Pearson Edexcel A Level Chemistry Year 1 and Year 2
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PEARSON EDEXCEL A LEVEL
CHEMISTRY
GRAHAM CURTIS
ANDREW HUNT
GRAHAM HILL
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In order to ensure that this resource offers high-quality support for the associated Pearson qualification,
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covers the teaching and learning content of the specification or part of a specification at which it is aimed.
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knowledge and understanding, in addition to preparation for assessment.
Endorsement does not cover any guidance on assessment activities or processes (e.g. practice questions or
advice on how to answer assessment questions), included in the resource nor does it prescribe any particular
approach to the teaching or delivery of a related course.
While the publishers have made every attempt to ensure that advice on the qualification and its assessment
is accurate, the official specification and associated assessment guidance materials are the only authoritative
source of information and should always be referred to for definitive guidance.
Pearson examiners have not contributed to any sections in this resource relevant to examination papers for
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Examiners will not use endorsed resources as a source of material for any assessment set by Pearson.
Endorsement of a resource does not mean that the resource is required to achieve this Pearson qualification,
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ISBN: 978 1 5104 6998 3
© Andrew Hunt, Graham Curtis, Graham Hill 2019
First published in 2019 by
Hodder Education,
An Hachette UK Company
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50 Victoria Embankment
London EC4Y 0DZ
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Year 2023 2022 2021 2020 2019
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A catalogue record for this title is available from the British Library
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Contents
Get the most from this book v
Introductionvii
Prior knowledge 1
1 Atomic structure and the Periodic Table 12
2 Bonding and structure 39
3 Redox I 83
4 Inorganic chemistry and the Periodic Table 97
5 Formulae, equations and amounts of substance 123
6.1 Introduction to organic chemistry 155
6.2 Hydrocarbons: alkanes and alkenes 177
6.3 Halogenoalkanes and alcohols 209
7 Modern analytical techniques I 233
8 Energetics I 246
9 Kinetics I 272
10 Equilibrium I 285
11 Equilibrium II 298
12 Acid–base equilibria 318
13.1 Lattice energy 349
13.2 Entropy 367
14 Redox II 387
15 Transition metals 418
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16 Kinetics II 450
17.1 Chirality 473
17.2 Carbonyl compounds 484
17.3 Carboxylic acids and their derivatives 500
18.1 Arenes – benzene compounds 519
18.2 Amines, amides, amino acids and proteins 541
18.3 Organic synthesis 568
19 Modern analytical techniques II 593
Appendix
A1 Mathematics in chemistry Year 1 626
A2 Preparing for the exam 641
Index647
The Periodic Table of Elements 656
Acknowledgements657
Answers for the ‘Test yourself ’ questions and activities found in this book are available online at
[Link]/EdexcelChemistry
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Get the most from this book
Welcome to the Pearson Edexcel A level Chemistry Student’s Book!
This book covers Year 1 and Year 2 of the Pearson Edexcel A level Chemistry
specification.
The following features have been included to help you get the most from
this book.
Tips
These highlight important facts,
common misconceptions and
signpost you towards other relevant
topics.
Key terms and formulae
These are highlighted in the text and definitions are given in the margin to
help you pick out and learn these important concepts.
Test yourself questions
These short questions, found
throughout each chapter, are useful
for checking your understanding as
you progress through a topic.
Examples
Examples of questions and
calculations feature full workings
and sample answers.
Get the most from this book v
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Activities and Core
practicals
These practical-based activities will help
consolidate your learning and test your
practical skills. Pearson Edexcel’s Core
practicals are clearly highlighted.
In this edition the authors describe many
important experimental procedures to
conform to recent changes in the
A level curriculum. Teachers should be
aware that, although there is enough
information to inform students of
techniques and many observations for
exam purposes, there is not enough
information for teachers to replicate
the experiments themselves, or
with students, without recourse to
CLEAPSS Hazcards or Laboratory
worksheets which have undergone a
risk assessment procedure.
Exam practice questions
You will find Exam practice questions at the end of every
chapter. These follow the style of the different types of
questions you might see in your examination and are
colour coded to highlight the level of difficulty. Quality
of extended response questions are marked with an
asterisk (*).
Dedicated chapters for developing your Maths and Preparing for your
exam are also included in this book.
vi Get the most from this book
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Introduction
This book is an extensively revised, restructured, updated and combined version of Pearson Edexcel Chemistry for AS and
Pearson Edexcel Chemistry for A2 by Graham Hill and Andrew Hunt. We have relied heavily on the contributions that
Graham Hill made to the original books and are most grateful that he has encouraged us to build on his work. The team
at Hodder Education, led initially by Hanneke Remsing and then by Emma Braithwaite, has made an extremely valuable
contribution to the development of the book and the website resources. In particular, we would like to thank Abigail
Woodman, the project manager, for her expert advice and encouragement. We are also grateful for the skilful work on
the print and electronic resources by Anne Trevillion.
At the end of each chapter we have added a ‘Chapter Summary’. This contains a brief bulleted list of the main points
covered in the Chapter to help with your revision.
We have grouped each set of ‘Exam practice’ questions broadly by difficulty. In general, a question with is straightforward
and based directly on the information, ideas and methods described in the chapter. Each problem-solving part of the
question typically only involves one step in the argument or calculation. A question with is a more demanding, but still
structured, question involving the application of ideas and methods to solve a problem with the help of data or information
from this chapter or elsewhere. Arguments and calculations typically involve more than one step. The questions marked
by are hard and they may well expect you to bring together ideas from different areas of the subject. In these harder
questions, you may have to structure an argument or work out the steps required to solve a problem. In the earlier chapters,
you may well decide not to attempt the questions with until you have gained wider experience and knowledge of
the subject.
Practical work is of particular importance in A Level chemistry. Each of the Core Practicals in the specification
features in the main chapters of this book, with an outline of the procedure and data for you to analyse and interpret.
Throughout the text there are references to Practical skills sheets, which can be accessed via [Link].
[Link]/EdexcelChemistry. Sheets 1 to 3 provide general guidance for the first year of your course, and sheets 11 to 15
for the second year of your course, and the remainder provide more detailed guidance for the Core Practicals.
1 Practical skills for advanced chemistry
2 Assessing hazards and risks
3 Researching and referencing
4 Making measurements
5 Identifying errors and estimating uncertainties
6 Measuring chemical amounts by titration
7 Analysing inorganic unknowns
8 Synthesising organic liquids
9 Analysing organic unknowns
10 Measuring enthalpy changes
11 Assessment of practical work
12 Overview of practical skills
13 Assessing hazards and risks
14 Researching and referencing
15 Identifying errors and estimating uncertainties
16 Finding the K a value for a weak acid
17 Measuring chemical amounts by titration
18 Investigating reaction orders and activation energies
19 Analysing inorganic unknowns
20 Analysing organic unknowns
21 Synthesis of an organic solid
Introduction vii
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You will need to refer to the Pearson Edexcel Data booklet when answering some of the questions in this book. This
will help you to become familiar with the booklet. This is important because you will need to use the booklet to find
information when answering some questions in the examinations. You can download the Data booklet from the Pearson
Edexcel website. It is part of the specification. The booklet includes the version of the Periodic Table that you use in the
examinations.
Andrew Hunt and Graham Curtis
February 2019
Photo credits: p. 1 Karina Baumgart – Fotolia; blueskies9 – Fotolia (inset); p. 3 image originally created by IBM Corporation; p. 5 Andrew Lambert Photography/
Science Photo Library (both); p. 6 theartofphoto – Fotolia; p. 10 Gayvoronskaya_Yana/Shutterstock; p. 12 t Science Source/Science Photo Library; b Sheila Terry/
Science Photo Library; p. 15 Jason Hawkes/Corbis via Getty Images; p. 16 Graham J. Hills/Science Photo Library; p. 23 Gilbert Iundt; Jean-Yves Ruszniewski/
TempSport/Corbis/VCG via Getty Images; p. 24 Dept. of Physics, Imperial College/Science Photo Library; p. 40 Philippe Plailly/Eurelios/Science Photo Library; p. 41
t marcel – Fotolia, b Monkey Business – Fotolia; p. 42 Andrew Lambert Photography/Science Photo Library; p. 44 Ruddy Gold/age fotostock/SuperStock;
p. 50 Andrew Lambert Photography/Science Photo Library; p. 60 Charles D. Winters/Science Photo Library; p. 61 nico99 – Fotolia; p. 66 marcaletourneux – Fotolia;
p. 70 jurra8 – Fotolia; p. 72 Stuart Franklin/Getty Images; p. 73 bl James King-Holmes/Science Photo Library, br Alfred Pasieka/Science Photo Library; p. 76 branex –
Fotolia; p. 83 Miredi – Fotolia; p. 86 Andrew Lambert Photography/Science Photo Library; p. 97 Martyn F. Chillmaid/Science Photo Library; p. 98 Andrew Lambert
Photography/Science Photo Library; p. 99 Lawrence Migdale/Science Photo Library; p. 101 Andrew Lambert Photography/Science Photo Library (all);
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Photo Library, bl Andrew Lambert Photography/Science Photo Library; p. 108 Javier Trueba/Msf/Science Photo Library; p. 109 l [Link] – Fotolia, r Alfred
Pasieka/Science Photo Library; p. 111 l Andrew Lambert Photography/Science Photo Library, c sciencephotos/Alamy, r Andrew Lambert Photography/Science Photo
Library; p. 112 Andrew Lambert Photography/Science Photo Library; p. 115 Andrew Lambert Photography/Science Photo Library (both); p. 117 Martyn F. Chillmaid/
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Fotolia; p. 178 Alvey & Towers Picture Library/Alamy; p. 181 Andrew Lambert Photography/Science Photo Library (all); p. 187 Tony Craddock/Science Photo Library;
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Images; p. 218 Agencja Fotograficzna Caro/Alamy; p. 219 Roger Job/Reporters/Science Photo Library; p. 225 Andrew Lambert Photography/Science Photo Library;
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Degginger/Alamy; p. 272 tl Clive Freeman, The Royal Institution/Science Photo Library, b Israel Sanchez EPA/REX/Shutterstock; p. 285 bl albinoni – Fotolia, br
Santi Rodríguez – Fotolia; p. 286 Andrew Lambert Photography/Science Photo Library; p. 298 yodiyim/Fotolia; p. 305 FOOD-micro/Fotolia; p. 309 Maximilian
Stock Ltd/Science Photo Library; p. 310 Andrew Lambert Photography/Science Photo Library; p. 313 tl Charles D. Winters/Science Photo Library, b Manfred Kage/
Science Photo Library; p. 318 tl toa555/Fotolia, b Science Photo Library; p. 323 Charles D. Winters/Science Photo Library; p. 327 nenetus/Fotolia; p. 332 Andrew
Lambert Photography/Science Photo Library; p. 336 Andrew Lambert Photography/Science Photo Library (all); p. 338 gabriffaldi/Fotolia; p. 340 JPC-PROD/Fotolia;
p. 342 Sebastian Kaulitzki/Fotolia; p. 349 Andrew Lambert Photography/Science Photo Library; p. 360 Vit Kovalcik/Fotolia; p. 367 cl maros_bauer/Fotolia, cr ju_skz/
Fotolia, bl ova/Fotolia; p. 368 t Anna Khomulo/Fotolia, c Ghen/Fotolia, b Science Photo Library; p. 369 Charles D. Winters/Science Photo Library (both); p. 371
WavebreakmediaMicro/Fotolia; p. 374 kalpis/Fotolia; p. 381 c Fuse/Thinkstock, r Mark Williamson/Science Photo Library; p. 382 marcel/Fotolia; p. 387 Stocktrek
Images/Getty images; p. 392 Andrew Lambert Photography/Science Photo Library (both); p. 396 Martyn F. Chillmaid/Science Photo Library; p. 402 design56/Fotolia;
p. 404 tr Andrew Lambert Photography/Science Photo Library, br Peticolas/Megna/Fundamental Photos/Science Photo Library; p. 410 mikanaka/Thinkstock;
p. 418 Andrew Lambert Photography/Science Photo Library; p. 423 sumnersgraphicsinc/Fotolia; p. 425 Andrew Lambert Photography/Science Photo Library; p. 426
Andrew Lambert Photography/Science Photo Library; p. 430 Andrew Lambert Photography/Science Photo Library; p. 435 Interfoto/Alamy; p. 436 Science Photo
Library; p. 443 Biosym Technologies, Inc./Science Photo Library; p. 450 Kadmy/Fotolia; p. 473 l mosinmax/Fotolia, r atoss/Fotolia; p. 474 full image/Fotolia; p. 476
c indigolotos/Fotolia, b James Watson;p. 485 Steve Gschmeissner/Science Photo Library; p. 487 Vesna Cvorovic/Fotolia; p. 493 Andrew Lambert Photography/Science
Photo Library; p. 494 Andrew Lambert Photography/Science Photo Library (both); p. 500 l skynet/Fotolia, r Debu55y/Fotolia; p. 501 Susan Wilkinson; p. 503 xeni4ka/
Thinkstock; p. 507 Andrew Lambert Photography/Science Photo Library; p. 513 tr Philippe Hallé/Thinkstock, br Sally and Richard Greenhill/Alamy; p. 519 Corbis
Super RF/Alamy; p. 526 Andrew Lambert Photography/Science Photo Library; p. 528 sashagrunge/Fotolia; p. 533 Martyn F. Chillmaid/Science Photo Library; p. 541
WavebreakmediaMicro/Fotolia; p. 543 tr Mediablitzimages/Alamy, br Nomadsoul1/Thinkstock; p. 545 Martyn F. Chillmaid/Science Photo Library; p. 558 tl Charles
D. Winters/Science Photo Library, cl Jeff Morgan 09/Alamy; p. 560 Ashley Cooper/Corbis via Getty Images; p. 562 Eye of Science/Science Photo Library; p. 568 James
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Inc./Alamy; p. 593 Geoff Tompkinson/Science Photo Library; p. 602 Food Collection/Alamy; p. 607 Vince Bevan/Alamy; p. 608 Roger Hutchings/Alamy; p. 614
Jerry Mason/Science Photo Library; p. 616 Michael Donne/Science Photo Library; p. 618 l ESA/ATG medialab, r STFC
b = bottom, c = centre, l = left, r = right, t = top
Acknowledgement
Data used for the mass spectra in Figures 7.4 and 7.6, the IR spectra on page 244, proton NMR spectrum in Figure 19.21 and the two IR spectra on page 625 come
from the SDBS of the National Institute of Advanced Industrial Science and Technology.
Although every effort has been made to ensure that website addresses are correct at time of going to press, Hodder Education cannot be held responsible for the
content of any website mentioned in this book. It is sometimes possible to find a relocated web page by typing in the address of the home page for a website in the
URL window of your browser.
viii Introduction
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Prior knowledge
1 Working like a chemist
Chemistry is about understanding the material world. Chemists develop
their explanations by observing the properties of substances and looking at
patterns of behaviour (Figure 1). They devise theories and models that can
be used in chemical analysis and synthesis.
Figure 1 Aspirin is probably the
commonest medicine in use. The bark
of willow trees was used to ease pain
for more than 2000 years. Early in the
twentieth century, chemists extracted the
active ingredient from willow bark. Their
understanding of patterns in the behaviour
of similar compounds enabled them to
synthesise aspirin.
Tip
This first chapter surveys the main themes of chemistry and indicates how you will be learning
more about chemistry during your A Level course. The chapters in this book build on what
you already know about chemistry. The text and ‘ Test yourself ’ questions in the early part of
each chapter can help you to check on what you have learned before and what you need to
understand at the start of each topic.
Looking for patterns in chemical behaviour
Part of being a chemist involves getting a feel for the way in which chemicals
behave. Chemists get to know chemicals just as people get to know their friends
and family. They look for patterns in behaviour and recognise that some of
the patterns are familiar. For example, the elements sodium and potassium are
both soft and stored under oil because they react so readily with air and water;
copper sulfate is blue, like other copper compounds. By understanding patterns,
chemists can design and make plastics like polythene and medicines like aspirin.
1 Working like a chemist 1
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Tip Test yourself
The Periodic Table links together Remind yourself of some patterns in the ways that chemicals behave.
many of the key patterns of behaviour
1 What happens when a more reactive metal (such as zinc) is added to
of elements. You will extend your
a solution in water of a compound of a less reactive metal (such as
knowledge of the Periodic Table in
copper sulfate)?
Chapter 1. You will also make a detailed
study of patterns in the properties of the 2 What forms at the negative electrode (cathode) during the electrolysis
elements and compounds in some of the of a solution of a salt?
Periodic Table groups in Chapter 4. 3 What happens on adding an acid (such as hydrochloric acid) to a
carbonate (such as calcium carbonate)?
4 What do sodium chloride, sodium bromide and sodium iodide look like?
Discovering the composition and structure
Tip of materials
Theories of structure and bonding are New materials exist only because chemists understand how atoms, ions and
key to understanding the properties molecules are arranged in different materials, and about the forces which
of materials. You will extend your hold these particles together. Thanks to this knowledge, people can enjoy
knowledge of these ideas when you fibres that breathe but are waterproof, plastic ropes that are 20 times stronger
study Chapter 2. Chapter 8 shows how than similar ropes of steel and metal alloys which can remember their shape.
measuring energy changes can provide
Understanding the structure and bonding of materials is a central theme in
evidence of the nature and strength of
modern chemistry. Fundamental to this is an understanding of how the atoms,
chemical bonds.
molecules or ions are arranged in different states of matter (Figure 2).
Particles in a solid are packed
close together in a regular way.
The particles do not move freely,
but vibrate about fixed positions.
The particles in a liquid are closely packed
but are free to move around, sliding past
each other.
In a gas the particles are spread out, so the densities of
gases are very low compared with solids and liquids.
The particles move rapidly in a random manner, colliding
with other particles and the walls of the container.
Pressure is caused by particles hitting the walls.
Lighter particles move faster than heavier ones.
Figure 2 The arrangements of particles in solids, liquids and gases.
2 Prior knowledge
469983_00_Chem_Y1-2_001-[Link] 2 13/04/19 10:15 PM
Explaining and controlling chemical changes Tip
Four key questions are at the heart of many chemical investigations.
Chapters 5 and 8 show you how
● How much? – How much of the reactants is needed to make a product, chemists answer the question ‘How
how much of the product is produced, and how much energy is needed? much?’. The questions ‘How fast?’
● How fast? – How can a reaction be controlled so that it goes at the right and ‘How far?’ are the focus of
speed: not too fast and not too slow? Chapters 9 and 10. Understanding how
● How far? – Do the chemicals react completely, or does the reaction stop reactions occur is a feature of organic
before all the reactants have turned into products? If it does, what can be chemistry and so the study of reaction
done to get as big a yield as possible? mechanisms is explored in the three
● How do reactions occur? – Which bonds between atoms break and parts of Chapter 6.
which new bonds form during a reaction?
Developing new techniques and skills
Chemistry involves doing things as well as gaining knowledge and
understanding about materials. Chemists use their thinking skills and
practical skills to solve problems. One of the frontiers of today’s chemistry
involves nanotechnology, in which chemists work with particles as small as
individual atoms (Figure 3).
Increasingly, chemists rely on modern instruments to explore structures Figure 3 In the 1990s, two scientists
and chemical changes. They also use information technology to store data, working for IBM cooled a nickel surface
search for information and to publish their findings. to −269 °C in a vacuum chamber. Then
they introduced a tiny amount of xenon so
Analysis and synthesis that some of the xenon atoms stuck to the
A vital task for chemists is to analyse materials and find out what they nickel surface. Using a special instrument
are made of. When chemists have analysed a substance, they use symbols called a scanning tunnelling microscope,
and formulae to show the elements it contains. Symbols are used to the scientists were able to move individual
represent the atoms in elements; formulae are used to represent the ions xenon atoms around on the nickel surface
and molecules in compounds. and construct the IBM logo. Each blue blob
is the image of a single xenon atom.
Analysis is involved in checking that water is safe to drink and that food
has not been contaminated. People may worry about pollution of the
environment, but without chemical analysis they would not know about the Tip
causes or the scale of any pollution.
You will be developing your practical
Chemists have devised many ingenious methods of analysis. Spectroscopy is skills and understanding of practical
especially important. At first spectroscopists just used visible light, but now chemistry during your A Level
they have found that they can find out much more by using other kinds of course. Most chapters in this book
radiation such as ultraviolet and infrared rays, radiowaves and microwaves. include activities and core practicals
with results and data to analyse.
Chemistry is also about making things. Chemists take simple chemicals
General guidance on practical
and join them together to make new substances. This is synthesis. On a
work can be accessed online at
large scale, the chemical industry converts raw materials from the earth, sea
[Link]/
and air into valuable new products. A well-known example is the Haber
EdexcelChemistry.
process which uses natural gas and air to make ammonia. Ammonia is the
chemical needed to make fertilisers, dyes and explosives. On a smaller scale,
chemical reactions produce the specialist chemicals used for perfumes, dyes
and medicines.
1 Working like a chemist 3
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Tip Linking theories and experiments
Scientists test their theories by doing experiments. In chemistry,
Chapter 7 includes an account of some
experiments often begin with careful observation of what happens as
of the modern instrumental techniques
chemicals react and change. Theories are more likely to be accepted
used by chemists. Organic reactions
if predictions made from them turn out to be correct when tested by
that are important in synthesis feature
experiment.
in all parts of Chapter 6. The study of
synthesis is a key feature of the organic One of the reasons why Mendeléev’s Periodic Table was so successful was
chemistry in the second half of your because he left gaps in his table for elements that had not yet been discovered
A Level course. and then made predictions about the properties of missing elements that
turned out to be accurate (Table 1).
Table 1 Mendeléev’s predictions for germanium in 1871 and the properties it was found
to have after its discovery in 1886.
Mendeléev’s predictions in 1871 Actual properties in 1886
Grey metal Pale grey metal
Density 5.5 g cm−3 Density 5.35 g cm−3
Relative atomic mass 73.4 Relative atomic mass 72.6
Tip
Melting point 800 °C Melting point 937 °C
Chemistry is a quantitative subject
which involves a variety of types of Formula of oxide GeO2 Ge forms GeO2
calculation. You will find many worked
examples in the chapters of this book
Studying chemistry is more than about ‘what we know’. It is also about
that will help you to solve quantitative
‘how we know’. For example, the study of atomic structure has provided
problems. The key mathematical ideas
evidence about the nature and properties of electrons, and this has led to an
and techniques involved are described
explanation of the properties of elements and the patterns in the Periodic
in Appendix A1.
Table in terms of the electron structures of atoms.
2 Elements
Everything is made of elements. Elements are the simplest chemical
substances which cannot be decomposed into simpler chemicals by heating
or using electricity. There are over 100 elements, but from their studies of
the stars, astronomers believe that about 90% of the Universe consists of just
one element, hydrogen. Another 9% is accounted for by helium, leaving only
1% for all the other elements.
Metals and non-metals
Most of the elements, nearly 90 of them, are metals. It is usually easy to
recognise a metal by its properties. Most metals are shiny, strong, bendable
and good conductors of electricity (Figure 4).
There are only 22 non-metal elements: this includes a few which are
solid at room temperature, such as carbon and sulfur, several gases, such
as hydrogen, oxygen, nitrogen and chlorine, and just one liquid, bromine
(Figure 5).
4 Prior knowledge
469983_00_Chem_Y1-2_001-[Link] 4 13/04/19 10:15 PM
Figure 4 Samples of metals: from left to right, copper, zinc, lead Figure 5 Samples of non-metals: sulfur, bromine, phosphorus
and silver. (behind), carbon and iodine (in front).
Atoms of elements Tip
Each element has its own kind of atom. An atom is the smallest particle of an
You will learn more about the properties
element. Atoms consist of protons, neutrons and electrons. Every atom has a
of metal and non-metal elements in
tiny nucleus surrounded by a cloud of electrons (Figure 6).
Chapter 4.
The mass of an atom is concentrated in the nucleus which consists of
protons and neutrons. The protons are positively charged and the neutrons
uncharged. All the atoms of a particular element have the same number of
protons in the nucleus. cloud of electrons
The electrons are negatively charged. The mass of an electron is so small protons
that it can often be ignored. In an atom the number of electrons equals the nucleus
neutrons
number of protons in the nucleus. So the total negative charge equals the
total positive charge and overall the atom is uncharged.
Test yourself Figure 6 Diagram of an atom showing a
nucleus surrounded by a cloud of electrons.
5 Give examples of substances which can be split into elements by This is not to scale. In reality the diameter of
heating or by using an electric current (electrolysis). an atom is about 100 000 times bigger than
6 Draw up a table to compare metal elements with non-metal elements the diameter of its nucleus.
using the following headings: Property; Metal; Non-metal.
Tip
3 Compounds You will learn more about atomic
Compounds form when two or more elements combine. Apart from the atoms structure in Chapter 1.
of the elements helium and neon, all elements can combine with other elements.
In order to explain the properties of compounds, chemists need to find out
how the atoms, molecules or ions are arranged (the structure) and what holds
them together (the bonding).
Compounds of non-metals with non-metals
Water, carbon dioxide, methane in natural gas, sugar and ethanol (‘alcohol’)
are examples of compounds of two or more non-metals. These compounds
of non-metals have molecular structures.
3 Compounds 5
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H The covalent bonds between the atoms in molecules are strong but the
attractive forces between molecules are weak. This means that molecular
H C compounds
H CH 4and vaporise easily. They may be gases, liquids or solids at
melt
H H room temperature and they do not conduct electricity.
H
H CH HC H CH 4 CH 4 Methane contains one carbon atom bonded to four hydrogen atoms. The
formula of the molecule is CH4. Figure 7 shows three ways of representing
H H a methane molecule.
Figure 7 Ways of representing a molecule Chemists have to analyse compounds to find their formulae. The results of
of methane. analysis give an empirical (experimental) formula. This shows the simplest
whole number ratio of the atoms of different elements in a compound, for
example CH4 for methane and CH3 for ethane.
Tip
More information is needed to work out the molecular formula of a
You will learn more about how chemists compound showing the numbers of atoms of the different elements in one
determine the formulae of compounds molecule of the compound. For example, CH4 is the molecular formula of
in Sections 5.2 and 5.3. methane but C2H6 is the molecular formula of ethane.
It is often possible to write the formula of non-metal compounds given how
many covalent bonds the atoms normally form (Table 2).
Table 2 Symbols, number of bonds and colour codes of some non-metals.
Element Symbol Number of bonds Colour in molecular
formed models
Carbon C 4 Black
O Nitrogen N 3 Blue
H H H 2O
Oxygen O 2 Red
Figure 8 Ways of representing a molecule of
Sulfur S 2 Yellow
water.
Hydrogen H 1 White
Chlorine CI 1 Green
O C O
Figure 9 Bonding in carbon dioxide Water is a compound of oxygen and hydrogen. Oxygen atoms form two
showing the double bonds between atoms. bonds and hydrogen atoms form one bond. So two hydrogen atoms can bond
to one oxygen atom (Figure 8) and the formula of water is H 2O.
There are double and even triple bonds between the atoms in some non-
metal compounds (Figure 9). Notice also that there is a colour code for the
atoms of different elements in molecular models – these colours are shown
in Table 2.
In practice, it is not possible to predict the formulae of all non-metal
compounds. For example, the simplified bonding rules in Table 2 cannot
account for the formulae of carbon monoxide, CO, sulfur dioxide, SO2, or
sulfur hexafluoride, SF6.
There are some compounds made up of non-metal elements in which the
Figure 10 Quartz crystal from Sentis, covalent bonding links all the atoms in a crystal together in a giant lattice.
Switzerland. Quartz is one of the Silicon dioxide, SiO2, is an important example which is found in many
commonest minerals of the Earth’s crust. igneous rocks (Figure 10). Compounds with covalent giant structures are
It consists of silicon dioxide, SiO2. hard and melt at high temperatures.
6 Prior knowledge
469983_00_Chem_Y1-2_001-[Link] 6 13/04/19 10:16 PM
Test yourself Tip
7 Draw the various ways of representing the following molecular You will learn more about the bonding
compounds in the style of Figure 7: in compounds of non-metals with non-
metals in Chapter 2.
a) hydrogen chloride b) carbon disulfide.
8 Name the elements present and work out the formula of the following
molecular compounds:
a) hydrogen sulfide b) dichlorine oxide
c) ammonia (hydrogen nitride).
Compounds of metals with non-metals
Common salt (sodium chloride), limestone (calcium carbonate) and copper
sulfate are all examples of compounds of metals with non-metals. These
metal/non-metal compounds consist of a giant structure of ions. An ion is an
atom, or a group of atoms, which has become electrically charged by the loss
or gain of one or more electrons. Generally metal atoms form positive ions
by losing electrons while non-metal atoms form negative ions by gaining
electrons. For example, sodium chloride consists of positive sodium ions,
Na+, and negative chloride ions, Cl− (Figure 11).
Na+
Cl–
space-filling model ball-and-stick model
Figure 11 A space-filling model and a ball-and-stick model showing the giant structure
of sodium chloride.
The strong ionic bonding between the ions means that such compounds melt
at much higher temperatures than the molecular compounds of non-metals.
They are solids at room temperature. They conduct electricity as molten liquids
but not as solids. Metal/non-metal compounds conduct electricity when heated
above their melting points because the ions are free to move in the liquid state.
The formula of sodium chloride is NaCl because the positive charge on one
Na+ ion is balanced by the negative charge on one Cl− ion. In a crystal of
sodium chloride there are equal numbers of sodium ions and chloride ions.
The formulae of all metal/non-metal (ionic) compounds can be worked out by
balancing the charges on positive and negative ions. For example, the formula of
potassium oxide is K2O. Here, two K+ ions balance the charge on one O2− ion.
Elements such as iron, which have two different ions (Fe2+ and Fe3+), have
two sets of compounds – iron(ii) compounds such as iron(ii) chloride, FeCl 2,
and iron(iii) compounds such as iron(iii) chloride, FeCl3.
3 Compounds 7
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Table 3 The names and formulae of some ionic compounds.
Name of compound Ions present Formula
Magnesium nitrate Mg2+ and NO3− Mg(NO3)2
Aluminium hydroxide Al3+ and OH− Al(OH)3
Zinc bromide Zn2+ and Br− ZnBr2
Lead(ii) nitrate Pb2+ and NO3− Pb(NO3)2
Calcium iodide Ca2+ and I− CaI2
Copper(ii) carbonate Cu2+ and CO32− CuCO3
Silver sulfate Ag+ and SO42− Ag2SO4
Table 3 shows the names and formulae of some ionic compounds. Notice
that the formula of magnesium nitrate is Mg(NO3)2. The brackets round
NO3− show that it is a single unit containing one nitrogen and three oxygen
atoms bonded together with a 1− charge. Other ions, such as OH−, SO42−
and CO32−, must also be treated as single units and put in brackets when
there are two or three of them in a formula.
Tip Test yourself
You will learn more about ionic crystals 9 This question concerns the substances ice, salt, sugar, copper, steel
and ionic bonding in Chapter 2. and limestone.
Which of these substances contain:
a) uncombined atoms
b) ions
c) molecules?
10 The structure of the main constituent in antifreeze is:
H H
H C C H
OH OH
What is:
a) its molecular formula
b) its empirical formula?
11 The formula of aluminium hydroxide must be written as Al(OH)3. Why
is AlOH3 wrong?
12 Write the formulae of the following ionic compounds given these charges
on ions: Al3+, Fe2+, Fe3+, K+, Pb2+, Zn2+, CO32−, O2−, OH−, SO42−:
a) potassium sulfate
b) aluminium oxide
c) lead carbonate
d) zinc hydroxide
e) iron(iii) sulfate.
8 Prior knowledge
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13 Which of the following compounds consist of molecules and which
consist of ions?
a) octane (C8H18) in petrol b) copper(i) oxide
c) concentrated sulfuric acid d) lithium fluoride
e) phosphorus trichloride
14 Compare non-metal (molecular) compounds with metal/non-metal
(ionic) compounds in:
a) melting temperatures and boiling temperatures
b) conduction of electricity as liquids.
4 Chemical changes
Burning, rusting and fermentation are all examples of chemical reactions.
Under the right conditions, chemical bonds break and new ones form. This
is what happens during a chemical reaction to create new chemicals.
Figure 12 shows a simple way of demonstrating that when hydrogen burns
the product is water. Hydrogen and oxygen (in the air) are both gases at room
temperature. When the gases react the changes give out so much energy that
there is a flame. Water condenses on cooling the steam that forms in the flame.
Figure 12 Demonstration that burning
hydrogen produces water.
to pump
ice and water
dry hydrogen
gas
a colourless liquid
condenses here
One way of describing what happens during a reaction is to write a word equation.
Writing word equations identifies the reactants (on the left) and products (on the
right), so it is a useful first step towards a balanced equation with symbols.
When hydrogen burns:
hydrogen(g) + oxygen(g) → water(l)
reactants product
When they are looking at this change, chemists imagine what is happening
to the molecules. The trick is to interpret the visible changes in terms of
theories about atoms and bonding. Models help to make the connection.
The hydrogen molecules and oxygen molecules consist of pairs of atoms.
They are diatomic molecules. Figure 13 shows how molecular models give a
picture of the reaction at an atomic level.
Figure 13 Model equation to show
+ hydrogen reacting with oxygen.
4 Chemical changes 9
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The formula of water is H2O. Each water molecule contains only one
oxygen atom. So one oxygen molecule can give rise to two water molecules,
provided that there are two hydrogen molecules available to supply all the
hydrogen atoms necessary.
There is the same number of atoms on both sides of the equation. The atoms
have simply been rearranged.
Chemists normally use symbols rather than models to describe reactions.
Symbols are much easier to write or type. State symbols added to a symbol
equation show whether the substances are solid, liquid, gases or dissolved
in water.
2H2(g) + O2(g) → 2H2O(l)
Modelling is increasingly important in modern chemistry but now the
Tip modelling is usually carried out with computers. In 2013 the Nobel prize
for chemistry was awarded to Martin Karplus, Michael Levitt and Arieh
You will learn more about writing
Warshel whose work, in the 1970s, laid the foundation for the powerful
equations for chemical reactions in
computer modelling programs that are used to understand and predict
Sections 3.2 and 4.1.
chemical processes.
Test yourself
15 a) Write a balanced symbol equation for the reaction of methane,
CH4, with oxygen.
b) Draw a diagram, similar to that shown in Figure 13, to show what
happens when methane burns in oxygen.
16 Write balanced equations, with state symbols, for the following word
equations:
a) hydrogen + chlorine → hydrogen chloride
b) zinc + hydrochloric acid (HCl) → zinc chloride + hydrogen
c) ethane + oxygen → carbon dioxide + water
d) iron + chlorine → iron(iii) chloride.
5 Acids, bases, alkalis and salts
Acids
Pure acids may be solids (such as citric, Figure 14, and tartaric acids), liquids
(such as sulfuric, nitric and ethanoic acids) or gases (such as hydrogen chloride
which becomes hydrochloric acid when it dissolves in water). All these acids
are compounds with characteristic properties:
● they form solutions in water with a pH below 7
● they change the colour of indicators such as litmus
Figure 14 Crystals of the solid acid citric
● they react with metals above hydrogen in the reactivity series forming
acid. This acid was first obtained as a pure
hydrogen plus an ionic metal compound called a salt
compound in 1784 when it was crystallised
from lemon juice. Fe(s) + 2HCl(aq) → FeCl 2(aq) + H2(g)
10 Prior knowledge
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● they react with metal oxides and metal hydroxides to form salts and water
CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)
● they react with carbonates to form salts, carbon dioxide and water
ZnCO3(s) + 2HCl(aq) → ZnCl 2(aq) + CO2(g) + H2O(l)
Bases and alkalis
Bases are ‘anti-acids’. They are the chemical opposites of acids. Alkalis are
bases which dissolve in water. The common laboratory alkalis are sodium
hydroxide, potassium hydroxide, calcium hydroxide and ammonia. Alkalis
form solutions with a pH above 7, so they change the colours of acid–base
indicators. Alkalis are useful because they neutralise acids.
Manufacturers produce powerful oven and drain cleaners containing sodium
hydroxide or potassium hydroxide because they can break down and remove
greasy dirt. These strong alkalis are highly ‘caustic’. They attack skin,
producing a chemical burn. Even dilute solutions of these alkalis can be
hazardous, especially if they get into your eyes (Section 4.3).
Test yourself
17 Write full balanced equations for the reactions of hydrochloric acid with:
a) zinc b) calcium oxide
c) potassium hydroxide d) nickel(ii) carbonate.
Salts
Salts are ionic compounds formed when an acid reacts with a base. In the
formula of a salt, the hydrogen of an acid is replaced by a metal ion. For
example, magnesium sulfate, MgSO4, is a salt of sulfuric acid, H2SO4.
Salts can be regarded as having two ‘parents’. They are related to a parent acid
and to a parent base. Hydrochloric acid, for example, gives rise to the salts
called chlorides, such as sodium chloride, calcium chloride and ammonium
chloride. The base sodium hydroxide gives rise to sodium salts, such as
sodium chloride, sodium sulfate and sodium nitrate.
Neutralisation is not the only way to make a salt. Some metal chlorides, for
example, are made by heating metals in a stream of chlorine. This is useful
for making anhydrous chlorides, such as aluminium chloride.
Test yourself
18 Name the salts formed from these pairs of acids and bases:
a) nitric acid and potassium hydroxide
b) hydrochloric acid and calcium hydroxide
c) sulfuric acid and copper(ii) oxide
d) ethanoic acid and sodium hydroxide.
5 Acids, bases, alkalis and salts 11
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Atomic structure and the
1 Periodic Table
1.1 Models of atomic structure
Early ideas about atoms
The idea that all substances are made of atoms is a very old one. It was
suggested by Greek philosophers, including Democritus, more than 2400
years ago (Figure 1.1).
Democritus was a philosopher whose idea was that if a lump of metal, such
as iron, was cut into smaller and smaller pieces, the end result would be
miniscule and invisible particles that could not be cut any smaller. Democritus
called these smallest particles of matter ‘atomos’ meaning ‘indivisible’. He
explained the properties of materials such as iron in terms of the shapes of the
atoms and the ‘hooks’ that he imagined joined them together.
Democritus was a great thinker but he did not do experiments and he
had no way to test his ideas. He, and other atomists of his time, failed
Figure 1.1 The Greek philosopher to convince everybody that the theory was correct. There were other
Democritus, who lived from 460 to 370 bce. competing theories and no convincing reasons to accept the idea of atoms
in preference to other ideas.
Modern atomic theory grew from work started about 2000 years after
Democritus, when scientists in Europe started to purify substances and to
carry out experiments with them. They found that many substances could
be broken down (decomposed) into simpler substances, which they called
elements. These elements could then be combined to make new compounds.
In the eighteenth century, chemists began to make accurate measurements
of the quantities of substances involved in reactions. To their surprise, they
found that the weights of elements which reacted were always in the same
proportions. So, for example, water always contained 1 part by weight of
hydrogen to 8 parts by weight of oxygen. And, black copper oxide always
contained 1 part by weight of oxygen to 4 parts by weight of copper.
At the start of the nineteenth century, John Dalton puzzled over these
results. He concluded that if elements were made of indivisible particles,
Figure 1.2 John Dalton was born in 1766 then everything made sense (Figure 1.2). Compounds, like copper oxide,
in the village of Eaglesfield in Cumbria. His were made of particles of copper and oxygen with different masses and these
father was a weaver. Dalton was always always combined in the same ratios. Dalton called the indivisible particles
curious and liked to study. When he was atoms in recognition of the ideas first proposed by Democritus.
only 12 years old, he started to teach
children in the village school. For most of Dalton began to publish his atomic theory in 1808. The main points in his
his life, he taught science and carried out theory were that:
experiments at the Presbyterian College in ● all elements are made up of indivisible particles called atoms
Manchester. ● all the atoms of a given element are identical and have the same mass
12 1 Atomic structure and the Periodic Table
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● the atoms of different elements have different masses
● atoms can combine to form molecules in compounds
● all the molecules of a given compound are identical.
Although some scientists were reluctant to accept Dalton’s ideas, his atomic
theory caught on because it could explain the results of many experiments.
Even today, Dalton’s atomic theory is still useful and very helpful. However,
research has since shown that atoms are not indivisible and that all atoms of
the same element are not identical.
Test yourself
1 Look at the five main points in Dalton’s atomic theory. Which of these
points:
a) are still correct
b) are now incorrect?
2 Look at the formulae below which Dalton used for water, carbon
dioxide and black copper oxide.
C
water carbon black copper
dioxide oxide
a) Write the formulae that are used today for these compounds.
b) What symbols did Dalton use for carbon, oxygen, hydrogen and
copper?
c) Which one of the formulae did Dalton get wrong?
Inside atoms
For much of the nineteenth century, scientists continued with the idea that
atoms were just as Dalton had described them: solid, indestructible particles
similar to tiny snooker balls. Then, between 1897 and 1932, scientists carried
out several series of experiments that revealed that atoms contain three
smaller particles: electrons, protons and neutrons.
The discovery of electrons
In 1897, J.J. Thomson was investigating the conduction of electricity by
gases in his laboratory at Cambridge. When he connected 15 000 volts across
the terminals of a tube containing air, the glass walls glowed bright green.
Rays travelling in straight lines from the negative terminal hit the glass and
made it glow. Experiments showed that a narrow beam of the rays could be
deflected by an electric field (Figure 1.3). When passed between charged
plates, the rays always bent towards the positive plate. This showed they were
negatively charged.
1.1 Models of atomic structure 13
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Figure 1.3 The effect of charged plates on fluorescent screen
a beam of electrons. which glows when
vacuum pump particles hit it
charged
plates
+ –
–
narrow beam before
plates were charged
+
deflected beam of
very high voltage rays after plates
(15 000 V) were charged
Further study showed that the rays consisted of tiny negative particles about
2000 times lighter than hydrogen atoms. This surprised Thomson. He had
discovered particles smaller than atoms. Thomson called the tiny negative
– ball of positive
–
charge particles electrons.
–
–
– – –
– Thomson obtained the same electrons with different gases in the tube and
– when the terminals were made of different substances. This suggested to him
– – – – that the atoms of all substances contain electrons. Thomson knew that atoms
–
–
–
had no electrical charge overall. So, the rest of the atom must have a positive
negative charge to balance the negative charge of the electrons.
– –
electrons
In 1904, Thomson published his model for the structure of atoms. He
Figure 1.4 Thomson’s plum pudding model suggested that atoms were tiny balls of positive material with electrons
for the structure of atoms. embedded in it like fruit in a Christmas pudding. As a result, Thomson’s idea
became known as the ‘plum pudding’ model of atomic structure (Figure 1.4).
alpha
gold foil Rutherford and the nuclear atom
particles Radioactivity was discovered by Henri Becquerel in Paris in 1896. Two
+
+ years later, Ernest Rutherford, in Manchester, showed that there were at least
+ two types of radiation given out by radioactive materials. He called these
+
alpha rays and beta rays.
+
+ At the time, Rutherford and his colleagues didn’t know exactly what alpha
+ + rays were. But they did know that alpha rays contained particles. These
+
alpha particles were small, heavy and positively charged. Rutherford and his
+
colleagues realised that they could use the alpha particles as tiny ‘bullets’ to
+
+ fire at atoms.
+
+
+
In 1909, two of Rutherford’s colleagues, Hans Geiger and Ernest Marsden,
directed narrow beams of positive alpha particles at very thin gold foil only
a few atoms thick (Figure 1.5). They expected the particles to pass straight
Figure 1.5 When positive alpha particles
through the foil or to be deflected slightly.
are directed at a very thin sheet of gold
foil, they emerge at different angles. Most The results showed that:
pass straight through the foil, some are
● most of the alpha particles went straight through the foil
deflected and a few appear to rebound
● some of the alpha particles were scattered (deflected) by the foil
from the foil.
● a few alpha particles rebounded from the foil.
14 1 Atomic structure and the Periodic Table
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Test yourself
3 Suggest explanations for these results of the Geiger–Marsden
experiment:
a) Most of the alpha particles passed straight through the foil.
b) Some alpha particles were deflected.
c) A few alpha particles rebounded from the foil.
4 a) What did the Geiger–Marsden experiment suggest about the size
of any positive and negative particles in the gold atoms.
b) Why did the results cast doubts on Thomson’s plum pudding model
for atomic structure?
5 Rutherford and his team published a series of papers about their
work, including a paper The Laws of Deflexion of α Particles through
Large Angles in a 1913 edition the Philosophical Magazine. Why is nucleus –
it important that scientists publish their experimental results and
theories?
–
Rutherford came up with a new model of the atom to explain the results + ++
++
of Geiger and Marsden’s experiment. In this model a very small positive –
nucleus is surrounded by a much larger region of empty space in which
–
electrons orbit the nucleus like planets orbiting the Sun (Figure 1.6).
Rutherford’s nuclear model quickly replaced Thomson’s plum pudding
model and it is still the basis of models of atomic structure used today. –
electrons
The work of Thomson, Rutherford and their colleagues showed that: Figure 1.6 Rutherford’s nuclear model for
● atoms have a small positive nucleus surrounded by a much larger region of the structure of atoms. Rutherford pictured
empty space in which there are tiny negative electrons (Figure 1.7) atoms as miniature solar systems with
● the positive charge of the nucleus is due to positive particles which electrons orbiting the nucleus like planets
Rutherford called protons around the Sun.
● protons are about 2000 times heavier than electrons
● the positive charge on one proton is equal in size, but opposite in sign, to
the negative charge on one electron
● atoms have equal numbers of protons and electrons, so the positive charges
on the protons cancel the negative charges on the electrons
● the smallest atoms are those of hydrogen with one proton and one electron.
The next smallest atoms are those of helium with two protons and two
electrons, then lithium atoms with three protons and three electrons, and
so on.
Chadwick and the discovery of neutrons
Although Rutherford was successful in explaining many aspects of atomic
structure, one big problem remained. If hydrogen atoms contain one proton
and helium atoms contain two protons, then the relative masses of hydrogen
Figure 1.7 If the nucleus of a hydrogen
and helium atoms should be one and two, respectively. But the mass of helium
atom were to be enlarged to the size of
atoms relative to hydrogen atoms is four and not two. It took the discovery of
a marble and put in the centre of the
isotopes and much further research before the problem was solved.
Wembley pitch, the atom’s one electron
In 1932, James Chadwick, in Cambridge, solved the mystery of the extra mass would be whizzing around somewhere in
in helium atoms. Chadwick studied the effects of bombarding a beryllium the stands.
1.1 Models of atomic structure 15
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target with alpha particles. This produced a new kind of radiation with no
electric charge but with enough energy to release protons when fired at
a material such as wax. In time, Chadwick was able to demonstrate that
there must be uncharged particles in the nuclei of atoms, as well as positively
charged protons. Chadwick called these particles neutrons. It was soon found
that neutrons had the same mass as protons.
The discovery of neutrons accounted for the relative masses of hydrogen
and helium atoms. Hydrogen atoms have one proton and no neutrons, so
a hydrogen atom has a relative mass of one unit, Helium atoms have two
protons and two neutrons, so a helium atom has a relative mass of four units.
This makes a helium atom four times as heavy as a hydrogen atom.
It is now understood that all atoms are made up from protons, neutrons and
electrons. The relative masses, relative charges and positions within atoms of
these sub-atomic particles are summarised in Table 1.1.
Tip Table 1.1 Relative masses, relative charges and positions in atoms of protons, neutrons
For a time, protons, neutrons and electrons.
and electrons were described as Particle Mass relative to that Charge relative to Position in the atom
‘fundamental’ or ‘elementary’ particles – of a proton that on a proton
that is particles not made up of anything Proton 1 +1 Nucleus
smaller or simple. Electrons are still
Neutron 1 0 Nucleus
thought to be fundamental particles but
protons and neutrons are now known to Electron 1 –1 Shells
be made up of quarks. 1840
Test yourself
6 Draw and label a diagram to show how Chadwick explained that the
mass of a helium atom is four times the mass of a hydrogen atom.
7 Summarise the development of atomic models in a table with the
models listed in the left-hand column and a brief note on the evidence
which gave rise to the models in the right-hand column.
1.2 Atomic number and mass number
All the atoms of a particular element have the same number of protons, and
atoms of different elements have different numbers of protons.
Hydrogen atoms are the simplest of all atoms – they have just one proton and
one electron. The next simplest are atoms of helium with two protons and
two electrons, then lithium with three protons, and so on. Large atoms have
large numbers of protons and electrons. For example, gold atoms (Figure 1.8)
have 79 protons and 79 electrons.
Figure 1.8 Photo of the surface of a The only atoms with one proton are those of hydrogen; the only atoms
gold crystal taken through an electron with two protons are those of helium; the only atoms with three protons
microscope. Each yellow blob is a are those of lithium, and so on. This means that the number of protons in
separate gold atom – the atoms have been an atom decides which element it is. Because of this, scientists have a special
magnified about 35 million times. name for the number of protons in the nucleus of an atom. They call it the
16 1 Atomic structure and the Periodic Table
469983_01_Chem_Y1-2_012-[Link] 16 13/04/19 10:15 PM
atomic number and use the symbol Z to represent it. So, hydrogen has an
atomic number of 1 (Z = 1), helium has an atomic number of 2 (Z = 2), and
Key terms
so on.
The atomic number of an atom is the
Protons do not account for all the mass of an atom – neutrons in the nucleus number of protons in its nucleus. The
also contribute. Therefore, the mass of an atom depends on the number of term ‘proton number’ is sometimes
protons plus neutrons. This number is called the mass number of the atom used for atomic number.
(symbol A). The mass number of an atom is the
Hydrogen atoms, with one proton and no neutrons, have a mass number number of protons plus neutrons in
of 1. Lithium atoms, with 3 protons and 4 neutrons, have a mass number its nucleus. Protons and neutrons are
of 7 and aluminium atoms, with 13 protons and 14 neutrons, have a mass sometimes called nucleons, so the term
number of 27. ‘nucleon number’ is an alternative to
mass number.
There is an agreed shorthand for showing the mass number and atomic
number of an atom. This is shown for a potassium atom, 39 19K, in Figure 1.9.
Ions can also be represented using this shorthand. For example, the potassium
39K+.
ion can be written as 19
K
39
Test yourself mass
number
8 Use Figure 1.8, and the information in the caption, to estimate the
diameter of a gold atom in nanometres.
9 How many protons, neutrons and electrons are there in the following
atoms and ions:
a) 94Be
235U
b) 39
19 K
19 F –
atomic
number 19
c) 92 d) 9 Figure 1.9 The mass number and atomic
e) 40 Ca2+? number can be shown with the symbol of
20
an atom.
10 Write symbols showing the mass number and atomic number for
these atoms and ions:
a) an atom of oxygen with 8 protons, 8 neutrons and 8 electrons
b) an atom of argon with 18 protons, 22 neutrons and 18 electrons
c) an ion of sodium with a 1+ charge and a nucleus of 11 protons
and 12 neutrons
d) an ion of sulfur with a 2− charge and a nucleus with 16 protons
and 16 neutrons.
1.3 Comparing the masses of
atoms – mass spectrometry
Individual atoms are far too small to be weighed, but in 1919 F.W. Aston
invented the mass spectrometer. This gave scientists an accurate method of
comparing the relative masses of atoms and molecules. Since its invention,
mass spectrometry has been developed into a sophisticated technique for
chemical analysis based on a variety of types of instrumentation.
A mass spectrometer separates atoms and molecules according to their mass,
and also shows the relative numbers of the different atoms and molecules
present. Figure 1.10 shows a schematic diagram of a mass spectrometer.
1.3 Comparing the masses of atoms – mass spectrometry 17
469983_01_Chem_Y1-2_012-[Link] 17 13/04/19 10:15 PM
Ionisation of the Mass analyser Ion detector giving an
Gaseous sample sample by separating ions by electrical signal which is
bombardment mass-to-charge converted to a digital
from inlet system with electrons or ratio, e.g. by magnetic response that is stored in
other methods field or time of flight a computer
Figure 1.10 A schematic diagram to show
the key features of a mass spectrometer.
Before atoms, or molecules, can be separated and detected in a mass
spectrometer, they must be converted to positive ions in the gaseous or
vapour state. This can be done in various ways. In some mass spectrometers,
a beam of high-energy electrons bombards the atoms or molecules of the
sample. This turns them into ions by knocking out one or more electrons.
e− + X → X+ + e− + e−
fast-moving atom in sample positive electron knocked slower-moving
electron vapour ion out of X electron
Inside a mass spectrometer there is a high vacuum. This allows ionised
atoms or molecules from the chemical being tested to be studied without
interference from atoms and molecules in the air.
Key term After ionisation, the charged species are separated to produce the mass
spectrum, which distinguishes the positive ions on the basis of their mass-
The mass-to-charge ratio (m/z) is to-charge ratios.
the ratio of the relative mass, m, of There are various types of mass spectrometer. They differ in the method
an ion to its charge, z, where z is the used to separate ions with different ratios of mass to charge. One type uses an
number of charges (1, 2 and so on). electric field to accelerate ions into a magnetic field, which then deflects the
Spectrometers usually operate so that ions onto the detector. A second type accelerates the ions and then separates
most ions produced have the value them by their flight time through a field-free region. A third type, the so-
of z = 1. called transmission quadrupole instrument, is now much the most common
because it is very reliable, compact and easy to use. It varies the fields in the
instrument in a subtle way to allow ions with a particular mass-to-charge
ratio to pass through to the detector at any one time.
Relative abundance
The output from the detector of a mass spectrometer is often presented as
a ‘stick diagram’. This shows the strength of the signal produced by ions
of varying mass-to-charge ratio. The scale on the vertical axis shows the
relative abundance of the ions. The horizontal axis shows the m/z values.
Each of the four peaks on the mass spectrum of lead in Figure 1.11 represents
a lead ion of different mass, and the heights of the peaks give the proportions
204 206 207 208 of the ions present.
Mass-to-charge ratio (m/z)
Figure 1.11 A mass spectrum of the Test yourself
element lead. The lead ions that produce
the peaks in the mass spectrum are all 1+ 11 Look carefully at Figure 1.11.
ions formed by ionising atoms in a lead a) How many different ions are detected in the mass spectrum of
vapour at very low pressure. The lead ions lead?
that form under these conditions are not b) What are the relative masses of these different ions?
the same as the stable lead ions normally
c) Make a rough estimate of the relative proportions of these
found in solid lead compounds or in
different ions in the sample of lead.
solutions.
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1.4 Isotopes and relative
isotopic masses
Mass spectrometer traces, like that in Figure 1.11, show that lead and most
other elements contain atoms that are not exactly alike. When atoms of
Relative abundance
these elements are ionised in a mass spectrometer, the ions separate and are
detected as two or more peaks with different values of m/z. This shows that
the atoms from which the ions formed must have different relative masses.
These atoms of the same element with different masses are called isotopes.
Look closely at Figure 1.12. It shows a mass spectrometer print out (mass
spectrum) for magnesium. The three peaks show that magnesium consists
of three isotopes with relative masses of 24, 25 and 26. These relative masses 23 24 25 26
are best described as relative isotopic masses because they give the relative Mass-to-charge ratio (m/z)
mass of particular isotopes. Figure 1.12 A mass spectrum for
Chemists originally measured the relative masses of atoms relative magnesium.
to hydrogen. Then, because of the existence of isotopes, it became
necessary to choose one particular isotope as the standard. Today, Key terms
the isotope carbon-12 (126C) is chosen as the standard and given a relative
mass of exactly 12. Isotopes are atoms of the same
The heights of the peaks in Figure 1.12 show the relative proportions of the element which have the same number
three isotopes. The isotope magnesium-24 has a mass number of 24 with of protons in the nucleus but a different
12 protons and 12 neutrons, whereas magnesium-25 has a mass number of number of neutrons. So isotopes have
25 with 12 protons and 13 neutrons. Table 1.2 summarises the important the same atomic number but different
similarities and differences in isotopes. mass numbers.
Relative isotopic mass is the mass of
Table 1.2 Similarities and differences in isotopes. one atom of an isotope relative to 121 th
of the mass of an atom of the isotope
Isotopes have the same Isotopes have different
carbon-12. The values are relative so
• number of protons • numbers of neutrons they do not have units.
• number of electrons • mass numbers
• atomic number • physical properties Relative atomic mass, Ar, is the
• chemical properties average mass of an atom of an element
1
relative to 12 th of the mass of an atom
of the isotope carbon-12. The values
Relative atomic masses are relative so they do not have units.
The relative atomic mass of an element is the average mass of an atom of the
element relative to one twelfth the mass of an atom of the isotope carbon-12.
The symbol for relative atomic mass is Ar, where ‘r’ stands for relative.
H=1
average mass of an atom of the element He=4
relative atomic mass = 1
H=1 H=1 H=1
12 × the mass of one atom of carbon-12
Using this scale, the relative atomic mass of hydrogen is 1.0, that of helium
is 4.0, and that of oxygen is 16.0. This can be written as: Ar(H) = 1.0,
Ar(He) = 4.0 and Ar(O) = 16.0, or simply H = 1.0, He = 4.0 and Cl = 35.5
for short (Figure 1.13). Figure 1.13 If atoms could be weighed, the
The values of relative atomic masses have no units because they are relative. scales would show that helium atoms are
The relative atomic masses of all elements are shown in the Periodic Table four times as heavy as hydrogen atoms.
on page 652.
1.4 Isotopes and relative isotopic masses 19
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The accurate relative atomic masses of most elements in tables of data are
35Cl 35Cl
not whole numbers. This is because these elements contain a mixture of
isotopes. For example, chlorine contains two isotopes, chlorine-35 and
chlorine-37, in the relative proportions of 3 : 1 (Figure 1.14). This is 34 , or
35Cl 37Cl
75%, chlorine-35 and 14 , or 25%, chlorine-37.
So, the average mass of a chlorine atom on the 12C scale is given by:
Figure 1.14 On average, for every four
chlorine atoms, three are chlorine-35 and (3 × 35) + (1 × 37)
one is chlorine-37. = 35.5
4
This is the relative atomic mass of chlorine.
Tip
The relative masses of individual isotopes are called relative isotopic masses,
whereas the relative masses of the atoms in an element (often containing a mixture of
isotopes) are called relative atomic masses.
Example
The mass spectrum of magnesium (Figure 1.12) shows that it consists of
three isotopes with these percentage abundances:
magnesium-24: 78.6%
magnesium-25: 10.1%
magnesium-26: 11.3%
Calculate the relative atomic mass of magnesium.
Notes on the method
The relative atomic mass of magnesium is an average value that takes
into account the relative masses of its isotopes and their relative
abundance. It is a ‘weighted’ average (Section A1.4).
The percentages show you how many atoms of each isotope are present
in a sample of 100 atoms.
Answer
The total relative mass of 100 atoms of magnesium
= (78.6 × 24) + (10.1 × 25) + (11.3 × 26) = 2432.7
The average relative mass of a magnesium atom = 2432.7 ÷ 100 = 24.3
(to three significant figures)
Tip
The values for Ar are average values for the mixture of isotopes found naturally. This
means that the values of relative atomic masses are often not whole numbers. In
calculations you should use Ar values to one decimal place, as in the Periodic Table on
page 652.
20 1 Atomic structure and the Periodic Table
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Test yourself
12 Look up the values of relative atomic masses in the Periodic
Table on page 652. How many times heavier (to the nearest whole
number) are:
a) C atoms than H atoms
b) Mg atoms than C atoms
c) S atoms than He atoms
d) C atoms than He atoms
e) Fe atoms than N atoms?
13 Silicon consists of three naturally occurring isotopes, 28
14 Si (93.0%),
29 Si (5.0%) and 30 Si (2.0%).
14 14
a) How many protons and neutrons are present in the nuclei of each
of these isotopes?
b) What is the relative atomic mass of silicon?
14 Neon has two isotopes with mass numbers of 20 and 22.
a) How do you think the boiling temperature of neon-20 compares
with that of neon-22? Explain your answer.
b) Neon in the air contains 90% neon-20 and 10% neon-22. What is
the relative atomic mass of neon in the air?
15 Why do isotopes have the same chemical properties, but different
physical properties?
Relative molecular and formula masses Key terms
Relative atomic masses can also be used to compare the masses of different
molecules. The relative masses of molecules are called relative molecular The relative molecular mass of an
masses (symbol Mr). element or compound is the sum of the
relative atomic masses of all the atoms
The relative molecular mass of an element or compound is the sum of the
in its molecular formula.
relative atomic masses of all the atoms in its molecular formula.
The relative formula mass of a
For oxygen, O2, Mr(O2) = 2 × Ar(O) = 2 × 16.0 = 32.0 compound is the sum of the relative
and for sulfuric acid, M r(H2SO4) = 2 × Ar(H) + Ar(S) + 4 × Ar(O) atomic masses of all the atoms in its
= (2 × 1.0) + 32.1 + (4 × 16.0) = 98.1 formula.
Metal compounds consist of giant structures of ions and not molecules. To
avoid the suggestion that their formulae represent molecules, chemists use
the term relative formula mass (symbol Mr), not relative molecular mass, Tip
for ionic compounds and for other compounds with giant structures such as
silicon dioxide, SiO2. Section A1.1 of Appendix A1 on
page 626 gives advice on how to
For magnesium nitrate, work out the value of maths equations
Mr(Mg(NO3)2) = Ar(Mg) + 2 × [Ar(N) + 3 × Ar(O)] with brackets and combinations of
multiplication and addition.
= 24.3 + 2 × (14.0 + 48.0) = 148.3
1.4 Isotopes and relative isotopic masses 21
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Mass spectrometers can also be used to study molecules (Chapter 7). After
Relative abundance
injecting a sample into the instrument and vaporising it, bombarding
43
electrons not only ionise the molecules but also break them into fragments.
Because of the high vacuum inside the mass spectrometer, it is possible to
29 study these molecular fragments and ions which do not normally exist. As a
15 58 result the mass spectrum consists of a ‘fragmentation pattern’ (Figure 1.15).
When analysing molecular compounds, the peak of the ion with the highest
0 10 20 30 40 50 60 mass is usually the whole molecule ionised. So the mass of this ‘parent ion’ or
Mass-to-charge ratio (m/z) ‘molecular ion’, M+, is the relative molecular mass of the compound.
Figure 1.15 The mass spectrum of a
hydrocarbon and its fragments. e – + M(g) ⎯⎯→ M+ + e− + e –
high-energy molecule in molecular ion electrons
electron sample
Test yourself
16 What is the relative molecular mass of:
a) chlorine, Cl2
b) sulfur, S8
c) ethanol, C2H5OH
d) tetrachloromethane, CCl4?
17 What is the relative formula mass of:
a) magnesium chloride, MgCl2
b) iron(iii) oxide, Fe2O3
c) hydrated copper(ii) sulfate, CuSO4.5H2O?
18 Look carefully at Figure 1.15.
a) What is the relative molecular mass of the hydrocarbon?
b) The fragment of the hydrocarbon with relative mass 15 is a CH3
group. What do you think the fragments are with relative masses
of 29 and 43?
c) Draw a possible structure for the hydrocarbon.
Notice that, by carefully interpreting the data from mass spectrometers,
chemists can deduce:
● the isotopic composition of elements
● the relative atomic masses of elements
● the relative molecular masses of compounds.
Chemists who separate and synthesise new compounds can also identify
the fragments in the mass spectra of these compounds. Then, by piecing
the fragments together, they can identify possible structures for the new
compounds.
The combination of gas chromatography and mass spectrometry is
particularly important in modern chemical analysis. Chromatography is first
used to separate the chemicals in an unknown mixture, such as polluted
water or similar compounds synthesised for possible use as drugs. Then mass
spectrometry is used to detect and identify the separated components.
22 1 Atomic structure and the Periodic Table
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Activity
Mass spectrometry in sport
Mass spectrometry provides an incredibly sensitive method of Great care is taken during sampling, transport, storage and
analysis in areas such as space research, medical research, analysis to ensure that the results of analysis will stand up in
monitoring pollutants in the environment and the detection of court.
illegal drugs in sport.
Figure 1.17 shows the molecular ion and the largest fragments
Detecting the use of anabolic steroids in sport in the mass spectrum of a banned chemical that is thought to
Since the 1980s, unscrupulous sportsmen and sportswomen be dihydrocodeine (C18H23O3N).
have tried to improve their performance by using anabolic
steroids. These drugs increase muscle size and strength, which 301
increases the chance of winning (Figure 1.16). But anabolic
steroids also have serious harmful effects on the body. Women
Relative abundance
develop masculine features and anyone using them may suffer
heart disease, liver cancer and depression leading to suicide.
258 270 284
0 250 260 270 280 290 300
Mass-to-charge ratio (m/z)
Figure 1.17 The molecular ion and the largest fragments in the mass
spectrum of a banned chemical.
1 What is the probable relative molecular mass of the banned
chemical on the mass spectrum?
2 Is the probable relative molecular mass consistent with that
Figure 1.16 Ben Johnson won the men’s 100 m race at the Olympic of dihydrocodeine, (C18H23O3N)? Explain your answer.
Games in 1992. Unfortunately, urine tests showed that he had used 3 What is the relative mass of the fragment lost from one
anabolic steroids – Johnson was stripped of his title and the gold molecule of the banned substance, leaving the fragment of
medal. relative mass 284?
Sporting bodies, such as the International Olympic Committee, 4 Dihydrocodeine contains a CH3O– group and an –OH group.
have banned the use of anabolic steroids in all sports and have What evidence does the mass spectrum provide for these
introduced a rigorous testing regime. The testing procedures two groups?
involve analysis of urine samples using mass spectrometry.
1.5 Evidence for the electronic
structure of atoms
In a mass spectrometer, a beam of electrons can be used to bombard the
sample, turning atoms (or molecules) into positive ions. The electrons in the
beam must have enough energy to knock electrons off atoms in the sample.
By varying the intensity of the beam, it is possible to measure the minimum
amount of energy needed to remove electrons from the atoms of an element.
From these measurements, scientists can predict the electron structures of
atoms.
1.5 Evidence for the electronic structure of atoms 23
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The energy needed to remove one electron from each atom in one mole of
Key terms gaseous atoms is known as the first ionisation energy. The product is one
mole of gaseous ions with one positive charge.
An ionisation energy is the energy
needed to remove one mole of So, the first ionisation energy of sodium is the energy required for the process
electrons from one mole of gaseous
Na(g) → Na+(g) + e – first ionisation energy = +496 kJ mol−1
atoms, or ions, of an element.
Atomic energy levels are the energies Ionisation energies like this are always endothermic. Energy is taken in by
of electrons in atoms. According to the reaction so the energy change is given a positive sign.
quantum theory, each electron in an Scientists can also determine ionisation energies by using a spectroscope to
atom has a definite energy. When study the light given out by atoms when heated in a flame (as in a flame test).
atoms gain or lose energy, the electrons The spectroscope shows up a series of bright lines (Figure 1.18). Heating the
jump from one energy level to another. atoms gives them energy which makes some of the electrons jump to higher
energy levels. Each line in the spectrum arises from the energy given out as
the electrons drop back from a higher energy level to a lower level.
Tip
The shells of electrons at fixed or
specific levels are sometimes called
quantum shells. The word ‘quantum’
is used to describe something related
to a fixed amount or a fixed level.
Figure 1.18 The line spectrum of hydrogen in the visible region of the electromagnetic
spectrum.
Using data from spectra, it is possible to measure the energy required to
remove electrons from ions with increasing charges. A succession of ionisation
energies is obtained. For example:
Na(g) → Na+(g) + e− first ionisation energy = +496 kJ mol−1
Na+(g) → Na2+(g) + e− second ionisation energy = +4563 kJ mol−1
Na2+(g) → Na 3+(g) + e− third ionisation energy = + 6913 kJ mol−1
There are 11 electrons in a sodium atom so there are 11 successive ionisation
energies for this element.
The successive ionisation energies for an element get bigger and bigger. This
is not surprising because, having removed one electron, it is more difficult to
remove a second electron from the positive ion formed.
The graph in Figure 1.19 provides evidence to support the theory that the
electrons in an atom are arranged in a series of levels or shells around the
nucleus.
Tip
Logarithms reduce the range of numbers that vary over several orders of magnitude.
Figure 1.19 uses logarithms which work like this: log 10 = 1, log 100 = 2, log 1000 = 3
and so on. A calculator can be used to find the values of the logarithms (log) of other
numbers.
24 1 Atomic structure and the Periodic Table
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Notice the big jumps in value between the first and second ionisation energies,
and between the ninth and tenth ionisation energies in Figure 1.19. This
Log ionisation energy
suggests that sodium atoms have one electron in an outer shell or energy level
furthest from the nucleus. This outer electron is relatively easily removed
because it is shielded from the full attraction of the positive nucleus by 10
inner electrons.
Below this outer single electron, sodium atoms appear to have eight electrons
in a second shell, all at roughly the same energy level. These eight electrons
are closer to the nucleus than the single outer electron.
0 5 10
Finally, sodium atoms have two inner electrons in a shell closest to the Number of electrons removed
nucleus. These two electrons feel the full attraction of the positive nucleus
Figure 1.19 Log ionisation energy against
and are hardest to remove with the most endothermic ionisation energies.
the number of electrons removed for
This electronic structure for a sodium atom can be represented in an energy level sodium. The values for the ionisation
diagram as in Figure 1.20. The electron arrangement in sodium can sometimes energies range from 496 kJ mol−1 to
be written simply as 2, 8, 1 (but see Section 1.6). 159 079 kJ mol−1. Plotting the logarithms of
these values makes it possible to fit them
In energy level diagrams such as that in Figure 1.20, the electrons are
on to the vertical axis, while still showing
represented by arrows. When an energy level is filled, the electrons are paired
where there are big jumps in the values.
up and in each of these pairs the electrons are spinning in opposite directions.
Chemists have found that paired electrons can only be stable when they spin
in opposite directions so that the magnetic attraction resulting from their
opposite spins can counteract the electrical repulsion from their negative
Key term
charges.
In energy level diagrams such as Figure 1.20, the opposite spins of the paired Shielding is an effect of inner electrons
electrons are shown by drawing the arrows in opposite directions. which reduces the pull of the nucleus
on the electrons in the outer shell of an
The quantum shells of electrons correspond to the periods of elements atom. Thanks to shielding, the electrons
in the Periodic Table. By noting where the first big jump comes in the in the outer shell are attracted by an
successive ionisation energies of an element, it is possible to predict the ‘effective nuclear charge’ which is less
group to which the element belongs. For example, the first big jump in the than the full charge on the nucleus.
successive ionisation energies for sodium comes after the first electron is
removed. This suggests that sodium has just one electron in its outermost
shell and, therefore, it must be in Group 1. Highest energy
level – electron
easily removed
Test yourself
Intermediate
19 Write equations to represent: energy level –
electrons harder
a) the second ionisation energy of calcium to remove
b) the third ionisation energy of aluminium.
Lowest energy
20 The successive ionisation energies of beryllium are 900, 1757, level – electrons
14 849 and 21 007 kJ mol−1. hardest to remove
a) What is the atomic number of beryllium? Figure 1.20 The energy levels of electrons
b) Why do successive ionisation energies always get more in a sodium atom.
endothermic?
c) Draw an energy level diagram for the electrons in beryllium, and
predict its electron structure.
d) To which group in the Periodic Table does beryllium belong?
1.5 Evidence for the electronic structure of atoms 25
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Activity
Evidence for sub-shells of electrons
By studying the first ionisation energies of successive elements 2 When you have plotted the points, draw lines from one point
in the Periodic Table, it is possible to compare how easy it to the next to show a pattern of peaks and troughs. Label
is to remove an electron from the highest energy level in the each point with the symbol of its corresponding element.
atoms of these elements. This provides us with evidence for the 3 a) Where do the alkali metals in Group 1 appear in the
arrangement of electrons in sub-shells. pattern?
b) Where do the noble gases in Group 0 appear in the
1 Refer to the data sheet for Chapter 1, ‘The first ionisation
pattern?
energies of successive elements in the Periodic Table’, which
4 What similarities do you notice in the pattern for elements in
you can access online at [Link]/
Period 2 (lithium to neon) with that for elements in Period 3
EdexcelChemistry. Using this data, plot a graph of the first
(sodium to argon)?
ionisation energy for the first 20 elements in the Periodic
5 Identify three sub-groups of points in both Period 2 and
Table. Put first ionisation energy on the vertical axis and
Period 3. How many elements are there in each sub-group?
atomic number on the horizontal axis.
1.6 Electrons in energy levels
From the study of ionisation energies and spectra, scientists have found that
the electrons in atoms are grouped together in energy levels or quantum
shells. The numbers 1, 2, 3, etc. are used to label these main shells, starting
nearest to the nucleus.
Each quantum shell can hold only a limited number of electrons:
● the n = 1 shell can hold 2 electrons
● the n = 2 shell can hold 8 electrons
● the n = 3 shell can hold 18 electrons
● the n = 4 shell can hold 32 electrons.
These main shells divide into sub-shells labelled s, p, d and f. The labels
s, p, d and f are left over from the early studies of the spectra of different
elements. These studies used the words ‘sharp’, ‘principal’, ‘diffuse’ and
‘fundamental’ to describe different lines in the spectra. The terms have no
special significance now.
The sub-shells are further divided into atomic orbitals (Figure 1.21). Each
Key term orbital is defined by its:
Atomic orbitals are the sub-divisions of ● energy level
the electron shells in atoms. The main ● shape
shells divide into sub-shells labelled s, ● direction in space.
p, d and f. The sub-shells are further
The shapes and directions in space of the atomic orbitals are found by
divided into atomic orbitals. An orbital
calculating the probability of finding an electron at any point in an atom. These
is a region in space around the nucleus
calculations are based on a theoretical model described by the Schrödinger
of an atom in which there is a 95%
wave equation. The one orbital in the first shell is spherical. It is an example
chance of finding an electron, or a pair
of an s orbital (1s). The four orbitals in the second shell are made up of one
of electrons with opposite spins.
s orbital (2s) and three dumbbell-shaped p orbitals. The three p orbitals (2px,
2py, 2pz) are arranged at right angles to each other along the x-, y- and z-axes
(Figure 1.22).
26 1 Atomic structure and the Periodic Table
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Energy
4f
4d
n=4
4p
4s 3d
3p
n=3
3s
2p
2s
n=2
1s
n=1
Figure 1.21 The energies of atomic orbitals in atoms. The terms ‘energy level’
and ‘orbital’ are often used interchangeably. In a free atom the orbitals in a
sub-shell have the same energy.
y y y y
z z z z
x x x x
boundary of sphere
within which there
nucleus is a greater than
at origin 95% chance of 2px 2py 2pz
finding an electron
s orbital p orbitals
Figure 1.22 The shapes of s and p atomic orbitals. The density of shading indicates the
probability of finding an electron at any point.
The electrons in an atom fill the energy levels according to a set of rules
which determine electron arrangements in atoms.
The three rules are:
● electrons go into the orbital with the lowest available energy level first
● each orbital can only contain at most two electrons (with opposite spins) Key term
● where there are two or more orbitals at the same energy, they fill singly
before the electrons pair up. The electronic configuration of an
element describes the number and
The application of these rules is illustrated for the atoms of four elements in arrangement of electrons in an atom
Figure 1.23. These descriptions of the arrangement of electrons in the atoms of the element. A shortened form
of elements are called electronic configurations. Chemists sometimes of electronic configuration uses the
use the term ‘auf bau principle’ for these rules from the German word symbol of the previous noble gas, in
meaning ‘build up’. This is a reminder that electron configurations build up square brackets, to stand for the inner
from the bottom. There are several common conventions for representing shells. So, using this convention, the
electron configurations in a shorthand way. Figure 1.24, for example, shows electronic configuration of sodium is
the electrons-in-boxes representations and the s, p, d, f notations for the [Ne]3s1.
electronic structures of beryllium, nitrogen and sodium.
1.6 Electrons in energy levels 27
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Energy
Energy
3d 3d
3p 3p
3s 3s
2p 2p
2s 2s
1s 1s
hydrogen, 1s1 carbon, 1s2 2s2 2p2
Energy
Energy
3d 3d
3p 3p
3s 3s
2p 2p
2s 2s
1s 1s
sodium, 1s2 2s2 2p6 3s1 sulfur, 1s2 2s2 2p6 3s2 3p4
Figure 1.23 Electrons in energy levels for four atoms to show the application of the building-up principle.
Element Electrons-in-boxes notation of electronic structure s,p,d,f electron
notation
1s 2s 2p 3s
Beryllium 1s2 2s2
Nitrogen 1s2 2s2 2p3
Sodium 1s2 2s2 2p6 3s1
Figure 1.24 Electrons-in-boxes representations and s, p, d, f notations for the electronic
structure of beryllium, nitrogen and sodium.
Test yourself
21 Sketch a graph of log ionisation energy against number of electrons
removed when all the electrons are successively removed from a
phosphorus atom. (Sketch the graph in the style of Figure 1.19.
There is no need to look up logarithms.)
22 Write out the electron structure in terms of shells (for sodium this
would be 2, 8, 1) for the atoms of following elements:
a) lithium b) oxygen
c) neon d) silicon.
28 1 Atomic structure and the Periodic Table
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23 Write the electronic sub-shell structure for the elements in
Question 22 – for sodium this would be 1s22s22p63s1.
24 Draw the electrons-in-boxes representations for the following
elements:
a) boron b) fluorine
c) phosphorus d) potassium.
25 Identify the elements with the following electron structures in their
outermost shells:
a) 1s2 b) 2s22p2
c) 3s2 d) 3s23p4.
The development of knowledge and understanding about electronic
structures illustrates how chemists use the results of their experiments, such
as the measurements of ionisation energies, to devise atomic models that they
can use to explain the properties of elements. It also illustrates the important
distinction between evidence and experimental data on the one hand, and
ideas, theories and explanations on the other.
In particular, ionisation energies and spectra have provided chemists with
evidence and information that has caused them to develop and modify their
models and theories about electron structure. Early ideas about electrons
arranged in shells have been developed to take in the evidence for sub-shells,
and then modified to include ideas about orbitals.
1.7 Electron structures and the
Periodic Table
The Periodic Table helps chemists to bring order and patterns to the vast
amount of information they have discovered about all the elements and their
compounds.
In the modern Periodic Table, elements are arranged in order of atomic
number. The horizontal rows in the table are called periods – each period Key terms
ends with a noble gas. The vertical columns in the table are called groups
which can be divided into four blocks – the s block, p block, d block and A period is a horizontal row of elements
f block – based on the electron structures of the elements (Figure 1.25). in the Periodic Table.
So, the modern arrangement of elements in the Periodic Table reflects the A group is a vertical column of
underlying electronic structures of the atoms, while the more sophisticated elements in the Periodic Table.
model of electron structure in terms of orbitals allows chemists to explain Elements in the same group have
the properties of elements more effectively. The four blocks in the Periodic similar properties because they have
Table are shown in different colours in Figure 1.25. the same outer electronic configuration.
● The s block comprises the reactive metals in Group 1 and Group 2 – such
as potassium, sodium, calcium and magnesium. In these metals, the
outermost electron is in an s orbital in the outer shell.
● The p block comprises the elements in Groups 3, 4, 5, 6, 7 and 0 on the
right of the Periodic Table. These elements include relatively unreactive
metals such as tin and lead, plus all the non-metals. In these elements, the
last electron added goes into a p orbital in the outer shell.
1.7 Electron structures and the Periodic Table 29
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0
1 2 3 4 5 6 7 He
Li C O Ne
Na Mg Cl
K Ca Ti Fe Cu Zn Br
p block
s block Ag
d block
Ba La Au
Fr Ac
f block
U
Figure 1.25 The s, p, d and f blocks in the Periodic Table.
● The d-block elements occupy a rectangle across Periods 4, 5, 6 and 7
Tip between Group 2 and Group 3. The d-block elements are all metals –
The International Union of Pure and including titanium, iron, copper and silver – in which the last electron
Applied Chemistry (IUPAC) now added goes into a d orbital. These metals are much less reactive than the
recommends that the groups in the s-block metals in Groups 1 and 2. Within the d block there are marked
Periodic Table should be numbered similarities across the periods, as well as the usual vertical similarities. The
from 1 to 18. Groups 1 and 2 are the d-block elements are sometimes loosely called ‘transition metals’.
same as before. Groups 3 to 12 are the ● The f-block elements occupy a low rectangle across Periods 6 and 7 within
vertical families of d-block elements. the d block, but they are usually placed below the main table to prevent
The groups traditionally numbered 3 to it becoming too wide to fit the page. Like the d-block elements, those in
7 and 0 then become Groups 13 to 18. the f block are all metals. Here, the last added electron is in an f orbital.
The f-block elements are often called the lanthanoids and actinoids, or
lanthanides and actinides, because they are the 14 elements immediately
4p following lanthanum, La, and actinium, Ac, in the Periodic Table. Another
orbitals in name used for the f-block elements is the ‘inner transition elements’.
the 4th shell
As the shells of electrons around the nuclei of atoms get further from the
3d
4s nucleus, they become closer in energy (see Figure 1.21). Therefore, the
difference in energy between the second and third shells is less than that
3p between the first and second. When the fourth shell is reached there is, in
fact, an overlap between the orbitals of highest energy in the third shell
orbitals in (the 3d orbital) and that of lowest energy in the fourth shell (the 4s orbital)
the 3rd shell
(Figure 1.26). As a result the orbitals that fill in the fourth period are the 4s,
3s 3d and 4p orbitals in that order. This accounts for the position of the d-block
Figure 1.26 The relative energy levels of elements in the Periodic Table.
orbitals in the third and fourth shells.
Tip
The 4s orbital fills before the 3d orbital because it has a lower energy. However, the
4s orbital is the outer orbital and it is the electrons in the 4s orbital that are lost
first when a d-block element ionises. Chromium and copper each only have one 4s
electron in their atoms. The explanation for the irregularities lies in the stability of half-
filled and filled sub-shells. So the electronic structure of chromium is [Ar]3d54s1 and
that of copper is [Ar]3d104s1.
Table 1.3 shows the electron configurations of four elements in the fourth
period. The rules for the order in which electrons fill orbitals still apply.
30 1 Atomic structure and the Periodic Table
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Table 1.3 Electron configurations of
Electronic structure
Element four elements in the fourth period. [Ar]
and symbol spdf notation Electrons-in-boxes notation represents the electronic configuration of
3d 4s 4p argon: 1s22s22p63s23p6.
Potassium
[Ar]4s1 [Ar]
K
Vanadium [Ar]3d34s2 [Ar]
V
Iron [Ar]3d64s2 [Ar]
Fe
Bromine [Ar]3d104s24p5 [Ar]
Br
Test yourself
26 Write the electronic sub-shell structure for the atoms of these
elements using spdf notation:
a) scandium b) manganese
c) zinc d) germanium.
27 Identify the elements with the following electron structures:
a) 1s22s22p63s23p64s2
b) 1s22s22p63s23p63d84s2
c) 1s22s22p63s23p63d104s24p2
28 Write the electronic sub-shell structure for these ions using spdf
notation:
a) Al3+ b) S2−
c) Zn2+ d) Br −
Groups
The elements in each group have similar properties because they have similar
Group1
electron structures. This important point is well illustrated by the alkali
metals in Group 1. Look at Figure 1.27 – notice that each alkali metal has The alkali metals
one s electron in its outer shell. This similarity in their electron structures
Lithium
explains why they have similar properties. Li
2, 1
Alkali metals: (1s2 2s1)
● are very reactive because they lose their single outer electron so easily
Sodium
● form ions with a charge of 1+ (Li+, Na+, K+, etc.) so the formulae of their Na
compounds are similar 2, 8, 1
● form very stable ions with an electron structure like that of a noble gas. (1s2 2s2 2p6 3s1)
The chemical properties of all other elements are also determined by their Potassium
K
electronic structures. Chemistry is largely about the electrons in the outer 2, 8, 8, 1
shells of atoms. The reactivity of an element depends on the number of (1s2 2s22p6 3s23p6 4s1)
electrons in the outer shell and how strongly they are held by the nuclear
charge. This is a fundamental feature of chemistry and an essential principle Figure 1.27 Electron structures of the first
which governs the way in which chemists think and work. three alkali metals.
1.7 Electron structures and the Periodic Table 31
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Test yourself
29 Why are sodium and potassium so alike?
30 Why are the noble gases so unreactive?
31 a) Write down the electron shell structures and sub-shell structures
of fluorine and chlorine in Group 7.
b) Why do you think fluorine and chlorine are so reactive with metals?
c) Why do the compounds of fluorine and chlorine with metals have
similar formulae?
1.8 Periodic properties
Modern versions of the Periodic Table are all based on the one suggested
by the Russian chemist Dmitri Mendeléev in 1869. When Mendeléev
arranged the elements in order of atomic mass, he saw repeating patterns in
their properties. A repeating pattern is a periodic pattern – hence the terms
‘periodic properties’ and ‘periodicity’.
Perhaps the most obvious repeating pattern in the Periodic Table is from metals
on the left, through elements with intermediate properties (called metalloids), to
non-metals on the right. Graphs of the physical properties of the elements – such
as melting temperatures, electrical conductivities and first ionisation energies –
against atomic number, also show repeating patterns. Using the models of
bonding between atoms and molecules, chemists can explain the properties
of elements and the repeating patterns in the Periodic Table.
Melting temperatures of the elements
Figure 1.28 shows the periodic pattern revealed by plotting the melting
temperatures of elements against atomic number.
Figure 1.28 Periodicity in the melting
temperatures of the elements. C
3000
Melting temperature/°C
2000
Si
Be
1000
Mg
Li Na
0
Ne Ar
–250
3 4 5 6 3 8 9 10 11 12 13 14 15 16 17 18
Atomic number
32 1 Atomic structure and the Periodic Table
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The melting temperature of an element depends on both its structure and the
type of bonding between its atoms. In metals, the bonding between atoms
is strong (Section 2.9), so their melting temperatures are usually high. The
more electrons each atom contributes from its outermost shell to the shared
delocalised electrons, the stronger the bonding and the higher the melting
temperature.
Therefore, melting temperatures rise from Group 1 to Group 2 to Group 3.
In Group 4, the elements carbon and silicon have giant covalent structures.
The bonds in these structures are strong and highly directional, so most of
the bonds must break before the solid melts. This means that the melting
temperatures of Group 4 elements are very high and at the peaks of the graph
in Figure 1.28.
The non-metal elements in Groups 5, 6, 7 and 0 form simple molecules. The
intermolecular forces between these simple molecules are weak, so these
elements have low melting temperatures (Section 2.3).
First ionisation energies of the elements
Figure 1.29 shows the clear periodic trend in the first ionisation energies of
the elements. The general trend is that first ionisation energies increase from
left to right across a period.
2500 He
Ne
2000
First ionisation energy/kJ mol –1
F
Ar
1500 N Group 0
H
O Cl
C P
1000 Be S
Mg Si
B Ca
500 Al Group 2
Li Na Group 1
K
0
1 5 10 15 20
Atomic number
Figure 1.29 Periodicity in the first ionisation energies of the elements.
The ionisation energy of an atom is determined by three atomic properties.
● The size of the positive nuclear charge. As the positive nuclear charge increases,
its attraction for outermost electrons increases and this tends to increase
the ionisation energy.
● The distance of the outermost electron from the nucleus. As this distance increases,
the attraction of the positive nucleus for the negative electron decreases
and this tends to reduce the ionisation energy.
1.8 Periodic properties 33
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● The shielding effect of electrons. Electrons in inner shells exert a repelling
effect on electrons in the outer shells of an atom. This reduces the pull of
the nucleus on the electrons in the outer shell. Thanks to shielding, the
‘effective nuclear charge’ attracting electrons in the outer shell is much
less than the full positive charge of the nucleus. As expected, the shielding
effect increases as the number of inner shells increases.
Moving from left to right across any period, the nuclear charge increases as
electrons are added to the same outer shell. The increasing nuclear charge
tends to pull the outer electrons closer to the nucleus. The shielding effect
of full inner shells is constant, the extra electrons in the same outer shell
do not shield each other well so shielding hardly changes across the period.
The increased nuclear charge and the reduced distance of the outer electrons
from the nucleus makes the outer electrons more difficult to remove and, in
general, the first ionisation energy increases.
But notice in Figure 1.29 that the rising trend in ionisation energies across
a period is not smooth. There is a 2-3-3 pattern, which reflects the way
in which electrons feed into s and p orbitals. The first ionisation energy
decreases from beryllium to boron and again from nitrogen to oxygen.
A beryllium atom loses one of the 2s2 electrons from its outer shell when it
ionises. The electronic configuration of boron is 2s22p1, so the electron lost
when a boron atom ionises is a 2p electron. The 2p electron is in a higher
energy sub-shell than a 2s electron, so it takes less energy to remove the
boron 2p electron, despite the increase in nuclear charge.
The electronic configuration of oxygen is 1s22s22p4. This means that one
of the paired 2p electrons is removed on ionisation. In a nitrogen atom the
electronic configuration is 1s22s22p3 and all three p electrons are unpaired.
Ionisation of nitrogen involves losing an unpaired electron. The repulsion
between the negative electrons is greater for the paired electrons in the
same sub-shell of an oxygen atom than between the unpaired electrons in
a nitrogen atom. As a result it is easier to ionise an oxygen atom despite the
increase in nuclear charge.
Test yourself
32 Why do the first ionisation energies of elements decrease with
increasing atomic number in every group of the Periodic Table?
34 1 Atomic structure and the Periodic Table
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Chapter summary
l The study of successive ionisation energies and
Chapter 1 Atomic structure and
spectra shows that electrons are grouped in energy
the Periodic Table levels, or quantum shells. These shells are divided
l The nucleus of an atom consists of protons (charge into sub-shells labelled s, p, d and f.
+1, relative mass 1) and neutrons (charge 0, relative l Each s sub-shell has one spherical s-orbital; each p
mass 1). sub-shell has three dumbbell-shaped p-orbitals.
l The atomic (proton) number gives the number of l Electrons in an atom fill energy levels according to
protons in the nucleus. The mass number is the rules: electrons go into the orbital with the lowest
number of protons plus the number of neutrons. available energy; each orbital can only contain at
l The isotopes of an element have the same atomic most two electrons; where there are two or more
number but different mass numbers. orbitals with the same energy, they fill singly
l Relative isotopic mass is the mass of one atom of before they pair up; electrons in the same orbital
an isotope relative to 1/12th of the mass of an atom have opposite spins.
of carbon-12. l In the Periodic Table, the elements are arranged in
l Mass spectrometry measures the relative mass and order of atomic number: the horizontal rows are
relative abundance of the isotopes of an element. periods and the vertical columns are groups.
The output is often presented as a ‘stick diagram’ l Electronic configuration determines the chemical
showing the relative abundance of the ions of properties of the elements, and the division of the
differing mass-to-charge ratios. Periodic Table into s, p and d blocks reflects the
l Data from a mass spectrum can be used to calculate underlying electron configurations of the atoms.
the relative atomic mass of the element, which l There are properties of elements that show
is the average mass of the atoms of the element repeating (periodic) patterns in the Periodic Table:
relative to 1/12th of the mass of an atom of these include atomic radii, melting and boiling
carbon-12. temperatures and first ionisation energies.
l The relative molecular mass of an element or l The melting temperature of an element depends on
compound is the sum of the relative atomic masses both its structure and the type of bonding between
of all the atoms in its molecular formula. its atoms.
l The mass spectrum of a molecular substance l The sizes of ionisation energies are related to the
consists of a fragmentation pattern in which number of protons in the nucleus, shielding and
the peak of the highest m/z value is that of the the electron sub-shell (orbital) from which the
‘molecular ion’, M+. The relative mass of this ion is electron is removed. These factors can account for
the relative molecular mass of the compound. the general rise of first ionisation energies across a
l An ionisation energy measures the energy needed period and the fall in first ionisation energies down
to remove one mole of electrons from one mole of a group.
gaseous atoms or ions.
Chapter summary 35
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Exam practice questions
Antimony has two main isotopes –
1 c) Copy and complete the electronic
antimony-121 and antimony-123. A configuration of an atom of 24Mg.
forensic scientist was asked to help a crime 1s2 _________(1)
investigation by analysing the antimony in a
bullet. This was found to contain 57.3% of 5 The table shows the melting temperatures of the
121Sb and 42.7% of 123Sb. elements in Period 3 of the Periodic Table.
a) State what is meant by the term ‘relative
atomic mass’. (3) Element Melting temperature/°C
b) Calculate the relative atomic mass of the Na 98
sample of antimony from the bullet. Write
Mg 649
your answer to an appropriate number of
significant figures. (3) Al 660
c) State one similarity and one difference, in Si 1410
terms of sub-atomic particles, between the
P 44
isotopes.(2)
S 119
This question concerns the following five
2
species: Cl −101
16 O2− 19 F− 20 23 Na 25Mg2+ Ar −189
8 9 10Ne 11 12
a) Identify two species that have the same The trend in the melting temperatures across
number of neutrons. (2) Period 3 and other periods is described as a
b) Identify two species that have the same periodic property.
ratio of neutrons to protons. (2) a) Give the general pattern in melting
c) Identify the species that does not have 10 temperatures across periods in the Periodic
electrons. (1) Table. (2)
b) Show that this general trend is related to the
3 a) Identify the elements with these electron
different types of elements. (1)
configurations as s-, p- or d-block
c) State what is meant by the term ‘periodic
elements.
property’. (2)
i) 1s22s22p63s2
d) State two other properties which can be
ii) 1s22s22p63s23p4
described as periodic in relation to the
iii) 1s22s22p63s23p63d64s2 (3)
Periodic Table. (2)
b) Give the electrons-in-boxes notation for the
electron configurations of: 6 The table shows the first and second ionisation
i) a nitrogen atom energies of lithium and sodium in Group 1 of
ii) a sodium ion the Periodic Table.
iii) a sulfide ion. (3)
Element First ionisation Second
4 The isotopes of magnesium, 24Mg, 25Mg and energy/kJ mol−1 ionisation
26Mg, can be separated by mass spectrometry.
energy/kJ mol−1
a) State what is meant by the term ‘isotope’.(2)
Lithium 520 7298
b) Copy and complete the table below to
show the composition of the 24Mg and Sodium 496 4563
26Mg isotopes. (2)
a) Write an equation, with state symbols, for
Protons Neutrons Electrons the second ionisation energy of sodium. (2)
24Mg b) Explain why the second ionisation energies
of lithium and sodium are larger than their
26Mg
first ionisation energies. (3)
36
1 Atomic structure and the Periodic Table
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c) Explain why the first and second ionisation ii) three lines with m/z values of 158,
energies of sodium are smaller than those of 160 and 162 with heights in the
lithium.(4) ratio 1 : 2 : 1. (3)
d) The first five successive ionisation energies, b) Chlorine consists of two isotopes
in kJ mol−1, of an element, X, in Period 3 chlorine-35 and chlorine-37. Explain why
of the Periodic Table are 578, 1817, 2745, the mass spectrum of chlorine includes:
11 578, 14 831. i) two lines with m/z values of 35 and 37
i) Identify element X. (1) with heights in the ratio 3 : 1 (2)
ii) Explain how you obtained your ii) three lines with m/z values of 70, 72
answer. (2) and 74 with heights in the
e) Predict which element in the Periodic Table ratio 9 : 6 : 1. (3)
has the highest first ionisation energy and c) Explain these features of the
explain your answer. (3) mass spectrum of dichloroethene,
C2H2Cl2 (Mr = 97):
7 The graph shows the first ionisation energies of
i) the absence of a peak at m/z = 97 (2)
the elements in Period 2 of the Periodic Table.
ii) the presence of three peaks at
1600 m/z values of 96, 98 and 100 with
First ionisation energy/kJ mol –1
1400 intensities in the ratio 9 : 6 : 1 (3)
1200 iii) the presence of two peaks at m/z values
1000
of 61 and 63 with intensities in the
ratio 3 : 1. (3)
800
600 9 This diagram shows the order in which sub-
400 shells are filled by electrons according to the
200 Aufbau principle which accounts for the
0 arrangement of elements in the modern form
Na Mg Al Si P S Cl Ar of the Periodic Table.
Element
1s
a) Describe and explain the general trend in
first ionisation energies from Na to Ar. (3) 2s 2p
b) Explain why aluminium, Al, has a lower first
3s 3p 3d
ionisation energy than magnesium, Mg. (2)
c) Explain why the ionisation energy decreases 4s 4p 4d 4f
from phosphorus, P, to sulfur, S. (2)
5s 5p 5d
d) Predict the value for the first ionisation
energy of potassium and explain your 6s 6p
answer. (2)
7s
e) The first five ionisation energies of an
element are 738, 1451, 7733, 10 541,
13 629 kJ mol−1. Explain why the element a) Show that this diagram accounts for the
cannot have an atomic number less position of the d-block elements in the
than 12. (3) Periodic Table. (2)
b) Give the electronic configuration of tin
8 a) Bromine consists of two isotopes
(atomic number 50). (1)
bromine-79 and bromine-81 which are
c) Explain why this diagram cannot account
equally abundant. Explain why the mass
for elements with atomic numbers greater
spectrum of bromine includes:
than 88. (2)
i) two lines with m/z values of 79 and 81
d) Predict the number of orbitals in the 4f sub-
with heights in the ratio 1 : 1 (3)
shell. Show how you decide on your answer.
(2)
37
Exam practice questions
469983_01_Chem_Y1-2_012-[Link] 37 13/04/19 10:15 PM
e) Comment on the relative energies of c) Explain why formulae OCl2 and FCl
the 4f and 5d orbitals given the electron are normally written as Cl2O and ClF,
configurations of the four elements with respectively. (1)
atomic numbers from 57 to 60. d) i) Describe the pattern in the boiling
temperatures of the chlorides of the
Lanthanum: [Xe]4f 05d16s2
elements in Periods 2 and 3. (2)
Cerium: [Xe]4f 25d06s2
ii) Explain the pattern you describe. (4)
Praseodymium: [Xe]4f 35d06s2
e) Phosphorus forms a second chloride, PCl5,
Neodymium: [Xe]4f 45d06s2
but nitrogen only forms the one chloride.
Show that this information, and the diagram Explain this difference in terms of the
above, can account for the position of the electron configurations of the atoms of
elements with atomic numbers 58–71 in phosphorus and nitrogen. (5)
the Periodic Table. (4)
11* Discuss the following statements using
10 The table below shows the groups, formulae examples to show the extent to which you
and boiling temperatures of chlorides for the think that they are true or false:
elements in Periods 2 and 3. a) The atomic number of an element is a better
a) Explain why are there no entries in the guide to its atomic structure and is more
table for Group 0. (2) useful in its classification than its relative
b) i) Describe the pattern shown by the atomic mass. (6)
formulae of the chlorides in Periods 2 b) The chemical properties of an element
and 3. (2) are largely determined by the number of
ii) Suggest an explanation for the pattern electrons in the outer shell of its atoms. (6)
you describe. (4)
Group
1 2 3 4 5 6 7
Period 2
Formula of chloride LiCl BeCl2 BCl3 CCl4 NCl3 OCl2 FCl
Boiling temperature of chloride/°C 1340 520 13 77 71 4 −101
Period 3
Formula of chloride NaCl MgCl2 AlCl3 SiCl4 PCl3 S2Cl2 Cl2
Boiling temperature of chloride/°C 1413 1412 423 58 76 136 −35
38
1 Atomic structure and the Periodic Table
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Bonding and structure
2
2.1 Investigating structure
and bonding
The word ‘structure’ has different levels of meaning in science. On a grand
scale, engineers design the structures of buildings and bridges; on the smallest
scale, chemists and physicists explore the inner structure of atoms. Not
surprisingly, scientists use different models and different theories to explain
the structure and properties of materials at these different levels.
Scientists have developed increasingly sophisticated models to account for
the structure, bonding and properties of materials as their knowledge has
increased. No single model can be used to explain the properties of elements
and compounds at all levels. Each has its advantages and its limitations and
particular models are more appropriate in different contexts.
In this topic, crystal structures are best explained using Dalton’s model of
atoms and ions as discreet, tiny spheres. Metallic, ionic and covalent bonding
are best explained using the model of electron shells.
The regular shapes of crystals suggest an underlying arrangement of the
atoms, ions or molecules in their structure. Until the early part of the
twentieth century, scientists could only guess at the arrangement of invisible
atoms in crystals. Then, Sir Lawrence Bragg (1890–1971) realised that X-rays
could be used to investigate crystal structures because their wavelengths are
about the same as the distances between atoms in a crystal.
A narrow beam of X-rays is directed at a crystal of the substance being
studied (Figure 2.1).
diffracted X-rays
lead shield
source of
crystal
X-rays
X-rays
narrow beam
of X-rays
X-ray film
Figure 2.1 Using X-rays to study the structure of atoms or ions in
a crystal.
2.1 Investigating structure and bonding 39
469983_02_Chem_Y1-2_039-[Link] 39 13/04/19 10:13 PM
The atoms or ions in the crystal scatter the X-rays producing a pattern of
diffracted rays. The diffracted X-rays were originally photographed using
X-ray film but can now be recorded electronically. From the diffraction
pattern produced, such as that shown in Figure 2.2, it is possible to deduce
the three-dimensional crystal structure by studying the pattern of dots.
Once the arrangement of the atoms or ions in a substance (the structure)
is known – and also how they are held together (the bonding) – then the
properties of a substance can be explained.
Figure 2.2 An X-ray diffraction pattern For example, copper is composed of closely packed atoms with freely moving
of lysozyme, a protein found in egg outer electrons. These electrons move through the structure when copper
white. Data like this is now stored in the is connected to a battery, so it is a good conductor of electricity. Atoms in
Worldwide Protein Data Bank, wwPDB. the closely packed structure can slide over each other and because of this
copper can be drawn into wires. These properties of copper lead to its use in
electrical wires and cables.
Notice how:
● the structure and bonding of copper determine its properties
● the properties of copper determine its uses.
The links between structure, bonding and properties help to explain the uses
of different materials. They explain why metals are used as conductors and
why graphite is used in pencils.
Two types of structure
Figure 2.3 Molecules in bromine liquid Broadly speaking there are two types of structure – giant structures and
and vapour. Many molecular elements simple molecular structures.
and compounds are liquids or gases at
room temperature because little energy Materials with giant structures form crystals in which all the atoms or ions
is needed to overcome the weak forces are linked by a network of strong bonding extending throughout the crystal.
between their molecules. This strong bonding results in giant structures with high melting and boiling
temperatures.
Substances with simple molecular structures consist of small groups of atoms.
The covalent bonds linking the atoms in the molecules (intramolecular forces)
are relatively strong, but the forces between molecules (intermolecular
forces) are weak. These weak intermolecular forces allow the molecules to
be separated easily. So molecular substances, such as bromine (Figure 2.3),
have low melting and boiling temperatures.
Key terms
Giant structures are crystal structures in which all the atoms or ions are linked by a
network of strong bonding extending throughout the crystal.
Simple molecular structures consist of groups of atoms held together by strong
covalent bonding within the molecules, but with weak forces of attraction between the
molecules.
Intermolecular forces are weak attractive forces between molecules.
40 2 Bonding and structure
469983_02_Chem_Y1-2_039-[Link] 40 13/04/19 10:13 PM
Test yourself
1 a) The melting temperatures and boiling temperatures of selected
elements are given in Table 2.1.
Use the data to decide whether they have giant or simple
molecular structures.
b) Choose those elements from Question 1(a) where no covalent
bonds are broken when the element melts.
Table 2.1 Melting temperatures and boiling temperatures of selected elements.
Element Melting temperature/K Boiling temperature/K
Boron 2573 2823
Fluorine 53 85 Figure 2.4 Crystals of rock salt (sodium
Silicon 1683 2628 chloride, NaCl).
Sulfur 386 718
Manganese 1517 2235
Iodine 387 457
2 Look at the crystals of rock salt in Figure 2.4.
a) What shape are most of the crystals of rock salt?
b) How do you think the ions are arranged in rock salt?
The main types of giant structures are ionic solids, giant covalent solids (these
include ceramics and glasses, as well as diamond and graphite) and metals.
All of these materials are solids that depend for their properties on three
types of strong bonding – ionic bonding, covalent bonding and metallic
bonding. Materials with specific properties can be chosen based on the type
of bonding present. The pylons in Figure 2.5 contain metals which conduct
electricity well and ceramics which don’t.
These three types of bonding – ionic, covalent and metallic – will be the main
focus in the following sections of this topic. For each type of bonding, its strength
depends on electrostatic attractions between positive and negative charges.
Figure 2.5 Metal cables in the electricity
grid supported by steel pylons – a reminder
that metals are strong, bendable and good
conductors of electricity. Ceramic insulators
between the conducting cables and the
pylons prevent the electric current leaking
away to earth.
2.1 Investigating structure and bonding 41
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2.2 Ionic bonding and structures
Atoms into ions
Compounds of metals with non-metals, such as sodium chloride and calcium
oxide, are composed of ions. When compounds form between metals and
non-metals, the metal atoms lose electrons and become positive ions (cations).
At the same time, the non-metal atoms gain electrons and become negative
ions (anions). For example, when sodium reacts with chlorine (Figure 2.6)
each sodium atom loses its one outer electron to form a sodium ion, Na+,
which has the same electron structure as the noble gas neon. Chlorine atoms
gain these electrons to form chloride ions, Cl−, with the same electronic
configuration as the noble gas argon (Figure 2.7).
In many cases, when atoms react to form ions, they gain or lose electrons in
such a way that the ions formed have the same electronic configuration as a
noble gas. This transfer of electrons involves redox (Section 3.2).
Figure 2.6 Hot sodium reacting with
Chemists describe diagrams like that in Figure 2.7 as dot-and-cross diagrams,
chlorine gas.
in which the electrons belonging to one reactant are shown as dots and those
belonging to the other reactant are shown as crosses. But remember, all
electrons are the same – dots and crosses are simply used to show which
electrons come from the metal and which come from the non-metal.
Dot-and-cross diagrams are useful because they provide a balance sheet for
keeping track of the electrons when ionic compounds form.
Na Cl Figure 2.8 shows simplified dot-and-cross diagrams for the formation of
sodium chloride and calcium fluoride in which only the outer shell electrons
are drawn.
–
sodium atom, Na chlorine atom, Cl
2,8,1 2,8,7 Na• + Cl Na+ + Cl
sodium atom chlorine atom sodium ion chloride ion
+ – (2,8,1) (2,8,7) (2,8) (2,8,8)
– –
Ca + F F Ca2+ + F F
Na Cl
calcium atom two fluorine atoms calcium ion two fluoride ions
(2,8,8,2) (2,7) (2,8,8) (2,8)
Figure 2.8 Dot-and-cross diagrams for the formation of sodium chloride and calcium
sodium ion, Na+ chloride ion, Cli–
2,8 2,8,8 fluoride showing only the electrons in the outer shells of the reacting atoms.
Figure 2.7 Formation of ions when sodium Test yourself
reacts with chlorine.
3 Draw dot-and-cross diagrams, showing only the outer electrons, for
the ions present in:
a) lithium fluoride b) magnesium chloride
c) lithium oxide d) calcium sulfide.
4 With the help of a Periodic Table, predict the charges on ions of each
of the following elements: caesium, strontium, gallium, selenium and
astatine.
5 Why do metals form positive ions, while non-metals form negative
ions?
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Ionic bonding Key terms
When metals react with non-metals, the ions produced form ionic crystals.
An ionic crystal is a giant lattice containing billions of positive and negative A lattice is a regular three-dimensional
ions packed together in a regular pattern. arrangement of atoms or ions in a
crystal.
Figure 2.9 shows how the ions are arranged in one layer of sodium chloride
(NaCl) and Figure 2.10 shows a three-dimensional model of its structure. Ionic bonding refers to the strong
electrostatic forces between oppositely
In the lattice, each Na+ ion is surrounded by Cl− ions, and each Cl− ion is charged ions in a lattice.
surrounded by Na+ ions. These oppositely charged ions attract each other.
Chloride ions repel other nearby chloride ions and sodium ions repel other
nearby sodium ions, but overall there are strong net electrostatic attractions
between ions in all directions throughout the lattice. These electrostatic
attractions between oppositely charged ions are described as ionic bonding.
Many other compounds have the same structure as sodium chloride including
the chlorides, bromides and iodides of Li, Na, K and the oxides and sulfides
of Mg, Ca, Sr and Ba.
Cl – Na+ Cl – Na+
The strength of the electrostatic attractions between ions depends on the
charges of the ions and their radii. Ions with high charges and small radii
Na+ Cl – Na+ Cl –
produce the strongest electrostatic attractions. So, in general, a Group 2
compound has stronger ionic bonding with a higher melting temperature
and lower solubility in water than a corresponding Group 1 compound. Cl – Na+ Cl – Na+
Tip
Na+ Cl – Na+ Cl –
In mathematical terms, the size of the electrostatic force, F, between two charges is
given by:
Q 1 × Q2
F ∝ Figure 2.9 The arrangement of ions in one
d2 layer of a sodium chloride crystal.
● The bigger the charges, Q1 and Q2, the stronger the force.
● The greater the distance, d, between the two charges, the smaller the force. This
has a big effect because it is the square of the distance that affects the force.
Test yourself
6 Look carefully at Figures 2.9 and 2.10.
a) How many Cl− ions surround one Na+ ion in a layer of the NaCl
crystal?
b) How many Cl− ions surround one Na+ ion in the three-dimensional
crystal?
c) How many Na+ ions surround one Cl− ion in the three-dimensional
crystal?
d) The structure of crystalline sodium chloride is described as 6 : 6
co-ordination. Why is this? Figure 2.10 A three-dimensional model
of the structure of sodium chloride. The
e) Use Figure 2.9 to explain that overall the attractive forces are
smaller red balls represent Na+ ions and
stronger than the repulsive forces in an ionic crystal.
the larger green balls represent Cl− ions.
2.2 Ionic bonding and structures 43
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Tip Properties of ionic compounds
Strong ionic bonding holds the ions firmly together in ionic compounds.
During electrolysis, positive ions
This explains the properties of ionic compounds. They:
gain electrons at the cathode; this is
reduction. At the same time, electrons ● are hard, brittle crystalline substances
are lost at the anode; this is oxidation ● have high melting and boiling temperatures
(see Chapter 3). ● are often soluble in water and other polar solvents, but insoluble in non-
polar solvents (Section 2.7)
● do not conduct electricity when solid, because their ions cannot move
away from fixed positions in the giant lattice
● conduct electricity when they are melted or dissolved in water, because
the charged ions are then free to move.
For example, when molten, sodium chloride conducts electricity. Ions from
the electrolyte move towards the electrodes. Positive sodium ions move
Key terms towards the negative terminal (cathode) while negative chloride ions move
towards the positive terminal (anode). When the sodium ions reach the
Electrolysis is the decomposition of a cathode, they gain electrons and become sodium atoms:
compound by electricity. The compound
which is decomposed is called an cathode (−): 2Na+(l) + 2e− → 2Na(l)
electrolyte and it is described as being When chloride ions reach the anode, they lose electrons. The chlorine atoms
electrolysed. formed then bond in pairs to become chlorine molecules:
anode (+): 2Cl−(l) → 2e− + Cl 2(g)
This process is described as electrolysis. It reverses the changes that happen
when an ionic compound such as sodium chloride forms from its elements
(Figure 2.6).
Tip
Electrolysis decomposes molten salts such as sodium chloride into their constituent
elements. Electrolysis of salts in aqueous solution is more complicated. Elements
such as oxygen (at the anode) or hydrogen (at the cathode) may be produced by the
decomposition of water, rather than simple decomposition of the salt.
Migration of ions
The movement of ions can be observed during the electrolysis of coloured
compounds. If a green solution of copper(ii) chromate(vi) is electrolysed
in a U-tube (Figure 2.11), the solution around the cathode turns blue and
the solution around the anode turns yellow. This is because blue Cu 2+(aq)
cations are attracted by the negative cathode and migrate towards it.
At the same time, yellow CrO42−(aq) ions are attracted by the positive
anode and migrate towards it. This movement provides evidence for the
existence of ions.
Ionic radii
Figure 2.11 The migration of coloured X-ray diffraction methods (Section 2.1) are used to study ionic compounds
ions during the electrolysis of copper(ii) and to measure the spacing between ions in crystals. From the diffraction
chromate(vi) solution. patterns, it is possible to calculate the radii of individual ions. The radius of
44 2 Bonding and structure
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the positive ion of an element is smaller than its atomic radius because it loses
electrons from its outer shell when turning into an ion. The radius of the
negative ion of an element is larger than its atomic radius because electrons
are added to the outer shell (Figure 2.12).
Na Na+ F F–
r atom = 0.191 nm r ion = 0.102 nm r atom = 0.085* nm r ion = 0.133 nm
Mg Mg 2+ O O2–
r atom = 0.160 nm r ion = 0.072 nm r atom = 0.090* nm r ion = 0.140 nm
Figure 2.12 Comparing the radii of atoms and ions. (*The values for fluorine and
oxygen atoms are estimates.)
Tip
● Atoms are neutral because the number of protons equals the number of electrons.
● Positive ions contain more protons than electrons; these cations are smaller than
the neutral atom.
● Negative ions contain more electrons than protons; these anions are larger than the
neutral atom.
Test yourself
7 The melting temperature of sodium fluoride is 993 °C, but that of
magnesium oxide is 2852 °C.
a) Write the formulae of these two compounds, showing charges on
the ions.
b) Suggest why the melting temperature of magnesium oxide is so
much higher than that of sodium fluoride.
8 Write equations for the reactions at the cathode and anode during
electrolysis of the following compounds:
a) molten potassium bromide
b) molten magnesium chloride.
9 A strip of wet filter paper is placed on a microscope slide and a
small crystal of potassium manganate(vii) is placed at the centre
of the paper. Leads from a 40 V DC power supply are attached to
the ends of the filter paper and the power supply is switched on.
After 30 minutes a purple colour is seen to have spread towards the
positive terminal.
Explain the movement of the purple colour.
2.2 Ionic bonding and structures 45
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Activity
Identifying and explaining the trends in ionic radii
Table 2.2 shows the radii of ions of the elements in Group 1, 4 a) All the ions of consecutive elements in the Periodic Table
and those of the consecutive elements from nitrogen to from N3− to Al3+ are described as ‘isoelectronic’. What do
aluminium in the Periodic Table. you think this means?
b) Describe the trend in ionic radii for the isoelectronic ions
Look carefully at the data in Table 2.2.
from N3− to Al3+.
1 Describe and explain the trend in ionic radii in Group 1 of c) Explain the trend in ionic radii for the isoelectronic ions
the Periodic Table. from N3− to Al3+.
2 Do you think this trend is repeated in other groups of the
Periodic Table? State ‘yes’ or ‘no’ and explain
your answer. Table 2.2 Ionic radii.
3 Use the s, p, d, f notation to describe the
Ions of Group 1 Li+ Na+ K+ Rb+ Cs+
electronic configuration of:
a) a nitride ion, N3− Radius/nm 0.074 0.102 0.138 0.149 0.170
b) a fluoride ion, F −
Ions of N to Al N3− O2− F− Na+ Mg2+ Al3+
c) a sodium ion, Na+
Radius/nm 0.171 0.140 0.133 0.102 0.072 0.053
d) an aluminium ion, Al3+.
2.3 Covalent bonding and structures
Ionic bonding always produces giant structures of ionic lattices. Ionic
compounds are, therefore, always solids at room temperature.
Covalent bonding can also produce giant structures, called giant covalent
lattices, but can also lead to simple molecular structures.
A covalent bond forms when atoms share electrons – a single covalent bond
Key term consists of a shared pair of electrons.
A covalent bond is the strong The atoms are held together by the strong electrostatic attraction between
electrostatic attraction between two the positive charges on their nuclei and the negative charge on the shared
nuclei and the shared pair of electrons electrons.
between them.
The electronic configuration of fluorine is 1s22s22p5 or more simply 2, 7,
with seven electrons in the outer shell. When two fluorine atoms combine to
form a molecule, they share a pair of electrons, one provided by each fluorine
atom. The electronic configuration of each atom in the molecule is then like
that of neon, the nearest noble gas (Figure 2.13).
Tip
Metallic bonding also involves the fluorine atoms fluorine molecule
sharing of electrons but, whereas the
electrons in metals are delocalised and
can move throughout the lattice, the F F F F
electron pair shared in a covalent bond
is fixed in position between the two
nuclei. These electrons are ‘localised’.
Figure 2.13 Covalent bonding in a fluorine molecule.
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Covalent bonds also link the atoms in non-metal compounds. Figure 2.14
shows the covalent bonding in methane.
H H
Dot-and-cross diagrams, showing only the electrons in outer shells, provide
a simple way of representing covalent bonding. An even simpler way of
showing the bonding in molecules represents each covalent bond as a line C
between two symbols. So, chemists write a fluorine molecule as F–F. This is
the structural formula, showing atoms and bonding. The molecular formula
of fluorine is F2. Three examples of both methods are shown in Figure 2.15. H H
methane molecule, CH4
Tip
Figure 2.14 Covalent bonding in methane.
The number of covalent bonds formed by a non-metal atom in Period 2 can be
predicted from its place in the Periodic Table. An atom of fluorine (Group 7) has
one electron fewer than a noble gas so it forms one covalent bond. An atom of
oxygen (Group 6) has two electrons fewer than a noble gas so it forms two covalent H H H H H H
bonds. An atom of nitrogen (Group 5) has three electrons fewer than a noble gas Cl Cl Cl Cl
Cl Cl H HO HO O H HN HNH NH H
so it forms three covalent bonds. An atom of carbon (Group 4) has four electrons
fewer than a noble gas so it forms four covalent bonds. Note that this only applies
H H H H H H
to the elements in Period 2; the situation for the elements in Period 3 and beyond
is more complex. Cl Cl Cl Cl Cl H H O
HO O H HN
H N HN H H
chlorine
chlorine
chlorine water
water
water ammonia
ammonia
ammonia
Multiple bonds Figure 2.15 Covalent bonds in three
One shared pair of electrons makes a single bond. Double bonds and triple molecules shown both as dot-and-cross
bonds are also possible with two or three shared pairs, respectively. diagrams and by using lines between the
symbols.
There is a double bond between the two oxygen atoms in an oxygen
molecule, and double bonds between both the oxygen atoms and the carbon
atom in carbon dioxide (Figure 2.16). With two electron pairs involved in
the bonding, there is a region of high electron density between the two
atoms joined by a double bond. Figure 2.17 shows two molecules which each
contain a triple bond.
H H
O O O C O C C N N H C C H
H H
oxygen carbon dioxide ethene nitrogen ethyne
H H
O O O C O C C N N H C C H
H H
Figure 2.17 Two molecules with triple covalent bonds.
Figure 2.16 Three molecules with double covalent bonds.
Tip
The carbon–carbon double bond in alkenes is considered in more detail in
Section 6.2.7.
Key term
Lone pairs of electrons A lone pair of electrons is pair of
In many molecules, there are atoms with outer shells that contain pairs electrons in the outer shell of one of the
of electrons which are not involved in the bonding between atoms in the atoms in a molecule or ion which is not
molecule. Chemists call these lone pairs of electrons (Figure 2.18). involved in bonding.
2.3 Covalent bonding and structures 47
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H H
– –
H O H N H H O Br
Figure 2.18 Molecules and ions with lone pairs of electrons.
Lone pairs of electrons:
Key term
● affect the shapes of molecules (Section 2.4)
A dative covalent bond is a bond ● are used to form dative covalent bonds
in which two atoms share a pair of ● are important in the chemical reactions of some compounds including
electrons, both the electrons being water and ammonia.
donated by one atom.
Dative covalent bonds
In a covalent bond, two atoms share a pair of electrons and each atom supplies
one electron to make up the pair. Sometimes, however, one atom provides both
the electrons and chemists call this a dative covalent bond. The word ‘dative’
means ‘giving’ and one atom gives both the electrons to make the covalent bond.
The other atom accepts the electron pair into a vacant orbital. Once formed,
there is no difference between a dative covalent bond and any other covalent
bond. An alternative name for a dative covalent bond is a co-ordinate bond.
An ammonia molecule has a lone pair and a hydrogen ion has a vacant
1s orbital. A dative covalent bond is formed when ammonia reacts with a
hydrogen ion to make an ammonium ion, NH4+ (Figure 2.19). A dative
bond is represented by an arrow in displayed formulae like that of NH4+.
The arrow points from the atom donating the electron pair to the atom
receiving them (Figure 2.19).
H H
+
+
H N H H N H
H H
Figure 2.19 Formation of an ammonium ion, NH4+.
Dative covalent bonding also accounts for the structure of Al 2Cl6 molecules.
Tip When solid aluminium chloride is heated, it sublimes (turns straight to
vapour) and Al 2Cl6 molecules are formed. These molecules contain two
When an acid dissolves in water,
dative covalent bonds formed when a lone pair on a chlorine atom is donated
aqueous hydrogen ions called oxonium
into the empty orbital on an aluminium atom (Figure 2.20).
ions are formed. A lone pair of electrons
on a water molecule forms a dative At higher temperatures these double molecules (dimers) split into AlCl3
covalent bond with a hydrogen ion from molecules.
an acid. The formula of the oxonium ion
is H3O+. It is often convenient to write
H+(aq) instead, but remember that the Cl Cl Cl Cl Cl Cl
hydrogen ion is hydrated.
H + Al Al Al Al
H +
Cl Cl Cl Cl Cl Cl
H O H
H O H
oxonium ion
Figure 2.20 An Al2Cl6 molecule shown as a dot-and-cross diagram and also using arrows
oxonium ion to represent the dative covalent bonds.
48 2 Bonding and structure
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0.200 0.200
Test yourself
10 Look at the electron density map of hydrogen, H2, in Figure 2.21.
a) What kind of bond exists between the hydrogen atoms in a
hydrogen molecule? H 0.150 H
b) What evidence does the electron density map provide for the
0.100
existence of a bond between the hydrogen atoms in a hydrogen
0.050
molecule?
11 Draw dot-and-cross diagrams showing all the electrons in:
Figure 2.21 An electron density map for
a) hydrogen chloride, HCl hydrogen, H2. The units for the contours
b) ammonia, NH3 . are electrons per 10−30 m3.
c) ammoniaum ion, NH4+.
12 Draw dot-and-cross diagrams using only the outer shell electrons
and also draw lines between symbols to show the covalent
bonding in:
a) hydrogen sulfide, H2S
b) ethane, C2H6
c) carbon disulfide, CS2
d) nitrogen trifluoride, NF3
e) phosphine, PH3.
13 Identify the atoms with lone pairs of electrons in the following
molecules and state the number of lone pairs:
a) ammonia
b) water
c) hydrogen fluoride
d) carbon dioxide.
14 a) In aqueous solution, acids donate H+ ions to water molecules
forming H3O+ ions. Draw a dot-and-cross diagram to show the
formation of an H3O+ ion.
b) Boron fluoride forms molecules with the formula BF3. Draw
a dot-and-cross diagram for BF3 and then explain why BF3
molecules readily react with other molecules.
Bond length and bond strength Key term
X-ray diffraction studies (Section 2.1) enable chemists to investigate
structures and to measure bond lengths in covalent substances in the solid Bond length is defined as the distance
phase. Microwave spectroscopy can be used to obtain values for bond lengths between the nuclei of two bonded
in molecules in the vapour phase. atoms in a molecule.
Bond length depends both on the size of the atoms involved and the number
of pairs of electrons shared (see Table 2.3).
Larger atoms form longer bonds because larger atoms have more electrons
which shield the nuclei and reduce the attraction for the electron cloud. For
instance, the length of the bond between hydrogen and the halogen atoms
increases down the group as the halogen atoms get larger.
2.3 Covalent bonding and structures 49
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Table 2.3 Some bond lengths and bond Single bonds are longer than double bonds, which are longer than triple
energies. bonds. The nuclei can remain closer together if the shared electron cloud
Bond Bond length/ Bond energy/
contains more electrons to overcome the repulsion of the nuclei.
nm kJ mol−1 The carbon–carbon single bond consists of one shared pair of electrons and
H−F 0.092 568 is longer than the carbon–carbon double bond which involves two shared
pairs. The triple bond in which three pairs are shared is shorter still. The
H−Cl 0.127 432
values for the carbon–carbon bonds in Table 2.3 are averages as these bonds
H−Br 0.141 366 occur in many different compounds.
H−I 0.161 298 The strength of a bond varies inversely with its length. A short bond is
C−C 0.154 347 stronger with a greater bond energy.
C=C 0.134 612 Bond energies are discussed in more detail in Section 8.7.
C C 0.120 838
Simple molecular structures
In most non-metal elements, atoms are joined together in small molecules
such as hydrogen (H2), nitrogen (N2), phosphorus (P4), sulfur (S8) and
Key term chlorine (Cl 2).
Most of the compounds of non-metals with other non-metals also have
The bond energy of a particular bond
simple molecular structures. This is true of simple compounds such as water,
is the energy required to break one
carbon dioxide, ammonia, methane and hydrogen chloride. It is also true of
mole of the bonds in a substance in the
the many thousands of carbon compounds (see Chapter 6.1).
gaseous state.
The covalent bonds holding atoms together within these simple molecular
structures are strong, so the molecules do not break up into atoms easily.
However, the forces between the individual molecules (intermolecular forces)
are weak, so it is quite easy to separate them. This means that molecular
substances are often liquids or gases at room temperature and that molecular
solids are usually easy to melt and evaporate (Figure 2.22).
Some non-metal elements including diamond and some compounds of non-
Tip metals including silicon dioxide consist of giant structures of atoms held
together by covalent bonding. These substances are hard and have high
Energy is needed to break bonds and
melting temperatures because the covalent bonds are strong. Giant covalent
energy is given out when bonds form.
structures are considered in Section 2.8.
Figure 2.22 The structure of iodine showing the arrangement of I2 molecules. The forces
between I2 molecules are so weak that iodine changes directly from solid to vapour on
only gentle warming; it sublimes easily.
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Properties of simple molecular substances
Tip
In general, simple molecular substances, whether elements or compounds:
Some molecular substances dissolve
● are usually gases, liquids or soft solids at room temperature
in water and react with it to form ions,
● have relatively low melting and boiling temperatures
so the ‘solution’ in water does conduct
● do not conduct electricity as solids, liquids or gases because they contain
electricity. For example, molecules of
neither ions nor free electrons to carry the electric charge
hydrogen chloride dissolve in water
● are usually more soluble in non-polar solvents (see Section 2.7), such
and react with it to form H+(aq) and
as hexane, than in water – and the solutions in hexane do not conduct
Cl−(aq) ions. The solution is known
electricity.
as hydrochloric acid and conducts
electricity because of the ions present.
Test yourself
15 Look at the structure of iodine shown in Figure 2.22. Describe the
arrangement of the molecules in solid iodine. Tip
16 The Cl−Cl and Br−Br bond lengths are 0.199 nm and 0.228 nm Ionic and covalent bonding both
respectively. depend on electrostatic attractions to
a) Explain why the Br−Br bond is longer. hold the ions and atoms together. But,
b) State which of these two bonds has the higher bond energy and whereas the electrostatic attraction by
explain your answer. an ion is the same in all directions, a
covalent bond between two atoms is
17 Explain why the O=O bond is shorter and stronger than the O−O
directional.
bond.
2.4 The shapes of molecules
and ions
Electron-pair repulsion theory Key term
X-ray diffraction studies provide very accurate evidence not only about
bond lengths but also about bond angles in molecules and in ions such as A bond angle is the angle between two
NH4+ which have covalent bonds. The results show that covalent bonds have covalent bonds in a molecule or giant
a definite direction and a definite length. For example, X-ray diffraction covalent structure.
studies show that all the C−H bond lengths in methane, CH4, are 0.109 nm
and all the H−C−H bond angles are 109.5° (Figure 2.23).
H
Tip
In 3D structures, such as methane in Figure 2.23, the two normal lines represent
C
covalent bonds in the plane of the paper. The solid wedge represents a bond coming
H
out of the paper towards the reader, while the hashed bond represents a bond going H
into the paper away from the reader.
H
Drawing 3D structures is difficult, so molecules are often represented with normal line Figure 2.23 All the bond angles in
bonds but still with an attempt at a 3D representation. See the methane structure in methane are 109.5° and all the C—H bond
Table 2.4. Section A1.8 in Appendix A1 discusses this further. lengths are 0.109 nm.
2.4 The shapes of molecules and ions 51
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Chemists have developed a very simple theory to explain and predict the
shapes and bond angles of simple molecules and ions containing covalently
bonded atoms. The theory is based on the repulsion of electron pairs in the
outermost shell of the central atom. It is called the electron-pair repulsion
theory. The theory says that electron pairs in the outer shell of atoms and
ions repel each other and get as far apart as possible.
If a methane molecule was shaped as a flat cross, the angle between carbon-
Tip hydrogen bonds would be 90°. By adopting the three-dimensional shape
Sometimes the abbreviation VSEPR known as a tetrahedron, shown in Figure 2.23, the bond angle increases to
is used for the electron-pair repulsion 109.5°, so there is maximum separation and therefore minimum repulsion
theory. This is short for ‘valence shell between the pairs of electrons. Consider the molecules of beryllium chloride
electron-pair repulsion’. The valence and boron trifluoride in Figure 2.24. In the BeCl 2 molecule, beryllium has
shell of an atom is its outer shell. only two pairs of electrons in its outer shell. In order to get as far apart
as possible, these two pairs of electrons must be on opposite sides of the
beryllium atom. The shape of the molecule is described as linear and the
Cl−Be−Cl bond angle is 180°.
F F F F
Cl Be Cl
B B
+ F
F
H
Cl Be Cl
H N H linear trigonal planar
H Figure 2.24 The shapes of molecules with two and three electron pairs around the central
atom.
The next simplest example of the electron-pair repulsion theory is shown by
H boron trifluoride, BF3. In the BF3 molecule, boron has three electron pairs
+ in its outer shell. This time, to get as far apart as possible, the three pairs
N must occupy the corners of a triangle around the boron atom. The shape
H H
of this molecule is described as trigonal planar and the F−B−F bond angles
H are 120°.
tetrahedral Now consider the ammonium ion, NH4+, in Figure 2.25. In this ion, nitrogen
+
Figure 2.25 The shape of the NH4 ion. has four electron pairs in its outer shell, each bonded to a hydrogen atom.
These electron pairs repel each other and get as far apart as possible. The
four hydrogen atoms are, therefore, at the corners of a tetrahedron. All the
Key term H−N−H bond angles are 109.5° and the shape of the NH4+ ion is tetrahedral,
exactly the same as methane.
Isoelectronic molecules and ions
The methane molecule, CH4, and the ammonium ion NH4+ have exactly the
have exactly the same number and
same number and arrangement of electrons; they are said to be isoelectronic.
arrangement of electrons.
Table 2.4 summarises the shapes of molecules with two, three, four, five and
six pairs of electrons, based on the electron-pair repulsion theory. In each
case, the electron pairs are repelled as far apart as possible. The table also
Tip shows the predicted bond angles for each molecule.
Ions and molecules which are
The electron-pair repulsion theory shows how chemists can make
isoelectronic have exactly the same
generalisations from their results and use these generalisations to make
shape.
predictions.
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Table 2.4 The shapes of molecules with two to six pairs of bonding electrons around the central atom.
Number of Shape Description Bond angle XMX Example of molecule in gas phase
electron pairs
2 X M X Linear 180° BeCl2 Cl Be Cl
X Cl
3 Trigonal planar 120° BCl3
M B
X X Cl Cl
X H
4 M Tetrahedral 109.5° CH4 C
X X H H
X H
X Cl
X Trigonal bipyramid Cl
5 X M (two triangle-based 90°, 120° and 180° PCl5 Cl P
X pyramids base to base) Cl
X Cl
X F
X X Octahedral F F
6 M (two square-based 90° and 180° SF6 S
X X pyramids base to base) F F
X F
Tip
The elements in Period 2 only have 2s and 2p orbitals available for bonding. The
maximum number of electrons these orbitals can contain is eight in four pairs, so no
Period 2 element can form more than four bonds. This eight-electron maximum is
sometimes called the octet rule.
Elements in Period 3 and beyond also have d orbitals available for bonding. Together
with s and p orbitals these d orbitals allow more than four bonds to be formed. So
elements after Period 2 are not constrained by the octet rule.
Test yourself Tip
18 Why are covalent bonds described as ‘directional bonds’? Learn the five basic shapes and
bond angles shown in Table 2.4.
19 Draw dot-and-cross diagrams of the following simple molecules,
For simplicity, the bonds may all
showing only electrons in the outer shell of all atoms.
be drawn as single lines (see
a) PF5 b) SiCl4 c) BCl3 Appendix A1.8, page 640).
20 Predict the shape and bond angle in the following molecules.
a) PF5 b) SiCl4 c) BCl3
21 Draw dot-and-cross diagrams of the following ions, then predict the
shape and give the bond angles.
a) PH4+ b) BH4− c) PF6 −
2.4 The shapes of molecules and ions 53
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Molecules and ions with lone pairs and
multiple bonds
Lone pairs
Some molecules, such as ammonia and water, contain non-bonding pairs (or
lone pairs) of electrons as well as bonding pairs (Figure 2.26).
H H H
H C H H N H H O
H
H lone pair
C H N H O
H H H H H H
109.5° 107° 104.5°
Figure 2.26 Shapes and bond angles in molecules with bonding pairs and lone pairs of
electrons.
Tip
Learn the shapes and bond angles shown in Figure 2.26. Each extra lone pair reduces
the bond angle by about 2.5°.
Ammonia and water are isoelectronic with methane. All have four pairs of
electrons in the outer shell of the central atom (see dot-and-cross diagrams
in Figure 2.26).
In methane, all four pairs of electrons are bonding pairs between the central
carbon atom and a hydrogen atom. In ammonia, three of the four pairs make
up N−H bonds as bonding pairs, but the fourth is a lone pair. Each of these
four electron pairs repels the others, so they form a tetrahedral shape around
the nitrogen atom. But the positions of the atoms in the NH3 molecule make
a shape which is pyramidal – a triangle-based pyramid – with a nitrogen
atom at the top and hydrogen atoms at the three corners of its base.
In water, there are also four pairs of electrons around the central atom – two
bonding pairs and two lone pairs. The shape formed by these electron pairs is
tetrahedral again, but the shape of the water molecule, H−O−H, is described
as V-shaped or bent.
Lone pairs of electrons are held closer to the central atom than the bonding
pairs. This means that they have a stronger repelling effect than bonding
pairs. Therefore, the strength of repulsion between electron pairs is:
lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair
This explains why the bond angle in ammonia, with one lone pair, is less
than that in methane; and why the bond angle in water, with two lone pairs,
is less than that in ammonia.
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Similar predictions about shapes and bond angles can be made for ions such
as H3O+, BF4− and NH2− (Figure 2.27).
Figure 2.27 Dot-and-cross diagrams and
F shapes of some ions.
+ – –
H O H F B F H N
H F H
H3O+ BF4– NH2–
F
+ – –
O B N
H H F F H
H F H
pyramidal tetrahedral bent (V-shaped)
Test yourself
22 Draw dot-and-cross diagrams of the following species, then predict
the shape and give the bond angles.
a) PH3 b) SF4
c) ICl2 −
d) XeF2
23 a) Draw a dot-and-cross diagram of the molecule SiH4, then predict
the shape and give the bond angle.
b) Write the formulae of two ions that are isoelectronic with SiH4.
Multiple bonds
The arrangement of the electrons in double bonds and triple bonds is
considered in more detail in Section 6.2.7. However, when it comes to
predicting molecular shapes, double bonds and triple bonds count as just
one centre of negative charge (electron-pair repulsion axis) and affect the
shapes of molecules and ions in a similar way to electrons in single bonds. So
all of these (single bonds, lone pairs, double bonds and triple bonds) can be
regarded as separate centres of negative charge when predicting the overall
shapes of molecules and ions (Figure 2.28).
Tip
O As a double bond is a greater centre
H O S O of electron density than a single bond,
H O
O O O C O there is slightly greater repulsion of
H H other bonding pairs by the electrons in
O double bonds than by those in single
H bonds. This increases the bond angles
O C O C O S
HO O around the double bond. For instance,
H
OH the H−C−O bond angle in methanal
linear trigonal planar tetrahedral (Figure 2.28) is found to be 121° rather
than the expected 120° in trigonal
Figure 2.28 The shapes of some molecules with multiple bonds:
planar molecules such as BCl3.
carbon dioxide, CO2; methanal, H2CO; and sulfuric acid, H2SO4.
2.4 The shapes of molecules and ions 55
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Test yourself
24 Draw dot-and-cross diagrams for the following molecules and ion and
predict their shapes.
a) HCN b) N2
c) SO2 d) SO42−
25 a) Draw a dot-and-cross diagram for CH3Cl.
b) What is the shape of the CH3Cl molecule?
c) The H−C−H bond angles in CH3Cl are 111°, whereas the Cl−C−H
bond angles are 108°.
Why do you think the Cl−C−H bond angles are smaller than the
H−C−H bond angles? (Hint: The C−H bond is much shorter than
the C−Cl bond.)
Key term 2.5 Polar bonds and polar molecules
Polar covalent bonds are bonds A spectrum of bonding
between atoms of different elements.
The electron pair in a covalent bond is shared equally if the two atoms joined
The shared electrons are drawn towards
by the bond are the same. The bonding is purely covalent. However, the
the atom with the stronger pull on the
electron pair in a covalent bond is not shared equally if the two atoms joined
electrons. The bonds have a positive
by the bond are different. The nucleus of one atom attracts the electrons
pole at one end and a negative pole at
more strongly than the nucleus of the other. This means that one end of the
the other.
bond has a slight excess of negative charge. This excess is represented by
the symbol δ−. At the other end of the bond, the electrons are less strongly
attracted and the charge cloud of electrons does not cancel the positive charge
on the nucleus. This end of the bond has a partial positive charge (δ+).
The bonding is covalent but the polar covalent bond has some separation
of charge (Figure 2.29).
H CI
Figure 2.29 A polar covalent bond in hydrogen chloride. Overall the molecule is
uncharged – it is not an ion – but the uneven distribution of electrons leads to partial
Tip charges at the ends of the covalent bond.
The δ symbol is the Greek letter
Compounds such as potassium fluoride and sodium chloride exist as
‘delta’. Chemists use this symbol for
giant lattices of spherical ions held together by electrostatic forces. This
a small quantity or change. They use
bonding is purely ionic. However, in ionic compounds where the cations
the symbols δ+ and δ− for the small
are small and highly charged, these cations distort the electron clouds of
charges at the ends of a polar bond.
the anions in a process called polarisation. This leads to an increase in the
They use the capital Greek ‘delta’, Δ, for
electron density in the space between the ions, some sharing of electrons
larger changes or differences.
and partial covalency. Polarisation is considered in more detail in Section
4.7 where the thermal stability of Group 1 and 2 carbonates and nitrates
is discussed.
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So the bonding in many compounds is neither purely ionic nor purely
covalent but lies somewhere on a bonding spectrum between these two
extremes (Figure 2.30).
Na+Cl– MgI2 H Cl Cl Cl Key term
Ionic bonding: Ionic bonding with Polar covalent Covalent bonding:
electron transfer polarisation of bonding between electrons evenly Electronegativity is the ability of an
from a reactive anions by small atoms with different shared between two
metal to a highly highly charged values of identical atoms
atom to attract the bonding electrons
electronegative cations causing electronegativity in a covalent bond. In a polar bond, the
non-metal partial covalency shared electrons are drawn towards the
more electronegative atom.
Figure 2.30 A bonding spectrum from purely ionic to purely covalent.
Electronegativity
Chemists use electronegativity values to predict the extent to which the
bonds between different atoms are polar. The stronger the pull of an atom
on the electrons it shares with other atoms, the higher its electronegativity.
Oxygen is more electronegative than hydrogen, so an O−H bond is polar
with a slight negative charge on the oxygen atom and a slight positive charge
on the hydrogen atom.
There are several scales of electronegativity which reflect the changes
in electronegativity in the Periodic Table (Figure 2.31), but that devised electronegativity
increases
by Linus Pauling (1901–1994) is the most commonly used. Pauling assigned
values on a scale from 0 to 4, with fluorine, the most electronegative
element, given the value 4.0. Figure 2.31 Trends in electronegativity for
Electronegativity is used to compare one element with another qualitatively, so s- and p-block elements.
when comparing elements it is enough to know the trends in electronegativity
values across and down the Periodic Table.
Highly electronegative elements, such as fluorine and oxygen, are at the
top right of the Periodic Table. The least electronegative elements, such as
caesium, are at the bottom left.
Electronegativity increases across a period. The nuclear charge increases but
the number of shielding electrons remains constant, so the attraction for the
shared electron pair increases.
Electronegativity decreases down a group. Although the nuclear charge
increases, there is an increase in the number of shielding electrons and
the shared electron pair is further from the nucleus so is attracted less
strongly.
The bigger the difference in the electronegativity of the elements forming
a bond, the more polar, and possibly more ionic, the bond. The bonding in
a compound becomes ionic when the difference in electronegativity is large
enough for the more electronegative element to remove electrons completely
from the other element. This happens in compounds such as sodium chloride,
magnesium oxide and calcium fluoride.
2.5 Polar bonds and polar molecules 57
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Activity
Interpreting electronegativity values
The version of the Periodic Table in Table 2.5 shows Pauling Table 2.5 Pauling electronegativity values.
electronegativity values for selected elements.
1 Why are electronegativity values for He and Ne not included in H
the table? 2.1
2 Identify two elements in the table that would combine to form
a compound with covalent bonds that are not polar.
3 Identify the two elements in the table that would form the Li Be B C N O F
most purely ionic compound. 1.0 1.5 2.0 2.5 3.0 3.5 4.0
4 What is the trend in the values of electronegativity from left to
right across a period? Na Mg Al Si P S Cl
5 a) Draw diagrams showing the electrons in the main shells for 0.9 1.2 1.5 1.8 2.1 2.5 3.0
lithium and fluorine.
b) Use your diagrams and the concept of shielding to explain K Ca Br Kr
why fluorine is much more electronegative than lithium. 0.8 1.0 2.8 3.0
6 What is the trend in the values of electronegativity down a group?
Rb Sr I Xe
7 a) Draw diagrams showing the electrons in the main shells for
fluorine and chlorine. 0.8 1.0 2.5 2.6
b) Use your diagrams and the concept of shielding
to explain why fluorine is more electronegative than chlorine.
Test yourself
26 a) Use Figure 2.31 and the electronegativity values in Table 2.5 to
predict the polarity of the bonds in these molecules: H2S, NO,
CCl4, ICl.
b) Put these bonds in order of polarity, with the most polar first:
C−I, C−H, C−Cl, C−O, C−F, C−Br.
27 Put these sets of compounds in order of the character of the
bonding, with the most ionic on the left and the most covalent on
the right:
a) Al2O3, Na2O, MgO, SiO2 b) LiI, NaI, KI, CsI.
28 Iron(iii) chloride can be prepared by passing dry chlorine over hot
iron. The iron(iii) chloride sublimes away from the metal surface and
can be collected where the vapour solidifies on a cold surface.
Iron(ii) chloride does not sublime and cannot be prepared in this way.
a) Write an equation for this preparation of iron(iii) chloride.
b) Explain why iron(iii) chloride easily turns to vapour despite being
a metal compound.
c) Explain why iron(ii) chloride does not sublime in the same way.
29 The ionic model of bonding involves the transfer of electrons from
metals to non-metals to form oppositely charged ions held together
by strong electrostatic forces.
Discuss the strengths and weaknesses of this model in explaining
the properties of metal compounds.
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Polar molecules Key term
The covalent bonds in hydrogen chloride are polar and, as there is only
one bond in each molecule, overall these are polar molecules. There are, Polar molecules contain polar bonds
however, molecules with polar bonds that are not polar overall. One example which do not cancel each other out, so
is the tetrahedral molecule tetrachloromethane (Figure 2.32). The four polar that the whole molecule is polar.
bonds in CCl4 are arranged symmetrically around the central carbon atom
so that overall they cancel each other out.
O
H H O C O
H Cl
H C Cl C
Cl Cl
H Cl
overall polar overall non-polar
Figure 2.32 Molecules with polar bonds. Note that in the examples on the left, the net
effect of all the polar bonds is a polar molecule; in the examples on the right the overall
effect is a non-polar molecule.
Tip H H
Make sure you consider the three- Cl C Cl C
dimensional structure of a molecule Cl Cl
when working out whether it has an H H
overall dipole. A flat representation of Figure 2.33 In the left–hand flat
dichloromethane could suggest that the representation, the polar bonds are drawn
effect of the two polar bonds cancel, opposite to each other, so it appears that their
but the 3D structure shows that this is effects would cancel. In the right–hand 3D
not the case and the molecule has an representation, the bonds are shown correctly
overall dipole (Figure 2.33). 109.5° apart in the tetrahedral molecule. So
the molecule does have an overall dipole.
Polar molecules are little electrical dipoles – they have a positive electric pole
and a negative electric pole. These two poles of opposite charge in a molecule
are called dipoles. Dipoles tend to line up in an electric field (Figure 2.34).
polar molecules
+ – + – –
+
+
–
+
–
– +
– –
+ +
–
– +
+
– +
electric field
Figure 2.34 Polar molecules in an electric field. The electrostatic forces tend to line
up the molecules with the field. Random movements due to the kinetic energy of the
molecules tend to disrupt the alignment of the molecules.
2.5 Polar bonds and polar molecules 59
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The bigger the dipole, the bigger the twisting effect – or dipole moment –
on a molecule in an electric field. By making measurements with a polar
substance between two electrodes it is possible to calculate dipole moments.
The units are debye units, named after the physical chemist Peter Debye
(1884–1966).
Table 2.6 Measures of the polarities of
some molecules. Tip
Molecule Dipole moment/ Dipole moment is a measure of the overall polarity of a molecule. Mathematically it
debye units is the product of the magnitude of the charge multiplied by the distance between the
HCl 1.08
charges. Where a molecule has several polar bonds, the overall dipole moment is the
vector addition of the individual bond dipole moments taking into account both their
H2O 1.94 size and direction.
CH3Cl 1.86
CHCl3 1.02 A thin stream of a polar liquid is attracted towards an object with an
electrostatic charge (Figure 2.35). This is because the polar molecules tend to
CCl4 0
move and rotate because the charge on one side of the molecules is attracted
CO2 0 to the opposite charge on the object.
Figure 2.35 A thin stream of water is bent by a nearby comb carrying an electrostatic
charge.
Test yourself
30 Consider the shapes of the following molecules and the polarity of
their bonds. Then, divide the molecules into two groups – polar and
non-polar: HBr, CHBr3, CBr4, CO2, SO2.
31 Account for the relative values of the dipole moments of the
molecules in Table 2.6.
32 Draw the structure of the molecule OF2 and use the symbols δ+ and
δ− to show the polarity of the atoms in the bonds.
Compare your answer with the water molecule in Figure 2.32.
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2.6 Intermolecular forces
The covalent bonds linking the atoms in the molecules (intramolecular forces)
are relatively strong, whereas the forces between molecules (intermolecular
forces) are weak. However, without these intermolecular forces there
would be no rivers or oceans and the DNA double helix would not exist.
Figure 2.36 shows another example of the effect of intermolecular forces.
Figure 2.36 Geckos can climb smooth
walls and hang from ceilings thanks to the
intricate design of their feet. Each of a
gecko’s toes is lined with microscopic hairs,
and each hair is further branched into finer
structures. Weak intermolecular forces
over the large surface area of the hairs are
strong enough to grip on any surface, but
weak enough to break as the gecko moves
by peeling its feet away.
Intermolecular forces, as their name states, are forces between molecules. It is
these forces which are overcome when a molecular substance melts or boils. The Key term
covalent bonds within a molecule are not broken when a substance melts or boils.
Intermolecular forces are weak
attractive forces between molecules.
Tip
If you are ever confused about whether bonds or intermolecular forces are breaking,
think what happens when water boils in a kettle. Vaporised water molecules come out
of the spout because the forces between the water molecules are broken. However,
hydrogen and oxygen gases are not formed! Boiling does not break the covalent bonds
between oxygen and hydrogen atoms inside the water molecules.
London forces
Key term
The Dutch physicist Johannes van der Waals (1837–1923) developed a theory
of intermolecular forces to explain why real gases behave in the way that they London forces are the intermolecular
do. If there were no attractions between molecules, it would be impossible to forces that exist between all molecules.
turn a gas into a liquid by cooling. For some gases, the attractive forces are so They arise from the attractions between
weak that they do not liquefy until very low temperatures are reached. The temporary instantaneous dipoles and
boiling temperature of hydrogen, for example, is −253 °C, just 20 degrees the fleeting dipoles they induce in
above absolute zero. neighbouring molecules.
It is not obvious why there are weak attractions between uncharged non-
polar molecules, such as those of iodine, hydrocarbons and the noble gases.
The German physicist who developed the theory to explain these forces was
Fritz London (1900–1954), so they are sometimes called London forces.
2.6 Intermolecular forces 61
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When non-polar atoms or molecules meet, there are fleeting repulsions and
attractions between the nuclei of the atoms and the surrounding clouds of
electrons. Temporary displacements of the electrons lead to temporary dipoles as
shown in Figure 2.37. These temporary instantaneous dipoles can induce dipoles
in neighbouring molecules – positive poles induce negative poles and vice versa.
The attractions between these instantaneous and induced dipoles are the weakest
kind of intermolecular force, but their presence gives molecules a tendency to
cohere. Intermolecular forces of this kind between small molecules are roughly
a hundred times weaker than the covalent bonds within the molecules.
Figure 2.37 The origins of temporary molecules meet:
there are temporary
induced dipoles. attractions and repulsions
between electrons and nuclei
two non-polar molecules: weak, short-lived
the centres of positive and attractions between
negative charge coincide temporary dipoles
Bigger molecules with a larger number of electrons have a higher polarisability
Key term and the possibility for temporary, induced dipoles is greater. This explains why
the boiling temperatures of the elements rise down Group 7 (the halogens) and
Polarisability is an indication of the
Group 0 (the noble gases) (see Figure 2.38). For the same reason, the boiling
extent to which the electron cloud in a
temperatures of alkanes increase with the increasing number of carbon atoms.
molecule (or an ion) can be distorted by
The chemistry of the alkanes is considered in Sections 6.2.2 and 6.2.3.
a nearby electric charge.
The shapes of molecules can also affect the overall size of London forces. The
attractions between long thin molecules are stronger than those between
short fat molecules. This is because the attractions between long thin
molecules can take effect over a larger surface area.
For a given volume, the minimum surface area is a sphere. Consider molecules
with similar total volume but different surface areas, such as isomers of
alkanes; the more branched the alkane, the more spherical and compact the
Test yourself molecule. This means that branched alkanes have a lower surface area of
contact and therefore weaker London forces than unbranched isomers (see
33 Explain how Figure 2.38 Activity: Intermolecular forces and the properties of alkanes).
illustrates the fact that the
200
strength of intermolecular
forces varies with the number Xe
of electrons in the molecules
Boiling temperature/K
of monatomic gases. Kr
34 a) Account for the states
100 Ar
of the halogens at room
temperature – chlorine is
a gas; bromine is a liquid;
Ne
while iodine is a solid.
He
b) Predict the state of the
0
element astatine at room 1 2 3 4 5
temperature and explain Period
your answer. Figure 2.38 The boiling temperatures of noble gases plotted against atomic number.
62 2 Bonding and structure
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Activity
Intermolecular forces and the properties of alkanes
Table 2.7 shows the melting and boiling temperatures of the lowest molecular mass
alkanes. Use these data when answering the questions in this activity. Explain the
patterns you find in terms of intermolecular forces.
Table 2.7 Melting and boiling temperatures of the lowest molecular mass alkanes.
Straight-chain alkanes Tm /K Tb /K
Methane CH4 91.1 109.1
Ethane CH3CH3 89.8 184.5
Propane CH3CH2CH3 83.4 231.0
Butane CH3(CH2)2CH3 134.7 272.6
Pentane CH3(CH2)3CH3 143.1 309.2
Hexane CH3(CH2)4CH3 178.1 342.1
Heptane CH3(CH2)5CH3 182.5 371.5
Octane CH3(CH2)6CH3 216.3 398.8
Nonane CH3(CH2)7CH3 222.1 423.9
Decane CH3(CH2)8CH3 243.4 447.2
Branched alkanes Tm /K Tb /K
2-Methylpropane (CH3)2CHCH3 113.7 261.4
2-Methylbutane (CH3)2CHCH2CH3 113.2 301.0
2-Methylpentane (CH3)2CH(CH2)2CH3 119.4 333.4
2-Methylhexane (CH3)2CH(CH2)3CH3 154.8 363.1
2-Methylheptane (CH3)2CH(CH2)4CH3 164.1 390.7
2,2-Dimethylpropane C(CH3)4 256.6 282.6
Boiling temperatures of the unbranched alkanes 6 What is the effect of chain branching on the boiling
Plot the boiling temperatures of unbranched alkanes against the temperatures of alkanes?
number of carbon atoms in the molecules for the range C1 to C10. 7 How do you account for this trend?
1 Which of these alkanes are gases at room temperature and Melting temperatures of the unbranched alkanes
pressure and which are liquids? On the same axes as your other graphs, plot the melting
2 What is the approximate increase in boiling temperature for temperatures of unbranched alkanes.
each −CH2− added to an alkane chain? 8 Identify one similarity and one difference between the plots
3 Estimate the boiling temperature for dodecane, C12H26. of melting temperatures and boiling temperatures.
4 What type of intermolecular forces act between alkane 9 Suggest an explanation for the pattern of melting
molecules? temperatures for alkanes with an odd number of carbon
5 What two features of alkane molecules account for the trend atoms compared to the alkanes with an even number of
in values shown by your graph? carbon atoms.
Boiling temperatures of branched alkanes 10 Polythene can be regarded as a long chain polymer of −(CH2)n−.
Add to your graph the points for three 2-methyl alkanes, and The value of n can be around 100 000. How do you account for
also one for a 2,2-dimethyl alkane. the strength of this material, which softens and melts in the range
100–150 °C?
2.6 Intermolecular forces 63
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Dipole–dipole interactions
Molecules with permanent dipoles attract each other a little more strongly
repulsion than non-polar molecules. The positive pole of one molecule tends to attract
the negative poles of others and vice versa (Figure 2.39).
attraction This contribution to intermolecular forces from permanent dipoles occurs in
addition to the London forces which act between all molecules.
Data such as boiling temperatures can be used to deduce the relative
contribution of each type of force to the overall intermolecular forces.
Figure 2.39 Forces between molecules
with permanent dipoles. Test yourself
35 Account for the difference in boiling temperature between the
following pairs of molecules by considering both London forces and
dipole–dipole attractions:
a) ethane which boils at −88 °C and fluoromethane which boils at
−78 °C
b) butane which boils at −0.5 °C and propanone, CH3COCH3, which
boils at 56 °C.
36 Iodine monochloride boils at 371 K and bromine boils at 332 K
although both molecules contain exactly the same number of
electrons. Explain why the boiling temperatures differ.
37 Three isomers with molecular formula C7H16 (heptane,
3-methylhexane and 2,2-dimethylpentane) have boiling temperatures
of 79.2 °C, 92.0 °C, and 98.4 °C, but not in that order.
a) Draw the structures of the three isomers.
b) Match the structures with their boiling temperatures and give
your reasons.
Tip
Despite its name, a hydrogen ‘bond’ Hydrogen bonding
is an intermolecular force and not a Hydrogen bonding is an extreme type of dipole–dipole attraction between
covalent bond. Hydrogen bonds are at molecules. It is much stronger than other types of intermolecular force, but
least 10 times weaker than covalent still at least 10 times weaker than covalent bonds. This strongest type of
bonds. They affect the physical intermolecular force acts in addition to London forces.
properties of many substances, but not
Hydrogen bonding affects molecules in which hydrogen is covalently bonded to
the way they react.
one of the three highly electronegative elements – fluorine, oxygen and nitrogen.
H
H H 180°
H O O
H H H
O O
H H H H H
O O
H H H
O O
180°
H
Figure 2.40 Hydrogen bonding in water.
64 2 Bonding and structure
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The highly electronegative atom attracts electrons strongly away from
hydrogen so the covalent bond between them is extremely polar and the
hydrogen atom is very electron deficient or δ+. A strong intermolecular
force then results between this δ+ hydrogen atom and a lone pair of electrons
on the δ− fluorine, oxygen or nitrogen atom of a nearby molecule.
The three atoms associated with a hydrogen bond are always in a straight
line. According to electron-pair repulsion theory, the bonding pair of
electrons in the covalent bond to hydrogen and the lone pair of electrons
on the fluorine, oxygen or nitrogen atom of the adjacent molecule keep as
far apart as possible to minimise the repulsion between them. So the angle
between the covalent bond to hydrogen and the hydrogen bond is 180° (see
Figures 2.40 and 2.41).
F F F
H H H H H
F F covalent bond
hydrogen bond
Figure 2.41 Hydrogen bonding in hydrogen fluoride.
The essential requirements for hydrogen bonding are:
Key term
● a hydrogen atom covalently bonded to a highly electronegative atom
● a lone pair of electrons on a second electronegative atom. Hydrogen bonding
In a water molecule there are two O−H bonds and two lone pairs on the A strong intermolecular force between
oxygen atom. This means that each water molecule can take part in up to a δ+ hydrogen atom covalently bonded
four hydrogen bonds, two via the hydrogen atoms and two others via the to fluorine, oxygen or nitrogen and a
lone pairs of electrons (see Figure 2.42). This helps to explain the three- lone pair of electrons on the δ− fluorine,
dimensional structure of ice (Figure 2.43). In liquid water, molecular motion oxygen or nitrogen atom of a nearby
means that not all possible hydrogen bonds are formed at all times. molecule.
oxygen
hydrogen
hydrogen bond
covalent bond
Figure 2.42 Molecules in ice are held together by hydrogen bonding. Each oxygen atom is
bonded to two hydrogen atoms by covalent bonds and two others by hydrogen bonds.
Hydrogen bonding accounts for:
● the relatively high boiling temperatures of ammonia, water and hydrogen
fluoride, which are out of line with those of the other hydrides in Groups
4, 5 and 6 (see Figure 2.44)
● the open structure (see Figure 2.43) and low density of ice (see Figure 2.45)
● the solubility of simple alcohols in water
● the pairing of bases in a DNA double helix.
2.6 Intermolecular forces 65
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400 H2O
300
Boiling temperature/K
H2Te
SbH3
NH 3
H2S H2 Se
SnH4
200
AsH3
PH3
SiH 4 GeH 4
CH4
100
0
2 3 4 5
oxygen hydrogen Period
Figure 2.43 The hydrogen bonding in ice holds Figure 2.44 Boiling temperatures for the hydrides of the elements in Groups 4,
the water molecules in an open structure. 5 and 6.
This structure collapses as the ice melts. The
molecules then get closer together so the
density rises to a maximum at 4 °C.
Test yourself
38 Draw diagrams to show hydrogen bonding between water molecules
and:
a) ammonia molecules in a solution of ammonia, NH3
b) ethanol molecules in a solution of ethanol, CH3CH2OH.
39 a) The boiling temperatures of the hydrogen halides are shown in
Table 2.8. Plot a graph showing how the boiling temperatures of
the hydrogen halides vary with the atomic number of the halogen.
b) Describe and explain the pattern shown by the graph with
Figure 2.45 An iceberg in Antarctica. Only reference to the types of intermolecular forces which act
about 10% of the ice is above the surface between the molecules.
of the sea because ice is less dense than
Table 2.8 Boiling temperatures of the hydrogen halides.
water at 0 °C.
Hydrogen halide Tb /K
Hydrogen fluoride 293
Hydrogen chloride 188
Hydrogen bromide 206
Hydrogen iodide 238
40 Explain the differences in Figure 2.44 between the plot for the
hydrides of Group 6 and the plot for the hydrides of Group 4.
41 Which types of intermolecular force hold the molecules together in:
a) hydrogen bromide, HBr
b) propane, CH3CH2CH3
c) methanol, CH3OH?
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2.7 Solutions and solubility
Patterns of solubility
As a rough-and-ready rule, ‘like dissolves like’. Water, which is highly polar,
dissolves many ionic compounds and compounds with −OH groups such Key terms
as alcohols and sugars. Non-polar solvents, such as cyclohexane, dissolve
hydrocarbons, molecular elements and molecular compounds. A solute is a chemical which dissolves
in a solvent to make a solution.
But there is a limit to the quantity of a substance, a solute, that can dissolve
in a solvent. A saturated solution is one which contains as much of the A saturated solution contains as much
dissolved solute as possible at a particular temperature. The solubility is at of the solute as possible at a particular
maximum in a saturated solution. temperature.
Solubility is a measure of the
Soluble or insoluble? concentration of a saturated solution
No chemicals are completely soluble and none are completely insoluble
of a solute at a specified temperature.
in water. Even so, chemists find it useful to use a rough classification of
Solubilities are commonly recorded as
solubility based on what they see on shaking a little of the solid with water
‘mol per 100 g water’ or ‘g per 100 g
in a test tube:
water’ at 25 °C (298 K).
● very soluble, like potassium nitrate – plenty of the solid dissolves quickly
● soluble, like copper(ii) sulfate – crystals visibly dissolve to a significant
extent Tip
● sparingly or slightly soluble, like calcium hydroxide – little solid seems to
The term ‘saturated’ is also used
dissolve but, in this case, the pH of the solution changes in organic chemistry to describe
● insoluble, like iron(iii) oxide – no sign that any of the material dissolves.
compounds which contain only single
A similar rough classification applies to gases dissolving in water. Ammonia bonds. In hydrogenation reactions,
and hydrogen chloride are very soluble; sulfur dioxide is soluble; carbon hydrogen adds across double or triple
dioxide is slightly soluble; helium is insoluble. bonds in unsaturated hydrocarbons.
The saturated compounds formed
Solubility and intermolecular forces contain as much hydrogen as possible.
Patterns of solubility for molecular solids are determined by intermolecular
forces. The dissolving of a molecular solute is shown in Figure 2.46. Three
interactions are involved:
● the intermolecular forces between solute molecules
● the intermolecular forces between solvent molecules
● the intermolecular forces between solute and solvent molecules.
Figure 2.46 A molecular substance
dissolves if the energy needed to break
intermolecular forces and to separate the
molecules of a solid solute
molecules in the solute and in the solvent
is about the same as the energy released
as the solute forms new intermolecular
forces with the solvent.
molecules of the solute
molecules of a liquid solvent dissolved in the solvent
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When all three types of force are about the same strength, the solute dissolves
Key terms freely in the solvent. So non-polar molecules, such as those of a hydrocarbon
wax, dissolve and mix freely with non-polar liquids such as cyclohexane.
A non-aqueous solvent is any solvent
In this example, cyclohexane is acting as a non-aqueous solvent. All the
other than water.
intermolecular forces involved are London forces.
Miscible liquids are those which mix
with each other – water and ethanol are
However, non-polar molecules, such as hydrocarbons, do not dissolve in water.
miscible; oil and water are immiscible.
The non-polar molecules can separate easily because their intermolecular
forces are relatively weak. But the stronger hydrogen bonding between water
molecules acts as a barrier which keeps out molecules that cannot, themselves,
form hydrogen bonds.
Polar organic molecules, such as halogenoalkanes, are also insoluble in water.
Again the weak dipole-dipole forces between the organic molecules allow
them to separate fairly easily. But, as they cannot form hydrogen bonds
with water, they cannot disturb the hydrogen bonding between the water
molecules and so remain separate from water.
Organic molecules that can form hydrogen bonds, such as alcohols, do
dissolve and mix with water. Ethanol molecules, for example, can break into
the hydrogen-bonded structure of water by forming new hydrogen bonds
between ethanol and water molecules.
The two liquids, ethanol (C2H5OH) and water, are miscible. Alcohols with
longer hydrocarbon chains do not mix with water so easily. The longer the
chain, the less the miscibility of the alcohol with water.
Solutions of ionic salts in water
It is not obvious why the charged ions in a crystal of sodium chloride can
separate and dissolve in water with only a small energy change. The high
melting point of a salt such as sodium chloride (801 °C) shows that a large
amount of energy is needed to separate the ions from a crystal.
The explanation of the solubility of some ionic salts in water is that the ions are
strongly hydrated by polar water molecules (Figure 2.47). The water molecules
cluster around the ions and bind to them. The energy released when the water
molecules bind to the ions is enough to compensate for the energy needed to
overcome the electrostatic attractions holding the ionic lattice together.
Figure 2.47 Sodium ions and chloride ionic crystal
ions leaving a crystal lattice and becoming lattice
hydrated as they dissolve in water. Here the hydrated
bonding between the ions and the polar – + – cation
water molecules is electrostatic attraction.
+ – + –
– + – +
+ – + – + –
– + – + – +
hydrated
polar water anion
molecule
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Other salts are insoluble in water because the hydration energy that would
be released when the ions are hydrated is not large enough to overcome the
Key term
electrostatic forces in the lattice of the crystals.
Hydration takes places when water
molecules bond to ions or add to
Tip molecules. Water molecules are polar
Not all reactions which occur are exothermic; some reactions do occur despite so they are attracted to both positive
being endothermic. Similarly, some substances dissolve in water even though the ions and negative ions.
overall enthalpy change of solution is endothermic. The reason for this is that other
factors are involved when a solution forms, including the disorder created as the
solute particles move apart into the solvent. The overall energy change which takes
these other factors into account is called the Gibbs free energy change and will be
considered in detail later.
Test yourself
42 Explain why methane gas is insoluble in water, but ammonia is freely
soluble.
43 Explain why iodine is soluble in a non-aqueous solvent such as
cyclohexane, but almost insoluble in water.
44 Explain why methanol is miscible with water whereas decan-1-ol is not.
45 Table 2.9 shows the solubility in water of several salts.
Table 2.9 Solubility in water of some Group 1 and Group 2 salts.
Salt Solubility in mol/100 g water
Barium sulfate 9.43 × 10−7
Caesium fluoride 3.84
Calcium hydroxide 1.53 × 10−3
Calcium sulfate 4.66 × 10−3
Lithium chloride 2.00
Lithium fluoride 5.09 × 10−3
Magnesium chloride 5.57 × 10−1
Magnesium sulfate 1.83 × 10−1
Potassium iodide 8.92 × 10−1
Use the data in Table 2.9 to classify the salts as very soluble, soluble,
slightly soluble or insoluble according to their solubility in water.
2.8 Giant covalent structures
A few non-metal elements – including carbon and silicon – consist of giant
structures of atoms held together by covalent bonding.
Some compounds of non-metals, such as silicon dioxide and boron nitride,
also exist as giant covalent structures. The covalent bonds in these structures
are strong, so giant covalent substances are very hard and have very high
melting temperatures.
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Test yourself
Si
46 Carbon dioxide is a gas at room temperature, whereas silicon dioxide
O (silicon(iv) oxide) is a solid with a very high melting point.
Explain this difference in terms of the structures of the two compounds.
47 Consider the structure of silicon dioxide (Figure 2.48) which
shows that each silicon atom is bonded to four oxygen atoms and
each oxygen atom is bonded to two silicon atoms. Confirm that,
overall, there are two oxygen atoms for every one silicon atom and,
Figure 2.48 Part of the giant structure of therefore, the (empirical) formula of silicon dioxide is SiO2. Drawing a
silicon dioxide in the mineral quartz. Silicon planar representation of Figure 2.48 may help.
atoms are arranged in the same way as
carbon atoms in diamond, but with an
oxygen atom between each pair of silicon Structure and bonding in different forms
atoms. Sandstone and sand consist mainly of carbon
of silicon dioxide. Carbon can exist in different solid forms – diamond, graphite, various
fullerenes and graphene. These solid forms of carbon are called allotropes –
Key term different forms of the same element in the same physical state.
Allotropes are different forms of the These forms of carbon illustrate the important connections between the
same element in the same physical structure and bonding of materials, their properties and hence their uses.
state.
All these forms of carbon are held together by strong covalent bonds with a
definite length and direction. Diamond, graphite and graphene are giant covalent
structures, whereas fullerenes, in comparison, are relatively simple molecules.
Diamond
Strong covalent bonds with a definite length and fixed direction help to
account for the rigid covalent structure of diamond (Figure 2.49). It is the
hardest naturally occurring substance with a high sublimation point. People
have always valued diamonds for their brilliance as gemstones. But diamonds
are also used industrially as abrasives for cutting and grinding hard materials
such as glass and stone (Figure 2.50).
Figure 2.49 Part of the giant covalent
structure in diamond – each carbon atom Figure 2.50 Diamonds that cannot be sold as gemstones are used in glass cutters and
is linked to four other atoms in a network diamond-studded saws. This photo shows an engraver using a diamond-studded wheel to
extending throughout the giant structure. make patterns on a glass.
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Diamond does not conduct electricity because the electrons in its covalent
bonds are fixed (localised) between pairs of atoms.
Diamond conducts thermal energy very well – five times better than copper.
This important property means that diamond-tipped cutting tools don’t
overheat. The rigidity of the strong covalent bonds in diamond means that,
as the atoms close to the tip of a cutting tool get hotter and move faster, the
vibrations move rapidly throughout the giant structure.
Graphite
Graphite is used to make crucibles for molten metals. It can withstand high
temperatures because it sublimes at the extremely high temperature of
3650 °C. For the same reason, graphite blocks are used to line the walls of
industrial furnaces.
This high sublimation temperature also suggests that graphite has a giant
structure with strong covalent bonds. This is confirmed by X-ray diffraction
studies which show that the atoms are held together in extended sheets
(layers) of atoms. Each layer contains billions and billions of carbon atoms
arranged in hexagons (Figure 2.51). Each carbon atom is held strongly in its
layer by strong covalent bonds to three other carbon atoms. So every layer is
a giant covalent structure. The distance between neighbouring carbon atoms
in the same layer is only 0.14 nm, but the distance between layers is 0.34 nm.
Figure 2.51 The giant covalent structure
of graphite. The layers are vast sheets of
carbon atoms piled on top of each other.
The bonding between atoms within the
layers is strong, but the bonding between
layers is relatively weak.
Each carbon atom in graphite uses three of its outer shell electrons to form
three normal covalent bonds with other carbon atoms. This accounts for
Key term
the trigonal arrangement of bonds around each atom and the hexagonal
Composites combine two or more
arrangement of the atoms within a layer.
materials to create a new material
The fourth outer shell electron on each carbon atom forms part of a cloud of which has the desirable properties of
delocalised electrons spread out over each layer. Because of these delocalised both its constituents. For example,
electrons, graphite conducts electricity well. This explains why graphite is plastic reinforced with graphite fibres
used for electrodes in industry and as the positive terminal in cells. combines the flexibility of the plastic
with the high tensile strength of
The covalent bonds between carbon atoms within the layers of graphite are
graphite.
so strong that many modern composites incorporate graphite fibres for
greater tensile strength (Figure 2.52).
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Unlike diamond, graphite is soft and feels greasy – this property leads to
the use of graphite as a lubricant. It has lubricating properties because the
bonding between the well-separated layers is relatively weak, allowing the
layers to slide over each other.
Fullerenes
At one time, chemists believed there were only two forms of crystalline
carbon. Then, in 1985, Harry Kroto and his research team at the University
of Sussex, working with teams led by Bob Curl and Richard Smalley in Texas,
discovered buckminsterfullerene, C60 – a black solid with a simple molecular
structure. Since 1985, several similar subtances have been prepared; these are
now known as ‘fullerenes’.
At the molecular level, the fullerenes mimic the geodesic (football-like)
dome invented by the American engineer Robert Buckminster Fuller
(Figure 2.53). Hence, the original name ‘buckminsterfullerene’ and the
nicknames ‘bucky balls’ and ‘footballene’.
Figure 2.52 Graphite fibres are used
to reinforce the shafts of broken bones,
badminton rackets and golf clubs, like
the one being used by Rory McIlroy in this
photo.
Figure 2.53 The structure of C60 is roughly spherical with each carbon atom bonded to
three nearest neighbours. Look carefully and see if you can count all 60 carbon atoms.
Other fullerenes have the formulae C32, C50 and C240.
Fullerenes are fundamentally different from diamond and graphite because
they are molecular forms of carbon, rather than infinite giant covalent
structures.
Fullerenes are black solids which are soluble in various solvents because of
their molecular structure. This has already led to the use of C60 in mascara
and printing ink.
The bonding at each carbon atom in fullerenes resembles that in graphite.
Three of the outer shell electrons are combined in covalent bonds with other
atoms, while the fourth electron is delocalised over the whole molecule. But,
unlike graphite which conducts, the fullerenes are good electrical insulators
because the delocalised electrons cannot move between molecules. However,
metals in Groups 1 and 2 can react with C60 to form superconducting systems
at very low temperatures. The reaction produces a rare type of salt in which
electrons transferred to the C60 move around the whole salt in the same way
as electrons move in a metal.
3Rb(s) + C60(s) → (Rb+)3C603−(s)
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At present, these superconducting fullerene–metal systems can conduct
only relatively small currents at very low temperatures, below −190 °C. The
priority is to develop superconductors that can work at higher temperatures
and carry much larger currents.
Now that chemists understand the structure of fullerenes, they are able to
produce fullerenes in the form of tubes as well as spheres. These ‘buckytubes’
or ‘carbon nanotubes’ are not only the narrowest tubes ever made, but also
the strongest and toughest weight for weight.
These carbon nanotubes have enormous potential in very diverse applications,
from the replacement of graphite fibres in golf clubs and fishing rods to their
use in medicine as vehicles for carrying drugs into specific body cells.
Graphene
Graphene is effectively a two-dimensional material although essentially a one–
atom thick layer of carbon atoms, the same as a single layer of graphite. Graphene,
first isolated in 2003 in Manchester by Andre Geim (Figure 2.54) and Kostya
Novoselov, is an exciting new materials with a huge number of possible uses.
It is the thinnest material known but is also one of the strongest. Graphene-
plastic composites are extremely strong but very light weight and so have
potential uses in aircraft and cars.
Graphene is as good a conductor of electricity as copper and is also a better
conductor of heat than any other material. Composites again allow the
possibility of plastics which conduct.
Graphene’s transparency, flexibility (see Figure 2.55) and conductivity also
raise the possibility of its use in touchscreens for mobile devices. It is also
being investigated for use in ultrasensitive chemical sensors and photocells.
Figure 2.54 Professor Andre Geim holding a model of graphene. Figure 2.55 Computer model of the molecular structure of
Working with Kostya Novoselov at the University of Manchester, graphene.
he isolated this single layer structure in 2003. They were jointly
awarded the Nobel Prize for Physics in 2010 for their work.
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Test yourself
48 a) Why does a zip fastener move more freely if it is rubbed with a
soft pencil?
b) Why should you use a pencil for this rather than oil?
c) Why is graphite sometimes mixed with the oil used to lubricate
the moving parts of machinery?
49 a) Name one element, other than carbon, which exists as different
allotropes and name the allotropes.
b) Explain what is meant by a ‘composite’. Illustrate your answer
with an example, such as reinforced concrete or fibre glass
(glass-reinforced plastic).
50 a) Why is diamond such a poor conductor of electrical energy, but a
good conductor of thermal energy?
Figure 2.56 Close packing of atoms in one
layer of a metal. b) Why does graphene conduct electricity but fullerenes do not?
second-layer atom
2.9 Metallic bonding and structures
Metals are very important and useful materials. Just look around you and
notice the uses of different metals – in vehicles, bridges, pipes, taps, radiators,
cutlery, pans, jewellery and ornaments. X-ray studies show that the atoms in
first-layer atom most metals are packed together as closely as possible. This arrangement is
Figure 2.57 Atoms in two layers of a metal called ‘close packing’.
crystal.
Figure 2.56 shows a model of a few atoms in one layer of a metal crystal.
Notice that each atom in the middle of the layer ‘touches’ six other atoms in
positive ion the same layer.
sea of When a second layer is placed on top of the first, atoms in the second layer
delocalised sink into the dips between atoms in the first layer (Figure 2.57).
electrons
This packing arrangement allows atoms in one layer to get as close as possible
to those in the next layer, so the structure of most metals is a giant lattice
of closely packed atoms in a regular pattern. In this giant lattice, electrons
Figure 2.58 Metallic bonding results from
in the outer shell of each metal atom are free to drift through the whole
the strong attractions between metal ions
structure. These electrons do not have fixed positions – they are described as
and the sea of delocalised electrons.
‘delocalised electrons’.
So, metallic bonding consists of positive ions with electrons moving around
and between them as a ‘sea’ of delocalised negative charge (Figure 2.58).
Key terms
The strong electrostatic attractions between the positive metal ions and the
Delocalised electrons are bonding ‘sea’ of delocalised electrons result in strong forces between the metal atoms.
electrons which are not fixed in a bond
between two atoms. They are free to The properties of metals
move and are shared by many atoms. In general, metals:
Metallic bonding is the strong ● have high melting and boiling temperatures
electrostatic attraction between metal ● have high densities
ions and the ‘sea’ of delocalised ● are good conductors of heat and electricity
electrons. ● are malleable – can be bent or hammered into different shapes.
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All the properties of metals can be explained and interpreted in terms of
their close-packed structure and delocalised electrons.
Tip
Metals in Group 1 have lower melting
● High melting and boiling temperatures – metal atoms are closely packed with
temperatures than other metals, for
strong forces of attraction between the positive ions and delocalised
example, sodium melts at 98 °C. An
electrons. So, it takes a lot of energy to move the positive ions away from
irregular piece of sodium melts to form
their positions in the giant lattice, allowing the metal to melt. It takes
a sphere as it reacts on the surface of
even more energy to separate individual atoms in the metal at the boiling
cold water. Atoms of Group 1 metals
temperature.
have only one electron in their outer
● High densities – the atoms are close packed with as little space between
energy level, so the delocalised cloud
them as possible.
contains only one electron per atom
● Good conductors of heat – when a metal is heated, energy is transferred to
and the metallic bonding is relatively
the electrons. The delocalised electrons in the heated region move around
weak. Magnesium (two outer electrons)
faster and conduct the heat (energy) rapidly to other parts of the metal.
and aluminium (three outer electrons)
● Good conductors of electricity – when a potential difference is applied across a
and the transition metals have more
metal, the delocalised electrons are attracted to the positive electrode and
delocalised electrons per atom, so the
flow through the metal. This flow of electrons is an electric current.
forces of attraction in the lattice and
● Malleable – the bonds between metal atoms are strong, but they are not
the resulting melting temperatures are
directional because the delocalised electrons can drift throughout the
higher than for sodium.
lattice and attract any of the positive ions. When a force is applied to a
metal, lines or layers of atoms can slide over each other. This is known
as ‘slip’. After slipping, the atoms settle into close-packed positions again.
Figure 2.59 shows the positions of atoms before and after slip. This is what
Tip
happens when a metal is bent or hammered into different shapes. It is often convenient to reduce the
malleability of a metal to make it
force harder. This can be achieved by adding
applied
here
other metals or carbon to the metal.
Atoms of different sizes in the lattice
disrupt the layers of atoms and make
it more difficult for layers to slide over
each other. These mixtures are called
alloys and have important engineering
a) b)
uses, e.g. the addition of a few percent
Figure 2.59 Positions of atoms in a metal a) before and b) after ‘slip’ has occurred.
of tungsten and molybdenum to iron
produces harder steel used for high
speed drill bits.
Test yourself
51 Why are the electrons in the outermost shell of metal atoms
described as ‘delocalised’?
52 Look carefully at Figures 2.56 and 2.57.
a) Choose one central atom in a layer. How many atoms in the
same layer touch this atom?
b) How many atoms in the layer above this first layer also touch this
atom?
c) In total, how many atoms touch a single metal atom in a close-
packed arrangement.
53 Explain why most metals:
a) have high densities
b) are good conductors of electricity.
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Test yourself
54 Name a metal or alloy and its particular use to illustrate each of the
following typical properties of metals:
a) shiny
b) conduct electricity
c) bend without breaking
d) high tensile strength.
55 Consider the patterns of metal properties in the Periodic Table.
Figure 2.60 Blacksmiths rely on the a) Which metals have:
malleability of metals to hammer and bend i) relatively high densities
them into useful shapes.
ii) relatively low densities?
b) Which metals have:
i) relatively high melting temperatures
ii) relatively low melting temperatures?
Activity
Choosing metals for different uses
Various properties of six metals are shown in Table 2.10.
Table 2.10 Some metals and their properties.
Metal Density/g cm−3 Tensile Melting Electrical Thermal Cost per
strength/ temperature/°C resistivity/ conductivity/ tonne/£
107 N m−2 10 −8ohm m J s−1 cm−1 K−1
Aluminium 2.7 8 660 2.5 2.4 960
Copper 8.9 33 1083 1.6 3.9 1 200
Iron 7.9 21 1535 8.9 0.8 130
Silver 10.5 25 962 1.5 4.2 250 000
Titanium 4.5 23 1660 43.0 0.2 27 000
Zinc 7.1 14 420 5.5 1.1 750
1 Use the information in Table 2.10 to explain the following 2 If the atoms in a metal pack closer together then the density
statements. should be higher, the bonds between atoms should be
a) Copper is used in most electrical wires and cables, but stronger and so the melting temperature should be higher.
high-tension cables in the National Grid are made of This suggests there should be a relationship between the
aluminium. density and melting temperature of a metal.
b) Bridges are built from steel which is mainly iron, even Use the data in the table to check if there is a relationship
though the tensile strength of iron is lower than that of between density and melting temperature. State ‘yes’ or ‘no’
some other metals. and explain your answer.
c) Metal gates and dustbins are made from steel coated with 3 The explanation of both electrical and thermal conductivity
zinc (galvanised). in metals uses the concept of delocalised electrons. This
d) Silver is no longer used to make our coins. suggests that there should be a relationship between the
e) Aircraft are now constructed from an aluminium/titanium electrical and thermal conductivities of metals.
alloy, rather than pure aluminium. Use the data in the table to check if there is a relationship.
f) The base of high-quality saucepans is copper rather than (Hint: electrical resistivity is the reciprocal of electrical
steel (iron). conductivity.) State ‘yes’ or ‘no’ and explain your answer.
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Activity
Structure, bonding and physical properties
The physical properties of a substance can be predicted a substance in terms of its structure and bonding and then use
knowing its structure and bonding. Similarly, structure and numerical data to predict the type of structure and bonding.
bonding can be deduced from physical properties. In this Solids exist in four types of structure; these types and some
activity you will predict the physical properties of properties are shown in Table 2.11.
Table 2.11 The four types of solid structure and some properties.
Type of structure Giant ionic lattice Giant metallic Simple molecular Giant covalent
lattice (covalent) lattice
Type of substance Compound of metal Metal element Non-metal element Non-metal element
and non-metal or compound of or compound of
non-metals non-metals
Attraction between particles Strong Strong Weak Strong
Melting temperature Mostly high
Electrical conductivity of solid Poor Poor Poor
Solubility in water
Example
1 Copy and complete the table by adding the missing 4 Explain why simple molecular solids are poor electrical
properties and examples. conductors. Give an example of a simple molecular
2 State why solid ionic compounds do not conduct electricity substance which conducts when dissolved in water and
and explain under what conditions ionic compounds can be explain why the solution conducts electricity.
electrolysed. 5 Name a giant covalent substance which does conduct
3 Transition metals have high melting temperatures. Give electricity and explain why it is a conductor.
an example of a group of metals with much lower melting 6 Table 2.12 gives some properties of substances A to H. Use
temperatures and suggest why these are different from this information to identify the type of bonding and structure
transition metals. of these substances. It is not expected that the actual
identity of each substance is deduced.
Table 2.12 Some properties of the substances A–H.
Melting Boiling Electrical conductivity Electrical conductivity Electrical conductivity
temperature/K temperature/K as solid as liquid in aqueous solution
A 918 1563 Poor Good Good
B 162 319 Poor Poor Poor
C 2345 3253 Poor Good Insoluble
D 302 942 Good Good Good
E 185 206 Poor Poor Good
F 1883 2503 Poor Poor Insoluble
G 1728 3003 Good Good Insoluble
H 279 353 Poor Poor Insoluble
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Chapter summary
Chapter 2 Bonding and structure planar (3 bp); tetrahedral (4 bp); trigonal-bipyramid
(5 bp); octahedral (6 bp).
l Material exists in two types of structure: giant and l Lone pairs (lp) of electrons are held closer to the
simple molecular. central atom than bonding pairs (bp), so lone pairs
l Giant structures are present in ionic solids,
have a stronger repelling effect than bonding pairs.
in covalently bonded solids, such as diamond The strength of repulsion between electron pairs
and silicon(IV) oxide, and in solid metals. In is: lp–lp > lp–bp > bp–bp. This repulsion reduces
these giant lattices, the atoms or ions are linked bond angles slightly.
throughout the crystal by a network of strong l Electronegativity is the ability of an atom to attract
bonding. the bonding electrons in a covalent bond. Ionic
l Simple molecular structures consist of groups of
and covalent bonding are the extremes of bonding
atoms held together by strong covalent bonding types. Electronegativity differences lead to bond
within the molecules, but with weak forces of polarity in bonds and molecules.
attraction between the molecules. These include l Intermolecular forces include interactions between
substances such as iodine and ice. temporary and induced dipoles (London forces) and
l There are three types of bonding: ionic, covalent
between permanent dipoles. Hydrogen bonding in
and metallic. molecules such as water gives rise to its anomalous
l Ionic bonding is the strong electrostatic attraction
properties.
between oppositely charged ions. This attraction l The physical properties of alkanes, alcohols and
is stronger between ions with larger charges and hydrogen halides depend on the type and strength
smaller radii. Positive ions form when metal atoms of their intermolecular forces.
lose electrons and negative ions form when non- l The solubility of ionic substances in water
metal atoms gain electrons. The physical properties involves the hydration of ions. The solubility of
of ionic compounds and the migration of ions molecular substances in water and in non-aqueous
provide evidence for the existence of ions. solvents depends on the relative strength of the
l Simple ions and their formation by the transfer of
intermolecular forces present.
electrons between atoms can be shown using dot- l Metallic bonding is the strong electrostatic
and-cross diagrams. attraction between metal ions and the delocalised
l A single covalent bond is the strong electrostatic
electrons. In general, metals have high melting
attraction between two nuclei and the shared and boiling temperatures, high densities and high
pair of electrons between them. Double bonds electrical conductivity.
and triple bonds have two and three shared pairs, l Allotropes of carbon with different structures
respectively. Multiple bonds are shorter and formed by carbon atoms include graphite, diamond
stronger than single bonds. Larger atoms form and graphene.
longer and therefore weaker bonds. l Data including melting and boiling temperatures,
l A dative covalent (coordinate) bond is one in
electrical conductivity and solubility in water
which two atoms share a pair of electrons, both can be used to predict the type of structure and
electrons being donated by one atom. bonding present in a substance.
l The shape of a simple molecule or ion is
l Information about the structure and bonding
determined by the repulsion between the electron of a substance can be used to predict its physical
pairs that surround a central atom. The shapes of properties.
molecules are linear (2 bonding pairs); trigonal
78 2 Bonding and structure
469983_02_Chem_Y1-2_039-[Link] 78 13/04/19 10:14 PM
Exam practice questions
1 The table shows the melting temperatures of 4 a) Draw dot-and-cross diagrams, with outer
the elements in Period 3 of the Periodic Table. shell electrons only, to show the bonding
in ammonia and water. (2)
Element Melting temperature/K
b) On mixing with water, ammonia reacts
Na 371 to form an alkaline solution containing
Mg 922 ammonium ions and hydroxide ions.
i) Write an equation for this reaction,
Al 933
including state symbols. (2)
Si 1683 ii) The ammonium ion has dative covalent
P 317 bonding. Explain the term ‘dative
covalent bonding’. (2)
S 386
iii) Draw a dot-and-cross diagram of the
Cl 172 ammonium ion and label the dative
Ar 84 covalent bond. (2)
a) Explain why the melting temperature 5 Draw dot-and-cross diagrams of the following
of sodium is much lower than that of molecules and ions. Predict the shape and give
magnesium. (3) the bond angle in each case.
b) Phosphorus and sulfur exist as molecules a) H3O+ (4)
of P4 and S8, respectively. Explain their b) IF5 (4)
difference in melting temperatures. (2) c) ClO3−(4)
c) State the type of structure and the nature d) PO43− (4)
of the bonding in each of the following 6 a) Phosphorus forms the chloride PCl3. Draw
elements: a dot-and-cross diagram for PCl3. (2)
i) aluminium ii) silicon b) Draw and name the shape of the PCl3
iii) chlorine. (6) molecule and give the bond angle. (3)
2 This question is about calcium and calcium c) Explain why PCl3 has this shape and this
oxide. angle. (3)
a) i) Describe the bonding in calcium. (3) d) Explain why PCl3 forms a stable compound
ii) Explain why calcium is a good with BCl3. (3)
conductor of electricity. (2) e) State the Cl−P−Cl bond angle and the
b) Draw dot-and-cross diagrams for the ions Cl−B−Cl bond angle in the compound
in calcium oxide showing all the electrons formed and explain your answer. (3)
and the ionic charges. (4) 7 a) State the types of intermolecular forces
c) State the conditions under which calcium present in:
oxide conducts electricity. Explain your i) propane ii) ethanol. (2)
answer. (6) b) Explain why the boiling temperature of
3 a) Using sodium chloride, hydrogen chloride propane (−42.2 °C) is lower than the boiling
and copper, explain what is meant by temperature of ethanol (78.5 °C). (2)
covalent, ionic and metallic bonding. (9) c) Glycerol (IUPAC name propane-1,2,3-triol)
b) Compare and explain the conduction of is a type of alcohol.
electricity by sodium chloride and copper H H H
in terms of structure and bonding. (3)
H C C H
C H
c) By considering their lattice structures,
explain why sodium chloride is brittle but OH OH OH
copper is malleable. (3)
79
Exam practice questions
469983_02_Chem_Y1-2_039-[Link] 79 13/04/19 10:14 PM
E
xplain why glycerol is very viscous and 13 When space travel was being pioneered, one of
predict whether the boiling temperature the first rocket fuels was hydrazine, H2NNH2.
of glycerol is higher or lower than that of a) Draw a dot-and-cross diagram to show the
ethanol, giving your reasons. (3) electron structure of a hydrazine molecule. (2)
8 Deduce the shapes of the molecules CF4, SF4 b) Predict the size of the H−N−H bond angle
and XeF4 and compare their overall polarity. (6) in a hydrazine molecule and explain your
reasoning. (3)
9 Explain why: c) Suggest a possible equation for the reaction
a) FCl is polar but F2 is not (2) which occurs when hydrazine vapour burns
b) SO2 is polar but CO2 is not (2) in oxygen. (2)
c) NCl3 is polar but BCl3 is not. (2) d) When 1.00 g of hydrazine burns in excess
10 a) Name the strongest of the intermolecular oxygen, 18.3 kJ of thermal energy is
forces between water molecules, and released. Calculate the enthalpy change of
describe the bonding with the help of a combustion of hydrazine. (2)
diagram. (3) 14 a) Draw a dot-and-cross diagram for
b) Explain why: methylamine, CH3NH2. (1)
i) the boiling temperature of water is b) Give approximate values for the H−C−H
higher than the boiling temperatures and C−N−H bond angles in methylamine.
of the other hydrides of Group 6 Explain your answer. (5)
elements c) Amines such as methylamine form hydrogen
ii) ice is less dense than water at 0 °C bonds with each other. Using displayed
iii) water and pentane are immiscible formulae, draw a diagram to show the
liquids hydrogen bond between two methylamine
iv) methoxymethane (CH3−O−CH3) boils molecules and give the bond angle around
at −24.8 °C but ethanol, an isomer of the shared hydrogen atom.(3)
methoxymethane, boils at 78.5 °C. (8) d) i) Write the formula of the compound
formed when methylamine reacts with
11 Diamond and graphite are described as allotropes.
hydrogen chloride. (1)
a) Explain what is meant by the term
ii) Give the C−N−H bond angle in this
‘allotropes’. (2)
product and explain your answer. (2)
b) State why fullerenes and graphene are
also allotropes together with diamond and 15 Predict three possible arrangements of bonds
graphite. (1) for the ICl3 molecule. By considering the
c) Graphite fibres are often used for the electron-pair repulsions in each of your
brushes (contacts) in electric motors. structures, suggest which is the most likely
i) Give two reasons why graphite fibres shape and justify your answer. (9)
are used in this way. (2) 16* Explain why the melting temperatures of the
ii) Give three reasons why diamonds Group 7 elements rise down the group, but the
would be unsuitable for this use. (3) melting temperatures of the Group 1 elements
12 The covalently bonded compound urea has fall down the group. (6)
the formula (NH2)2C=O. Urea is commonly 17 Thin streams of some liquids are attracted
used as a fertiliser in most of Europe, whereas towards a charged rod, but with other liquids
ionic ammonium nitrate, NH4NO3, is the most there is no effect.
popular fertiliser in the UK. a) Explain why some liquids are attracted
a) Draw a dot-and-cross diagram for urea. (2) while others are not. (2)
b) Describe the arrangement of atoms b) Predict which of the following liquids
i) around the carbon atom in urea are deflected towards a charged rod and
ii) around a nitrogen atom in urea. (2) explain your predictions: water, hexane,
c) Suggest two advantages of using urea as a bromoethane, tetrachloromethane. (4)
fertiliser compared with ammonium nitrate.(2) c) Why are the affected liquids always attracted
towards the charged rod and not repelled? (2)
80
2 Bonding and structure
469983_02_Chem_Y1-2_039-[Link] 80 13/04/19 10:14 PM
Substance Melting Boiling Electrical Electrical Electrical
temperature/°C temperature/°C conductivity conductivity conductivity
as solid as liquid in water
A −113 101 Poor Poor Insoluble
B 1510 2370 Good Good Insoluble
C 1020 1660 Poor Good Good
D 1883 2503 Poor Poor Insoluble
E −88 −67 Poor Poor Good
F 2072 2980 Poor Good Insoluble
G −39 357 Good Good Insoluble
18 a) Use the information in the table above b) State the type of intermolecular forces
to deduce the bonding (ionic, covalent or between the molecules of H2S and between
metallic) in the following substances and molecules of H2Se. (1)
whether they exist as giant lattices or small c) Use your graph to estimate the value
molecules or neither. (7) ΔHvaporisation for water, assuming that the
b) Given that the seven substances are aluminium only intermolecular forces in water are the
oxide, 1-bromobutane, hydrogen bromide, same as in the other hydrides in Group 6. (1)
manganese, mercury, silicon dioxide and d) Use your graph and the answer to c) to
sodium bromide, deduce the letter of each. (7) deduce a value for the strength of hydrogen
bonding in water.(1)
19 Explain each of the following: 21* Consider the following three molecules:
a) Sodium has a higher melting temperature
than potassium. (4) H H
b) Magnesium oxide has a higher melting H Cl
temperature than magnesium chloride. (3) Cl C C Cl C C
Cl H
c) The boiling temperature of chlorine is
H H
238 K, but temperatures in excess of 1300 K
are needed to form chlorine atoms from 1,2-dichloroethane E-1,2-dichloroethene
chlorine molecules. (4)
d) When aluminium chloride is heated, it Cl Cl
sublimes at 451 K to form vapour which C C
contains Al2Cl6 molecules. (4) H H
20 The enthalpy change of vaporisation of Z-1,2-dichloroethene
a liquid is a measure of the strength of its (IUPAC names of organic molecules such as
intermolecular forces. The table shows values these are studied in Chapter 6.1.)
for the enthalpy change of vaporisation of the
Deduce whether each molecule has an overall
hydrides of Group 6 elements.
dipole and justify your answer. (6)
Compound ΔHvaporisation/ 22 Hydrogen reacts with sodium to form sodium
kJ mol−1 hydride, an ionic compound which has the
H2 O 40.7
same lattice structure as sodium chloride.
a) i) Write an equation, including state
H2 S 18.7 symbols, for the formation of sodium
H2Se 19.3 hydride from its elements. (2)
H2Te 23.2
ii) Draw dot-and-cross diagrams for the
ions in sodium hydride showing the
a) Plot a graph of ΔHvaporisation against molar outer electrons and the ionic charges. (2)
mass for the four compounds. (4)
81
Exam practice questions
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b) Electrolysis of sodium hydride dissolved in 24 The boiling temperatures of five compounds
molten sodium chloride produces hydrogen. are shown in the table.
State at which electrode hydrogen is
discharged and write a half-equation for the Name Formula Boiling
formation of hydrogen in this electrolysis. (2) temperature/°C
c) Unlike sodium chloride, sodium hydride Water H2O 100
does not simply dissolve in water, but reacts Methanol CH3OH 65
with water to form a strongly alkaline
solution. Ammonia NH3 −33
Write an equation for the reaction of the Ethanamide CH3CONH2 221
hydride ion with water and state the role of Ethanoic acid CH3COOH 118
the hydride ion in this reaction. (2)
d) Magnesium hydride has been suggested a) Explain why the boiling temperature of
as a medium for the storage of hydrogen. water is higher than that of methanol and
Comment on the likely bonding in much higher than that of ammonia. (6)
magnesium hydride and give two possible b) Consider the structures of ethanamide and
ways to release hydrogen from magnesium ethanoic acid and predict why the boiling
hydride. (4) temperature of ethanamide is higher than
23 a) Benzenecarboxylic acid (C6H5COOH) is that of ethanoic acid. (4)
almost insoluble in cold water; only 2.9 g NOTE: Part (b) requires study beyond
dissolves in 1 dm3 water at 25 °C. Year 1.
However about 70 g of the acid will dissolve
25 The electronegativity value of tin is 1.8. Tin
in 1 dm3 of tetrachloromethane at 25 °C. In
reacts with fluorine to form a fluoride which
this solution, the solute particles are dimers
contains 61.0% of tin and which has a melting
of benzenecarboxylic acid.
temperature of 705 °C. Tin also reacts with
i) Explain why, at 25 °C,
iodine to form an iodide which contains 19.0%
benzenecarboxylic acid is much more
of tin and which has a melting temperature
soluble in tetrachloromethane than in
of 144 °C. Use other electronegativity values
water. (2)
from Table 2.5 to help you discuss the bonding
ii) Draw a structure for the dimer of
types in these two tin halides. Explain why the
benzenecarboxylic acid formed in
compounds are different. (9)
tetrachloromethane and suggest how
and why it forms. (4)
b) A conical flask contains 2.90 g of
benzenecarboxylic acid. Aqueous sodium
hydroxide solution is added from a burette
and the mixture shaken until a colourless
solution is formed.
i) Write an equation for the reaction
of sodium hydroxide with
benzenecarboxylic acid. (1)
ii) Calculate the volume of
0.500 mol dm−3 sodium hydroxide
needed to react exactly with 2.90 g of
benzenecarboxylic acid. (3)
iii) Explain why the organic product of the
reaction is very soluble in water. (2)
82
2 Bonding and structure
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Redox I
3
3.1 Oxidation and reduction
METAL EXTRACTION Oxidation and reduction reactions are very common. Chemists have devised
oxygen is removed from a number of ways of recognising and describing what happens during changes
the oxide (reduction)
of this kind.
ore Burning is perhaps the commonest example of oxidation. Another example
iron oxide iron metal is rusting, which converts iron to a form of iron oxide. At its simplest,
rust
oxidation involves adding oxygen to an element or compound.
CORROSION Reduction is the opposite of oxidation. Metal oxides are reduced during the
the metal combines extraction of metals from their ores. In a blast furnace, for example, carbon
with oxygen (oxidation)
monoxide takes the oxygen away from iron oxide to leave metallic iron
Figure 3.1 The cycle of extraction and (Figure 3.1).
corrosion for iron.
Tip Test yourself
The elements oxygen and hydrogen can 1 In terms of gain or loss of oxygen, which element or compound is
be regarded as chemical opposites oxidised and which is reduced in the reaction of:
in oxidation and reduction reactions.
a) steam with hot magnesium
Older definitions also defined oxidation
at the loss of hydrogen and reduction b) copper(ii) oxide with hydrogen
as the gain of hydrogen. Defining c) aluminium with iron(iii) oxide
oxidation and reduction in terms of the d) carbon dioxide with carbon to form carbon monoxide?
loss or gain of hydrogen is now rarely
used, except in organic chemistry.
3.2 Equations to explain oxidation
and reduction reactions
Balanced symbol equations
Chemists write equations, with symbols and formulae, to describe and
explain what happens during reactions. Writing a word equation is a useful
first step towards a balanced chemical equation with symbols. This is because
it is not possible to write an equation without first knowing the identity of
the reactants and products. Figure 3.2 shows sparks from a sparkler – these
Figure 3.2 When sparklers burn, bits of sparks are bits of burning magnesium. When magnesium burns in air, it
magnesium react with oxygen in the air. reacts with oxygen to form white magnesium oxide.
3.2 Equations to explain oxidation and reduction reactions 83
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Example
Write a balanced equation for the reaction of magnesium with oxygen.
Notes on the method
There are four key steps in writing an equation. Before writing the
equation you must identify the reactants and the products.
Oxygen, hydrogen, nitrogen and the halogens are diatomic molecules with
two atoms in their molecules – so they are written as O2, H2, N2, F2, Cl2,
Br2 and I2. There are some other molecular non-metals but most other
elements are shown as single atoms.
Never change a formula to make an equation balance. The formula of
magnesium oxide is always MgO and never MgO2 or Mg2O, or anything
else for that matter.
Answer
Step 1: Write a word equation for the reaction.
magnesium + oxygen → magnesium oxide
Step 2: Write symbols for the elements and formulae for the compounds
in the word equation.
Mg + O2 → MgO
Step 3: Balance the equation by putting numbers in front of the symbols
and formulae, so that the number of each type of atom is the
same on both sides of the equation.
2Mg + O2 → 2MgO
Step 4: Add state symbols to show the state of each substance in the
equation. Use (s) for solid, (l) for liquid, (g) for gas and (aq) for an
aqueous solution (a substance dissolved in water).
2Mg(s) + O2(g) → 2MgO(s)
Balanced chemical equations are important because they:
Test yourself
● identify the reactants and products with their formulae
2 Write balanced equations, ● indicate relative numbers of atoms and molecules in the reaction
with state symbols, for these ● make it possible to calculate the amounts of the chemical substances
reactions: involved (Chapter 5).
a) steam with hot magnesium
b) copper(ii) oxide with Electron transfer
hydrogen The compound formed when magnesium burns in air is ionic. It is made up
c) aluminium with iron(iii) oxide of magnesium ions, Mg2+, and oxide ions, O2−. During the reaction, each
d) carbon dioxide with carbon magnesium atom gives up two electrons, turning into a magnesium ion:
to form carbon monoxide. 2Mg → 2Mg2+ + 4e−
Oxygen takes up the electrons from the magnesium producing oxide ions:
O2 + 4e− → 2O2−
In this way, electrons transfer from magnesium atoms to oxygen atoms,
forming ions from atoms.
84 3 Redox I
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Magnesium atoms also turn into ions when they react with other non- Mg Mg2+ + 2e
_
metals such as chlorine, bromine and sulfur. In all its reactions with
non-metals, magnesium loses electrons and its atoms become positive ions.
_ _
The atoms of the non-metal gain electrons and become negative ions. So all Cl2 + 2e 2Cl
the reactions involve electron transfer (Figure 3.3). Magnesium is oxidised as
it loses electrons; the non-metal is reduced as it gains electrons. Reduction Figure 3.3 Electron transfer in the reaction
and oxidation go together in these redox reactions. of magnesium with chlorine.
Table 3.1 The charges on common ions.
Positive ions (cations) Negative ions (anions)
Charge Cation Symbol Charge Anion Symbol
1+ Sodium Na+ 1− Chloride Cl−
Potassium K+ Bromide Br−
Silver Ag+ Iodide I− Tip
Copper(i) Cu+ Hydroxide OH− You may find the mnemonic, oil rig
Hydrogen H+ Nitrite NO2−
Ammonium NH4+ Nitrate NO3−
helpful when thinking about redox in
terms of the loss or gain of electrons.
2+ Magnesium Mg2+ 2− Oxide O2−
Calcium Ca2+ Sulfide S2− Oxidation Reduction
Zinc Zn2+ Sulfite SO32− Is Is
Copper(ii) Cu2+ Sulfate SO42−
Iron(ii) Fe2+ Carbonate CO32− Loss Gain
3+ Aluminium Al3+ 3− Nitride N3−
Iron(iii) Fe3+ Phosphate PO43−
Notice from Table 3.1 that:
Key terms
● metal ions are always positive
+
● non-metal ions are negative except hydrogen, H , and ammonium, NH4
+ A redox reaction is a reaction that
● some metals can form more than one ion – this is characteristic of metals involves reduction and oxidation.
in the d block such as copper and iron An oxoanion is an ion with the general
● some non-metal ions are compound ions containing more than one kind of
formula is X xOyz− (where X represents
atom, including oxoanions such as the sulfate, nitrate and phosphate ions. any element while O represents
an oxygen atom). Metals and non-
Ionic half-equations metal elements form oxoanions. The
A half-equation is used to describe either the gain or the loss of electrons oxoanions of non-metals shown in
during a redox process. Half-equations help to show what is happening Table 3.1 are particularly common.
during a reaction. Two half-equations combine to give the overall balanced A half-equation is an ionic equation
equation. used to describe either the gain, or
the loss, of electrons during a redox
Zinc metal can reduce copper ions to copper. This happens when pieces
reaction.
of zinc are added to a solution of copper(ii) sulfate (Figure 3.4). In this
example of a displacement reaction the more reactive metal, zinc, Displacement reactions are redox
displaces the less reactive metal, copper. The reaction can be shown as two reactions which can be used to
half-equations: compare the relative strengths of
metals as reducing agents and non-
electron gain (reduction): Cu2+(aq) + 2e− → Cu(s) metals as oxidising agents. A more
electron loss (oxidation): Zn(s) → Zn 2+(aq) + 2e− reactive metal displaces a less reactive
metal from one of its salts.
3.2 Equations to explain oxidation and reduction reactions 85
469983_03_Chem_Y1-2_083-[Link] 85 13/04/19 10:13 PM
Note that the sulfate ions in copper(ii) sulfate are not shown. They are left out
of the half-equation because they are the same before and after the reaction.
For this reason they are called spectator ions. For clarity and simplicity,
chemists often omit spectator ions from equations.
Adding together the two half-equations leads to the full ionic equation.
The electrons must cancel in order that the electrons gained on one side
of the equation equal the electrons lost on the other side. It is sometimes
necessary to multiply one of the half-equations by a factor of two or three
such that the electrons gained and lost are equal.
Cu2+(aq) + 2e− → Cu(s)
Zn(s) → Zn 2+(aq) + 2e−
Cu2+(aq) + Zn(s) → Cu(s) + Zn 2+(aq)
Test yourself
3 With the help of Table 3.1, write the separate ionic half-equations for
Figure 3.4 Zinc displacing copper from the reactions of:
copper(ii) sulfate solution. The copper a) sodium with chlorine
appears as a reddish solid. Colourless zinc
b) zinc with oxygen
ions replace the copper ions in solution.
c) calcium with bromine.
Key terms 4 Write the ionic half-equations and the full ionic equation for the
reaction of zinc with silver nitrate solution.
Spectator ions are ions which are present
in solution but take no part in a reaction.
Ionic equations describe chemical
3.3 Oxidation numbers
changes by showing only the reacting Chemists use oxidation numbers to keep track of the electrons transferred
ions and any other reacting atoms or shared during chemical changes. With the help of oxidation numbers
or molecules, while leaving out the it becomes much easier to recognise redox reactions. Oxidation numbers
spectator ions (Section 4.1). also provide a useful way of organising the chemistry of elements such as
chlorine, which can be oxidised or reduced to varying degrees. Chemists
Oxidation originally meant combination
base the names of inorganic compounds on oxidation numbers.
with oxygen, but the term now covers
all reactions in which atoms, molecules
or ions lose electrons. The definition is Oxidation numbers and ions
extended to cover molecules, as well as Oxidation numbers show how many electrons are gained or lost by an element
atoms and ions, by defining oxidation when atoms turn into ions and vice versa. In Figure 3.5, movement up the
as a change which makes the oxidation diagram involves the loss of electrons and a shift to more positive oxidation
number of an element more positive, or numbers – this is oxidation. Movement down the diagram involves the
less negative. gain of electrons and a shift to less positive, or more negative, oxidation
numbers – this is reduction.
Reduction originally meant removal
of oxygen or addition of hydrogen, but The oxidation number of all uncombined elements is zero. In a simple ion,
the term now covers all reactions in the oxidation number of the element is the charge on the ion. For example,
which atoms, molecules or ions gain in calcium chloride the metal is present as the Ca 2+ ion and the oxidation
electrons. The definition is extended to number of calcium is +2.
cover molecules, as well as atoms and
Oxidation numbers distinguish between the compounds of elements such as
ions, by defining reduction as a change
iron that can exist in more than one oxidation state. In iron(ii) chloride the
which makes the oxidation number of an
Roman number ‘ii’ shows that iron is in oxidation state +2. Iron atoms lose two
element more negative, or less positive.
electrons when they react with hydrogen chloride to make iron(ii) chloride.
86 3 Redox I
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+3 Fe3+
Tip
It is ambiguous just to state that
+2 Mg2+ Fe2+
oxidation numbers get higher or lower
because both positive and negative
+1 Na+ numbers are involved. To be clear, state
Oxidation number
that during oxidation the oxidation
reduction
oxidation
0 Mg Fe Na O2 Cl2 number of an element gets more
positive, or less negative; while during
reduction the oxidation number of an
–1 Cl–
element gets less positive, or more
negative.
–2 O2–
Figure 3.5 Oxidation numbers of atoms and ions.
Oxidation number rules
1 The oxidation number of uncombined elements is zero.
2 In ions made of just one atom the oxidation number of the element
is the charge on the ion.
3 The sum of the oxidation numbers in a neutral compound is zero.
4 The sum of the oxidation numbers for an ion is the charge on the ion.
5 Some elements have fixed oxidation numbers in all their compounds.
Metals Non-metals
Group 1 metals +1 hydrogen +1
(e.g. Li, Na, K) (except in metal hydrides, H–)
Group 2 metals +2 fluorine –1
(e.g. Mg, Ca, Ba)
aluminium +3 oxygen –2
(except in peroxides, O22–, and
compounds with fluorine)
chlorine –1 NH4+ MnO4–
(except in compounds with –3 +1 +7 –2
oxygen and fluorine)
Figure 3.6 Oxidation number rules.
SO42– Cr2 O72–
+6 –2 +6 –2
Figure 3.7 Oxidation numbers in ions
With the help of the rules in Figure 3.6, it is possible to extend the use of with more than one atom. Note the use
oxidation numbers to ions consisting of more than one atom. The charge of 2− for the electric charge on a sulfate
on an ion, such as the sulfate ion, is the sum of the oxidation numbers of the ion (number first for ionic charges) but the
atoms. The normal oxidation state of oxygen is −2. There are four oxygen use of −2 to refer to the oxidation state of
atoms (four at −2) in the sulfate ion, so the oxidation state of sulfur must be oxygen in the ion (plus or minus first for
+6 to give an overall charge on the ion of −2 (Figure 3.7). oxidation states in ions and molecules).
3.3 Oxidation numbers 87
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Care is necessary when assigning oxidation numbers in ions where there are
covalent bonds between same element. The formula of the peroxide ion,
O22−, for example is:
[O-O]2−
In this ion the oxidation number of each oxygen atom is −1 (and not −2 as
usual). This follows from the rule that the sum of the oxidation numbers
sulfur at +6
adds up to the charge on the ion.
H2 SO4 Oxidation numbers and molecules
The rules in Figure 3.6 make it possible to apply the definitions of oxidation
two hydrogens at +1 four oxygens at –2
and reduction to molecules. In most molecules, the oxidation state of an
Figure 3.8 Oxidation numbers of the
atom corresponds to the number of electrons from that atom that are shared
elements in sulfuric acid.
in covalent bonds (Figure 3.8).
Where two atoms are linked by covalent bonds, the more electronegative
atom (Section 2.5) has the negative oxidation state. Fluorine always has a
negative oxidation state of −1 because it is the most electronegative of all
atoms. Oxygen normally has a negative oxidation state (−2) but it has a
positive oxidation state (+1) when combined with fluorine.
Tip
Oxidation numbers are written with the + or − in front of the number: +1, +2 or −1, −2.
Key term
This is to make it quite clear that when dealing with molecules these numbers do not
Oxidation states are the states of refer to electric charges, unlike charges on ions such as Ca2+ or N3−. Molecules are not
oxidation, or reduction, shown by an charged. The sum of the oxidation states for all the atoms in a molecule is zero.
element in its chemistry. The states are
labelled with the oxidation numbers of
the element in that state. Test yourself
5 What is the oxidation number of:
a) aluminium in aluminium b) nitrogen in magnesium
+5 BrO3–
oxide, Al2O3 nitride, Mg3N2
+4 c) nitrogen in barium nitrate, d) nitrogen in the ammonium
Ba(NO3)2 ion, NH4+?
+3
6 Are these elements oxidised or reduced when they react to form these
+2 compounds?
+1 BrO– a) calcium to calcium bromide b) chlorine to lithium chloride
c) chlorine to chlorine dioxide d) sulfur to hydrogen sulfide
0 Br2
e) sulfur to sulfuric acid
–1 Br – HBr
Figure 3.9 Bromine is reduced when it Oxidation numbers and the chemistry
reacts to form bromide ions. A reaction
turning bromine into BrO− ions involves of elements
oxidation of bromine. The conversion of Oxidation numbers help to make sense of the chemistry of an element such
BrO− ions to BrO3− ions involves further as bromine (see Figure 3.9). The compounds of an element can be classified
oxidation of bromine. according to their oxidation states.
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The oxidation numbers of the elements lithium to chlorine in their oxides
reveal a periodic pattern when plotted against atomic number (Figure 3.10).
The most positive oxidation number for each element corresponds to the
number of electrons in the outer shell of the atoms.
+7
+6
+5
+4
+3
+2
Oxidation number
+1
0
Li Be B C N O F Ne Na Mg Al Si P S Cl
–1
–2
–3
–4
–5
–6 Oxidation numbers in oxides
Oxidation numbers in hydrides
Figure 3.10 Oxidation numbers of elements in their oxides and hydrides. Note that some
elements form oxides in a variety of oxidation states.
Test yourself
7 This question refers to Figure 3.10.
a) Give the formula of the oxide of lithium.
b) Give the formulae of the two oxides of carbon.
c) What are the oxidation states of nitrogen in these oxides: NO, N2O,
NO2, N2O3, N2O5?
d) Give the formulae of the hydrides of nitrogen and phosphorus.
e) Why is there only one element with a negative oxidation number in
an oxide?
Oxidation numbers and the names
of compounds
The names of inorganic compounds are becoming increasingly systematic,
but chemists still use a mixture of names. Most prefer the name ‘copper
sulfate’ for the blue crystals with the formula CuSO4.5H 2O. This is
hydrated copper(ii) sulfate. Its fully systematic name, tetraaquocopper(ii)
tetraoxosulfate(vi)-1-water, is rarely used. This fully systematic name has
much more to say about the arrangement of atoms, molecules and ions
in the blue crystals but it is too cumbersome for normal use. The fully
systematic name also shows the oxidation states of copper and sulfur in the
compound.
3.3 Oxidation numbers 89
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Here are some of the basic rules for naming common inorganic compounds:
Tip
● the ending ‘-ide’ shows that a compound contains just the two elements
In practice, the oxidation number for
mentioned in the name. The more electronegative element comes second –
metals is always given in names where
for example, sodium sulfide, Na 2S, carbon dioxide, CO2, and magnesium
it can vary, but the oxidation number
nitride, Mg3N2
for the element in common oxoanions is
● the Roman numbers in names indicate the oxidation numbers of the
less often given. So the name iron(iii)
elements – for example iron(ii) sulfate, FeSO4, and iron(iii) sulfate,
nitrate is regularly used rather than
Fe2(SO4)3
iron(iii) nitrate(v); similarly copper(ii)
● the traditional names of oxoacids end in ‘-ic’ or ‘-ous’ as in sulfuric acid,
sulfate is much more common than
H2SO4, and sulfurous acid, H2SO3, as well as in nitric acid, HNO3, and
copper(ii) sulfate(vi).
nitrous acid, HNO2. The ‘-ic’ ending is for the acid in which the central
atom has the higher oxidation number
● the corresponding traditional endings for the salts of oxoacids are ‘-ate’
Test yourself and ‘-ite’ as in sulfate, SO42−, and sulfite, SO32−, and in nitrate, NO3−, and
nitrite, NO2−
8 Write the formulae of the ● the more systematic names for oxoacids and oxosalts use oxidation numbers
compounds: as in sulfate(vi) for sulfate, SO42−, sulfate(iv) for sulfite, SO32−, as well as
a) tin(ii) oxide nitrate(v) for nitrate and nitrate(iii) for nitrite.
b) tin(iv) oxide When in doubt, chemists give the name and the formula. In some cases, they
c) iron(iii) nitrate(v) may give two names – the systematic name and the traditional name.
d) potassium sulfate(vi).
3.4 Recognising redox reactions
Oxidation numbers help us to identify redox reactions. In the equation for
any redox reaction, at least one element changes to a more positive oxidation
state, while another changes to a less positive oxidation state. A reaction is
not a redox reaction if there are no changes of oxidation state.
Oxidising and reducing agents
An agent is someone or something which gets things done. In spy stories,
the main players are secret agents with a mission to make a change. In redox
reactions, the chemicals with a mission are the oxidising and reducing agents.
The term ‘oxidising agent’ (or oxidant) describes chemical reagents which
can oxidise other atoms, molecules or ions by taking electrons away from
them. Common oxidising agents are oxygen, chlorine, nitric acid, potassium
manganate(vii), potassium dichromate(vi) and hydrogen peroxide.
The term ‘reducing agent’ (or reductant) describes chemical reagents which
can reduce other atoms, molecules or ions by giving them electrons. Common
reducing agents are hydrogen, sulfur dioxide and zinc or iron in acid.
It is easy to get into a mental tangle when using these terms. An oxidising
agent reacts by removing electrons from the reducing agent. The oxidising
agent gains electrons and so is reduced, the reducing agent loses electrons
and so is oxidised (Figure 3.11).
Tip
Fluorine is a very powerful oxidising agent but it is much too reactive and dangerous
for normal use.
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Figure 3.11 Magnesium is oxidised by
+2 Mg2+
loss of electrons. It is oxidised by chlorine,
Magnesium so chlorine is the oxidising agent. At the
Oxidation number
+1 is oxidised – same time chlorine gains electrons and is
it loses The oxidising agent which
electrons takes the electrons reduced by magnesium. Magnesium is the
–
2e from magnesium reducing agent.
0 Mg Cl 2
The reducing agent Chlorine is reduced
which gives electrons – it gains electrons
to chlorine
–1 2Cl
–
Disproportionation reactions
In some reactions, the same element both increases and decreases its
oxidation number. In other words, some of the element is oxidised while the
rest of it is reduced. This is called disproportionation. One example is the
decomposition of hydrogen peroxide to oxygen and water.
2H2O2(aq) → 2H2O + O2
−1 −2 0
Half of the oxygen in hydrogen peroxide is reduced from the −1 to the −2
state in water, while the other half is oxidised from the −1 to the 0 state in
oxygen gas.
Another example of a disproportionation reaction takes place on warming
copper(i) oxide with dilute sulfuric acid. The reaction does not produce a Key term
solution of copper(i) sulfate. Instead, the products are a solution of copper(ii)
A disproportionation reaction involves
sulfate and a precipitate of copper metal.
an element in a single species being
Cu2O(s) + H2SO4(aq) → CuSO4(aq) + Cu(s) + H2O(l) simultaneously oxidised and reduced.
+1 +2 0
Half of the copper(i) is oxidised to copper(ii), while the rest is reduced to
copper(0). Reactions of this kind are important in the chemistry of the
halogens (Section 4.11).
Test yourself
9 Use oxidation numbers to show that these are disproportionation
reactions.
a) 2CO(g) → C(s) + CO2(g)
b) 3K2MnO4(aq) + 2H2O(l) → MnO2(s) + 2KMnO4(aq) + 4KOH(aq)
c) 2Ca(OH)2(s) + 2Cl2(aq) → CaCl2(aq) + Ca(ClO)2(aq) + 2H2O(l)
3.5 Balancing redox equations
Balancing redox equations using half-equations
Half-equations can help to balance equations for redox reactions because the
electrons lost when an atom, molecule or ion is oxidised in one half-equation
have to equal the electrons gained by the reduction of another species in the
second half-equation.
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The half-equations in Section 3.2 only included atoms, ions and electrons.
There are also redox half-equations for the reactions of molecules or ions
that include oxygen such as the nitrate ion, hydrogen peroxide and sulfur
dioxide. These chemicals often react in acid solutions and the half-equations
are balanced by including hydrogen ions and water molecules.
Tip
A redox half-equation must include the substance being oxidised or reduced plus
electrons. It may also include hydrogen ions (in acid solution) and water molecules.
Example
An acidic solution of chloric(i) acid, HOCl, is reduced to chloride ions as it
oxidises iodide ions to iodine. What is the balanced equation for the reaction?
Notes on the method
You have to know, or be told, what the reactants and products are before
you can write the equation for a reaction. You can then follow the five
steps illustrated in this example. In acid solution you can only use water
molecules or hydrogen ions to balance the half-equations for O and H.
You do not need to keep rewriting the equations so long as you leave spaces
on each side to add the water molecules, hydrogen ions and electrons.
Answer
Step 1: Write down the given information about the half-equations, then
balance the atoms being oxidised and reduced.
HOCl(aq) → Cl−(aq)
2I−(aq) → I2(aq)
Step 2: Balance the hydrogen and oxygen atoms by adding H2O and/or H+
(in acid solution).
HOCl(aq) + H+(aq) → Cl−(aq) + H2O(l)
2I−(aq) → I2(aq)
Step 3: Balance the electric charges by adding electrons.
HOCl(aq) + H+(aq) + 2e− → Cl−(aq) + H2O(l)
2I−(aq) → I2(aq) + 2e−
Test yourself Step 4: If necessary, multiply one half-equation so that the numbers of
electrons in each are the same, then add them, cancelling the
10 Use half-equations to write
electrons (in this example the numbers of electrons in each are
balanced ionic equations for already the same).
these redox reactions.
HOCl(aq) + H+(aq) + 2e− → Cl−(aq) + H2O(l)
a) H2O2 with Fe2+ to give H2O
and Fe3+ 2I−(aq) → I2(aq) + 2e−
b) SO32− and Cl2 to give HOCl(aq) + H+(aq) + 2I−(aq) → Cl−(aq) + H2O(l) + I2(aq)
SO42− and Cl−.
Step 5: If necessary, simplify the equation by cancelling molecules or ions
c) the disproportionation of that appear on both sides of the equation (not needed in this
IO − into I− and IO3− example).
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Balancing redox equations using oxidation
numbers
Oxidation numbers offer an alternative way to balance redox equations. This is
because the total decrease in oxidation number for the element that is reduced
must equal the total increase in oxidation number for the element that is oxidised.
Example
What is the balanced equation for the reaction of concentrated sulfuric acid
with hydrogen bromide? The products are bromine, sulfur dioxide and water.
Notes on the method
You have to know, or be told, what the reactants and products are before
you can write the equation for a reaction. You can then follow the five
steps illustrated in this example.
Answer
Step 1: Write down the formulae for the atoms, molecules and ions
involved in the reaction.
HBr + H2SO4 → Br2 + SO2 + H2O
Step 2: Identify the elements which change in oxidation number and the
extent of change.
In this example only bromine and sulfur show changes of oxidation
state.
Step 3: Balance so that the total increase in oxidation number of one
element equals the total decrease of the other element.
In this example, the increase of +1 in the oxidation number of two
bromine atoms (from −1 to 0) balances the −2 decrease of one
sulfur atom (from +6 to +4).
2HBr + H2SO4 → Br2 + SO2 + H2O
Step 4: Balance for oxygen and hydrogen.
In this example, the four hydrogen atoms on the left of the
equation join with the two remaining oxygen atoms to form two
water molecules.
2HBr + H2SO4 → Br2 + SO2 + 2H2O
Step 5: Add the state symbols.
2HBr(g) + H2SO4(l) → Br2(l) + SO2(g) + 2H2O(l)
Test yourself
11 Use oxidation numbers to write the full ionic equation for each of
these redox reactions.
State which element is oxidised and which is reduced in each
example.
a) Fe with Br2 to give FeBr3
b) F2 with H2O to give HF and O2
c) IO3− and H+ with I− to give I2, and H2O
d) S2O32− and I2 to give S4O62− and I−
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Activity
Preparing a sample of an oxide of nitrogen
Lead(ii) nitrate decomposes on heating to form three products. 1 Give the oxidation states of all the elements in:
These are lead(ii) oxide, nitrogen dioxide and oxygen. Nitrogen a) lead(ii) nitrate b) lead(ii) oxide
dioxide, NO2, is a brown gas. Some of the nitrogen dioxide c) nitrogen dioxide d) N2O4
molecules pair up to form N2O4. Cooling the gas mixture e) oxygen gas.
condenses the N2O4 as a greenish liquid (Figure 3.12). 2 Identify as oxidation, reduction or neither, the formation of:
a) lead(ii) oxide from lead(ii) nitrate
lead(ii) nitrate (20 g) b) nitrogen dioxide from lead(ii) nitrate
c) oxygen from lead(ii) nitrate
d) N2O4 from NO2.
3 Write balanced equations for:
a) the decomposition of lead(ii) nitrate
heat
b) the formation of N2O4 from NO2.
freezing
mixture
4 Calculate the theoretical mass of N2O4 that could be
(ice + salt) collected by condensing the gases given off when 20 g
Figure 3.12 Heating liquid N2O4 lead(ii) nitrate decomposes (Section 5.4).
lead(ii) nitrate in a fume 5 Suggest reasons why the mass of N2O4 collected by heating
cupboard to collect a 20 g lead(ii) nitrate in the apparatus in Figure 3.12 is less
sample of N2O4.
than your answer to Question 4.
Chapter summary
Chapter 3 Redox I numbers is the charge on the ion; the sum of the
oxidation numbers in a neutral compound is zero.
l A redox reaction involves both reduction and l Some elements have fixed oxidation numbers in all
oxidation. their compounds.
l Oxidation is a term that covers all reactions in
l Oxidation is a change which makes the oxidation
which atoms or ions lose electrons. number of an element more positive, or less negative.
l Reduction covers all reactions in which atoms or
l Reduction is a change that makes the oxidation of
ions gain electrons. an element more negative, or less positive.
l In a redox reaction, the oxidising agent gains
l Metal elements, in general, form positive ions
electrons while the reducing agent loses electrons. when their atoms react by loss of electrons with an
l An ionic equation describes a chemical change
increase in oxidation number.
by showing only the reacting ions and any other l Non-metal elements, in general, form negative ions
reacting atoms or molecules, while leaving out the when their atoms or molecules react by gain of
spectator ions. electrons with a decrease in oxidation number.
l A half-equation is an ionic equation used to
l The compounds of an element can be classified
describe either the gain, or loss, of electrons during according to their oxidation states.
a redox reaction. l Roman numerals are used in chemical names to
l The two ionic half-equations for a redox reaction
show the oxidation numbers of elements in the
can be combined to give the full ionic equation. compound. The oxidation number of a metal is
l Oxidation numbers extend redox ideas to cover
always given in the name if it can vary. It is less
reactions involving molecules as well as atoms and ions. usual to give the oxidation number of the element
l The oxidation numbers of uncombined elements
in common oxoanions (such as sulfate or nitrate).
are zero; in simple ions the charge on the ion gives l A disproportionation reaction involves an
the oxidation number of the element; for ions with element in a single atom, molecule or ion being
more than one atom, the sum of the oxidation simultaneously oxidised and reduced.
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Exam practice questions
These are incomplete half-equations for
1 8 Rewrite each of these full ionic equations as
changes involving reduction in solution. two half-equations. In each case state which
Complete and balance the half-equations. element has been oxidised and which has been
a) H+(aq) → H2(g) (1) reduced.
b) Fe3+(aq) → Fe2+(aq) (1) a) 2Fe2+(aq) + Br2(aq)
c) H2O2(aq) + 2H+(aq) → 2H2O(l) (1) → 2Fe3+(aq) + 2Br−(aq) (3)
b) 2I−(aq) + Cl2(aq) → I2(aq) + 2Cl−(aq) (3)
These are incomplete half-equations for
2
c) Zn(s) + 2V3+(aq) → Zn2+(aq) + 2V2+(aq) (2)
changes involving oxidation in solution.
Complete and balance the half-equations. 9 Complete these redox equations so that they
a) Mg(s) → Mg2+(aq) (1) are balanced:
b) Sn2+(aq) → Sn4+(aq) (1) a) CuO(s) + NH3(g)
c) I−(aq) → I2(aq) (1) → N2(g) + H2O(l) + Cu(s) (2)
b) KI(s) + H2SO4(l)
Select a reduction from Question 1 and an
3
→ K2SO4(s) + I2(s) + H2S(g) + H2O(l) (2)
oxidation from Question 2 and combine them
c) NaIO3(aq) + NaI(aq) + H2SO4(aq)
to give the full ionic equation for these reactions:
→ I2(aq) + H2O(l) + Na2SO4(aq) (2)
a) iron(iii) ions with tin(ii) ions (2)
d) Cu(s) + HNO3(aq)
b) magnesium with dilute hydrochloric acid (2)
→ Cu(NO3)2(aq) + NO2(g) + H2O(l) (2)
c) hydrogen peroxide with iodide ions. (2)
Identify the atoms, molecules or ions that are
10
State the oxidation numbers of chlorine
4
oxidised and reduced in each of these reactions,
in these ions: Cl−, ClO−, ClO2−, ClO3−,
stating the changes in oxidation number. In each
ClO4−. (3)
case, write a full balanced equation for the reaction.
State the oxidation numbers of nitrogen in
5 a) Hydrogen bromide gas reacts with sulfuric
these molecules: N2, NH3, N2H4, HNO3, acid to form bromine and sulfur dioxide. (3)
HNO2, NH2OH, NF3. (4) b) An acidic solution of manganate(vii) ions,
MnO4−, reacts with aqueous iron(ii) ions. (3)
Write half-equations for these changes in
6
c) A sample of sodium chromate(vi) is made
solution – in each case state whether the
by the reaction of a chromium(iii) salt with
process is an example of oxidation or of
hydrogen peroxide in alkaline solution. (3)
reduction:
d) An acidic solution of dichromate(vi) ions,
a) cobalt(ii) ions turning into cobalt(iii)
Cr2O72−, is used to test for sulfur dioxide.
ions (2)
When SO2 is present, the solution turns
b) sulfur dioxide molecules in acid solution
green as Cr3+ ions form as well as sulfate
turning into hydrogen sulfide molecules (2)
ions. (3)
c) hydroxide ions turning into oxygen and
e) The reaction of gaseous hydrazine, N2H4,
water molecules (2)
with gaseous dinitrogen tetroxide used
d) hydrogen molecules turning into hydrogen
to propel rockets in spacecraft producing
ions. (2)
nitrogen and steam in the exhaust gases. (3)
Identify the element that disproportionates in
7
11 Briefly state four different definitions of the
each of these reactions by giving the oxidation
terms ‘oxidation’ and ‘reduction’. (4)
states of the element before and after reaction:
Discuss the application of your definitions to
a) 2H2O2(aq) → 2H2O(l) + O2(g) (2)
these examples:
b) Cl2(aq) + 2NaOH(aq)
a) the reaction of hydrogen sulfide gas with
→ NaCl(aq) + NaClO(aq) + H2O(l) (2)
moist sulfur dioxide to form sulfur and
c) 3MnO42−(aq) + 4H+(aq)
water (4)
→ 2MnO4−(aq) + MnO2(s) + 2H2O(l) (2)
95
Exam practice questions
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b) the reaction of hydrogen gas with hot relationship, if any, between the oxidation
sodium metal to form sodium hydride (4) numbers of sulfur in these two ions and the
c) the reaction of hydrogen peroxide with number of electrons used in bonding. (4)
potassium iodide in acid solution to form b)* A Level textbooks usually state that the
water and iodine (4) oxidation number of sulfur in the thiosulfate
d) the decomposition of hydrogen peroxide to ion is +2. However, some chemists suggest
form oxygen and water. (4) that that the two sulfur atoms in the ion have
different oxidation states: −2 and +6. Similarly,
The diagram shows four oxoanions of sulfur.
12
the oxidation number in the tetrathionate ion
O O O O is normally stated O to be +2.5 in textbooks but
in other sources the sulfur atoms are considered
–O S O– –S S O– –O S S S S O– –O S O–
to be divided between the +6 and −1 states.
O O O Discuss the application
O of the oxidation
number rules to:
sulfite ion thiosulfate ion tetrathionate ion sulfate ion
• the thiosulfate ion
O O O O • the tetrathionate ion. (6)
O– –S S O– –O S S S S O– –O S O– c)* Adding dilute hydrochloric acid to a
solution of sodium thiosulfate produces a
O O O O precipitate of sulfur and a solution of sulfur
on thiosulfate ion tetrathionate ion sulfate ion dioxide. Discuss the type of reaction taking
place in terms of the alternative assignments
a) State the oxidation states of sulfur in the of oxidation numbers suggested in (b). (6)
sulfite and the sulfate ions. Comment on the
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Inorganic chemistry
4 and the Periodic Table
4.1 Types of inorganic reaction
Chemists classify the chemical reactions of inorganic elements and compounds
in order to make sense of many thousands of reactions. By identifying
the characteristics of similar reactions, chemists can predict how different
substances will behave. Recognising and understanding the different types
Tip of reaction makes it possible to explain observations and to interpret the
Redox reactions are described in detail results of chemical tests.
in Chapter 3. This section covers other
types of inorganic reaction that are Thermal decomposition
important in the study of groups in the Compounds such as metal carbonates and metal nitrates may decompose
Periodic Table. when heated in a Bunsen flame. Green copper(ii) carbonate, for example,
breaks up on heating to form black copper(ii) oxide and carbon dioxide. This
Key term is an example of thermal decomposition.
CuCO3(s) → CuO(s) + CO2(g)
Thermal decomposition is the name
for a reaction in which a compound Other metal carbonates, except those of most Group 1 metals, also decompose
decomposes on heating. on heating. Carbonates of metals below copper in the reactivity series are so
unstable that they cannot exist at room temperature.
Hydrated compounds like blue copper(ii) sulfate, CuSO 4.5H 2 O, contain
water as part of their structure. They also decompose on heating.
Fairly gentle heating causes most hydrates to give off water vapour,
which often condenses to water on the cooler parts of the apparatus
(Figure 4.1).
heat
CuSO4.5H2O(s) ——→ CuSO4(s) + 5H2O(g)
blue hydrated white anhydrous
copper(ii) sulfate
copper(ii) sulfate
Test yourself
1 Some thermal decomposition reactions are also redox reactions.
Use oxidation numbers to decide whether or not these examples of
thermal decomposition are also redox reactions.
a) 2KClO3(s) → 2KCl(s) + 3O2(g)
Figure 4.1 Copper(ii) sulfate crystals b) 2Cu(NO3)2(s) → 2CuO(s) + 4NO2(g) + O2(g)
decompose on heating to anhydrous c) Ca(HCO3)2(s) → CaO(s) + 2CO2(g) + H2O(l)
copper(ii) sulfate. The water driven off d) (NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(l)
condenses on the cool part of the tube.
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Reactions of acids and alkalis
Acids and alkalis are commonly used as chemical reagents. Dilute hydrochloric
acid is a convenient strong acid. Sodium hydroxide solution is often chosen
as a strong base.
Key term
Acids do not simply mix with water when they dissolve in it – they react
Ionic equations describe chemical with it to produce aqueous hydrogen ions, H+(aq).
equations by showing only the reacting
→ H+(aq) + Cl−(aq)
water
HCl(g) ——
ions in solution while leaving out the
spectator ions. Ionic equations are All acids produce H+(aq) ions with water and this is why dilute acids all
balanced both for atoms and for charges. react in a similar way. The typical reactions of dilute acids in water are
the reactions of aqueous hydrogen ions. This makes it possible to rewrite
Spectator ions are ions which are present
the equations for the reactions of acids as ionic equations leaving out
in solution but take no part in a reaction.
the spectator ions.
Acids reacting with metals
Acids react with the more reactive metals to form hydrogen gas plus an ionic
metal compound called a salt (Figure 4.2).
Mg(s) + 2HCl(aq) → MgCl 2(aq) + H2(g)
The equation can be rewritten to show the ions in solution.
Mg(s) + 2H+(aq) + 2Cl−(aq) → Mg2+(aq) + 2Cl−(aq) + H2(g)
spectator ion spectator ion
unchanged
The chloride ions are spectator ions so they can be left out of the ionic equation.
Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)
Acids reacting with metal oxides and hydroxides
Figure 4.2 Bubbles of hydrogen forming All alkalis dissolve in water to produce hydroxide ions, OH−. Sodium
as magnesium ribbon reacts with dilute hydroxide (Na+OH−) and potassium hydroxide (K+OH−) contain hydroxide
hydrochloric acid. The magnesium atoms ions in the solid as well as in solution.
turn into ions and pass into solution where
they mix with chloride ions to give a dilute During a neutralisation reaction, an acid reacts with metal hydroxide, or
solution of the salt magnesium chloride. metal oxide, to form a salt. Mixing the right amounts of dilute hydrochloric
acid and sodium hydroxide solution, for example, produces a neutral solution
of sodium chloride.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
This equation can also be rewritten to show the ions in solution:
H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) → Na+(aq) + Cl−(aq) + H2O(l)
spectator ions spectator ions
unchanged
The sodium ions and chloride ions do not react. By cancelling the spectator
ions, the equation simplifies to the ionic equation.
H+(aq) + OH−(aq) → H2O(l)
This is true of all reactions between aqueous solutions of acids and alkalis. It
shows that acids and alkalis neutralise each other because hydrogen ions react
with hydroxide ions to form water.
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Acids reacting with carbonates
Crystals of calcium carbonate consist of calcium and carbonate ions. The
reaction with hydrochloric acid is used to test for the presence of this mineral
in rocks. The acid fizzes by producing carbon dioxide when dripped onto a
mineral made of calcium carbonate.
Ca2+CO32−(s) + 2H+(aq) + 2Cl−(aq) → Ca2+(aq) + 2Cl−(aq) + CO2(g) + H2O(l)
In this case, the Ca2+ and 2Cl− are spectator ions, so the ionic equation is:
CO32−(s) + 2H+(aq) → CO2(g) + H2O(l)
Test yourself
2 Give the names and symbols of the ions formed when these acids
dissolve in water:
a) nitric acid, HNO3
b) sulfuric acid, H2SO4.
3 Write full balanced equations for the reactions of:
a) zinc with sulfuric acid
b) calcium oxide with nitric acid Key term
c) sodium carbonate with hydrochloric acid.
An ionic precipitation reaction is
4 Rewrite the equations in Question 3 as ionic equations.
a reaction which produces a solid
precipitate on mixing two solutions
containing ions.
Ionic precipitation reactions
The simplest examples of ionic precipitation reactions involve mixing two
solutions of soluble ionic compounds. The positive ions from one compound
combine with the negative ions of the other to form an insoluble precipitate.
When ionic compounds dissolve in water, the ions move away from the
crystals and each ion becomes surrounded by water molecules. So a solution
of potassium iodide, KI, in water, for example, contains separate K+(aq) ions
and I−(aq) ions mixed up with water molecules.
On mixing solutions of potassium iodide and lead(ii) nitrate, there are
two possible new combinations of ions: lead ions with iodide ions, and
potassium ions with nitrate ions. Lead(ii) iodide is insoluble, so it precipitates
(Figure 4.3). Potassium nitrate is soluble so the potassium and nitrate ions
stay in solution.
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
Rewriting the equation in terms of aqueous ions, gives:
Pb2+(aq) + 2NO3−(aq) + 2K+(aq) + 2I−(aq) → PbI2(s) + 2K+(aq) + 2NO3−(aq)
spectator ions spectator ions
Cancelling the spectator ions, leads to the simpler, ionic equation for the reaction.
Pb2+(aq) + 2I−(aq) → PbI2(s)
Figure 4.3 A precipitate of lead(ii) iodide
The solubility rules in Table 4.1 can be used to predict whether or not a forms on adding a solution of potassium
precipitate forms on mixing two solutions. iodide to a solution of lead(ii) nitrate.
4.1 Types of inorganic reaction 99
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Table 4.1 Solubilities of acids, bases and salts.
Type of compound Soluble in water Insoluble in water
Acids All common acids
Metal hydroxides and Soluble hydroxides and carbonates are All other metal oxides and hydroxides.
carbonates alkalis. They include the hydroxides of All other carbonates.
sodium and potassium, and also the
carbonates of sodium and potassium.
Calcium hydroxide is slightly soluble.
Salts All nitrates.
All chlorides… … except silver chloride and lead chloride.
All sulfates … … except barium sulfate and lead sulfate.
(Calcium sulfate and silver sulfate are
slightly soluble.)
All sodium, potassium and ammonium salts.
Test yourself
5 Use Table 4.1 to decide whether or not a precipitate forms on mixing
solutions of these pairs of substances. If yes, state the name and
formula of the precipitate and give the ionic equation for the reaction.
a) zinc sulfate and barium nitrate
b) potassium nitrate and copper(ii) sulfate
c) sodium carbonate and calcium chloride
d) lead(ii) nitrate and sodium chloride
e) sodium hydroxide and copper(ii) sulfate
6 Classify these reactions as redox, acid–alkali, precipitation or thermal
decomposition:
a) CaCl2(aq) + K 2SO4(aq) → CaSO4(s) + 2KCl(aq)
b) CaCO3(s) → CaO(s) + CO2(g)
c) Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g)
d) Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)
4.2 Group 1, the alkali metals
Tip Inorganic chemistry is the study of the hundred or so elements and their
compounds. The amount of information can be bewildering, hence the
The detailed study of Group 1 chemistry importance of the Periodic Table which helps to identify patterns in all the
is not required for examinations. However, facts about properties and reactions.
there are Group 1 compounds that you
need to know about because they are The Group 1 elements are better known as the alkali metals. These elements
important chemical reagents. You also are more alike than the elements in any other group. Their compounds are
have to know the results of flame tests widely used as chemical reagents.
for some Group 1 compounds. You are
expected to be able to compare some The elements
properties of the carbonates and nitrates The metals are soft and easily cut with a knife. They are shiny when freshly
of Group 2 elements with those of cut but quickly become dull in air as they react with moisture and oxygen
Group 1 elements. (Figures 4.4, 4.5 and 4.6). Laboratory specimens are kept in oil to protect
them from the air.
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Figure 4.4 A sample of lithium metal. Figure 4.5 A sample of sodium metal. Figure 4.6 A sample of potassium
Lithium compounds such as lithium Sodium metal is a powerful reducing metal. Potassium ions are an
carbonate are used in drugs for treating agent used to extract titanium and some essential nutrient for plants and an
mental illness. Other lithium compounds other metals. Sodium vapour is used in ingredient of NPK fertilisers. There
are important reagents for organic street lights. Sodium hydroxide is the most are many other important potassium
synthesis. important industrial alkali. Many other sodium compounds.
compounds are important commercially.
Key term
Atoms of the elements A trend describes the way in which a
The Group 1 elements have similar chemical properties because their atoms property increases or decreases along a
have similar electron structures with one electron in the outer s orbital series of elements or compounds. In the
(Figure 4.7 and Section 1.7). Even so, there are trends in properties down Periodic Table the term can describe the
the group from lithium to caesium. variations of a property down a group or
The atoms change down the group: the charge on the nucleus increases; across a period.
also the number of filled inner shells increases and so the atomic radius
increases. The number of electrons in the inner shells is always one less
than the number of protons in the nucleus. So the shielding effect of the
Li
inner electrons means that the effective nuclear charge attracting the outer Na
electron is 1+. Down the group, the outer electrons get further and further
away from the same effective nuclear charge and so they are held less
strongly (Figure 4.8 and Section 1.8).
Figure 4.7 Diagrams to represent the electron
Test yourself configurations of lithium and sodium.
7 The electronic configuration of lithium can be shortened to [He]2s1.
Using this style, give the electron configurations of:
Li Li+
a) sodium
b) potassium.
8 The first ionisation energies of the alkali metals get smaller down the
group. Why is this? Na Na+
9 Explain why the ions of Group 1 metals are smaller than their atoms.
Oxidation states K K+
When the atoms of alkali metals react, they lose their single s electron
from the outer shell, turning into ions with a single positive charge: Li+,
Na+, K+ and so on. So the only oxidation state of these elements in their
compounds is +1. Figure 4.8 Relative sizes of the atoms and
ions of Group 1 elements.
4.2 Group 1, the alkali metals 101
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Reactions of the elements
The alkali metals are powerful reducing agents because they react by giving
up electrons to form M+ ions.
All the metals react with water to form hydroxides, MOH, and hydrogen
(where M is Li, Na, K and so on). The rate and violence of the reaction
increases down the group. Lithium reacts steadily with cold water giving
off a steady stream of bubbles of hydrogen. Sodium melts to form a shiny
bead which skates around on the surface of the water. The reaction with
potassium is so violent that the hydrogen catches fire and burns with a violet
flame as the metal rapidly reacts.
Key term 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)
All the metals react vigorously with chlorine to form colourless, ionic
Bases are ‘anti-acids’ – they are the chlorides, M+Cl−. The chlorides are soluble in water, forming colourless
chemical opposites of acids. Acids solutions.
donate (give up) hydrogen ions; bases
(such as the oxide ion, hydroxide ion and 2K(s) + Cl 2(g) → 2KCl(s)
carbonate ion) accept hydrogen ions.
Test yourself
Tip 10 Write balanced equations and use oxidation numbers to show that
sodium acts as a reducing agent when it reacts with:
It is the hydroxide ion which is the base
and which makes the solutions alkaline, a) water
not the metal ions. b) chlorine.
4.3 Properties of the compounds
of Group 1 metals
The hydroxides
The hydroxides are all white solids, commonly supplied as pellets or flakes
(Figure 4.9). These are soluble in water, forming alkaline solutions, although
the solubility of these hydroxides increases down the group. The hydroxides
are strong bases because they are fully ionised in water, giving solutions
containing hydroxide ions.
The carbonates
The carbonates are all white with the general formula M 2CO3. They are
Figure 4.9 Sodium hydroxide, NaOH, is unusual metal carbonates in that they dissolve in water. Solutions of these
deliquescent, which means that it picks up carbonates are alkaline because the carbonate ions remove H+ ions from
water from moist air and then dissolves in water molecules to form hydrogencarbonate ions and hydroxide ions. It is
it. Sodium hydroxide is a strong base – it the hydroxide ions that make the solution alkaline.
dissolves in water to form a highly alkaline
CO32−(aq) + H2O(l) → HCO3−(aq) + OH−(aq)
solution. The traditional name for the alkali
is ‘caustic soda’. Sodium hydroxide is Another unusual feature of Group 1 carbonates is that most of them do not
highly corrosive and more hazardous to the decompose on heating. The exception is lithium carbonate, which breaks
skin and eyes than many acids. down to the oxide and carbon dioxide when hot (see Section 4.7).
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The nitrates Test yourself
The nitrates of Group 1 metals are white crystalline solids with the formula
MNO3. They are very soluble in water. 11 Write full and ionic equations
for the reaction of:
The crystals of sodium and potassium nitrates are much harder to decompose
a) potassium hydroxide with
on heating than most other metal nitrates. On heating, these nitrates first
dilute sulfuric acid
melt and then on stronger heating start to decompose, giving off oxygen.
They only decompose as far as the nitrite. b) aqueous sodium carbonate
with dilute nitric acid.
2KNO3(s) → 2KNO2(s) + O2(g) 12 Write balanced equations for
Lithium nitrate is the exception. It behaves like most other metal nitrates, the thermal decomposition of:
decomposing to the oxide, nitrogen dioxide and oxygen (see Section 4.7). a) lithium carbonate
b) lithium nitrate.
Sodium and potassium compounds as
chemical reagents
The compounds of sodium and potassium are widely used as chemical
reagents. One reason for this is that the ions of alkali metals are unreactive.
So they act as spectator ions which take no part in reactions when the
reagents are used.
A second reason is that most sodium and potassium compounds are soluble
in water, including their hydroxides and carbonates. Most other metal
hydroxides and carbonates are insoluble so not available in aqueous solution.
A third reason is that the ions of alkali metals are colourless in aqueous solution
so they do not hide or interfere with colour changes. Sodium or potassium
compounds are coloured only if the negative ion is coloured. Potassium
chromate(vi), for example, is yellow because CrO42− ions are yellow.
Flame colours
Flame colours help to detect some metal ions (Figures 4.10 and 4.11, and
Table 4.2). They are particularly useful in identifying Group 1 metal ions,
which are otherwise very similar.
Ionic compounds such as sodium chloride do not burn during a flame test.
The energy from the flame excites the outer electrons of sodium ions, raising
them to higher energy levels. The atoms then emit the characteristic yellow
light as the electrons drop back to lower energy levels (Section 1.5).
flame
nichrome wire
crystals on wire
concentrated crystals to be tested
hydrochloric acid
Bunsen burner
Figure 4.10 Procedure for a flame test. Chlorides evaporate more easily and so colour flames more strongly than less
volatile compounds. Concentrated hydrochloric acid converts involatile compounds such as carbonates to chlorides.
4.3 Properties of the compounds of Group 1 metals 103
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Table 4.2 Flame colours of Group 1 metal compounds.
Metal ion Colour
Lithium Bright red
Sodium Bright yellow
Potassium Lilac
Tip
Some chemical reagents contain traces of sodium compounds as impurities. The
sodium flame colour is so strong that it can easily obscure the colour of other flames,
especially the pale lilac flame from potassium compounds.
Figure 4.11 The flame colour from a Test yourself
potassium salt.
13 Classify these reactions of Group 1 elements and compounds as
acid–base, thermal decomposition or redox.
a) 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)
b) 2KNO3(s) → 2KNO2(s) + O2(g)
c) 2Li(s) + Br2(l) → 2LiBr(s)
d) Li2CO3(s) → Li2O(s) + CO2(g)
4.4 Group 2, the alkaline earth
Figure 4.12 Stalactites and stalagmites in
metals
a limestone cave. Stalactites are formed Group 2 elements belong to the family of alkaline earth metals. Many
by calcium carbonate, dissolved in ground of the compounds of these elements occur as minerals in rocks – hence the
water, dripping through into caves. The name ‘earth metals’. Chalk, marble and limestone, for example, are forms of
calcium carbonate then precipitates out calcium carbonate (Figure 4.12). Dolomite consists of a mixture of calcium and
of the water, forming rock. Stalagmites magnesium carbonates. Fluorspar is a form of calcium fluoride which is mined as
are formed when the water falls from the the ornamental mineral (Figure 4.13). These Group 2 compounds are insoluble,
stalactite onto the cave floor. unlike the equivalent Group 1 compounds. They do not dissolve in water.
The elements
The Group 2 metals are harder and denser than Group 1 metals and they
have higher melting temperatures (Figure 4.14). In air, the surface of the
metals is covered with a layer of oxide.
Figure 4.13 A sample of fluorite mined in Figure 4.14 Samples of the Group 2 metals. From the left: beryllium, magnesium, calcium,
Weardale, County Durham. strontium and barium.
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The first member of the group, beryllium (Be), is a strong metal with a Table 4.3 The shortened forms of the
high melting temperature, but its density is much less than that of transition electron configurations of Group 2 metals.
metals such as iron. The element makes useful alloys with other metals. Metal Electronic
configuration
Magnesium is manufactured by the electrolysis of molten magnesium chloride
Beryllium, Be [He]2s2
from sea water or from salt deposits. The low density of the metal helps to
make light alloys, especially with aluminium. These alloys, which are strong Magnesium, Mg [Ne]3s2
for their weight, are especially useful for car and aircraft manufacture. Calcium, Ca [Ar]4s2
Barium is a soft, silvery-white metal. It is so reactive with air and moisture Strontium, Sr [Kr]5s2
that it is generally stored under oil, like the alkali metals. Barium, Ba [Xe]6s2
Atoms and ions
Like the atoms of the alkali metals, the Group 2 atoms change down the
group: the charge on the nucleus increases, and the number of filled inner
shells also increases (Table 4.3 and Figure 4.15).
The increasing number of filled inner shells means that atomic and ionic radii
increase down the group (Figure 4.16). For each element, the 2+ ion is smaller
Mg
than the atom because of the loss of the outer shell of electrons. The tendency
to react and form ions increases down the group.
The first and second ionisation energies decrease down the group
(Figure 4.17). The shielding effect of the inner electrons means that the
effective nuclear charge attracting the outer electron is 2+. Down the group
the outer s electrons get further and further away from the same effective
nuclear charge, and so they are held less strongly and the ionisation energies
Ca
decrease. This trend helps to account for the increasing reactivity of the
elements down the group.
Li+ Be2+
Figure 4.15 Diagrams to represent the
Na+ Mg2+
electron configurations of magnesium and
Be2+
calcium atoms.
Sum of first two ionisation
2500
Mg2+
energies/kJ mol –1
2000
K+ Ca2+ Ca 2+ Sr 2+
Ba2+
1500
1000
Rb+ Sr2+
500
0
0 10 20 30 40 50 60
Cs+ Ba2+ Atomic number
Figure 4.17 Graph showing the trend in the
Figure 4.16 Comparison of the trend in sum of the first two ionisation energies of
ionic radii of Group 1 and Group 2 metals. Group 2 metals: M(g) → M2+(g) + 2e−.
The removal of a third electron to form a 3+ ion takes much more energy
because the third electron has to be removed against the attraction of a much
larger effective nuclear charge. This means that it is never energetically
favourable for the metals to form M3+ ions.
4.4 Group 2, the alkaline earth metals 105
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Tip Test yourself
Always check carefully that you are 14 Write the full electron configurations of the atoms and ions of Mg, Ca
using the right term when answering and Sr showing the numbers of s, p and d electrons (Section 1.6).
questions about the radii of atoms and
15 Explain why Group 2 ions in any period are smaller than the Group 1
ions. Be sure to distinguish atomic radii
ions in that period.
from ionic radii.
16 Explain why the radii of Group 2 metal ions increase down the group.
Oxidation states
All the Group 2 metals have similar chemical properties because they have
similar electron structures with two electrons in an outer s orbital. When the
metal atoms react to form ions, they lose the two outer electrons, giving ions
with a 2+ charge: Mg2+, Ca 2+, Sr2+ and Ba2+. So these elements exist in the
+2 oxidation state in all their compounds.
4.5 Reactions of the Group 2
elements
Group 2 metals are reducing agents. Apart from beryllium, they readily
give up their two s electrons to form M 2+ ions (where M represents Mg,
Ca, Sr or Ba).
M → M 2+ + 2e−
Tip
Note that beryllium is not a typical Reactions with oxygen
Group 2 element and so the coverage
Apart from beryllium, the Group 2 metals burn in oxygen on heating to
of its chemical properties in this
form white, ionic oxides, consisting of M 2+ ions and O2− ions.
chapter is limited. The small size of the
beryllium ion (electronic configuration Magnesium burns very brightly in air with an intense white flame and for
1s2) means that it has a much higher this reason, magnesium powder is used in fireworks and flares. The reaction
polarising power than other Group 2 produces the white solid magnesium oxide, MgO (Section 3.2).
ions (Section 4.7). The polarising
Calcium also burns brightly in air but with a red flame forming the white
power of the ions of a metal determine
solid calcium oxide, CaO. Strontium reacts in a similar way.
to a large extent the type of bonding
between the element and non-metals Barium burns in excess air or oxygen with a green flame to form a peroxide,
such as oxygen and chlorine and hence BaO2, which contains the peroxide ion, O22−.
the chemical characteristics of the
compounds. Reactions with water
The metals Mg to Ba in Group 2 react with water. The reactions are not as
vigorous as the reactions of the Group 1 metals, but, as in Group 1, the rate
of reaction increases down the group.
Magnesium reacts very slowly with cold water producing the hydroxide, Mg(OH)2,
and hydrogen. This metal reacts much more rapidly on heating in steam.
Mg(s) + H2O(g) → MgO(s) + H2(g)
Calcium reacts with cold water to produce hydrogen and calcium
hydroxide. Initially the Ca(OH)2 formed dissolves, but the solubility is
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low so that, as more is formed, the solution soon becomes saturated and
a white precipitate appears.
Ca(s) + 2H2O(l) → Ca(OH)2(s) + H2(g)
Barium reacts even faster with cold water and its hydroxide is more soluble.
Reactions with chlorine
All the metals, including beryllium, react with chlorine on heating to form
white chlorides, MCl 2.
Be(s) + Cl 2(g) → BeCl 2(s)
Test yourself
17 What is the oxidation state of oxygen in barium peroxide?
18 Write balanced equations for the reactions of:
a) strontium with oxygen
b) barium with oxygen
c) strontium with chlorine
d) barium with water.
4.6 Properties of the compounds
of Group 2 metals
The oxides
Key term
Apart from beryllium oxide, the oxides of Group 2 metals are basic oxides.
They react with acids to form salts. A basic oxide is a metal oxide which
CaO(s) + 2HNO3(aq) → Ca(NO3)2(aq) + H2O(l) reacts with acids to form salts and
water. It is the oxide ion which acts as
Magnesium oxide is a white solid. In water it turns to magnesium hydroxide, a base by taking a hydrogen ion from
which is slightly soluble. Magnesium oxide has a high melting temperature the acid. Basic oxides which dissolve in
and is used as a heat-resistant ceramic to line furnaces. water are alkalis.
Calcium oxide is a white solid made by heating calcium carbonate. Calcium
oxide reacts very vigorously with cold water, hence its traditional name
‘quicklime’. The product is calcium hydroxide.
The hydroxides
The hydroxides of the elements Mg to Ba are:
● similar in that they all have the formula M(OH)2 and are, to some degree,
soluble in water forming alkaline solutions
● different in that their solubility increases down the group.
Magnesium hydroxide is the active ingredient in milk of magnesia, used as
an antacid and laxative. It is insoluble in water.
Calcium hydroxide is slightly soluble in water forming an alkaline solution,
usually called limewater. The limewater test for carbon dioxide works
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because a solution of calcium hydroxide reacts with the gas forming a white,
Tip insoluble precipitate of calcium carbonate.
There is no simple explanation for the
Barium hydroxide is the most soluble of the hydroxides. It is sometimes
trends in solubility of alkaline earth
used as an alkali in chemical analysis. It has the advantage over sodium and
metal compounds down the group. If
potassium hydroxides in that it cannot be contaminated by its carbonate
the negative ion is small (as in the oxide
because barium carbonate is insoluble in water.
and hydroxide), then the compounds of
the metal with the smallest ions, Mg 2+,
are least soluble. If the negative ion is Test yourself
relatively large (as in the sulfates and 19 Write balanced equations for the reactions:
carbonates), then the metal compounds
a) magnesium oxide with dilute hydrochloric acid
of the metal with the largest ions, Ba2+,
are least soluble. In other words, the b) calcium oxide with water
rule of thumb is that these compounds c) limewater with carbon dioxide.
tend to be insoluble if both ions are
small, or both ions are big.
The carbonates
The carbonates of Group 2 metals (Mg to Ba) are:
● similar in that they all have the formula MCO3, are insoluble in water,
react with dilute acids and decompose on heating to give the oxide and
carbon dioxide
CaCO3(s) → CaO(s) + CO2(g)
● different in that they become more difficult to decompose down the
group – in other words, they become more thermally stable (Section 4.7).
The nitrates
The nitrates of Group 2 metals (Mg to Ba) are:
● similar in that they all have the formula M(NO3)2, are colourless crystalline
solids, are very soluble in water and decompose to the oxide on heating
2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g)
● different in that they become more difficult to decompose down the group
(Section 4.7).
The sulfates
The sulfates are:
● similar in that they are all colourless solids with the formula MSO4
● different in that they become less soluble down the group.
Epsom salts consist of hydrated magnesium sulfate, MgSO4.7H2O, which is
a laxative.
Figure 4.18 A geologist in the Cave of A hydrated form of calcium sulfate occurs naturally as gypsum (Figure
Crystals (Cueva de los Cristales) in Naica 4.18) which is produced on a large scale by the process that removes sulfur
Mine, Chihuahua, Mexico. The crystals are dioxide from the flue gases of coal-fired power stations. Plaster of Paris is the
the largest known in the world, and are main ingredient of building plasters and much is used to make plasterboard.
formed of the selenite form of gypsum The white powder is made by heating gypsum in kilns to remove most of
(hydrated calcium sulfate). the water of crystallisation. Stirring plaster of Paris with water produces a
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paste which soon sets as it turns back into interlocking grains of gypsum.
Plaster makes good moulds because it expands slightly as it sets so that it fills
every crevice.
Barium occurs naturally in minerals such as witherite, BaCO3 and as baryte,
BaSO4 (Figure 4.19). Barium sulfate absorbs X-rays strongly so it is the main
ingredient of ‘barium meals’ used to diagnose disorders of the stomach or
intestines. Soluble barium compounds are toxic but barium sulfate is very
insoluble so it is not absorbed into the bloodstream from the gut. X-rays
cannot pass through the ‘barium meal’, which therefore creates a shadow on
the X-ray film (Figure 4.20).
Tip
Magnesium compounds do not give
a colour when heated in a flame.
The flame test colour for magnesium
compounds is not ‘bright white’.
Figure 4.19 Sample of baryte (barium Figure 4.20 Coloured X-ray photograph of
sulfate) from Poland. the healthy stomach of a patient who has
taken a barium meal.
A soluble barium salt can be used to test for sulfate ions because barium Table 4.4 Flame colours of Group 2 metal
sulfate is insoluble, even when the solution is acidic. Adding a solution of compounds.
barium nitrate or barium chloride to an acidified solution produces a white Metal ion Colour
precipitate only if sulfate ions are present.
Beryllium No colour
Ba2+(aq) + SO42−(aq) → BaSO4(s)
Magnesium No colour
white precipitate
Calcium Brick red
Flame colours Strontium Bright red
Flame tests help to identify compounds of calcium, strontium and barium
Barium Pale green
(Table 4.4).
4.7 Thermal stability of the
carbonates and nitrates Key term
Whenever chemists use the term ‘stability’ they are making comparisons. Thermal stability is an indication
For the Group 2 carbonates, the question is which is more stable – the of the ease with which compounds
metal carbonate, or a mixture of the metal oxide and carbon dioxide? decompose on heating. Compounds
are stable if they do not tend to
The carbonates and nitrates of Group 1 and 2 elements are ionic. Chemists
decompose into their elements or into
explain differences in the thermal stabilities of their carbonates and nitrates
other compounds. A compound which
in terms of two factors:
is stable at room temperature and
● the charge on the metal ions pressure may become more or less
● the size of the metal ions. stable as conditions change.
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Table 4.5 The temperature at which Group Group 2 carbonates and nitrates are generally less stable than the corresponding
2 carbonates begin to decompose. Group 1 compounds. This suggests that the larger the charge on the metal ion
Carbonate Decomposition and the smaller the metal ion, the less stable the compounds.
temperature/°C
The carbonates become more stable down both Group 1 and Group 2.
MgCO3 540 This helps to confirm that, the larger the metal ion, the more stable the
CaCO3 900 compounds.
SrCO3 1280 Beryllium carbonate is so unstable that it does not exist. Table 4.5 shows
the temperatures at which the carbonates of Group 2 metals begin to
BaCO3 1360
decompose. The values indicate that magnesium carbonate is the least stable.
It decomposes easily to the oxide and carbon dioxide when heated with a
Key term Bunsen flame. Barium carbonate is the most stable.
Trends in the thermal stability of carbonates and nitrates can be explained in
Polarising power gives an indication of terms of the polarising power of the metal ions. The larger the charge on
the extent to which a positive ion is able an ion and the smaller its radius, the greater its charge density. The greater the
to distort the electron cloud around a charge density, the greater the polarising power of the ion. A metal ion with
neighbouring negative ion. The larger the a higher polarising power attracts the bonding electrons in neighbouring
charge on a positive ion and the smaller ions more strongly. This pull on the electrons of an ion, such as a carbonate
its size, the greater its polarising power. ion, distorts the bonding and makes it easier to break up the negative ion into
an oxide ion and carbon dioxide (Figure 4.21).
Down Group 2, for example, the charge on the metal ion is always 2+.
However the increasing number of full, inner shells means that the ionic
CO 32–
radii increase down the group. As a result, the charge density of the ions
decreases down the group. So the trend in polarising power is Mg 2+ >
CO 32– M2+ CO 32– Ca 2+ > Sr2+ >Ba 2+. This means that the thermal stability of carbonates and
nitrates increases down the group.
CO 32–
Test yourself
20 a) Draw and label a diagram of a simple apparatus to investigate
the trend in the thermal stability of Group 2 carbonates.
O2–
b) Describe the observations you would expect to make with the
carbonates of magnesium, calcium and barium.
O2– M2+ O2–
21 Write equations for:
O2– a) the thermal decomposition of magnesium carbonate
Figure 4.21 Decomposition of a Group 2 b) the reaction of magnesium carbonate with hydrochloric acid
carbonate to a Group 2 oxide. The smaller c) the thermal decomposition of calcium nitrate
the metal ion, the less stable the carbonate. d) the reaction of barium nitrate solution with zinc sulfate solution.
22 With the help of oxidation numbers, identify the elements that
are oxidised and reduced during the thermal decomposition of
magnesium nitrate.
23 Why does calcium carbonate decompose on heating strongly in a
Bunsen flame while potassium carbonate does not?
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4.8 Group 7, the halogens
Fluorine, chlorine, bromine and iodine belong to the family of halogens. All
four are reactive non-metals – fluorine and chlorine extremely so. The elements
are hazardous because they are so reactive. For the same reason, they are never
found free in nature. However, they do occur as compounds with metals. Many
of the compounds of Group 7 elements are salts – hence the name ‘halo-gen’
meaning ‘salt-former’. Halogen compounds are important economically as the
ingredients of plastics, pharmaceuticals, anaesthetics and dyestuffs.
The elements
Under laboratory conditions chlorine is a yellow-green gas (Figure 4.22), bromine
is a dark red liquid (Figure 4.23), while iodine is a dark grey solid (Figure 4.24).
Figure 4.22 Chlorine gas. Figure 4.23 Bromine is a dark red liquid at Figure 4.24 Iodine is a lustrous grey-black
room temperature. It is very volatile, giving off solid at room temperature. It sublimes
a choking, orange vapour. For this reason, it when gently warmed to give a purple
should always be studied in a fume cupboard. vapour.
Fluorine is much too dangerous to be used in normal laboratories. Astatine,
the final member of the group is the rarest naturally occurring element. It is
highly radioactive – its most stable isotope has a half-life of just over 8 hours.
All the halogens consist of diatomic molecules, X 2, linked by a single covalent
bond. They are all volatile. Intermolecular forces (London forces) increase Table 4.6 Shortened form of the electron
down the group as the numbers of electrons in the molecules increase (Section configurations of the halogens.
2.6). The larger molecules are more polarisable than the smaller molecules, so
Halogen Electronic configuration
melting temperatures and boiling temperatures rise down the group.
Fluorine, F [He]2s22p5
The halogens have similar chemical properties because they all have seven electrons
in the outer shell – one fewer than the next noble gas in Group 8 (Table 4.6). Chlorine, Cl [Ne]3s23p5
Bromine, Br [Ar]3d104s24p5
Fluorine is the most electronegative of all elements. Its oxidation state is −1 in
all its compounds. Uses of fluorine include the manufacture of a wide range of Iodine, I [Kr]4d105s25p5
compounds consisting of only carbon and fluorine (fluorocarbons). The most
familiar of these is the very slippery, non-stick polymer, poly(tetrafluorethene).
4.8 Group 7, the halogens 111
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Chlorine reacts directly with most elements. In its compounds, chlorine is usually
present in the −1 oxidation state, but it can be oxidised to positive oxidation states
by oxygen and fluorine. Most chlorine is used in the production of polymers
such as PVC. Water companies use chlorine to kill bacteria in drinking water,
while another important use of the element is to bleach paper and textiles.
Bromine, like the other halogens, is a reactive element but it is a less powerful
oxidising agent than chlorine. Bromine is used to make a range of products
including flame retardants, medicines and dyes.
Iodine is also an oxidising agent but it is less powerful than bromine. Iodine and
its compounds are used to make many products including medicines, dyes and
catalysts. In many regions, sodium iodide is added to table salt to supplement
iodine in the diet and to drinking water to prevent goitre – a swelling of the
thyroid gland in the neck. Iodine is needed in our diet so that the thyroid gland
can make the hormone thyroxine, which regulates growth and metabolism.
Tip
Test yourself
In exams it is important to avoid losing
marks by careless use of chemical 24 Predict the state of the following at room temperature and pressure,
language. For example, you must giving your reasons:
distinguish the element ‘chlorine’ from a) fluorine b) astatine.
the ‘chloride’ ion in its compounds.
25 Write down the full electron configurations of:
a) a chlorine atom b) a chloride ion
c) a bromine atom d) a bromide ion.
26 Explain why:
a) the atomic radii of halogen atoms increase down the group
b) the ionic radii of halides are larger than their corresponding
atomic radii.
27 Draw dot-and-cross diagrams, showing just the electrons in the outer
shells of the atom, to describe the bonding in:
a) a fluorine molecule
b) a molecule of hydrogen bromide
c) a molecule of iodine monochloride, ICl.
Solutions of the halogens
The halogens dissolve freely in hydrocarbon solvents, such as cyclohexane.
When dissolved in cyclohexane, the solutions have a very similar colour
Figure 4.25 Iodine dissolved in water to the free halogen vapours. So iodine in cyclohexane, for example, has an
(bottom layer). There is some solid iodine at attractive violet colour (Figure 4.25).
the bottom of the beaker. Iodine does not
dissolve well in water, so the brown colour The halogens are less soluble in water than in organic solvents. Aqueous
of the solution is faint. Iodine does dissolve chlorine and bromine are useful reagents. Their colours are similar
well in cyclohexane, giving a strong purple to the colours of their vapours. These elements also react with water
colour. (Section 4.11).
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Iodine does not dissolve in water, but it does dissolve in aqueous potassium
iodide. Iodine dissolves in this way because iodine molecules react with Test yourself
iodide ions to form triiodide ions, I3−. 28 Why are the halogens more
I2(s) + I−(aq) → I3 −(aq) soluble in cyclohexane than
in water?
A reagent labelled ‘iodine solution’ is normally I2(s) in KI(aq). The I3−(aq)
ion is a yellow-brown colour which explains why aqueous iodine looks
quite different from a solution of iodine in a non-polar solvent such as
cyclohexane. The aqueous solution is yellow when very dilute but dark
orange-brown when more concentrated.
4.9 Reactions of the Group 7 elements
The halogens are all oxidising agents. The reactions of the halogens show a
clear trend in their reactivity as oxidising agents down the group. Fluorine is
the most powerful oxidising agent and iodine the least powerful.
Halogen atoms are highly electronegative (Section 2.5), although the
electronegativity decreases down the group. They form ionic compounds or
compounds with polar bonding.
Reactions of halogens with metals
Chlorine and bromine react with s-block metals to form ionic halides in
which the halogen atoms gain one electron to fill the outermost p sub-shell.
Iodine also reacts with metals to form iodides, but because of the polarisability
of the large iodide ion, those iodides formed with small cations such as Li+, Test yourself
or highly charged cations such as Al3+, are essentially covalent (Section
2.5). The halogens also react with most metals in the d block. Hot iron, for 29 Write balanced equations for
example, burns brightly in chlorine, forming iron(iii) chloride (Figure 4.26). these reactions, and show
The reaction with bromine is similar but much less exothermic. Iron(iii) the changes in oxidation
iodide does not exist because iodide ions reduce iron(iii) ions to iron(ii) ions. states:
So, heating iron with iodine vapour produces iron(ii) iodide. a) bromine with magnesium
b) chlorine with iron
c) iodine with iron.
30 In Figure 4.26, explain the
drying agent reasons for:
a) carrying out the reaction
specimen tube
in a fume cupboard
or small bottle b) drying the chlorine gas
c) using iron wool instead of
iron wool
small lumps of iron
dry
chlorine d) collecting the product in a
gas specimen tube
combustion tube e) allowing excess gas to
heat
escape through a tube
Figure 4.26 The laboratory apparatus for making anhydrous iron(iii) chloride. This
with a drying agent.
reaction must be carried out in a fume cupboard.
4.9 Reactions of the Group 7 elements 113
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Reactions of halogens with non-metals
Chlorine reacts with most non-metals to form molecular chlorides. Hot silicon,
for example, reacts to form silicon tetrachloride, SiCl4(l), and phosphorus
produces phosphorus trichloride, PCl3(l) or phosphorus pentachloride, PCl5(s),
depending on whether the supply of the gas is limited or in excess. However,
chlorine does not react directly with carbon, oxygen or nitrogen.
Hydrogen burns in chlorine to produce the colourless, acidic gas hydrogen
chloride, HCl. Igniting a mixture of chlorine and hydrogen gases leads to a
violent explosion.
Bromine also oxidises non-metals such as sulfur and hydrogen on heating,
forming molecular bromides. A mixture of bromine vapour and hydrogen
gas reacts smoothly with a pale bluish flame.
H2(g) + Br2(g) → 2HBr(g)
Iodine oxidises hydrogen on heating to form hydrogen iodide. Unlike the
reactions of chlorine and bromine, this is a reversible reaction.
Tip
H2(g) + I2(g) ⇋ 2HI(g)
Be careful to distinguish hydrogen
chloride gas, HCl(g) from hydrochloric
acid, HCl(aq). Hydrochloric acid is a Reactions of halogens with aqueous Fe2+ ions
solution of hydrogen chloride in water. Chlorine and bromine can oxidise iron(ii) ions in solution to iron(iii) ions.
Even concentrated hydrochloric acid is Iodine is such a weak oxidising agent that it cannot oxidise iron(ii) compounds.
a solution.
2Fe2+(aq) + Cl 2(aq) → 2Fe3+(aq) + 2Cl−(aq)
Test yourself
31 Show that the reactions of chlorine, bromine and iodine with
hydrogen illustrate a trend in reactivity down Group 7.
32 Predict the formula of the product and vigour of the reaction when
fluorine reacts with hydrogen.
33 Explain, in terms of structure and bonding, why silicon tetrachloride
is a liquid.
34 Write equations for the reactions and use oxidation numbers to show
that:
a) phosphorus is oxidised when it reacts with chlorine
b) chlorine is reduced when it displaces iodine from a solution of
potassium iodide.
35 Write ionic half-equations and the overall ionic equation for the
reaction of bromine with aqueous iron(ii) ions.
4.10 Halogens in oxidation state −1
Tip
Halide ions
Remember that halide ions are
Halide ions are the ions of the halogen elements in oxidation state −1.
colourless. It is the halogen molecules
They include the fluoride (F−), chloride (Cl−), bromide (Br−) and iodide
that are coloured.
(I−) ions.
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Figure 4.27 Precipitates of silver chloride, silver bromide and silver iodide formed by Key term
adding silver nitrate solution to solutions of the halide ions.
Displacement reactions are redox
Silver nitrate solution can be used to distinguish between halides (Figure 4.27). reactions which can be used to
Silver fluoride is soluble, so there is no precipitate on adding silver nitrate compare the relative strengths of
to a solution of fluoride ions. The other silver halides are insoluble – adding metals as reducing agents and non-
silver nitrate to a solution of one of these halide ions produces a precipitate metals as oxidising agents. A more
(Table 4.7). For example: reactive element displaces a less
Ag+(aq) + Cl−(aq) → AgCl(s) reactive element from one of its salts.
Table 4.7 Properties of silver halides.
Silver halide Observations when aqueous Effect of adding aqueous
silver nitrate is added to a ammonia to a precipitate of
solution of the halide the silver halide
Silver chloride, White precipitate quickly Precipitate dissolves easily
AgCl turns purple–grey in sunlight in dilute ammonia to form a
colourless solution
Silver bromide, Creamy coloured precipitate Precipitate dissolves in
AgBr concentrated aqueous ammonia
to form a colourless solution
Silver iodide, Yellow precipitate Precipitate does not dissolve in
AgI ammonia solution
In Group 7, a more reactive halogen oxidises the ions of a less reactive
halogen. So chlorine displaces bromine from a bromide, while bromine
reacts with a solution of an iodide to produce iodine (Figure 4.28). This is
an example of a displacement reaction. Bromine has a stronger tendency
than iodine to gain electrons and turn into ions. Figure 4.28 Chlorine bubbling through
potassium bromide solution. The more
Br2(aq) + 2I−(aq) → 2Br−(aq) + I2(s)
reactive chlorine displaces bromine. The
In Group 7, a more reactive halogen displaces a less reaction halogen. aqueous solution of bromine is orange.
4.10 Halogens in oxidation state −1 115
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6 SO3 H2SO4 SO42 Reactions of halides with concentrated sulfuric acid
5 Warming solid sodium chloride with concentrated sulfuric acid produces
4 SO2 H2SO3 SO3 2 hydrogen chloride gas. This colourless gas produces white fumes in moist
Oxidation number
air. The acid–base reaction between sodium chloride and sulfuric acid can
3
be used to make hydrogen chloride.
2
NaCl(s) + H2SO4(l) → HCl(g) + NaHSO4(s)
1
0 S This type of reaction cannot be used to make pure hydrogen bromide
because bromide ions are oxidised to bromine by concentrated sulfuric acid.
1
However the reaction with potassium bromide is used to convert alcohols to
2 S2 H2S bromoalkanes (Section 6.3.8).
Figure 4.29 The oxidation states of sulfur. Iodide ions are such strong reducing agents that they reduce sulfur from the
+6 state in H2SO4 to sulfur and hydrogen sulfide (Figure 4.29). This reaction
is so rapid that little or no hydrogen iodide is formed. These reactions of
halide ions with sulfuric acid show that there is a trend in the strength of the
halide ions as reducing agents (Table 4.8).
Table 4.8 Reactions of concentrated sulfuric acid with sodium halides.
Tip Reaction Observations and products Type of reaction
NaCl + Colourless acidic gas forms Acid–base reaction. No
Hydrogen iodide can be made by warming
concentrated H2SO4 that fumes in moist air redox.
potassium iodide with concentrated (HCl). A white solid remains
phosphoric acid, H3PO4. Phosphoric acid (NaHSO4).
is not an oxidising agent.
NaBr + Orange vapour (Br2) mixed Mainly a redox reaction
concentrated H2SO4 with a colourless, acidic gas in which bromide
(SO2). The solid product is ions are oxidised to
NaHSO4. Some HBr is also bromine. Sulfur is
formed. reduced from the +6 to
the +4 state. Also an
acid–base reaction to
form HBr.
NaI + A dark solid forms which A redox reaction in
concentrated H2SO4 gives off a purple vapour on which iodide ions are
Test yourself warming (I2). Some yellow oxidised to iodine.
solid may be seen (S) and Sulfur is reduced from
36 Draw a diagram of laboratory there is a bad-egg smell the +6 to the 0 and −2
(H2S). states.
apparatus for collecting several
large test tubes full of hydrogen
chloride gas from the reaction The trend in the power of the halide ions to act as reducing agents is I− >
of sodium chloride with Br− > Cl−. Chlorine is the strongest oxidising agent of these halogens, so it
concentrated sulfuric acid. has the greatest tendency to form negative ions. Conversely, chloride ions are
37 With the help of oxidation reluctant to give up their electrons and turn back into chlorine molecules. So
numbers, write a balanced the chloride ion is the weakest reducing agent.
equation for the redox reaction
Iodine is the weakest oxidising agent, so it has the least tendency to form
of sodium bromide with
negative ions. Conversely, the iodide ion is the strongest reducing agent,
concentrated sulfuric acid.
being most ready to give up electrons and turn back into iodine molecules.
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Hydrogen halides
H+
The hydrogen halides are compounds of hydrogen with the halogens. transferred
They are all colourless, molecular compounds with the formula HX, HCl + H2O Cl – (aq) +
+
H3O (aq)
where X stands for Cl, Br or I. The bonds between hydrogen and the oxonium
halogens are polar. ion
Figure 4.30 The reaction of hydrogen
Hydrogen chloride, hydrogen bromide and hydrogen iodide are similar in chloride with water. Hydrogen chloride is a
that they are: strong acid which is fully ionised in aqueous
● colourless gases at room temperature which fume in moist air solution.
● very soluble in water, forming acidic solutions (hydrochloric, hydrobromic
and hydriodic acids) which ionise completely in water (Figure 4.30)
● strong acids, so they ionise completely in water.
Mixing any of the hydrogen halides with ammonia produces a white
smoke of an ammonium salt (Figure 4.31). Ammonia molecules turn into
ammonium ions, NH4+, in this reaction. The ammonia is acting as a base
by accepting hydrogen ions from the hydrogen halides.
NH3(g) + HCl(g) → NH4Cl(s)
Test yourself
38 Write ionic equations for the reactions of silver nitrate solution with:
a) potassium iodide solution
b) sodium bromide solution.
39 Describe the colour changes on adding:
a) a solution of chlorine in water to aqueous sodium bromide Figure 4.31 Fumes of ammonium
b) a solution of bromine in water to aqueous potassium iodide. chloride forming as gases escaping from
40 Put the chloride, bromide and iodide ions in order of their strength concentrated ammonia solution and
as reducing agents, with the strongest reducing agent first. Justify concentrated hydrochloric acid mix and
your answer. react. Ammonium chloride is a white solid,
while hydrogen chloride and ammonia are
41 Explain why the compound of hydrogen and fluorine is a liquid at
colourless gases.
room temperature on a cool day, when the other hydrogen halides
are gases.
42 a) S
how that the reaction of ammonia with hydrogen bromide gas
involves proton transfer.
b) Explain why the product of the reaction is a solid.
4.11 Halogens in oxidation states
+1 and +5
Chlorine oxoanions form when chlorine reacts with water and alkalis. Key term
When chlorine dissolves in water, it reacts reversibly, forming a mixture An oxoanion is an ion with the general
of weak chloric(i) acid and strong hydrochloric acid. This is an example of formula is X xOyz− (where X represents
a disproportionation reaction (Section 3.4). any element while O represents an
H2O(l) + Cl 2(g) → HClO(aq) + HCl(aq) oxygen atom).
4.11 Halogens in oxidation states +1 and +5 117
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7 CIO4 KCIO4 Bromine reacts in a similar way but to a much lesser extent. Iodine is insoluble
in water and hardly reacts at all.
5 CIO3 KCIO3 When chlorine dissolves in potassium (or sodium) hydroxide solution
at room temperature it produces chlorate(i) and chloride ions
(Figure 4.32).
3 CIO2 KCIO2
Cl 2(g) + 2OH−(aq) → ClO−(aq) + Cl−(aq) + H2O(l)
1 CIO KOCI The active ingredient in household bleach is sodium chlorate(i), made by
dissolving chlorine in sodium hydroxide solution.
0 CI2
1 CI HCI On heating, the chlorate(i) ions disproportionate to chlorate(v) and
Figure 4.32 The oxidation states of chlorine. chloride ions.
3ClO−(aq) → ClO3−(aq) + 2Cl−(aq)
+1 +5 −1
The overall equation for the reaction of chlorine with hot potassium
hydroxide is:
3Cl 2(g) + 6OH−(aq) → ClO3−(aq) + 5Cl−(aq) + 3H2O(l)
Bromine and iodine react in a similar way to chlorine with alkalis. The
BrO− and IO− ions are less stable, so they disproportionate at a lower
temperature. A hot solution of iodine in potassium hydroxide produces a
solution containing potassium iodate(v) and potassium iodide.
Test yourself
43 a) Write a balanced, ionic equation for the reaction of iodine with hot
aqueous hydroxide ions.
b) Use oxidation numbers to show that this is a disproportionation
reaction.
Activity
Water treatment
100%
At very low concentrations, chlorine is used to disinfect tap water. It forms
chloric(i) acid, HClO, when it reacts with water. Chloric(i) acid is a powerful
oxidising agent and a weak acid. It is an effective disinfectant because,
un-ionised HClO
Percentage of
unlike ClO— ions, the molecule can pass through the cell walls of bacteria.
Once inside the bacterium, the HClO molecules break the cell open and kill
the organism by oxidising and chlorinating molecules which make up its
structure.
Chloric(i) acid is a weak acid. It is only very slightly ionised in solution. The
concentration of un-ionised HClO in a solution depends on the pH, as shown 0%
4 5 6 7 8 9 10 11
in Figure 4.33. pH
Swimming pools can be sterilised with much higher concentrations of Figure 4.33 Graph to show how the concentration of
chloric(i) acid, HClO, varies over a range of pH values at
chlorine compounds which react to produce chloric(i) acid when they
20 °C.
dissolve in water. Swimming pool managers have to check the pH of the
water carefully – they aim to keep the pH in the range 7.2–7.8 (Figure 4.34).
118 4 Inorganic chemistry and the Periodic Table
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1 Write an equation to show the formation of chloric(i) acid when
chlorine reacts with water.
2 Swimming pools used to be treated with chlorine gas from
cylinders containing the liquefied gas under pressure. Now they
are usually treated with chemicals that are supplied as solids
and that produce chloric(i) acid when added to water. Suggest
reasons for this change.
3 Explain why sodium chlorate(i) produces chloric(i) acid when
added to water at pH 7–8.
4 Explain why the pH of swimming pool water must not be allowed
to rise above 7.8.
5 Suggest reasons why pool water must not become acidic, even
though this would increase the concentration of
un-ionised HClO. Figure 4.34 Chlorine compounds treat the water in swimming pools.
6 Nitrogen compounds, including ammonia, urea and proteins,
react with HClO to form chloramines, which are irritating to skin and eyes. Different
chloramines can react to form nitrogen and hydrogen chloride, which gets rid of the
problem. However, if there is excess HClO, another reaction produces nitrogen trichloride,
which is responsible for the so-called ‘swimming pool smell’. Nitrogen trichloride is very
irritating to skin and eyes. Write equations to show:
a) the formation of the chloramine, NH2Cl, from ammonia and chloric(i) acid
b) the removal of chloramines by reaction of NH2Cl with NHCl2
c) the formation of nitrogen trichloride from chloramine, NH2Cl.
7 Explain why swimming pools do not smell of chlorine if they are properly maintained.
Chapter summary
l Group 2 elements become more reactive down the
Chapter 4 Inorganic chemistry and
group as the tendency to lose electrons and form
the Periodic Table positive ions increases.
l Elements in compounds of some Group 1 and l The metals react: with oxygen to form ionic oxides,
Group 2 elements can be recognised by their flame MO; with chlorine to form ionic chlorides, MCl2;
colours. and with water to form hydroxide, M(OH)2.
l The energy from the flame excites outer electrons l The oxides of the elements Mg to Ba react with
in the atoms, raising them to higher energy water to turn into hydroxides; they are basic oxides
levels. The atoms then emit radiation as the that react with acids to form salts.
electrons drop back to lower energy levels. This l The solubility of the hydroxides of the elements
gives rise to the characteristic colours if the size Mg to Ba increases down the group. The
of the energy jumps means that the frequency of hydroxides react with acids to form salts.
the radiation corresponds to colours in the visible l The sulfates of the elements Mg to Ba become less
spectrum. soluble down the group.
l The first and second ionisation energies of the l The carbonates and nitrates of elements in
elements decrease down Group 2, because the Groups 1 and 2 become more thermally stable
charge on the nucleus increases but the number of down each group. In Group 1, only lithium
shielding electrons in inner shells also increases so carbonate decomposes on heating with a flame.
that the two outer electrons get further and further Only lithium nitrate decomposes on heating to the
from the same effective nuclear charge. oxide, nitrogen dioxide and oxygen; the nitrates to
Na and K decompose to the nitrite and oxygen.
Chapter summary 119
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l The trend in thermal stability of carbonates and l Halide ions in solution can be recognised by the
nitrates can be explained in terms of the size and colour of the precipitates formed with silver nitrate
charge of the metal ions. solution, and the solubility of the precipitates in
l Polarising power gives an indication of the extent ammonia solution.
to which a positive ion is able to distort the electron l The trend of increasing reducing ability of the halides
cloud of a neighbouring negative ion. The larger from Cl– to Br– is illustrated by the reactions of solid
the charge on a positive ion, and the smaller its size, Group 1 halides with concentrated sulfuric acid.
the greater its polarising power. l The hydrogen halides are colourless gases that react
l The halogens consist of diatomic molecules. with ammonia gas to form a white smoke of the
London forces between the molecules increase ammonium halide.
down the group as the numbers of electrons l Formation of positive oxidation states of halogens
in the molecules increase. Also, the larger can be illustrated by the disproportionation
molecules are more polarisable than the smaller reactions of chlorine with water, with cold aqueous
molecules. As a result, the melting and boiling alkali and with hot alkali.
temperatures of the elements increase down l Trends in the properties of halogens from Cl
the group. to I can be used to make predictions about the
l The trend of decreasing reactivity down the properties of fluorine and astatine.
group can be illustrated by the redox reactions of l Characteristic chemical reactions are used
solutions of halogens with aqueous halide ions. to identify halide ions as well as carbonate,
l Halogens oxidise Group 1 and Group 2 metals to hydrogencarbonate, sulfate and ammonium ions
form ionic halides, MX and MX 2 respectively. (see Chapter 5 in this book).
Exam practice questions
The elements Mg to Ba in Group 2, and their
1 for the reaction and show that it is a
compounds, can be used to show the trends in disproportionation reaction. (3)
properties down a group of the Periodic Table. Identify the following salts and account for the
3
State and explain the trend down the group in: observations.
a) atomic radius (3)
a) X is a white solid which colours a flame
b) first ionisation energy (3)
c) thermal stability of the carbonates. (3) bright yellow. No precipitate forms on
mixing a solution of X with sodium
2 a) Write an equation for the reaction of
hydroxide solution. On heating, X gives
chlorine with aqueous sodium hydroxide,
and use this example to state what is meant off a colourless gas that relights a glowing
by disproportionation. (2) splint. (4)
b) On heating, chlorate(i) ions in solution b) Y is a white, crystalline solid which colours
disproportionate to chlorate(v) ions and a flame green. Adding dilute nitric acid
chloride ions. Write an ionic equation for followed by silver nitrate solution to a
this reaction. (2) solution of Y produces a white
c) On heating to just above its melting precipitate. (2)
temperature, KClO3 reacts to form KClO4 c) Z is a white crystalline solid which colours
and KCl. Write a balanced equation a flame lilac. Z is soluble in water – the
120
4 Inorganic chemistry and the Periodic Table
469983_04_Chem_Y1-2_097-[Link] 120 13/04/19 10:13 PM
solution does not change the colour of bromide and sodium iodide. Include in your
indicators. Mixing the solution with answer any additional tests that help to confirm
a solution of silver nitrate produces a the interpretation of observations:
cream precipitate that is insoluble in a) aqueous chlorine (6)
dilute aqueous ammonia but soluble in b) aqueous silver nitrate (6)
concentrated aqueous ammonia. A solution c) concentrated sulfuric acid. (6)
of Z turns orange on adding aqueous
State the trend in the power of chloride,
8
chlorine. (4)
bromide and iodide ions as reducing agents.
Radium is a highly radioactive element which
4 Describe how you could demonstrate the
is below barium in the Periodic Table. Use your trend by carrying out experiments in the
knowledge of the chemistry of the elements laboratory. Include in your description the
Mg to Ba in Group 2 to predict properties of main observations that illustrate the differences
radium and its compounds. Include in your between the ions. Include equations for the
predictions a description of the following reactions. (6)
changes, equations for any chemical changes
9 a) Hydrogen fluoride is manufactured by heating
and the appearance of the products:
concentrated sulfuric acid with fluorite, CaF2,
a) the reaction of radium with oxygen (3)
obtained by purifying the mineral fluorspar.
b) the reaction of radium with water (3)
Hydrogen fluoride leaves the kiln as a gas.
c) the reaction of radium oxide with water (3)
It can be condensed to a liquid and purified
d) the reaction of radium hydroxide with
by fractional [Link] other product
dilute hydrochloric acid (3)
formed in the kiln is solid calcium sulfate.
e) the solubility of radium sulfate in water (2)
i) Write an equation for the reaction in
f ) the effect of heating radium nitrate. (3)
the kiln. (1)
5 This generalisation is sometimes stated for the ii) Explain why it is much easier to
ionic compounds of Group 2 elements: condense hydrogen fluoride to a liquid
‘For Group 2 compounds with small anions than the other hydrogen halides. (3)
solubility in water increases down the group; b) Hydrogen chloride was traditionally made
for compounds with large anions solubility by adding concentrated sulfuric acid to
decreases down the group.’ sodium chloride. Today, the main, industrial
a) Discuss, with the help of examples, whether source of hydrogen chloride is as a co-
or not this generalisation can be justified.(6) product of processes in the petrochemical
b) Assess whether magnesium fluoride is likely industry, such as the chlorination of
to be more or less soluble than barium hydrocarbons. A small amount of the
fluoride. (2) gas is made by the reaction between the
hydrogen and chlorine produced during
Astatine, At, is the element below iodine in
6 the manufacture of sodium hydroxide
Group 7. Predict, giving your reasons: during the electrolysis of aqueous sodium
a) the physical state of astatine at room chloride. This is used to make the purest
temperature (3) hydrochloric acid.
b) the effect of bubbling chorine though an i) Explain why the chlorination of
aqueous solution of sodium astatide (2) hydrocarbons is a source of hydrogen
c) whether or not hydrogen astatide forms on chloride and suggest reasons why 90%
adding concentration sulfuric acid to solid of the gas is now produced by the
sodium astatide (3) petrochemical industry. (3)
d) the colour of silver astatide and its solubility ii) Explain why the manufacture of
in concentrated ammonia solution. (2) sodium hydroxide also produces
Describe the observations and write equations
7 hydrogen and chlorine. Give reasons
to explain how each of the following reagents why the gases made in this way are
can be used to distinguish between sodium very pure. (2)
121
Exam practice questions
469983_04_Chem_Y1-2_097-[Link] 121 13/04/19 10:13 PM
iii) Explain why the reaction of hydrogen reacts with excess potassium iodide solution
gas with chlorine gas has to be very to form four moles of iodine, I2. Determine
carefully controlled when it is used to the value of x and write equations for the two
manufacture hydrochloric acid. (1) reactions. Identify the molecules or ions that
iv) Give an example to show why it is are oxidised or reduced in the reactions. (6)
sometimes important for chemists to
A mixture of two solid potassium halides was
12
use very pure hydrochloric acid. (1)
investigated in two ways.
c) Hydrogen bromide is manufactured
A A sample of the mixture was dissolved in
on a much smaller scale than hydrogen
water to make a concentrated solution.
chloride. Explain why hydrogen bromide
Chlorine gas was bubbled through the
is not made by adding concentrated
solution until there was no further reaction.
sulfuric acid to a bromide, such as sodium
A dark grey precipitate formed and orange-
bromide. (2)
brown fumes appeared above the solution.
The table at the bottom of the page compares
10 B A 0.214 g sample of the mixture was
properties of the chlorides of Group 2 elements. dissolved in water. An excess of silver nitrate
a) Assess whether or not the data justifies the solution was added. The precipitate that
claim that the bonding and structure of formed was separated washed and dried. The
beryllium chloride differ from the bonding mass of precipitate was 0.317 g. Next, the
and structure of the chlorides of other precipitate was treated with concentrated
elements in Group 2. (2) ammonia solution. Some of the precipitate
b) Explain the differences identified in (a). (4) dissolved. After separating, washing and
c) In the gas phase, beryllium chloride is drying again, the mass of precipitate reduced
molecular. Explain why beryllium chloride to 0.176 g.
molecules have a linear shape. (3) a) Deduce what you can about the
d) In the solid state, BeCl2 molecules two halides in the mixture from the
polymerise to make long chains. Explain, observations in A. (3)
with the help of a diagram, how these b) Write ionic equations for the
chains can form by beryllium forming reactions that led to the observations
four bonds with chlorine atoms arranged described. (2)
tetrahedrally around each Be atom. (4) c) Use the data from B to determine the
percentage by mass of the two potassium
Solid iodine reacts with excess liquid chlorine
11
halides in the original mixture. Show how
at a low temperature. A yellow solid remains
you arrive at your answer. (6)
after evaporating the excess chlorine. The
formula of the solid is IClx. One mole of IClx
Compound Melting temperature/°C Boiling temperature/°C Solubility in mol/100 g
water
Beryllium chloride, BeCl2 450 520 0.19
Magnesium chloride, MgCl2 714 1412 0.56
Calcium chloride, CaCl2 782 2000 0.54
Strontium chloride, SrCl2 911 1250 0.01
Barium chloride, BaCl2 963 1560 0.15
122
4 Inorganic chemistry and the Periodic Table
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Formulae, equations and amounts
5 of substance
5.1 Understanding chemical quantities
Chemists often need to measure how much of a particular chemical there is
in a sample. Analysts in pharmaceutical companies, for example, test samples
of tablets and medicines to check that they contain the right amount of a
drug. Industrial chemists measure the amounts of substances they need for
chemical processes. Laboratory chemists calculate the yield of product they
expect when they mix measured quantities of the reactants for a synthesis. In
these, and many other contexts, chemists have to be able to measure amounts
of substances accurately.
Key terms
Chemical amounts
Amount of substance is a physical In chemistry, the amount of a substance is measured in moles. Chemists
quantity (symbol n) which is measured use the unit ‘mole’ to measure an amount of substance containing a standard
in the unit mole (symbol mol). number of atoms, molecules or ions. The word ‘mole’ entered the language of
Relative atomic mass, Ar, is the chemistry at the end of the nineteenth century. It is based on the Latin word
average mass of an element relative for a heap or a pile.
1
to 12 th of the mass of an atom of the When chemists are determining formulae or working with equations,
isotope carbon-12. The values are they need to measure amounts in moles. Chemists have balances to
relative so they do not have units. measure masses and graduated containers to measure volumes, but there is
Molar mass is the mass of one mole no instrument for measuring chemical amounts directly. Instead, chemists
of a chemical – the unit is g mol−1. As must first measure the masses or volumes of substances and then calculate
always with molar amounts, the symbol the chemical amounts.
or formula of the chemical must be
specified. Molar masses
The key to working with chemical amounts in moles is to know the relative
masses of different atoms. The accurate method for determining relative
atomic masses involves the use of a mass spectrometer (Section 1.3).
One mole of an element has a mass that is equal to its relative atomic mass
in grams. So, the mass of one mole of carbon is 12.0 g and the mass of one mole
of copper is 63.5 g. These masses of one mole are called molar masses (symbol
M). So, the molar mass of carbon, M(C) = 12.0 g mol−1 and the molar mass of
copper, M(Cu) = 63.5 g mol−1 (Figure 5.1).
Similarly the molar mass of the molecules of an element or a compound
is numerically equal to its relative molecular mass. So, the molar mass of
oxygen molecules, M(O2) = 32.0 g mol−1 and the molar mass of sulfuric acid,
M(H2SO4) = 98.1 g mol−1. Likewise, the molar mass of an ionic compound
is numerically equal to its relative formula mass. The molar mass of sodium
chloride, NaCl, is therefore 58.5 g mol−1 (Figure 5.2).
5.1 Understanding chemical quantities 123
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iron 55.8 g
carbon 12.0 g hydrated cobalt nitrate iron(iii) chloride
sulfur 32.1 g
Co(NO3)2.6H2O = 290.9 g FeCl3 = 162.3 g
sodium choride
NaCl = 58.5 g
mercury 200.6 g potassium iodide potassium
KI = 166.0 g manganate(vii)
copper 63.5 g aluminium 27.0 g hydrated copper(ii) sulfate KMnO4 = 158.0 g
CuSO4.5H2O = 249.6 g
Figure 5.1 One mole amounts of copper, carbon, iron, Figure 5.2 One mole amounts of some ionic compounds.
aluminium, mercury and sulfur.
Amount in moles
The mole is the SI unit for amount of substance. The name of the quantity
Tip is ‘mole’. Its unit is ‘mol’. So,
Section A1.1 of Appendix A1 gives 12 g of carbon contains 1 mol of carbon atoms
advice on how to work out the value
of maths equations with brackets and 24 g of carbon contains 2 mol of carbon atoms
combinations of multiplication and 240 g of carbon contains 20 mol of carbon atoms.
addition.
Notice that:
mass of substance/g
Key term amount of substance/mol =
molar mass/g mol−1
The term species is a useful collective It is important to be precise about the chemical species involved when measuring
noun used by chemists to refer amounts in moles. In calcium chloride, CaCl2, for example, there are two chloride
generally to atoms, molecules or ions. ions, Cl−, combined with each calcium ion, Ca2+. So in one mole of calcium
chloride there is one mole of calcium ions and two moles of chloride ions.
Tip The Avogadro constant
Relative atomic masses show that one atom of carbon is 12 times heavier than
Section A1.5 of Appendix A1 gives
one atom of hydrogen. This means that 12 g of carbon contains the same number
help with substituting values into
of atoms as 1 g of hydrogen. Similarly, one atom of oxygen is 16 times as heavy
mathematical formulae.
as one atom of hydrogen, so 16 g of oxygen also contains the same number of
atoms as 1 g of hydrogen.
In fact, the molar mass of every element (1 g of hydrogen, 12 g carbon, 16 g
Key term oxygen, and so on) contains the same number of atoms. This number is called the
The Avogadro constant is the number
Avogadro constant, after the Italian scientist Amedeo Avogadro. Experiments
of atoms, molecules or ions in one mole
show that the Avogadro constant, L, is 6.02 × 1023 mol−1. Written out in full this
of a substance. The constant has the
is 602 000 000 000 000 000 000 000 atoms, molecules or ions per mole.
unit mol−1. The Avogadro constant is the number of atoms, molecules or formula units in
one mole of any substance. So, one mole of oxygen (O2) contains 6.02 × 1023 O2
molecules and two moles of oxygen (O2) contains 2 × 6.02 × 1023 O2 molecules.
The number of atoms, molecules or formula units
= amount of chemical/mol × the Avogadro constant/mol−1
124 5 Formulae, equations and amounts of substance
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Again, it is vital to specify the chemical species concerned in calculating
the amount of a substance or the number of particles in a sample of a
substance. For example, 2 g of hydrogen contains 2 mol of hydrogen (H)
atoms (12.04 × 1023 atoms) but only 1 mol of hydrogen (H 2) molecules
(6.02 × 1023 molecules).
Test yourself
1 What is the amount, in moles, of:
a) 20.05 g of calcium atoms
b) 3.995 g of bromine atoms
c) 159.8 g of bromine molecules
d) 6.41 g of sulfur dioxide molecules
e) 10.0 g of sodium hydroxide?
2 What is the mass of:
a) 0.1 mol of iodine atoms
b) 0.25 mol of chlorine molecules
c) 2.0 mol of water molecules
d) 0.01 mol of ammonium chloride, NH4Cl
e) 0.125 mol of sulfate ions, SO42−?
3 How many moles of:
a) sodium ions are there in 1 mol of sodium carbonate, Na2CO3
b) bromide ions are there in 0.5 mol of barium bromide, BaBr2
c) nitrogen atoms are there in 2 mol of ammonium nitrate, NH4NO3?
4 Use the Avogadro constant to calculate:
a) the number of chloride ions in 0.5 mol of sodium chloride, NaCl
b) the number of oxygen atoms in 2 mol of oxygen molecules, O2
c) the number of sulfate ions in 3 mol of aluminium sulfate, Al2(SO4)3.
5.2 Finding empirical formulae Key term
Although the formulae of most compounds can be predicted, the only sure
way of determining the formula of a substance is by experiment. This has An empirical formula shows the simplest
been done for all common compounds and their formulae can be checked whole number ratio of the atoms of
in tables of data. different elements in a compound; for
example, CH4 for methane and CH3 for
‘Empirical’ evidence is information based on experience or experiment, so
ethane.
chemists use the term empirical formulae for formulae calculated from the
results of experiments.
An experiment to find an empirical formula involves measuring the
masses of elements which combine in the compound. From these masses,
Tip
it is possible to calculate the number of moles of atoms which react, and Section A1.4 of Appendix A1 gives
hence the ratio of atoms which react. This gives an empirical formula help with calculations involving ratio
which shows the simplest whole number ratio for the atoms of different and proportion.
elements in a compound.
5.2 Finding empirical formulae 125
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Example
Analysis of 20.1 g of an iron bromide sample showed that it contained 3.80 g
of iron and 16.3 g of bromine. What is its empirical formula?
Notes on the method
The molar masses of the elements come from the Periodic Table (see page 652).
Convert the masses in grams to amounts in moles by dividing by the molar
masses of the atoms of the elements.
Divide the amounts by the smaller of the amounts to find the simplest
whole number ratio.
Answer
iron bromine
Combined masses 3.80 g 16.3 g
Molar mass 55.8 g mol−1 79.9 g mol−1
3.80 g 16.3 g
Combined moles of atoms = 0.0681 mol = 0.204 mol
55.8 g mol−1 79.9 g mol−1
0.0681 0.204
Ratio of combined atoms = 1.00 = 2.996 = 3.00
0.0681 0.0681
Simplest whole number ratio of atoms is 1 : 3
So, the empirical formula is FeBr3.
Key term Percentage composition
Sometimes, the results of an analysis of a compound show the percentages
Percentage composition is the of the different elements, rather than their masses. This is the percentage
percentage by mass of each of the composition of the compound. The empirical formula of the compound
elements in a sample of a compound. can be calculated from these results.
Example
What is the empirical formula of copper pyrites which has the analysis
34.6% copper, 30.5% iron and 34.9% sulfur by mass?
Notes on the method
Follow the procedure in the example for finding an empirical formula. The percentages,
in effect, show the combining masses in a 100 g sample of the compound.
Answer
copper iron sulfur
Combining masses 34.6 g 30.5 g 34.9 g
Molar masses of elements 63.5 g mol−1 55.8 g mol−1 32.1 g mol−1
34.6 g 30.5 g 34.9 g
Amounts combined
63.5 g mol−1 55.8 g mol−1 32.1 g mol−1
= 0.545 mol = 0.546 mol = 1.09 mol
Simplest whole number ratio of amounts is 1 : 1 : 2
The empirical formula is CuFeS2.
126 5 Formulae, equations and amounts of substance
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Test yourself
5 What is the empirical formula of the compound in which:
a) 0.60 g carbon combines with 0.20 g hydrogen
b) 1.02 g vanadium combines with 2.84 g chlorine
c) 1.38 g sodium combines with 0.96 g sulfur and 1.92 g oxygen?
6 What is the empirical formula of the compound in which the
percentages by mass of the elements present are:
a) 2.04% hydrogen, 32.65% sulfur and 65.31% oxygen
b) 52.18% carbon, 13.04% hydrogen and 34.78% oxygen?
Activity
Finding the formula of red
red copper oxide excess natural
copper oxide gas burning
A group of students investigated the combustion tube
formula of red copper oxide by reducing
it to copper using natural gas as shown in
natural
Figure 5.3. gas
The experiment was carried out five
times, starting with different amounts of
red copper oxide. The results are shown strong heat
in Table 5.1. Figure 5.3 Reducing red copper oxide by heating in natural gas.
Table 5.1 b) Enter a formula in column 5 to find the amount of copper
Experiment Mass of red Mass of copper in moles in the oxide.
number copper oxide/g in the oxide/g c) Enter a formula in column 6 to find the amount of oxygen
in moles in the oxide.
1 1.43 1.27
01.22 Edexcel Chemistry for AS
5 From the spreadsheet, plot a line graph of amount of
2 2.14 Barking
1.90 Dog Art copper (y-axis) against amount of oxygen (x-axis). Print
3 2.86 2.54 red copper oxide by heatingout
Figure 5.3 Reducing in your graph.
natural gas. If you cannot plot graphs directly from the
spreadsheet, draw the graph by hand.
4 3.55 3.27
6 Which of the points should be disregarded in drawing the
5 4.29 3.81 line of best fit?
7 a) What, from your graph, is the average value of the ratio:
1 Look at Figure 5.3. What safety precautions should the amount of copper/mol
?
students take during the experiments? amount of oxygen/mol
2 What steps should the students take to ensure that all the b) How much copper, in moles, combines with one mole of
copper oxide is reduced to copper? oxygen in red copper oxide?
3 Start a spreadsheet program on a computer and open up c) What is the formula of red copper oxide?
a new spreadsheet for your results. Enter the experiment 8 Give reasons why the students could claim that their answer
numbers and the masses of copper oxide and copper in the for the formula of the oxide was valid?
first three columns of your spreadsheet, as in Table 5.1. 9 Write a word equation, and then a balanced equation, for the
4 a) Enter a formula in column 4 to work out the mass of reduction of red copper oxide to copper using methane (CH4)
oxygen in the red copper oxide used. in natural gas. (Hint: The only solid product is copper.)
5.2 Finding empirical formulae 127
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Key terms 5.3 Amounts of gases
The gas laws describe the behaviour of Gases and the gas laws
gases and are summarised by the ideal The Irish chemist Robert Boyle (1627–91) was one of the first people to
gas equation. investigate the effect of pressure on the volume of gases. About a hundred
Ideal gases are gases which obey the years after, in the late 18th century, the hot air balloon flights of the
ideal gas equation. In practice, real Montgolfier brothers stimulated scientists to study the behaviour of gases.
gases deviate to at least some extent Two of these scientists were French: Joseph Gay-Lussac (1778–1850) and
from ideal behaviour. Jacques Charles (1746–1823). They were particularly interested in the
variation of the volumes of gases with temperature. Jacques Charles put his
SI units are the internationally agreed theories to the test and in 1783 made the first ascent in a hydrogen balloon.
units for measurement in science. Meanwhile, the Italian scientist Amedeo Avogadro (1776–1856) proposed
Pressure is defined as force per unit area. the law that equal volumes of all gases, at the same temperature and pressure,
The SI unit of pressure is the pascal (Pa) contain the same number of molecules.
which is a pressure of one newton per These scientists discovered the gas laws that show how the volume, V, of a
square metre (1 N m−2). The pascal is a sample of gas depends on three things:
very small unit and so pressures are often
quoted at kilopascals (kPa). Atmospheric ● the temperature, T
pressure is measured in bar: ● the pressure, p
● the amount of gas in moles, n.
1 bar = 100 000 Pa = 100 kPa
Volume is the amount of space taken
Real and ideal gases
up by a sample. The SI unit of volume
Scientists use the concept of an ‘ideal gas’ which obeys the gas laws perfectly.
is the cubic metre (m3). Chemists
In practice, real gases do not obey the laws under all conditions. Under
generally measure volumes in cubic
laboratory conditions, however, there are gases which are close to behaving
decimetres (dm3) or cubic centimetres
like an ideal gas. These are the gases which, at room temperature, are well
(cm3): 1 dm = 10 cm and so 1 dm3 =
above their boiling points, such as helium, nitrogen, oxygen and hydrogen.
10 cm × 10 cm × 10 cm = 1000 cm3;
1 dm3 is the same volume as a litre; Chemists generally find that the gas laws predict the behaviour of real gases
1 m = 10 dm so 1 m3 = 103 dm3 = 106 cm3. accurately enough to make them a useful practical guide, but it is important
to bear in mind that gases such as ammonia, butane, sulfur dioxide and
The kelvin is the SI unit of temperature on
carbon dioxide can show marked deviations from ideal behaviour. These
the absolute, or Kelvin, temperature scale.
are the gases which boil only a little below room temperature and can be
On this scale, absolute zero is 0 K, water
liquefied just by raising the pressure.
freezes at 273 K and boils at 373 K.
The ideal gas equation
The behaviour of an ideal gas can be summed up by combining the gas laws
Test yourself
into a single equation called the ideal gas equation:
7 What are the values of these
pV = nRT
temperatures on the Kelvin
scale? When SI units are used, the pressure is measured in pascals, Pa, the volume
a) boiling temperature of in cubic metres, m 3, and the temperature in kelvin, K. R, in the ideal gas
nitrogen, −196 °C equation, is the gas constant. R has the value 8.31 J mol−1 K−1 if all quantities
are in SI units.
b) boiling temperature of
butane, −0.5 °C
c) melting temperature of
Measuring molar masses of gases
sucrose, 186 °C In the days before mass spectrometry (Section 1.3) chemists used the ideal
gas equation to measure the molar masses of gases and of other substances
d) melting temperature of iron,
that evaporate easily. The method is accurate enough to determine the
1540 °C
molecular formula of elements and compounds.
128 5 Formulae, equations and amounts of substance
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One practical approach is to inject a weighed sample of a liquid into a syringe
heated in an oven. Measurements taken include the volume of vapour,
Key term
the temperature of the vapour and its pressure (the atmospheric pressure).
The molecular formula gives the actual
Measurements are converted to SI units and then substituted in the ideal gas
number of atoms of each element in a
equation to find the amount in moles, n.
molecule.
Example
A 0.124 g sample of a liquid with the empirical formula C3H7 evaporates
to give 45.0 cm3 vapour at 100 °C and a pressure of 100 kPa. What is Tip
the molecular formula of the liquid?
Including the units at every stage of the
calculation is a useful check that the
Notes on the method
steps have been carried out correctly.
Convert all units to SI units: 1 cm3 = 10−6 m3.
The units should cancel to give the
Substitute in the equation pV = nRT to find n (the amount in moles). expected units for the answer (in this
pV example, mol). To check the units in
The equation rearranges to give: n =
RT the ideal gas equation you need to
The molar mass can then be calculated by dividing the mass of the sample remember that a pressure of 1 Pa =
in grams by the amount in moles, giving an answer with the units g mol−1.
1 N m−2 and that an energy transfer of
1 J = 1 N m (force times distance).
Answer
For the sample of liquid:
● pressure = 100 000 N m−2
● volume = 45.0 × 10−6 m3 Tip
● temperature = 373 K
Section A1.5 of Appendix A1 gives
The gas constant = 8.31 J mol−1 K−1 help with rearranging mathematical
equations.
pV 100 000 Pa × 45.0 × 10−6 m3
n= = = 1.45 × 10−3 mol
RT 8.31 J mol−1 K−1 × 373 K
Mass of the sample = 0.124 g
The amount of substance in the sample = 1.45 × 10−3 mol
0.124 g
Therefore, the molar mass of the liquid = = 85.5 g mol−1
1.45 × 10−3 mol
A molecular formula is always a simple multiple of the empirical formula
(Section 6.1.3).
The relative mass of the empirical formula of the liquid,
Mr (C3H7) = (3 × 12.0) + (7 × 1.0) = 43.0
Even though the vapour of the compound does not behave as an ideal gas,
the result is accurate enough to show that the molecular formula is twice
the empirical formula. The molecular formula of the compound is C6H14.
Test yourself
8 The mass of 200 cm3 of a gaseous hydrocarbon is 0.356 g at 298 K
and 100 kPa. What is the molar mass of the gas?
9 A 0.163 g sample of a liquid evaporates to give 65.0 cm3 of vapour at
101 °C and 100 kPa. What is the molar mass of the liquid?
5.3 Amounts of gases 129
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Key term The molar volume of gases
The ideal gas equation shows that, at a fixed temperature and fixed pressure,
The molar volume of a gas is the the volume of a gas depends only on the amount of gas in moles; the type
volume of 1 mol of the gas under stated or the formula of the gas does not matter. This is only strictly true for ideal
conditions. At room temperature and gases but is nevertheless useful when dealing with real gases.
atmospheric pressure, the molar volume
Substituting in the ideal gas equation makes it possible to calculate the volume of
of all gases is 24.0 dm3 mol−1.
one mole of gas under any conditions. This shows that the volume of one mole
of any gas occupies about 24 dm3 (24 000 cm3) at a typical room temperature of
around 16 °C (298 K) and at atmospheric pressure (100 kPa). This volume of one
mole of gas is called the molar volume under the stated conditions.
So, 1 mole of oxygen (O2) and 1 mole of carbon dioxide (CO2) each occupy 24 dm3
at room temperature. Therefore 2 moles of O2 occupy 48 dm3 and 0.5 moles of O2
occupy 12 dm3 at room temperature. Notice from these simple calculations that:
volume of gas/cm 3 = amount of gas/mol × molar volume/cm 3 mol−1
So, under laboratory conditions at room temperature:
volume of gas/cm 3 = amount of gas/mol × 24 000/cm 3 mol−1
Activity
Measuring the molar volumes of three gases Table 5.2 Results recorded at room temperature and pressure.
The syringe shown in Figure 5.4 is used in an experiment to Mass/g
measure the molar volume of several gases. The procedure is
outlined in steps A—I. Sample results are given in Table 5.2. Syringe + cap + nail (step D) 142.213
50 cm3 plastic syringe Syringe + cap + nail + carbon dioxide 142.302
Syringe + cap + nail + methane 142.247
Syringe + cap + nail + butane 142.322
nail to hold the plunger
at the 50 cm3 mark
Figure 5.4 Plastic syringe with nail to lock the plunger at the 50 cm3 mark.
A Remove the nail. Fill the syringe to the 50 cm3 mark. Seal 1 Explain the purpose of step A.
the syringe with a syringe cap. Check that the plunger 2 At the end of step C, what is in the syringe and why is it
returns to the 50 cm3 mark after pushing in the plunger by necessary to lock the plunger with the nail?
10 cm3 and releasing, and after pulling out the plunger by 3 Why is the syringe weighed in step B with the plunger
10 cm3 and releasing. pulled out, rather than weighing the empty syringe with
B Push in the plunger to empty the syringe. Block the nozzle with the plunger pushed in?
a syringe cap. 4 How might a bag be filled with a dry sample of carbon
C Pull out the plunger. Lock it at the 50 cm3 mark with the nail. dioxide if the gas is not available from a cylinder? Why
D Measure and record the mass of the syringe, syringe cap must the gas be dry?
and nail using a three-place balance. 5 Use the results in Table 5.2 to determine the molar volumes
E Remove the syringe cap and the nail from the plunger. Push of the three gases.
in the plunger completely. 6 Why is it necessary to use a three-place balance to measure
F Draw 50 cm3 gas into the syringe from a plastic bag the masses?
containing one of the gases. 7 What are the main sources of measurement uncertainty in
G Seal the syringe again and use the nail to lock the syringe. this experiment?
H Measure and record the mass of the syringe, cap and nail. 8 How might the procedure be modified to reduce the
I Flush out the gas and repeat the procedure with another gas. measurement uncertainty in the results?
130 5 Formulae, equations and amounts of substance
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5.4 Calculations from equations Test yourself
An equation is more than a useful shorthand for describing what happens 10 What is the amount, in
during a reaction. In industry, in medicine and anywhere that chemists moles, of each gas at room
make products from reactants, it is vitally important to know the amounts of temperature and pressure?
reactants that are needed for a chemical process and the amount of product
a) 240 000 cm3 chlorine
that can be obtained. Chemists can calculate these amounts using equations.
b) 48 cm3 hydrogen
c) 3.0 dm3 ammonia
Calculating the masses of reactants and products
11 What is the volume in cm3 of
There are four key steps in solving problems using equations.
each amount of gas at room
Step 1: Write the balanced equation for the reaction. temperature and pressure?
Step 2: Write down the amounts in moles of the relevant reactants and a) 2.0 mol nitrogen
products in the equation. b) 0.00020 mol neon
Step 3: Convert these amounts in moles of the relevant reactants and c) 0.125 mol carbon dioxide
products to masses.
Step 4: Scale the masses to the quantities required.
Example
What mass of iron can be obtained from 1.0 kg of iron(iii) oxide (iron ore)?
Notes on the method
Only do the calculation for the substances in the equation that affect the
answer. In this instance, the CO and CO2 can be ignored.
In Step 2, you obtain the numbers of moles from the numbers in front of
the formulae for the substances. The number is ‘1’ if there is no number
in front of the formula.
Look up the relative atomic masses of the elements in the Periodic Table
so that you can work out the molar masses.
The proportions are the same whether the mass of iron oxide is 1.0 g or
1.0 kg (1000 g).
Answer
Step 1: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
Test yourself
Step 2: 1 mol Fe2O3 → 2 mol Fe
12 What mass of calcium
Step 3: M(Fe2O3) = (2 × 55.8 g mol−1) + (3 × 16.0 g mol−1) oxide, CaO, forms when 25 g
calcium carbonate, CaCO3,
= 111.6 + 48.0 = 159.6 g mol−1
decomposes on heating?
So, 159.6 g Fe2O3 → 2 × 55.8 g Fe = 111.6 g Fe
13 What mass of sulfur
Step 4: 159.6 g Fe2O3 → 111.6 g Fe combines with 8.0 g copper
to form copper(i) sulfide,
111.6
1.0 g Fe2O3 → g Fe Cu2S?
159.6
14 What mass of sulfur is
= 0.70 g Fe (giving the answer to two significant figures)
needed to produce 1.0 kg of
Scaling up, 1.0 kg of iron(iii) oxide produces 0.70 kg of iron. sulfuric acid, H2SO4?
5.4 Calculations from equations 131
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Measuring the volumes of gases in reactions
The apparatus in Figure 5.5 can be used to measure the reacting volumes of
dry ammonia gas (NH3) and dry hydrogen chloride gas (HCl).
Figure 5.5 Measuring the reacting volumes dry ammonia 3-way tap dry hydrogen chloride
of ammonia and hydrogen chloride.
100
100
75
50
25
25
50
75
Syringe A Syringe B
When 30 cm 3 of ammonia gas and 50 cm3 of hydrogen chloride gas are
mixed, ammonium chloride (NH4Cl) forms as a white solid. The volume of
this solid is insignificant compared to the volume of the gases. The volume
of gas remaining is 20 cm3, which turns out to be excess hydrogen chloride.
So,
30 cm3 of NH3 reacts with 30 cm3 of HCl
1 cm3 of NH3 reacts with 1 cm3 of HCl
and 24 dm3 of NH3 reacts with 24 dm3 of HCl.
This shows that 1 mol of NH3 reacts with 1 mol of HCl.
Notice that the ratio of the reacting volumes of these gases is the same as
the ratio of the reacting amounts in moles shown in the equation for the
reaction. This is always the case when gases react.
NH3(g) + HCl(g) → NH4Cl(s)
1 mol 1 mol
1 volume 1 volume
Gas volume calculations
Gas volume calculations are straightforward when all the relevant substances
are gases. In these cases, the ratio of the gas volumes in the reaction is the
same as the ratio of the numbers of moles in the equation. This is the case
because the volume of a gas, under given conditions of temperature and
pressure, depends only on the amount of the gas and not on the type of gas.
Tip Example
Remember that you cannot ignore What volume of oxygen reacts with 60 cm3 methane and what volume
the volume of water in a gas volume of carbon dioxide is produced if all volumes are measured at the same
calculation if the temperature is above temperature and pressure?
100 °C and the water is in the gaseous
state. Notes on the method
Write the balanced equation.
Note that below 100 °C the water formed condenses to an insignificant
volume of liquid.
Apply the rule that the ratios of gas volumes are the same as the ratio
of the amounts in moles if measured under the same conditions of
temperature and pressure.
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Answer
The equation for the reaction is:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
1 mol 2 mol 1 mol
1 volume 2 volumes 1 volume
So, 60 cm3 methane reacts with 120 cm3 oxygen to produce 60 cm3
carbon dioxide.
The other approach to gas volume calculations is also based on the fact that
the volume of a gas, under given conditions, depends only on the amount of
gas in moles. It is possible to determine the molar volume of a gas given the
equation for a reaction that forms the gas.
Core practical 1
Measuring the molar volume of a gas before reaction
gas syringe
A group of students carried out an experiment to find the
volume of hydrogen produced when magnesium reacts with
excess dilute hydrochloric acid. One of the students drew
the diagram in Figure 5.6 to describe the method used. Each length of magnesium
ribbon
student used a different, measured, length of ribbon.
5 cm depth of
Results dilute
The results are shown in Table 5.3. during reaction hydrochloric acid
Mass of 25.0 cm of clean magnesium ribbon = 0.200 g
Questions
1 Explain the apparatus and technique used by the students to Figure 5.6 Apparatus for measuring the volume of gas formed when
start the reaction and collect the gas given off. a metal reacts with an acid.
2 Plot a graph of the volume of gas given off against the length of
magnesium ribbon used. Draw a line of best fit. Table 5.3
3 Calculate the length of magnesium ribbon that gives 1.0 × Length of magnesium Volume of hydrogen
10−3 mol of the metal. ribbon/cm gas collected/cm3
4 Read off from the line on the graph the volume of gas formed 1.0 7.5
when 1.0 × 10−3 mol metal reacts with excess dilute 2.0 16.5
hydrochloric acid. 3.0 24.5
5 Write the equation for the reaction of magnesium with
4.0 31.0
hydrochloric acid.
5.0 39.5
6 According to the equation, how much hydrogen, in moles, is
6.0 47.5
formed when 1.0 × 10−3 mol magnesium reacts?
7 Use the results to calculate the molar volume of hydrogen.
8 How can the ideal gas equation be used to evaluate the Tip
accuracy of the experiment? What other measurements would
Refer to Practical skills sheets 1, 3, 4 and 5, which you
be needed?
can access online at [Link]/
9 Suggest likely sources of measurement uncertainty that might
EdexcelChemistry.
have affected the results.
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Test yourself
15 Assuming that all gas volumes are measured 16 What volume of gas forms at room temperature
under the same conditions of temperature and and pressure when 0.654 g of zinc reacts with
pressure, what volume of oxygen is needed to excess dilute hydrochloric acid?
react with 50 cm3 ethane, C2H6, when it burns,
and what volume of carbon dioxide forms?
Key terms 5.5 Amounts in solutions
The concentration of a solution shows how much solute is dissolved in a
Concentration/g dm−3
certain volume of solution. It can be measured in grams per cubic decimetre
mass of solute/g
= (g dm−3) but in chemistry it is more useful to measure concentrations in
volume of solution/dm3
moles per cubic decimetre (mol dm−3). For example, a solution of sodium
Concentration/mol dm−3
amount of solute/mol hydroxide containing 1.0 mol dm−3 has one mole of sodium hydroxide
= (40.0 g of NaOH) in 1.0 dm3 (1000 cm3) of solution.
volume of solution/dm3
Example
Tip A car battery contains 2350 g of sulfuric acid (H2SO4) in 6.0 dm3 of the
Concentrations are measured in moles battery liquid. What is the concentration of sulfuric acid in:
per dm3 of solution − not per dm3 of a) g dm−3 b) mol dm−3?
water used to make up the solution.
This is because there are small volume
Notes on the method
changes when solutes dissolve in water. Divide the mass in grams of solute by the volume in dm3 to find the
concentration in g dm−3.
Divide the mass of solute by its molar mass to find its amount in moles.
Divide the amount in moles of solute by the volume in dm3 to find the
concentration in mol dm−3.
Answer
mass of solute/g
a) Concentration of the acid/g dm−3 =
volume of solution/dm3
2350 g
=
6.0 dm3
= 392 g dm−3
b) M(H2SO4) = 98.1 g mol−1
2350 g
So, amount of H2SO4 in the battery = = 24.0 mol
98.1 g mol−1
amount of solute/mol 24.0 mol
Concentration = =
volume of solution/dm3 6.0 dm3
= 4.0 mol dm−3
When ionic compounds dissolve, the ions separate in the solution.
For example:
aq
CaCl 2(s) ⎯→ Ca 2+(aq) + 2Cl−(aq)
So, if the concentration of CaCl 2 is 0.1 mol dm−3, then the concentration of
Ca 2+ is also 0.1 mol dm−3, but the concentration of Cl− is 0.2 mol dm−3.
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Test yourself
17 What is the concentration, in mol dm−3, of a solution containing:
a) 4.25 g silver nitrate, AgNO3, in 500 cm3 solution
b) 4.0 g sodium hydroxide, NaOH, in 250 cm3 of solution
c) 20.75 g potassium iodide, KI, in 200 cm3 of solution?
18 What mass of solute is present in:
a) 50 cm3 of 2.0 mol dm−3 sulfuric acid
b) 100 cm3 of 0.010 mol dm−3 potassium manganate(vii), KMnO4
c) 250 cm3 of 0.20 mol dm−3 sodium carbonate, Na2CO3?
5.6 Solutions for quantitative analysis
Quantitative analysis involves techniques which answer the question ‘How
much?’ In many laboratories, quantitative analysis is based on instrumental
techniques such as chromatography and spectroscopy (see Chapter 7).
Accurate chemical analysis generally involves preparing a solution of an
unknown sample. It may then be necessary to dilute the solution before Key term
analysing it by titration or by some instrumental method. Titrations are an
A titration is a volumetric analysis
important procedure for checking and calibrating instrumental methods. In a
technique for finding the concentrations
titration, the analyst finds the volume of the sample solution that reacts with a
of solutions and for investigating the
certain volume of a reference solution with an accurately known concentration.
amounts of chemicals involved in
Titrations are widely used because they are quick, convenient, accurate and easy
reactions.
to automate.
Many laboratories have automatic instruments for carrying out titrations
(Figure 5.7), but the principle is exactly the same as in titrations where the
volumes are measured with a traditional burette and pipette. Volumetric
titrations with the kinds of glassware used in school and college laboratories
are widely used in the food, pharmaceutical and other industries.
Figure 5.7 A scientist in Nigeria adjusting an
automatic titration device. This is being used
to check that a pharmaceutical product
contains the right amount of folic acid.
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Test yourself
19 Why might the results of the following examples of quantitative
analysis be important and why must the results be accurate:
a) the concentration of sugars in urine
b) the concentration of alcohol in blood
c) the percentage by mass of haematite (iron ore) in a rock sample
d) the concentration of nitrogen oxides in the air?
Pipettes, burettes and graduated flasks make it possible to measure out
volumes of solutions very precisely during a titration. There are correct
techniques for using all this glassware which must be followed carefully for
accurate results.
Standard solutions
Any titration involves two solutions. Typically, a measured volume of one
solution is run into a flask from a pipette. Then the second solution is added,
bit by bit, from a burette until the colour change of an indicator, or the
change in a signal from an instrument, shows that the reaction is complete.
The procedure only gives accurate results if the reaction between the two
solutions is rapid and proceeds exactly as described by the chemical equation.
So long as these conditions apply, titrations can be used to study acid–base
and other types of reactions.
Standard solutions make volumetric analysis possible. The direct way of
Key terms preparing a standard solution is to dissolve a known mass of a chemical in
water and then to make the volume of solution up to a definite volume in a
A standard solution is a solution with graduated flask.
an accurately known concentration.
This method for preparing a standard solution is only appropriate with a
A primary standard is a chemical which chemical that:
can be weighed out accurately to make
● is very pure
up a standard solution.
● does not gain or lose mass when in the air
● has a relatively high molar mass so that weighing errors are minimised.
Chemicals that meet these criteria are called primary standards.
A titration with a primary standard can be used to measure the
concentration of a solution.
Test yourself
20 S
uggest a reason why sodium hydroxide cannot be used as a
primary standard (see Figure 4.9).
21 S
uggest a reason why anhydrous sodium carbonate can be used as
a primary standard but hydrated sodium carbonate cannot.
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Core practical 2
Preparation of a standard solution from a solid acid
A standard solution of a solid acid was prepared in a graduated flask using the
procedure illustrated in Figure 5.8. The acid used was a potassium salt of
benzene-1,2-dicarboxylic acid. The traditional name for the salt is potassium
hydrogenphthalate, which is often referred to as KHP.
glass rod
Weigh solid into sample Dissolve solute in small
tube then tip into beaker amount of solvent,
and reweigh warming if necessary
stirring rod Transfer to
standard flask
glass rod
wash
Stopper and Carefully make bottle Rinse all solution
mix well up to the mark into flask with
on the flask more solvent
Figure 5.8 Using a standard flask to prepare a solution with a specified concentration.
1 The formula of KHP is KHC8H4O4. What mass of KHP is graduation mark. The contents are then mixed well before
needed to prepare a 0.10 mol dm−3 solution in a 250 cm3 finally adding water dropwise until the meniscus just rests on
graduated flask? the mark. What are the reasons for following this procedure?
2 Suggest a reason why KHP is a better primary standard to use 5 Calculate the concentration of the standard solution made
than the oxalic acid (H2C2O4.2H2O) which is also available as a by the procedure in Figure 5.8 when the readings from the
pure solid. balance when weighing out the solid are as follows and the
3 a) Why is the solution poured down a glass rod as the volume of the graduated flask is 250.0 cm3.
liquids are transferred from the beaker to the graduated
Mass of weighing bottle plus sample of KHP = 20.216 g
flask?
b) What other steps must be taken to ensure that every drop Mass of weighing bottle after tipping KHP into the beaker
of the solution is transferred to the graduated flask? = 14.855 g
4 After transferring the solution from the beaker, the graduated
flask is filled with water to within about 1 cm of the
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Diluting a solution quantitatively
Quantitative dilution is an important procedure in analysis. Two common
reasons for carrying out dilutions are:
● to make a solution with the concentration needed for a particular
experiment from a standard solution
● to dilute an unknown sample for analysis to give a concentration suitable
for titration.
The procedure for dilution is to take a measured volume of the more
concentrated solution with a pipette (or burette) and run it into a graduated
flask. The flask is then carefully filled to the mark with pure water.
The key to calculating the volumes to use when diluting a solution is to
remember that the amount, in moles, of the chemical dissolved in the diluted
solution is equal to the amount, in moles, of the chemical taken from the
concentrated solution. If c is the concentration in mol dm−3 and V is the
volume in dm 3, then we can write the following expressions.
The amount, in moles, of the chemical taken from the concentrated
solution = c AVA
The amount, in moles, of the same chemical in the diluted solution = c BV B
These two amounts are the same, so cAVA = c BV B
Example
An analyst requires a 0.10 mol dm−3 solution of sodium hydroxide,
NaOH(aq). The analyst has a 250 cm3 graduated flask and a supply
of 0.50 mol dm−3 sodium hydroxide solution. What volume of the
concentrated solution should be measured into the graduated flask?
Notes on the method
Use the relationship cAVA = cBVB
cBVB
This can be rearranged to show that: VA = cA
Answer
cA = 0.50 mol dm−3 cB = 0.10 mol dm−3
VA = to be calculated VB = 250 cm3 = 0.25 dm3
cBVB 0.10 mol dm−3 × 0.25 dm3
VA = =
cA 0.50 mol dm−3
= 0.050 dm3 = 50.0 cm3
Pipetting 50.0 cm3 of the concentrated solution into the 250 cm3
graduated flask and making up to the mark with pure water gives the
required dilution after thorough mixing.
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Test yourself
22 How would you prepare:
a) a 0.0500 mol dm−3 solution of HCl(aq) given a 1000 cm3
graduated flask and a 1.00 mol dm−3 solution of the acid
b) a 0.010 mol dm−3 solution of NaOH(aq) given a 500 cm3 graduated
flask and a 0.50 mol dm−3 solution of the alkali?
23 What is the concentration of the solution produced when making up
to the mark with pure water and mixing:
a) 10.0 cm3 of a 0.010 mol dm−3 solution of AgNO3(aq) in a 100 cm3
graduated flask
b) 50.0 cm3 of a 2.00 mol dm−3 solution of nitric acid in a 250 cm3
graduated flask?
5.7 Titration principles
A titration involves two solutions. A measured volume of one solution is run
into a flask. The second solution is then added, bit by bit, from a burette until
the reaction is complete (Figure 5.9).
safety filler
burette
pipette
solution of
substance A
solution of
substance B
volume VB of substance B
conical flask concentration cB in mol dm–3
mean titre = VA
Figure 5.9 The apparatus used for a titration based on a reaction between two chemicals
in solution, A and B.
Some titrations are used to investigate reactions. In these experiments the
concentrations of both solutions are known and the aim is to determine the
equation for the reaction.
More often, titrations are used to measure the concentration of an unknown
solution, knowing the equation for the reaction and using a second solution
of known concentration.
In general, nA moles of A react with nB moles of B.
nA A + nBB → products
The concentration of solution B in the flask is c B and the concentration of
solution A in the burette is c A. Both are measured in mol dm−3. The analyst uses
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a pipette to run a volume V B of solution B into the flask. Then solution A is
Key terms added from the burette until an indicator shows that the reaction is complete.
This is the end-point of the titration.
The end-point in a titration is the point
at which a colour change shows that In a well planned titration the colour change observed at the end-point
enough of the solution in the burette corresponds exactly with the point when the amount in moles of the reactant
has been added to react with the added from a burette is just enough to react exactly with all of the measured
chemical in the flask. amount of chemical in the flask as shown by the balanced equation. This is
the equivalence point.
The equivalence point during a titration
is reached when the amount of reactant At the end-point, the volume added is the titre, VA. The analyst should
added from a burette is just enough repeat the titration enough times to achieve consistent results.
to react exactly with all the measured
amount of chemical in the flask. Titration calculations
In the laboratory, volumes of solutions are normally measured in cm3, but
they should be converted to dm 3 in calculations so that they are consistent
with the units used for concentrations.
The amount, in moles, of B in the flask at the start = c B × V B
Tip The amount, in moles, of A added from the burette = c A × VA
Instead of trying to remember a formula The ratio of these amounts must be the same as the ratio of the reacting
for working out titration calculations, amounts nA and nB. This means that:
it is better to work through the cA × VA nA
calculation, step by step, as shown in c B × V B = nB
the example in Section 5.8. In any titration, all but one of the values in this relationship are known. The
one unknown is calculated from the results, so this formula can be used
to analyse titration results. It is generally better, however, to work out the
results, step by step, as shown in the worked examples in this chapter.
Analysing solutions
In titrations designed to analyse solutions, the equation for the reaction is
given so that the ratio nA /nB is known. The concentration of one of the
solutions is also known. The volumes VA and V B are measured during the
titration. Substituting all the known quantities in the titration formula allows
the concentration of the unknown solution to be calculated.
Investigating reactions
In titrations to investigate reactions, the problem is to determine the ratio nA/nB.
The concentrations cA and cB are known and the volumes VA and VB are measured
during the titration. So the ratio nA/nB can be calculated from the formula.
5.8 Acid–base titrations
Coloured indicators can be used to detect the end-points of acid–base reactions.
These are chemicals which change colour as the pH varies. Typically, an
indicator completes its colour change over a range of about two pH units as
shown in Table 5.4. In any acid–base titration, there is a sudden change of pH
at the end-point. The chosen indicator must, therefore, complete its colour
change within the range of pH values spanned at the end-point.
140 5 Formulae, equations and amounts of substance
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Some acids and alkalis are fully ionised in solution. These are strong acids
and strong alkalis. For a titration of a strong acid with a strong alkali, the pH
Key terms
jumps from around pH 3 to pH 10 at the end-point. Most common indicators A strong acid is one that is fully ionised
change colour sharply within this range. when it dissolves in water. Hydrochloric
Other acids are only slightly ionised in solution. These are weak acids. acid is an example of a strong acid.
During a titration of a weak acid with a strong alkali, the jump is from about A weak acid is one that is only slightly
pH 6 to pH 10. So the indicator must be chosen with care so that it changes ionised when it dissolves in water.
colour in this range. Ethanoic acid is an example of a weak
Table 5.4 Some common indicators and the pH range over which they change colour. acid.
Indicator Colour change pH range over which the
low pH–high pH colour change occurs
Methyl orange Red–yellow 3.2−4.2
Tip
You will learn more about indicators and
Methyl red Yellow–red 4.8−6.0
why they change colour over different
Bromothymol blue Yellow–blue 6.0−7.6 pH ranges later in your Advanced
Phenolphthalein Colourless–red 8.2−10.0 chemistry course.
Example
Calcium hydroxide is an alkali that is only slightly soluble in water. Its
solubility, at a given temperature, can be determined by titration of a
saturated solution of the alkali with a standard solution of hydrochloric acid,
as shown in Figure 5.10. Work out the solubility of Ca(OH)2 in moles per dm3,
and in grams per dm3, given that the volume VA of acid added from the burette
at the end-point was 23.50 cm3.
solution B: safety filler
saturated solution of
calcium hydroxide at
20 °C concentration
cB to be measured
25.00cm3
pipette solution A:
cA = 0.0500 mol dm–3 hydrochloric acid
solution B with 2 drops
phenolphthalein indicator
VB = 25.00 cm3
VB = 25.00cm3
mean titre VA = 23.50 cm3
Figure 5.10 A titration to determine the solubility of calcium hydroxide.
Notes on the method
Both the volume and concentration of the acid are known, so the first step
is to work out the amount in moles of acid added from the burette.
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Next look at the balanced equation to see how much Ca(OH)2 this amount
of acid reacts with.
Finally work out the concentration of calcium hydroxide in the saturated solution.
Answer
Step 1: Work out the amount of acid added from the burette.
The concentration of the acid, cA = 0.0500 mol dm−3
23.50
Amount of HCl(aq) added from the burette = dm3 × 0.0500 mol dm−3
1000
= 0.001 175 mol
Step 2: Use the equation for the titration reaction to find the amount of alkali in the flask.
Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)
So 1 mol of the alkali reacts with 2 mol of the acid.
Hence the amount of calcium hydroxide in the flask = 0.5 × 0.001 175 mol = 0.0 005 875 mol
Step 3: Work out the concentration of the saturated solution.
The 0.0 005 875 mol of alkali is dissolved in 25.0 cm3 of saturated solution.
So the concentration of saturated calcium hydroxide solution
= 0.0 005 875 mol ÷ 0.0250 dm3 = 0.0235 mol dm−3
The molar mass of calcium hydroxide is 74.1 g mol−1.
So the concentration of saturated calcium hydroxide solution
= 0.0235 mol dm−3 × 74.1 g mol−1 = 1.74 g dm−3
Test yourself
24 Suggest why methyl orange is distinctly orange when the pH is 3.7.
25 A 25.0 cm3 sample of nitric acid was neutralised by 18.0 cm3
of 0.150 mol dm−3 sodium hydroxide solution. Calculate the
concentration of the nitric acid.
26 A 2.65 g sample of anhydrous sodium carbonate was dissolved in
water and the solution made up to 250 cm3. In a titration, 25.0 cm3
of this solution was added to a flask and the end-point was
reached after adding 22.5 cm3 of hydrochloric acid. Calculate the
concentration of the hydrochloric acid.
27 A 41.0 g sample of the acid H3PO3 was dissolved in water and the
volume of solution was made up to 1 dm3. 20.0 cm3 of this solution
was required to react with 25.0 cm3 of 0.800 mol dm−3 sodium
hydroxide solution. What is the equation for the reaction?
5.9 Evaluating results
Accuracy of data is determined by how close a measured quantity is to the
Key term correct value. In chemical analysis the correct value is often not known and
so chemists need to estimate the measurement uncertainty.
Measurements are accurate if they are
precise and free from bias. Every time an analyst carries out a titration, there is some uncertainty
in the result. It is important to be able to assess measurement uncertainty
(Figure 5.11). Key decisions are based on the results of chemical analysis in
healthcare, in the food industry, in law enforcement and in many other areas
of life. It is important that the people making these decisions understand the
extent to which they can rely on the data from analysis.
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Figure 5.11 Sources of uncertainty in
B It is difficult to determine The material used to
volumetric analysis.
EX 20°C
accurately the volume of prepare a standard
0 liquid in a burette if the solution may not be
meniscus lies between 100% pure.
two graduation marks. KHP
10 PURITY
99.5%
A 250 cm3 volumetric A burette is calibrated
flask may actually contain by the manufacturer for
250.3 cm3 when filled to use at 20 °C. When it is
23 °C
the calibration mark due used in the laboratory
to permitted variation in the the temperature may be 23 °C.
manufacture of the flask. This difference in
temperature causes
a small difference in the
B 250ml actual volume of liquid
20°C
in the burette when it is
filled to a calibration mark.
It is difficult to make an The display on a
exact judgement of the laboratory balance
end-point of a titration only shows the mass to
(the exact point at which a certain number of
the colour of the indicator decimal places.
changes).
Random errors in titrations
Every time an analyst carries out a titration, there are small differences in the
results. This is not because the analyst has made mistakes but because there
are factors that are impossible to control. Unavoidable random errors arise
in judging when the bottom of the meniscus is level with the graduation on
a pipette, in judging the colour change at the end-point and when taking
the reading from a burette scale. If these random errors are small, then the Key terms
results will be close together – in other words, they are precise (Figure 5.12).
The precision of a set of results can be judged from the range in a number of Measurements are precise if repeat
repeated titrations. measurements have values that are
close together. Precise measurements
have a small random error.
Systematic errors in titrations
Bias arises from systematic errors
Systematic errors mean that the results differ from the true value by the same
which affect all the measurements in
amount each time. The measurement is always too high or too low, so it
the same way, making them all higher
is biased in one way or the other (Figure 5.12). One source of systematic
or lower than the true value. Systematic
error is the tolerance allowed in the manufacture of graduated glassware. The
errors do not average out.
tolerance for grade B 250 cm3 graduated flasks is ±0.3 cm3. This means that
5.9 Evaluating results 143
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when an analyst chooses to use a particular flask, the volume of solution may
be as little as 249.7 cm3 or as much as 250.3 cm3 when it is filled correctly to
the graduation mark. This introduces a systematic error when using this same
flask to make up solutions for a series of titrations.
Similar tolerances are allowed for pipettes and burettes: for a class B 25 cm3
pipette the tolerance is ±0.06 cm3, while for a class B 50 cm3 burette the
tolerance is ±0.1 cm3.
Systematic errors can be allowed for by calibrating the measuring instruments.
It is possible to calibrate pipettes and burettes by using them to measure out
pure water and then weighing the water with an accurate balance.
A darts player is practising throwing darts at a board. The aim is
to get all the darts close together near the centre of the board.
The results of some of the attempts are shown below.
1st attempt: The shots are quite widely scattered and some have
not even hit the board. The shots show poor precision as they are
quite widely scattered. There is also a bias in where the shots
have landed – they are grouped in the top right-hand corner, not
near the centre of the board.
2nd attempt: The precision has improved as the shots are now
bias
more closely grouped. However, there is still a bias, as the group
of shots is offset from the centre of the board.
3rd attempt: The player has improved to reduce the bias – all the
shots are now on the board and scattered round the centre.
Unfortunately the precision is poor as the shots are quite widely
scattered.
Some time later: The shots are precise and unbiased – they are
all grouped close together in the centre of the board.
Figure 5.12 Throwing darts at the bullseye of a dartboard illustrates the notions of
precision and bias. Reliable players throw precisely and without bias so that their darts
hit the centre of the board accurately.
Tip
Test yourself
Refer to Practical skills sheet 5,
‘Identifying errors and estimating 28 Identify examples of random and systematic error when:
uncertainties’, to find out how to a) using a pipette
estimate measurement errors and b) using a burette
calculate overall measurement
c) making up a standard solution in a graduated flask.
uncertainties.
144 5 Formulae, equations and amounts of substance
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Core practical 3
Tip
Finding the concentration of a solution of hydrochloric acid
A student carried out a titration to determine the concentration of a solution of Refer to Practical skills sheets 5 and 6,
hydrochloric acid. He used the apparatus shown in Figure 5.13 and followed the which you can access online at
instructions numbered A–F. The results are shown in Table 5.5. [Link]/
EdexcelChemistry.
5 Identifying errors and estimating
safety filler uncertainties
6 Measuring chemical amounts by
titration.
25.00 cm3
pipette dilute hydrochloric acid
standard solution of
sodium carbonate
standard solution of sodium
carbonate with three drops
of indicator
conical flask
Figure 5.13
Instructions
A Wash out the pipette, burette and conical flask with pure 1 Explain briefly the reasons for carrying out each of the
(deionised or distilled) water. steps A–F.
B Rinse the burette with a little of the solution of hydrochloric 2 Describe what the student should do to ‘allow the
acid, then fill the burette, remembering to run out some of pipette to drain adequately’ in step C.
the solution through the tap. 3 The standard solution of sodium carbonate was
C Rinse the 25.00 cm3 pipette with the standard solution of prepared with 2.920 g anhydrous Na2CO3 in a 500 cm3
sodium carbonate. Fill the pipette to the mark and run out graduated flask. Calculate the concentration of the
the measured alkali into a clean conical flask, allowing the sodium carbonate solution.
pipette to drain adequately. 4 How should the student read the burette in order to
D Add three drops of methyl orange indicator. justify recording results to the nearest 0.05 cm3?
E Carry out one rough and then accurate titrations to give two 5 Use the titration results in Table 5.5 to calculate the
titres that are within 0.10 cm3 of each other. In the accurate concentration of the dilute hydrochloric acid.
titrations the colour change at the end-point should be 6 The glassware used for the titration was all grade B
caused by adding one drop of acid. apparatus. Estimate the total uncertainty in your
F Each time, record the initial and final burette readings. Take calculated result.
the burette readings to the nearest half-scale division.
Results
Burette: Solution of hydrochloric acid to be standardised
Pipette: Standard solution of sodium carbonate
Indicator: Methyl orange
Table 5.5 Titration results.
Rough Accurate 1 Accurate 2 Accurate 3
Final burette reading 28.0 24.00 25.70 26.50
Initial burette reading 5.0 1.55 3.30 4.15
Titre/cm 3 23.0 22.45 22.40 22.35
5.9 Evaluating results 145
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5.10 Yields and atom economies
If there are no losses during a chemical reaction, the starting reactants are
converted to the required products. Often, some reactants are added in excess
to ensure that the most valuable reactant is converted to as much product as
possible. Then the reactant that is not in excess is the substance that limits the
maximum yield that is possible.
Converting all of the limiting reagent to the desired product gives a 100%
Key terms
yield. But few reactions are so efficient and many give low yields. There are
A limiting reagent is a substance which various reasons why yields are not 100%:
is present in an amount which limits the ● the reactants may not be totally pure
theoretical yield. ● some of the product may be lost during transfer of the chemicals from one
Yield calculations are used to assess container to another, when the product is separated and purified
● there may be side reactions in which the reactants form different products
the efficiency of a chemical process.
● some of the reactants may not react because the reaction is so slow
The actual yield is the mass of
product obtained from a reaction. The (Chapter 9) or because it comes to equilibrium (Chapter 10).
theoretical yield is the mass of product
obtained if the reaction goes according
to the equation. Example
actual yield A modern gas-fuelled lime kiln produces 500 kg of calcium oxide,
percentage yield = × 100%
theoretical yield CaO (quicklime), from 1000 kg of crushed calcium carbonate,
CaCO3(limestone).
What is the percentage yield of calcium oxide? Give your answer
to 2 significant figures.
Notes on the method
Start by writing the balanced equation for the reaction.
Use the method for calculating reactant and product masses in Section 5.4.
Answer
The equation for the reaction involved is:
CaCO3(s) → CaO(s) + CO2(g)
From the equation, 1 mole CaCO3 → 1 mole CaO
[40.1 + 12 + (3 × 16)] g CaCO3 → (40.1 + 16) g CaO
So 100.1 g CaCO3 → 56.1 g CaO
Thus 1 g CaCO3 → 56.1 g CaO
100.1 56.1
Theoretical yield from 1000 kg CaCO3 = × 1000 kg CaO
100.1
= 560 kg
The actual yield of CaO = 500 kg CaO
500 kg
Percentage yield = × 100 = 89%
560 kg
146 5 Formulae, equations and amounts of substance
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Test yourself
29 1000 kg of pure iron(iii) oxide, Fe2O3, was reduced to iron and
630 kg of iron was obtained. What are the theoretical and
percentage yields of iron?
30 500 kg of calcium oxide (quicklime) was reacted with water to
produce calcium hydroxide, Ca(OH)2 (slaked lime). 620 kg of calcium
hydroxide was produced. Calculate the theoretical and percentage
yields.
Atom economy
The yield in a laboratory or industrial process focuses on the desired product.
But many atoms in the reactants do not end up in the desired product. This
can lead to a huge waste of material. For example, when calcium carbonate
(limestone) is decomposed to produce calcium oxide (quicklime), part of the
calcium carbonate is lost as carbon dioxide in the atmosphere. Figure 5.14 The production of ibuprofen
is an excellent example of atom economy.
The waste in many reactions has led scientists and industrialists to use the
Ibuprofen is an important medicine which
term atom economy in calculating the overall efficiency of a chemical
reduces swelling and pain. In the 1960s,
process (Figure 5.14). The atom economy of a reaction is the molar mass of
Boots made ibuprofen in five steps with
the desired product expressed as a percentage of the sum of the molar masses
an atom economy of only 40%. When
of all the products as shown in the equation for the reaction.
the patent expired, another company
molar mass of the desired product developed a new process requiring just two
atom economy = × 100%
sum of the molar masses of all the products steps with an atom economy of 100%.
Example Key terms
Titanium is manufactured by heating titanium(iv) chloride with magnesium. Atom economy is a measure of how
The equation for the reaction at 1200 °C is: efficiently a chemical reaction converts
TiCl4(g) + 2Mg(l) → Ti(s) + 2MgCl2(l) the atoms in its reactants to atoms
What is the atom economy of this process? in the product. The atom economy
for a reaction is calculated from
Answer the balanced equation to show the
Molar mass of all products = M(Ti) + 2M(MgCl2) percentage of the mass of the atoms in
the reactants that is converted to the
= 47.9 g mol−1 + 190.6 g mol−1 = 238.5 g mol−1
desired product.
Molar mass of desired product = 47.9 g mol−1
Therefore: atom economy = 47.9 ÷ 238.5 × 100% = 20.1%
Almost 80% of the reactants are ‘wasted’ in the manufacture of titanium
by the process described above, because magnesium and chlorine atoms are
lost as magnesium chloride. If society is to use raw materials as efficiently
as possible, chemists must look for high atom economies as well as high
percentage yields, particularly in industrial processes.
5.10 Yields and atom economies 147
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Test yourself
31 Calculate the atom economy for:
a) the conversion of nitrogen (N2) to ammonia (NH3) in the Haber
process:
N2(g) + 3H2(g) → 2NH3(g)
b) the fermentation of glucose (C6H12O6) to ethanol (C2H5OH)
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)
c) the manufacture of tin (Sn) from tinstone (SnO2)
SnO2(s) + 2C(s) → Sn(s) + 2CO(g)
5.11 Tests and observations
in inorganic chemistry
Formulae and equations are not only used to solve quantitative problems.
They are also important in qualitative analysis.
Qualitative analysis answers the question ‘What is it?’ In Advanced chemistry
courses, this question is often answered by careful observation of the changes
during test-tube experiments and flame tests. These changes include gases
bubbling off, different smells, precipitates forming, solids dissolving,
temperatures changing or new colours appearing.
The skill is knowing what to look for. Some visible changes are much more
significant than others and a capable analyst can spot the important changes
and know what they mean. Good chemists have a ‘feel’ for the way in which
chemicals behave and recognise characteristic patterns of behaviour. With
experience they know what to look for when making observations.
Success also depends on good techniques when mixing chemicals, heating
mixtures and testing for gases.
In inorganic chemistry most observations can be explained in terms of a
number of types of reaction (see also Chapter 3 and Section 4.1).
Ionic precipitation reactions
This type of reaction can be used to test for negative ions (anions). Adding
a solution of silver nitrate to a halide produces a precipitate that can be used
to distinguish chlorides, bromides and iodides. Adding a soluble barium salt
(nitrate or chloride) to a solution of a sulfate produces a white precipitate of
insoluble barium sulfate.
Acid–base reactions
Acids and alkalis are commonly used in chemical tests. Dilute hydrochloric
acid is a convenient strong acid. Sodium hydroxide solution is often chosen
as a strong base.
148 5 Formulae, equations and amounts of substance
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Adding dilute hydrochloric acid to a carbonate, for example, adds hydrogen
ions to the carbonate ions, CO32−, turning them into carbonic acid molecules,
H2CO3, which immediately decompose into carbon dioxide and water.
2H+(aq) + CO32−(aq) → H2CO3(aq) → H2O(l) + CO2(g)
Testing with limewater can then identify the gas given off, confirming that
the compound tested is a carbonate.
Hydrogencarbonate ions react in a similar way to carbonate ions.
H+(aq) + HCO3−(aq) → H2CO3(aq) → H2O(l) + CO2(g)
The strong base, sodium hydroxide, is used to test for ammonium ions in
salts such as ammonium chloride, NH4Cl. Warming an ammonium salt with
a solution of sodium hydroxide produces an alkaline gas with a pungent
smell. This gas is ammonia, which is formed when hydroxide ions remove
hydrogen ions from ammonium ions.
NH4+(aq) + OH−(aq) → NH3(g) + H2O(l)
Redox reactions
Common oxidising agents used in inorganic tests include chlorine, bromine
and acidic solutions of iron(iii) ions, manganate(vii) ions or dichromate(vi) ions.
Some reagents change colour when oxidised, which makes them useful for
detecting oxidising agents. In particular, a colourless solution of iodide ions
turns to a yellow–brown colour when oxidised. This can be a very sensitive
test if starch is present because starch gives an intense blue–black colour with
low concentrations of iodine. This is the basis of using starch–iodide paper
to test for chlorine and other oxidising gases. The oxidation of iodide ions
by chlorine or bromine is a redox reaction in which one halogen displaces
another (Section 4.10).
Common inorganic reducing agents are metals (in the presence of acid or
alkali), sulfur dioxide and iron(ii) ions.
Some reagents change colour when reduced. In particular, dichromate(vi)
ions in acid change from orange to green. This is the basis of a test for sulfur
dioxide gas.
Test yourself
32 For each test, identify the type of chemical reaction taking place,
name the products and write a balanced equation for the reaction:
a) testing for iodide ions with silver nitrate solution
b) adding dilute hydrochloric acid to magnesium carbonate
c) testing for sulfate ions with barium chloride
d) strongly heating a sample of potassium nitrate
e) using concentrated ammonia solution to detect hydrogen chloride
f) adding chlorine to a solution of potassium bromide
g) warming ammonium chloride with aqueous sodium hydroxide.
5.11 Tests and observations in inorganic chemistry 149
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Core practical 7 (part 1)
Tip
Analysis of inorganic unknowns Analysis of organic unknowns is covered
A series of tests was carried out on two unknown inorganic salts labelled X and Y.
in Core practical 7 (part 2) in Chapter 7.
The tests and observations were recorded as in Table 5.6.
Table 5.6 Tests and observations on two inorganic unknowns.
Test Observations with compound X Observations with compound Y
1 Carry out a flame test on Lilac coloured flame Lilac coloured flame
the salt
2 Heat a sample of the salt Melts to a colourless liquid. It gives off a Melts to a clear liquid but no gas is given
first gently and then more colourless gas that relights a glowing splint. off. In time the hot liquid turns red. There
strongly. On strong heating the gas is tinged with are slight traces of a purple vapour.
Identify any gases purple. In time the liquid turns red. Molten Y resembles the liquid formed on
evolved. decomposing X.
3 Allow the residue from On cooling, the liquid crystallises to a On cooling, the liquid crystallises to
test 2 to cool, then colourless (white) solid. The cold crystals a colourless (white) solid. The solid reacts
add a few drops of react immediately with concentrated sulfuric with concentrated sulfuric acid in the
concentrated sulfuric acid. There are traces of a fuming, acidic gas. same way as the residue after heating X.
acid. Warm gently and There is a smell of bad eggs. On warming a
then more strongly. purple vapour can be seen.
Identify any gases
evolved.
4 Make separate aqueous Both X and Y dissolve in water. The solutions are colourless. There is no change at first
solutions of X and Y. Mix when the solutions are mixed. On adding dilute sulfuric acid the solutions turn dark brown.
the two solutions and Specks of a grey solid separate from the solution.
then add dilute sulfuric
acid.
1 Describe in outline the procedure for carrying out a flame test on an unknown salt.
2 What precautions have to be taken to avoid contamination, and why are they
Tip
necessary? Refer to Practical skills sheet 7,
3 Describe in outline the procedure for the gas tests mentioned in Table 5.6. ‘Analysing inorganic unknowns’, which
4 What can be deduced from the results of the flame tests in Table 5.6? you can access online at
5 Suggest explanations for the observations on heating X and Y, including equations for [Link]/
any reactions. EdexcelChemistry.
6 What can be deduced from the results of Test 3?
7 Explain the observations in Test 4 and write an equation for the reaction which took
place on adding acid.
8 Describe two further tests that could be carried out to confirm the conclusions
based on these observations. What are the expected results of these tests?
150 5 Formulae, equations and amounts of substance
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Chapter summary
l Masses of substances involved in a reaction can be
Chapter 5 Formulae, equations
calculated from the balanced equation using the
and amounts of substance concepts amount of substance and molar mass.
l Amount of substance is a physical quantity which is l The amounts of reactants and products involved
measured in the unit ‘mole’ (symbol mol). in a reaction can be determined experimentally by
l Molar mass is the mass of one mole of a specified measuring the masses, gas volumes, or volumes in
chemical species. The unit is g mol‒1. solution and then converting the results to amounts
l Amount of substance/mol = mass of in moles and finding the ratios.
substance/g ÷ molar mass/g mol‒1. l The molar volume of a gas is the volume of 1 mol
l Mass of substance/g = amount of substance/mol × of gas under stated conditions of temperature and
molar mass/g mol‒1. pressure.
l The Avogadro constant is the number of specified l Concentration/g dm
−3
atoms, molecules or ions in one mole of a = mass of solute/g ÷ volume of solution/dm 3.
substance. The constant has the units mol‒1. l Concentration/mol dm
−3
l An empirical formula shows the simple whole = amount of solute/mol ÷ volume of solution/dm3.
number ratio of the atoms of the different elements l A standard solution is a solution with an accurately
in a compound. known concentration.
l Percentage composition is the percentage by mass l Standard solutions can be prepared by weighing a
of each element in a sample of a compound. sample of a primary standard and then dissolving it
l An empirical formula is determined by converting in water and making up the volume to a specified
the masses of elements combined in a sample of the value in a graduated flask.
substance to amounts in moles and then finding the l Standard solutions are used in titrations to
simplest ratio. determine the concentrations of acids and alkalis.
l The molecular formula of a substance gives the l The endpoint of a titration is the point at which
actual number of atoms of each element in a the colour change of an indicator (such as methyl
molecule of the element or compound. orange or phenolphthalein) shows that enough of
l A molecular formula is always a simple multiple of the solution in the burette has been added to react
the empirical formula, so the molecular formula with the measured volume of solution in the flask.
of a substance can be determined by finding the l Appropriate experimental design and care with
molar mass of the substance. measurements help to reduce percentage error and
l The molar mass of gases and volatile liquids can be percentage uncertainty in the results of quantitative
determined using the gas equation: experiments.
pV = nRT l Two quantities used to assess the efficiency of
Measuring the volume of a known mass, m, of gas, processes to make chemical products are percentage
or vapour, at a measured pressure and temperature yield and percentage atom economy.
allows the amount, n, of substance in the sample l Observations of changes during test-tube reactions
to be calculated and hence the value of the molar can be related to the full and ionic equations
mass, m/n. When substituting in the gas equation (including state symbols). Examples include
all quantities must be in SI units. displacement reactions (redox), the reactions
l Balanced chemical equations show the ratios of the of acids with metals, oxides, hydroxide and
amounts, in moles, of the reactants and products carbonates, and precipitation reactions. (See also
taking part in reactions. Chapter 4 in this book.)
Chapter summary 151
469983_05_Chem_Y1-2_123-[Link] 151 13/04/19 10:11 PM
Exam practice questions
1 Complete these equations so that they are b) adding dilute hydrochloric acid to
balanced. magnesium carbonate (3)
a) Cu2S(s) + O2(g) c) testing for sulfate ions with barium
→ CuO(s) + SO2(g) (1) chloride (3)
b) FeS(s) + O2(g) + SiO2(s) d) heating a sample of zinc carbonate (3)
→ FeSiO3(s) + SO2(g) (3) e) adding zinc metal to a solution of
c) Fe(NO3)3(s) copper(ii) sulfate. (3)
→ Fe2O3(s) + NO2(g) + O2(g) (3) 7 The concentration of cholesterol (C27H46O) in a
2 a) Calculate the number of molecules present in patient’s blood was found to be 6.0 mmol dm−3.
4.0 g of oxygen, O2. (O = 16) (3) a) Calculate the concentration of cholesterol
b) Calculate the number of ions present in in mol dm−3. (1000 mmol = 1 mol) (1)
9.4 g of potassium oxide, K2O. (K = 39.1, b) Calculate the concentration of cholesterol
O = 16.0) (Avogadro constant = 6.02 × in g dm−3. (2)
1023 mol−1) (3) c) Calculate the mass of cholesterol in 10 cm3
of the patient’s blood. (1)
3 One cubic decimetre of tap water was found
8 a) Ammonium sulfate was prepared by adding
to contain 0.112 mg of iron(iii) ions (Fe3+)
ammonia solution to 25 cm3 of 2.0 mol dm−3
and 12.40 mg of nitrate ions (NO3−).
sulfuric acid.
a) Calculate the masses of Fe3+ and NO3−
in grams. (1) 2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq)
b) Calculate the amounts in moles of Fe3+ i) Calculate the volume of 2.0 mol dm−3
and NO3−. (2) ammonia solution needed to just
c) Calculate the numbers of Fe3+ and NO3− neutralise the sulfuric acid. (1)
ions. (2) ii) Describe a test to check that enough
ammonia had been added to neutralise
4 a) D
etermine the empirical formula of a all the acid without contaminating the
substance X with this percentage composition: solution.(2)
C = 42.9%, H = 2.36%, N = 16.7% and b) Iron(ii) sulfate, FeSO4, was dissolved in
O = 38.1%. (4) the solution of ammonium sulfate
b) Mass spectrometry shows that the relative solution to produce the double salt
molecular mass of X is 168. Determine the ammonium iron(ii) sulfate hexahydrate,
molecular formula of X. (2) (NH4)2SO4.FeSO4.6H2O.
5 For each of the following equations, state the i) Calculate the mass of iron(ii) sulfate added
type of reaction which it represents. to the ammonium sulfate solution. (3)
a) Ca(NO3)2(aq) + K2CO3(aq) ii) The double salt was crystallised from the
→ CaCO3(s) + 2KNO3(aq) solution. The percentage yield was 50%.
b) Mg(s) + 2HCl(aq) Calculate the mass of ammonium iron(ii)
→ MgCl2(aq) + H2(g) sulfate hexahydrate that was obtained.
c) 2Zn(NO3)2(s) (H = 1.0, N = 14.0, Fe = 55.8, S = 32.1,
→ 2ZnO(s) + 4NO2(g) + O2(g) O = 16.0) (4)
d) H2SO4(aq) + 2NaOH(aq) 9 a) A compound Z is a compound of carbon,
→ Na2SO4(aq) + 2H2O(l) (4) hydrogen and oxygen only. Analysis of a
6 For each of the following tests, identify the sample of the compound shows that it is
type of chemical reaction taking place, made up of 54.5% by mass of carbon and
name the products and write a balanced 9.10% by mass of hydrogen. Determine the
equation for the reaction: empirical formula of Z. (4)
a) testing for iodide ions with silver nitrate b) When 0.270 g of Z is heated to 100 ºC
solution (3) it vaporises to produce 100 cm3 gas at a
152
5 Formulae, equations and amounts of substance
469983_05_Chem_Y1-2_123-[Link] 152 13/04/19 10:11 PM
pressure of 95.0 kPa. Determine the molar ii) the reaction of ethene, C2H4, with
mass and molecular formula of Z. (5) hydrogen bromide to make bromoethane,
C2H5Br, as the only product. (3)
10 Excess calcium reacted vigorously with
c) Explain why both yield and atom economy
25.0 cm3 of 1.00 mol dm−3 hydrochloric acid
have to be considered when selecting a process
producing calcium chloride solution and
for manufacturing a chemical product. (6)
hydrogen.
a) Write a balanced equation, with state An analyst investigates an impure sample
13 a)
symbols, for the reaction. (3) of sodium sulfate. The impurities are
b) Draw a labelled diagram showing how the unreactive. A 0.250 g sample of the
hydrogen gas could be collected during the Na2SO4 is dissolved in water. Excess
reaction. (2) barium chloride is added to the solution
c) Suggest a procedure for obtaining clean, to precipitate all the sulfate ions as barium
dry crystals of hydrated calcium chloride, sulfate, BaSO4. The mass of the pure, dry,
CaCl2.6H2O, from the solution formed. (4) precipitated barium sulfate is 0.141 g.
d) Calculate the maximum possible yield Calculate the percentage purity of the
of hydrated calcium chloride, CaCl2.6H2O, sample of sodium sulfate.(5)
from the reaction. (Ca = 40.1, H = 1.0, b) A 0.500 g sample of steel consisting of iron
Cl = 35.5, O = 16.0) (4) alloyed with carbon and silicon gave off
e) Give two reasons why the actual yield of 191 cm3 hydrogen gas when it reacted with
hydrated calcium chloride is much less than excess hydrochloric acid. The gas volume was
the mass calculated in part (d). (2) measured at room temperature and pressure.
11 A carbonate of metal M has the formula Calculate the percentage of iron in the steel.
M2CO3. In a titration, a 0.245 g sample of (Fe = 55.8, molar volume of a gas at room
M2CO3 was found to neutralise 23.6 cm3 of temperature and pressure = 24.0 dm3 mol−1)(5)
0.150 mol dm−3 hydrochloric acid. Follow these
steps to identify the metal M. 14 1.576 g of ethanedioic acid crystals,
a) Write the equation for the reaction of (COOH)2.nH2O, was dissolved in water and
M2CO3 with hydrochloric acid. (1) made up to 250 cm3. In a titration, 25.0 cm3 of
b) Calculate the amount, in moles, of the acid solution reacted exactly with 15.6 cm3
hydrochloric acid needed to react with the of 0.160 mol dm−3 sodium hydroxide solution.
sample of the metal carbonate. (1) Show by calculation that this data confirms
c) Use the equation to calculate the amount, in that n = 2 in the formula for the acid. (8)
moles, of M2CO3 in the sample. (1)
15 A sample of sodium carbonate crystals,
d) Use your answer to part (c) and the mass of
Na2CO3.10H2O, had lost part of its water
the sample to calculate the relative formula
of crystallisation on exposure to air. 2.696 g
mass of M2CO3. (2)
of the crystals were dissolved in water and
e) Calculate the relative atomic mass of
made up to 250 cm3 in a graduated flask. In
metal M. (1)
a series of titrations, 20.0 cm3 portions of the
f) Identify the metal M. (1)
solution were titrated with 0.10 mol dm−3
12 a) The reaction of ammonia, NH3, with hydrochloric acid, giving the results shown in
sodium chlorate(i), NaOCl, produces the the table.
rocket fuel hydrazine, N2H4, together with
sodium chloride and water. Determine the Titration number 1 (rough) 2 3
atom economy for the process. (4) Final burette 22.00 23.00 22.15
reading/cm3
b) Calculate the atom economies for each of
these processes for making bromoethane: Initial burette 1.00 2.35 1.60
reading/cm3
i) the reaction of ethane, C2H6 with
bromine to form bromoethane, C2H5Br, Determine the percentage of loss of mass from
and hydrogen bromide (4) the crystals from the titration results. (10)
153
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16 Egg shells contain calcium carbonate. It is react with the egg shell. The mean titre was
possible to determine the percentage of calcium 24.30 cm3.
carbonate in an egg shell by titration. Calcium a) Give two reasons why titrating the calcium
carbonate reacts with acids but it is not possible carbonate in an egg shell with hydrochloric
to titrate this directly with an acid from a burette. acid from a burette is not possible. (2)
b) Use the data about the titration to calculate
An analyst adds 40.00 cm3 of 1.200 mol dm−3
the amount of excess hydrochloric acid, in
hydrochloric acid (an excess) to a 1.510 g
moles, left over after reaction with the
sample of the crushed shell. When the reaction
egg shell. (3)
with calcium carbonate in the egg shell is
c) Hence calculate how much hydrochloric
complete, all the solution is transferred to a
acid, in moles, reacted with calcium
250 cm3 graduated flask. Water is then added to
carbonate in the sample of egg shell. (2)
the mark and the diluted solution is well mixed.
d) Use the results to calculate the percentage of
Next the analyst titrates separate 25.0 cm3 calcium carbonate in the egg shell. (5)
portions of the diluted solution with a e) The procedure used in this analysis is
0.100 mol dm−3 solution of sodium hydroxide called a ‘back titration’. Explain what you
to determine the amount of acid that did not understand by this term. (3)
154
5 Formulae, equations and amounts of substance
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Introduction to organic chemistry
6.1
6.1.1 Carbon – a special element
Carbon is an amazing element. The number of compounds containing carbon
is well over ten million. This is far more than the number of compounds of
all the other elements put together. Most compounds containing carbon also
contain hydrogen. The main sources of these compounds are organic – living
or once-living materials in animals and plants. Because of this, the term
‘organic chemistry’ is used to describe the branch of chemistry concerned
with the study of compounds containing C–H bonds. This covers most of
the compounds of carbon. Simple carbon compounds which don’t contain
C–H bonds, such as carbon dioxide and carbonates, are usually included in
the study of inorganic chemistry.
Organic compounds in their millions make up the cells in our bodies, the
food we eat, the clothes we wear, the plastic or wooden objects we use and
much of the world around us (Figure 6.1.1).
Figure 6.1.1 From the cells in the people’s
bodies and the fibres in their clothes, to
the plastics in the plates and the food on
the plates, almost everything in this photo
of a summer party consists of organic
chemicals.
There are two main reasons why carbon can form so many compounds.
The first reason is that carbon atoms have an exceptional ability to form
chains, branched chains and rings of varying size. No other element can
form long chains of its atoms in the same way as carbon.
The second reason why carbon can form so many compounds is the relative
inertness and unreactive nature of the C–C and C–H bonds, because of their
relatively high bond enthalpies (Section 8.7).
Figure 6.1.2 Sky divers can use their arms Figure 6.1.2 shows sky divers forming four links to each other. Like carbon
and legs to form four links to one another. atoms, they can form chains and rings, although carbon atoms can do it in
three dimensions also.
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When carbon atoms form a chain or a ring linked by single covalent bonds,
no more than two of the bonds on each atom are used. This leaves at least
two other bonds on each carbon atom which can bond with other atoms.
Carbon often forms bonds with hydrogen, oxygen, nitrogen and halogen
atoms (Figure 6.1.3).
Figure 6.1.3 The structure of ethanol, A knowledge of organic chemicals enables chemists to extract, synthesise and
CH3CH2OH. Ethanol is commonly called manufacture a wide range of important products including fuels, plastics,
‘alcohol’. Its structure shows two carbon medicines, anaesthetics and antibiotics.
atoms linked to each other and to hydrogen
and oxygen atoms by covalent bonds. Big molecules
Most molecules in living things are molecules of carbon compounds, so
biochemistry and molecular biology are important applications of organic
chemistry. The compounds found in living cells include carbohydrates, fats,
proteins and nucleic acids. The molecules of these compounds are large –
some of them are very large. For example, cellulose, the carbohydrate in
cotton, is a natural polymer made of very long chains of glucose units linked
together. Its relative molecular mass is about one million.
Organic chemists can synthesise other long-chain molecules by linking together
thousands of small molecules to make polymers. These synthetic polymers include
polythene (Figure 6.1.4), PVC (polyvinylchloride), polystyrene and nylon.
With so many organic compounds to study and understand, a way of
simplifying and organising this knowledge is needed.
Chemists have found a method of classifying organic compounds into families
or series, each of which has a distinctive group of atoms called a functional
group (Section 6.1.2). Examples of these families include hydrocarbons such
as alkanes and alkenes (Chapter 6.2) and compounds where other elements
are also present such halogenoalkanes and alcohols (Chapter 6.3). A family
of similar compounds with the same functional group is sometimes called
a homologous series and can be represented by a general formula. For
Figure 6.1.4 A short section of a polythene instance, alkanes can be represented by the general formula, CnH2n+2; alkenes
molecule. have the general formula CnH2n and halogenoalkanes containing one halogen
atom (X) have the general formula CnH2n+1X.
Key terms
A functional group is the group Test yourself
of atoms which gives an organic 1 What is organic chemistry?
compound its characteristic properties
2 State two reasons why carbon can form so many compounds.
and reactions.
3 The table shows some mean bond
A hydrocarbon is a compound of enthalpies. Use the data to answer Bond Mean bond
hydrogen and carbon only. enthalpy/kJ mol−1
the following questions.
H–H 436
A homologous series is a family of a)
How does the strength of the
compounds which all contain the same Cl–Cl 243
single C–C bond compare with
functional group and each member other single bonds between two Br–Br 193
of the series contains one –CH2 – unit atoms of the same non-metal? I–I 151
more than the previous member. b) How does the relative strength of C–C 347
A general formula represents all the C–C bond affect the number N–N 158
members of a homologous series. of carbon compounds? O–O 144
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6.1.2 Functional groups
H H
Ethane (CH3CH3) and ethanol (CH3CH2–OH) have very different properties
despite their similar structures (Figure 6.1.5). Ethane is a gas at room H C C H
temperature, ethanol is a liquid; ethane does not react with phosphorus(v)
chloride, but ethanol reacts vigorously, forming hydrogen chloride gas, H H
which is seen as misty fumes. Clearly, the –OH group in ethanol has a big ethane
effect on its properties. H H
The –OH group in ethanol, called the hydroxyl group, is an example of a
H C C OH
functional group – the group of atoms which gives an organic compound its
characteristic properties. The functional group in a molecule is responsible H H
for most of its reactions, while the hydrocarbon chain which makes up the ethanol
rest of the organic compound is relatively unreactive (Figure 6.1.6). Figure 6.1.5 Structures of ethane and
these bonds this active group is
ethanol.
are unreactive found in all alcohols
these bonds
are reactive
the number of
carbon and hydrogen atoms
does not have much effect
on the chemistry of alcohols
Figure 6.1.6 The structure of ethanol showing the reactive functional group and the
unreactive hydrocarbon skeleton.
Tip
Figure 6.1.6 shows ethanol in its correct three-dimensional representation. The
tetrahedral arrangement of bonds around each carbon atom is clear. Figure 6.1.5,
however, shows ethanol in a planar (flat) representation. This is much easier to draw,
but it is important to remember that the real molecule is not flat and the H—C—H bond
angles are not 90° or 180° but are 109.5° (see Section 2.4).
Functional groups, such as –OH, have more or less the same effect whatever
the size of the hydrocarbon skeleton to which they are attached.
This makes the study of organic compounds much simpler because all
molecules containing the same functional group have similar chemical
properties. Their physical properties are similar, but vary depending on the
length of the carbon chain attached to the functional group. In this respect,
molecules with the same functional group can be regarded as a chemical
family like a group of elements in the Periodic Table.
Ethanol is a member of the series of compounds called alcohols, all of which
contain the –OH functional group. Ethene, CH 2=CH2, is a member of the
series of compounds called alkenes which contain the C=C functional
group. The functional groups and homologous series of organic compounds
met in Year 1 of the course are shown in Table 6.1.1.
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Table 6.1.1 Common functional groups and their series of compounds.
Functional group Name of the series Example
of compounds
–OH Alcohols CH3CH2OH ethanol
Alkenes H H2C=CH2 ethene
C C C
O
C C Alkynes HC≡CH ethyne
OH
–Hal
C O C
Halogenoalkanes CH3CH2Cl chloroethane
O
Ethers CH3OCH3 methoxymethane
C O C
H
C CH Aldehydes
C CH3CHO ethanal
C C C (–CHO) O
O
H
C C C KetonesOH CH3COCH3 propanone
C OOH
O C
C O C O
O
Carboxylic acids CH3COOH ethanoic acid
OH
C O C (–COOH)
O
–NH2 Primary amines CH3CH2NH2 ethylamine
alcohol functional group carboxylic Some organic molecules have two or more functional groups. Lactic acid
– a hydroxy group acid group in sour milk, for example, has both an –OH group and a –COOH group
(Figure 6.1.7). In its reactions, lactic acid sometimes acts like an alcohol,
H OH sometimes like an acid and sometimes it shows the properties of both
3 2 1
O types of compound.
H C C C
O H
H H Test yourself
chain of 4 a) Why do all alkenes have similar chemical properties?
three carbon atoms
b) Why is there a gradual change in the physical properties of
Figure 6.1.7 The structure of lactic acid alkenes from gaseous ethene (C2H4) to liquid hex-1-ene (C6H12)?
(2-hydroxypropanoic acid). (See Section 2.6.)
5 Identify the functional groups in the following compounds and the
series to which they belong:
a) CH3CH2CH2OH
b) CH3CH2CHO
c) CH3CH2I
d) CH2=CHCH2Cl.
6 Molecules of amino acids contain the primary amine and the
carboxylic acid functional groups. Draw the structure of the amino
acid molecule, 2-aminoethanoic acid (common name glycine), which
contains these two groups bonded to the same carbon atom.
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6.1.3 Empirical, molecular and
structural formulae
Empirical formulae
The term ‘empirical formula’ was introduced in Section 5.2. The empirical Key term
formula of a compound is the formula found by experiment. In Section
5.2, we used the combined masses of elements in a compound to calculate its The empirical formula is the simplest
empirical formula. The formulae we obtain by this method show the simplest whole number ratio of the atoms of
whole number ratio of the atoms of different elements in a compound. each element in a compound.
Example
0.150 g of a liquid was analysed and found to contain 0.060 g of carbon,
0.010 g of hydrogen and 0.080 g of oxygen. What is the empirical formula
of the liquid?
Notes on the method
The molar masses of the elements come from a table of data.
Convert the masses in grams to amounts in moles by dividing by the molar
masses of the atoms of the elements.
Divide the amounts by the smallest of the amounts to find the simplest
whole number ratio.
Answer
C H O
Masses of elements combined/g 0.060 0.010 0.080
Molar masses/g mol−1 12.0 1.0 16.0
Amounts of elements combined 0.060 g 0.010 g 0.080 g
12.0 g mol−1 1.0 g mol−1 16.0 g mol−1
Figure 6.1.8 Many compounds have the
Ratio of moles of elements = 0.005 mol = 0.010 mol = 0.005 mol empirical formula CH2O. These include
Simplest whole number ratio 1 2 1 ethanoic acid (in vinegar) with molecular
formula C2H4O2, lactic acid (in milk or
So, the empirical formula of the compound is CH2O. This empirical formula athletes’ muscles) with molecular formula
can represent many different compounds. Three possibilities are shown C3H6O3 and glucose (in sugar/glucose
in Figure 6.1.8. tablets) with molecular formula C6H12O6.
Molecular formulae Key term
The molecular formula of a compound shows the actual number of atoms
of each element in one molecule. The molecular formula of ammonia is NH3 The molecular formula gives the actual
and that of ethanol is C2H6O. The term ‘molecular formula’ only applies to number of atoms of each element in a
substances that consist of molecules. molecule.
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For molecular compounds, the relative molecular mass shows whether or
not the molecular formula is the same as the empirical formula. A molecular
formula is always a simple multiple of the empirical formula.
For example, analysis shows that the empirical formula of octane in petrol
is C4H9, but the mass spectrum of octane shows that its relative molecular
mass is 114.
The relative mass of the empirical formula is given by:
Mr(C4H9) = (4 × 12.0) + (9 × 1.0)
= 57.0
The relative molecular mass is twice this value – so, the molecular formula
is twice the empirical formula.
∴ the molecular formula of octane = C8H18
Although the empirical and molecular formulae of organic compounds can
be determined by analysis in the way shown above, the modern way to find
molecular formulae is by mass spectrometry (Section 7.1).
Test yourself
7 A compound containing only carbon, hydrogen and oxygen was
analysed. It consisted of 38.7% carbon and 9.68% hydrogen by mass.
a) What percentage by mass of oxygen does it contain?
b) What is its empirical formula?
c) The relative molecular mass of the compound is 62. What is its
molecular formula?
8 A sample of a hydrocarbon was burned completely in oxygen. All the
carbon in the sample was converted to 1.69 g of carbon dioxide, and
all the hydrogen was converted to 0.346 g of water.
a) What is the percentage of carbon in carbon dioxide?
b) What is the mass of carbon in 1.69 g of carbon dioxide?
c) What is the percentage of hydrogen in water?
d) What is the mass of hydrogen in 0.346 g of water?
e) Use the masses of carbon and hydrogen from parts (b) and (d) to
calculate the empirical formula of the hydrocarbon.
f) The relative molecular mass of the hydrocarbon is found to be 26.
Deduce its molecular formula.
9 A hydrocarbon which consists of 82.8% by mass of carbon has an
approximate relative molecular mass of 55.
a) What is its empirical formula?
b) What is its molecular formula?
10 The three compounds shown in Figure 6.1.8 all have the empirical
formula CH2O. Explain how it is possible for different compounds to
have the same empirical formula.
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Structural formulae
The molecular formula of a compound gives the numbers of atoms of each
element in one molecule, but it does not show how the atoms are arranged.
To understand the properties of a compound, we need to know its structural
formula – this shows which atoms, or groups of atoms, are attached to each
other. For example, the molecular formula of ethanol is C2H6O – but this
does not show how the two carbon atoms, six hydrogen atoms and one
oxygen atom are arranged. Its structural formula, however, is written as H H
CH3CH2OH. This shows that ethanol has a CH3 group attached to a CH2
H C C O H
group which, in turn, is attached to an OH group.
Often, it is clearer to draw a full structural formula showing all the atoms H H
and all the bonds (Figure 6.1.9). This type of formula is called a displayed Figure 6.1.9 The displayed formula
formula. of ethanol.
Sometimes a skeletal formula is used – this shows only the carbon–carbon
bonds and functional groups in a compound (Table 6.1.2).
Table 6.1.2 Alternative formulae for propane and ethanol.
Molecular formula Structural formula Displayed formula Skeletal formula
H H H
C3H8 CH3CH2CH3 H C C C H
H H H Key terms
H H
A structural formula shows in minimal
C2H6O CH3CH2OH H C C O H OH
detail which atoms, or groups of atoms,
are attached to each other in one
H H
molecule of a compound.
A displayed formula shows all the
Skeletal formulae are outline formulae only – they provide a useful shorthand atoms and all the bonds between them
for large and complex molecules. However, skeletal formulae need careful in one molecule of a compound.
study because they show the hydrocarbon part of a molecule as nothing
A skeletal formula shows the functional
more than lines for the bonds between carbon atoms and for the bonds from
groups fully, but the hydrocarbon part
carbon atoms to functional groups. The symbols for carbon and hydrogen
of a molecule simply as lines between
atoms in the carbon skeleton are omitted. In contrast, functional groups are
carbon atoms, omitting the symbols for
shown in full.
carbon and hydrogen atoms.
Tip
Molecular formulae should not normally be used for describing a particular compound
because several different structures may be represented by the same molecular
formula. Displayed formulae are clear and unambiguous, but can be time-consuming
to draw. Skeletal formulae are the simplest to draw and, with experience, may be the
formula of choice, but initially structural formulae should be used as these are clear,
unambiguous and easy to understand.
6.1.3 Empirical, molecular and structural formulae 161
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Activity
The alkanes – an important series of organic compounds
Methane (CH4), ethane (C2H6), propane (C3H8) and butane (C4H10) are the first four
members of the homologous series of alkanes. Most of their empirical, molecular,
structural and displayed formulae are shown in Table 6.1.3.
Table 6.1.3 Empirical, molecular, structural and displayed formulae of the first four alkanes. H H H H
Name Methane Ethane Propane Butane
H C H H C C C
Empirical formula CH4 C3H8 C2H5
H H H H
Molecular formula CH4 C2H6 C4H10
Structural formula CH3CH3 CH3CH2CH3
Displayed formula H H H H HH H H H H H H
H C HH C H H C H C CC CH C H H C C C C H
H H H H HH H H H H H H
1 Copy and complete the table by adding the missing formulae. except the end carbon atoms, each of which has 1 extra
H H H H HH H
2 The names of all alkanes end in -ane. The names of the first H hydrogen atom.)
four alkanes in Table 6.1.3 do not follow a logical system. b) What is the value of y in terms of n?
H C H C CC CC CH C H
All other straight-chain alkanes are named using a Greek c) Write the general formula for alkanes in terms of C, H
numerical prefix for the numberHof carbon atoms
H HH HH H in one H and n.
molecule, with the ending -ane. So, C5H12 is pentane and 5 a) Draw the skeletal formula of butane, C4H10.
C7H16 is heptane. The prefixes are the same as those used for b) Why is it not possible to draw a skeletal formula of methane?
geometrical figures (pentagon, etc.). 6 a) Use a molecular model kit to construct a model of
What is the name for: propane.
a) CH3CH2CH2CH2CH2CH3 b) Connect one more carbon atom to the carbon chain in
b) CH3CH2CH2CH2CH2CH2CH2CH3? your model of propane to produce butane.
3 Which of the following molecular formulae are alkanes? c) Connect a carbon atom to a different place on the carbon
C2H2 C3H8 C4H8 C8H18 C10H20 chain in your model of propane to produce an alkane
4 It is possible to write a general formula for alkanes in the which is not butane.
form of CxHy. d) Draw the skeletal formula of this alternative structure of
a) Suppose x equals n. If an alkane has n carbon atoms, C4H10.
how many hydrogen atoms will it have? (Hint: In long- e) How many alternative structures of molecular formula
chain alkanes, every carbon atom has 2 hydrogen atoms, C5H12 can you make? Draw a skeletal formula of each one.
6.1.4 Naming simple organic
compounds
The International Union of Pure and Applied Chemistry (IUPAC) is the
recognised authority for naming chemical compounds. IUPAC has developed
systematic names based on a set of rules. These IUPAC rules make it possible
to work out the structure of a compound from its name and to work out
its name from its structure. The names of organic compounds are based on
the longest chain or main ring of carbon atoms in the carbon skeleton. The
IUPAC names of the first ten unbranched alkanes are shown in Table 6.1.4.
162 6.1 Introduction to organic chemistry
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Naming alkanes Table 6.1.4 Names and molecular formulae
of the first ten alkanes with unbranched
In naming an alkane, it is important to follow the IUPAC rules.
chains of carbon atoms.
1 Look for the longest unbranched chain of carbon atoms in the carbon
Number of Molecular Name
skeleton of the molecule and name that part of the compound. So: carbon formula
CH3CH2CH2CH2CH3 is pentane atoms
5 4 3 2 1 1 CH4 Methane
CH3CH2CH2CH CH3 is pentane with a CH3 group attached
2 C2H6 Ethane
CH3
7 6 5 4 3 2 1 3 C3H8 Propane
3 CH2
CH CH 2 CH CH CH 2 CH 3 is heptane with one CH3 and one
CH3CH2 group attached 4 C4H10 Butane
CH2 CH3
5 C5H12 Pentane
CH3 6 C6H14 Hexane
2 Identify the alkyl groups attached to the longest unbranched chain. The 7 C7H16 Heptane
simplest alkyl group is the methyl group, CH3, which is methane with one 8 C8H18 Octane
hydrogen atom removed. Alkyl groups are alkane molecules minus one
9 C9H20 Nonane
hydrogen atom (Table 6.1.5). So:
5 4 3 2 1 10 C10H22 Decane
CH3CH2CH2CH CH3 has a methyl side group
CH3
7 6 5 4 3 2 1
CH3 CH2 CH 2 CH CH CH 2 CH 3 h
as an ethyl side group and a Table 6.1.5 The structures of alkyl groups.
methyl side group Alkyl group Formula
CH2 CH3
Methyl CH3–
CH3
Ethyl CH3CH2–
3 Number the carbon atoms in the main chain to identify which carbon
Propyl CH3CH2CH2–
atoms the side groups are attached to.
Butyl CH3CH2CH2CH2–
4 Name the compound using the name of the longest unbranched chain,
prefixed by the names of the side groups and the numbers of the carbon
atoms to which they are attached. The numbering of the carbon atoms
can be from either the left or the right to give the name with the lowest
numbers. So:
5 4 3 2 1
CH3CH2CH2CH CH3 is 2-methylpentane – not 4-methylpentane Tip
CH3 When writing names, use a comma
7 6 5 4 3 2 1
CH3 CH2 CH 2 CH CH CH 2 CH 3 is 4-ethyl-3-methylheptane – between two numbers, but a hyphen
not 4-ethyl-5-methylheptane between a number and letter.
CH2 CH3
CH3
5 When there is more than one type of side group, they should be arranged
alphabetically. So:
7 6 5 4 3 2 1
3 CH2
CH CH 2 CH CH CH 2 CH 3 is 4-ethyl-3-methylheptane –
not 3-methyl-4-ethylheptane
CH2 CH3
CH3
6.1.4 Naming simple organic compounds 163
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6 When there are two or more of the same side group, add the prefix ‘di’,
‘tri’, ‘tetra’ and so on. So:
5 4 3 2 1
3 CH2 CH CH CH 3
CH is 2,3-dimethylpentane –
not 3,4-dimethylpentane and
CH3 CH3
not 2-methyl-3-methylpentane
Using the prefix ‘di’, ‘tri’, ‘tetra’ and so on does not change the alphabetical
order of the side groups, so:
7 6 5 4 3 2 1
CH3 CH2 CH2 CH CH CH CH3 is 4-ethyl-2,3-dimethylheptane.
CH2 CH3 CH3
CH3
Beware that the longest carbon chain may involve a side group:
3 4 5 6
CH
3 CH CH CH 2 CH 3 is 3,4-dimethylhexane (numbering from
either end of the six-carbon chain) –
2 CH CH3
2
not 2-ethyl-3-methylpentane
1 CH3
Tip Test yourself
Make sure you understand that names 11 Name the following alkanes.
which include 1-methyl or 2-ethyl
a) CH3(CH2)6CH3 b) CH 3 CH CH CH CH 2 CH 3
must be wrong (unless the compound
contains a ring). CH 3 CH 3 CH 3
c) CH 3 CH 2 CH 2 CHCH
CH 3 d) CH 3
CH 2 CH 3 CH 2 C CH 3
CH 3 CH 3
12 Draw the displayed formulae of the following alkanes:
a) 2-methylpropane b) 2,3-dimethylbutane.
13 Draw the structural formulae and the skeletal formulae of the
following alkanes:
a) 3,3,4-trimethylheptane b) 2-methylbutane
c) 3-ethyl-2-methyl-5,5-dipropyldecane.
Naming alkenes
Ethene (CH2=CH2) and propene (CH3CH=CH2) are the first two members
of the homologous series of alkenes with the functional group C=C .
Alkenes are named using the same general rules as alkanes, with the
suffix -ene instead of -ane, sometimes prefixed by a number to indicate the
position of the double bond in the chain.
With ethene and propene there is no need to number the carbon atoms because
the double bond must be between carbon atoms 1 and 2. But with a chain of
four or more carbon atoms, the double bond may be in more than one position.
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Thus, the molecule CH2=CHCH2CH3 is named but-1-ene. Although the
double bond links carbon atoms 1 and 2, the lower number 1 is used to
Tip
indicate the start of the bond. When more than one double bond
is present, the letter ‘a’ is included
Using the same rule, CH3CH=CHCH3 is named but-2-ene.
between the consonants ‘t’ and ‘d’,
The methods of naming organic compounds with other functional groups as in butadiene, or more fully
will be explained as they arise. buta-1,3-diene, H2C=CH—CH=CH2.
Tip
Ethene was shown in Table 6.1.1 as H2C=CH2, but is shown at the start of this section
as CH2=CH2. Both representations are accepted as correct structural formulae,
although H2C=CH2 is preferred by some because it shows the double bond between
carbon atoms more clearly.
Test yourself
14 Name the following alkenes:
a) CH 3 C CH 2
CH 3
b) CH 3CH 2 CH 2 CH CH CH 3
c) CH 3 CH 2 C C CH 3
CH 3 CH 3
15 Draw the structural formulae of the following alkenes:
a) 2-methylbut-2-ene
b) 3,4-dimethylpent-1-ene.
16 Draw the skeletal formulae of the following alkenes:
a) 2-methylbut-2-ene
b) 3,4-dimethylpent-1-ene.
6.1.5 Isomerism
Another reason why carbon forms so many compounds is that it is sometimes
possible to join the same atoms together in different ways. Consider, for example,
Key term
the molecular formula C4H10. You probably realise already that this could be
butane – but there is another compound, 2-methylpropane, which also has Structural isomers are compounds
the molecular formula C4H10. Both are shown in Figure 6.1.10. Compounds with the same molecular formula but
like butane and 2-methylpropane, which have the same molecular formula but different structural formulae.
different structural formulae, are called structural isomers.
There are two structural isomers of C4H10, three structural isomers of C5H12
and five structural isomers of C6H14. Table 6.1.6 shows that the number of
structural isomers of the alkanes increases very quickly.
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Table 6.1.6 The number of structural It is useful to divide structural isomers into three different types – chain
isomers of the alkanes. isomers, position isomers and functional group isomers.
Number of Number of ● Chain isomers have different chains of carbon atoms (Figure 6.1.10).
carbons isomers ● Position isomers have different positions of the same functional group
1 1 (Figure 6.1.11).
● Functional group isomers have different functional groups (Figure 6.1.12).
2 1
3 1 H
4 2 H H H H H H C H H
5 3
H C C C C H H C C C H
6 5
H H H H H H H
7 9
butane 2-methylpropane
8 18
Figure 6.1.10 Structural isomers of C4H10.
9 35
10 75 H
11 159 H H H H O H
12 355
H C C C O H H C C C H
13 802
H H H H H H
14 1 858
propan-1-ol propan-2-ol
15 4 347
Figure 6.1.11 Propan-1-ol and propan-2-ol are both alcohols like ethanol, CH3CH2OH. All
20 366 319 alcohols contain the —OH group. In propan-1-ol and propan-2-ol the —OH group is in a
25 36 797 588 different position on the carbon chain.
30 4 111 846 763 H H H H H H
40 62 491 178 805 831
or H C C C O H H C O C C H
62 481 801 147 341
opinions differ! H H H H H H
propan-1-ol methoxyethane
(an alcohol) (an ether)
Figure 6.1.12 Propan-1-ol is an alcohol with the —OH functional group. Methoxyethane
is an ether with the C—O—C functional group. Both these compounds have the same
molecular formula, C3H8O.
Notice that the word describing the type of structural isomer (chain, position
and functional group) tells you how the isomers differ from each other.
Test yourself
17 a) Draw displayed formulae of the three structural isomers with the
molecular formula C5H12 and name them.
b) What type of structural isomerism is shown by the three isomers
in part (a)?
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18 a) Draw skeletal formulae of the five structural isomers with the
molecular formula C6H14.
b) Identify the position isomers in your answer to part (a).
19 a) Draw structural formulae of the alkene structural isomers with
molecular formula C4H8 and name them.
b) Draw skeletal formulae of the two functional group isomers of
C4H8 which are not alkenes.
6.1.6 Types of organic reaction Key terms
Almost all organic molecules will burn, so most organic molecules can be
Addition is a reaction in which two
oxidised in redox reactions. Apart from redox, the type of reaction depends
molecules add together to form a single
on the structure of the molecule.
product.
If a molecule contains a double bond, it is said to be unsaturated and another
Substitution is a reaction in which one
molecule can join on to it in an addition reaction.
atom or group is replaced by another
If a molecule is saturated, that is it contains only single bonds, then it can atom or group.
react in two possible ways: Elimination is a reaction which
● An atom or group can be replaced in a substitution reaction. produces an unsaturated product by
● Adjacent atoms or groups can be removed to form an unsaturated molecule loss of atoms or groups from adjacent
in an elimination reaction. carbon atoms.
Both saturated and unsaturated molecules undergo hydrolysis reactions Hydrolysis is a reaction in which a
which may involve substitution or addition steps. compound splits apart in a reaction
involving water.
Addition – adding bits to molecules
Addition reactions are characteristic of unsaturated compounds with double
bonds. During an addition reaction, two molecules add together to form
a single product. Ethene, for example, reacts with bromine to form the
colourless addition product 1,2-dibromoethane (Section 6.2.9).
All types of unsaturated molecules can be saturated by reaction with
hydrogen in the presence of a catalyst in a hydrogenation reaction. Addition
of hydrogen can also be described as reduction.
Hydrogen adds to C=C double bonds in alkenes such as propene in the
presence of a platinum or palladium catalyst at room temperature, or on
heating to 150 °C in the presence of a nickel catalyst (Figure 6.1.13).
Addition to alkenes is considered in more detail in Section 6.2.9.
H H
CH3 H Ni catalyst
Tip
C C + H2 CH 3 C C H
150i°C The atom economy of an addition
H H
H H reaction is always 100% because only
one product is formed.
propane
Figure 6.1.13 The hydrogenation of propene.
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H H Polymerisation – making long chains
C C Under the right conditions, small molecules with double bonds can add
to each other and join up in long chains to form polymers. This is called
H H addition polymerisation (Section 6.2.12).
Figure 6.1.14 The repeating unit of
For instance, ethene molecules can join together to form poly(ethene), a
poly(ethene).
substance commonly known as polythene (Figure 6.1.14).
Elimination – splitting bits off from molecules
Key term An elimination reaction splits off a simple molecule, such as water or
hydrogen chloride, from a larger molecule leaving a double bond.
Addition polymerisation is an addition
Examples of elimination reactions are:
reaction in which small molecules,
called monomers, join together forming ● the removal of a hydrogen halide from a halogenoalkane in alkaline
a giant molecule, called a polymer. conditions to produce an alkene (Section 6.3.4)
CH3CH2CH2CH2Br + OH− → CH3CH2CH=CH2 + H2O + Br−
1-bromobutane but-1-ene
● the removal of water from an alcohol in acidic conditions to produce an
Tip alkene (Section 6.3.8).
Addition reactions convert unsaturated conc. H SO
2 4
compounds into saturated. CH3CH2CH2OH → CH3CH=CH2 + H2O
Elimination reactions convert saturated propan-1-ol propene
compounds into unsaturated.
Substitution – replacing one or more atoms
by others
Substitution reactions replace an atom or a group of atoms by another atom
or group of atoms. An example is the reaction of butan-1-ol with hydrogen
bromide to make 1-bromobutane (Section 6.3.4 Activity).
CH3CH2CH2CH2OH + HBr → CH3CH2CH2CH2Br + H2O
butan-1-ol 1-bromobutane
Substitution reactions are characteristic of halogenoalkanes (Section 6.3.4).
Other examples of substitution reactions include the replacement of hydrogen
atoms in alkanes by chlorine or bromine atoms in the presence of ultraviolet
light (Section 6.2.3).
UV light
CH4 + Cl 2 → CH3Cl + HCl
methane chloromethane
Hydrolysis – splitting apart with water
The word ‘hydrolysis’ comes from two other words – ‘hydro’ related to water
and ‘lysis’ meaning splitting. So, the term hydrolysis is used to describe any
reaction in which water causes another compound to split apart. Hydrolysis
reactions are often catalysed by acids or alkalis. They are often substitution
reactions in which the chemical attack is by nucleophiles such as water
molecules or hydroxide ions.
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An example of hydrolysis is the reaction of halogenoalkanes with water. In this
case, bonds in both the water and halogenoalkanes split to form alcohols and
hydrogen halides (Section 6.3.4). The reaction is very slow in the absence of alkali.
CH3CHBrCH3 + H2O → CH3CHOHCH3 + HBr
2-bromopropane propan-2-ol
Test yourself
20 Classify the following reactions as redox, addition, substitution,
elimination or hydrolysis.
a) C2H5OH → C2H4 + H2O
b) CH3COOCH2CH3 + H2O → CH3COOH + C2H5OH
c) + H2
d) CH3CHBrCH2CH3 + OH− → CH3CH(OH)CH2CH3 + Br −
e) Br
+ OH – + H2O + Br –
21 Give the structures and names of the products of the addition
reactions of but-2-ene with:
a) hydrogen b) chlorine
c) hydrogen bromide.
22 Write equations for the elimination reactions which occur when:
a) 2-bromopropane reacts with a hot solution of potassium
hydroxide in ethanol
b) butan-1-ol is dehydrated by passing over a hot catalyst.
23 Draw a structure to represent the addition polymer PTFE,
poly(tetrafluoroethene), formed from tetrafluoroethene.
24 Write equations for:
a) the substitution reaction in which ethane reacts with bromine to
form bromoethane and one other product
b) the substitution reaction in which 1-bromopropane reacts with
sodium hydroxide to form propan-1-ol and one other product
c) the hydrolysis reaction in which the ester ethyl ethanoate,
CH3COOCH2CH3, reacts with water to form ethanoic acid and one
other product.
6.1.7 Introduction to the mechanisms
of organic reactions
Reactions between ionic compounds in solution are almost instantaneous, the
ions simply collide (see Section 9.4). However, when organic compounds react
the covalent bonds in the molecules require energy to break, and a sequence
of bond breaking and bond forming steps can occur as reactants turn into
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products. This sequence is called a mechanism and chemists have used great
ingenuity to work out mechanisms. They began with simple reactions but
are now using their knowledge to explain what happens during industrial
processes and in living cells, where enzymes control biochemical processes.
There are two ways in which a bond can break – homolytically or heterolytically.
Tip Homolytic bond breaking
A covalent bond involves a shared pair of electrons. If the bond breaks
The use of curly half-arrows is not
homolytically then each atom keeps one electron – this is ‘equal splitting’
expected in this mechanism.
(homolytic fission) (Figure 6.1.15).
This type of splitting produces fragments with unpaired electrons. A
fragment of this type is called a free radical. Free radicals (often simply
Cl Cl Cl + Cl
called radicals) exist for a short time during a reaction, but quickly react to
chlorine chlorine atoms with
molecule unpaired electrons form new products. So, free radicals are intermediates which form during a
reaction but then disappear as the reaction is completed.
Figure 6.1.15 Homolytic bond breaking.
The covalent bond breaks and the atoms Chemists use a single dot to represent the unpaired electron on a radical.
separate, each taking one of the shared Other paired electrons in the outer shells are not usually shown.
pair of electrons.
Free radicals often occur in reactions taking place in the gas phase or in a
non-polar solvent. Ultraviolet light can speed up free-radical reactions.
Key term Examples of free-radical processes include the thermal cracking of
hydrocarbons (Section 6.2.4), the burning of petrol and other alkanes
A free radical is a species with an
(Section 6.2.3) and the substitution reactions of alkanes with halogens
unpaired electron.
(Section 6.2.3). Free-radical reactions are important high in the atmosphere
where gases are exposed to intense ultraviolet radiation from the Sun. The
reactions which form and destroy the ozone layer are free-radical reactions.
Heterolytic bond breaking
When a covalent bond breaks heterolytically, one atom takes both of the
electrons from the bond, the other atom takes none (Figure 6.1.16).
H H
H C Br H C+ + Br –
H H
Tip
The prefix ‘homo’ means ‘the same’ or H H
‘similar’. Chemical terms which include
this prefix include homolytic fission, H C Br H C+ + Br –
homogeneous catalyst and homologous
series. H H
Figure 6.1.16 Heterolytic bond breaking. Note the use of a curly arrow to show what
The prefix ‘hetero’ means ‘different’.
happens to the electrons as the bond breaks. A curly arrow shows the movement of
Chemical terms which include this
a pair of electrons. The covalent bond breaks and the atoms separate, with one atom
prefix include heterolytic fission and
taking both electrons in the shared pair. The arrow starts from the pair of electrons that is
heterogeneous catalyst.
moving. The head of the arrow points to where the electron pair will be after the change.
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Heterolytic bond breaking (heterolytic fission) produces ionic intermediates,
such as CH3+ and Br− in reactions like that shown in Figure 6.1.16. This type
of bond breaking is favoured when reactions take place in polar solvents such
as water. Often, the bond which breaks is already polar (Section 2.5) with a
δ+ end and a δ− end, like the C–Br bond in Figure 6.1.17.
H H Key terms
H C Br H C H + Br – Nucleophiles are electron pair donors.
– They are negative ions or molecules
HO
H HO with a lone pair of electrons that attack
Figure 6.1.17 A nucleophile attacking a δ+ carbon leading to heterolytic bond breaking. positive ions or positive centres in
molecules.
Some of the reagents which initiate reactions seek out the δ+ end of polar
bonds. These are called nucleophiles and will always have a lone pair of Electrophiles are electron pair
electrons to donate. acceptors. They are positive ions or
molecules with a vacant orbital that
Other reagents seek out the δ− end of polar bonds or the electron dense
attack negative ions or negative centres
regions in molecules. These are called electrophiles.
in molecules.
Nucleophiles
Nucleophiles are molecules or ions with a lone pair of electrons which can H
form a new covalent bond (Figure 6.1.18). They are electron-pair donors. –
Nucleophiles are reagents which attack molecules that have a partial positive H O H O
charge, δ+, so they seek out positive charges – they are ‘nucleus loving’. hydroxide ion water molecule
The substitution reactions of halogenoalkanes involve nucleophiles
(Section 6.3.4). H N H
–
C N H
Electrophiles
cyanide ion ammonia
Electrophiles are molecules or ions that attack negative ions or parts of molecule
molecules which are rich in electrons with negative centres, δ−. They are
Figure 6.1.18 Examples of nucleophiles.
‘electron-loving’ reagents. Electrophiles form a new bond by accepting a pair
of electrons from the molecule or ion attacked during a reaction.
An example of an electrophile is the H atom at the δ+ end of the H–Br bond
in hydrogen bromide. See, for example, the electrophilic addition reactions
of alkenes (Section 6.2.10).
Test yourself
25 Write equations (without curly arrows) to show how: c) an ammonia molecule reacts with water to
a) a bromine molecule breaks homolytically form an ammonium ion and a hydroxide ion.
b) a hydrogen bromide molecule breaks 27 In each of the following examples decide whether
heterolytically the reagent attacking the carbon compound is a
free radical, a nucleophile or an electrophile:
c) a C–H bond in a methane molecule breaks
homolytically. a) CH3CH2I + H2O → CH3CH2OH + HI
26 Write equations including curly arrows to show how: b) CH2=CH2 + HBr → CH3CH2Br
a) a bromide ion reacts with CH3+ to form c) CH4 + Cl • → • CH3 + HCl
bromomethane d) CH3CH2Br + CN− → CH3CH2CN + Br −
b) a hydroxide ion reacts with a hydrogen ion to
form water
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6.1.8 Investigating reaction
mechanisms
The techniques which chemists use to study reactions are becoming more
and more sophisticated. With the help of laser beams and spectroscopy, it
is now possible to follow extremely fast reactions and to watch molecules
breaking apart and rearranging themselves in fractions of a second. This
section outlines other ways in which reaction mechanisms can be studied.
Labelling with isotopes
Chemists can use isotopes as markers to track what happens to particular
atoms during a chemical change. They replace atoms of the normal isotope
of an element in a molecule with a different isotope (Section 1.4).
Isotopes of an element have identical chemical properties, so it is possible
to use them to follow what happens during a change without altering the
normal course of a reaction. Radioactive isotopes can usually be tracked
fairly easily and conveniently by detecting their radiation. The fate of
non-radioactive isotopes can be followed by analysing samples with a mass
spectrometer (Section 7.1).
Trapping intermediates
An important key to understanding reaction mechanisms was the realisation
that most reactions do not take place in one step, as implied by the balanced
equation. Instead, most reactions involve a series of steps. In the course
of these mechanisms, atoms, molecules and ions, which do not appear in
the balanced equation, exist as intermediates as chemicals change from the
reactants to the products.
Activity
Investigating the mechanism of a hydrolysis reaction
Alcohols react with carboxylic acids to form esters. Hydrolysis 2 How do atoms of the oxygen-18 and oxygen-16 isotopes
splits esters back to the alcohol and acid. Isotopic labelling differ?
has been used to investigate the mechanism of this hydrolysis 3 Why do oxygen-18 and oxygen-16 isotopes have the same
reaction (Figure 6.1.19). The researchers used water labelled chemical properties?
with oxygen-18 instead of the normal oxygen-16 isotope. After 4 What method of analysis can be used to distinguish ethanoic
hydrolysis with H218O, they found that the heavier oxygen atoms acid molecules with 18O atoms from those with 16O atoms?
from the water ended up in the acid and not in the alcohol. In (Neither isotope of oxygen is radioactive.)
this way they were able to identify exactly which bond breaks 5 Look closely at Figure 6.1.19. Which bond in the ester
during hydrolysis of the ester. breaks during the reaction?
6 Where would the oxygen-18 atoms have appeared if the
1 Why is the reaction of an ester with water described as
mechanism involved breaking the other C—O bond in the ester?
‘hydrolysis’?
O O
18
CH3 C + H2 O CH 3 C + C2H5 OH
O C2H5 18
OH
Figure 6.1.19 Use of labelling to investigate bond breaking during the hydrolysis
of an ester.
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Chemists can often use spectroscopy (Chapter 7) to detect intermediates
which exist for only a short time during a reaction. Intermediates can also
be detected chemically. This can be done by adding chemicals to trap the
intermediates by reacting with them. This gives rise to products that would
never be formed without first producing the intermediate. In this way,
chemists showed that the addition of bromine to alkenes must be a two-step
reaction (Section 6.2.10).
Studying reaction rates
Chemists have learned a great deal about reaction mechanisms by studying
the rates of chemical reactions. This is illustrated by the mechanisms of
substitution reactions of halogenoalkanes (Section 6.3.4). What these
mechanisms show is that the effects of changing the concentrations of the
reactants are not the same for primary and for tertiary halogenoalkanes. Taken
with other evidence, this suggests that the two types of halogenoalkane react
by different mechanisms.
Studying the shapes of molecules
Studying the shapes of molecules can also give clues to the details of reaction
mechanisms. Part of the evidence that helped to confirm the two-step
mechanism for electrophilic addition to alkenes (Section 6.2.10) came from
studies of the isomers which form when bromine adds to compounds such
as cyclohexene.
H H
C C
H H
H H H H
C C
C C H H
H H H H
H H + Br2 C C
C C
Br Br
H H
C C
H H
H H
C C
H H
H H
C C
H Br
H H
C C
Br H
Figure 6.1.20 Two possible products when bromine adds to cyclohexene forming
1,2-dibromocyclohexane.
There are two possible isomers when bromine adds to cyclohexene. One has
both bromine atoms on the same side of the ring of carbon atoms and the
other has the bromine atoms on opposite sides. It turns out that the main
product of the reaction is the isomer with the bromine atoms on opposite
sides of the ring (trans), which is the lower structure in Figure 6.1.20. This
suggests that the bromine does not add directly to alkene molecules as Br2
molecules, but a two-stage mechanism via a positively charged carbocation
occurs.
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Chapter summary
Chapter 6.1 Introduction to formulae. Structural isomerism occurs in three
ways: different arrangements of the carbon atoms
organic chemistry in the chain, different functional groups and
l The element carbon can form many millions of different positions of the functional groups on the
compounds. Carbon can form long chains and carbon chain.
rings of atoms in a way no other element can. l Stereoisomerism occurs when molecules with the
C–C and C–H bonds are also relatively strong and same molecular formula and the same structural
unreactive. formula have a different spatial arrangement of
l The term ‘organic chemistry’ is used to describe bonds.
the branch of chemistry concerned with the study l Stereoisomerism, as illustrated by E/Z isomerism,
of compounds containing C–H bonds. is covered in Section 6.2.8. Optical isomerism,
l A hydrocarbon is a compound of hydrogen and another form of stereoisomerism is covered in
carbon only. Section 17.1.1.
l Organic compounds are classified into families l Reactions in organic chemistry are often classified
with similar chemical properties called by terms such as addition, where two molecules
homologous series. Each member of a series has combine; elimination, where a small molecule
a distinctive group of atoms called a functional is removed to leave a double bond; substitution,
group. Members contain one –CH2– unit more where one functional group is replaced by
than the previous member and can be represented another; and hydrolysis, where a compound splits
by a general formula. apart in a reaction involving water.
l The empirical formula of an organic compound, l The sequence in which covalent bonds break
the simplest whole number ratio of the atoms of and form in an organic reaction is called a
each element in a compound, can be determined reaction mechanism. A curly arrow is used in
by analysis. a mechanism to show the movement of a pair
l The molecular formula gives the actual number of of electrons in the breaking or the forming of a
atoms of each element in a molecule. covalent bond.
l The structural formula shows how the atoms are l Covalent bonds may break homolytically, which
arranged. If all the bonds are shown, this is called is where the atoms separate and each takes one of
a displayed formula. A skeletal formula shows the shared pair of electrons to form radicals, or
only the carbon–carbon bonds and the functional heterolytically, where one atom takes both of the
groups in a compound. electrons from the bond and the other atom takes
l International Union of Pure and Applied none.
Chemistry (IUPAC) rules are used to name l Nucleophiles are electron pair donors. They
compounds. Names are based on the longest are negative ions or molecules with a lone pair
unbranched chain of carbon atoms in the carbon of electrons that attack positive ions or positive
skeleton of the molecule, with a suffix or prefix centres in molecules.
added to indicate the functional group and l Electrophiles are electron pair acceptors. They
numbers used to show its position. are positive ions or molecules with a vacant
l Structural isomers are compounds with the orbital that accept electrons from electron rich
same molecular formula but different structural centres.
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Exam practice questions
1 a) Carbon is able to form a vast number of i) State what is meant by the term
chemical compounds. Give two reasons for ‘hydrocarbon’. (1)
this. (2) ii) One molecule of dodecane contains
b) Petrol is a mixture of hydrocarbons 12 carbon atoms. Deduce the molecular
containing between 6 and 10 carbon atoms. formula of dodecane. (1)
Some of these hydrocarbons are structural iii) Give the empirical formula of
isomers. dodecane. (1)
i) Explain the term ‘structural isomers’.(2) b) Decane, C10H22, is a straight-chain alkane.
ii) Some of the hydrocarbons in petrol are It reacts with chlorine in a free-radical
alkanes. Write the general formula of reaction to form the compound C10H21Cl.
the alkanes and the molecular formula i) Explain the term ‘free radical’. (2)
of an alkane that could be present in ii) Write an equation for the formation of
petrol. (2) chlorine free radicals from Cl2. (1)
c) Petrol also contains cycloalkanes. Draw iii) State the type of bond fission involved
the structure of cyclohexane and write the in the formation of chlorine free
general formula of cycloalkanes. (2) radicals. (1)
iv) Deduce how many structural isomers
The compound X below, containing two
2
can be produced when decane reacts
functional groups, can be extracted from oil of
with chlorine to form C10H21Cl. (1)
violets.
v) Draw the structural formula of one of
CH 3 CH 2 CH 2 OH these structural isomers and give its
C C IUPAC name. (2)
H H
Identify the following conversions as addition,
4
X
elimination, substitution, oxidation, reduction,
a) State the empirical and molecular formula hydrolysis or polymerisation reactions. (Note
of X and draw its skeletal formula. Explain that a reaction may belong to more than one
the term ‘functional group’ and name the category.)
functional groups present in X. (7) a) 2-iodobutane to but-2-ene (1)
b) X reacts with hydrogen and a nickel catalyst b) butane to 1-bromobutane (1)
in the gas phase to produce compound Y c) but-1-ene to 1,2-dichlorobutane (1)
with the formula CH3(CH2)3CH2OH. d) butanal to butanoic acid (1)
i) Give the IUPAC name of Y. (1) e) 1-bromobutane to butan-1-ol (1)
ii) Write an equation, including state f) buta-1,3-diene to synthetic rubber. (1)
symbols, for the reaction of X with
hydrogen to form Y. (2) 5 a) i) Explain the term ‘electrophile’,
iii) Y has two structural isomers which are giving an example. (2)
also position isomers. Name these two ii) Use symbols to describe the mechanism
position isomers of Y. (2) of the electrophilic addition of
iv) Y also has structural isomers with hydrogen bromide to ethene. Show any
different functional groups. Write the relevant dipoles. (5)
structural formula of one of these b) i) Explain the term ‘nucleophile’, giving
isomers. (1) an example. (2)
ii) Use symbols to describe the mechanism
3 a) Crude oil is a mixture of many of nucleophilic substitution during
hydrocarbons. Using fractional distillation the reaction of hydroxide ions with
it can be separated into fractions that can bromoethane. Show any relevant
be refined to produce hydrocarbons such as dipoles. (5)
dodecane.
175
Exam practice questions
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6 a)* Discuss the circumstances that favour The reaction of ethanol with 50% sulfuric
8
homolytic bond breaking and those that acid and potassium bromide produces
favour heterolytic bond breaking during 1-bromobutane as the main product.
organic reactions. (6) Side reactions also form ethene and some
b) Explain why it is helpful for chemists to ethoxyethane, C2H5–O–C2H5. The first step
classify reagents as free radicals, electrophiles in all the reactions is for ethanol to gain a
or nucleophiles. (4) hydrogen ion, H+, from the acid.
a) Draw a diagram, with a curly arrow, to show
Complete combustion of 0.292 g of a
7
ethanol forming a bond with H+. (2)
compound containing carbon, hydrogen and
b) The main product forms by reaction of the
oxygen formed 408 cm3 of carbon dioxide
intermediate from part (a) with bromide
and 0.308 g of water. (Molar volume of a gas is
ions. Draw a diagram with curly arrows to
24.0 dm3 mol−1.)
show that this is a nucleophilic substitution
a) Calculate the empirical formula of the
reaction. (2)
compound. (6)
c) Draw a diagram with curly arrows to show
b) The molecular formula of the compound
how an elimination reaction turns the
is the same as its empirical formula. Explain
intermediate from part (a) into an alkene.(2)
why the molecule must either be cyclic or
d) Identify the nucleophile that reacts with
contain a double bond. Draw the structural
the intermediate from part (a) to form
formula of one cyclic compound and the
ethoxyethane and explain how it acts as a
structural formulae of three unsaturated
nucleophile. (2)
compounds that are functional group
e) Ethoxyethane is unaffected by acidified
isomers with this molecular formula. (5)
potassium dichromate(vi).
i) Draw a functional group isomer of
ethoxyethane that is also unaffected by
acidified potassium dichromate(vi). (1)
ii) Describe how the isomer from part
(e)(i) and ethoxyethane could be
distinguished by a physical test and by a
chemical test. (4)
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6.1 Introduction to organic chemistry
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Hydrocarbons: alkanes
6.2 and alkenes
CH2 6.2.1 Types of hydrocarbon
H2C CH
or Hydrocarbons make up the majority of crude oil – the source of most
H2C CH fuels and the main raw material for the chemical industry. Hydrocarbons are
CH2 compounds containing hydrogen and carbon only. The most common and
cyclohexene C6 H10 most important hydrocarbons are alkanes and alkenes.
Figure 6.2.1 Structures of the aliphatic There are two types of hydrocarbon.
hydrocarbon cyclohexene, which is a
● Aliphatic hydrocarbons are those with chains of carbon atoms which
cycloalkene.
may be branched or unbranched and with rings that are not aromatic.
Key terms Alkanes and alkenes such as cyclohexene (Figure 6.2.1) are examples of
aliphatic compounds.
Hydrocarbons are compounds of ● Aromatic hydrocarbons, such as benzene, methylbenzene and
carbon and hydrogen only. naphthalene, are ring compounds in which there are delocalised electrons
Aliphatic hydrocarbons have branched (Figure 6.2.2). They are called aromatic because of their smells (aromas).
or unbranched chains of carbon atoms These hydrocarbons are sometimes called arenes.
or rings of carbon atoms.
H
Aromatic hydrocarbons or arenes
contain rings of carbon atoms in which H C H
there are delocalised electrons. C C
Saturated compounds have only C C
single bonds between atoms in their H C H
molecules.
H
Figure 6.2.2 Representations of the structure of benzene. At one time, chemists thought that
the ring structure in benzene had three double and three single bonds. X-ray studies have
shown that all six bonds in the ring are identical and that each carbon atom contributes one
electron to a cloud of delocalised electrons. This has led to the third structure with a ring inside
a hexagon.
6.2.2 Alkanes
Alkanes are the hydrocarbons which make up most of crude oil and natural gas.
Alkanes form a series of organic compounds with the general formula CnH2n+2.
Alkanes are saturated compounds – they have only single bonds between
the atoms in their molecules. The term ‘saturated’ is also used for compounds
with saturated hydrocarbon chains, such as saturated fats and fatty acids
in food (Figure 6.2.3). If eaten in excess, saturated fats lead to high levels
Figure 6.2.3 The saturated fats in foods such of cholesterol in the blood which causes furring and blocking of the arteries.
as beef burgers and doughnuts contain alkyl Unfortunately, not all unsaturated fats are good for health – trans fats should also
groups with long chains of carbon atoms. be avoided (see Section 6.2.9).
6.2.2 Alkanes 177
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Physical properties
Alkanes are composed of simple molecules which are held together by
only weak intermolecular forces (Section 2.6). As the molecules get larger,
the intermolecular forces increase, and therefore the melting and boiling
temperatures rise as the number of carbon atoms per molecule increases.
At room temperature and pressure, alkanes in the range C1 to C4 are gases, those
in the range C5 to C17 are liquids, while those from C18 upwards are solids.
Tip Test yourself
All combustion reactions are exothermic 1 a) Write the molecular formulae and names of the first four members
(Section 8.2). So an alkane plus oxygen of the alkanes.
is unstable with respect to the products
b) What is the difference in molecular formula from one alkane to the
of oxidation which are at a lower energy.
next?
But fuels do not spontaneously ignite
without the input of sufficient activation 2 Explain why the general formula of the alkanes is CnH2n+2.
energy (Section 9.4). The combustion 3 Write the molecular formulae of the first liquid alkane and of the first
reaction does not take place and the solid alkane at room temperature.
rate of the reaction is effectively zero
without, for example, a spark to provide
this activation energy. So although 6.2.3 Chemical reactions
energetics might predict that a reaction
should occur, kinetics prevents it of the alkanes
happening at room temperature. This The bond enthalpies of C−C and C−H bonds are relatively high, so the bonds
kinetic stability is extremely fortunate in alkanes are difficult to break. In addition, these bonds are non-polar (Section
because, without it, no fuels could be 2.5). This means that alkanes are very unreactive with ionic reagents in water –
stored for future use. such as acids, alkalis, oxidising agents and reducing agents. There are, however,
three important reactions of alkanes involving homolytic bond breaking and
free radicals (Section 6.1.7). These three important reactions are combustion
(burning), halogenation and cracking (Section 6.1.7).
Reaction with oxygen – combustion
Many common fuels consist mainly of alkanes. Natural gas is mainly
methane, Calor Gas® (Figure 6.2.4) is mainly propane and Gaz® is mainly
butane. In a plentiful supply of air or oxygen, the alkanes are completely
oxidised to carbon dioxide and water. The reaction is highly exothermic.
C4H10(g) + 6 12 O2(g) → 4CO2(g) + 5H2O(l) ΔcH 1 −2876 kJ mol−1
butane
If the air is in short supply, the products include soot (carbon) and highly
toxic carbon monoxide as well as carbon dioxide (see Section 6.2.5).
Alkanes are kinetically stable in the air (oxygen), but they are energetically
(thermodynamically) unstable with respect to the products of oxidation.
The combustion of alkanes involves a free-radical mechanism, which occurs
rapidly in the gas phase. This means that liquid and solid alkanes must vaporise
before they burn and it explains why less volatile alkanes burn less easily.
The burning of alkanes is immensely important in any advanced,
Figure 6.2.4 Red Calor Gas® cylinders
technological society. It is used to generate energy of one kind or another
contain propane for use as a fuel.
178 6.2 Hydrocarbons: alkanes and alkenes
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in power stations, furnaces, domestic heaters, cookers, candles and vehicles.
Unfortunately this mass burning of alkanes is now accepted to be a major
cause of global warming and the increased greenhouse effect (Section 6.2.5).
Reactions with chlorine and bromine
(halogenation) Key terms
Alkanes react with chlorine and bromine, either on heating or on exposure
A substitution reaction is one in which
to ultraviolet light. During the reactions, hydrogen atoms in the alkane
an atom, or group of atoms, is replaced
molecules are replaced (substituted) by halogen atoms. These are described
by another atom, or group of atoms.
as substitution reactions.
Initiation – the step which produces
Any of the hydrogen atoms in an alkane may be replaced, and the reaction free radicals from molecules
can continue until all the hydrogen atoms have been substituted by halogen
atoms. Consequently, the product is a mixture of compounds.
In strong sunlight, methane and chlorine react explosively. The initial
products are chloromethane, CH3Cl, and hydrogen chloride (Figure 6.2.5).
sunlight
+ +
sunlight
CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)
Figure 6.2.5 The equation and models representing the initial substitution
reaction of methane with chlorine.
The reaction involves breaking some bonds – for which energy must be
supplied – and making new bonds – when energy is released. Possible
reaction mechanisms can be tested using bond enthalpies (energies) and this
leads to a probable reaction mechanism.
The reaction between methane and chlorine does not occur in the dark
because the molecules do not have enough energy for bonds to break when
they collide. But in ultraviolet light, the energy provided by absorbed
photons is 400 kJ mol−1. This is enough to cause homolytic fission of chlorine
molecules into free radicals:
Cl 2 → Cl• + Cl• ΔH = +242 kJ mol−1
But this is not enough for the homolytic fission of methane, which requires
435 kJ mol−1:
CH4 → CH3• + H• ΔH = +435 kJ mol−1
and definitely not enough for the heterolytic fission of either chlorine or
methane:
Cl 2 → Cl+ + Cl− ΔH = +1130 kJ mol−1
CH4 → H+ + CH3− ΔH = +1700 kJ mol−1
These figures suggest that ultraviolet light starts the reaction by splitting chlorine
molecules into chlorine atoms (free radicals). This stage is called initiation.
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The chlorine atoms, each with an unpaired electron, are highly reactive.
Key terms They remove hydrogen atoms from methane molecules to form hydrogen
chloride and a new free radical, CH3•
Free-radical substitution is the
replacement of hydrogen atoms in Cl• + CH4 → HCl + CH3• ΔH = +4 kJ mol−1
a molecule by halogen atoms in a
The CH3• free radical now reacts with a chlorine molecule to form
reaction which involves free radicals.
chloromethane, CH3Cl, and generate another chlorine free radical:
Free-radical chain reactions involve
three stages: CH3• + Cl 2 → CH3Cl + Cl• ΔH = −97k kJ mol−1
Initiation – the step which produces The new Cl• free radical can react with another CH4 molecule and the last
free radicals from molecules. two reactions can be repeated again and again until either all the Cl 2 or all
the CH4 is used up. These two repeated reactions create a chain reaction and
Propagation – steps which form are described as propagation stages.
products and more free radicals.
Propagation ends when two free radicals combine. This is the termination
Termination – steps which remove free stage of the reaction, which is very exothermic. There are several possible
radicals by turning them into molecules. termination steps:
A chain reaction occurs when a product
Cl• + Cl• → Cl 2 ΔH = −242 kJ mol−1
in a reaction can react with a starting
material so the reaction continues. CH3• + Cl• → CH3Cl ΔH = −339 kJ mol−1
CH3• + CH3• → CH3CH3 ΔH = −346 kJ mol−1
The three stages in the free-radical substitution of methane with chlorine
(initiation, propagation and termination) are summarised in Figure 6.2.6.
light
Stage 1 Initiation Cl Cl Cl + Cl
Stage 2 Propagation Cl + CH4 HCl + CH3
CH3 + Cl2 CH3Cl + Cl
Stage 3 Termination Cl + Cl Cl2
CH3 + Cl CH3Cl
CH3 + CH3 CH3CH3
Figure 6.2.6 Stages in the free-radical substitution of methane with chlorine.
The chlorine radical formed in the second propagation step (Figure 6.2.6)
reacts with another methane molecule so the two propagation steps repeat and
keep on repeating. This is called a chain reaction and can lead to explosions
if chlorine and methane mixtures are exposed to sunlight.
The number of radicals present at any one time is quite small. Each
Tip propagation step uses a radical and then forms a radical, so the number of
radicals remains fairly constant during the reaction. The likelihood of two
Adding the two propagation steps
radicals colliding is relatively low, so the amount formed of a termination
together gives the overall equation for
product such as ethane is small. But the fact that any ethane is formed at all
that radical substitution reaction.
confirms that the proposed mechanism is correct.
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In the first propagation step, a chlorine radical extracts a hydrogen atom to
form hydrogen chloride. Chloromethane, CH3Cl, the product of this first
substitution, still contains three hydrogen atoms, so further substitution can
take place. Further pairs of propagation steps produce dichloromethane,
CH2Cl2, then trichloromethane, CHCl3, then finally tetrachloromethane,
CCl4, so when chlorine and methane react a mixture of products is formed.
To reduce the likelihood of further substitution, an excess of methane can
be used but this cannot prevent the formation of a mixture of products.
Therefore, radical substitution has limited use in the synthesis of chloroalkanes.
Alternative methods of making chloroalkanes are considered in Section 6.3.3.
Other halogen elements such as bromine also react with methane in radical
substitution reactions (Figure 6.2.7).
0s 10 s 16 s
Figure 6.2.7 The effect of light on a mixture of bromine and hexane after 0, 10 and 16 seconds.
Test yourself
4 To what extent is a series of organic compounds, b) Write an equation for the substitution reaction
such as the alkanes, comparable to a group of which occurs when chloromethane reacts with
elements in the Periodic Table? chlorine to form dichloromethane. Write a
5 a) Write an equation for the complete combustion of mechanism for this substitution and label each
propane in Calor Gas®. step.
b) What are the products formed when propane c) Write overall equations for the two further
burns in a poor supply of oxygen? substitution reactions which occur when
dichloromethane reacts with an excess of
c) Why is it important for gas water heaters to be
chlorine and name the products.
serviced regularly?
d) How could a pure sample of dichloromethane be
6 The boiling temperatures of propane and butane
obtained from the mixture of products formed?
are −42 °C and −0.5 °C respectively. Why is it wise
for campers to use Calor Gas®(propane) rather 8 a) Why does a mixture of bromine in hexane
than Gaz®(butane) for cooking during the winter? remain orange in the dark, but fade and become
colourless in sunlight?
7 a) In the substitution of methane with chlorine,
chloromethane can be formed both in a b) Write an equation for the reaction in part (a).
propagation step and in a termination step. Write c) Why can acidic fumes be detected above the
an equation for each step and explain which of solution once the colour has faded?
the two is more likely.
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6.2.4 Fuels from crude oil
Crude oil, also known as petroleum, is arguably the most important naturally
occurring raw material. It provides a very large proportion of our energy
needs, and is the source of most of our organic chemicals – including plastics,
fibres, drugs and pesticides.
Crude oil is a complex mixture of hydrocarbons, most of which are alkanes.
Crude oil has no uses in its raw form. The challenge for refineries is to
produce the various oil products in the proportions required by industrial
and domestic users.
Generally, crude oil contains too much of the high boiling fractions with
larger molecules, and not enough of the low boiling fractions with the smaller
molecules needed for fuels such as petrol. In order to satisfy the demand for
very different products, crude oil undergoes three main processes – fractional
distillation, cracking and reforming.
Fractional distillation
Fractional distillation is the first stage in refining crude oil (Figure 6.2.8).
This produces fuels and lubricants, as well as feedstocks for the petrochemical
industry. The continuous process operates on a large scale, separating crude
oil into different fractions.
Figure 6.2.8 The fractional distillation of 20 °C
gas C1 – C4
crude oil.
naphthai/gasoline C5 – C10
kerosene C10 – C16
distillation
fractional
diesel oil
crude C14 – C20
oil heavy diesel oil
furnace
400 °C
lubricating oil C20 – C50
distillation
vacuum
feed to catalytic
cracker
fuel oil C20 – C 70
fuel oil for sale or
combustion on refinery
bitumen >C70
A furnace heats the crude oil to about 400 °C. The oil then flows into a
fractionating tower containing 40 or so horizontal ‘trays’ pierced with small
holes.
The column is hotter at the bottom and cooler at the top. Rising vapour
condenses when it reaches the tray with liquid at a temperature just below its
boiling temperature. Condensing vapour releases energy. This heats the liquid on
the tray and evaporates the more volatile compounds in the mixture on the tray.
With a series of trays, the outcome is that hydrocarbons with small molecules and
low boiling temperatures rise to the top of the column, while larger molecules
stay at the bottom. Fractions are drawn off from the column at various levels.
182 6.2 Hydrocarbons: alkanes and alkenes
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Some components of crude oil have boiling temperatures too high for them
to vaporise at the furnace temperature and atmospheric pressure. Lowering
the pressure in a separate vacuum distillation column reduces the boiling
temperatures of these hydrocarbons and this makes it possible to separate
them.
Fuel fractions
Petrol is a blend of hydrocarbons based on the gasoline fraction (hydrocarbons
with 5–10 carbon atoms), whereas jet fuel is produced from the kerosene (or
paraffin) fraction (hydrocarbons with 10–16 carbon atoms). Fuel for diesel
engines is made from diesel oil (hydrocarbons with 14–20 carbon atoms).
All these fuels must be refined to remove sulfur compounds, which
would cause air pollution when they burn. In addition, petrol must be
blended carefully if modern engines are to start reliably and run smoothly
(Figure 6.2.9). The proportion of volatile hydrocarbons added to petrol
is higher in winter to help cold-starting, but lower in summer to prevent
vapour forming too readily.
valves
spark plug
compressed
fuel and air
piston
cooling water
crankshaft
Key term
Octane number is a measure of the
performance of a fuel by comparison
Figure 6.2.9 The working parts of a cylinder in an internal combustion engine that runs on
with 2,2,4-trimethylpentane, which is
petrol. Sparks from the plugs cause the compressed fuel and air to ignite. This produces
given the number 100, and heptane,
more gas molecules, increasing the pressure and forcing the piston down. The product
which is given the number 0. Most UK
gases are then allowed to escape and, as the pressure falls, the piston rises ready for the
petrol has an octane number of 95.
next ignition.
For smooth running, petrol must burn smoothly in the engines of vehicles
and not in fits and starts. To ensure smooth combustion, companies produce Tip
fuel with a high octane number by increasing the proportions of branched Straight-chain alkane does not mean
alkanes and arenes, or blending-in oxygen compounds. The three main literally straight but means unbranched.
methods used to increase the octane number of fuels are: There are no side chains attached to
● cracking – which makes smaller molecules and converts straight-chain the carbon skeleton.
hydrocarbons to branched and cyclic hydrocarbons The C−C−C bond angles in the carbon
● reforming – which turns straight-chain alkanes into branched-chain or
skeleton are all about 109.5° (see
cyclic alkanes or arenes such as benzene and methylbenzene and turns Section 2.4).
cyclic alkanes into arenes
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● adding ethanol and ethers such as ETBE (initials based on its older name
ethyl tertiary butyl ether (Figure 6.2.10). Its correct IUPAC name is
2-ethoxy-2-methylpropane). In the USA and Brazil, a mixture of 10%
ethanol and 90% petrol (gasoline), known as gasohol, is commonly used
(see Section 6.2.6).
CH3
CH3 CH2 O C CH 3
CH3
Figure 6.2.10 Structure of the ether ETBE. Adding ETBE is one of the methods used
to raise the octane number of gasoline from as low as 70 to 120, the required level for
premium petrol.
Cracking
Fractional distillation of crude oil produces a larger supply of the heavier
fractions than needed but a lower supply of the fractions most in demand for
use as fuels such as petrol. In order to supply more of the smaller molecules,
a process called cracking is used to convert heavier fractions, such as diesel
oil and fuel oil, into more useful hydrocarbon fuels by breaking up large
molecules into smaller ones.
Cracking converts long-chain alkanes with 12 or more carbon atoms into
smaller, more useful molecules in a mixture of branched alkanes, cycloalkanes,
alkenes and branched alkenes. When conducted at high temperatures in
the presence of steam, a higher proportion of alkenes is produced. When
conducted in the presence of a catalyst (catalytic cracking), higher yields of
branched and cyclic alkanes are produced.
The catalyst is a synthetic sodium aluminium silicate belonging to a class
of compounds called ‘zeolites’. A zeolite has a three-dimensional structure
(Figure 6.2.11) similar in structure to clay, in which the silicon, aluminium
and oxygen atoms form tunnels and cavities into which small molecules can
fit. Cracking takes place on the surface of the catalyst at about 500 °C.
Figure 6.2.11 A model of the structure of a zeolite crystal.
Synthetic zeolites make excellent catalysts because they can be developed
with active sites to favour the shapes and sizes of those molecules which react
to give the desired products.
184 6.2 Hydrocarbons: alkanes and alkenes
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Reforming
Reforming converts straight-chain alkanes into branched-chain or cyclic
alkanes or arenes such as benzene and methylbenzene, and turns cyclic
alkanes into arenes (Figure 6.2.12). Hydrogen is a valuable by-product of the
process – it can be used in processes elsewhere at the refinery.
The catalyst for this process is often one or more of the precious metals such
as platinum and rhodium supported on an inert material such as aluminium
oxide. The process operates at about 500 °C.
Figure 6.2.12 Examples of reforming.
Pt catalyst
500 °C
high pressure
octane 2,5-dimethylhexane
CH3
H C H
Pt catalyst C C
CH3CH2CH2CH2CH2CH2CH3 + 4H2
500 °C C C
high pressure H C H
H
heptane methylbenzene
H H H
H H
C H C H
H C C H Pt catalyst C C
+ 3H2
H C C H 500 °C C C
C high pressure H C H
H H
H H H
cyclohexane benzene
Test yourself
9 a) Why is crude oil so important? b)
Use skeletal formulae to show how
b)
Why should we try to conserve our reserves of catalytic cracking converts decane into
crude oil? 2,3-dimethylpentane and propene.
10 a)
Why do you think that ethanol and ETBE can c)
Use structural formulae to show how reforming
raise the octane number of petrol? converts hexane into cyclohexane.
b)
What other methods are used to increase the d)
Use structural formulae to show how reforming
octane number of petrol? converts octane into 1,4-dimethylbenzene and
hydrogen.
11 a) Why is cracking important?
e)
Use molecular formulae to show how an
b)
What conditions are used for catalytic
alkane with 16 carbons can be cracked to form
cracking?
molecules of 2,2,4-trimethylpentane, propene
12 a)
Use displayed formulae to show how catalytic and ethene.
cracking converts hexane into butane and
ethene.
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6.2.5 Combustion and air pollution
The complete combustion of fuels containing alkanes (Section 6.2.3) provides
energy to heat homes and to power vehicles but also releases carbon dioxide
into the atmosphere. This carbon dioxide is enhancing the greenhouse effect
and is responsible for global warming and climate change. However, motor
vehicles and power stations also produce other pollutants during combustion
and these can affect the quality of the air we breathe.
Air pollution from motor vehicles
Engines that burn petrol or diesel fuels can pollute the air for three main reasons:
● they do not burn the fuel completely
● the fuel contains impurities
● they run at such a high temperature that nitrogen and oxygen in the air
can react.
Although a controlled quantity of air and fuel enters the cylinder of an
engine (Figure 6.2.9) before being compressed and burned, some of the fuel
Tip may not burn completely and some may not burn at all.
Oxygen always combines with hydrogen
in preference to carbon. So when Incomplete combustion
alkanes are sparked in a limited If the supply of air is insufficient, alkanes burn to form water together with
supply of air, if there is not enough either carbon monoxide or carbon.
oxygen present to produce water, no
CH4(g) + 112 O2(g) → CO(g) + 2H2O(l)
combustion occurs at all.
C8H18(l) + 4 12 O2(g) → 8C(s) + 9H2O(l)
Carbon monoxide is a toxic gas that combines strongly with haemoglobin so
that blood can carry less oxygen. In low doses this puts a strain on the heart;
in higher doses, it kills.
Carbon particles are also produced when combustion is incomplete, especially
Tip in diesel engines. The soot formed includes fine particles and nanoparticles
PM-10s, Particulate Matter with a which can penetrate deep into the lungs. This causes short-term symptoms
diameter of under 10 μm (1 micrometre such as coughing and headache but long-term exposure is suspected of
(μm) = 10−6 metre) is of particular leading to more serious health problems including heart disease and lung
concern as it can penetrate so deeply cancer. Modern diesel engines use a diesel particulate filter to capture these
into the lungs. Levels of PM-10s are carbon particles which are then automatically burned to remove them from
monitored daily and under European the filter.
Union air quality laws, levels of PM-10s
Unburned hydrocarbons which may enter the atmosphere from incomplete
must not exceed 75 μg m−3 on more
combustion or evaporation of fuel include benzene, which is carcinogenic.
than 35 days in a year.
Careful monitoring of the air : fuel ratio is used to minimise the release of
unused fuel.
Impurities in the fuel
Sulfur compounds are the main impurities in crude oil and must be removed
before the petroleum products are used as fuels. This is to ensure that sulfur
dioxide emissions are reduced as much as possible as these can lead to acid
rain. Removal of sulfur is also necessary to prevent damage which sulfur
causes to the catalyst in catalytic converters.
186 6.2 Hydrocarbons: alkanes and alkenes
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Crude oil is mixed with hydrogen and passed over a hot catalyst. Any sulfur
compounds react with hydrogen to form hydrogen sulfide and a hydrocarbon.
For instance, ethanethiol, C2H5SH, is converted into ethane.
C2H5SH(l) + H2(g) → C2H6(g) + H2S(g)
The hydrogen sulfide produced is then oxidised to sulfur in two steps.
2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(l)
2H2S(g) + SO2(g) → 3S(s) + 2H2O(l)
Huge quantities of sulfur are obtained in this way and most of it is used to Tip
make sulfuric acid.
The same reaction happens during a
thunderstorm where lightning provides
Formation of oxides of nitrogen the energy needed to split nitrogen
When petrol vapour burns in an internal combustion engine, the temperature molecules into atoms. The rain
can rise to 2800 °C. At this temperature, sufficient activation energy is which falls during a thunderstorm is,
available for nitrogen to react with oxygen and form nitrogen monoxide. therefore, a very dilute solution of nitric
N2(g) + O2(g) → 2NO(g) acid, but it also produces nitrates in the
soil and these promote plant growth.
Reaction of NO with more oxygen produces nitrogen dioxide.
2NO(g) + O2(g) → 2NO2(g)
These oxides of nitrogen, NO and NO2, sometimes referred to as NOx,
react with water and more oxygen to form nitric acid, which leads to the
production of acid rain.
4NO2(g) + 2H2O(l) + O2(g) → 4HNO3(l)
In bright sunshine, nitrogen dioxide molecules break down into nitrogen
monoxide and oxygen radicals. These oxygen atoms combine with oxygen
molecules to form ozone. Ozone itself is a serious pollutant, but it can lead to
further harm in still, sunny weather near cities when it mixes with unburned
hydrocarbons. The reaction of ozone with hydrocarbons forms a complex
mixture of irritant chemicals that in, the absence of any wind, builds up to Figure 6.2.13 Photochemical smog over
create a photochemical smog (Figure 6.2.13). Hong Kong, China.
Catalytic converters
Catalytic converters improve air quality by removing the pollutants that
would otherwise be released from car exhausts. In the presence of the catalyst,
carbon monoxide and unburned hydrocarbons react with nitrogen oxides
to form carbon dioxide and water. The converter contains a honeycomb of
ceramic material coated with a thin layer of metals such as rhodium, platinum
or palladium. The large surface area increases the rate of reaction (Section 9.4)
so that 90% of the pollutant gases are removed in a fraction of a second.
2CO(g) + 2NO(g) → 2CO2(g) + N2(g)
C8H18(g) + 25NO(g) → 8CO2(g) + 12 12 N2(g) + 9H2O(g)
Tip
These are redox reactions (Section 3.4). In the first example, the toxic reducing agent,
CO, is oxidised to CO2 by a harmful oxidising agent (NO), which is itself reduced to
harmless nitrogen.
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Test yourself
13 Write equations for:
a) the complete combustion of butane
b) the incomplete combustion of pentane to form only gaseous
products
c) the incomplete combustion of hexane to form products which
include a solid product
d) the complete combustion of ethanethiol.
14 The flame on a Bunsen burner changes as the air hole is closed.
Describe the change and explain why a Bunsen burner should not be
used for heating if the air hole is closed.
15 a) Acid rain can contain sulfur compounds or nitrogen compounds.
Explain how each type of acid rain is formed.
b) Explain how a catalytic converter removes pollutant gases from
car exhaust fumes.
6.2.6 Alternative fuels
Fossil fuels, such as petroleum and natural gas, were formed millions of years
ago by the anaerobic decomposition of remains of organisms that settled
at the bottom of the sea. These fuels are considered to be non-renewable
because supplies of them are being consumed at a much faster rate than they
are being replenished. Therefore, supplies of fossil fuels will eventually run
out and alternative supplies of energy must be used to provide the world’s
energy needs.
The combustion of fossil fuels produces carbon dioxide, so concerns both
about supplies of fossil fuels and also the enhanced greenhouse effect are
encouraging people and governments to reduce their CO2 emissions and
seek a more sustainable development. This involves planning to live within
the means of the environment in order that the Earth’s natural resources are
not destroyed, but remain available for future generations.
Any attempt to reduce CO2 emissions means seeking alternatives to fossil
fuels. These alternatives include nuclear, solar, wind and wave power, but
also include renewable energy supplies such as biofuels – bioethanol and
biodiesel.
Crops take in carbon dioxide from the air as they photosynthesise,
making sugars and vegetable oils that can be used to produce biofuels.
When the biofuels burn, the carbon dioxide that is taken up during
Key term photosynthesis is returned to the air. This analysis suggests that the use
of biofuels should have no overall effect on the level of carbon dioxide
A process is termed carbon neutral if in the atmosphere. Because of this, biofuels are sometimes described as
the carbon dioxide (or other greenhouse carbon neutral.
gases) released is balanced by actions
However, this analysis ignores the carbon dioxide released from fossil fuels
which remove an equivalent amount of
during the mechanical planting, harvesting and processing of the crop, and in
carbon dioxide from the atmosphere.
the manufacture of fertilisers applied to the crop during the growing season.
188 6.2 Hydrocarbons: alkanes and alkenes
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The principal biofuels now being used are bioethanol and biodiesel.
Bioethanol is manufactured by fermenting carbohydrates such as starch and
sugar in crops like sugar cane. Countries, such as Brazil, with limited natural
oil supplies use fuels consisting of bioethanol on its own or in mixtures
(Figure 6.2.14).
Fermentation converts starch to glucose, and then glucose to ethanol and
carbon dioxide. The process is catalysed by enzymes in yeast.
C6H12O6(aq) → 2CO2(g) + 2C2H5OH(aq)
glucose carbon dioxide ethanol
Figure 6.2.14 Brazil is the leading country
Biodiesel is produced by extracting and processing the oils from crops such to use bioethanol. This photo shows a
as rapeseed. petrol station in São Paulo. The ethanol
Studies of the overall impact of biofuels have produced conflicting results. fuel (‘álcool’) is cheaper than the petrol
Some have even suggested that more energy is expended in making the (‘gasolina’).
bioethanol than is available from burning the fuel. A more optimistic study
has suggested that ethanol from maize in the USA does reduce greenhouse
gas emissions, but only by about 12% compared to petrol.
Reduction in CO2 emissions is more favourable for biodiesel from vegetable
oil. Biodiesel from soya beans can reduce emissions by 41% compared with
conventional diesel fuel. This is mainly because energy is not needed for
distillation during the production of the fuel. In addition, far fewer fertilisers
and pesticides are used in growing soya beans.
Instead of making bioethanol from maize, a better solution is to produce
ethanol from non-food sources, such as woody plants and agricultural wastes
including straw. Using micro-organisms to break down cellulose to sugars does
not compete with food supplies and makes it possible to process large amounts
of waste. However, this approach is still at the research and development stage.
The conditions for producing bioethanol in Brazil are more favourable. The
ethanol is manufactured by fermenting sugars from sugar cane. The refineries that
make bioethanol in Brazil have the advantage that they can meet all their energy
needs for heating and electricity by burning the sugar-cane waste. However,
there are several negative aspects of this industry in Brazil where, among other
things, large-scale deforestation has been carried out to make way for sugar cane
plantations. Also, when land use is switched from food crops to biofuel crops,
food prices rise and food production is displaced. This causes cropland expansion
elsewhere, threatening tropical forests and causing loss of biodiversity.
Test yourself
16 Why are governments around the world becoming increasingly
concerned about our use of fossil fuels and the need for sustainable
development?
17 Suggest three ways in which our use of fossil fuels might be
reduced.
18 Biofuels are sometimes described as ‘carbon neutral’. Why is this
and to what extent is it true?
19 Why does the use of bioethanol in Brazil have a smaller carbon
footprint compared to the use of bioethanol in the USA?
6.2.6 Alternative fuels 189
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6.2.7 Alkenes
The alkenes form a series of organic compounds in which the functional
Key term group is a carbon–carbon double bond, C=C. Because of this, their molecules
have two atoms of hydrogen fewer than the corresponding alkane, and their
Unsaturated compounds contain one
general formula is CnH2n. As they have less than the maximum content of
or more double or triple bonds between
hydrogen, they are described as being unsaturated. The term is applied
atoms in their molecules.
to alkenes and also used describe unsaturated fats which have C=C double
bonds in their hydrocarbon chains.
Tip Names and structures
The name of an alkene is based on the name of the corresponding alkane
The general formula CnH2n applies
with the ending -ene in place of -ane (Figure 6.2.15).
to cyclic alkanes as well as alkenes.
Both types of hydrocarbon have 4 3
two hydrogen atoms fewer than the H H CH3 CH2 2 1
H
corresponding alkane. C C C C
H H H H
ethene but-1-ene
4 1
CH 3 H CH3 3 2 CH3
C C C C
H H H H
propene but-2-ene
Figure 6.2.15 The names and structures of the four simplest alkenes.
Where necessary, a number in the name shows the position of the double
bond, as in the structural isomers but-1-ene and but-2-ene. Counting starts
from the end of the chain that gives the lowest possible number in the name.
This number indicates the first of the two atoms connected by the double
bond. In but-1-ene, for example, the double bond is between carbon atoms
numbered 1 and 2.
Physical properties
Like the alkanes, the melting and boiling temperatures of alkenes increase as
the number of carbon atoms in the molecules increases. Ethene, propene and
the butenes are gases at room temperature. Alkenes, like other hydrocarbons,
do not mix with or dissolve in water.
The double bond in alkenes
Chemists have extended the theory of atomic orbitals (Section 1.6) to
describe the distribution of electrons in molecules. This molecular orbital
theory is helpful in discussing the bonding and reactivity of alkenes.
Molecular orbitals result when atomic orbitals overlap, forming bonds between
atoms. The shape of a molecular orbital shows the regions in space where there is
a high probability of finding electrons. A sigma (σ) bond is a single covalent bond
formed by a pair of electrons in an orbital in a molecule with the electron density
concentrated between two nuclei. Free rotation is possible around single bonds.
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Sigma bonds can form by the overlap of two s orbitals, an s orbital and a
p orbital, or two p orbitals (Figure 6.2.16).
Figure 6.2.16 Examples of sigma bonds in
molecules.
1s 1s
hydrogen atoms hydrogen molecule, H2
1s 3p
hydrogen chlorine hydrogen chloride, HCl
atom atom
A pi (π) bond is the type of bond found in molecules with double and triple
Tip
bonds. The bonding electrons are in a π orbital formed by the sideways overlap
of two atomic p orbitals. In a π bond, the electron density is concentrated In order for a π bond to form, two
in two regions – one above and the other below the plane of the molecule, atomic p orbitals must overlap
on either side of the line between the nuclei of the two atoms joined by the sideways. If there is any rotation of the
bond (Figure 6.2.17). Rotation around a double bond is restricted because of carbon–carbon bond, the p orbitals can
these regions of high electron density. no longer overlap sideways.
H H H H Figure 6.2.17 The π bond in ethene.
C C C C
H H H H
p orbital orbital
6.2.8 E/Z isomerism Key terms
X-ray diffraction studies show that the ethene molecule is planar (Figure 6.2.15) E/Z isomerism occurs where there is
with the three atoms around each carbon atom arranged trigonally at restricted rotation about a bond and
approximately 120°. However, the CH2 groups in ethene cannot be rotated also different groups are attached to the
around the carbon–carbon double bond. In ethane and other alkanes, it is carbon atoms at each end of the bond.
possible to rotate the whole molecule around a single σ bond, because this
Stereoisomerism occurs when
does not affect the overlap of orbitals (Figure 6.2.18). But with a double bond,
molecules with the same molecular
rotation would involve breaking the π bond and this requires more energy
formula and the same structural
than the molecules possess at normal temperatures. So, free rotation is not
formula have a different spatial
possible around the carbon–carbon double bond in alkenes and this gives rise
arrangement of bonds.
to E/Z isomerism, which is a form of stereoisomerism.
H H H CH 3 Figure 6.2.18 Free rotation can occur
These two structures are the
same compound because the around a single σ bond.
C C ends of the molecule can rotate C C
CH3 CH3 freely around the single bond. CH 3 H
H H H H
E/Z isomerism involves molecules with
● restricted rotation about a bond
● different groups attached to the carbon atoms at each end of the bond.
Restricted rotation usually involves C=C double bonds but can also involve
single bonds in cyclic compounds (see the Activity on page 192: Renaming
cis–trans as E/Z isomers).
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Activity
Renaming cis–trans as E/Z isomers
Cis–trans isomerism arises in compounds containing carbon– This is where the cis–trans naming system breaks down and it
carbon double bonds when there are two different groups on becomes necessary to use the E/Z naming system. In fact, the
both carbon atoms of the double bond. Different isomers result E/Z naming system was developed in order to name complex
because of restricted rotation around the double bond. alkenes in naturally occurring materials, such as the red pigment
in tomatoes. One molecule of this pigment has 13 C=C bonds.
The existence of a ring structure in a molecule can also restrict
rotation and give rise to cis and trans isomers. Look at the The E/Z naming system
structure of trans-1,2-dichlorocyclobutane in Figure 6.2.19. Its cis ● Look at the atoms bonded to the first carbon atom in the
isomer is cis-1,2-dichlorocyclobutane. C=C bond. The atom with the highest atomic number takes
priority.
H Cl ● If two atoms with the same atomic number, but in different
C C groups, are attached to the first carbon atom, then the next
H H
bonded atom is taken into account. Thus, CH3CH2− has
H H
C C priority over CH3−.
● This consideration is then repeated with the second carbon
H Cl
atom in the C=C bond.
Figure 6.2.19 The structure of trans-1,2-dichlorocyclobutane.
4 a) What is the order of priority among Br, C in CH3, Cl and H?
1 Draw the displayed formula of cis-1,2-dichlorocyclobutane. b) What is the priority among CH3−, CH3CH2−, CH3O− and
2 There is another pair of cis–trans isomers named HOCH2−?
dichlorocyclobutane and a separate structural isomer.
Name and draw displayed formulae for these three 5 Look again at 2-bromo-1-chloroprop-1-ene in Figure 6.2.20.
molecules. a) What is the priority between H− and Cl− attached to the
first carbon atom in the double bond?
Now look at the isomer of 2-bromo-1-chloroprop-1-ene in b) What is the priority between Br− and CH3− attached to the
Figure 6.2.20. second carbon atom in the double bond?
H CH3 If the two groups of highest priority are on the same side
C C of the double bond, the isomer is designated Z- (from the
German ‘zusammen’ meaning ‘together’), and if the two
Cl Br groups of highest priority are on opposite sides of the
Figure 6.2.20 An isomer of 2-bromo-1-chloroprop-1-ene. double bond, the isomer is designated E- (from the German
‘entgegen’ meaning ‘opposite’).
3 Draw the other cis–trans isomer of 2-bromo-1-chloroprop-1- 6 Draw the displayed structure of Z-2-bromo-1-chloroprop-1-ene.
ene. 7 Use the E/Z system to name the cis and trans isomers of:
But, which of these isomers is cis and which is trans? The rule a) but-2-ene
normally used to name the isomers as cis or trans is: b) 1,2-dichlorocyclobutane.
● in the cis isomer, similar groups are on the same side of the 8 Name the two compounds in Figure 6.2.21 using the cis–trans
double bond system and also using E/Z. Comment on your answers.
● in the trans isomer, similar groups are on opposite sides of the
Cl H Cl Br
double bond.
C C C C
In 2-bromo-1-chloroprop-1-ene, there are four different groups on
H Cl H Cl
the atoms joined by the double bond. So the normal rule, which a) b)
requires one group to be the same on both carbon atoms, cannot
Figure 6.2.21
be used.
192 6.2 Hydrocarbons: alkanes and alkenes
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The simplest type of E/Z isomerism occurs when at least one group on each
carbon is the same. This is called cis-trans isomerism.
In the cis isomer, similar groups are on the same side of the double bond
(in Latin, cis means ‘on the same side’). In the trans isomer, similar groups
are on opposite sides of the double bond (in Latin, trans means ‘opposite’ or
‘across’). But-2-ene can exist as a cis isomer where the methyl groups are on
the same side or a trans isomer where the methyl groups are on opposite sides
(Figure 6.2.22).
H H These two structures CH 3 H
are different compounds
C C C C
because the double bond
CH3 CH3 stops rotation. H CH 3
melting temperature = – 139°C melting temperature = – 106°C
cis-but-2-ene trans-but-2-ene
Figure 6.2.22 Cis and trans isomers of but-2-ene – different compounds with different
melting temperatures, boiling temperatures and densities.
Naming isomers becomes more complicated for examples such as 2-bromo-
1-chloroprop-1-ene (Figure 6.2.20), where no substituents on the double
bond are the same. For these molecules it is necessary to use the E/Z naming
system (see Activity: Renaming cis–trans as E/Z isomers).
Test yourself
20 Why do you think the bond angles around each carbon atom in
ethene are approximately 120°?
21 Why is rotation about a carbon–carbon double bond restricted?
22 Which of the following unsaturated compounds have cis–trans
isomers: but-1-ene, 1,1-dichloropropene, pent-2-ene, buta-1,3-diene?
23 a) Draw and name the displayed structures of the three isomers of
C2H2Br2.
b) What types of isomerism do they show?
6.2.9 Chemical reactions
of the alkenes
Combustion
Alkenes are hydrocarbons so like alkanes will burn in air and oxygen. The flame
when an alkene burns is more smoky than that of an alkane with the same number
of carbon atoms because the alkene contains a higher percentage of carbon.
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) ΔcH 1 = −1411 kJ mol−1
Alkenes are rarely burned as fuels because they react in other more useful
ways. The characteristic reactions of alkenes are addition reactions (Section
6.1.6), in which small molecules such as H 2, Cl 2 and HBr add across the
double bond to form a single product.
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Tip Addition of hydrogen
Hydrogen adds to C=C double bonds in alkenes such as propene at room
When describing an organic reaction,
temperature in the presence of a platinum or palladium catalyst, or on heating
always write an equation, name the
to 150 °C in the presence of a nickel catalyst (Figure 6.2.23).
reagents and the products, and state
the conditions (temperature, pressure, H H
catalysts). CH3 H Ni catalyst
C C + H2 CH 3 C C H
150 °C
H H
H H
propene propane
Figure 6.2.23 Addition of hydrogen to propene.
This process is known as ‘catalytic hydrogenation’ and is used in the
manufacture of margarine from unsaturated vegetable oils in palm seeds and
sunflower seeds. The advantage of using a solid metal catalyst is that it can be
held in the reaction vessel as the reactants flow in and the products flow out.
There is no difficulty in separating the products from the catalyst.
Vegetable oils are liquids containing carbon–carbon double bonds, nearly all
of which are cis-double bonds. During hydrogenation, some of these double
bonds are converted to carbon–carbon single bonds by addition of hydrogen.
The change in structure turns oily unsaturated liquids into soft saturated
fatty solids like margarine. However, research in the 1960s started to show
that saturated fats contribute to heart disease.
Fats that have been partially saturated by hydrogenation often contain trans-
fats and, during the 1990s, evidence began to suggest that these trans-fats
could also lead to high cholesterol levels in the blood and so also cause heart
disease. In some parts of the world, trans-fats such as E-octadec-9-enoic acid
(Figure 6.2.24) were banned by law. Elsewhere manufacturers of margarine
and vegetable fat spreads agreed voluntarily to remove artificial trans-fats
from their products. In UK, the agreed date for their removal was the end
of 2012. A new strategy adopted by the manufacturers is to hydrogenate
a proportion of the vegetable oils completely. This removes all the double
bonds, hence the trans-fats. They then blend this product with untreated oil
to make a spread of the correct texture, but with no trans-fats (Figure 6.2.25).
COOH
Figure 6.2.24 E-octadec-9-enoic acid is the main trans-fat found in hydrogenated
vegetable oils.
Figure 6.2.25 Manufacturers of vegetable
fat spreads, such as Bertolli, often claim
that their products contain virtually
no trans-fats, and that they are rich in
unsaturated fat and low in saturated fats
compared to butter. These spreads are
claimed to be a healthier option than
butter.
194 6.2 Hydrocarbons: alkanes and alkenes
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Addition of halogens Tip
Chlorine and bromine add rapidly to alkenes at room temperature
The reaction of bromine water (aqueous
(Figure 6.2.26). The products are dichloroalkanes and dibromoalkanes and
bromine, Br 2(aq)) with hydrocarbons
the reaction proceeds by an electrophilic addition mechanism (Section 6.2.10).
is a useful test for alkenes and the
Fluorine reacts explosively with small alkenes, such as ethene and propene,
carbon–carbon double bond.
but the reaction of iodine with alkenes is slow.
Unsaturated hydrocarbons with a C=C
H H bond, such as ethene and cyclohexene,
CH 3 H 3 2 1
C C + Br2 CH 3 C C H
quickly decolourise yellow/orange
H H bromine water, producing a colourless
Br Br mixture (see Section 6.2.10).
propene 1,2-dibromopropane Saturated hydrocarbons, such as
ethane and cyclohexane, have no
Figure 6.2.26 Addition of bromine to propene.
reaction with bromine water and
the yellow/orange colour of bromine
Addition of hydrogen halides remains.
Hydrogen halides react readily with alkenes at room temperature, forming
halogenoalkanes. For example, hydrogen bromide reacts with ethene in the
gas phase to form bromoethane (Figure 6.2.27). The reaction follows an
electrophilic addition mechanism (Section 6.2.10).
The other hydrogen halides, HCl and HI, react in a similar way.
H H Figure 6.2.27 Addition of hydrogen
H H bromide to ethene.
C C + H Br H C C H
H H
H Br
ethene bromoethane
Addition of steam
Alkenes react with steam in the presence of an acid catalyst to produce
alcohols. The direct catalytic hydration of ethene, for example, produces
ethanol in a reversible reaction between ethene and steam (Figure 6.2.28).
The phosphoric acid catalyst is adsorbed on silica and the conditions used are
570 K and a pressure of 6.5 MPa (65 atmospheres) (see Section 10.4 Activity:
The manufacture of ethanol).
H H Figure 6.2.28 Hydration of ethene to
H H produce ethanol.
C C + H2O H C C H
H H
H OH
ethene
ethanol
Oxidation by potassium manganate(vii)
Potassium manganate(vii) oxidises alkenes – the products depend on the
conditions. A dilute, acidified solution of potassium manganate(vii) converts
an alkene to a diol at room temperature (Figure 6.2.29). At the same time,
purple manganate(vii) ions, MnO4−, are reduced to very pale pink Mn 2+
ions, and the purple colour disappears if there is excess alkene. So, like the
6.2.9 Chemical reactions of the alkenes 195
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reaction with aqueous bromine (Section 6.2.10), this reaction with dilute
acidified potassium manganate(vii) solution can be used to distinguish
between unsaturated and saturated hydrocarbons.
H H
H H
Tip C C + [O] + H2O
dilute acid
H C C H
MnO4–(aq)
The symbol [O] represents the oxidising H H
agent and can be used in simplified OH OH
equations for such reactions. The ethene ethane-1,2-diol
mechanism for this reaction is not
Figure 6.2.29 The reaction of ethene with dilute acidified manganate(vii) ions producing
required at A Level.
ethane-1,2-diol.
Test yourself
24 a) Write the structures and names of the products and the
conditions for the reactions when propene reacts with:
i) hydrogen ii) chlorine.
b) Write the structures and names of the products and the
conditions for the reactions when but-2-ene reacts with:
i) hydrogen chloride ii) steam.
25 State what you would see when a few cm3 of acidified potassium
manganate(vii) solution is added to a gas jar of propene and the
mixture is shaken. Write an equation for the reaction and name the
product.
26 Catalytic hydrogenation is sometimes used in the manufacture of
spreads, such as ‘Flora™’, from vegetable oils.
a) What is meant by the term ‘catalytic hydrogenation’?
b) E xplain the terms ‘saturated’ and ‘unsaturated’ as applied
to organic compounds such as those in low-fat spreads and
vegetable oils.
c) Why are some unsaturated fats, such as olive oil and sunflower
oil, thought to be healthier foods than more saturated fats, such
as cream, and which type of unsaturated fats are now known to
be unhealthy?
6.2.10 Mechanism for electrophilic
addition to alkenes
Most of the reactions of alkenes are electrophilic addition reactions.
Electrophiles (Section 6.1.7) attack the electron-rich region of the double
bond in alkenes and, in particular, the exposed π bond. Electrophiles that
add to alkenes include hydrogen bromide, bromine and water in the presence
of H+ ions.
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The addition of hydrogen bromide to ethene
Hydrogen bromide molecules are polar (Section 2.5). The hydrogen atom,
with its δ+ charge at one end of the molecule, acts as an electrophile
(Figure 6.2.30).
Tip
H H H H
H H step 1 step 2
The δ+ hydrogen in HBr acts as an
C C H C C H H C C H electrophile. Because of the electron-
+
H H rich double bond, ethene donates
H H – H Br a pair of electrons and so acts as a
Br
Br nucleophile. The reaction, however,
is always described as electrophilic
Figure 6.2.30 Electrophilic addition of hydrogen bromide to ethene. The reaction takes
addition of the inorganic species to the
place in two steps.
organic compound.
In the first step of the reaction, an HBr molecule approaches an ethene
molecule. The δ+ hydrogen end of the HBr is attracted towards the electron-
rich double bond. As the HBr molecule gets even closer, heterolytic fission Tip
of the π bond occurs. The electrons in the π bond form a covalent bond
to the hydrogen atom and, at the same time, heterolytic fission of the HBr A curly arrow represents the movement
bond also occurs. Electrons in the H−Br bond are taken over by the bromine of a pair of electrons. In step 1 each
atom, producing a Br− ion. curly arrow starts from the bond that is
breaking.
The other product of step 1 is the highly reactive carbocation CH3CH2+.
This reacts immediately with the bromide ion; a pair of electrons from Br− In step 2, the curly arrow starts from a
is used to form a covalent bond with the electron-deficient carbon of the lone pair on the bromide ion.
carbocation. This rapid second step forms bromoethane, CH3CH2Br.
The addition of bromine to ethene Key term
The addition of bromine to ethene is also an electrophilic addition. Bromine A carbocation is a reactive species
molecules are not polar, but they become polarised as they approach the which contains a carbon atom which
electron-rich region of the double bond. Electrons in the double bond repel has a positive charge.
electrons in the bromine molecule, so the end of the bromine molecule
(nearer the double bond) becomes δ+ and electrophilic (Figure 6.2.31). In
this case, the product is 1,2-dibromoethane.
Tip
H H H H
H H step 1 step 2
Always show relevant dipoles in these
C C H C C H H C C H mechanisms to make it clear that the
+
H H electrophilic δ+ end of a molecule is
Br Br Br Br
–
attracted to the electron-rich double
Br
Br bond.
Figure 6.2.31 Electrophilic addition of bromine to ethene.
The reaction of bromine water (aqueous bromine, Br2(aq)) with hydrocarbons
is a useful test for alkenes (see Section 6.2.9).
When bromine water is shaken with ethene, bromine molecules react
with the electron-rich region of the C=C bond forming an intermediate
carbocation, the same as with liquid bromine. But in aqueous solution, this
carbocation can react either with Br− ions or with water molecules in the
bromine water to form a mixture of 1,2-dibromoethane and 2-bromoethanol
6.2.10 Mechanism for electrophilic addition to alkenes 197
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(Figure 6.2.32). There are many more water molecules than bromide ions
present, so water molecules are more likely to collide with the carbocation.
The major product of the reaction of ethene with bromine water is, therefore,
2-bromoethanol.
H H
H C C Br
Br–
H H p2
H H ste Br H
step 1 1,2-dibromoethane
C C H C C+
ste
H H Br H p3
H H
Br H2O
H C C OH + H+
Br
Br H
2-bromoethanol
Figure 6.2.32 Reaction of bromine water with ethene producing 1,2-dibromoethane and
2-bromoethanol.
When bromine is added to water, some of it reacts with the water to form
a mixture of hydrobromic acid and bromic(i) acid. Hydrobromic acid is a
strong acid and ionises immediately. In comparison, bromic(i) acid is a weak
acid which remains un-ionised as polar HOδ−−Brδ+ molecules.
Br2(l) + H2O(l) ⇋ H+(aq) + Br−(aq) + HOBr(aq)
The reaction of ethene with bromine water can be represented by the
addition of HOBr to ethene (Figure 6.2.33).
H H H H
C C + Br2 + H 2O H C C H + HBr
H H Br OH
ethene 2-bromoethanol
Figure 6.2.33 Formation of 2-bromoethanol when bromine water reacts with ethene.
These mechanisms for electrophilic additions are not simply hypotheses
or good ideas. They are supported by significant experimental evidence to
help our understanding of the reactions at a molecular level. This evidence
includes data from studies of the kinetics of the reactions, as well as the
structures of the products formed. For instance, if the addition of HBr
is carried out in the presence of NaCl then, in addition to the expected
product, CH 3CH 2 Br, some chloroethane, CH 3CH 2Cl, is also obtained.
Similarly, if the addition of Br2 is carried out in the presence of NaCl
then 1-bromo-2-chloroethane, CH 2 BrCH 2Cl, is obtained as well as
CH 2 BrCH 2 Br.
The formation of these extra products can only be explained using the two-
step mechanism proposed. The highly reactive carbocations CH3CH2+ and
CH2BrCH2+ formed during the slow first step can be attacked either by
Br− or Cl− ions in the rapid second step of the mechanism, so two products
are formed.
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6.2.11 Addition to unsymmetrical
alkenes
When a molecule such as HBr or HCl adds to an unsymmetrical alkene, such
as propene, there are two possible products. Although it might be expected
that equal amounts of both products would be formed, in fact much more of
one product is usually produced (Figure 6.2.34).
H H
H3C C C H
major
Br H
2- bromopropane
H3C H
C C + HBr
H H
H H
minor
H3C C C H
H Br
1- bromopropane
Figure 6.2.34 Possible products when hydrogen bromide adds to propene.
Analysis of the products of the reaction between hydrogen bromide and propene
shows that much more 2-bromopropane is produced than 1-bromopropane.
This suggests that the hydrogen atom from HBr adds mainly to the carbon
atom of the double bond which already has more hydrogen atoms attached
to it. This pattern is usually called Markovnikov’s rule because it was first
reported by the Russian chemist Vladimir Markovnikov who studied many
alkene addition reactions during the 1860s.
Modern theories can explain which of the two products is more likely to
form by considering the mechanism of the reaction and the relative stability Key terms
of the intermediates (carbocations) formed.
Intermediates are atoms, molecules,
The stability of a carbocation depends on the number of alkyl groups attached,
ions or free-radicals which do not
because these alkyl groups exert an inductive effect on the carbocation.
appear in the overall equation for a
Alkyl groups have a small tendency to push electrons towards any carbon reaction, but which are formed during
atom to which they are bonded; they are said to have a positive inductive one step of a reaction and then used up
effect. in the next.
One consequence of this is that any carbon atom with a positive charge is The inductive effect describes the way
more stable the more alkyl groups are attached to it. in which electrons are either pushed
towards or pulled away from a carbon
A primary carbocation has one alkyl group attached to C+. atom by the atoms or groups to which it
A secondary carbocation has two alkyl groups attached to C+ and is more is bonded.
stable than a primary carbocation. A primary carbon is attached to one
A tertiary carbocation has three alkyl groups attached to C+ and is more other carbon. A secondary carbon
stable than a secondary or a primary carbocation. is attached to two others. A tertiary
carbon is attached to three others.
6.2.11 Addition to unsymmetrical alkenes 199
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Figure 6.2.35 shows the possible carbocation structures of C4H9+ and their
relative stabilities.
Primary carbocations Secondary carbocation Tertiary carbocation
+ + +
CH3 CH2 CH 2 CH 2 CH3 CH 2 CH CH 3 CH3 C CH 3
+ CH3
CH3 CH CH 2
CH3
increasing stability
Figure 6.2.35 Isomeric primary, secondary and tertiary carbocations, C4H9+.
In the mechanism for electrophilic addition of HBr to propene, there are
Tip two possible carbocation intermediates (Figure 6.2.36).
Halogen atoms are very electronegative
and draw electrons in a bond to inductive effect
themselves; halogen atoms are said to H H H H
have a negative inductive effect.
H3C C C H H3C C C H
+
H3C H HBr H Br H
more stable carbocation 2-bromopropane
C C major product
H H
HBr H H H H
H3C C C H H3C C C H
+
H H Br
less stable carbocation 1-bromopropane
minor product
Figure 6.2.36 The formation of primary and secondary carbocations in the reaction of
propene with HBr.
The secondary carbocation with its positive charge in the middle of the
carbon chain is slightly more stable than the primary carbocation with its
charge at the end of the chain. The secondary carbocation has two alkyl
groups pushing electrons towards the positively charged carbon atom. By
contrast, the primary carbocation only has one alkyl group pushing electrons.
This extra inductive effect helps to stabilise the secondary ion slightly more
by reducing its positive charge. Because of this extra stability, the secondary
carbocation is more likely to form. Subsequent rapid attack by bromide ions
then leads to the formation of the major product, 2-bromopropane.
Tip All electrophilic additions proceed via carbocation intermediates, so major and
The stability of a carbocation depends minor products will be formed where the alkene and the molecule added are
on the number of alkyl groups attached both unsymmetrical. For instance, the hydration of propene produces propan-
and not on the size of the alkyl group. 2-ol as the major product (Figure 6.2.37). Propan-2-ol is used as a solvent and
also used to make propanone, an important compound in the plastics industry.
CH3 CH CH 2 + H2O CH3 CH CH 3
OH
Figure 6.2.37 The hydration of propene to form propan-2-ol.
200 6.2 Hydrocarbons: alkanes and alkenes
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Test yourself
27 a) Draw a mechanism for the reaction of bromine ii) explain which carbocation is the more stable
with but-2-ene and name the product. iii) write the displayed formula and name of the
b) Write an equation for the reaction of bromine major product.
water with but-2-ene showing the structure of 29 When propene reacts with bromine in the
the major product. Explain why this product is presence of sodium chloride, a mixture of
different from that obtained in part (a). products is formed. Explain why the mixture
28 Consider the reactions of: contains:
a) hydrogen chloride with but-1-ene a) more 1-bromo-2-chloropropane than
b) hydrogen bromide with 2-methylpropene. 2-bromo-1-chloropropane
In each case: b) 1,2-dibromopropane but not
1,2-dichloropropane.
i) draw the structure of the two possible
carbocation intermediates
6.2.12 Addition polymers
from alkenes
If the conditions are right, molecules of ethene undergo addition reactions
with each other to form polythene – or more correctly, poly(ethene). Two
different kinds of polythene are manufactured – low-density polythene and
high-density polythene.
Low-density polythene (LDPE) is manufactured by heating ethene at high
pressures and high temperatures with special substances called initiators.
These initiators are often peroxides. The weak O−O single bond easily breaks
homolytically to form free radicals that initiate (start) the reaction (Figure
6.2.38). After this initiation step, the polymerisation proceeds in a series of
propagation steps: a radical reacts with ethene and a new radical is formed in
each step. The polymer chain grows stepwise until termination steps occur.
The polyethene produced has very long chains but it also has lots of branches
(which result from complex collisions between radicals). The branches prevent
the molecules from packing closely and this results in low-density material.
Initiation
R O O R R O O R
Propagation
H H H H
R O + C C R O C C
H H H H
H H H H H H H H
R O C C + C C R O C C C C
H H H H H H H H
Figure 6.2.38 Molecules of ethene can undergo stepwise addition reactions with each other.
6.2.12 Addition polymers from alkenes 201
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Activity
A more sustainable future for polymers
Manufactured goods, such as polymers, are produced at a 1 Suggest three significant steps that should be taken to
heavy cost to society and the environment. In general, their reduce the consumption of crude oil.
production uses up scarce natural resources, consumes Large amounts of non-renewable fossil fuels are used in
non-renewable fuels and energy, and disrupts wildlife and the extracting crude oil, in transporting it to refineries, in processing
countryside. at the refineries and then in the production of specific products
Today’s industrialists and manufacturers are expected to assess such as plastic bags, bottles, toys and fibres.
the life cycles of their products more closely, in order to find After production, various polymer products are moved to shops
ways in which industry and society can contribute to a more where they are sold, used and then thrown away in landfill sites.
sustainable use of materials. This life cycle assessment (LCA) This is very wasteful.
is part of the legislation designed to protect the environment.
2 What are local councils and the government now doing to
In terms of energy and materials, manufactured goods such as reduce landfill waste?
computers, clothes and cars go through a life cycle with three 3 Why is this approach so important?
distinct phases:
Most plastic waste can be melted and then remoulded. Recycling
● birth – raw materials and energy are used to make the goods would seem an obvious way forward, but there are problems in
● life – chemicals and energy are needed to maintain and use
sorting and separating different types of polymer, particularly
the goods when some products include more than one type of polymer.
● death – energy, and possibly space, is needed to recycle the
Polymers are classified into seven types for recycling purposes
goods or dispose of them as waste. (see Table 6.2.1). The symbol of each type is clearly visible on
Here are some issues and questions that are being raised plastic objects.
about the life cycle of polymers from the extraction of crude Sorting plastic waste into these different types is both difficult
oil, through the production and use of commercial polymer and costly, and until recently has been done by hand. Automatic
products, to their eventual disposal. methods have now been developed which use optical techniques,
Only about 4% of crude oil is used to make polymers. Most such as the use of infrared cameras. These are set to detect
crude oil is used to provide fuel for transport, heating homes specific readings that are unique for each material. Other materials
and industry. Although ‘fracking’ may provide further resources, are then ejected from the conveyor belt using a jet of air.
if crude oil continues to be used at the present rate, known 4 Suggest any difficulties that an optical detector might have
reserves are unlikely to last much into the next century. with a supply of household plastic waste.
• recycling
• reuse
• energy recovery
mining the ore • greater
efficiency
• less use of
materials
metals, glass and
polymers in use
extracting oil • design for long life landfill site
Figure 6.2.39 Processes and products can be redesigned to slow down the rate at which valuable Figure 6.2.40 A plastic bag that
resources are transformed into waste. is compostable.
202 6.2 Hydrocarbons: alkanes and alkenes
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Table 6.2.1 Types of plastic for recycling. produced. Modern incinerators use very high temperatures
Polymer Examples of applications Symbol to ensure the breakdown of any such gases into harmless
types products. Also, before the flue gases are released into the
atmosphere they are cleaned to remove any pollutants.
PETE or PET
Bottles for water and fizzy 1 6 a) Is incineration better than landfill? Explain your answer.
polyethylene
drinks b) Why are people often opposed to incinerators in their area?
terephthalate PET
c) What environmental problems does incineration make worse?
HDPE d) Gases including SO2, NOx and HCl are produced in an
Bottles for milk, juices, 2
high density incinerator. How do chemists ensure that they are removed
detergents
polyethylene HDPE from the waste gases and do not enter the atmosphere?
When plastic bags were first produced and started to replace
PVC
Clingfilm, pipes, window 3 paper bags in the 1950s, they were hailed as a great success.
polyvinyl
and door frames Chemists had developed a new material that would not
chloride PVC
disintegrate when it got wet as a paper bags did. However the
LDPE invention was just too good and its lack of any weakness has
Carrier bags, bin liners and 4
low density led to the problem of plastic waste today.
packaging films
polyethylene LDPE
So chemists have now developed different materials to solve
the waste problem. One solution has been to incorporate weak
PP Margarine tubs, 5 links in the addition polymer chain. Condensation polymers,
polypropylene microwaveable meal trays such as starch or cellulose, contain polar functional groups
PP
which can be attacked by nucleophiles such as water. You
will study condensation polymers in the second year of the
PS In expanded form, used as 6 A Level course. This hydrolysis breaks the polymer chain, so
polystyrene packaging and insulation
PS incorporating a few per cent of starch weakens the polymer and
allows degradation into dust.
Types that do not fall into
Other any of the above categories 7 Perhaps a better solution is to produce bags (Figure 6.2.40)
– or mixtures of polymers made entirely from condensation polymers such as PLA
OTHER
(polylactic acid) (Figure 6.2.41). These bags are completely
biodegradeable and when used to wrap waste material will
In some cases, it is possible to decompose the polymers to compost in months, rather than tens of years for traditional
provide a feedstock for cracking. The polymers are shredded polythene bags. However, as with biofuels, growing the raw
and heated in the absence of air in a process called pyrolysis. material for them competes with food production.
The vapours produced are cracked and the products distilled to
produce liquid fuel, including diesel and kerosene. 7 Why do you think that traditional polythene bags are still
used? What is likely to bring an end to their use?
5 a) Suggest where the energy comes from for the pyrolysis of
polymers at high temperature. CH3
b) Why are polyethylene or polypropylene used for pyrolysis
O CH C
but PVC is not?
O n
Some polymers that are difficult to recycle can be burned and
used as fuel. Incineration reduces the volume of the waste Figure 6.2.41 The repeat unit of PLA, which despite its common
by a huge amount, but care is needed as toxic gases may be name is actually a polyester made from lactic acid.
6.2.12 Addition polymers from alkenes 203
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High-density polythene (HDPE) is manufactured at relatively low pressure
Key term and low temperature with a special catalyst. This produces extra long chains
with very little branching, so the chains pack closer.
Addition polymerisation is an addition
reaction in which small molecules, Processes like these are called ‘addition polymerisations’. During addition
called monomers, join together forming polymerisation, small molecules like ethene, known as monomers, add to
a giant molecule, called a polymer. each other to form a giant molecule called a polymer. Addition polymers may
contain several thousands of monomer units. A polymer can be represented
by a repeat unit (Table 6.2.2). These units ignore the RO− group present on
H H
H H each end of a chain as they make up a negligibly tiny proportion of the whole
n C C C C polymer molecule. An overall equation for the polymerisation of ethene is
shown in Figure 6.2.42.
H H H H
n
Figure 6.2.42 An overall equation for the Polythene is the most important addition polymer. Two other useful
polymerisation of ethene. polymers are poly(propene) and poly(chloroethene) or PVC. These are also
manufactured by addition polymerisation. Polythene, polypropene and PVC
are soft, flexible and slightly elastic – because of this, they are often called
Tip plastics. One of their major advantages, apart from relatively low cost, is that
The abbreviation PVC comes from the they are almost chemically inert. This lack of reactivity, as well as being a
original name for the polymer, which benefit, can also be a problem where disposal of these plastics is concerned
was polyvinyl chloride. (see Activity: A more sustainable future for polymers). Some concern has
been also expressed about the use of PVC to wrap food (Figure 6.2.43)
because of the transfer to the food of compounds called plasticisers used in
the film.
The monomers, polymer structures, properties and uses of polythene,
polypropene and PVC are shown in Table 6.2.2.
Table 6.2.2 Some important addition polymers.
Monomer Polymer repeat unit Properties Uses
Ethene Polythene Light, flexible Plastic bags and
Figure 6.2.43 Clingfilm is usually made (poly(ethene))
H H Easily moulded bottles
H H
from PVC although alternatives such as H H H
HH H
H Transparent Beakers
HH
H C HC
HH
H HH
H
C
HH C
LDPE are sometimes used. H
C C C Good insulator Insulation for
CC CC CC C
C
H
CC CC H CC H
CC H
H H
C
H H
n Resistant to water, cables
H H HH H n
H
HH H
HH HH HH n acids and alkalis Joint replacements
H nnn
Propene Polypropene Tough Fibre for ropes and
H H (poly(propene))
H H Easily moulded carpets
HH C HC H
H
HH H H Easily coloured Crates
HH HH
H HH
H
C
HH C
H
C
C CC C
C CC Very resistant to Toys
CC
H C
C CC
CH3 CC
C
HC
CC CH3 n
H water, acids and
HH CH
CH3 CH33 HCH3CHCH33 n
HH
H CH
CH33 HH
H CH
CH3 3nn n alkalis
n
Chloroethene PVC Tough Guttering and
H H (poly(chloroethene))
H H Rigid or flexible window frames
H H H H
HH
H C HC
HH
H HH
H
HH
C
HH
H
C Very resistant to Insulation for
H
C
H
C
CC CC Cl CC
C
CC
H
CC
C
C
Cl water, acids and cables
CC CC C n
HH Cl
Cl H
HCl Cl
Cl n alkalis Waterproof clothing
H
HH Cl
ClCl HH ClCl n
H nnn Flooring
Clingfilm
204 6.2 Hydrocarbons: alkanes and alkenes
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Test yourself
30 CF2=CF2 is the monomer for the polymer PTFE. b) Why is polythene so useful?
a) What is the systematic name for CF2=CF2? c) What is the major disadvantage of such
b) Draw a section of the PTFE polymer composed plastics?
of three monomer units. 32 a) Why is only a small amount of initiator needed
c) The name ‘PTFE’ contains four key letters from to make polymer chains of several thousand
the full name of the polymer. What is the full units long and why should the initiator not be
name? described as a catalyst?
d) State one use for PTFE and the property of the b) Using displayed formulae, write an equation for
polymer on which this use depends. the polymerisation of propene.
31 a) What conditions are used to manufacture low-
density polythene?
Chapter summary
l Halogen radicals can also react with the first
Chapter 6.2 Hydrocarbons:
product causing further substitution and a mixture
alkanes and alkenes of products.
l Alkanes are saturated compounds containing only l Crude oil is a complex mixture of hydrocarbons,
single bonds. The general formula for alkanes is mainly alkanes. The mixture is separated by
Cn H2n+2. fractional distillation. This produces too much
l C−C and C−H bonds are strong and of the high boiling point fractions with larger
non-polar, so alkanes are unreactive but do molecules and not enough of the low boiling
react with oxygen (combustion) and halogens point fractions needed for fuels such as petrol. To
(halogenation). satisfy this imbalance, the heavy fractions undergo
l In a plentiful supply of air, complete combustion cracking, which splits large molecules into smaller
occurs. Alkanes are oxidised to carbon dioxide and ones, and reforming, which turns straight-chain
water in a highly exothermic reaction. If the air hydrocarbons into branched-chain alkanes and
supply is limited, incomplete combustion occurs cyclic hydrocarbons, which burn more efficiently.
and the products include carbon particles and toxic l Sulfur impurities in the fuel combust to form
carbon monoxide. sulfur dioxide, which causes acid rain, so sulfur
l Alkanes react with chlorine and bromine, either compounds are removed before the fuel is burned.
on heating or on exposure to ultraviolet light. l In internal combustion engines, the high
Hydrogen atoms are replaced (substituted) by temperature provides sufficient activation energy
halogen atoms. This involves homolytic fission for nitrogen to react with oxygen and form oxides
of covalent bonds, which produces free radicals, of nitrogen. These also cause acid rain. Catalytic
species with an unpaired electron. converters remove carbon monoxide and oxides of
l Free-radical substitution involves three stages: nitrogen from car exhausts.
an initiation step, which produces radicals from l Fossil fuels are non-renewable and will run out.
molecules; propagation steps, which form products Alternative renewable fuels include ethanol, formed
and more radicals; and termination steps when by fermentation, and biodiesel, obtained from
radicals combine. vegetable oils. These fuels are described as carbon
l Halogen radicals react in the first propagation step neutral because the amount of carbon dioxide
but are re-formed in the second step and react removed from the atmosphere in their formation
again. This leads to a chain reaction. balances the amount released when they burn.
Chapter summary 205
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l Alkenes are unsaturated compounds with the l In electrophilic additions, the electron-rich C=C
functional group C=C. The general formula for reacts with an electrophile (electron pair acceptor)
alkenes is Cn H2n. to form a carbocation. This reacts immediately
l The C=C double bond contains a π bond in which with an electron pair donor.
electron density is concentrated above and below l Unsymmetrical alkenes react with hydrogen
the plane of the molecule. This restricts rotation of halides to give two isomeric products. The major
the bond and can lead to E/Z stereoisomerism. product is formed via the more stable carbocation.
l Alkenes undergo addition reactions. Alkenes react Carbocation stability increases with the number
with hydrogen in the presence of metal catalysts to of alkyl groups attached to the positive carbon.
form alkanes. Margarine is manufactured by this Tertiary carbocations are the most stable and
catalytic hydrogenation of unsaturated vegetable oils. primary ones are the least stable.
l Alkenes react with halogens and hydrogen halides to l Alkenes form polymers by addition polymerisation.
form halogenoalkanes, with steam to form alcohols Although polymers are useful materials, concern
and with acidified KMnO4 to form diols. is increasing about the effect of waste polymers
l Alkenes decolourise bromine water; this tests for a on the environment. Chemists are developing
C=C bond. biodegradable polymers to limit the problems
caused by polymer disposal.
Exam practice questions
The diagram shows three important reactions
1 iii) Give a common use of
of ethene. CH2OHCH2OH. (1)
c) i) Give the conditions used in Reaction
H2/Ni
CH3CH3 CH2 D CH2 CH2OHCH2OH
3 to convert ethene to high-density
Reaction 1 ethene Reaction 2 poly(ethene). (2)
ii) Draw a section of the poly(ethene)
structure consisting of three
Reaction 3 monomer units.(1)
d) State four properties of poly(ethene) which
poly(ethene) make it particularly suitable for making
plastic bags. (2)
a) i) Give the conditions of temperature
and pressure used in Reaction 1. (2) Crude oil is an important source of materials
2
ii) Reaction 1 is used to convert unsaturated for the petrochemical [Link] products
alkenes to saturated alkanes. State what are obtained from crude oil by fractional
is meant by the terms ‘unsaturated’ and distillation, followed by processes involving
‘saturated’ in this context. (3) cracking and reforming.
iii) Explain why saturated and unsaturated a)* Explain how crude oil is separated into
chemicals are important to dieticians.(3) fractions by fractional distillation. (6)
b) i) Identify the chemicals used in Reaction 2 b) i) State what is meant by ‘cracking’. (3)
to produce CH2OHCH2OH. (2) ii) Dodecane, C12H26, can be cracked
ii) A common name for CH2OHCH2OH into ethene and a straight-chain
is ethylene glycol. Give its systematic alkane so that the molar ratio of ethene
name. (1) to the straight-chain alkane is 2 : 1.
206
6.2 Hydrocarbons: alkanes and alkenes
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Write an equation for this reaction and Heptane can undergo isomerisation to
5
name the straight-chain alkane. (2) produce branched-chain alkanes such as
iii) Heat alone can be used to crack 2-methylhexane and 2,3-dimethylpentane.
alkanes, but oil companies normally use a) State what is meant by the term
catalysts as well. Give two reasons why ‘isomerisation’. (1)
oil companies use catalysts. (2) b) Write an equation using structural formulae
c) Straight-chain alkanes such as heptane can for the isomerisation of heptane to
be reformed into cyclic compounds. 2-methylhexane. (1)
Write an equation to show how heptane c) Write an equation using skeletal formulae
can be reformed into methylcyclohexane. (2) for the isomerisation of heptane to
d) Oxygen-containing compounds are added 2,3-dimethylpentane. (1)
to some brands of petrol to improve their d) Draw and name the structure of one isomer
performance. In Formula One racing cars, of heptane with a name ending in butane. (2)
2-methylpropan-2-ol is usually added to e) The boiling temperature of hexane is 69 °C
the petrol. and that of heptane is 99 °C. Explain why the
i) Draw the structural formula of boiling temperature of heptane is higher than
2-methylpropan-2-ol. (1) that of hexane. (2)
ii) State why compounds like f ) The boiling temperature of
2-methylpropan-2-ol improve a racing 2,3-dimethylpentane is 84 °C. Predict the
car’s performance on the track. (1) boiling temperature of 2-methylhexane and
3 a) Write equations using molecular formulae for: explain your prediction. (2)
i) the complete combustion of pentane(1) 6 Two isomeric hydrocarbons contain 85.7%
ii) the incomplete combustion of hexane carbon by mass and have Mr = 84.0
to form a gaseous product (1) Isomer X decolourises an acidified solution of
iii) the incomplete combustion of heptane potassium manganate(vii) but isomer Y has no
to form a solid product. (1) effect.
b) State and explain a difference you would When isomer Y reacts with chlorine in the
see in the flames of pentane and of pentene presence of UV light, only one monochloro-
burning under identical conditions. (2) substituted product is formed.
4 a) The reaction between propene, There is also only one monochloro-substituted
CH3CH=CH2, and hydrogen bromide isomer of X, but this is not formed by reaction
involves electrophilic addition and gives of X with chlorine.
CH3CHBrCH3 as the major product. a) Deduce the molecular formula of isomers
i) Explain the term ‘electrophilic X and Y. (4)
addition’. (3) b) Draw the structure of X and of Y and of
ii) Give the name of the compound their monochloro-substituted
CH3CHBrCH3. (1) compounds. (4)
iii) Draw displayed structures to show the c) Draw and name the compound formed
mechanism of the reaction between when X reacts with chlorine. (2)
propene and hydrogen bromide. (4) Propene and but-2-ene are used in the
7
iv) Explain why CH3CHBrCH3 is petrochemical industry to produce
the major product rather than important polymers.
CH3CH2CH2Br. (3) a) i) Explain the term ‘polymer’. (3)
b) Name the major product formed when ii) Poly(propene) does not have a sharp
propene reacts with iodine(i) chloride (ICl) melting temperature, but softens over a
and explain your answer. (3) wide temperature range. Explain why
c) Write an equation for the reaction of this is. (2)
propene with bromine water to form the iii) Draw a section of the polymer formed
major product. (1) from but-2-ene showing two
repeat units. (1)
207
Exam practice questions
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iv) Give two ways in which chemists Bond Bond enthalpy/kJ mol−1
contribute to a more sustainable use of C−H 412
materials such as poly(propene). (2)
C−C 348
b) But-2-ene can be converted to buta-1,3-diene
by a process called dehydrogenation. Buta-1, C−Br 276
3-diene is used to make synthetic rubber. H−Br 366
i) Explain the term ‘dehydrogenation’. (1) Br−Br 193
ii) Draw the structure of buta-1,3-diene. (1)
iii) Write an equation for the a) Use the bond enthalpy data above to
dehydrogenation of but-2-ene to form calculate the enthalpy change for this
buta-1,3-diene. (2) reaction. (5)
8 A student studied the amount of unsaturation b) The mechanism of the reaction occurs in a
in a sample of vegetable oil by reacting samples series of steps. Write equations for the two
of the oil with bromine water using the steps which, when added together, give the
following practical instructions. overall equation above. Name this
• Using a measuring cylinder, add 1 cm3 of type of step. (3)
vegetable oil to a conical flask. Add 20 cm3
10 Chlorine reacts with an excess of propane in
of inert organic solvent to the flask.
the presence of UV light to form a mixture of
• Fill a burette with bromine water and take
two isomers of chloropropane and very little
the burette reading.
dichloropropane.
• Add the bromine water slowly from the
a) Draw displayed structures of the two
burette to the solution in the conical flask.
chloropropane isomers. (2)
Shake the flask vigorously until the bromine
b) Explain why little dichloropropane is formed
colour disappears.
under these conditions (2)
• Add more bromine water, shaking the flask
c) At low temperatures, the proportion of
after each addition, until the bromine water
each chloropropane formed depends
is no longer decolourised.
on the relative stabilities of the alkyl
• Record the burette reading.
radicals formed during the reaction. The
• Repeat the experiment.
relative stabilities of radicals are similar
The student obtained the following results.
to the relative stabilities of the equivalent
Experiment 1 2 3 carbocations. Predict the proportion of each
Volume of bromine chloropropane formed at low temperatures.
8.3 7.8 9.0
water used/cm3 Explain your answer. (2)
d) At higher temperatures, the proportion of
a) Explain why it was necessary to shake the each chloropropane in the product mixture is
flask after each addition of bromine water. (2) found to be as expected statistically. State this
b) Describe what the student saw in the proportion. Explain your answer, stating any
conical flask at the point where the bromine assumptions you have made. (2)
water was not decolourised. (2) e) State why the relative stabilities of the
c) The student’s results were not concordant. intermediates is less important as the
Give improvements to the experimental temperature rises. (1)
method that could give more concordant
titres. (3) 11 A family of three drive their car about 8000
d) The volumes of bromine water used were miles each year using 1000 dm3 of petrol.
relatively small. Give improvements to the Assuming that petrol consists of octane, C8H18,
experimental method that would give with a density of 0.8 g cm−3, calculate the mass
larger titres. (2) of carbon dioxide that their travel adds to the
atmosphere in one year. (6)
9 Ethane reacts with bromine in the gas phase to
form bromoethane and hydrogen bromide.
C2H6(g) + Br2(g) → C2H5Br(g) + HBr(g)
208
6.2 Hydrocarbons: alkanes and alkenes
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Halogenoalkanes and alcohols
6.3
6.3.1 Halogenoalkanes
Halogenoalkanes are important to organic chemists both in research and
in industry. Many halogenoalkanes are reactive compounds which can be
converted into other more valuable products (Figure 6.3.1). This makes them
useful as intermediates in synthesis – the production of one compound from
another.
In the structure of a halogenoalkane, one or more of the hydrogen atoms in
an alkane molecule is replaced by a halogen atom, for example:
CH3F fluoromethane
CH3CH2Cl chloroethane
CH3CH2CH2Br 1-bromopropane
CH3CH2CH2CH2I 1-iodobutane
The bond between carbon and the electronegative halogen atom is polar.
This polarity affects the physical properties of halogenoalkanes (Section
6.3.2) and also their chemical properties (Section 6.3.4).
IUPAC rules name halogenoalkanes by placing the prefix fluoro, chloro,
Figure 6.3.1 The Gore-tex® membrane in bromo or iodo before the name of the parent alkane.
this waterproof jacket contains the fluoro
Where necessary, the position of the halogen group is noted by including
compound poly(tetrafluorethene),
the number of the carbon to which it is attached – numbering either from
–(CF2–CF2)n –.
left or right to give the lower number, for example, 1-iodobutane not
4-iodobutane.
If more than one halogen atom is present then the position of both must be
given. If different types of halogen are present, as in CFCs (Section 6.3.5),
the halogens are listed in alphabetical order.
CH3CCl3 1,1,1-trichloroethane
Key term BrCH2CH2Br 1,2-dibromoethane
CBrClF2 bromochlorodifluoromethane
A primary halogenoalkane has the
The terms primary, secondary and tertiary are used with halogenoalkanes
halogen atom bonded to a carbon
and other organic compounds to show the positions of functional groups
at the end of the chain. A secondary
(Figure 6.3.2).
halogenoalkane has the halogen atom
bonded to a carbon in the middle of the
chain but not at a branch. A tertiary
Tip
halogenoalkane has the halogen atom The terms ‘primary’, ‘secondary’ and ‘tertiary’ have a different meaning when applied
bonded to a carbon at a branch in the to amines which are derived from the ammonia molecule with one (primary), two
chain. (secondary) or three (tertiary) hydrogen atoms replaced by alkyl or aryl groups.
6.3.1 Halogenoalkanes 209
469983_6.3_Chem_Y1-2_209-[Link] 209 13/04/19 10:09 PM
H
H C H
H H H H H H H H H H
H C C C C I H C C C C H H C C C H
H H H H H H Br H H Cl H
1-iodobutane 2-bromobutane 2-chloro-2-methylpropane
(primary) (secondary) ( tertiary)
Figure 6.3.2 Names and displayed formulae of three halogenoalkanes.
6.3.2 Physical properties
of halogenoalkanes
Chloromethane, bromomethane and chloroethane are gases at room
temperature, but most other halogenoalkanes are colourless liquids.
Their boiling points are higher than the parent alkane, because the polarity
of the carbon–halogen bond leads to stronger intermolecular forces.
However, these dipole–dipole forces are much weaker than the hydrogen
bonding that alcohols can form with water so, unlike alcohols, even small
halogenoalkanes do not mix with water.
Test yourself
1 Draw the structures of the following compounds, 3 Which of the following molecules are polar and
and identify them as primary, secondary or tertiary: which are non-polar: CHCl3, CH2Cl2, CHCl3 and CCl4?
a) 1-iodopropane 4 The boiling temperatures of 1-chlorobutane,
b) 2-chloro-2-methylbutane 1-bromobutane and 1-iodobutane are 352 K, 375 K
and 404 K respectively. Suggest an explanation for
c) 3-bromopentane
the trend in values.
d) 1-bromo-2-chloropropane.
5 The boiling temperatures of the isomers
2 Name the following compounds. 1-bromobutane, 2-bromobutane and 2-bromo-
a) CH CH CH 2 CH 2 Cl 2-methylpropane are 375 K, 364 K and 346 K
3
respectively. Suggest an explanation for the
Cl differences in boiling temperatures of the primary,
b) Cl F secondary and tertiary compounds.
6 Explain why, despite containing a polar carbon–
Cl C C F
halogen bond, halogenoalkanes are immiscible with
F Cl water.
c) Br
210 6.3 Halogenoalkanes and alcohols
469983_6.3_Chem_Y1-2_209-[Link] 210 13/04/19 10:09 PM
6.3.3 The preparation
of halogenoalkanes
Three types of reaction can be used to produce organic molecules containing
halogen atoms.
From alkanes
Reaction of alkanes with chlorine or bromine, either on heating or on
exposure to ultraviolet light, leads to free-radical substitution in which
halogen atoms replace hydrogen atoms (Section 6.2.3).
CH4(g) + Cl 2(g) → CH3Cl(g) + HCl(g)
From alkenes
Reaction of alkenes with hydrogen halides at room temperature produces
halogenoalkane molecules containing one halogen atom (Section 6.2.9).
CH3CH=CHCH3(g) + HBr → CH3CH2CHBrCH3
Alkenes also react with halogen elements to form halogenoalkane molecules
containing two halogen atoms.
CH3CH=CH2 + Br2 → CH3CHBrCH2Br
From alcohols
Alcohols react with phosphorus halides or hydrogen halides to form
halogenoalkanes.
C2H5OH(l) + PCl5(s) → C2H5Cl(l) + POCl3(l) + HCl(g)
CH3CH2CH2OH(l) + HCl(aq) → CH3CH2CH2Cl(l) + H2O(l)
On a laboratory scale, the usual preparative methods are based on the reactions
of alcohols with a hydrogen halide, or with a phosphorus halide (Section 6.3.8).
Test yourself
7 Name the five halogenoalkanes produced in the c) Write an equation for a preparation of
reactions in Section 6.3.3. 1-bromobutane that is more efficient than those
8 a) Write an equation to show the formation of in parts (a) and (b).
1-bromobutane from butane. Give a necessary 9 Give the reagents and name the three types of
condition for the reaction and explain why reaction used to make halogenoalkanes as shown
1-bromobutane is not the only organic product. in the scheme below.
b) Write an equation for a possible preparation alkenes
of 1-bromobutane from but-1-ene. Explain why
a low yield of 1-bromobutane is obtained in alcohols halogenoalkanes
this reaction.
alkanes
6.3.3 The preparation of halogenoalkanes 211
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Key terms 6.3.4 Chemical reactions
Nucleophiles are electron pair donors of halogenoalkanes
(Section 6.1.7). The polarity of the carbon–halogen bond means that the carbon atom in the
Substitution is a reaction in which one bond is slightly electron deficient. This δ+ atom is, therefore, vulnerable to
atom or group is replaced by another attack by nucleophiles. The two important reactions of halogenoalkanes
atom or group. are substitution and elimination (Section 6.1.6).
Elimination is a reaction which Substitution by reaction with water
produces an unsaturated product by Cold water slowly hydrolyses halogenoalkanes, replacing the halogen atoms
loss of atoms or groups from adjacent with an –OH group to form alcohols.
carbon atoms.
CH3CH2CH2I(l) + H2O(l) → CH3CH2CH2OH(l) + H+(aq) + I−(aq)
Hydrolysis is a reaction in which a
compound splits apart in a reaction The nucleophile in this nucleophilic substitution reaction is the water
involving water. molecule. The rate of hydrolysis of different halogenoalkanes can be
compared by carrying out the reaction in the presence of silver ions. The
halogen atoms in halogenoalkanes are covalently bonded to carbon. They
cannot react with silver ions and so give no precipitate of a silver halide.
Hydrolysis releases halide ions, which immediately react with silver ions to
form precipitates of silver halides (Section 4.10).
Halogenoalkanes do not mix with water or aqueous solutions. For this
reason, the reaction is carried out in the presence of ethanol, which can
dissolve the halogenoalkane and also mix with the aqueous silver nitrate.
Core practical 4
Investigation of the rates of hydrolysis of halogenoalkanes
A student studied the hydrolysis of 1-chlorobutane, 1-bromobutane and 1-iodobutane
to investigate the effect of the halogen atom on the rate of hydrolysis. A similar
method was also used to compare the rates of hydrolysis of three isomeric
bromoalkanes: one primary, one secondary and one tertiary.
The student followed the instructions labelled A to D. The results are shown below
the method on page 213.
The effect of the halogen atom on the rate of hydrolysis
A Set up three labelled test tubes as shown in Table 6.3.1. Stand the tubes in a water bath
at about 60 °C. Put a tube containing 5 cm3 silver nitrate solution in the same beaker.
Leave the tubes for about 10 minutes to allow them to reach the temperature of the
water bath.
Table 6.3.1
Tube 1 Tube 2 Tube 3
1 cm3 ethanol 1 cm3 ethanol 1 cm3 ethanol
2 drops 1-chlorobutane 2 drops 1-bromobutane 2 drops 1-iodobutane
B Note the time. Quickly add 1 cm3 of the warm silver nitrate solution to each of the
three tubes. Shake the tubes to mix the contents. Replace them in the water bath and
observe for the next five minutes or so.
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Results
Table 6.3.2
Tube 1 Tube 2 Tube 3
No precipitate Cream precipitate appears A heavy yellow precipitate
after 5 minutes. between 2 and 3 minutes. is visible after 1 minute.
1 What was the purpose of adding ethanol to the tubes?
2 In what ways did the experiment try to ensure that the only factor that affected the
rate of reaction was the halogen atom?
3 Suggest any improvements to the method to make the comparison fairer.
4 Identify the precipitates and explain why they formed slowly and not immediately.
5 Which halogenoalkane hydrolysed fastest and which slowest?
6 What is the order of polarity of the carbon—halogen bonds? What are the bond
energies of the three carbon—halogen bonds? Does the rate of hydrolysis correlate
better with bond polarity or bond strength?
The effect of the structure of the carbon skeleton on the rate of hydrolysis
C Set up three labelled test tubes as shown in Table 6.3.3. This time there is no need
to warm the tubes.
Table 6.3.3
Tube 1 Tube 2 Tube 3
1 cm3 ethanol 1 cm3 ethanol 1 cm3 ethanol
2 drops 2 drops 2 drops
1-bromobutane 2-bromobutane 2-bromo-2-
methylpropane
D Note the time. Quickly add 1 cm3 of silver nitrate solution to each of the three tubes.
Observe the tubes for the next five minutes or so.
Results
Table 6.3.4
Tube 1 Tube 2 Tube 3
Cream precipitate Cream precipitate Cream precipitate
appears between appears between appears in
2 and 3 minutes. 1 and 2 minutes. under a minute.
7 Suggest any improvements to the method to make the comparison as fair as possible.
8 What might account for the differing rates of hydrolysis of the three compounds?
Substitution by reaction with hydroxide ions
Replacement of the halogen atom of a halogenoalkane by an –OH group
is much quicker with an aqueous solution of an alkali, such as potassium or
sodium hydroxide (Figure 6.3.3). Heating increases the rate of reaction even
further.
H H H H H H H H
– heat
H C C C C Br + OH (aq) H C C C C OH + Br – (aq)
H H H H H H H H
1-bromobutane butan-1-ol
Figure 6.3.3 Reaction of a halogenoalkane with an alkali on heating.
6.3.4 Chemical reactions of halogenoalkanes 213
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The nucleophile in this nucleophilic substitution reaction is the hydroxide
ion. The bromide ion released into the solution will react with silver ions to
form a cream precipitate of silver bromide.
The red curly arrows in the mechanism (Figure 6.3.4) show the movement
of electron pairs.
Tip H H
–
Take care when using C3H7 – in structures C3H7 C Br C3H7 C OH + Br
as it does not show whether the alkyl
group attached is CH3CH2CH2 – or H H
–
HO
(CH3)2CH–.
Figure 6.3.4 A nucleophilic substitution mechanism using hydroxide ions.
Hydrolysing halogenoalkanes makes it possible to distinguish between
chloro-, bromo- and iodo- compounds (Figure 6.3.5).
Figure 6.3.5 Heating the compound with acidify with add a few drops of
an alkali releases halide ions. Acidifying dilute nitric acid silver nitrate solution
with nitric acid and then adding silver
nitrate produces a precipitate of the silver
halide. NaOH(aq) AgCl(s): white
plus a drop AgBr(s): cream
of the hot AgI(s): yellow
halogenoalkane water
Key terms
heat
hydrolysis acidification precipitation
Heating under reflux means heating with
a condenser placed vertically in the flask.
A reflux condenser is fitted vertically in Test yourself
a flask to prevent vapour escaping while
10 Explain the use of the terms ‘nucleophile’, ‘substitution’ and
a liquid is being heated. Vapour from
‘hydrolysis’ to describe the reaction of halogenoalkanes with water.
the boiling reaction mixture condenses
and flows back into the flask. 11 Write equations for the two reactions which take place when
2-bromobutane reacts with water in the presence of silver ions, Ag+.
12 Refer to Figure 6.3.5 in answering this question.
a) Why is hydrolysis necessary before testing with silver nitrate?
b) Why must nitric acid be added before the silver nitrate solution?
water out
c) Write the equations for the three reactions that take place when
condenser detecting bromide ions in 1-bromobutane by this method.
13 Suggest why the polymer, PTFE, –(CF2–CF2)n – is unaffected by
prolonged exposure to boiling water or hot alkali.
water in
Substitution by reaction with cyanide ions
When a halogenoalkane is heated under reflux with a solution of potassium
reaction cyanide in ethanol, the halogen atom is replaced by the CN group and a
mixture compound called a nitrile is formed. Use of a reflux condenser (Figure 6.3.6)
Figure 6.3.6 Apparatus for heating under ensures that the volatile substances, the halogenoalkane and ethanol, condense
reflux. and drip back into the reaction mixture. This nucleophilic substitution
214 6.3 Halogenoalkanes and alcohols
469983_6.3_Chem_Y1-2_209-[Link] 214 13/04/19 10:09 PM
reaction, where the cyanide ion acts as a nucleophile, is useful in synthesis as a
way of increasing the length of the carbon chain in a molecule (Figure 6.3.7).
Tip
It is necessary to state that an excess of
For example, the two-carbon chain in bromoethane becomes three in
ammonia is used in the formation of a
propanenitrile.
primary amine from a halogenoalkane. For
CH3CH2Br + CN− → CH3CH2CN + Br− each reaction in organic chemistry, you
bromoethane propanenitrile should learn not only the reagents, but
also the reaction conditions. The same
–
reagents may give different products
CH3CH2 Br CH3CH2 CN + Br
under different reaction conditions.
–
NC
Figure 6.3.7 A nucleophilic substitution mechanism using cyanide ions.
H H H H
Substitution by reaction with ammonia H C C C C Br + 2NH3
Warming a halogenoalkane with a concentrated solution of ammonia in H H H H
ethanol in a sealed tube produces a primary amine. ethanol
heat solution under
The ammonia molecule acts as the nucleophile. In the presence of excess pressure
ammonia, the other product is an ammonium salt (Figure 6.3.8).
H H H H
The mechanism for this reaction, a nucleophilic substitution, occurs in two
steps (Figure 6.3.9). H C C C C NH2 + NH4 Br
+ –
H H H H
H H H
+ Figure 6.3.8 Reaction of 1-bromobutane
–
C3H7 C Br C3H7 C N H + Br with ammonia to make butylamine
(1-aminobutane).
H H H
NH3
H H H H
+ +
C3H7 C N H C3H7 C N H + NH4
H H H
NH3
Figure 6.3.9 A nucleophilic substitution mechanism using ammonia.
The reactivity of ammonia as a nucleophile depends on the lone pair of
Tip
electrons on its nitrogen atom. A problem in this case is that there is also Hydrogenation of a C ≡ N bond in the
a lone pair on the nitrogen atom of the primary amine formed and this is presence of a nickel catalyst forms a
even more reactive because of the inductive effect of the alkyl group. So primary amine. This reaction produces
the primary amine can also react with the halogenoalkane and this can lead a pure product. The alternative
to a mixture of further products. Fortunately, it is possible to limit further preparation of amines by substitution
reaction by using an excess of ammonia, so that there is a much greater in halogenoalkanes using ammonia can
chance of ammonia – rather than the amine – acting as nucleophile with the lead to an impure product because of
halogenoalkane molecules. the possibility of further substitution.
6.3.4 Chemical reactions of halogenoalkanes 215
469983_6.3_Chem_Y1-2_209-[Link] 215 13/04/19 10:09 PM
Test yourself
14 a) Write an equation for the formation of butanenitrile in a reaction
using potassium cyanide.
b) Name the organic starting material and state the conditions needed.
c) Aqueous solutions of potassium cyanide are alkaline. Name
another product in the reaction to form butanenitrile if aqueous
conditions are used.
15 Refer to the mechanism for the reaction of ammonia with
1-bromobutane in answering this question.
a) Describe the behaviour of the ammonia molecule in the first step.
b) Describe the behaviour of the ammonia molecule in the second step.
16 Write an overall equation for the reaction of an excess of ammonia
with bromoethane.
17 Suggest the structure of the organic product when ethylamine reacts
with bromoethane. Can this product also react with bromoethane? If
so, what is formed?
Elimination reactions
In the reaction of aqueous potassium hydroxide with a halogenoalkane
(Figure 6.3.3), the hydroxide ion acts as a nucleophile and brings about
substitution. But in an alternative reaction in ethanolic solution, the
hydroxide ion acts as a base and brings about elimination of a hydrogen
halide to form an alkene instead of substitution. The mechanism shown in
Figure 6.3.10 is not required for your specification.
CH3CHBrCH3 + OH− → CH3CH=CH2 + H2O + Br−
–
HO H2O
H H H H H
H
H C C C H C C C H
H
H Br H H
–
Br
Figure 6.3.10 Elimination of hydrogen bromide from 2-bromopropane on heating with
a solution of potassium hydroxide in ethanol.
favoured by The hydroxide ion provides both the electrons needed to form a new bond
warm aqueous alcohol
KOH to the hydrogen atom. The C–H bond breaks and the pair of electrons from
substitution
that bond forms a second bond between the two carbon atoms. At the same
halogenoalkane time, the C–Br bond breaks heterolytically. In this case, both electrons in the
elimination bond leave with the bromine atom, which is set free as a bromide ion.
favoured by alkene
hot ethanolic Although changing the reaction conditions can favour substitution or
KOH
elimination, the result of these reactions is usually a mixture of products
Figure 6.3.11 Alternative reactions (Figure 6.3.11). Also, elimination happens more readily with secondary
of a halogenoalkane with solutions of or tertiary halogenoalkanes and substitution more readily with primary
hydroxide ions. halogenoalkanes.
216 6.3 Halogenoalkanes and alcohols
469983_6.3_Chem_Y1-2_209-[Link] 216 13/04/19 10:09 PM
Activity
The preparation of 1-bromobutane from butan-1-ol
Figure 6.3.12 shows the steps Carrying out Separating the product
the reaction from the reaction mixture
in synthesising a pure sample of
1-bromobutane from butan-1-ol. The
alcohol reacts with hydrogen bromide
formed from sodium bromide and heat
under
50% sulfuric acid. reflux
1 a) Identify three aspects of this distil off
preparation that might be impure product
hazardous. butan-1-ol mixed reaction mixture
after refluxing impure product
b) What steps would you take to with sodium heat
bromide and 50%
reduce the risks from these sulfuric acid
hazards?
2 Write an equation for the reaction heat Purifying the product
of sodium bromide with 50%
sulfuric acid to form hydrogen Drying the product
bromide. concentrated washing with
3 Explain why 50% sulfuric acid hydrochloric HCl(aq) to remove
anhydrous acid unchanged butan-1-ol
is used, and not concentrated organic
sodium layer from 1-bromobutane then with NaHCO3(aq)
sulfuric acid. sulfate separating to remove acids
4 The reaction mixture is heated (a drying funnel
agent)
for about 40 minutes but even
after this time some of the alcohol
does not turn into the product. Final purification
and identification
Suggest a reason why.
5 Explain why the reaction flask is final distillation
fitted with a reflux condenser. and measurement
6 After heating the reaction mixture of boiling temperature
for some time, the flask contains
a mixture of chemicals including
anti-bumping
1-bromobutane, unchanged granule 1-bromobutane
(fraction boiling
butan-1-ol, hydrogen bromide heat between 100 ºC and 103 ºC)
and unchanged sodium bromide.
Which of these chemicals is likely Figure 6.3.12 Steps in the synthesis of 1-bromobutane from butan-1-ol.
to distil over and collect in the
measuring cylinder when separating the impure product? 11 When this synthesis was carried out, the yield was 6.8 g
7 Why are there two layers in the separating funnel when the of 1-bromobutane from 7.5 cm3 butan-1-ol. The density of
product is shaken with aqueous reagents? butan-1-ol is 0.81 g cm—3. Calculate the percentage yield.
8 Suggest a reason why shaking the product with hydrochloric 12 Suggest three reasons why the percentage yield is below
acid helps to remove unchanged butan-1-ol from the impure 100%.
product.
9 Why is aqueous sodium hydrogencarbonate used in the
Tip
separating funnel, and not aqueous sodium hydroxide, to For the differences between hazards and risks,
remove acidic impurities? refer to Practical skills sheet 2, ‘Assessing hazards
10 Explain the term ‘fraction’ to describe the sample of and risks’, which you can access online at www.
product collected during the final distillation. [Link]/EdexcelChemistry.
6.3.4 Chemical reactions of halogenoalkanes 217
469983_6.3_Chem_Y1-2_209-[Link] 217 13/04/19 10:09 PM
Test yourself
18 Give the name and structure of the main organic product when
2-bromopropane reacts on heating with:
a) an aqueous solution of sodium hydroxide.
b) a solution of potassium hydroxide in ethanol.
19 Outline mechanisms to show how both but-1-ene and but-2-ene
can be formed when 2-bromobutane reacts with a hot solution of
potassium hydroxide in ethanol.
20 Consider the mechanisms for the competing substitution and
elimination reactions of haloalkanes with hydroxide ions and suggest
why elimination is favoured at higher temperatures.
6.3.5 The uses of halogenoalkanes
and their impacts
The discovery, production and use of halogenoalkanes during the last
century have led to dramatic examples of the way in which science-based
technology can provide us with products we value, and which make our lives
more comfortable and safer. Yet, at the same time, these new technologies
can turn out to have unintended and undesirable consequences.
One particular class of unreactive halogenoalkanes is notorious because of the
damaging effect they have had on the Earth’s ozone layer. These are called
chlorofluorocarbons (CFCs) and were developed in the early twentieth century,
supposedly as safe alternatives to toxic refrigerants such as ammonia and sulfur
dioxide. As CFCs are also powerful greenhouse gases, their use is doubly damaging.
There are now increasing restrictions on the uses of halogenoalkanes because
of concerns about their hazards to health, their persistence in the environment
Figure 6.3.13 A recycling plant which and their effect on the ozone layer (Figure 6.3.13).
recovers CFCs from the coolant systems
Halogenoalkanes are used as:
of old refrigerators and freezers. The CFCs
are removed from the systems as gases, ● solvents, for example CH2Cl 2
liquefied and then chemically destroyed. ● refrigerants, for example CF3CH 2F, which has replaced the CFC CF2Cl 2
● fire extinguishers and fire retardants, for example C3HF7, which has
replaced halons such as CBr2ClF.
Chlorofluorocarbons (CFCs) are compounds containing just the elements
chlorine, fluorine and carbon, such as CCl3F, CCl2F2 and CCl2FCClF2.
They contain no hydrogen. CFCs have some desirable properties – they are
unreactive, do not burn and are not toxic. It is also possible to make CFCs with
different boiling temperatures to suit different applications. These properties
made CFCs ideal as the working fluid in refrigerators and air-conditioning
units. They can also act as the blowing agents to make the bubbles in expanded
plastics and insulating foams. CFCs were once valued as good solvents for dry
cleaning and for removing grease from electronic equipment.
218 6.3 Halogenoalkanes and alcohols
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The manufacture of CFCs and their general use was banned by the Montreal
Protocol in 1987 because of their effects on the ozone layer.
CFCs can also contribute to global warming. In 1987, governments also
agreed to ban the production of bromofluoroalkanes, which were then in
common use as halon fire extinguishers in homes, on aircraft and on ships,
as well as in a wide range of electronic equipment.
Since the Montreal Protocol, the concentration of CFCs in the atmosphere
has been falling, after peaking in 1994. The Antarctic ozone ‘hole’ has begun
to decrease but complete recovery of the ozone layer to pre-1980 levels is not
predicted until 2060.
These bans set chemists the challenge of finding replacement chemicals
with similar properties to CFCs, but without the harmful impacts on the
environment. The chemists’ first step was to try the effect of adding hydrogen
to make hydrochlorofluorocarbons (HCFCs). These are much less stable in
the lower atmosphere and were expected to break down before reaching the
ozone layer. But these were still found to deplete the ozone layer and have
been banned after 2020. More recently, new chemicals have been made that
contain no chlorine. These are hydrofluorocarbons (HFCs), which survive
for an even shorter time in the lower atmosphere. HFCs have no effect on
the ozone layer, but they are greenhouse gases.
6.3.6 Alcohol names and structures
Compounds which contain the –OH functional group are called alcohols.
Ethanol is the best known member of this family because it is easily produced
by fermentation and is the alcohol in beer, wine and spirits. Because it is so
common, the person in the street will often say ‘alcohol’ when they mean
ethanol. But to the chemist, there are many alcohols with different properties
and uses. Alcohols are useful solvents in the home, in laboratories and in Figure 6.3.14 This fuel is a blend of
industry and so called ‘bioethanol’ is an important fuel (Figure 6.3.14). conventional petrol and 85% bioethanol.
Understanding the properties of the –OH functional group in alcohols helps Using a mixture of the two fuels produces
to make sense of the reactions of some important biological compounds, less carbon in the vehicle exhaust
especially carbohydrates such as sugars and starch. emissions than using petrol alone.
Alcohols are compounds with the formula R–OH, where R represents an
alkyl group. The hydroxy group, –OH, is the functional group which gives
alcohols their characteristic reactions.
Tip
The name of an alcohol ends in –ol.
The simplest alcohols are:
So if the name of any compound ends
● methanol CH3OH in -ol, for example, cholesterol or
● ethanol
CH3CH2OH paracetamol, there must be an alcohol
● propan-1-ol CH3CH2CH2OH. group present, even if the rest of the
molecule is complicated.
IUPAC rules name alcohols by replacing the ‘e’ at the end of the corresponding
alkane with ‘-ol’ – so ethane becomes ethanol.
Where necessary, the position of the –OH group is noted by including
the number of the carbon to which it is attached – numbering either
from left or right to give the lower number, e.g. propan-1-ol not
propan-3-ol.
6.3.6 Alcohol names and structures 219
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The terms primary, secondary and tertiary are used with alcohols
Key terms and other organic compounds to show the positions of functional groups
(Figure 6.3.15).
A primary alcohol has the —OH group
at the end of the chain. A secondary
alcohol has the —OH group in the CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH CH 3
middle of the chain but not at a branch. butan-1-ol
A tertiary alcohol has the —OH group at a primary alcohol OH
butan-2-ol
a branch in the chain. a secondary alcohol
CH 3
CH3 CH 3 C CH 3
H3C CH CH2 OH OH
2-methylpropan-2-ol
2-methylpropan-1-ol
a tertiary alcohol
a primary alcohol
Figure 6.3.15 The names and structures of the four isomeric alcohols with the formula
C4H9OH.
Tip
The final ‘e’ is not removed when naming an alcohol containing more than one —OH
group. In these cases the ending diol or triol is added after the numbers which show the
positions of the —OH groups, for example, ethane-1,2-diol and propane-1,2,3-triol below.
HO CH2 CH2 OH HO CH2 CH CH2 OH
OH
ethane-1,2-diol propane-1,2,3-triol
Test yourself
21 Draw the structural formulae of these alcohols, and state whether
they are primary, secondary or tertiary compounds:
a) propan-1-ol b) propan-2-ol
c) 2-methylbutan-2-ol d) 3-methylbutan-2-ol.
22 Draw the skeletal formula and name one isomer with the formula
C5H11OH that is:
a) a primary alcohol b) a secondary alcohol
c) a tertiary alcohol.
23 Name the following alcohols and classify each OH group as primary,
secondary or tertiary:
a) CH3 b) OH c)
OH
CH2
HO
CH3 CH CH CH3
OH
220 6.3 Halogenoalkanes and alcohols
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6.3.7 Physical properties of alcohols
Although methane and ethane are gases at room temperature, the simplest
alcohols, methanol and ethanol, are liquids under the same conditions. Alcohols
are much less volatile than hydrocarbons that have roughly the same molar mass.
This is because the hydrogen bonding between –OH groups in alcohols is much
stronger than the London forces between alkanes (Section 2.6). For the same
reason, alcohols with short hydrocarbon chains also mix completely with water.
If two or three –OH groups are present, the intermolecular hydrogen
bonding becomes even stronger. This can lead to very viscous liquids such as
propane-1,2,3-triol (commonly known as glycerol) or solids such as glucose.
Test yourself
24 Refer to the data sheet for Chapter 6.3, ‘Melting and boiling
temperatures of some alkanes and alcohols’, which you can access
online at [Link]/EdexcelChemistry. Use data
from this data sheet to show that alcohols are less volatile than
alkanes with similar molar masses.
25 a) Draw a diagram to show the hydrogen bonding between a
methanol molecule and a water molecule.
b) Explain why hydrogen bonding accounts for the fact that
methanol, at room temperature, is a liquid that mixes freely with
water, while ethane is a gas which is insoluble in water.
26 A half-full bottle of propan-1-ol is stoppered and shaken for a few
seconds. When the shaking is stopped, the bubbles of air escape from
the liquid very quickly. By contrast, after a half-full bottle of propane-
1,2,3-triol is shaken in a similar way, the bubbles of air rise very slowly.
Suggest why there is a difference in behaviour of the two liquids.
6.3.8 Chemical properties of alcohols
Alcohols are much more reactive than alkanes because the C–O and O–H
bonds in the molecules are polar (see Section 2.5).
Combustion Tip
Alcohols burn in a plentiful supply of air with a clean, pale blue flame. Methanol
and ethanol are both common fuels (see Section 6.2.6) and fuel additives. When balancing equations for the
combustion of alcohols, don’t forget the
CH3CH2OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) oxygen atom in the alcohol.
Tip
It is often convenient to write an equation for the combustion of one mole of alcohol
and use fractions of moles of oxygen, for instance:
CH3OH(l) + 23 O2(g) → CO2(g) + 2H2O(l)
6.3.8 Chemical properties of alcohols 221
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Tip Substitution of the –OH group by a halogen atom
The –OH group in an alcohol can be replaced by a halogen atom. This is an
The reaction with phosphorus(v)
example of a substitution reaction. Several reagents can be used, including
chloride also produces hydrogen
phosphorus halides or some hydrogen halides.
chloride gas, which is seen as misty
fumes. This observation makes the Chloroalkanes are best prepared from alcohols using phosphorus(v) chloride.
reaction a useful test for the presence
C3H7OH(l) + PCl5(s) → C3H7Cl(l) + POCl3(l) + HCl(g)
of the —OH group in molecules.
Tertiary chloroalkanes can be prepared by reacting tertiary alcohols with
concentrated hydrochloric acid. For instance, a tertiary chloroalkane can be
made at room temperature by reacting the alcohol, 2-methylpropan-2-ol,
Tip with concentrated hydrochloric acid (see Core practical 6 on page 223). The
Beware that C3H7OH is an ambiguous
reaction of concentrated hydrochloric acid with primary alcohols is very
formula as it represents both
slow and not suitable for synthesis.
propan-1- ol and propan-2-ol and Bromoalkanes are conveniently prepared from alcohols using hydrogen
does not specify which. Both alcohols bromide. For example, butan-1-ol reacts with hydrogen bromide on heating
react in the same way with PCl5, so this in the presence of sulfuric acid to form 1-bromobutane. It is usual to make
formula is suitable. However, the hydrogen bromide in the reaction flask by mixing 50% sulfuric acid with
propan-1-ol, CH3CH2CH2OH, and sodium or potassium bromide (see the Activity in Section 6.3.4, page 217).
propan-2-ol, CH3CH(OH)CH3, react in
different ways with oxidising agents, so
C4H9OH(l) + HBr(aq) → C4H9Br(l) + H2O(l)
more precise structures should be used Bromoalkanes can also be made from alcohols using phosphorus(v) bromide
in those cases. or phosphorus(iii) bromide.
3C3H7OH(l) + PBr3(s) → 3C3H7Br(l) + H3PO3(l)
Iodoalkanes can be made by reacting alcohols with phosphorus(iii) iodide,
but, as phosphorus(iii) iodide is unstable, it is made in the reaction flask
using a mixture of red phosphorus and iodine. Once formed, it reacts with
the alcohol.
3C3H7OH(l) + PI3(s) → 3C3H7I(l) + H3PO3(l)
Hydrogen iodide is unstable with respect to its elements and also easily
oxidised to iodine, so cannot be used to prepare iodoalkanes.
Test yourself
27 Write equations for:
a) the complete combustion of propan-1-ol in a plentiful supply of air
b) the incomplete combustion of butan-1-ol in a limited supply of air
to form carbon and water.
Tip 28 Write an equation for the formation of 2-bromobutane from
butan-2-ol and hydrogen bromide.
Bromide ions are protonated by
29 Describe a test to confirm that hydrogen chloride is formed when
concentrated sulfuric acid to form HBr.
PCl5 reacts with an alcohol.
But bromide ions are also oxidised
by concentrated sulfuric acid, so the 30 Write an equation for the formation 1-iodobutane from butan-1-ol.
reaction mixture turns orange because of 31 Why is it not possible to convert an alcohol into an iodoalkane using
the formation of bromine (Section 4.10). a mixture of potassium iodide and concentrated sulfuric acid?
222 6.3 Halogenoalkanes and alcohols
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Core practical 6
Chlorination of 2-methylpropan-2-ol with concentrated hydrochloric acid
A sample of 2-chloro-2-methylpropane is prepared using the procedure outlined in steps A–K.
Preparation Questions
A Add about 10 cm3 of 2-methylpropan-2-ol to a measuring 1 Write an equation for the reaction showing the structure of
cylinder. Weigh the cylinder and contents. both organic compounds.
B Pour the alcohol into a 250 cm3 conical flask and reweigh the 2 What are the main hazards associated with use of
measuring cylinder. Add, a little at a time, a total of 35 cm3 of 2-methylpropan-2-ol and concentrated hydrochloric acid,
concentrated hydrochloric acid. Stopper and swirl the flask to and what precautions should be taken? (Refer to Practical
mix the contents. skills sheet 2, ‘Assessing hazards and risks’, which you
C After adding all the acid, leave the stoppered flask to stand can access online at [Link]/
for about 20 minutes in a fume cupboard. Shake the mixture EdexcelChemistry.)
carefully from time to time releasing the pressure after 3 Suggest a reason why the preparation can be carried out at
each shaking. During this time, set up a clean distillation room temperature.
apparatus in a fume cupboard or in a well-ventilated 4 List the steps taken to purify the product and identify the
laboratory. impurities removed at each stage.
5 Why is sodium hydrogencarbonate used in the purification
Separation and purification process instead of a stronger alkali such as sodium
D Add about 3 g of powdered anhydrous calcium chloride to the hydroxide?
flask. Shake to dissolve. 6 Describe how the pressure is released in Stage F.
E Pour the contents of the flask into a tap funnel. Allow the two 7 Which layer contains the chloroalkane in Stage G?
layers to separate. Run off the lower layer. 8 What is the plug of cotton wool for in Stage I?
F Neutralise your product in the tap funnel by adding a solution of 9 Why is the distillation an example of fractional distillation?
sodium hydrogencarbonate, 2 cm3 at a time. Stopper and shake 10 The density of 2-methylpropan-2-ol is 0.786 g cm−3. In
after each addition. Take care to release the pressure. Continue an experiment, 0.75 g of 2-chloro-2-methylpropane was
until no more carbon dioxide forms when you add alkali. obtained.
G Allow the layers to separate and run off the lower layer. Pour a) Calculate the theoretical yield of product.
your product into a small, dry conical flask. b) Hence, calculate the percentage yield.
H Add 2–3 small spatula measures of anhydrous sodium
sulfate to the flask. Swirl the mixture. Stopper and allow to thermometer
stand for about 5 minutes.
screw cap adaptor
I Carefully pour your product through a small funnel
fitted with a plug of cotton wool into the pear-shaped water out
distillation flask. Add a few anti-bumping granules. tube to
Weigh the receiving flask and reassemble the sink
distillation apparatus (Figure 6.3.16). impure product
gauze on small gap receiver with
J Distil the product. Heat gently. Continue heating just tripod adaptor with
water in
strongly enough to distil the liquid at a rate of not more heat
vent
than about 2 drops per second. Collect the fraction that
boils between 48 °C and 52 °C.
K Reweigh the receiving flask to determine the yield. Figure 6.3.16 Distillation apparatus.
Tip Tip
Bumping is violent boiling which shakes the apparatus and can throw liquid from the For practical guidance, refer to
container in which it is being heated. Adding a few fragments of porous pottery or Practical skills sheet 8, ‘Synthesising
some jagged anti-bumping granules cuts the risk of bumping by helping the bubbles of organic liquids’, which you can access
vapour to form smoothly as the liquid boils. online at [Link]/
EdexcelChemistry.
6.3.8 Chemical properties of alcohols 223
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Key terms Oxidation by acidified potassium dichromate(vi)
All alcohols react in a similar way with phosphorus halides because they all
An aldehyde contains the functional contain the –OH functional group. However, different classes of alcohol
group —CHO as in propanal. form different products when reacted with oxidising agents such as acidified
A ketone contains the functional group potassium dichromate(vi). A primary alcohol is oxidised to an aldehyde and
C=O as in propanone. then to a carboxylic acid. A secondary alcohol is oxidised to a ketone.
There is no reaction with a tertiary alcohol.
A carboxylic acid contains the
functional group —COOH as in CH3
O O
propanoic acid (Figure 6.3.17).
CH3 CH2 C C O CH3 CH2 C
H CH3 O H
propanal propanone propanoic acid
an aldehyde a ketone a carboxylic acid
Figure 6.3.17 Propanal, propanone and propanoic acid.
Oxidation of a primary alcohol
This takes place in two steps. In the first step, an aldehyde is formed
(Figure 6.3.18). If this is the required product, the apparatus used is that
shown in Figure 6.3.19. This arrangement allows the aldehyde to distil
off as soon as it forms and prevents further oxidation of the aldehyde.
H H H H H
H
H C C C
H OH H C C C + 2H+ + 2e–
O
H H H H H
propan-1-ol propanal
an aldehyde
Figure 6.3.18 Oxidation of propan-1-ol to propanal by acidified K 2Cr 2O7. The oxidising
agent takes away the electrons (Section 3.2).
This oxidation can also be represented as a simplified equation, where [O]
represents the oxidising agent:
CH3CH2CH2OH + [O] → CH3CH2CHO + H2O
water out
tube to
sink
receiver with
excess propan-1-ol + water in adaptor with
sodium dichromate(VI) heat vent
+ dilute sulfuric acid
propanal
Figure 6.3.19 Apparatus used to oxidise a primary alcohol to an aldehyde. The aldehyde
distils off as it forms.
224 6.3 Halogenoalkanes and alcohols
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In the second stage, the aldehyde is oxidised further to a carboxylic acid
(Figure 6.3.20). If a carboxylic acid is the required product, the primary
alcohol is heated with the oxidising agent and a reflux condenser is used
(Figure 6.3.21). The reflux condenser ensures that any volatile aldehyde
reflux
condenses and flows back into the flask, where excess oxidising agent ensures condenser
complete conversion into the carboxylic acid.
H H H H
O O
H C C C + H2O H C C C + 2H+ + 2e–
H OH
H H H H
propan-1-ol with
propanal propanoic acid
excess potassium
dichromate(VI)
Figure 6.3.20 Oxidation of propanal to propanoic acid. Oxidation is completed when a and sulfuric acid heat
primary alcohol is heated with acidified potassium dichromate(vi) in the apparatus shown
in Figure 6.3.21. This converts the alcohol first to an aldehyde, then to a carboxylic acid. Figure 6.3.21 Apparatus used to oxidise
a primary alcohol to a carboxylic acid. The
A simplified equation for the second step is reflux condenser ensures that any volatile
CH3CH2CHO + [O] → CH3CH2COOH aldehyde condenses and flows back into
the flask, where excess oxidising agent
Oxidation of a secondary alcohol ensures complete conversion.
This produces a ketone (Figure 6.3.22).
H H H H H
+ –
H C C C
H H H C C C H + 2H + 2e
H OH H H O H
propan-2-ol propanone
a ketone
Figure 6.3.22 Oxidation of propan-2-ol produces propanone, a ketone.
Distinguishing between types of alcohol
An acidified solution of potassium dichromate(vi) is orange. It turns green
on warming with primary or secondary alcohols as the orange colour of
Cr2O72−(aq) turns to the green colour of Cr3+(aq).
Tertiary alcohols do not change the colour of acidified potassium
dichromate(vi), so can easily be distinguished from primary and secondary
alcohols which do (Figure 6.3.23).
Figure 6.3.23 The result of warming
three alcohols with an acidic solution of
potassium dichromate(vi). Dichromate(vi)
ions are reduced to green chromium(iii)
ions if there is a reaction. Tertiary
alcohols do not react with potassium
dichromate(vi).
6.3.8 Chemical properties of alcohols 225
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However, this reaction does not distinguish between primary and secondary
Tip alcohols so a further test is needed. This is carried out on the oxidation
Mild oxidising agents are used to make products, the aldehydes and ketones, formed by the alcohols.
sure there is no oxidation of ketones at
Aldehydes formed from primary alcohols can easily be oxidised further to
all. Strong oxidising agents may cause
carboxylic acids. Ketones are not easily oxidised.
some oxidation of ketones by breaking
carbon–carbon bonds. So, using mild oxidising agents, such as Fehling’s solution or Benedict’s
solution, it is possible to distinguish between aldehydes and ketones and,
hence, between the alcohols which formed them.
Distinguishing between aldehydes and ketones
Fehling’s reagent does not keep, so it is made when required by mixing
two solutions. One solution is copper(ii) sulfate in water. The other is a
solution of 2,3-dihydroxybutanedioate (tartrate) ions in strong alkali. The
2,3-dihydroxybutanedioate ions form a complex with copper(ii) ions so that
they do not precipitate as copper(ii) hydroxide with the alkali.
Benedict’s solution is similar to Fehling’s solution but is more stable. It is less
strongly alkaline and does not react so reliably with all aldehydes.
Aldehydes reduce the copper(ii) ions in Fehling’s, or Benedict’s, solution
to copper(i), which then precipitates in the alkaline conditions to give an
orange-brown precipitate of copper(i) oxide (Figure 6.3.24).
Tip
Oxidation of primary or secondary
alcohols occurs by loss of a hydrogen
atom from the carbon atom to which the
—OH group is attached. Primary alcohols
have two of these hydrogen atoms
so can be oxidised via aldehydes to
carboxylic acids in two steps. Secondary
alcohols contain one such hydrogen so
can be oxidised to ketones in one step.
Tertiary alcohols have no hydrogen on
this atom so are not oxidised, except
by powerful oxidising agents such as
concentrated nitric acid which can break
C—C bonds.
Figure 6.3.24 Fehling’s reagent is used to test for aldehydes. The reagent has a blue colour as
it contains copper(ii) ions. The test tube in the middle contains Fehling’s reagent that has been
reduced by an aldehyde, to form an orange-brown precipitate of copper(i) oxide. The test tubes
on the left and right contain Fehling’s reagent and ketones. Ketones do not react with Fehling’s
reagent, hence the colour is unchanged.
Tip
Infrared spectroscopy can be used to detect the functional groups in organic
molecules. It is an analytical tool that can be used to show the change in functional
groups when alcohols are oxidised (Section 7.2).
226 6.3 Halogenoalkanes and alcohols
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Core practical 5
Tip
The oxidation of propan-1-ol Refer to Practical skills sheet 8,
A student oxidised the primary alcohol, propan-1-ol, to see how the choice of conditions affected ‘Synthesising organic liquids’.
the product formed according to the following instructions. The products obtained by each
method were then tested and the results of the tests are shown below.
Part 1: Complete oxidation of propan-1-ol
A Add 5 cm3 water to a boiling tube. Wearing gloves, add 6 g of sodium dichromate(vi).
water out
Shake and stir to dissolve.
B Put 1.5 cm3 propan-1-ol into a small pear-shaped flask. Add 5 cm3 water and a couple condenser
of anti-bumping granules. Fit a reflux condenser and clamp the apparatus ready for
heating as in Figure 6.3.25.
water in
C Taking great care, add 2 cm3 of concentrated sulfuric acid down the condenser. Add it
drop by drop from a dropping pipette.
D Then, while the mixture is still warm, begin to add the solution of sodium dichromate(vi)
down the condenser. Add this solution drop by drop too – just fast enough to keep the
mixture boiling without any further heating.
reaction
E When you have added all the sodium dichromate(vi) solution, heat the mixture gently mixture anti-bumping
granules
with a small flame for 10 minutes.
F Stop heating and rearrange the apparatus for Figure 6.3.25 Apparatus for
distillation as in Figure 6.3.26. heating under reflux.
G Distil about 3 cm3 into a small flask.
Part 2: Partial oxidation of ethanol thermometer
Now repeat the oxidation as follows using half as much of
the oxidising agent as in part 1. Also allow the product to screw cap
adaptor
distil off as it forms. water out
H Add 10 cm3 of dilute sulfuric acid to a pear-shaped flask. tube to
reaction mixture
Wearing gloves, add 3 g sodium dichromate(vi) together after heating sink
with 2 or 3 anti-bumping granules. under reflux
gauze on small receiver with
I Add 1.5 cm3 propan-1-ol a few drops at a time. Shake gap
tripod adaptor with
to mix the contents until all the solid has dissolved. water in vent
heat
J Set up the apparatus for distillation as in Figure
6.3.26. Then gently distil 2–3 cm3 of liquid into a ice-water
small flask cooled in a beaker of iced water. Figure 6.3.26 Apparatus for distillation.
Tests and results
Table 6.3.5
Test Observations with product Observations with product
from Part 1 from Part 2
1 Carefully, note the smell of the Acrid smell. Fruitier smell.
product.
2 Add a solution of sodium carbonate Neutralises a significant quantity of the Only neutralises a drop or two of
to 1 cm3 of the distillate. carbonate solution, giving off bubbles of sodium carbonate solution.
Note how much of the solution you gas.
need to neutralise the sample.
3 Add a few drops of the product to No reaction. Reagent turns from blue to
freshly made Fehling’s solution. Heat green before the main colour is
the mixture in a hot water bath. Look a red–orange precipitate.
for colour changes and the formation
of a precipitate.
6.3.8 Chemical properties of alcohols 227
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Questions
1 Show that the results indicate that two distinct oxidation 5 a) By considering the structures of propan-1-ol and the two
products have been formed and identify the two oxidation products, list the three compounds in order of
products. increasing boiling point and explain your order.
2 Write an equation for the reaction of sodium carbonate with b) Hence discuss the use of the condenser in the preparation
the oxidation product from Part 1. of these two compounds from propan-1-ol.
3 Name the red–orange precipitate formed in Test 3 and the 6 What precautions should be taken with the dropping pipette after
oxidation product formed at the same time. it has been used to add concentrated sulfuric acid in Stage C?
4 Write half-equations for the oxidation of propan-1-ol 7 What particular precaution should be taken during Stage F?
a) to the product formed in Part 1 8 State where the clamps should be placed on the apparatus
b) to the product formed in Part 2. in Parts 1 and 2.
Key term
Dehydration by concentrated phosphoric acid
When a substance is dehydrated it Alcohols can be dehydrated to alkenes by heating with concentrated acids
loses water. This can be by the loss of such as phosphoric or sulfuric. Concentrated phosphoric acid is preferred as
water molecules from crystals such as it gives a purer product. This is because, unlike concentrated sulfuric acid, it
CuSO4.5H2O or by the removal of an H is not also an oxidising agent and so leads to fewer side reactions.
atom and an —OH group from adjacent
atoms leading to the formation of a
double bond. CH2 CH2
H2C CHOH conc. H3PO4 H2C CH
An alcohol such as cyclohexanol + H2O
H2C CH2 H2C CH
can be heated with concentrated
CH2 CH2
phosphoric acid and the alkene formed,
cyclohexene, distilled off cyclohexanol cyclohexene
(Figure 6.3.27). This is an elimination Figure 6.3.27 Dehydration of cyclohexanol to cyclohexene.
reaction (see also Section 6.3.4).
Tip Test yourself
Although knowledge of the mechanism 32 Use oxidation numbers to show that it is the chromium that is
for this reaction is not expected, it reduced when Cr2O72−(aq) ions turn into Cr3+(aq) ions.
is one of several examples where 33 Predict the products, if any, of oxidising the following alcohols with
removal of an —OH group occurs in acidified potassium dichromate(vi):
acid conditions. Protonation of the —OH a) butan-1-ol when the product is distilled off immediately
group occurs first so that it is actually a
b) butan-1-ol when the reagents are heated under reflux for some time
water molecule which is lost. H+ is also
c) butan-2-ol
lost from an adjacent carbon, so overall
the acid acts as a catalyst. d) 2-methylbutan-2-ol.
34 Write both half equations and simplified overall equations using [O],
Tip to show the oxidation of different alcohols to form:
a) propanal
There are two common elimination
reactions which form alkenes: b) butanoic acid
c) cyclohexanone.
● the removal of water from an alcohol
under acid conditions 35 a) Write an equation for the dehydration of propan-2-ol using
● the removal of a hydrogen halide
concentrated phosphoric acid and name the type of reaction.
from a halogenoalkane under b) Dehydration of butan-2-ol forms several products. Draw and name
alkaline conditions. each product.
228 6.3 Halogenoalkanes and alcohols
469983_6.3_Chem_Y1-2_209-[Link] 228 13/04/19 10:09 PM
Chapter summary
l The –OH group in an alcohol can be replaced by
Chapter 6.3 Halogenoalkanes
a halogen atom. Chloroalkanes are best prepared
and alcohols using phosphorus(v) chloride. Bromoalkanes use
l Halogenoalkanes can be classified as primary, hydrogen bromide, which is made in the reaction
secondary or tertiary. They can be prepared from flask by mixing 50% sulfuric acid with potassium
alkanes by free-radical substitution, from alkenes bromide. Iodoalkanes can be made using red
by electrophilic addition and from alcohols by phosphorus and iodine.
reaction with PCl5 or HX. l Chlorination of 2-methylpropan-2-ol with
l The polar carbon–halogen bond means that concentrated hydrochloric acid (Core practical 6)
the carbon atom in the bond is δ+ and can be involves the use of a separating funnel, drying the
attacked by nucleophiles (electron pair donors) liquid with an anhydrous salt and determining the
in nucleophilic substitution reactions and in boiling temperature during distillation.
elimination reactions. l Whereas all alcohols react in a similar way with
l Nucleophilic substitution with water (hydrolysis) phosphorus halides or hydrogen halides, different
or with hydroxide ions from aqueous potassium classes of alcohol form different products when
hydroxide forms alcohols. The rate of reaction is reacted with oxidising agents such as acidified
greater with halogenoalkanes containing the larger potassium dichromate(vi). A primary alcohol is
halogens (weaker C–X bond) and with tertiary oxidised to an aldehyde and then to a carboxylic
halogenoalkanes (Core practical 4). acid. A secondary alcohol is oxidised to a ketone.
l Nucleophilic substitution with cyanide ions from There is no reaction with a tertiary alcohol.
potassium cyanide forms nitriles and the carbon l To prepare an aldehyde from a primary alcohol and
chain length is increased by one. prevent further oxidation, the aldehyde is distilled
l Nucleophilic substitution with ammonia molecules away from the oxidising agent as soon as it forms.
forms amines. Using an excess of ammonia To produce a carboxylic acid, the reaction mixture
improves the yield of primary amines and limits is heated under reflux (Core practical 5).
further substitution. l Aldehydes can be distinguished from ketones
l In ethanolic solution, the hydroxide ion from using mild oxidising agents. Aldehydes reduce the
potassium hydroxide acts as a base and produces copper(ii) ions in Benedict’s or Fehling’s solutions
alkenes in an elimination reaction. to form a red-brown precipitate of copper(i) oxide.
l Alcohols can be classified as primary, secondary or Ketones have no effect on Benedict’s or Fehling’s
tertiary. solutions.
l Alcohols burn in a plentiful supply of air with a l Alcohols can be dehydrated to alkenes by heating
clean blue flame. Methanol and ethanol are both with concentrated phosphoric acid. This is an
common fuels and fuel additives. elimination reaction.
Chapter summary 229
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Exam practice questions
Consider the following reaction scheme:
1 Propylamine can be formed either in a two-
4
step synthesis starting from bromoethane
CH3 CH2CH2CH2OH X CH3 CH2CH2COOH or in a one-step synthesis starting from
W Y 1-bromopropane.
a) For each step, give the reagents and
CH3 CH2CH2CH2I conditions and write an equation. (10)
Z b) Identify a disadvantage of each synthetic
route. (2)
a) State the reagents and conditions needed
to convert compound W directly to Consider the following synthesis of butanoic
5
compound Y. (3) acid:
b) Draw the structure of X and state how the C4H9Br → C4H9OH → CH3CH2CH2COOH
conditions you have given in part (a) could A B
be modified to produce compound X a) Draw displayed formulae for compounds A
instead of Y.(2) and B. (2)
c) Give the reagents and conditions b) For each step, give reagents and conditions
for converting compound W into and name the type of reaction involved. (6)
compound Z. (2)
Consider the six reactions shown below.
6
A few drops of a halogenoalkane were added
2
1
to 2 cm3 of ethanol in a test tube and 5 cm3 of
CH3 CH2CH2Br CH3 CH2CH2OH
aqueous silver nitrate was then added. The test
2
tube was then placed in a water bath and after
5
a few minutes a cream precipitate had formed. 3
This precipitate was soluble in concentrated 6 4
ammonia.
a) Explain why ethanol was used in this CH3 CH CH2
experiment. (2)
b) State why the test tube containing the a) For each reaction give the necessary reagent
mixture was warmed in a water bath. (1) and conditions. (11)
c) Give the formula of the precipitate. b) One of the six reactions does not show
Explain your answer. (2) formation of the major product. Identify
d) Deduce the type of halogenoalkane. (1) which reaction, give the major product and
explain why the product shown is not the
Draw the structure of the organic product of
3 major product. (3)
reacting:
a) 2-bromo-2-methylpropane under reflux Isomers of C4H9Br include CH3CH2CH2CH2Br
7
with a hot solution of potassium hydroxide and (CH3)3CBr.
in ethanol (1) Both react with aqueous potassium hydroxide
b) 1-iodopropane with an aqueous solution of to form alcohols, but the reaction mechanism is
potassium hydroxide (1) different in each case.
c) 1-bromopropane with excess ammonia in a) Draw the structure of the intermediate
ethanol (1) or transition state formed during each
d) butane-1,4-diol with acidified potassium reaction and explain why the two routes are
dichromate(vi) under reflux conditions (1) different. (4)
e) cyclohexanol with phosphorus(v)
chloride. (1)
230
6.3 Halogenoalkanes and alcohols
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b) State how the rate of the reaction would Identify three improvements that should be
change if CH3CH2CH2CH2I were used made to the arrangement of apparatus and give
instead. Explain your answer. (2) a reason for each improvement you suggest.
c) Draw the alternative product formed You may assume suitable clamps are used. (6)
in each case if hot ethanolic potassium
10* A sample of cyclohexene was prepared by the
hydroxide were used instead. (2)
dehydration of cyclohexanol using concentrated
8 This question is about 2-bromo-3-methylbutane. phosphoric acid. The reaction mixture was
a) Give the name and formula of the product distilled and the distillate obtained contained
of heating 2-bromo-3-methylbutane under both cyclohexene and water.
reflux with aqueous potassium hydroxide. Describe how acidic impurities can be
Name the type of reaction taking place. (2) removed from the distillate and a sample of dry
b) Give the name and formula of the cyclohexene obtained.
two isomers that form on heating The density of cyclohexene is 0.81 g cm–3. (6)
2-bromo-3-methylbutane under reflux
with alcoholic potassium hydroxide. Name 11 a) 11 cm3 of propan-2-ol (density 0.78 g cm−3)
the type of reaction taking place. (5) are oxidised with sodium dichromate(vi)
c) Give the name and formula of the product in dilute sulfuric acid. After separating and
of heating 2-bromo-3-methylbutane under purifying the product, this preparation
reflux with alcoholic ammonia. Name the produces an 80% yield. Name the product,
type of reaction taking place. (3) state how it is separated and calculate the
d) Give the name and formula of the product mass of pure product obtained. (4)
of heating 2-bromo-3-methylbutane under b) 9.25 cm3 of butan-1-ol (density 0.81 g cm−3 )
reflux with potassium cyanide in ethanol.(2) are heated with an excess of red phosphorus
and iodine. After separating and purifying the
9 A student attempted to prepare ethanal by product, this preparation produces an 85%
oxidation of ethanol. The diagram shows the yield. Name the product and calculate the
apparatus set up by the student to collect the mass of pure product obtained. (4)
ethanal by distillation and to measure its boiling
12 Pentan-1-ol reacts with sodium to form
point.
compound P (C5H11ONa) and hydrogen gas
according to the equation:
2CH3CH2CH2CH2CH2OH + 2Na
→ 2CH3CH2CH2CH2CH2ONa + H2
water in
Reaction of P with bromomethane forms
ethanol mixed
compound Q (C6H14O) which has a much
with acidified lower boiling point than pentan-1-ol.
potassium
dichromate(VI)
Reaction of P with bromoethane forms a
water similar compound R (C7H16O) and in a
out
heat competing reaction also forms ethene.
beaker a) Give a structural formula for Q. (1)
b) i) Name and outline a mechanism for the
reaction between P and bromomethane. (3)
231
Exam practice questions
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ii) Explain why Q has a much lower i) Give a skeletal formula for S. (1)
boiling point than pentan-1-ol. (2) ii) Explain how S is formed and name the
c) i) Give a structural formula for R. (1) type of reaction. (2)
ii) Explain how P can react with iii) Explain why a similar sequence of reactions
bromoethane to form ethene and name starting from 2-bromopentan-1-ol does not
the type of reaction involved. (3) form a cyclic compound. (2)
d) Consider the reaction sequence to form the iv) State with a reason whether the boiling
cyclic compound S (C5H10O) below. point of S will be higher or lower than
pentan-1-ol. (2)
Na – +
Br CH2 CH2 CH2CH2CH2 OH Br CH2 CH2CH2CH2 CH2 O Na
heat
C5H10O + NaBr
232
6.3 Halogenoalkanes and alcohols
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Modern analytical techniques I
7
7.1 Mass spectra of organic
compounds
In modern chemistry, the use of instrumental techniques such as mass
Tip spectrometry for analysis is more important than chemical analysis. Although
Tests and observations in inorganic the instruments may be expensive to purchase, analysis using them is quick to
chemistry are described in Section 5.11. perform and extremely accurate. Chemical analysis also destroys the sample
by reacting it; instrumental analysis uses a very small sample and, in most
cases, does not destroy it.
Mass spectrometry is an accurate technique for determining relative atomic
masses (Section 1.4). Mass spectrometry can also help to determine the
relative molecular masses and molecular structures of organic compounds.
In this way, it can be used to identify unknown compounds. The technique
is extremely sensitive and requires very small samples, which can be as small
as one nanogram (10−9g).
Inside a mass spectrometer (Figure 7.1) there is a very high vacuum so that it is
possible to produce and study ionised molecules and fragments of molecules.
The molecular fragments could not exist other than in a high vacuum.
Figure 7.1 A mass spectrometer used for
analysis. The sample is fed into the bottom
left of the instrument where it is vaporised
and ionised. The ions are accelerated along
the U-shaped glass tube and steered by
electric and magnetic fields to reach the
gold-coloured detector on the bottom right.
This part of the instrument measures about
70 cm × 50 cm.
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A beam of high-energy electrons bombards the molecules of the sample
(Figure 7.2). This turns them into ions by knocking out one or more electrons.
high-energy
e–
electron e–
e–
+
M M
molecule in sample cation with an unpaired electron
(radical cation)
Figure 7.2 High-energy electrons ionising a molecule in a mass spectrometer. Knocking
out one electron leaves a positive ion, shown by the symbol M+.
Bombarding molecules with high-energy electrons not only ionises them,
but may also split them into fragments (Figure 7.3). As a result, the mass
spectrum consists of a ‘fragmentation pattern’.
e– e– positively
high- charged
energy m1+ fragment
electron (detected)
ionisation fragmentation
m1.m2 m1. m2+
molecule in sample (M) positive
shown as made up of ion M+
two parts
e– m2 uncharged
fragment
(not detected)
Figure 7.3 Ionisation and fragmentation of a single molecule m1.m2 which fragments
into two parts m1+ and m2. Only charged species show up in the mass spectrum because
electric and magnetic fields have no effect on neutral fragments. So, in this case, the
instrument only detects m1+.
Molecules break up more readily at weak bonds, or at bonds which give rise
to more stable fragments. It turns out that positive ions with the charge on a
secondary or tertiary carbon atom are more stable than ions with the charge
on a primary carbon atom. Species such as CH3CO+ are also more stable
where a bond breaks adjacent to the C=O double bond.
After ionisation and fragmentation, the charged species are accelerated
and deflected by electric and magnetic fields. The extent of the deflection
Key term depends on the ratio of the mass of the fragment to its charge, its mass-to-
charge ratio (m/z). The number of charges (1, 2 and so on) is called z, but
The mass-to-charge ratio (m/z) is the in most of the examples you will meet z = 1.
ratio of the relative mass of an ion to its
The positive ions finally reach a detector where they cause a small current.
charge.
This is amplified and the signal fed to a computer.
234 7 Modern analytical techniques I
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The output from the detector of a mass spectrometer is often present as
a ‘stick diagram’ (see Figure 7.4). This shows the strength of the signal
Key term
produced by ions of varying mass to charge ratio. The scale on the vertical
The molecular ion (also called parent
axis shows the relative abundance of the ions. The horizontal axis shows
ion) is the positive ion, M+, formed
the m/z values.
when a molecule loses one electron.
When analysing molecular compounds, the peak with the largest m/z value This species is also a radical since one
is usually the ionised whole molecule. So the mass of this ‘molecular ion’ or electron has been lost from a pair in
‘parent ion’, M, is the relative molecular mass of the compound. the molecule, leaving one unpaired
electron.
Tip
Recently, alternative less expensive mass spectrometers have been developed.
These involve alternative ionisation techniques (such as electrospray ionisation)
and alternative mass analysers which use quadrupoles instead of magnetic fields or
measure time of flight. (Your examination will not test these alternative methods.)
Tip
The existence of isotopes shows up in mass spectra. For instance, bromine
Take care when describing the peak for
exists as the isotopes 79Br and 81Br in almost equal amounts. The spectrum
the molecular ion. Do not simply use
of bromoethane (Figure 7.4) shows two molecular ion peaks at m/z = 108
the word highest to describe the peak
and 110 of almost equal abundance. Fragmentation of either molecular ion
because highest could be confused
(Figure 7.5) first involves breaking of the C−Br bond to produce the fragment
with tallest, which applies to the most
ion C2H5+ with m/z = 29. Further successive loss of hydrogen atoms forms
abundant ion.
ions which produce the peaks at m/z = 28, 27 and 26.
100 Figure 7.4 The mass spectrum of
bromoethane. There are two molecular
80 ion peaks because of the presence of two
Relative abundance/%
isotopes of bromine.
60
40
20
0
10 20 30 40 50 60 70 80 90 100 110
Mass-to-charge ratio (m/z)
loss of 79Br Figure 7.5 Fragmentation of the molecular
[C2H579Br]+
ions of bromoethane.
m/z = 108 [C2H3]+
[C2H5]+ loss of H [C2H4]+ loss of H
m/z = 28 m/z = 27
m/z = 29
[C2H581Br]+
m/z = 110 loss of 81Br
Chemists study mass spectra in order to gain insight into the structure of
molecules. They identify the fragments from their relative masses, and then
piece together likely structures, sometimes with the help of evidence from
other methods of analysis, such as infrared spectroscopy. The example on
page 236 shows use of mass spectrometry to identify a compound.
7.1 Mass spectra of organic compounds 235
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Example
The mass spectrum in Figure 7.6 is known to be that of ethyl propanoate,
CH3CH2COOCH2CH3 or methyl butanoate, CH3CH2CH2COOCH3. Use the
spectrum to identify which of the two isomers produced the spectrum.
100
80
Relative abundance/%
60
40
20
0
10 20 30 40 50 60 70 80 90 100
Mass-to-charge ratio (m/z)
Figure 7.6 Mass spectrum of an isomer of C5H10O2.
Answer
The spectrum shows a molecular ion peak at m/z = 102. Both isomers
have molecular formula C5H10O2 so both would have a molecular ion peak
at m/z = 102.
The other main peaks shown are at m/z = 57 and m/z = 29.
Ethyl propanoate can produce a peak at m/z = 57 for the fragment ion
CH3CH2CO+ and a peak at m/z = 29 for the fragment ion CH3CH2+. Both
of these peaks correspond to fragment ions formed by breaking of the
C − C bonds adjacent to the C = O group.
Methyl butanoate is unlikely to produce a peak at m/z = 57. Its major
fragment peaks are likely to be at m/z = 71 for the ion CH3CH2CH2CO+ or
at m/z = 43 for CH3CH2CH2+. Neither of these values correspond to major
peaks in the given spectrum.
The evidence indicates that the compound is ethyl propanoate.
Test yourself
1 The mass spectrum of butane, C4H10, is shown on 100 R
the right.
a) i) W
hich peak in the mass spectrum of butane 80
Relative abundance/%
corresponds to the molecular ion?
60
ii) What is the relative mass of this ion?
Q
b) Suggest the identity of the fragments labelled P, 40
Q and R.
c) Suggest a reason why the peak at m/z = 15 is 20
S
relatively weak.
P
d) Use symbols to show one way in which the 0
10 20 30 40 50 60
parent ion of butane could fragment.
Mass-to-charge ratio (m/z)
236 7 Modern analytical techniques I
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7.2 Infrared spectroscopy Test yourself
Spectroscopy is a term which covers a range of practical techniques 2 Put the following stages in order
for studying the composition, structure and bonding of compounds. during mass spectrometry:
Spectroscopic techniques are now the essential ‘eyes’ of chemistry and have acceleration, fragmentation,
many uses in detection and measurement. ionisation, deflection according
The range of techniques available covers many parts of the electromagnetic to m/z, detection, vaporisation.
spectrum – including the infrared, visible and ultraviolet regions. The 3 Give two reasons why it is
instruments used are called either spectroscopes, which emphasises the use important to have a high
of techniques for making observations, or spectrometers, which emphasises vacuum inside a mass
the importance of measurements. spectrometer.
Infrared spectroscopy (or IR spectroscopy) is an analytical technique used to 4 Propan-1-ol and propan-2-ol are
identify functional groups in organic molecules. Infrared radiation from a position isomers. Suggest which
glowing lamp or fire makes us feel warm. This is because infrared frequencies has a major peak in its mass
correspond to the natural frequencies of vibrating atoms in molecules. Our spectrum at m/z = 31 and predict
skin warms up as the molecules absorb infrared and vibrate faster. a major fragmentation peak in
the spectrum of the other isomer.
Most compounds absorb infrared radiation. A sample in a spectrometer Explain your answers.
(Figure 7.7) absorbs infrared radiation at wavelengths which correspond
to the natural frequencies at which vibrating bonds in the molecules bend
and stretch. However, it is only molecules that change their polarity as they
vibrate which interact with IR. The absorptions are detected, analysed and
the absorption spectrum displayed by a computer or printed (Figure 7.8).
The absorption spectrum is a plot of transmittance against wavenumber.
Key terms
An absorption spectrum is a plot
showing how strongly a chemical
absorbs radiation over a range of
frequencies.
Transmittance on the vertical axis
of infrared spectra measures the
percentage of radiation which passes
through the sample. The troughs appear
at those wavenumbers where the
compound absorbs strongly. Chemists
often refer to these dips in the line
as ‘peaks’ because they indicate high
levels of absorption.
Infrared wavenumbers range from
400 to 4500 cm−1. The wavenumber
Figure 7.7 Using an infrared spectrometer. The instrument covers a range of infrared
is the number of waves in 1 cm.
wavelengths and a detector records how strongly the sample absorbs at each wavelength.
Spectroscopists find the numbers more
Wherever the sample absorbs, there is a dip in the intensity of the radiation transmitted
convenient than wavelengths.
which shows up as a dip in the plot of the spectrum.
7.2 Infrared spectroscopy 237
469983_07_Chem_Y1-2_233-[Link] 237 13/04/19 10:09 PM
radiation sample detector computer printer
source
Figure 7.8 Essential features of a modern single-beam spectrometer.
Bonds vibrate in particular ways and absorb radiation at specific wavelengths.
This means that it is possible to look at an infrared spectrum and identify
particular functional groups. Figure 7.9 shows, on the left, a diatomic
molecule stretching and, on the right, a V-shaped molecule bending.
Figure 7.9 Bond vibrations give rise
to absorptions in the infrared region.
Vibrations of molecules which cause
a fluctuating polarity interact with
electromagnetic waves.
Tip
Only molecules which change polarity as
O C O O C O
they vibrate will absorb IR. Polar molecules
such as CO always absorb. Non-polar symmetrical stretch asymmetrical stretch
molecules such as N2 or O2 never absorb, no change in dipole net dipole changes
but some non-polar molecules such as CO2 does not absorb IR IR is absorbed
will absorb as some stretching or bending Figure 7.10 Symmetrical and asymmetrical
vibrations can cause a change in polarity stretching vibrations of carbon dioxide.
(Figure 7.10).
Spectroscopists have found that it is possible to correlate absorptions in the
region 4000 to 1500 cm−1 with the stretching or bending vibrations of particular
bonds. As a result, infrared spectra give valuable clues about the presence of
functional groups in organic molecules. The important correlations between
different bonds and observed absorptions are shown in Figure 7.11.
Wavenumber ranges
4000 cm–1 2500 cm–1 1900 cm–1 1500 cm–1 400 cm–1
C H C C C C C O
O H C N C O C X
N H
single bond triple bond double bond single bond
stretching stretching stretching stretching and
vibrations vibrations vibrations bending vibrations
Figure 7.11 The main regions of the infrared spectrum and important correlations
between bonds and observed absorptions.
238 7 Modern analytical techniques I
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However, the strength of a particular bond varies in different molecules
because of the effect of different atoms or groups next to the bond. Data
on bond strengths are usually given as mean bond enthalpies which take
into account the different environments. For the same reasons, infrared
absorption wavenumbers of particular bonds are usually quoted as a range of
values. In some cases more specific ranges are listed and this enables different
Tip
molecules to be identified (see the infrared spectroscopy data sheet in the Hydrogen bonding broadens the
Pearson Edexcel Data booklet). For example, the C=O stretching vibrations absorption peaks of −OH groups
distinguish between aldehydes (1740 to 1720 cm−1) and ketones (1720 to in alcohols, and even more so in
1700 cm−1), and the O−H stretching vibrations distinguish between alcohols carboxylic acids, where the O−H
and phenols (3750 to 3200 cm−1) and carboxylic acids (3300 to 2500 cm−1). absorption also overlaps with the C−H
absorption.
The example shows the use of infrared spectra to identify compounds.
Example
Compounds P and Q are isomers with molecular formula C4H10O.
P has an absorption peak in its infrared spectrum at 3355 cm−1. Q has
an absorption peak at 3337 cm−1
When P was heated with acidified potassium dichromate(vi), the colour of
the mixture changed from orange to green. The organic compound formed
was distilled off and was found to have an absorption peak in its infrared
spectrum at 1718 cm−1.
When Q was heated with acidified potassium dichromate(vi), the orange
colour did not change.
Identify compounds P and Q.
Answer
The absorption peaks at 3355 and 3337 cm−1 show the presence of the
O−H functional group, so both compounds are alcohols.
Reaction with acidified potassium dichromate(vi) oxidised P. The
absorption at 1718 cm−1 in the spectrum of the oxidation product shows
the presence of a ketone C=O bond, rather than an aldehyde C=O bond
which would have absorbed between 1740 and 1720 cm−1.
Therefore, the oxidation product must have been butanone,
CH3CH2COCH3, and P must be butan-2-ol, CH3CH2CH(OH)CH3.
When Q was heated with the oxidising agent, no reaction occurred.
So Q must be a tertiary alcohol and is, therefore, 2-methylpropan-2-ol,
(CH3)3COH.
Molecules with several atoms can vibrate in many ways because the vibrations
of one bond affect others close to it. The region between 1500 cm−1 and
400 cm−1 contains absorptions for some single bond stretching vibrations as
well as many bending vibrations. This leads to a very complex pattern in
which it is difficult to identify individual absorptions.
7.2 Infrared spectroscopy 239
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However, this complexity is useful because the unique absorption pattern
Key term here can be used as a ‘fingerprint’ to identify a particular compound, and so
this is called the fingerprint region of the spectrum. The complex pattern
The fingerprint region is the complex
for an unknown compound can be compared with recorded infrared spectra
region of the spectrum below
in a database. An exact match will identify the unknown compound.
1500 cm−1 which contains many single
bond stretching and bending vibrations Infrared spectroscopy is an analytical tool that can be used to monitor the progress
and is unique to each molecule. of an organic synthesis. Comparing the spectrum of the final product with the
known spectrum in a database can be used to check if the product is pure.
Tip
Test yourself
Infrared spectra are unique in giving
information about the absence of 5 Why do the vibrations of O−H, C−O and C=O bonds show up strongly
functional groups. If a characteristic in infrared spectra, while C−C vibrations do not?
absorption is not present in the 6 Figure 7.12 shows the infrared spectra of ethanol, ethanal and
spectrum, then the functional group ethanoic acid.
which would cause it cannot be present a) Which vibrations give rise to the peaks marked with the letters A–G?
in the molecule.
b) Which spectrum belongs to which compound?
c) Why do two of the spectra have broad peaks at wavenumbers
between 3000 and 3500 cm−1?
7 The infrared spectrum of a sample of propanal prepared by oxidation
of propan-1-ol contained a weak absorption at 3437 cm−1. Suggest
two possible reasons for the appearance of this absorption.
8 Suggest reasons why it is better to use infrared spectroscopy to check
the purity of a liquid product from a synthesis than to measure its
boiling temperature.
Figure 7.12 Infrared spectra for three a b
organic compounds.
Transmittance/%
Transmittance/%
B
A
D
4000 2000 1000 600 4000 2000 1000 600
Wavenumber/cm–1 Wavenumber/cm–1
c
T ransmittance/%
F
E
G
4000 2000 1000 600
Wavenumber/cm–1
240 7 Modern analytical techniques I
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Core practical 7 (part 2) Tip
Analysis of inorganic unknowns is
Analysis of some organic unknowns
covered in Core practical 7 (part 1), in
Bottles of six organic compounds have lost their original labels so have been
Chapter 5.
re-labelled with the letters P, Q, R, S, T and U. The compounds are known to be
hexane, hex-1-ene, hexan-2-ol, hexanal, hexanoic acid and 2-bromohexane.
A student carried out a series of test-tube reactions to identify each compound. The
tests are described below.
To interpret the results, you will need to refer to the data sheet ‘Characteristic
reactions of organic functional groups’, which you can access online at
[Link]/EdexcelChemistry. For Question 11, you will also need to
refer to the infrared spectroscopy data sheet from the Pearson Edexcel Data booklet.
Test A
Each of the six compounds was warmed in a separate test tube with acidified
potassium dichromate(vi) solution. Compounds R and U turned the colour of the Tip
solution from orange to green but the other four compounds had no effect.
For practical guidance, refer to Practical
1 Identify which of the compounds could be R or U. skills sheet 9, ‘Analysing organic
2 State how the test tubes were warmed and give a reason for this method. unknowns’, which you can access
Test B online at [Link]/
Samples of R and U were separately added to test tubes containing Fehling’s solution. EdexcelChemistry.
U gave a positive test but R did not.
3 Describe how Fehling’s test was carried out, including any precautions necessary.
4 State what was observed to indicate a positive Fehling’s test and hence identify R
and U.
Test C
Sodium hydrogencarbonate solution was added to compounds P, Q, S and T. Q gave a
positive result but P, S and T did not react.
5 Describe what was observed when Q reacted with sodium hydrogencarbonate solution.
6 Identify Q and write an equation for its reaction with sodium hydrogencarbonate
solution.
Test D
Bromine water was added drop-wise to samples of P, S and T. Only P reacted.
7 Describe what was seen when P reacted with bromine water.
8 Identify P and write an equation for the reaction of P with bromine water to form the
major product.
Test E
Samples of S and T were warmed with aqueous silver nitrate in ethanol. Only T gave a
positive test.
9 Describe what was seen when T reacted with aqueous silver nitrate in ethanol.
10 Identify S and T and write equations for the reactions occurring during Test E.
11 The infrared spectra of the compounds were compared.
a) Give the wavenumber of one absorption each in the spectra of hex-1-ene,
hexan-2-ol, hexanal and hexanoic acid which could be used to identify these
compounds.
b) Suggest how hexane and 2-bromohexane could be positively identified using
their infrared spectra but without referring to one single absorption.
7.2 Infrared spectroscopy 241
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Chapter summary
Chapter 7 Modern analytical or groups next to the bond. Infrared absorption
wavenumbers of particular bonds are quoted as a
techniques I range of values. In some cases, more specific ranges
l In modern chemistry, instrumental techniques for are listed and this enables different molecules
analysis are replacing chemical analysis. Although to be identified. For example, C–H stretching
the instruments may be expensive, using them is vibrations in alkanes (2962–2853 cm−1) and alkenes
quick and accurate and, in most cases, does not (3095–3010 cm−1) differ from those in aldehydes
destroy the sample. (2900–2820 cm−1 and 2775–2700 cm−1) and the
l Mass spectrometry is an accurate technique for O−H stretching vibrations can distinguish between
determining relative atomic masses and also the alcohols and phenols (3750 to 3200 cm−1) and
relative molecular mass of an organic compound carboxylic acids (3300 to 2500 cm−1).
from the molecular ion peak. The molecular ion, l Molecules of organic compounds can produce very
M+, is formed when a molecule loses one electron. complex patterns in which it is difficult to identify
l Bombarding molecules with high-energy electrons individual absorptions. However, this complexity
not only ionises them, but may also split them is useful because the unique absorption pattern
into fragments. A mass spectrum contains a can be used as a ‘fingerprint’. The region between
‘fragmentation pattern’ and analysis of the m/z of 1500 cm−1 and 400 cm−1 is called the fingerprint
these fragments can suggest possible structures of region of the spectrum. The pattern in this region
simple organic compounds. for an unknown compound can be compared with
l Infrared (IR) spectroscopy identifies functional spectra in a database. An exact match will identify
groups in organic molecules. A sample in a the unknown compound.
spectrometer absorbs infrared radiation at l Analysis of unknown organic compounds and
wavelengths that correspond to the natural identification of functional groups by test-tube
frequencies at which vibrating bonds in the reactions is covered in Core practical 7 (part 2).
molecules bend and stretch. The data sheet for Chapter 7 gives details of the
l The strength of a particular bond varies in different characteristic reactions of organic functional
molecules because of the effect of different atoms groups.
242 7 Modern analytical techniques I
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Exam practice questions
Where relevant, use the infrared spectroscopy data sheet from the Pearson Edexcel
Data booklet to help you answer these questions.
The mass spectrum of ethanol is shown below.
1
a) Identify which of the numbered peaks corresponds to each of these
positive ions from ethanol molecules in a low pressure mass spectrometer:
C2H5+, CH2OH+, C2H5O+, C2H5OH+, C2H3+. (3)
b) Write an equation to represent the formation of the molecular ion. (2)
c) i) Write an equation to show how the molecular ion fragments to
give CH2OH+. (2)
ii) Why does the other chemical species formed during this
fragmentation process not show up in the mass spectrum? (2)
100 3
80
Relative abundance/%
60
40 4
1 2
20 5
0
0 10 20 30 40 50
Mass-to-charge ratio (m/z)
2* Oxidation of an alcohol with formula C4H9OH gives a product with the
infrared spectrum shown below. Use the infrared spectroscopy data sheet from
the Pearson Edexcel Data booklet to interpret the spectrum. State the reagents
and conditions used to carry out the oxidation of the alcohol, giving your
reasons, and suggest two possible structures for the alcohol. (6)
100
Transmittance/%
80
60
40
20
0
4000 3500 3000 2500 2000 1500 1000 500
Wavenumber/cm–1
3 The existence of isotopes shows up in the mass spectrum of organic
compounds. The existence of the two chlorine isotopes 35Cl and 37Cl can be
detected in the spectra of chloroalkanes.
a) i) Explain why the mass spectrum of chloroethane has peaks with m/z
values of 64 and 66. (1)
243
Exam practice questions
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ii) State why the ion with an m/z value of 64 is about three times as
abundant as the ion with the m/z value of 66. (1)
b) Explain these statements about the mass spectrum of dichloroethene.
i) Although the Mr of dichloroethene is 97 there is no peak in the
mass spectrum at m/z = 97. (2)
ii) It includes three peaks at m/z values of 96, 98 and 100 with
intensities in the ratio 9 : 6 : 1. (3)
iii) It includes two peaks at m/z values of 61 and 63 with intensities
in the ratio of 3 : 1. (2)
4 High resolution mass spectrometers can measure relative molecular masses
to four decimal places. The Mr of a compound was found to be 72.0625.
Compounds with molecular formulae C5H12, C4H8O and C3H4O2 all
have Mr values of 72 to the nearest whole number.
a) Use the precise relative atomic masses given to identify the correct
molecular formula for this compound. (2)
Element Relative atomic mass
hydrogen 1.0079
carbon 12.0107
oxygen 15.9994
b) When added to aqueous sodium carbonate, the compound reacted
to give an effervescence. Deduce a structure for the compound and
explain your answer. (2)
Three isomeric compounds with molecular formula C3H6O were studied
5
using infrared spectroscopy and mass spectrometry.
Compounds A and B both had absorptions in their IR spectra at about
1720 cm−1. The IR spectrum of C is shown below.
100
Transmittance/%
50
0
4000 3000 2000 1500 1000 500
Wavenumber/cm–1
The mass spectrum of A had major peaks at m/z = 58, 43 and 15, B had
major peaks at m/z = 58 and 29 and C had major peaks at m/z = 58, 57
and 31.
Deduce structures for the three compounds and explain your answer. (9)
244
Modern analytical techniques I
Exam practice questions
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Five compounds were studied using infrared spectroscopy: hexan-1-ol,
6
hexan-3-one, hexanoic acid, hex-1-ene and 1-chlorohexane. IR spectra of
four of these compounds are shown below.
A 100 B 100
Transmittance/%
Transmittance/%
50 50
0 0
4000 3000 2000 1500 1000 500 4000 3000 2000 1500 1000 500
Wavenumber/cm–1 Wavenumber/cm–1
C 100 D 100
Transmittance/%
Transmittance/%
50 50
0 0
4000 3000 2000 1500 1000 500 4000 3000 2000 1500 1000 500
Wavenumber/cm–1 Wavenumber/cm–1
a) Use the infrared spectroscopy data sheet from the Pearson Edexcel
Data booklet to identify which spectrum corresponds to which
compound. (4)
b) Give reagents and conditions for the conversion of:
i) 1-chlorohexane into hexan-1-ol and name the mechanism of the
reaction (3)
ii) hexan-1-ol into hexanal. State how you could use IR spectroscopy
to show that the reaction was complete. (3)
7 An organic compound X contains the elements carbon, hydrogen and
oxygen only. Analysis showed that it contains 54.5% carbon and 9.1%
hydrogen by mass. The mass spectrum showed a molecular ion peak at
m/z = 88 and a fragmentation peak at m/z = 43. IR peaks were observed
at 3408 cm−1 and 1709 cm−1.
When X was heated under reflux with acidified potassium dichromate(vi),
a product Y was formed. The IR spectrum of Y contained a broad peak at
3087 cm−1.
Deduce the structure of compounds X and Y and explain your
deductions. (11)
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Exam practice questions
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Energetics I
8
8.1 Energy changes
Thermochemistry is the study of energy changes in chemistry. With the
help of thermochemistry, chemists can decide whether or not reactions are
likely to occur and explain the stability of compounds. Energy changes from
surroundings chemical reactions are also of great practical importance. The energy changes
during burning are crucial to the fuel and food industries. The prices of fuels
are closely related to their energy values and dieticians give advice related to
system
their knowledge of energy-providing foods
In thermochemistry, the term ‘system’ is important and it has a precise
meaning. It describes just the material or the mixture of chemicals being
Figure 8.1 A system and its surroundings. studied. Everything around the system is called the surroundings (Figure 8.1).
The surroundings include the apparatus, the air in the laboratory – in theory
everything else in the Universe.
In a closed system like that in Figure 8.1, the system cannot exchange matter
with its surroundings because the flask is closed with a bung. It can, however,
Key term exchange energy with the surroundings. If the bung is removed, the system
is described as ‘open’. An open system can exchange both energy and matter
An enthalpy change, ΔH, is the overall with its surroundings.
energy exchanged with the surroundings
when a change happens at constant Whenever a change occurs in a system, there is almost always an energy
pressure and the final temperature is change involving transfer of energy between the system and its surroundings.
the same as the starting temperature. The energy transferred between a system and its surroundings is described
as an enthalpy change when the change happens at constant pressure. The
symbol for an enthalpy change is ΔH and its units are kJ mol−1.
Tip
Scientists use the capital Greek letter ‘delta’, Δ, for a change or difference in a
physical quantity. So, ΔH means change in enthalpy and ΔT means change in
temperature.
8.2 Enthalpy changes
Exothermic changes
Exothermic changes give out energy that often just heats up the surroundings.
Tip Burning is an obvious exothermic chemical reaction. Respiration is another
Remember: in an exothermic change, exothermic reaction in which foods are oxidised to provide energy for living
energy leaves the system, just as things to grow, move and keep warm. Hot packs used in self-warming
people leave a building by the exit. drinks and in treating painful rheumatic conditions also involve exothermic
reactions (Figure 8.2).
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Figure 8.2 A self-warming can of coffee –
pressing the bulb on the bottom of the can
starts an exothermic reaction in a sealed
compartment. The energy released heats
the coffee.
Figure 8.3 shows what happens in the exothermic reaction between calcium
oxide and water. This reaction can be used in hot packs.
Figure 8.3 The exothermic reaction that
takes place in some hot packs.
calcium
oxide
calcium
hydroxide
solution
water
calcium
hydroxide
solid
reactants at products at
room temperature room temperature
and pressure energy and pressure
given out
When one mole of solid calcium oxide reacts with water to form calcium reactants:
hydroxide solution, 1067 kJ of energy are given out. The system loses energy CaO(s) + H2O(l)
by heating the surroundings. This loss of energy from the system means that
Energy
ΔH is negative. The enthalpy change is often written alongside the equation ∆H = –1067 kJ
for the reaction as in this example: product:
Ca(OH)2(aq)
CaO(s) + H2O(l) → Ca(OH)2(aq) ΔH = −1067 kJ mol−1
The energy changes in chemical reactions can be summarised in enthalpy Course of reaction
level diagrams. Figure 8.4 An enthalpy level diagram for
Figure 8.4 shows the enthalpy level diagram for the reaction of calcium the reaction of calcium oxide with water.
oxide with water. Energy is lost to the surroundings and, therefore, the
products are at a lower energy level than the reactants. For this and all other
exothermic reactions, ΔH is negative.
Tip
Arrows in an energy level diagram should be single-headed – pointing down or up.
Never draw double-headed arrows. Activation energy is not shown in an enthalpy level
diagram, but is shown in reaction profile diagrams (Section 9.4).
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Endothermic changes
Endothermic changes take in energy from the surroundings. They are the
opposite of exothermic changes. Melting and vaporisation are endothermic
changes of state. Photosynthesis is an endothermic chemical change. During
photosynthesis, plants take in energy from the Sun in order to convert carbon
dioxide and water to glucose. Figure 8.5 illustrates the use of an endothermic
reaction in a cold pack.
An enthalpy level diagram shows that the system has more energy after an
endothermic reaction than it had at the start. So, for endothermic reactions,
Figure 8.5 Twist a cold pack and it gets
the enthalpy change, ΔH, is positive and the products are at a higher energy
cold enough to reduce the pain of a sports
level than the reactants (Figure 8.6).
injury. When chemicals in the cold pack
react, they take in energy and the pack
gets cold. This, in turn, cools the sprained Test yourself
or bruised area and helps to reduce painful
1 Which of the following changes are exothermic and which are
swelling.
endothermic?
a) melting ice b) burning wood
products:
C6H12O6(s) + 6O2(g)
c) condensing steam d) metabolising sugar
e) subliming iodine
Energy
∆H = +2802 kJ 2 When 1.00 mol of carbon (as graphite) burns completely, 394 kJ of
energy is given out.
reactants: a) Write an equation for the reaction including state symbols and
6CO2(g) + 6H2O(l) show the value of the enthalpy change.
Course of reaction b) Draw an enthalpy level diagram for the reaction including the
Figure 8.6 An enthalpy level diagram for enthalpy change.
photosynthesis. 3 When 0.200 g of methane, CH4 (natural gas), burns completely, it gives
out 11.0 kJ.
a) Write an equation for the reaction when methane burns completely.
Tip
b) Calculate the molar mass of methane.
Chemists measure changes in c) Calculate the energy change when 1.00 mol of methane burns
enthalpy. Energy level diagrams show completely.
the difference in enthalpy between
d) Draw an enthalpy level diagram for the reaction showing the value
the reactants and the products. It is
of the enthalpy change.
not possible to put a scale on these
diagrams to show the absolute levels of
energy in a system.
8.3 Measuring enthalpy changes
The energy given out or taken in during many chemical reactions can be
measured and this makes it possible to calculate enthalpy changes.
Enthalpy changes from burning fuels
Figure 8.7 shows the simple apparatus that can be used to measure the energy
given out from a liquid fuel like methylated spirit (meths). Meths is ethanol
that is made undrinkable by adding some methanol which is toxic.
The results can be used to calculate the energy given out when one mole of
the fuel burns. This is the enthalpy of combustion of the fuel.
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The copper can is acting as a calorimeter. This name means ‘calorie-measurer’
and is based on the energy unit called the ‘calorie’.
metal can
Tip (calorimeter)
The calorie is the energy need to raise the temperature of 1 g water by 1 °C. measured
1 calorie = 4.18 J. The food industry often uses the ‘large calorie’, which is one volume
thousand times larger. 1 Cal = 4.18 kJ. of water
Typically, a calorimeter is insulated from its surroundings and contains
water. The energy from the reaction heats up the water and the rest of the
liquid burner meths
apparatus. An accurate thermometer measures the temperature rise.
The energy transferred to a material can be calculated using this expression: Figure 8.7 Measuring the enthalpy change
when meths is burned. Wear eye protection
energy transferred/J
if you try this experiment and remember
= mass/g × specific heat capacity/J g−1 K−1 × temperature rise/K
that liquid fuels are highly flammable.
If Q represents the energy transferred, this can be written as:
Q = mcΔT
Key term
The specific heat capacity of water is 4.18 J g−1 K−1. This means that:
4.18 J raises the temperature of 1 g of water by 1 K The specific heat capacity of a
material, c, is the energy needed to
m × 4.18 J raises the temperature of a mass m, in grams, of water by 1 K raise the temperature of 1 g of the
and material by 1 K.
m × 4.18 J g−1 K−1 × ΔT raises the temperature of a mass m, in grams, of For water c = 4.18 J g−1 K−1.
water by ΔT
Example Tip
Table 8.1 shows the results from an experiment to measure the energy Temperatures in thermodynamics are
given out by burning meths using the apparatus shown in Figure 8.7. Use measured on the Kelvin scale. However
the results to work out the enthalpy of combustion of meths. the size of a temperature change is the
same on the Celsius and Kelvin scales.
Table 8.1 Results from an experiment to measure the energy given out by burning
A temperature change of 1 °C is the
meths (ethanol).
same as a change of 1 K.
Mass of burner + meths at start of experiment = 271.80 g
Mass of burner + meths at end of experiment = 271.30 g
Volume of water in can = 250 cm3
Rise in temperature of water = 10.0 °C
= 10.0 K
Notes on the method
This calculation assumes that all the energy given out from the burning
meths heats up the water.
The density of water is 1.0 g cm−3. So the mass of 100 cm3 water is 100 g.
● From the mass of water in the can and its temperature rise, work out
the energy transferred.
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● From the loss in mass of the liquid burner, find the mass of meths
which burned.
● Calculate the energy given out per gram of fuel.
● Multiply by the molar mass of meths (ethanol) to calculate the energy
given out per mole of fuel
● Give an answer to an appropriate number of significant figures (see
Section A1.4 in Appendix A1).
Answer
From the data in the table:
● mass of water in the can = 250 g
● temperature rise = 10.0 K
Energy transferred to the water in the can
= 250 g × 4.18 J g−1 K−1 × 10.0 K = 10 450 J
Mass of meths burned = 0.50 g
Energy given out per gram of meths that burned = 10 450 J ÷ 0.50 g
= 20 900 J g−1
Molar mass of meths (ethanol, C2H6O) = 46.0 g mol−1
Energy given out when one mole of ethanol (meths) burns
= 20 900 J g−1 × 46.0 g mol−1
= 961 400 J mol−1 = 961.4 kJ mol−1
The enthalpy change of combustion of a fuel is given the symbol ΔcH.
There are many sources of error in this crude method of measuring
enthalpy changes and so the data should not be quoted to more than two
Tip significant figures.
In calculations with several steps it is Therefore, from these results the value for of enthalpy change of this
better not to use your calculator at each exothermic reaction is given by:
stage. The danger is that you introduce ΔcH [ethanol] = −960 kJ mol−1
‘rounding errors’ at every step. You will
In summary:
get a more accurate answer if you work
out the answer at the end. C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = −960 kJ mol−1
Assumptions and errors in thermochemical
experiments
In the experiment in the example, the calculation is based on the assumption
that all the energy from the flame heats the water. In practice much of the
energy heats the metal can and the surrounding air. In addition, the flame is
affected by draughts and sometimes the fuel burns incompletely, leaving soot
on the bottom of the metal can. So, the assumption is clearly flawed. The
result is certainly inaccurate. The major sources of error (loss of energy, flame
disturbance and incomplete combustion) all reduce the energy transferred to
the water. This leads to a result that is less exothermic than the true value.
Accurate values for energy changes during combustion are obtained using
a bomb calorimeter (Figure 8.8). The apparatus is specially designed to
ensure that the sample burns completely and that energy losses are avoided.
A measured amount of the sample burns in excess oxygen under pressure.
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There is enough oxygen to ensure that all carbon in the compound is fully
oxidised to carbon dioxide and no carbon monoxide or soot are produced.
The energy change is measured at constant volume so a correction is needed
to calculate the enthalpy change at constant pressure.
Figure 8.8 A bomb calorimeter.
thermometer
insulating lids
water bomb calorimeter
oxygen under pressure
electrically heated small dish containing
wire to ignite sample sample under test
stirrer
insulating
air jacket
Enthalpy changes in solution thermometer –10 to 50°C
Enthalpy changes for reactions in solution can be measured using insulated
plastic containers such as polystyrene cups as calorimeters (Figure 8.9).
Polystyrene is an excellent insulator and it has a negligible specific heat capacity. polystyrene cup and lid
If the reaction is exothermic, the energy released cannot escape to the
surroundings, so it heats up the solution. If the reaction is endothermic, no reaction mixture
energy can enter from the surroundings, so the solution cools. If the solutions are
dilute, it is sufficiently accurate to calculate the enthalpy changes by assuming
that the solutions have the same density and specific heat capacity as water. Figure 8.9 Measuring the enthalpy
The temperature changes are often quite small and so it is important to use change of a reaction in solution.
a thermometer that is graduated in tenths of a degree.
Example
When 4.00 g of ammonium nitrate (NH4NO3) dissolves in 100 cm3 of
water, the temperature falls by 3.0 °C. Calculate the enthalpy change per
mole when NH4NO3 dissolves in water under these conditions.
Notes on the method
Calculate the energy change by assuming that the solution has the same
specific heat capacity (4.18 J g−1 K−1) and density (1.00 g cm−3) as water.
Work out the amount in moles of ammonium nitrate added.
Divide the energy change by the amount to determine the energy change
in kJ mol−1.
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Answer
Energy taken in from the solution
= mass × specific heat capacity × temperature change
= 100 g × 4.18 J g−1 K−1 × 3.0 K
= 1254 J
mass of NH4NO3 4.00 g
Amount of NH4NO3 = =
molar mass of NH4NO3 80.0 g mol−1
= 0.050 mol
1254 J
Energy taken in per mole of NH4NO3 = = 25 080 J mol−1
0.050 mol
= 25 kJ mol−1 (2 significant figures)
The reaction is endothermic, so the enthalpy change for the system is
positive.
+aq
NH4NO3(s) NH4NO3(aq) ΔH = +25 kJ mol−1
Test yourself
4 Burning butane, C4H10, from a Camping Gaz®container raised the
temperature of 200 g water from 18.0 °C to 28.0 °C. The Gaz®
container was weighed before and after, and the loss in mass was
0.29 g. Estimate the molar enthalpy change of combustion of butane.
5 On adding 25 cm3 of 1.0 mol dm−3 nitric acid to 25 cm3 of 1.0 mol dm−3
potassium hydroxide in a plastic cup, the temperature rise is 6.5 °C.
a) Write an equation for the reaction.
b) Calculate the enthalpy change for the neutralisation reaction per
mole of nitric acid.
6 On adding excess powdered zinc to 25 cm3 of 0.20 mol dm−3 copper(ii)
sulfate solution, the temperature rises by 9.5 °C.
a) Write an equation for the reaction.
b) Calculate the enthalpy change of the reaction for the molar
amounts in the equation.
8.4 Standard enthalpy changes
The values of enthalpy changes vary with changes in temperature, pressure or
concentration. This means that the conditions have to be carefully specified
Key term for the standard enthalpy changes listed in data tables.
The standard enthalpy change of The standard conditions are:
a reaction, Δr H 1, is the energy
● a pressure of 100 kPa (this is the approximate pressure of the atmosphere at
transferred when the molar quantities
sea level)
of reactants as stated in the equation
● a stated temperature that is usually 298 K (25 °C)
react under standard conditions.
● substances in their standard (most stable) state at 100 kPa pressure and the
stated temperature
● solutions with a concentration of 1 mol dm .
−3
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Activity
Measuring and evaluating the enthalpy change for the reaction of zinc
with copper(ii) sulfate solution
Two students decided to measure the enthalpy change for the reaction between zinc and
copper(ii) sulfate solution.
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
The method which they used is shown in Figure 8.10 and their results are shown in Table 8.2.
After adding the zinc, it took a little while for the temperature to reach a peak and then the
mixture began to cool.
0–50 °C thermometer excess
powdered zinc
50cm3
0.25mol dm–3
CuSO4(aq)
Measure the temperature At 3.0 minutes add excess Continue stirring and record
every 30s for 2.5 minutes. powdered zinc and stir. the temperature every 30 s
for a further 6 minutes.
Figure 8.10 Measuring the enthalpy change for the reaction of zinc with copper(ii) sulfate solution.
Table 8.2
Time/min Temperature Time/min Temperature Time/min Temperature
/°C /°C /°C
0 24.1 3.5 34.2 6.5 33.7
0.5 24.0 4.0 34.8 7.0 33.6
1.0 24.1 4.5 35.0 7.5 33.5
1.5 24.1 5.0 34.6 8.0 33.4
2.0 24.2 5.5 34.2 8.5 33.2
2.5 24.1 6.0 33.9 9.0 33.1
3.0 −
Temperature/°C
ΔT
1 Plot a graph of temperature (vertically) against time (horizontally) using the results
in Table 8.2.
2 Extrapolate the graph backwards from 9 minutes to 3 minutes, as in Figure 8.11.
This gives an estimate of the maximum temperature if all the zinc had reacted at
once and there was no loss of energy to the surroundings.
a) What is the estimated maximum temperature at 3 minutes? 0 3 6 9
b) What is the temperature rise, ΔT, for the reaction? Time/minutes
3 Calculate the energy given out during the reaction using the equation: Figure 8.11 Estimating the maximum
temperature of the mixture when zinc reacts
energy transferred = mass × specific heat capacity × temperature change with copper(ii) sulfate solution.
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Assume that:
● all the energy is transferred to the solution in the polystyrene cup
● the density of the solution is 1.0 g cm−3
● the specific heat capacity of the solution is 4.18 J g−1 K−1.
4 How many moles of each chemical reacted?
a) CuSO4 b) Zn
5 What is the enthalpy change of the reaction, Δ r H, for the amounts of Zn and CuSO4
in the equation? (State the value of Δ r H in kJ mol−1 with the correct sign.)
6 With the help of Practical skills sheet 5, which you can access online at
[Link]/EdexcelChemistry, copy and complete
Table 8.3 for the various measurements in the experiment.
Table 8.3 The values, uncertainties and percentage uncertainties of measurements in the
experiment.
Measurement Value Uncertainty Percentage
uncertainty
Concentration of copper(ii) sulfate
solution
Volume of copper(ii) sulfate solution
measured from a 100 cm3 measuring
cylinder
Temperature rise, ΔT, estimated from
the difference in two readings taken
with a 0–50 °C thermometer
7 What is the total percentage uncertainty in the experiment?
8 What is the total uncertainty in the value you have calculated for the enthalpy change?
9 Write a value for the enthalpy change showing the uncertainty using the symbol ±.
10 What are the main sources of error in the measurements and procedure for the
experiment?
11 Look critically at the procedures in the experiment and suggest improvements to
minimise errors and increase the repeatability of the result.
Any enthalpy change measured under standard conditions is described as
a standard enthalpy change and given the symbol ΔH 1298 or simply ΔH 1,
pronounced ‘delta H standard’ (Figure 8.12).
Figure 8.12 The symbol for a standard the capital Greek letter standard state symbol
enthalpy change. delta means ‘change of’
ΔrH 1298 the temperature at which the
value is given, usually 298 K
(this is often omitted)
the type of change
r = reaction
c = combustion H = enthalpy
f = formation
In thermochemistry, it is important to specify the states of the substances and,
therefore, to include state symbols in equations. So, ΔH 1 for the reaction:
2H2(g) + O2(g) → 2H2O(l)
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must relate to hydrogen gas, oxygen gas and liquid water (not steam). The states
of elements and compounds must also be the most stable at the given temperature
(often 298 K) and 100 kPa. Thus, ΔH 1 measurements involving carbon refer to
graphite, which is energetically more stable than diamond.
Standard enthalpy changes of combustion
The standard enthalpy change of combustion of an element or
compound, ΔcH 1, is the enthalpy change when one mole of the substance Key term
burns completely in oxygen under standard conditions. The substance and
The standard enthalpy change of
the products of burning must be in their stable (standard) states. For a carbon
combustion of a substance, ΔcH 1 is
compound, complete combustion means that all the carbon burns to carbon
the enthalpy change when one mole
dioxide and there is no soot or carbon monoxide. If the substance contains
of the substance burns completely in
hydrogen, the water formed must end up as liquid and not as a gas.
oxygen under standard conditions.
Values of enthalpies of combustion are much easier to measure than many
other enthalpy changes. They can be calculated from measurements taken
with a bomb calorimeter (Section 8.3).
Chemists use two ways to summarise standard enthalpy changes of
combustion. One way is to write the equation with the enthalpy change
alongside it. So, for the standard enthalpy change of combustion of carbon,
they write:
C(graphite) + O2(g) → CO2(g) Δ cH 1 = −394 kJ mol−1 Tip
The other way is to use a shorthand form. For the standard enthalpy change Remember that all combustion reactions
of combustion of methane, this is written as: are exothermic, so ΔcH 1 values are
Δ cH 1 [CH4(g)] = −890 kJ mol−1 always negative.
Standard enthalpy changes of formation
The standard enthalpy change of formation of a compound, ΔH 1f, is the
enthalpy change when one mole of the compound forms from its elements. Key term
The elements and the compound formed must be in their stable standard
states. The more stable state of an element is chosen where there are allotropes The standard enthalpy change of
(different forms in the same state) such as graphite and diamond. formation of a compound, Δ fH 1 , is
the enthalpy change when one mole of
As with standard enthalpies of combustion, there are two ways of representing the compound forms from its elements
standard enthalpy changes of formation. under standard conditions with the
One way is to write the equation with the enthalpy change alongside it. For elements and the compound in their
the standard enthalpy change of formation of water this is: standard (stable) states.
H2(g) + 12 O2(g) → H2O(l) Δ f H 1 = −286 kJ mol−1
The other way is to use shorthand. For the standard enthalpy change of
formation of ethanol this is:
Δ f H 1 [C2H5OH(l)] = −277 kJ mol−1
Like all thermochemical quantities, the precise definition of the standard
enthalpy change of formation is important. Books of data tabulate values for
standard enthalpies of formation. These tables are very useful because they make
it possible to calculate the enthalpy changes for many reactions (Section 8.5).
Unfortunately, it is difficult to measure some enthalpy changes of formation
directly. For example, it is impossible to convert carbon, hydrogen and
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oxygen straight to ethanol under any conditions. Because of this, chemists
have had to find an indirect method of measuring the standard enthalpy
change of formation of ethanol and other compounds (Section 8.5).
One important consequence of the definition of standard enthalpy changes
of formation is that, for an element, Δ f H 1 = 0 kJ mol−1 because there is no
change, and therefore no enthalpy change, when an element forms from
itself. In other words, the standard enthalpy change of formation of an
element is zero. So:
Δ f H 1 [Cu(s)] = 0 and Δ f H 1 [O2(g)] = 0
Key term
Standard enthalpy changes of neutralisation
Many chemical reactions happen in solution. Chemists define standard
The standard enthalpy change of enthalpy changes for changes in solution including the standard enthalpy
neutralisation is the enthalpy change change of neutralisation. This is usually defined as the enthalpy change
when the acid and alkali in the equation per mole of water formed.
for the reaction neutralise each other
under standard conditions to form one Example
mole of water.
When 50.0 cm3 of 2.00 mol dm−3 hydrochloric acid is mixed with 50.0 cm3
of 2.00 mol dm−3 sodium hydroxide in a calorimeter at 25 °C and 100 kPa,
the temperature rises by 13.7 °C. Calculate the enthalpy change for the
neutralisation reaction.
Notes on the method
Write the equation for the reaction.
Assume that the dilute solution has the same specific heat capacity
(4.18 J g−1 K−1) and density (1 g cm−3) as water.
Calculate the energy change in the calorimeter, taking care to use the
total mass of water.
Next calculate the amount of hydrochloric acid that reacted.
Divide the energy change by the amount in moles to determine the
enthalpy change per mole of water formed.
Give the answer to an appropriate number of significant figures.
Answer
The equation for the reaction is:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
This shows that 1 mol acid reacts to form 1 mol water.
Energy given out and used to heat 100 cm3 of solution
= 100 g × 4.18 J g−1 K−1 × 13.7 K = 5727 J
Amount of HCl used = amount of NaOH used
= 50.0 dm3 × 2.00 mol dm−3 = 0.100 mol
1000
Energy given out per mole of acid = 5727 J = 57 270 J mol−1
0.100 mol
For this neutralisation reaction ΔH = −57.3 kJ mol−1
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Now, as the concentrations of the two solutions were effectively
1.0 mol dm−3 immediately after mixing, the temperature was 25 °C and
the pressure 100 kPa, the value of the enthalpy change of neutralisation
has been obtained under standard conditions.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
ΔnH 1 = −57.3 kJ mol−1
This is the standard enthalpy change of neutralisation for this reaction.
Test yourself
7 By writing a balanced equation, show that the standard enthalpy
change of formation of carbon dioxide is the same as the standard
enthalpy change of combustion of carbon (graphite).
8 Write equations for the reactions for which the enthalpy change is the
standard enthalpy change of formation of:
a) aluminium oxide, Al2O3(s)
b) hydrogen chloride, HCl(g)
c) propane, C3H8(g).
9 The temperature change on mixing 25 cm3 of 1.0 mol dm−3 hydrochloric
acid with 25 cm3 of 1.0 mol dm−3 potassium hydroxide is 6.5 °C. What
is the temperature on mixing 50 cm3 each of the same two solutions?
8.5 Hess’s Law and the indirect
determination of enthalpy changes
The enthalpy change of a reaction is the same whether the reaction happens
in one step or in a series of steps. As long as the reactants and products are Key term
the same, the overall enthalpy change is the same whether the reactants are
Hess’s Law states that the enthalpy
converted to products directly or through two or more intermediates. This is
change in converting reactants to
Hess’s Law. In Figure 8.13 the enthalpy change for Route 1 and the overall
products is the same regardless of the
enthalpy change for Route 2 are the same.
route taken, provided the initial and
final conditions are the same.
Route 1
∆H this way ...
∆H1
A D
∆H2 Route 2 ∆H4
...is the same
as ∆H this way
B C
∆H3
Figure 8.13 A diagram to illustrate Hess’s Law: ΔH1 = ΔH2 + ΔH3 + ΔH4.
8.5 Hess’s Law and the indirect determination of enthalpy changes 257
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Hess’s Law is a chemical version of the law of conservation of energy. Suppose
the enthalpy change for Route 1 in Figure 8.13 were more exothermic than the
total enthalpy change for Route 2. It would then be possible to go round the
cycle in Figure 8.13 from A to D direct and back to A via C and B, ending up
with the same starting chemical but with a net release of energy. This would
contravene the law of conservation of energy.
Hess’s Law is also an example of a mathematical model. This is shown by
the precise quantitative relationship between ΔH1, ΔH 2, ΔH3 and ΔH4 in
Figure 8.13. Using Hess’s Law it is possible to bring together data and calculate
enthalpy changes which cannot be measured directly by experiment. So,
Hess’s Law can be used to calculate:
● standard enthalpy changes of formation from standard enthalpy changes of
combustion and
● standard enthalpy changes of reaction from standard enthalpy changes of
formation.
Tip
The standard enthalpy changes for other reactions can be calculated using standard
enthalpy changes of combustion. However, it is the determination of standard enthalpy
changes of formation that is particularly important because these are the values that
are used in many thermochemical calculations.
Enthalpy changes of formation from enthalpy
change of combustion
Figure 8.14 shows the form of the energy cycle which we can use to calculate
enthalpy changes of formation from enthalpy changes of combustion.
∆ H1 ∆ f H 1 [compound]
elements compound
Route 1 oxygen oxygen
∆H this way...
∆ H2 sum of
∆ H3 ∆ c H [compound]
1
∆ c H 1 [elements]
combustion Route 2
products ... is the same
as ∆H this way.
Figure 8.14 An energy cycle for calculating standard enthalpy changes of formation from
standard enthalpy changes of combustion: ΔH2 = ΔH1 + ΔH3.
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Example
Calculate the standard enthalpy change of formation of propane, C3H8, at
298 K given the following standard enthalpy changes of combustion.
propane, Δ cH 1 [C3H8(g)] = −2219 kJ mol−1
carbon, Δ cH 1 [C(graphite)] = −394 kJ mol−1
hydrogen Δ cH 1 [H2(g)] = −286 kJ mol−1
Notes on the method
Draw up an energy cycle using the model in Figure 8.15. Use Hess’s Law
to produce an equation linking the relevant enthalpy changes.
Pay careful attention to the signs.
Put the value and sign for a quantity in brackets when adding or
subtracting enthalpy values.
Answer
An energy cycle linking the formation of propane with its combustion and the
combustion of its constituent elements is shown in Figure 8.15. Sometimes,
energy cycles like the one in Figure 8.15 are called Hess cycles.
∆H1
3C(s) + 4H2(g) C3H8(g)
+ 5O2(g) + 5O2(g)
∆H2 ∆H3
3CO2(g) + 4H2O(I)
Figure 8.15 An energy cycle for the combustion of propane and its
constituent elements.
According to Hess’s Law:
ΔH2 = ΔH1 + ΔH3
ΔH2 = 3 × Δ cH 1 [C(graphite)] + 4 × Δ cH 1 [H2(g)]
= 3 × (−394 kJ mol−1) + 4 × (−286 kJ mol−1) = −2326 kJ mol−1
ΔH1 = Δ f H 1 [C3H8(g)]
ΔH3 = Δ cH 1 [C3H8(g)] = −2219 kJ mol−1
Hence:
(−2323 kJ mol−1) = Δ f H 1 [C3H8(g)] + (−2220 kJ mol−1)
Δ f H 1 [C3H8(g)] = (−2326 kJ mol−1) − (−2219 kJ mol−1)
= −2326 kJ mol−1 + 2219 kJ mol−1
= −107 kJ mol−1
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Test yourself
10 Use the values of standard enthalpy changes of combustion below
to calculate the standard enthalpy change of formation of methanol,
CH3OH.
Δ cH 1 [CH3OH(l)] = −726 kJ mol−1
Δ cH 1 [C(graphite)] = −394 kJ mol−1
Δ cH 1 [H2(g)] = −286 kJ mol−1
11
Why is it useful to have standard enthalpy changes of combustion which
can be used to calculate standard enthalpy changes of formation?
Enthalpy changes of reaction from enthalpy
changes of formation
Data books contain tables of standard enthalpies of formation for both
inorganic and organic compounds. The great value of this data is that it
allows chemists to calculate the standard enthalpy change for any reaction
involving the substances listed in the tables.
The standard enthalpy change of a reaction is the enthalpy change when the
amounts shown in the chemical equation react. Like other standard quantities
in thermochemistry, the standard enthalpy change of reaction is defined at
100 kPa pressure with the reactants and products in their normal stable states
at a particular temperature, usually 298 K. The concentration of any solution
is 1.0 mol dm−3.
Thanks to Hess’s Law it is easy to calculate the standard enthalpy change of
a reaction from tabulated values of standard enthalpy changes of formation
(Figure 8.16).
Figure 8.16 An energy cycle for calculating Route 1
standard enthalpies of reaction from ∆H this way...
∆rH 1
standard enthalpies of formation. reactants products
∆ H1 sum of ∆ H2 sum of
∆ f H 1 [reactants] ∆ f H 1 [products]
Tip
Route 2
Take care! Enthalpy changes for ... is the same
as ∆H this way.
reactions can also be calculated from
elements
enthalpy changes of combustion. The
relationship is then: According to Hess’s Law:
Δ r H 1 = sum of Δ cH 1 [reactants] ΔH1 + Δ r H 1 = ΔH 2
− sum of Δ cH 1 [products] Rearranging gives:
So do not learn these formulae parrot Δ r H 1 = ΔH 2 − ΔH1
fashion. Check with the Hess cycle each
time. So:
Δ r H 1 = sum of Δ f H 1 [products] − sum of Δ f H 1 [reactants]
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Example
Calculate the standard enthalpy change for the reduction of iron(iii) oxide
by carbon monoxide.
Δ f H 1 [Fe2O3] = −824 kJ mol−1
Δ f H 1 [CO] = −110 kJ mol−1
Δ f H 1 [CO2] = −394 kJ mol−1
Notes on the method
Write the balanced equation for the reaction and then draw an energy
cycle (Hess cycle) using the model in Figure 8.17.
Remember that, by definition, Δ f H 1 [element] = 0 kJ mol−1.
Pay careful attention to the signs. Put the value and sign for a quantity in
brackets when adding or subtracting enthalpy values. Tip
Reversing the direction of a reaction in
Answer an energy cycle reverses the sign of ΔH.
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
∆ rH
1
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
sum of ∆ fH 1[reactants] sum of ∆ fH 1[products]
= ∆ fH 1[Fe2O3(s)] = 2∆ fH 1 [Fe(s)]
+ 3∆ fH 1[CO(g)] + 3∆ fH 1[CO2(g)]
2Fe(s) + 3O2(g) + 3C(graphite)
Figure 8.17 An energy cycle (Hess cycle) for calculating the enthalpy change of
reaction between iron(iii) oxide and carbon monoxide.
Applying Hess’s Law to Figure 8.17:
Δ r H 1 = sum of Δ f H 1 [products] − sum of Δ f H 1 [reactants]
Δ r H 1 = {2 × Δ f H 1 [Fe] + 3 × Δ f H 1 [CO2]} − {Δ f H 1 [Fe2O3] + 3 × Δ f H 1 [CO]}
= {0 + (3 × −394 kJ mol−1)} − {(−824 kJ mol−1) + (3 × −110 kJ mol−1)}
= −1182 kJ mol−1 + 824 kJ mol−1 + 330 kJ mol−1
Δ r H 1 = −28 kJ mol−1
Test yourself
12 The standard enthalpy change of formation of sucrose (sugar),
C12H22O11, is −2226 kJ mol−1. Write the balanced equation for which
the standard enthalpy change of reaction is −2226 kJ mol−1.
13 When calculating standard enthalpy changes for reactions involving
water at 298 K, why is it important to specify that the H2O is present
as water and not as steam?
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Test yourself
14 Calculate the standard enthalpy change for the reaction of one
mole of hydrazine, N2H4(l) with oxygen, O2, to form nitrogen, N2, and
water, H2O.
Δ f H 1 [N2H4(l)] = +51 kJ mol−1
Δ f H 1 [H2O(l)] = −286 kJ mol−1
Core practical 8
Applying Hess’s Law to find an enthalpy change that cannot be measured directly
Two students decided to determine the enthalpy change for the hydration of
magnesium sulfate to give crystals of the hydrated salt.
MgSO4(s) + 7H2O(l) → MgSO4.7H2O(s)
It is not possible to measure this enthalpy change directly because of the difficulty of
controlling the temperature and measuring the temperatures of solids. The students
were given the Hess’s Law cycle in Figure 8.18 which shows that it is possible to
determine the required enthalpy change at room temperature.
∆H1
MgSO4(s) + 7H2O(I) MgSO4.7H2O(s)
+93H2O(I) +93 H2O(I)
∆H2 ∆H3
MgSO4(aq, 100H2O)
Figure 8.18 An energy cycle (Hess cycle) for calculating the enthalpy change of the reaction.
Figure 8.19 shows the procedure that the students used for determining ΔH2. They used
a 0–50 °C thermometer with 0.2 °C graduations. They then used exactly the same
procedure to determine ΔH3 using hydrated magnesium sulfate in place of the
anhydrous salt and a little less water.
weighed sample tube stir and record the
+ MgSO4(s) (0.025 mol) highest temperature
reached
record the initial
temperature
45.0 g water MgSO4(aq,100H2O) reweigh the empty
(2.5 mol) (0.025 mol) sample tube
Figure 8.19 Outline of a procedure for measuring the enthalpy change when anhydrous
magnesium sulfate reacts with and dissolves in a measured amount of water.
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Results
Table 8.4 Results for the anhydrous and for the hydrated salt.
Results for anhydrous salt Results for hydrated salt
Mass of empty sample tube 15.92 g Mass of sample tube + MgSO4.7H2O(s) 19.58 g
Mass of sample tube + MgSO4(s) 12.91 g Mass of empty sample tube 13.42 g
Mass of polystyrene cup + water 47.10 g Mass of polystyrene cup + water 44.21 g
Mass of empty cup 2.10 g Mass of empty cup 2.36 g
Mass of water 45.00 g Mass of water 41.85 g
Temperature of the solution after reaction 35.4 °C Temperature of the solution after reaction 23.4 °C
Starting temperature of the water 24.1 °C Starting temperature of the acid 24.8 °C
1 Show that the ratio of the amount of water, in moles, to the amount of MgSO4 in Figure 8.19 is 100 : 1.
2 Explain why the hydrated salt was added to less water (as shown in Table 8.4) so that in both
reactions the mixture at the end was the same and equivalent to MgSO4(aq, 100H2O).
3 a) For the anhydrous salt, work out the mass of salt added and the temperature change.
b) Calculate the energy change on adding the anhydrous salt to excess water and hence determine ΔH2.
4 a) For the hydrated salt, work out the mass of salt added and the temperature change.
b) Calculate the energy change on adding the hydrated salt to excess water and hence determine ΔH3.
5 Write an expression connecting ΔH1, ΔH2 and ΔH3.
6 Calculate ΔH1 giving your answer to the number of significant figures justified by the data. Account for
your choice of number of significant figures.
7 Evaluate the results of the experiment by calculating the standard enthalpy change for the hydration
reaction using standard enthalpy changes of formation.
Δ f H 1 [MgSO4(s)] = −1285 kJ mol−1 Tip
Δ f H 1 [H2O(l)] = −286 kJ mol−1 Refer to Practical skills sheets 5 and
10, which you can access online at
Δ f H 1 [MgSO4.7H2O(s)] = −3389 kJ mol−1 [Link]/
Compare and comment on the two values. EdexcelChemistry:
8 The students measured the masses with a balance reading to two decimal places. 5 Identifying errors and estimating
Would they have reduced the overall error in their results by using a balance reading uncertainties
to three decimal places? 10 Measuring enthalpy changes.
9 Suggest two ways of modifying the procedure shown in Figure 8.19 that would have
improved the accuracy of the temperature changes measured by the students.
8.6 Enthalpy changes and the
direction of change
Strike a match and it catches fire and burns. Put a spark to petrol and it burns
furiously. These are two exothermic reactions which, once started, tend to
‘go’. They are examples of the many exothermic reactions which just keep
going once they have started. In general, chemists expect that a reaction will
go if it is exothermic.
What this means is that reactions which give out energy to their surroundings
are the ones which happen. This ties in with the common experience that
change happens in the direction in which energy is spread around and
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dissipated in the surroundings. So the sign of ΔH is a guide to the likely
direction of change, but it is not a totally reliable guide for three main reasons.
● The direction of change may depend on the conditions of temperature
and pressure. One example is the condensation of a vapour such as steam.
Steam condenses to water below 100 °C and energy is given out. This is
an exothermic change.
H2O(g) → H2O(l) ΔH = −44 kJ mol−1
At temperatures above 100 °C, the change goes in the opposite direction
and this process is endothermic.
● There are some examples of endothermic reactions which occur readily
under normal conditions. So some reactions for which ΔH is positive
can happen. One example of this is the reaction of citric acid solution
with sodium hydrogencarbonate. The mixture fizzes vigorously and
cools rapidly. This suggests that there are other factors that determine the
direction of change.
● Some exothermic reactions never occur because the rate of reaction is so
slow and the mixture of reactants is effectively inert. For example, the
change from diamond to graphite is exothermic, but diamonds do not
suddenly turn into black flakes.
8.7 Enthalpy changes and bonding
During reactions, the bonds in reactants break and then new bonds form in
the products. For example, when hydrogen reacts with oxygen:
2H2(g) + O2(g) → 2H2O(g)
Tip Bonds in the H2 and O2 molecules break to form H and O atoms (Figure 8.20).
New bonds then form between the H and O atoms to produce water, H2O.
Bond breaking is endothermic.
4 H (g) + 2 O (g)
Bond making is exothermic.
hydrogen and
Energy is needed to break the bonds oxygen atoms
between atoms. So, energy must be
released when the reverse occurs and a
bond forms.
energy needed to break one
mole of O O bonds
+
energy given out when
energy needed to break two four moles of O H
moles of H H bonds bonds are formed
2H2(g) + O2(g)
hydrogen and
oxygen molecules
2H2O(g)
energy released during
reaction to form two water molecules
moles of water in steam
Figure 8.20 An energy level diagram for the reaction between hydrogen and oxygen.
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Chemical reactions involve bond breaking followed by bond making. This
means that the enthalpy change of a reaction is the energy difference between
bond-breaking and bond-making processes.
When hydrogen and oxygen react, more energy is released in making four
new O–H bonds in the two H2O molecules than in breaking the bonds in
two H2 molecules and one O2 molecule, so the overall reaction is exothermic
Key terms
(Figure 8.20). The bond enthalpy of a particular bond
A definite quantity of energy known as the bond enthalpy (or bond energy) is the energy required to break one
can be associated with each type of bond. This energy is absorbed when mole of the bonds in a substance in the
the bond is broken and given out when the bond is formed. In measuring gaseous state.
and using bond enthalpies, chemists distinguish between the terms bond The mean bond enthalpy of the X–Y bond
enthalpy and mean (average) bond enthalpy. is the mean value of the bond enthalpy
Bond enthalpies are precise values for specific bonds in compounds (for values for the X–Y bond averaged across a
example the C–Cl bond in CH3Cl). wide range of compounds.
Mean bond enthalpies are average values for one kind of bond in different
compounds (for example an average value for the C–Cl bond in all Tip
compounds). Mean bond enthalpies take into account the fact that:
The symbol for bond enthalpy is
● the bond enthalpy for a specific covalent bond varies slightly from one E, so the C–H bond enthalpy is
compound to another (for example the O–H bond has a slightly different written as E(C–H) = 413 kJ mol−1.
bond enthalpy in H2O and C2H5OH) Values for mean bond enthalpies are
● successive bond enthalpies are not the same in compounds such as water
given in the data sheet for Chapter
and methane. (The energy needed to break the first O–H bond in H–O–H(g) 8, which you can access online at
is 498 kJ mol−1, but the energy needed to break the second O–H bond in [Link]/
OH(g) is 428 kJ mol−1.) EdexcelChemistry.
Using bond enthalpies
The most important use of mean bond enthalpies is in estimating the
enthalpy changes in chemical reactions involving molecular substances with
covalent bonds. These estimates are particularly helpful when experimental
measurements cannot be made, as in the following worked example.
Example
Use mean bond enthalpies to estimate the enthalpy of formation of
hydrazine, N2H4.
Note on the method
Write the equation for the reaction showing all the atoms and bonds in
the molecules. This makes it easier to count the number of bonds broken
and formed.
Answer
H H
N N + 2H H N N
H H
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Table 8.5
Bonds broken Total energy Bonds formed Total energy
(endothermic) change/ change/
kJ mol –1 kJ mol –1
1 × N≡N +945 1 × N–N −158
2 × H–H +(2 × 436) 4 × N–H −(4 × 391)
Δ r H = +945 kJ mol−1 + (2 × 436 kJ mol−1)
− 158 kJ mol−1 − (4 × 391 kJ mol−1)
= +1817 kJ mol−1 − 1722 kJ mol−1 = +95 kJ mol−1
For many reactions, the values of ΔH estimated from mean bond enthalpies
agree closely with experimental values. However, there are limitations to
the use of bond energy data in this way, and significant differences between
the values of ΔH estimated from bond enthalpies and those obtained by
experiment do occur. These differences usually arise:
● eitherfrom variations in the strength of one kind of bond in different
molecules (and mean bond enthalpies should not be used)
● or when one of the reactants or products is not in the gaseous state as bond
enthalpy calculations assume.
Unknown bond enthalpies can be calculated given the enthalpy change for
a reaction involving the compound which includes the bond, together with
other relevant bond enthalpies.
Example
Calculate a value for the bond enthalpy for the O–O known bond enthalpy values from the data sheet for
bond in the gas dimethyl peroxide, CH3OOCH3, given Chapter 8, which you can access online at
that the standard enthalpy of combustion, [Link]/EdexcelChemistry.
Δ c H 1 [CH3OOCH3(g)] = −1460 kJ mol−1. Let the unknown bond enthalpy term be x. Equate
Note on the method the known enthalpy change for the reaction with the
enthalpy change calculated from bond enthalpies,
Write the equation for the reaction, showing the
including the unknown value. Then rearrange the
molecules and the bonds, so that you can count the
equation to find the value of x.
number of bonds broken and formed. Look up the
Answer
H H
H C O O C H + 2.5 O O 2 O C O + 3 H O H
H H
Table 8.6
Bonds broken Total energy change/ Bonds formed Total energy change/
(endothermic) kJ mol –1 kJ mol –1
6 × C–H +(6 × 413) = 2478 6 × O–H −(6 × 464) = 2784
2 × C–O +(2 × 336) = 672 4 × C=O −(4 × 805) = 3220
2.5 × O=O +(2.5 × 498) = 1245
1 × O–O +x
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Δ c H 1 [CH3OOCH3(g)] = −1460 kJ mol−1
= +(2478 + 672 + 1245 + x) kJ mol−1 − (2784 + 3220) kJ mol−1
−1460 kJ mol−1 = +4395 kJ mol−1 + x − 6004 kJ mol−1
x = −1460 kJ mol−1 − 4395 kJ mol−1 + 6004 kJ mol−1 = 149 kJ mol−1
E(O–O) in dimethyl peroxide = + 149 kJ mol−1
Test yourself
Where relevant, refer to the data sheet for Chapter 8, ‘Mean bond enthalpies and bond lengths’, which you can
access online at [Link]/EdexcelChemistry, to help you answer these questions.
15 a)
Look back at Figure 8.20 and write out the 18 a)
Make a table to show the mean bond
equation enthalpies and bond lengths of the C – C, C=C
2H2(g) + O2(g) → 2H2O(g) and C ≡ C bonds.
showing all the bonds between atoms in the b) What generalisations can you make based on
molecules. your table?
b) Refer to the data sheet of mean bond 19
Use mean bond enthalpies to estimate the
enthalpies and calculate: enthalpy change when ethene, H2C=CH2(g),
i) the energy needed to break one mole of reacts with H2(g) to form ethane, CH3 –CH3(g).
O=O bonds plus two moles of H–H bonds 20 a)
Which are likely to give a more accurate
ii) the energy given out when four moles of answer to a calculation of the enthalpy change
O–H bonds are formed in two moles of for a reaction – mean bond enthalpies or
water (steam) molecules enthalpies of formation?
iii) the energy released during the reaction to b) Give a reason for your answer to part (a).
form two moles of water (steam). 21 Look carefully at the mean bond enthalpies for
16
Look up the bond enthalpies for the H–H, Cl–Cl hydrogen and the halogens (fluorine, chlorine,
and H–Cl bonds. bromine and iodine).
a) Calculate the overall enthalpy change for this a) Write an equation for the reaction of hydrogen
reaction. with chlorine.
H2(g) + Cl2(g) → 2HCl(g) b) Explain which bond (H–H or Cl–Cl) you think
b) Draw an energy level diagram for the reaction will break first in the reaction.
(similar to Figure 8.20). c) How would you expect the reaction of fluorine
17 a)
Calculate the average of the successive bond with hydrogen to compare with the reaction of
enthalpies for the two O–H bonds in water chlorine with hydrogen?
mentioned in Section 8.7.
b)
Compare your answer with the mean bond
enthalpy of the O–H bond given in the table of
mean bond enthalpies.
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Chapter summary
l Standard enthalpy changes are measured under
Chapter 8 Energetics I
standard conditions: a pressure of 100 kPa and a
l Enthalpy changes are energy changes measured at specified temperature (often 298 K).
constant pressure. l Standard enthalpy changes for reactions
l The enthalpy change is negative for an exothermic
are precisely defined and include standard
reaction and positive for an endothermic reaction. enthalpy changes of combustion, formation and
This can be represented using enthalpy level neutralisation.
diagrams. l Hess’s Law states that the enthalpy change in
l Enthalpy changes of combustion can be estimated
converting reactants to products is the same
by using the energy given out on burning a regardless of the route taken, provided that the
measured mass of fuel to heat a measured mass of initial and final conditions are the same.
water. The change in temperature of the water is l Enthalpy cycles drawn using Hess’s law make it
recorded. possible to determine enthalpy changes that cannot
l There can be significant errors in this type of
be measured directly, for example calculating
experiment because the assumption that all the enthalpy changes of formation from measured
energy from the flame heats the water is not enthalpy changes of combustion.
justified. l Hess’s law cycles can also be used to calculate
l Enthalpy changes in solution can be measured
enthalpy changes of reaction from tables of data
by mixing measured amounts of reactants in an giving enthalpy changes of formation.
insulated container and measuring the temperature l Bond breaking is endothermic; bond making is
change. exothermic.
l It is assumed that the specific heat capacity of
l The bond enthalpy of a particular bond is the
the solutions is the same as that of water. Errors energy required to break one mole of the bonds in a
arise because it is difficult to measure the small substance in the gaseous state.
temperature changes accurately. l The mean bond enthalpy of a bond is the mean
l Values of the enthalpy changes can be determined
value of the bond enthalpy values for the bond
from such experiments using the relationship: averaged across a range of compounds.
energy transferred/J l Bond enthalpies can be used to estimate the
= mass/g × specific heat capacity/J g−1 K−1 enthalpy changes for reactions between molecular
× temperature change/K
substances in the gaseous state. The values obtained
(Q = mcΔT) are approximate if mean bond enthalpies are used in
the calculation.
268 8 Energetics I
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Exam practice questions
1 a) Draw and label a diagram of the apparatus e) Calculate the standard enthalpy change of
that you could use to determine the the reaction. Show the correct sign and
enthalpy change of the reaction between units. (3)
powdered magnesium and excess copper(ii) f) Explain the number of significant figures in
sulfate solution.(3) your answer. (1)
b) When excess powdered magnesium was added
4 Hydrogen reacts with oxygen to form water or
to 50 cm3 of 0.040 mol dm-3 copper(ii) sulfate
steam depending on the conditions.
solution, the temperature rose by 5.0 °C.
i) Write a balanced equation with state H2(g) + 12 O2(g) → H2O(l)
symbols for the reaction. (1) ΔH 1 = −286 kJ mol–1
ii) Calculate the energy transferred to H2(g) + 2 O2(g) → H2O(g)
1
the copper(ii) sulfate solution. (2) ΔH 1 = −242 kJ mol–1
iii) State the assumptions made in your a) On the same diagram draw energy level
calculation in part (ii). (2) diagrams to represent these two changes (3)
iv) Calculate the enthalpy change for b) Use the diagram in (a) to determine the
the reaction shown in your equation enthalpy change for water turning to
in part (i). (3) steam. (2)
H2O(l) → H2O(g)
2 A butane gas burner is used to heat water.
The standard enthalpy of combustion of butane 5 a) Give a definition of the term ‘standard
gas, Δ c H 1 = −2876 kJ mol−1. The specific heat enthalpy change of combustion’. (3)
capacity of water = 4.18 J g− K−1. b) Write an equation for the change for which
a) Calculate the energy needed to heat 500 g the enthalpy change is the standard enthalpy
of water from 20 °C to its boiling point. (1) of combustion of propane. (2)
b) Calculate how much butane, in moles, must c) Give a definition of the term ‘standard
burn to supply the energy needed. (1) enthalpy change of formation’. (3)
c) Calculate the volume of butane gas needed, d) Write an equation for the change for which
measured under conditions such that its the enthalpy change is the standard enthalpy
molar volume is 24.0 dm3 mol−1. (1) change of formation of propanal. (2)
d) State the assumptions made in answering
When gypsum(CaSO4.2H2O(s)) is heated very
6
this question. (2)
strongly, it decomposes forming the anhydrous
3 An excess of solid sodium hydrogencarbonate salt (anhydrite) and water.
was added to 50 cm3 of 1.0 mol dm−3 ethanoic a) Write a balanced equation with state
acid in an insulated polystyrene container symbols for the decomposition of
under standard conditions. The temperature fell gypsum. (2)
by 8.0 °C. b) Predict whether the decomposition is
a) Complete the following equation for the exothermic or endothermic. Explain your
reaction. answer. (2)
CH3COOH(aq) + NaHCO3(s) c) Explain why the enthalpy change for the
→ ____+ ____ + ____(2) decomposition of gypsum cannot be
b) Explain why the NaHCO3 was added in measured directly. (1)
small portions. (1) d) Using the following values, calculate
c) Calculate the energy change during the the standard enthalpy change for the
reaction. (Assume that the specific heat decomposition. (5)
capacity of the solution is 4.18 J g−1 K−1 and Δ f H 1 [CaSO4.2H2O(s)] = −2023 kJ mol–1
its density is 1.0 g cm−3. Ignore the mass of
sodium hydrogencarbonate.) (2) Δ f H 1 [CaSO4(s)] = −1434 kJ mol–1
d) Calculate the amount of ethanoic acid, in Δ f H 1 [H2O(l)] = −286 kJ mol–1
moles, that reacted. (1)
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Exam practice questions
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7 Tin is manufactured by heating tinstone, SnO2, heated the can and water for a few minutes
at high temperatures with coke (carbon). the temperature of the water had risen by
14.8 °C. The mass of the alcohol that had
There are two possible reactions for the process.
burned was 0.898 g.
Reaction 1: SnO2(s) + C(s) → Sn(s) + CO2(g) i) Calculate the energy transferred to the
Reaction 2: SnO2(s) + 2C(s) → Sn(s) + 2CO(g) water. (The specific heat capacity of
water is 4.18 J g−1 K−1.) (2)
a) Calculate the standard enthalpy change for
ii) Calculate an experimental value
each of the possible reactions using the data
for the standard enthalpy change of
below.
combustion. Give your answer to an
Δ f H 1 [SnO2(s)] = −581 kJ mol–1 appropriate number of significant
figures. (3)
Δ f H 1 [CO2(g)] = −394 kJ mol–1
iii) Compare your answers to (a)(iii) and
Δ f H 1 [CO(g)] = −110 kJ mol–1 (6) (b)(ii). Give reasons to explain the
difference in the values. (2)
b) Use your calculations to explain which of
the reactions would be most economic for 9 Butane (Camping Gaz®), C4H10, burns readily
industry.(2) on a camp cooker. The equation for the
reaction is:
The Hess cycle below can be used to calculate
8
the standard enthalpy change of combustion of C4H10(g) + 6 12 O2(g) → 4CO2(g) + 5H2O(g)
ethanol, Δ c H 1 , using standard enthalpy changes a) Rewrite the equation showing all the
of formation. covalent bonds between atoms in the
reactants and products. (4)
∆ cH 1 b) Identify the bonds broken in the reactants
C2H5OH(I) + 3O2(g)
and the bonds formed in the products
during the reaction. (2)
∆H 1 ∆H 1
1
2 c) Use the following mean bond enthalpies
to calculate the enthalpy change of the
reaction. (4)
E(C–C) = 347 kJ mol −1
a) i) Copy and complete the cycle by filling
in the empty boxes. (2) E(C–H) = 413 kJ mol−1
ii) Give a definition of the term ‘standard
enthalpy change of formation’ of a E(O=O) = 498 kJ mol−1
compound. (3) E(C=O) = 805 kJ mol−1
iii) Use the Hess cycle to calculate the
standard enthalpy change of combustion E(H–O) = 464 kJ mol−1
of ethanol, Δ c H 1 . Use this data: d) Give two reasons why the value calculated
in part (c) is not the same as the standard
Δ f H 1 [CO2(g)] = −394 kJ mol–1 enthalpy change of the reaction calculated
Δ f H 1 [H2O(l)] = −286 kJ mol–1 at 298 K. (2)
Δ f H 1 [C2H5OH(l)] = −277 kJ mol–1 10 A student suggested that ethane might react
(6) with bromine vapour in two different ways in
b) A student carried out an experiment, using bright sunlight.
the apparatus shown in Figure 8.7 on Reaction 1: C2H6(g) + Br2(g)
page 249, to estimate the standard enthalpy → C2H5Br(g) + HBr(g)
change of combustion of ethanol. The Reaction 2: C2H6(g) + Br2(g) → 2CH3Br(g)
apparatus was surrounded with a screen to a) Use the following mean bond enthalpies to
reduce draughts. The student added 150 g calculate the enthalpy changes for the two
water to the metal can. After the burner had possible reactions.
270
8 Energetics I
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E(C–C) = 347 kJ mol−1 Octane is typical of the hydrocarbons found in
12
E(Br–Br) = 193 kJ mol−1 petrol made from crude oil.
E(C–H) = 413 kJ mol−1 Most methanol is still made from the methane
in natural gas but a growing amount of
E(H–Br) = 366 kJ mol−1 methanol is being made from other resources.
E(C–Br) = 290 kJ mol−1 (6) Methanol can be produced from anything that
is, or ever was, a plant. This includes biomass,
b) Assess the data, and your calculation, to agricultural and timber waste, solid municipal
account for the fact that Reaction 1 is the waste, landfill gas, industrial waste and pollution
one that is found to occur. (2) and a number of other feedstocks.
c) Give two reasons why your calculated
enthalpy changes may not agree with In some instances it is important to choose
the accurately determined experimental a fuel based on the energy released per gram
values. (2) when the fuel burns.
Hex-1-ene reacts with hydrogen gas to form
11 It can also be important to compare the energy
hexane. Hex-1-ene and hexane are liquids released per cm3 of fuel.
under standard conditions. Fuel Density/g cm –3 ΔcH 1 (298 K)
a) i) Write an equation of the reaction of /kJ mol –1
hex-1-ene with hydrogen and state the Methanol 0.793 −726
catalyst used for the reaction. (2) Octane 0.703 −5470
ii) Draw a Hess’s Law cycle to show how
the standard enthalpy change for the a) i) Give an example where it might be
reaction of hex-1-ene with hydrogen important to choose a fuel giving the
can be calculated from these enthalpy higher energy per gram. (1)
changes of combustion: ii) Calculate the standard enthalpy change of
combustion per gram for methanol and
Δ c H 1 [C6H12(l)] = −4003 kJ mol−1 octane and comment on the values. (2)
Δ c H 1 [H2(g)] = −286 kJ mol−1 b) i) Give an example where it might be
important to choose a fuel giving the
Δ c H 1 [C6H14(g)] = −4163 kJ mol−1(3) higher energy per cm3. (1)
iii) Use your cycle to calculate ΔH 1reaction.(3) ii) Calculate the standard enthalpy change
b) The table below shows the enthalpy change of combustion per cm3 for methanol
for three other alkenes with hydrogen. and octane. (2)
c)* Discuss the advantages and disadvantages
Reaction Standard enthalpy of octane and methanol as fuels, taking
change of reaction into account your answers to parts (a) and
/kJ mol−1
(b) and the information at the start of this
propene + hydrogen −125
question. (6)
→ propane
but-1-ene + hydrogen −126
→ butane
pent-1-ene + hydrogen −126
→ pentane
Explain why the values for the enthalpy
change for the reaction of these alkenes
with hydrogen are so similar. (3)
271
Exam practice questions
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Kinetics I
9
9.1 Reaction rates
The study of rates of reaction is important because it helps chemists to control
reactions both in the laboratory and on a large scale in industry. Chemists
have a model for explaining the effects of the various factors that affect the
rates of reactions. This model helps them to understand what happens to
atoms, molecules and ions during chemical changes.
In the chemical industry, manufacturers aim to get the best possible yield in
the shortest time. The development of new catalysts to speed up reactions
is, therefore, one of the frontier aspects of modern chemistry (Figure 9.1).
The aim is to make manufacturing processes more efficient so that they use
less energy and produce little or no harmful waste. The need for greater
Figure 9.1 Computer graphics showing a efficiency in chemical processes is now more pressing than ever as people
molecule of methanol (with green carbon become more aware of the harm that waste chemicals can do to our health
atom) passing through a channel in the and to the environment.
synthetic zeolite catalyst. This catalyst is
The study of reaction rates is called chemical kinetics which is important
used to make a new fuel from methanol.
in many other fields. The study of rates of reaction helped environmental
Chemists carry out research to understand
scientists, for example, to explain why CFCs and other chemicals are
reactions on an atomic scale so that they
destroying the ozone layer in the upper atmosphere. Pharmacologists who
can develop more effective catalysts.
study the chemistry of drugs must study the speed at which they change to
other chemicals or break down in the human body. Then the pharmacists
who formulate and supply medicines need to know about the rate at which
Key term the chemicals slowly degrade in the bottle or pack. For many medicines, the
shelf life is the time for which they can be stored before the concentration of
Chemical kinetics is the study of the the active ingredient has dropped by 10%.
rates of chemical reactions.
Chemical reactions happen at a variety of speeds (Figure 9.2). Ionic precipitation
reactions are very fast and explosions are even faster. However, the rusting of
iron and other corrosion processes are slow and may continue for years.
Figure 9.2 Firefighters have to know how
to slow down and stop burning. Water cools
the burning materials as it evaporates and
the steam produced can help to keep out
the air.
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Test yourself
1 How would you slow down or stop these reactions:
a) iron corroding
b) toast burning
c) milk turning sour?
2 How would you speed up these reactions:
a) fermentation in dough to make the bread rise
b) solid fuel burning in a stove
c) epoxy glue (adhesive) setting
d) the conversion of chemicals in engine exhausts to harmless gases?
9.2 Measuring reaction rates
Balanced chemical equations give no information about how quickly the
reactions occur. In order to get this information, chemists have to do
experiments to measure the rates of reactions under various conditions.
The amounts of the reactants and products change during any chemical
reaction – products form as reactants disappear. The rates at which these Key term
changes happen give a measure of the rate of reaction.
The rate of reaction is found by
The rate of the reaction between magnesium and hydrochloric acid
measuring the rate of formation of
Mg(s) + 2HCl(aq) → MgCl 2(aq) + H2(g) a product or the rate of removal of
a reactant. The usual procedure for
can be measured by:
finding the rate is to measure some
● the rate of loss of magnesium property of the reaction mixture, such
● the rate of loss of hydrochloric acid as its volume, and to see how this
● the rate of formation of magnesium chloride property varies with time.
● the rate of formation of hydrogen.
In this example, it is probably easiest to measure the rate of formation of
hydrogen by collecting the gas and recording its volume with time (Figure 9.3).
Figure 9.3 Collecting and measuring the
gas produced when magnesium reacts with
measuring
cylinder acid. A gas syringe can be used instead of
a measuring cylinder full of water.
acid
metal
water
Chemists design their rate experiments to measure a property which changes
with the amount or concentration of a reactant or product. Then:
change recorded in the property
rate of reaction =
time for the change
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In most chemical reactions the rate changes with time. The graph in
Test yourself Figure 9.4 is a plot of the results from a study of the reaction of magnesium
3 In an experiment to study the with dilute hydrochloric acid. The graph is steepest at the start, when the
reaction of magnesium with reaction is at its fastest. As the reaction continues it slows down, until it
dilute hydrochloric acid, 48 cm3 finally stops. This happens because one of the reactants is being used up until
of hydrogen forms in 10 s at none of it is left. The gradient at any point on a graph showing amount or
room temperature. Calculate concentration plotted against time measures the rate of reaction (Figure 9.5).
the average rate of formation of:
Product concentration/mol dm –3
a) hydrogen in cm3 s−1 A C
80
Volume of hydrogen/cm3
b) hydrogen in mol s−1 (Section
5.3) 60
c) the rates of appearance or 40 rate at time t
disappearance of the other =
AB
mol dm –3 s –1
20 AC
product and the reactants
B
in mol s−1. 0
0 100 200 300 400
Time/s
Figure 9.4 Volume of hydrogen t Time/s
plotted against time for the Figure 9.5 Graph showing the concentration of a
reaction of magnesium with product plotted against time. The gradient at any
hydrochloric acid. point measures the rate of reaction at that time.
A useful way of studying the effect of changing the conditions on the rate
of a reaction is to find a way of measuring the rate just after mixing the
reactants. Figure 9.6 is a graph for two different sets of conditions. When
one of the reactants was more concentrated, line A was produced. Near the
start, it took tA seconds to produce x mol of product. When the same reactant
was less concentrated, the results gave line B. This time, near the start it took
t B seconds to produce x mol of product. The reaction was slower when the
concentration was lower, so it took longer to produce x mol of product.
Figure 9.6 Formation of the same amount
(x mol) of product starting with different A
Amount of product
concentrations of one of the reactants. B
0
0 tA tB Time
The average rate of formation of product on line A = x
tA
The average rate of formation of product on line B = x
tB
1
If x is kept the same, it follows that the average rate near the start ∝ t
This means that it is possible to arrive at a measure of the initial rate of a
reaction by measuring how long the reaction takes to produce a small fixed
amount of product, or use up a small fixed amount of reactant.
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9.3 What factors affect reaction rates?
Concentration
In general, the higher the concentration of the reactants, the faster the
reaction. For gas reactions, a change in pressure has the same effect as
changing the concentration – a higher pressure compresses a mixture of
gases and increases their concentration. So, in a mixture of reacting gases,
the higher the pressure, the faster the reaction.
Activity
Investigation of the effect of concentration on the rate of
a reaction
Figure 9.7 illustrates an investigation of the effect of concentration on the rate at which
thiosulfate ions in solution react with hydrogen ions to form a precipitate of sulfur.
S2O32− (aq) + 2H+(aq) → S(s) + SO2(aq) + H2O(l)
The observer records the time taken for the sulfur
precipitate to obscure the cross on the paper under
the flask. In this example, the quantity x in Figure 9.6 look down at cross
is the amount of sulfur needed to hide the cross on from above
the paper. This is the same each time – so, the rate
of reaction is proportional to 1/t. The results of the
investigation are shown in Table 9.1. time = t seconds
1 Which factors must be kept constant in this sodium thiosulfate
solution with acid cloudy
investigation to ensure that the results are valid? added to start the liquid
Explain your answer. reaction
2 How would you prepare 50 cm3 of a solution of cross
cross
sodium thiosulfate solution with a concentration of invisible
white paper
0.12 mol dm−3 from a solution with a concentration
Figure 9.7 Investigating the effect of the concentration of thiosulfate ions on the
of 0.15 mol dm−3? rate of reaction in acid solution. The hydrogen ion concentration is the same in
3 Suggest why the mixture in the flask should be each experiment.
poured into a container of saturated sodium
carbonate solution after each experiment.
Table 9.1 Results of the investigation in Figure 9.7.
4 The pale yellow precipitate of sulfur often sticks to
the flask forming a thin film on the glass surface. Experiment Concentration of Time, t, for the Rate of reaction,
1
thiosulfate ions/ cross to be /s−1
Why is it important to thoroughly clean the flask mol dm−3 obscured/s
t
after each experiment?
1 0.15 43 0.023
5 Calculate the value for the rate of reaction
2 0.12 55
when the concentration of thiosulfate ions is
0.12 mol dm−3. 3 0.09 66 0.015
6 Plot a graph to show how the rate of reaction 4 0.06 105 0.0095
varies with the concentration of thiosulfate ions. 5 0.03 243 0.0041
7 What is the relationship between reaction rate and
concentration of thiosulfate for this reaction
according to your graph?
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Surface area of solids
Breaking a solid into smaller pieces increases the surface area in contact with
a liquid or gas. This speeds up any reaction happening at the surface of the
solid. This effect also applies to reactions between liquids which do not mix.
Shaking breaks up one liquid into droplets which are then dispersed in the
other liquid, thereby increasing the surface area for reaction.
Activity
cotton wool plug
Investigating the effect of
surface area on the rate of a
reaction 40 cm3 of about 20g folded
Figure 9.8 illustrates an investigation of the 2.0 mol dm 3 marble chips paper
nitric acid
rate of reaction of lumps of calcium carbonate
(marble) with dilute nitric acid. The results
are given in Table 9.2. Both sets of results
were obtained using 20 g of marble chips and top pan
balance
40 cm3 of 2.0 mol dm−3 nitric acid. The marble
was in excess.
Figure 9.8 Apparatus for comparing the reaction rate of calcium carbonate with nitric acid.
1 a) E xplain why all the equipment and
chemicals were placed together on the
balance throughout the experiment, as Table 9.2 Results of experiments to compare the reaction rate of calcium carbonate
shown in Figure 9.8. with nitric acid using the same mass of larger and smaller marble chips.
b) Why was a cotton wool plug placed in the Time/s Mass of carbon dioxide formed/g
neck of the flask? Small marble chips Large marble chips
2 Plot the two sets of results on the same axes.
30 0.45 0.18
3 Work out the initial rates of the two reactions
60 0.85 0.38
by drawing tangents and determining the
90 1.13 0.47
gradients.
4 a) After what time did the reaction stop for 120 1.31 0.75
each set of results? 180 1.48 1.05
b) Why did the reaction stop? 240 1.54 1.25
5 Why was the same mass of carbon dioxide 300 1.56 1.38
formed in both sets of results? 360 1.58 1.47
6 For a given mass of marble, how is surface area 420 1.59 1.53
related to particle size? 480 1.60 1.57
7 What is the effect on this reaction of changing 540 1.60 1.59
the surface area of the solid?
600 1.60 1.60
8 Sketch on your graph the results you would
expect if you repeated the experiment with 20 g
small marble chips and 40 cm3 of 1.0 mol dm−3 nitric acid.
9 a) Use the equation for the reaction to calculate the theoretical mass of carbon
dioxide formed when 40 cm3 of 2.0 mol dm−3 nitric acid reacts completely with
calcium carbonate.
b) Suggest reasons for the difference between the actual and the theoretical mass
of carbon dioxide formed.
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Temperature 0.06
Raising the temperature is a very effective way of increasing the rate of a reaction.
Rate of reaction / arbitrary units
0.05
For a number of chemical reactions at around room temperature, a 10 °C rise in
temperature roughly doubles the rate of reaction (see Figure 9.9).
0.04
Bunsen burners, hot-plates and heating mantles are common items of
equipment in laboratories because it is often convenient to speed up reactions 0.03
by heating the reactants. For the same reason, many industrial processes are
0.02
carried out at high temperatures.
0.01
Catalysts
Catalysts have an astonishing ability to speed up the rates of some chemical 0 10 20 30 40 50 60
Temperature/°C
reactions without themselves changing permanently. Very small quantities
of active catalysts can speed reactions to produce many times their own mass Figure 9.9 The effect of
of chemicals. temperature on the rate of
decomposition of thiosulfate ions to
Catalysts work by removing or lowering the barriers preventing reaction – form sulfur.
they bring reactants together in a way that makes a reaction more likely.
Some catalysts such as nickel metal can catalyse many different reactions.
However, catalysts can also be extraordinarily selective – a catalyst may Key terms
increase the rate of only one very specific reaction. Enzymes, the catalysts in
A catalyst speeds up the rate of
living cells, are especially selective.
a chemical reaction without itself
Catalysts change the mechanisms of reactions, but they are not reactants changing to a different substance. The
and they do not appear in the overall chemical equation. In theory, catalysts catalyst can often be recovered at the
can be used over and over again, but in practice there is some loss of catalyst. end of the reaction. A small amount of
Sometimes catalysts become contaminated, sometimes they are hard to recover catalyst can be effective.
completely from the products and sometimes the catalyst changes its state, such
The mechanism of a reaction is a
as from lumps to a fine powder, which means that it is no longer useable.
description of how a reaction takes
Most industrial processes involve passing a mixture of gases over a solid place showing, step by step, the bonds
catalyst. Such catalysts are described as heterogeneous catalysts because which break and the new bonds which
the reactants and catalyst are in different phases. The gas molecules are form as reactants turn into products.
briefly held onto the surface of the solid, where the atoms of the catalyst help
A phase is one of the three states of
them to react; then the product molecules break free and are carried away in
matter – solid, liquid or gas. Chemical
the flow of gas.
systems often have more than one
One of the targets in the modern chemical industry is to develop catalysts phase. Each phase is distinct but need
that make manufacturing processes more efficient, so that they produce less not be pure. For example, a solid in
waste and use less energy. A novel catalyst can make possible a new route equilibrium with its saturated solution
for making a chemical product that has a higher atom economy. Developing is a two-phase system. In the reactor
a new catalyst can also make it possible to carry out a reaction at a lower for ammonia manufacture, the mixture
temperature or at a lower pressure. This saves fuel. The cost of fuel is one of of nitrogen, hydrogen and ammonia
the factors that determines the profitability of large-scale chemical processes. gases make up one phase with the iron
catalyst being a separate solid phase.
Test yourself A heterogeneous catalyst is one that is
in a different phase from the reactants.
4 a) Use the Haber process for making ammonia to explain what is Generally a heterogeneous catalyst is a
meant by a heterogeneous catalyst. solid while the reactants are gases, or
b) Suggest advantages of using heterogeneous catalysts in industry. in solution.
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9.4 Collision theory
Scientists can explain the factors that affect reaction rates. They use a model
which gives a picture of what happens to atoms, molecules and ions as they
react. This model works best for gases but it can also be applied to reactions
in solution.
Gas molecules in motion
The model that scientists use to explain the behaviour of gases assumes that
the molecules in a gas are in rapid random motion and colliding with each
other. They call this particles-in-motion model the ‘kinetic theory’. The
name comes from a Greek word for movement.
The kinetic theory makes a number of assumptions about the molecules of a
gas. Applying Newton’s laws of motion to the collection of particles leads to
equations that can describe the properties of gases very accurately.
The assumptions of the kinetic theory model are that:
● gas pressure results from the collisions of the molecules with the walls of
the container
● there is no loss of energy in the elastic collisions between the molecules
and the walls of any container
● the molecules are so far apart that the volume of the molecules can be
neglected in comparison with the total volume of gas
● the molecules do not attract each other
● the average kinetic energy of molecules is proportional to their temperature
on the Kelvin scale.
A gas that behaves exactly as this model predicts, obeying the gas laws, is
called an ‘ideal gas’ (Section 5.3). Real gases do not behave exactly like this.
The assumptions built into the model help to explain why real gases approach
ideal behaviour at high temperatures and low pressures:
● at high temperatures, the molecules are moving so fast that any small
attractive forces between them can be ignored
● at low pressures the volumes are so big that the space taken up by the
molecules is insignificant.
This kinetic theory also helps to explain why real gases deviate from ideal
gas behaviour as they get nearer to becoming a liquid. As a gas liquefies, the
molecules get very close together and the volume of the molecules cannot
be ignored. Also, gases could not liquefy unless there were some attractive
(intermolecular) forces between the molecules to hold them together.
The Dutch physicist Johannes van der Waals (1837–1923) developed his
theory of intermolecular forces (Section 2.6) by studying the behaviour of
real gases and their deviations from the ideal gas behaviour.
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The Maxwell–Boltzmann distribution Key term
Two physicists used the kinetic theory to explore the distribution of energies
among the molecules in gases. They worked out the proportion of molecules The Maxwell–Boltzmann distribution
with a given energy at a particular temperature. The two physicists were shows the spread of molecular kinetic
James Maxwell (1831–1879) in Britain and Ludwig Boltzmann (1844–1906) energies for a gas at a particular
in Austria. temperature. It shows that there is a
wide spread of molecular energies and
Figure 9.10 shows the distribution of energies for the molecules of a gas so the molecules are moving at different
under two sets of conditions. This Maxwell–Boltzmann distribution speeds.
helps to explain the effects of temperature changes and catalysts on the rates
of reactions.
300 K
310 K
Number of molecules
with kinetic energy E
Kinetic energy E
Figure 9.10 The Maxwell–Boltzmann distribution of kinetic energies of the molecules of a
gas at two temperatures. The area under the curve gives the total number of molecules.
This area does not change as the temperature rises, so the peak height falls as the
temperature rises and the curve spreads to the right.
Explaining the effects of concentration,
pressure and surface area on reaction rates
In any reaction mixture the billions of atoms, molecules or ions are forever
bumping into each other. When they collide there is a chance that they will react.
Raising the pressure of a gas means that the reacting particles are closer
together. There are more frequent collisions and, therefore, the reaction
goes faster. Increasing the concentration of reactants in solution has a similar
effect (Figure 9.11).
lower concentration higher concentration Figure 9.11 Raising the pressure, or
concentration, means that the reacting
atoms, molecules or ions are closer
together. There are more frequent collisions
and the reaction is faster.
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In a reaction of a solid with either a liquid or a gas, the reaction is faster if
the solid is broken up into smaller pieces. Crushing the solid increases its
surface area – collisions can be more frequent and the rate of reaction is faster
(Figure 9.12).
smaller surface area larger surface area
Figure 9.12 Breaking a solid into smaller pieces increases the surface area exposed
to reacting chemicals in a gas or in solution. Note that this diagram shows the solid
fragments and the molecules on different scales. In reactions of this kind, the fragments
of solid are generally huge compared to the size of the molecules or ions.
Key terms
The activation energy is the height Explaining the effects of temperature
of the energy barrier separating on reaction rates
reactants and products during a It is not enough simply for the molecules to collide. Most collisions do not result
chemical reaction. It is the minimum in a reaction. Molecules simply bounce off each other if there is not enough
energy needed for a reaction between energy in the collision to break bonds. Molecules may also fail to react if they
the amounts, in moles, shown in the are not angled correctly as they collide. Molecules are in such rapid motion that
equation for the reaction. if every collision led to a reaction, most reactions would be explosive.
A transition state is the state of the Chemists use the term activation energy to describe the minimum energy
reacting atoms, molecules or ions when needed in a collision between molecules if they are to react. Activation energies
they are at the top of the activation account for the fact that reactions go much more slowly than would be expected
energy barrier for a reaction step. if every collision between atoms and molecules led to a reaction. Only a very
Transition states exist for such a brief small proportion of collisions bring about chemical change. Molecules can
moment that they cannot be detected only react if they collide with enough energy for bonds to stretch and then
or isolated. break so that new bonds can form. At around room temperature, only a minute
A reaction profile is a graph which proportion of molecules have enough energy to react.
shows how the total enthalpy (energy)
Figure 9.13 shows that the energy of the colliding molecules is taken in to
of the atoms, molecules or ions
stretch bonds as a transition state forms. Then, as old bonds break and new
changes during the progress of a
bonds form, the energy is released to create products. The net energy change
reaction from reactants to products.
is the enthalpy change for the reaction.
Figure 9.13 Reaction profile showing the
activation energy for a reaction. transition state
Energy
activation energy
reactants
products
Progress of reaction
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The shaded areas in Figure 9.14 show the small proportions of molecules having
enough energy to overcome the activation energy for a reaction at 300 K and
310 K. This area is larger at the higher temperature. So, at a higher temperature,
more molecules have enough energy to react when they collide and the reaction
goes faster. Also, when the molecules are moving faster, they collide more often.
300 K number of molecules with
energy greater than the
Number of molecules with
310 K activation energy at 310 K
kinetic energy E
number of molecules with
energy greater than the
activation energy at 300 K
activation
energy
Kinetic energy E
Figure 9.14 The Maxwell–Boltzmann distribution of kinetic energies in the molecules
of a gas at 300 K and 310 K. The area under each curve is a measure of the number of
molecules. At 310 K, more molecules have enough energy to react when they collide with
other molecules.
Test yourself
5 Two factors explain why reactions go faster when the temperature
rises. Identify these two factors in terms of the energy of molecules,
atoms and ions.
Explaining the effects of catalysts on
reaction rates
A catalyst works by providing an alternative pathway for the reaction with
a lower activation energy. Lowering the activation energy increases the
proportion of molecules with enough energy to react (Figure 9.15).
A catalyst changes the mechanism of a reaction and makes a reaction more
productive by increasing the yield of the desired product and reducing waste.
Number of molecules with
activation
energy with
kinetic energy E
a catalyst
activation
energy without
a catalyst
Kinetic energy E
Figure 9.15 Distribution of molecular energies in a gas, showing how the proportion of
molecules able to react increases when a catalyst lowers the activation energy.
9.4 Collision theory 281
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One of the ways in which a catalyst can change the mechanism of a reaction
Key term is to combine with the reactants to form an intermediate. The intermediate
is a stage in the transition from reactants to products – it breaks down to give
Intermediates in reactions are atoms,
the products and the catalyst is released. This frees the catalyst to interact
molecules and ions which do not
with further reactant molecules and the reaction continues (Figure 9.16).
appear in the balanced equation but
which are formed during one step of a
a)
reaction, then used up in the next step.
activation
energy without
catalyst
Energy
reactants ∆H
products
Progress of reaction
b)
activation alternative pathway
energy with
catalyst
Energy
reactants ∆H
products
Progress of reaction
Figure 9.16 Reaction profiles for a reaction a) without a catalyst and b) with
a catalyst. The dip in the curve of the pathway with a catalyst shows where an
unstable intermediate forms.
Test yourself
6 Which parts of Figure 9.16 b show:
a) the formation of an intermediate
b) a transition state?
7 a)
Why is a match or spark needed to light a Bunsen burner?
b) Why does the gas keep burning once it has been lit?
8 Suggest a reason why catalysts are often specific for a particular
reaction.
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Chapter summary
l Increasing the pressure of a gas mixture, or raising
Chapter 9 Kinetics I
the concentration of a solution, means that there are
l Chemical kinetics is the study of the rates of more frequent collisions between molecules and so
chemical reactions. the reaction goes faster.
l The rate of a reaction is found by measuring the
l The Maxwell–Boltzman distribution shows the
rate of formation of a product or the rate of removal spread of molecular energies for a gas at a particular
of a reactant. temperature. The area under the curve represents
l The rate of a reaction at a particular time can be
the number of molecules. At a higher temperature
determined by calculating the gradient of a tangent there is a wider spread of energies; the distribution
drawn on a concentration–time graph. shifts to the right and the peak height is lower.
l Factors that affect the rates of reaction include: the
l Only a small proportion of molecular collisions
concentration of solutions, the pressure of gases, the lead to reaction; these are the collisions involving
surface area of solids, temperature and catalysts. enough energy to break bonds.
l Many industrial processes involve passing a mixture
l The activation energy for a reaction is the height of
of gases over a solid (heterogeneous catalyst), which the energy barrier separating reactants and products,
provides a surface on which the molecules can as represented by a reaction profile.
react. l At a higher temperature, there are more molecules
l Catalysts help to make industrial processes more
with enough energy to overcome the activation
efficient by making it possible to carry out reactions energy when they collide, so the reaction is faster.
under milder conditions and by providing reaction l A catalyst provides an alternative pathway for a
pathways that create less waste and use less energy. reaction with a lower activation energy. Lowering
l Collision theory is used to explain how the various
the activation energy increases the proportion
factors affect reaction rates in terms of the chance of molecules with enough energy to react at a
that there will be a reaction when fast-moving particular temperature.
molecules bump into each other.
Chapter summary 283
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Exam practice questions
Hydrogen peroxide solution, H2O2(aq),
1 c) Explain why a small increase in temperature
decomposes very slowly at room temperature can lead to a large increase in the rate of a
releasing oxygen. The reaction is catalysed reaction. (3)
by manganese(iv) oxide. The table shows the Sketch the reaction profile for a reaction taking
3
volume of oxygen collected at regular intervals place with a catalyst given that:
when one measure of MnO2 powder is added to • the reaction is endothermic
50 cm3 of a hydrogen peroxide solution at 20 °C. • the activation energy for the formation of
a) Describe the apparatus that can be used an intermediate is higher than the activation
to obtain the results in the table. Explain energy for the conversion of the intermediate
how the volume of gas can be measured to the products. (3)
accurately from the moment that the
catalyst is added to the hydrogen peroxide 4 a) Sketch a graph, with labelled axes, to show
solution. (4) the Maxwell–Boltzmann distribution of
b) Write an equation for the reaction. (1) the energies of the reactant molecules. (2)
c) Plot a graph of the results on axes with a b) Label the energy axis to show a likely
vertical scale showing the volume of oxygen position for the activation energy of a
up to 100 cm3. (2) reaction without a catalyst and with a
catalyst.(2)
Time/s Volume of oxygen/cm3 c) Use your graph to explain why a catalyst
0 0 speeds up the rate of a reaction. (3)
20 10
Give reasons to account for these observations:
5
40 20
a) A mixture of oxygen and hydrogen gas does
60 26
not react at room temperature. The mixture
80 32
explodes if ignited with a spark or if a little
100 35 powdered platinum is added. (4)
120 38 b) Nitrogen and oxygen do not usually react
140 39 in the air but they do combine to give
160 40 nitrogen monoxide in the cylinder of a car
180 40 engine. (4)
c) There is a danger of explosions caused by
d) Explain the shape of your graph. (3) dust in flour mills and coal mines. (4)
e) On the same axes, sketch the graphs you
6* Some chemical reactions happen fast; others
would expect if, in separate experiments, all
are very slow. Give reasons for these differences.
the conditions are the same except that:
Illustrate your answer with examples. Use
i) the temperature is raised to 40 °C
graphs and diagrams to enhance your
ii) the volume of hydrogen peroxide
explanations. (6)
solution is 100 cm3
iii) manganese(iv) oxide granules are used 7* Two important scientific models are the kinetic
in place of powder theory of gases and the collision theory of
iv) the concentration of the hydrogen reaction rates. Identify key features of scientific
peroxide solution is halved. (8) models and show that they are illustrated by
these examples. Discuss the importance and
2 a) Explain why most collisions in a reaction
limitations of these examples of modelling. (6)
mixture do not result in a reaction. (2)
b) Give a way to increase the collision
frequency between molecules in a gas
without changing the temperature. (1)
284
9 Kinetics I
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Equilibrium I
10
10.1 Reversible changes
The study of reversible reactions helps chemists to answer the questions
‘How far?’ and ‘In which direction?’ – questions they need to answer when
trying to make new chemicals in laboratories and in industry.
Some changes go in only one direction – like baking bread. Once baked in an
oven, there is no way to reverse the process and split a loaf back into flour, water
and yeast. Burning a fuel, such as natural gas or petrol, is another example of a
one-way process. Once these fuels have burned in air to make carbon dioxide
and water, it is impossible to simply reverse the reaction to turn the products
back to natural gas and petrol. The combustion of fuels is an irreversible process.
Key term
Many other reactions involve reversible changes. Haemoglobin, for example,
A reversible change is a process combines with oxygen as red blood cells flow through the lungs, but then releases
which can be reversed by altering the the oxygen for respiration as blood flows in the capillaries throughout the rest of
conditions. the body. Another example is the reaction of water and dissolved carbon dioxide
with the calcium carbonate of limestone. This reaction erodes limestone rock
(Figure 10.1). The reaction is reversed in caves as stalactites form (Figure 10.2).
Another example of a reversible reaction is the basis of a simple laboratory
test for water. Hydrated cobalt(ii) chloride is pink and so is a solution of the
salt in water. Heating filter paper soaked in the solution in an oven makes
Figure 10.1 Eroded limestone rock near Malham in the Yorkshire Dales. Figure 10.2 Stalactites and stalagmites in a cave. Stalactites and
The cracks in the limestone rock have been widened by natural chemical stalagmites form in limestone caves because the reaction of carbon
erosion. Rainwater made acid with dissolved carbon dioxide reacts with dioxide and water with calcium carbonate is reversible. The reverse
the calcium carbonate in limestone as the water flows over it. reaction reforms solid calcium carbonate.
10.1 Reversible changes 285
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the paper turn blue, because water is driven off from the solution leaving
anhydrous cobalt(ii) chloride on the paper.
CoCl 2.6H2O(s) → CoCl 2(s) + 6H2O(g)
pink blue
The blue paper provides a sensitive test for water as it turns pink again if
exposed to water or water vapour. At room temperature water rehydrates the
blue salt (Figure 10.3).
CoCl 2(s) + 6H2O(l) → CoCl 2.6H2O(s)
blue pink
The reaction of ammonia with hydrogen chloride is an example of a reaction in
Figure 10.3 Using cobalt chloride paper to
which the direction of change depends on the temperature. At room temperature,
test for water.
the two gases combine to make a white smoke of ammonium chloride.
NH3(g) + HCl(g) → NH4Cl(s)
white smoke
Heating reverses the reaction and ammonium chloride decomposes at high
damp red
litmus paper
temperatures to give hydrogen chloride and ammonia.
damp blue NH4Cl(s) → NH3(g) + HCl(g)
litmus paper
The apparatus in Figure 10.4 can be used to show that ammonium chloride
glass wool decomposes into two gases on heating. Ammonia gas diffuses through the glass
wool faster than hydrogen chloride. After a short time, the alkaline ammonia
rises above the plug of glass wool and turns the red litmus blue. A while later both
ammonium strips of litmus paper turn red as the acid hydrogen chloride arrives. A smoke of
chloride ammonium chloride appears above the tube when both gases meet and cool.
heat Changing the temperature is not the only way to alter the direction of change.
Hot iron, for example, reacts with steam to make iron(iii) oxide and hydrogen.
Figure 10.4 Investigating the thermal
Supplying plenty of steam and ‘sweeping away’ the hydrogen means that the
decomposition of ammonium chloride.
reaction continues until all the iron changes to its oxide (Figure 10.5).
3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)
Figure 10.5 The forward reaction goes iron
when the concentration of steam is high
and the hydrogen is swept away, keeping steam hydrogen
its concentration low.
Tip heat
In an equation the chemicals on the Altering the conditions brings about the reverse reaction. A stream of hydrogen
left-hand side are the reactants – those reduces all the iron(iii) oxide to iron, so long as the flow of hydrogen sweeps
on the right are the products. The away the steam that has formed (Figure 10.6).
‘left-to-right’ reaction is the ‘forward’
reaction and the ‘right-to-left’ reaction 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)
is the backward reaction.
iron oxide
Figure 10.6 The reverse, or backward,
reaction goes when the concentration of hydrogen steam
hydrogen is high and the steam is swept
away, keeping its concentration low.
heat
286 10 Equilibrium I
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Test yourself
1 Write balanced equations, with state symbols, for the reactions when:
a) rainwater and carbon dioxide erode calcium carbonate in limestone
to form soluble calcium hydrogencarbonate
b) aqueous calcium hydrogencarbonate decomposes to reform
calcium carbonate as the solution drips from the roof of a cave.
2 How can the following changes be reversed, either by changing the
temperature or by changing the concentration of a reactant or product:
a) freezing water to ice
b) changing blue litmus to its red form
c) converting blue copper(ii) sulfate to its white form?
3 Explain why wet washing does not dry if kept in a plastic laundry bag,
but does dry if hung out on a line.
10.2 Reaching an equilibrium state
Reversible changes often reach a state of balance, or equilibrium. What is
special about chemical equilibria is that nothing appears to be happening,
but at a molecular level there is ceaseless change.
When chemists ask the question ‘How far?’, they want to know what the state
of a reaction will be when it reaches equilibrium. At equilibrium, the reaction
shown by an equation may be well to the right (mostly new products), well to
the left (mostly unchanged reactants) or somewhere in between.
Balance points exist in most reversible reactions where neither the forward
nor the reverse reaction is complete. Reactants and products are present
together and the reactions appear to have stopped – this is the state of
chemical equilibrium.
One way to study the approach to equilibrium is to watch what happens on shaking
a small crystal of iodine in a test tube with cyclohexane and a solution of potassium
iodide, KI(aq). The liquid cyclohexane and the aqueous solution do not mix.
Iodine freely dissolves in cyclohexane, which is a non-polar solvent (Section
2.7). The non-polar iodine molecules mix with the cyclohexane molecules –
there is no reaction. The solution is a purple–violet colour, the same colour as
iodine vapour. Iodine hardly dissolves in water but it does dissolve in a solution
of potassium iodide. The solution is yellow, orange or brown depending on
the concentration. In the solution, iodine molecules, I2, react with iodide
ions, I−, to form triiodide ions, I3−.
Tip
The symbol ⇋ represents a reversible reaction at equilibrium. In theory it is only possible
to achieve a state of equilibrium in a closed system (Section 8.1).
Figure 10.7 is a study of changes which can be summed up by this equilibrium:
I2(in cyclohexane) + I−(aq) ⇋ I3−(aq)
10.2 Reaching an equilibrium state 287
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potassium iodide
cyclohexane solution tube A1 tube A2 tube A3
iodine equilibrium
most iodine still distribution
dissolved in
in cyclohexane of iodine
cyclohexane
between
iodine cyclohexane and
potassium some iodine potassium
small dissolved in iodide in potassium iodide solution
iodine cyclohexane solution iodide solution
crystal
shake very shake
potassium iodide gently well
solution cyclohexane tube B1 tube B2 tube B3
equilibrium
some iodine distribution
cyclohexane dissolved in of iodine
cyclohexane between
cyclohexane and
small iodine iodine potassium
iodine dissolved in dissolved in most iodine iodide solution
crystal in potassium potassium still in
iodide solution iodide solution potassium
iodide solution
Figure 10.7 Two approaches to the same equilibrium state. Note that the tubes labelled A1, A2 and A3 are the same tube at three
different stages. The same is true for the tubes labelled B1, B2 and B3.
The graphs in Figure 10.8 show how the iodine concentration in the two
layers changes with shaking. After a little while, no further change seems
to take place – tubes A3 and B3 look just the same. Both contain the same
equilibrium system.
This demonstration shows two important features of equilibrium processes:
● at equilibrium the concentration of reactants and products does not change
● the same equilibrium state can be reached from either the ‘reactant side’ or
the ‘product side’ of the equation.
Figure 10.8 Change in concentration of
Concentration
iodine with time in the mixtures shown in
of iodine
in cyclohexane
Figure 10.7.
Concentration
of iodine
in cyclohexane
in KI(aq)
in KI(aq)
tube tube tube Time
A1 A2 A3
tube tube tube Time
A1 A2 A3
Concentration
of iodine
in cyclohexane
Concentration
of iodine
in cyclohexane
in KI(aq)
in KI(aq)
tube tube tube Time
B1 B2 B3
tube tube tube Time
288 10 Equilibrium I B1 B2 B3
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10.3 Dynamic equilibrium Key term
Figure 10.9 shows what is happening at a molecular level – not what your eye
In dynamic equilibrium, the forward
can see – in the equilibrium involving iodine moving between cyclohexane
and backward reactions continue, but
and a solution of potassium iodide. Consider tube A1 in Figure 10.7. All
at equal rates, so that the overall effect
the iodine molecules start in the upper cyclohexane layer. On shaking, some
is no change. At the molecular level,
move into the aqueous layer. At first, molecules can move in only one
there is continuous movement. At the
direction (the forward reaction). The forward reaction begins to slow down
macroscopic level, nothing appears to
as the concentration in the upper layer falls.
be happening.
Figure 10.9 Iodine molecules reaching
dynamic equilibrium between cyclohexane
and a solution of potassium iodide. The
formation of I3− ions in the aqueous layer
is ignored in this diagram.
Once there is some iodine in the aqueous layer, the reverse process can begin
with iodine returning to the cyclohexane layer. This backward reaction Test yourself
starts slowly but speeds up as the concentration of iodine in the aqueous layer 4 Under what conditions are
increases. these in equilibrium:
In time, both the forward and backward reactions happen at the same rate. a) water and ice
Movement of iodine between the two layers continues but overall there is b) water and steam
no change. In tube A3 in Figure 10.7 each layer is gaining and losing iodine c) copper(ii) sulfate crystals
molecules at the same rate. This is an example of dynamic equilibrium. and copper(ii) sulfate
solution?
10.4 Factors affecting equilibria 5 Draw a diagram to represent
the movement of particles
Changing the conditions can disturb a system at equilibrium. At equilibrium between a crystal and a
the rate of the forward and backward reactions is the same. Anything which saturated solution of the solid
changes the rates can shift the balance. in a solvent.
Predicting the direction of change
Le Chatelier’s principle is a qualitative guide to the effect of changes in
concentration, pressure or temperature on a system at equilibrium. The
principle was suggested as a general rule by the French physical chemist
Henri Le Chatelier (1850–1936).
The principle states that when the conditions of a system at equilibrium
change, the system responds by trying to counteract the change.
Changing the concentration
Table 10.1 shows the effects of changing the concentration in the generalised
equilibrium system
A+B⇋C+D
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Table 10.1 Effects of changing the concentration in the generalised equilibrium system.
Tip
Disturbance How does the equilibrium The result
Note that Le Chatelier’s principle is a
mixture respond?
guide based on lots of observations of
reversible processes at equilibrium. The Concentration System moves to the right – some Less B and more C and D
of A increases. A is removed by reaction with B. in the new equilibrium.
principle does not explain the effect
of changing conditions on systems at Concentration System moves to the left – some Less C and more A and B
equilibrium. of D increases. of the added D is removed by in the new equilibrium.
reaction with C.
Concentration System moves to the right to There is more C and
of D decreases. make up for the loss of D. less A and B in the new
equilibrium.
The reaction of bromine with water can be used to make predictions based
on Le Chatelier’s principle. A solution of bromine in water is a yellow–
orange colour because it contains bromine molecules in the equilibrium:
Br2(aq) + H2O(l) ⇋ HOBr(aq) + Br−(aq) + H+(aq)
orange colourless
Adding alkali turns the solution almost colourless (Figure 10.10). Hydroxide
ions in the alkali react with hydrogen ions, removing them from the
equilibrium. As the hydrogen ion concentration falls, the equilibrium
shifts to the right, converting orange bromine molecules to colourless
molecules and ions. Lowering the hydrogen ion concentration slows down
the backward reaction, while the forward reaction goes on as before. The
position of equilibrium shifts until, once again, the rates of the forward and
alkali
mainly backward reactions are the same.
mainly
Br2(aq) + H2O(l) HOBr(aq) +
acid Br–(aq) + H+(aq) Adding acid increases the concentration of hydrogen ions – this speeds up
the backward reaction and makes the solution turn orange–yellow again.
The equilibrium shifts to the left reducing the hydrogen ion concentration
Figure 10.10 The visible effects of adding
and increasing the bromine concentration until, once again, the forward and
alkali and acid to a solution of bromine in
backward reactions are in balance.
water.
Test yourself
6 Write an ionic equation for the reversible reaction of silver(i) ions with
iron(ii) ions to form silver atoms and iron(iii) ions. Make a table similar to
Table 10.1 to show how Le Chatelier’s principle applies to this equilibrium.
7 Yellow chromate(vi) ions, CrO42−(aq), react with aqueous hydrogen
ions, H+(aq), to form orange dichromate(vi) ions, Cr2O72−(aq), and
water molecules. The reaction is reversible. Write an equation for
the system at equilibrium. Predict how the colour of a solution of
chromate(vi) ions changes:
a) on adding acid
b) followed by adding hydroxide ions (OH−), which neutralise hydrogen
ions (Section 4.1).
8 Heating limestone, CaCO3, in a closed furnace produces an equilibrium
mixture of calcium carbonate with calcium oxide, CaO, and carbon
dioxide gas. Heating the solid in an open furnace decomposes the
solid completely into the oxide. How do you account for this difference?
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Activity
An equilibrium involving iodine and chlorine
Iodine reacts with chlorine gas to form a brown liquid called 3 Write an equation, with state symbols, for the reversible
iodine monochloride, ICl. Iodine monochloride reacts reversibly reaction of the brown liquid to produce the yellow solid.
with chlorine to form iodine trichloride, ICl3 which is a yellow solid. 4 At what stage in the demonstration is there a dynamic
Figure 10.11 shows the steps of a demonstration of these equilibrium between the brown liquid, chlorine and the
reactions. The steps are outlined above the diagrams. The yellow solid?
observations are described below the diagrams. 5 Explain the conditions under which more of the yellow solid
forms.
1 What are the hazards involved in this demonstration and 6 Explain the conditions under which most of the yellow solid
what steps must be taken to reduce the risks? disappears.
2 Write an equation, with state symbols, for the reaction that
produces the brown liquid.
1 Chlorine is passed over 2 The chlorine supply is stopped 3 More chlorine is passed through
iodine crystals. and the U-tube stoppered. the U-tube.
Cl2 Cl2
I2
A brown liquid forms. There is also a More yellow solid forms and the thick
little yellow solid. Chlorine gas stays brown liquid disappears.
in the tube.
6 The tube is inverted again. 5 More chlorine is passed through 4 The chlorine supply is disconnected
the U-tube. and the U-tube is inverted.
Cl2
The brown liquid reappears as the The yellow solid reappears as the Chlorine gas ‘falls out’ of the tube. The
yellow solid disappears. thick brown liquid disappears. brown liquid reappears as the yellow
solid disappears.
Figure 10.11 A demonstration to show how changing conditions can alter the position of an equilibrium.
10.4 Factors affecting equilibria 291
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Changing the pressure and temperature
Many industrial processes happen in the gas phase. High pressures and high
temperatures are often needed, even when there is a catalyst. One important
example of this is the Haber process used to make ammonia. The equilibrium
system involved is:
N2(g) + 3H2(g) ⇋ 2NH3(g) ΔH = −92.4 kJ mol−1
Test yourself The reaction takes place in a reactor packed with an iron catalyst. Adding a
catalyst does not affect the position of equilibrium. A catalyst simply speeds
9 E xplain why the conditions
up both the forward and the back reactions by the same amount, so it shortens
used in the industrial Haber
the time taken to reach equilibrium.
process are a compromise.
10 Methanol is manufactured Le Chatelier’s principle helps to explain the conditions chosen for the Haber
from carbon dioxide and process. There are 4 mol of gases on the left-hand side of the equation but
hydrogen in the presence of only 2 mol on the right. Increasing the pressure makes the equilibrium shift
a catalyst that consists of a from left to right, because this reduces the number of molecules and tends to
mixture of copper and zinc reduce the pressure. So, increasing the pressure increases the proportion of
oxides coated onto pellets ammonia at equilibrium.
of aluminium oxide. For this The reaction is exothermic from left to right and, therefore, endothermic from
reaction ΔH1 = −91 kJ mol−1. right to left. Le Chatelier’s principle predicts that raising the temperature makes
a) Write a balanced equation the system shift in the endothermic direction which takes in energy (tending
for the reaction. to lower the temperature). So, raising the temperature lowers the proportion of
b) Suggest a reason for ammonia at equilibrium.
coating the catalyst onto In industry, the conditions chosen are a compromise between the need to
the surface of pellets on convert as much nitrogen and hydrogen to ammonia as possible in the reactor
an inert material. (high pressure and low temperature) and the need to produce ammonia fast
c) Suggest reasons why the enough (high pressure, high temperature and the presence of a catalyst).
process is carried out
at 550 K and 100 times
In practice, a wide range of conditions are used, depending on the design of
atmospheric pressure.
the reactor. The temperatures used vary from 600 to 700 K. Pressures range
from 50 to 100 times atmospheric pressure.
Activity
The manufacture of ethanol
Ethanol is manufactured from ethene and steam in the condensed from the gases leaving the reactor. At this stage the
presence of a catalyst. ethanol formed contains a high proportion of water.
C2H4 (g) + H2O(g) ⇋ C2H5OH(g) ΔH = −45 kJ mol−1 1 What is the main source of ethene for industrial processes?
2 The ratio of water : ethene supplied to the reactor is
The catalyst in the reactor is phosphoric acid held as a thin film
0.6 mol : 1 mol. Suggest the factors that determine this ratio.
coating the surface of finely divided solid silicon dioxide. The
3 Suggest the factors that determine the choice of 500 K as
catalyst absorbs water under pressure. This dilutes the catalyst
the operating temperature for the process.
and may lead to it draining away from the solid support.
4 Suggest the factors that determine the choice of 60−70
The process is carried out at about 500 K with a pressure in times atmospheric pressure as the operating pressure for
the range 60−70 times atmospheric pressure. If the pressure is the process.
too high, the ethene starts to polymerise. 5 In practice, the process converts 95% of the ethene to
ethanol. Suggest how this is achieved.
Only about 5% of the ethene is converted to ethanol as the
6 Suggest the method that is used to concentrate the ethanol
mixture of reactants passes through the reactor. The product is
produced in the process.
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10.5 Equilibrium constants
Chemists have discovered a law that can be used to predict the amounts of
reactants and products when a reversible reaction reaches a state of dynamic
equilibrium. The equilibrium law has been established by experiment. It is a
quantitative law for predicting the amounts of reactants and products present
when a reaction reaches a state of dynamic equilibrium.
In general, for a reversible reaction at equilibrium:
[C]c[D]d
aA + bB ⇋ cC + dD Kc =
[A]a[B]b
This is the form for the equilibrium constant, Kc, when the concentrations
of the reactants and products are measured in moles per cubic decimetre.
Key term
[A], [B] and so on are the equilibrium concentrations, sometimes written as A shorthand for concentration in
[A]eqm and [B]eqm to make this clear. mol dm−3 is to write the formula
The concentrations of the chemicals on the right-hand side of the equation of a chemical in square brackets.
appear on the top line of the expression. The concentrations of reactants on For example: [A] represents the
the left appear on the bottom line. Each concentration term is raised to the concentration of A in mol dm−3.
power of the number in front of its formula in the equation.
Tip
In Year 1 you only have to be able to write the expression for Kc based on the
balanced equation for the reversible reaction. You will learn to apply the equilibrium
law quantitatively in Year 2 of your A Level course.
Equilibrium constants and balanced equations
An equilibrium constant always applies to a particular chemical equation and
can be deduced directly from the equation.
There are two common ways of writing the reaction of sulfur dioxide with
oxygen. As a result, there are two forms for the equilibrium constant, which
have different values. So long as the matching equation and equilibrium constant
are used, the predictions based on the equilibrium law are the same.
For:
2SO2(g) + O2(g) ⇋ 2SO3(g) Equation 1
[SO3(g)]2
Kc =
[SO2(g)]2[O2(g)]
But for:
1
SO2(g) + 2O2(g) ⇋ SO3(g) Equation 2
[SO3(g)]
Kc = 1
[SO2(g)][O2(g)] /2
So it is important to write the balanced equation and the equilibrium
constant together.
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Reversing the equation also changes the form of the equilibrium constant
because the concentration terms for the chemicals on the right-hand side of
the equation always appear on the top of the expression for Kc.
So for:
2SO3(g) ⇋ 2SO2(g) + O2(g) Equation 3
[SO2(g)]2[O2(g)]
Kc =
[SO3(g)]2
Test yourself
11 Write the balanced equation and the expression for Kc for these
reversible reactions:
a) hydrogen gas with iodine gas to form hydrogen iodide gas
b) nitrogen monoxide gas with oxygen gas to form nitrogen dioxide
gas
c) nitrogen gas with hydrogen gas to form ammonia gas.
Heterogeneous equilibria
In an equilibrium mixture of sulfur dioxide, oxygen and sulfur trioxide all
Key terms three substances are gases. They are all in the same gaseous phase. This is an
example of a homogeneous equilibrium.
A homogeneous equilibrium is an
equilibrium in which all the substances In many equilibrium systems the substances involved are not all in the
involved are in the same phase. same phase. An example is the equilibrium state involving two solids and
A heterogeneous equilibrium is
a gas formed on heating calcium carbonate in a closed container. In this
an equilibrium system in which the
system there are two solid phases and a gas phase. This is an example of a
substances involved are in more than
heterogeneous equilibrium.
one phase. CaCO3(s) ⇋ CaO(s) + CO2(g)
The concentrations of solids do not appear in the expression for the
equilibrium constant. Pure solids have, in effect, a constant ‘concentration’.
So Kc = [CO2(g)]
The same applies to heterogeneous systems which have a separate pure liquid
phase as one of the reactants or products.
Test yourself
12 Write the expression for Kc for these equilibria:
a) 3Fe(s) + 4H2O(g) ⇋ Fe3O4(s) + 4H2(g)
b) H2(g) + S(l) ⇋ H2S(g)
c) Ag+(aq) + Fe2+(aq) ⇋ Fe3+(aq) + Ag(s)
13 Write the balanced equations for the equilibria to which these
expressions for Kc apply:
[HI(g)]2 [H2(g)][CO2(g)]
a) Kc = b) Kc =
[H2(g)][I2(g)] [H2O(g)][CO(g)]
[Cr2+(aq)]²[Fe2+(aq)]
c) Kc =
[Cr3+(aq)]²
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Qualitative predictions based on the
equilibrium law
The equilibrium law makes it possible to explain qualitatively the effect
of changing the concentration of one of the chemicals in an equilibrium
mixture. This is an alternative approach to applying Le Chatelier’s principle.
An example is the equilibrium in an aqueous solution of bromine (Section Tip
10.4):
In dilute solution, the water is in such
Br2(aq) + H2O(l) ⇋ HOBr(aq) + Br−(aq) + H+(aq) large excess that the value of [H2O(l)]
At equilibrium: is effectively constant. As a result, it
does not appear in the equilibrium law
[HOBr(aq)][Br−(aq)][H+(aq)]
Kc = expression.
[Br2(aq)]
where these are equilibrium concentrations.
Tip
Adding a few drops of alkali neutralises H+(aq) on the right-hand side
of the equation. This reduces the value of [H+(aq)] and briefly upsets the Changing the concentrations does
equilibrium so that for an instant after adding alkali: not alter the value of the equilibrium
constant so long as the temperature
[HOBr(aq)][Br−(aq)][H+(aq)] stays constant.
Kc > [Br2(aq)]
The system restores equilibrium as the forward reaction predominates and
bromine molecules react with water to produce more of the products. There Test yourself
is very soon a new equilibrium. Once again: 14 Calculate the concentration
[HOBr(aq)][Br−(aq)][H+(aq)] of water in water (in
[Br2(aq)] = Kc
mol dm−3) to show that it
but now with new values for the various equilibrium concentrations. is reasonable to regard the
concentration of water as
Chemists sometimes say that adding alkali makes the ‘position of equilibrium
a constant when writing
shift to the right’. The effect is visible because the orange colour of the
the expression for Kc for
bromine molecules fades with the formation of more colourless molecules
equilibria in dilute aqueous
and ions on the right-hand side of the equation. This is as Le Chatelier’s
solution.
principle predicts (Section 10.4). The advantage of using Kc is that it makes
quantitative predictions possible.
10.5 Equilibrium constants 295
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Chapter summary
Chapter 10 Equilibrium I l Equilibrium constants make it possible to
describe the behaviour of equilibrium mixtures
l Many reactions are reversible; the direction of quantitatively.
change can be altered by changing the conditions. l For the reaction: aA + bB ⇋ cC + dD the
l Reversible reactions can reach a state of dynamic
equilibrium law expression is:
equilibrium such that the forward and backward
[C]c[D]d
reactions continue but at equal rates, so that there Kc =
[A]a[B]b
is no overall change to the concentrations of
l The value of Kc is constant at a particular
reactants and products.
l Changing the conditions of concentration, pressure
temperature.
l The value of Kc changes when the temperature
or temperature can disturb a system at equilibrium;
generally, when the conditions change, the system changes.
l In a homogeneous equilibrium, all the substances
responds by trying to counteract the change.
l The position of equilibrium is not affected by
involved are in the same
adding a catalyst. phase – all in the gas phase or in solution, for
l In industrial processes, the chosen conditions are
example.
l In a heterogeneous equilibrium, the substances are
often a compromise between the need to convert
as much of the reactants to product in the reactor in more than one phase.
l In heterogeneous equilibria with a solid in
(equilibrium factors) and the need to produce the
product fast enough (kinetic factors). equilibrium with a gas mixture, or solution, the
concentrations of the solid do not appear in the
expression for Kc.
Exam practice questions
1 Carbon dioxide is dissolved in water under c) Explain why lots of bubbles of gas form
pressure to make sparkling mineral water. In a when a bottle of sparkling mineral water is
bottle of mineral water, there is an equilibrium opened. (2)
between carbon dioxide dissolved in the drink, d) Less than 1% of the dissolved carbon
CO2(aq), and carbon dioxide in the gas above dioxide reacts with water. It forms
the drink, CO2(g). hydrogencarbonate ions:
a) Write an equation to represent the CO2(g) + H2O(l) ⇋ HCO3−(aq) + H+(aq)
equilibrium between carbon dioxide gas Use this equation to explain why carbon
and carbon dioxide in solution. (1) dioxide is much more soluble in sodium
b) Use this example to explain the term hydroxide solution than in water. (2)
‘dynamic equilibrium’. (2)
296
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Haemoglobin is a large molecule in red blood
2 of sulfur dioxide to sulfur trioxide in the
cells that can be represented by the symbol Hb. reactor. (6)
Haemoglobin carries oxygen in the blood from c) A mixture of sulfur dioxide and air is fed into
our lungs to the cells in our bodies: the reactor. Give a reason why equal amounts
Hb(aq) + 4O2(g) ⇋ HbO8(aq) of sulfur dioxide and oxygen are present in
the mixture fed into the reactor. (3)
a) Give a reason why haemoglobin takes up d)* Assess the factors which mean that, in
oxygen as blood passes through the blood practice, the process is carried out at
vessels in the lungs. (2) 700 K and at atmospheric pressure. (6)
b) Give a reason why haemoglobin releases
oxygen as blood passes through the blood Write equations and the expressions for Kc for
6
vessels between cells in muscles.(2) these reversible reactions:
c) Haemoglobin molecules are affected by the a) the reaction of hydrogen with chlorine to
presence of carbon [Link] molecules hold make hydrogen chloride. (3)
onto oxygen less strongly if the carbon dioxide b) the reaction of ammonia with oxygen to
concentration is higher. Explain why this helps form nitrogen monoxide and steam. (3)
the blood deliver oxygen to cells in muscles.(2) c) the decomposition of solid NH4HS to
ammonia gas and hydrogen sulfide gas. (3)
For each of the following equilibria predict the
3
effects, if any, of: 7 Hydrogen is manufactured from methane and
i) raising the pressure steam. In the first stage of the process, methane
ii) raising the temperature. is mixed with a large excess of steam and passed
a) H2(g) + I2(g) ⇋ 2HI(g) (3) through a reactor containing a nickel [Link]
This reaction is slightly exothermic. reversible reaction produces carbon monoxide and
b) NaCl(s) + aq ⇋ NaCl(aq) (3) hydrogen. For this reaction ΔH = +210 kJ mol−1.
The enthalpy change of solution of NaCl is In the second part of the process, the carbon
slightly positive and there is a slight decrease monoxide gas reacts with steam to form carbon
in volume when salt dissolves in water. dioxide and more hydrogen. At 700 K the
c) C(graphite) ⇋ C(diamond) equilibrium constant for this reaction Kc = 5.1.
ΔH = +1.9 kJ mol−1 (4) At 1100 °C the value of Kc = 1.0.
The density of diamond is 3.5 g cm−3. The Finally, the carbon dioxide is removed from the gas
density of graphite is 2.3 g cm−3. mixture using a molecular sieve made of a zeolite.
4 a) Explain why iodine is much more soluble in a) State the conditions that favour the
aqueous potassium iodide solution than in formation of products in the first part
water. (3) of the process. Explain your answer. (6)
b) A solution of iodine in KI(aq) is dark yellow- b) i) Write the equation and expression for
brown. Explain why, on adding sodium Kc for the reaction used in the second
thiosulfate solution, the solution gradually stage of the process.(2)
loses its colour, turning paler and paler ii) Use the two values of Kc to determine
yellow until finally becoming colourless. (5) whether the reaction is exothermic or
endothermic, giving your reasons. (3)
Sulfuric acid is manufactured by the contact
5 iii) Give reasons why the reaction is
process which involves the reversible reaction carried out at 650 K in the presence of
of sulfur dioxide with oxygen from the air an iron catalyst. (2)
to form sulfur trioxide. For this reaction c) Explain why zeolites can be used as
ΔH = −198 kJ mol−1. The reaction takes place molecular sieves. (2)
in the presence of a vanadium(v) oxide catalyst d) Give reasons why the carbon dioxide made
which does not work unless it is hot. in this process is captured and stored. (3)
a) Write a balanced equation for the reaction e) Explain why hydrogen is manufactured on a
involved in the contact process. (1) large scale. (2)
b) State the conditions that theory suggests
should favour the maximum conversion
297
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Equilibrium II
11
11.1 Reversible reactions
and dynamic equilibrium
All chemical reactions tend towards a state of dynamic equilibrium. An
understanding of equilibrium ideas helps to explain changes in the natural
environment, the biochemistry of living things and the conditions used in
the chemical industry to manufacture new products (Figure 11.1).
Tip
The first two sections of this chapter, and parts of Section 11.5, revisit ideas first
introduced in Chapter 10. In Chapter 10 the treatment of equilibrium was qualitative.
Chapter 11 builds on what you already know and shows you how to apply the equilibrium
Figure 11.1 Red blood cells flowing law quantitatively. Some of the ‘Test yourself’ questions are also designed to help revise
through a blood vessel magnified ×3000. ideas from the first year of the A Level course.
The protein haemoglobin has just the
right properties to take up oxygen in the Reversible reactions reach equilibrium when neither the forward change
lungs and release it to cells throughout nor the backward change is complete, but both changes are still going on at
the body. The position of equilibrium of equal rates. They cancel each other out and there is no overall change. This
this reversible process varies with the is dynamic equilibrium (see Chapter 10).
concentration of oxygen. Under given conditions the same equilibrium state can be reached either by
starting with the chemicals on one side of the equation for a reaction or by
starting with the chemicals on the other side. Figures 11.2 and 11.3 illustrate
this for the reversible reaction between hydrogen and iodine:
H2(g) + I2(g) ⇋ 2HI(g)
[H2(g)] = [I2(g)] [HI(g)]
Concentration
Concentration
[HI(g)]
[H2(g)] = [I2(g)]
Time Time
Figure 11.2 Reaching an equilibrium Figure 11.3 Reaching the same equilibrium
state by the reaction of equal amounts of state by decomposing HI(g) under the
hydrogen gas and iodine gas. same conditions as for Figure 11.2.
298 11 Equilibrium II
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Test yourself
1 Which of these statements are true, and which are false, for a
reversible reaction that is at equilibrium?
a) The concentrations of the reactants and products are constant.
b) The concentrations of the reactants and the products are equal.
c) The rate of formation of products is equal to the rate of formation
of reactants.
2 Give examples of dynamic equilibrium involving:
a) a solid and a liquid b) a solid and a solution
c) two solutions d) a chemical change.
3 Describe what is happening to the molecules in the gas mixtures from
time zero to the time at which each mixture reaches equilibrium, as
described by:
a) Figure 11.2 b) Figure 11.3.
Activity
Testing the equilibrium law
The reversible reaction involving hydrogen, iodine and hydrogen Once the tubes had reached equilibrium they were rapidly
iodide has been used to test the equilibrium law experimentally. cooled to stop the reactions. Then the contents of the
In a series of six experiments, samples of the chemicals were tubes were analysed to find the compositions of the
sealed in reaction tubes and then heated at 731 K until the equilibrium mixture. The results for the six tubes are shown
mixtures reached equilibrium. Four of the tubes started with in Table 11.1.
different mixtures of hydrogen and iodine. Two of the tubes
started with just hydrogen iodide.
Table 11.1
Tube Initial concentrations/10 −2 mol dm−3 Equilibrium concentrations/10 −2 mol dm−3
[H2(g)] [I 2(g)] [HI(g)] [H2(g)]eqm [I 2(g)]eqm [HI(g)]eqm
1 2.40 1.38 0 1.14 0.12 2.52
2 2.40 1.68 0 0.92 0.20 2.96
3 2.44 1.98 0 0.77 0.31 3.34
4 2.46 1.76 0 0.92 0.22 3.08
5 0 0 3.04 0.345 0.345 2.35
6 0 0 7.58 0.86 0.86 5.86
1 Write the equation for the reversible reaction to form 4 For each of the tubes, work out the value of:
hydrogen iodide from hydrogen and iodine. [HI(g)]eqm
a)
2 Show that the equilibrium concentration of: [H2(g)]eqm[I2(g)]eqm
a) hydrogen in tube 1 is as expected, given the value of [HI(g)]2eqm
[I2(g)]eqm b) .
[H2(g)]eqm[I2(g)]eqm
b) hydrogen iodide in tube 2 is as expected, given the value
Enter your values in a table and comment on the results.
of [I2(g)]eqm.
5 What is the value of Kc for the reaction of hydrogen with
3 Explain why [H2(g)]eqm = [I2(g)]eqm for tubes 5 and 6.
iodine at 731 K?
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11.2 The equilibrium law
The equilibrium law has been established by experiment. It is a quantitative
law for predicting the amounts of reactants and products when a reversible
reaction reaches a state of dynamic equilibrium.
In general, for a reversible reaction at equilibrium:
aA + bB ⇋ cC + dD
[C]c[D]d
Kc =
[A]a[B]b
This is the form for the equilibrium constant, Kc, when the concentrations
of the reactants and products are measured in moles per cubic decimetre.
[A], [B] and so on are the equilibrium concentrations, sometimes written as
[A]eqm and [B]eqm to make this clear.
The concentrations of the chemicals on the right-hand side of the equation
appear on the top line of the expression. The concentrations of reactants on
the left appear on the bottom line. Each concentration term is raised to the
power of the number in front of its formula in the equation.
Tip Equilibrium constants and balanced equations
An equilibrium constant always applies to a particular chemical equation and
Equilibrium constants are only constant
can be deduced directly from the equation.
at a particular temperature.
There can be different ways of writing the equation for a reversible reaction
The form of the expression for an
at equilibrium. As a result, there are different forms for the equilibrium
equilibrium constant can be deduced from
constant, each with a different value. So long as the matching equation and
the balanced chemical equation (unlike
equilibrium constant are used in any calculation, the predictions based on
rate equations – see Section 16.3).
the equilibrium law are the same.
For example, for this equilibrium involving dinitrogen oxide, oxygen and
nitrogen monoxide:
2N2O(g) + O2(g) ⇋ 4NO(g)
Tip [NO(g)]4
Kc =
[N2O(g)]2[O2]
It is important to write the balanced But for this equilibrium:
equation and the equilibrium constant
together. 1
N2O(g) + 2 O2(g) ⇋ 2NO(g)
[NO(g)]2
Kc = 1
[N2O(g)][O2] 2
Reversing the equation also changes the form of the equilibrium constant
because the concentration terms for the chemicals on the right-hand side of
the equation always appear on the top of the expression for Kc.
So, for this equilibrium:
4NO(g) ⇋ 2N2O(g) + O2(g)
[N2O(g)]2[O2]
Kc =
[NO(g)]4
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In an equilibrium mixture of dinitrogen oxide, oxygen and nitrogen
monoxide, all three substances are gases. They are all in the same gaseous
phase. This is an example of a homogeneous equilibrium.
Test yourself
4 Write the expression for Kc for each equation and state the units of
the equilibrium constant.
a) N2(g) + O2(g) ⇋ 2NO(g)
1 1
b) N2(g) + O2(g) ⇋ NO(g)
2 2
c) N2(g) + 3H2(g) ⇋ 2NH3(g)
d) 2NH3(g) ⇋ N2(g) + 3H2(g)
5 Explain what is meant by the term ‘homogeneous’.
Finding equilibrium constants by experiment
The strategy for determining the value of an equilibrium constant involves
three main steps:
Step 1: Mix measured quantities of reactants and/or products. Then allow
the mixture to reach equilibrium under steady conditions.
Step 2: Analyse the mixture to find the equilibrium concentration of one of
the chemicals at equilibrium.
Step 3: Use the equation for the reaction and the information from steps 1
and 2 to work out the values for the equilibrium concentrations of
all the atoms, molecules or ions. Then substitute these values into
the expression for Kc.
The challenge when investigating reactions at equilibrium is to find ways
to measure equilibrium concentrations without upsetting the equilibrium.
Many methods of analysis use up the chemical being analysed. Analytical
methods of this kind are generally unsuitable because, as Le Chatelier’s
principle shows (Section 10.4), the position of equilibrium shifts whenever
one of the reactants or products is removed from the equilibrium mixture.
There are two main ways to measure equilibrium concentrations. One way
is to find a method for ‘freezing’ the reaction and thus slowing down the rate
so much that it is possible to measure one of the equilibrium concentrations
by titration. The most obvious way to slow down the rate and ‘freeze’ the
equilibrium is by cooling. Other possibilities are to dilute the equilibrium
mixture or to remove a catalyst.
The second way of measuring equilibrium concentrations is to use an instrument
that responds to a property of the mixture that varies with concentration.
Well-established methods for doing this include measuring the pH with a pH
meter or measuring the intensity of a colour with a colorimeter.
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Measuring Kc for a reaction used to
make an ester
The equilibrium between ethanoic acid, ethanol, ethyl ethanoate and water
(Section 17.3.3) is one of the few reaction systems (other than acid–base
equilibria) that lends itself to study in an advanced chemistry course:
CH3COOH(l) + C2H5OH(l) ⇋ CH3COOC2H5(l) + H2O(l)
This esterification reaction is very, very slow at room temperature in the
absence of a catalyst. In the presence of an acid catalyst the reaction mixture
reaches equilibrium in about 48 hours.
Tip
Diluting the equilibrium mixture means that the reaction is slow enough
The equilibrium mixtures for the reaction to find the equilibrium concentration of ethanoic acid by titration without
of ethanoic acid with ethanol includes the position of equilibrium shifting perceptibly in the time taken for the
water from the dilute hydrochloric acid titration.
as well as any added water.
The procedure follows these three steps.
Step 1: Mix measured quantities of chemicals and allow the mixture to
reach equilibrium.
Precisely measured quantities of the chemicals are added to sample tubes.
The masses of the components of the mixture can be found by weighing.
The sample tubes are tightly stoppered to avoid loss by evaporation and set
aside at constant temperature for 48 hours.
Some of the tubes at first contain just ethanol, ethanoic acid and hydrochloric
acid. Others start with only ethyl ethanoate, water and hydrochloric acid.
Working in this way shows that it is possible to reach equilibrium from either
side of the equation.
Step 2: Analyse the mixture to find the equilibrium concentration of the
acid.
Each equilibrium mixture is transferred quantitatively to a flask and diluted
with water. Titration with a standard solution of sodium hydroxide determines
the total amount of acid in the sample at equilibrium: both hydrochloric acid
and ethanoic acid.
Step 3: Use the equation for the reaction and the information from steps 1
and 2 to work out the values for all the equilibrium concentrations.
Tip
Some of the sodium hydroxide used in the titration reacts with the hydrochloric
In this equilibrium system water is acid. Since the amount of HCl(aq) does not change as the reactants reach
present in relatively small amounts as a equilibrium, it is possible to work out how much of the titre was used to
reactant and not just as a solvent. The neutralise it, knowing how much HCl(aq) was added at the start. The
concentration of water is a variable and remainder of the alkali added during the titration reacts with ethanoic acid.
appears in the expression for Kc. Hence the amount of ethanoic acid at equilibrium can be calculated. The
other equilibrium concentrations can be found given the starting amounts of
chemicals present and the equation for the reaction.
302 11 Equilibrium II
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Activity
Analysing the results of an experiment to measure Kc
This activity is based on two sets of results from an experiment 1 Explain why the amount of hydrochloric acid in the mixture
carried out by students to measure the value of Kc for the did not change as the mixture of reactants reached
equilibrium between ethanoic acid, ethanol, ethyl ethanoate equilibrium.
and water. 2 Confirm, by calculation, that the mass of water in the
hydrochloric acid added to the reaction mixture was 4.81 g.
The approach to the calculation can be illustrated from the
3 Show that initially 0.322 mol of water was present.
results for one of the samples which initially contained only the
4 Use the titration result to confirm that the total acid (as
ester but no added ethanoic acid or ethanol. The sample was
H+ ions) in the equilibrium mixture was 0.0392 mol. Hence
initially made up of the following:
show that the amount of ethanoic acid at equilibrium was
● ethyl ethanoate (ester), 3.64 g (0.0413 mol) 0.0292 mol.
● water, 0.99 g 5 Explain why the answer to Question 3 shows that the
● 5.0 cm3 of 2.00 mol dm −3 HCl(aq) containing 0.010 mol HCl. equilibrium amounts of ethanol, ethyl ethanoate and water
were as shown in Table 11.2.
Mass of added hydrochloric acid = 5.17 g
6 Write the expression for Kc for the equilibrium reaction; then
This contained 4.81 g water use the values in the table to calculate the value of Kc.
7 Explain why it is not necessary to know the volume of the
So, the total mass of water at the start = 0.99 g + 4.81 g = 5.80 g
reaction mixture to calculate Kc and why the equilibrium
= 0.322 mol H2O
constant has no units.
Titration of the equilibrium mixture found that 39.20 cm3 of
8 Titration of the equilibrium mixture in another sample tube
1.00 mol dm−3 sodium hydroxide neutralised the total acid
from the experiment required 41.30 cm3 of 1.00 mol dm−3
present. Both HCl and CH3COOH react 1 : 1 with NaOH, so
NaOH(aq). Calculate a value for Kc given that, at the start,
this shows that the total amount of acid at equilibrium
the tube contained 4.51 g ethyl ethanoate with 5.0 cm3 of
= 0.0392 mol
2.00 mol dm−3 HCl(aq) but no added water other than the
So, taking away the amount of hydrochloric acid added at the water in the dilute acid.
start, this shows that the amount of ethanoic acid at equilibrium 9 Compare and comment on the two values for Kc.
= 0.0292 mol. The results are summarised in Table 11.2.
Table 11.2
Reaction CH3COOH(l) + C2H5OH(l) ⇋ CH3COOC2H5(l) + H2O(l)
Initial amount/mol 0.000 0.000 0.0413 0.322
Measured amount at 0.0292
equilibrium/mol
Other equilibrium amounts 0.0292 0.0121 0.293
calculated from the starting
amounts and the equation/mol
Calculating equilibrium constants
Experimental results can be used to calculate the values for equilibrium
constants along similar lines to the determination of the value of Kc for
the formation of ethyl ethanoate. Once a value of Kc is known, it can be
used to calculate the concentrations of the reactants and products in specific
equilibrium mixture.
11.2 The equilibrium law 303
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Example
1.00 mol of NOCl gas was enclosed in a 0.50 dm3 flask at 298 K. The
amount of NO gas in the flask at equilibrium was found to be 0.33 mol.
Calculate the value of Kc for this reaction:
2NOCl(g) ⇋ 2NO(g) + Cl2(g)
Note on the method
Write down the equation. Underneath write first the initial amounts, then
write the given amount at equilibrium. Next calculate the equilibrium
amounts not given, taking into account the numbers of moles of each
substance shown in the equation for the reaction.
Calculate the equilibrium concentrations given the volume of the solution.
Substitute the values and units in the expression for Kc.
Answer
Equation: 2NOCl(g) ⇋ 2NO(g) + Cl2(g)
Initial amounts/mol: 1.00 0 0
Equilibrium amount 0.33
given/mol:
Equilibrium amounts (1.00 − 0.33) (0.33 ÷ 2)
calculated/mol:
Equilibrium 0.67 ÷ 0.5 0.33 ÷ 0.5 (0.33 ÷ 2) ÷
concentrations/mol dm−3: = 1.34 = 0.66 0.5 = 0.33
[NO(g)]2[Cl2(g)] (0.66 mol dm−3)2(0.33 mol dm−3)
Kc = =
[NOCl(g)]2 (1.34 mol dm−3)2
Hence Kc = 0.080 mol dm−3
Test yourself
6 On mixing 1.68 mol PCl5(g) with 0.36 mol PCl3(g) in a 2.0 dm3 container,
and allowing the mixture to reach equilibrium, the amount of PCl5 in the
equilibrium mixture was 1.44 mol. Calculate Kc for the reaction:
PCl5(g) ⇋ PCl3(g) + Cl2(g)
7 Consider the equilibrium between sulfur dioxide, oxygen and sulfur trioxide:
2SO2(g) + O2(g) ⇋ 2SO3(g) Kc = 1.6 × 106 dm3 mol –1
a) Show that the units for the equilibrium constant, Kc, for the
equation are dm3 mol –1.
b) What is the value of Kc for this equation at the same temperature?
1
SO2(g) + O2(g) ⇋ SO3(g)
2
c) What is the value of Kc for this equation at the same temperature?
2SO3(g) ⇋ 2SO2(g) + O2(g)
8 Kc = 170 dm3 mol –1 at 298 K for the equilibrium system:
2NO2(g) ⇋ N2O4(g). If a 5 dm3 flask contains 1.0 × 10 –3 mol of NO2
and 7.5 × 10 –4 mol N2O4, is the system at equilibrium? Is there any
tendency for the concentration of NO2 to change and, if so, does it
tend to increase or decrease?
304 11 Equilibrium II
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Heterogeneous equilibria
In many equilibrium systems the substances involved are not all in the same
phase. An example is the equilibrium state formed when steam is heated with
coke (carbon) in a closed container. This is an example of a heterogeneous
equilibrium.
H2O(g) + C(s) ⇋ H2(g) + CO(g)
The concentrations of solids do not appear in the expression for the
equilibrium constant. Pure solids have, in effect, a constant ‘concentration’
so their values are incorporated into the value for the equilibrium constant.
Hence for the reaction of steam with carbon:
[H2(g)][CO(g)]
Kc =
[H2O(g)]
The same applies to heterogeneous systems which have a separate liquid
phase as one of the reactants or products.
Another example is the equilibrium state between solid calcium carbonate
and a dilute solution containing dissolved carbon dioxide and calcium
hydrogencarbonate.
CaCO3(s) + CO2(aq) + H2O(l) ⇋ Ca 2+(aq) + 2HCO3 –(aq)
This example illustrates another general rule. The Kc expression for dilute Tip
solutions does not include a concentration term for water. There is so much Remember that in dilute solution, the
water present that its concentration is effectively constant. water is in such large excess that the
So the expression for the equilibrium constant becomes: value of [H2O(l)] is effectively constant.
As a result, it does not appear in the
[Ca 2+(aq)][HCO3−(aq)]2 equilibrium law expression.
Kc =
[CO2(aq)]
Test yourself
9 Explain what is meant by the term ‘heterogeneous’.
10 Explain why the bottle shown in Figure 11.4 contained a
heterogeneous equilibrium before the top was unscrewed.
11 Write the expression for Kc for each equation and state the units of
the equilibrium constant.
a) NH4HS(s) ⇋ H2S(g) + NH3(g)
b) Pb2+(aq) + Sn(s) ⇋ Pb(s) + Sn2+(aq)
c) BiCl3(aq) + H2O(l) ⇋ BiOCl(s) + 2HCl(aq)
12 In the natural world, where is it possible to find solid calcium
carbonate and a dilute solution containing dissolved carbon
dioxide and calcium hydrogencarbonate close to a state of dynamic
equilibrium?
13 Calculate the concentration of water in water (in mol dm –3) to show
that it is reasonable to regard the concentration of water as a
constant when writing the expression for Kc for equilibria in dilute
Figure 11.4 Pouring fizzy water from a
aqueous solution.
glass bottle.
11.2 The equilibrium law 305
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The extent and direction of change
Chemists can use equilibrium constants to predict quantitatively the direction
and extent of chemical change.
Table 11.3 shows that if the value of an equilibrium constant is large, then
the position of equilibrium is over to the right-hand side of the equation.
Conversely, if the value of an equilibrium constant is small, then the position
of equilibrium is over to the left-hand side of the equation. If the value of
Kc is close to 1, then there are significant quantities of both reactants and
products present at equilibrium.
Table 11.3 Relating the value of Kc to the direction and extent of change.
Direction and extent of change Value of Kc
Reaction does not go Kc < 1 × 10–10
Reaction reaches an equilibrium in which the reactants Kc ≈ 0.01
predominate
Roughly equal amounts of reactants and products at equilibrium Kc = 1
Reaction reaches an equilibrium in which the products Kc ≈ 100
predominate
Reaction goes to completion Kc > 1 × 1010
It is very important to keep in mind that the equilibrium constant gives
no information about the time it takes for a reaction mixture to reach
equilibrium. The system may reach equilibrium rapidly or slowly. The value
of Kc for the reaction of hydrogen with chlorine to make hydrogen chloride,
for example, is about 1 × 1031 at room temperature, but in the absence of a
catalyst, ultraviolet light or a flame there is no reaction.
Test yourself
14 What can you conclude about the direction and extent of change in
each of these examples?
a) Zn(s) + Cu2+(aq) ⇋ Zn2+(aq) + Cu(s) Kc = 1 × 1037 at 298 K
b) 2HBr(g) ⇋ H2(g) + Br2(g) Kc = 1 × 10 –10 at 298 K
c) N2(g) + 3H2(g) ⇋ 2NH3(g) Kc = 2.2 at 623 K
15 In general, if the equilibrium constant for a forward reaction is large,
Key term what is the size of the equilibrium constant for the reverse of the
same reaction?
Pressure is defined as force per unit
area. The SI unit of pressure is the
pascal (Pa), which is a pressure of one 11.3 Gaseous equilibria
newton per square metre (1 N m –2). The
Many important industrial processes involve reversible reactions between
pascal is a very small unit, so pressures
gases. Applying the equilibrium law to these reactions helps to determine
are often quoted in kilopascals, kPa.
the optimal conditions for manufacturing chemicals. When it comes to
Standard atmospheric pressure is equal
gas reactions it is often easier to measure the pressure rather than the
to 101.3 kPa.
concentration and to use a modified form of the equilibrium law.
306 11 Equilibrium II
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Gas mixtures and partial pressures Key term
In any mixture of gases the total pressure of the mixture can be ‘shared out’
between the gases. The contribution each gas makes to the total pressure is The partial pressure of a gas is a
its partial pressure. It is possible to calculate a partial pressure for each gas measure of its concentration in a
in the mixture. In a mixture of gases A, B and C, the sum of the three partial mixture of gases. It is the pressure that
pressures equals the total pressure. the gas would exert if it were the only
gas present in the container.
pA + pB + pC = ptotal
Partial pressures are a useful alternative to concentrations when studying
mixtures of gases and gas reactions.
In gas mixtures it is the amounts (in moles) of gas molecules that matter and
not the chemical nature of the molecules. As a result the molar volume of
a gas is the same for all gases under the same conditions of temperature and
pressure. The gas laws show that this means that the total pressure is shared
between the gases simply according to their mole fractions in the mixture.
In a mixture of nA moles of A with nB moles of B and nC moles of C, the
total amount in moles is (nA + nB + nC). The mole fractions (symbol X) are
given by the following:
n nB nC
X A = n + nA + n X B = XC = n + n + n
A B C nA + nB + n C A B C
So, the mole fraction of A is the fraction of the total number of molecules
which are molecules of A.
The sum of all the mole fractions is 1, so X A + X B + XC = 1.
On this basis the partial pressures of three gases A, B and C in a gas mixture
with total pressure p are:
pA = X A p, pB = X Bp and p C = XC p
The partial pressure for each gas is the pressure it would exert if it was the
only gas in the container under the same conditions. The partial pressure
of a gas is proportional to the concentration of the gas in the mixture. This
makes it possible to work in partial pressures when applying the equilibrium
law to gas reactions.
Test yourself
16 A 20 mol sample of a gas mixture contains 15.6 mol nitrogen and
4.4 mol oxygen.
a) Calculate the mole fractions of the two gases in the mixture.
b) Calculate the partial pressures of each of the two gases if the
total pressure is 1 atm.
17 A mixture of 22 g propane gas and 11 g 2-methylpropane gas is
compressed into an aerosol can to give a total pressure of 1.5 atm.
a) What are the mole fractions of the two gases?
b) Calculate the partial pressures of each of the two gases.
11.3 Gaseous equilibria 307
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11.4 Kp
Kp is the symbol for the equilibrium constant for an equilibrium involving
Tip gases when the concentrations are measured by partial pressures. The rules
Do not use square brackets when for writing equilibrium expressions are the same for Kp as for Kc, with partial
writing Kp expressions. In the context pressures replacing concentrations. This is shown in Table 11.4.
of the equilibrium law, square brackets
Table 11.4 Examples of equilibrium expressions for Kp. Note that when writing an
signify concentrations in mol dm−3.
expression for Kp for a heterogeneous reaction the same rules apply as for Kc. The
expression does not include terms for any separate pure solid phases.
Equilibrium Kp Units of Kp
H2(g) + I2(g) ⇋ 2HI(g) (pHI )2 no units
Kp = p × p
H2 I2
N2(g) + 3H2(g) ⇋ 2NH3(g) (pNH3)2 atm–2
Kp =
pN2 × (pH2)3
N2O4(g) ⇋ 2NO2(g) (pNO )2 atm
Kp = p 2
N2O4
HCl(g) + LiH(s) ⇋ H2(g) + LiCl(s) pH no units
Kp = p 2
HCl
Tip Example
Changing the total pressure or the An experimental study of the equilibrium between N2(g), H2(g) and NH3(g)
composition of the gas mixture has no found that one equilibrium mixture contained 2.15 mol of N2(g), 6.75 mol
effect on the value of Kp as long as the of H2(g) and 1.41 mol of NH3(g) at a total pressure 10.0 atm. Calculate
temperature stays constant. the value for Kp under the conditions that the measurements were taken.
Notes on the method
First work out the mole fractions of the gases.
Multiply the total pressure by the mole fractions to get the partial pressures.
Check that the sum of the partial pressures equals the total pressure.
Finally substitute in the expression for Kp and give the units.
Answer
Total number of moles = 2.15 mol + 6.75 mol + 1.41 mol = 10.31 mol
2.15 mol
Mole fraction of N2(g) = = 0.208
10.31 mol
6.75 mol = 0.655
Mole fraction of H2(g) =
10.31 mol
1.41 mol
Mole fraction of NH3(g) = = 0.137
10.31 mol
Partial pressure of N2(g) = 0.208 × 10 atm = 2.08 atm
Partial pressure of H2(g) = 0.655 × 10 atm = 6.55 atm
Partial pressure of NH3(g) = 0.137 × 10 atm = 1.37 atm
Check: the total pressure = 2.08 atm + 6.55 atm + 1.37 atm = 10.0 atm
For the equilibrium: N2(g) + 3H2(g) ⇋ 2NH3(g)
(pNH )2
3
Kp =
pN × (pH2)3
2
(1.37)2
= = 3.21 × 10 –3 atm –2
2.08 × (6.55)3
308 11 Equilibrium II
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Test yourself
18 Write the expression for Kp for each equilibrium. Give the units with
pressures measured in atmospheres, atm.
a) 2SO2(g) + O2(g) ⇋ 2SO3(g)
b) 4NH3(g) + 3O2(g) ⇋ 2N2(g) + 6H2O(g)
c) CaCO3(s) ⇋ CaO(s) + CO2(g)
19 Calculate Kp at 330 K for this equilibrium mixture:
N2O4(g) ⇋ 2NO2(g) Tip
At this temperature, a sample of the gas mixture at 1.20 atm
Equilibrium constants for gaseous
pressure consists of 8.1 mol N2O4(g) and 3.8 mol NO2(g).
equilibria are not always given in the
20 Calculate Kp for this reversible reaction at 1000 K: form of Kp. Kc is sometimes used
C2H6(g) ⇋ C2H4(g) + H2(g) instead and there is no objection to
At this temperature, starting with just 5 mol ethane yields an this. However, it is generally easier to
equilibrium mixture containing 1.8 mol ethene at 1.80 atm pressure. measure gas pressures.
11.5 Factors affecting systems
at equilibrium
The chemical industry is being reinvented to make it more sustainable. It is
no longer acceptable to operate processes that make inefficient use of valuable
resources. Chemists and chemical engineers are devising new methods by
applying the theoretical ideas and models that explain how fast reactions go,
in which direction and how far (Figure 11.5).
Figure 11.5 This vast catalytic cracker in Germany is used to make ethene from natural
gas or oil. Controlling chemical reactions carried out on such a large scale requires
precise application of chemical principles.
11.5 Factors affecting systems at equilibrium 309
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The effect of changing concentrations on
systems at equilibrium
The equilibrium law makes it possible to explain the effect of changing the
concentration of one of the chemicals in an equilibrium mixture.
An example is the equilibrium in solution involving chromate(vi) and
dichromate(vi) ions in water (Figure 11.6):
2CrO42−(aq) + 2H+(aq) ⇋ Cr2O72−(aq) + H2O(l)
yellow orange
[Cr2O72−(aq)]
At equilibrium: Kc =
[CrO42−(aq)]2 [H+(aq)]2
where these are equilibrium concentrations.
Adding a few drops of concentrated acid increases the concentration of
H+(aq) on the left-hand side of the equation.
This briefly upsets the equilibrium. For an instant after adding acid:
Figure 11.6 On the left, a yellow solution [Cr2O72−(aq)]
of chromate(vi) ions in water. On the right, < Kc
the solution has turned orange as more [CrO42−(aq)]2 [H+(aq)]2
dichromate(vi) ions form after adding a The system restores equilibrium as chromate(vi) ions react with hydrogen
few drops of strong acid. ions to produce more of the products. There is very soon a new equilibrium.
Once again:
[Cr2O72−(aq)]
= Kc
[CrO42−(aq)]2 [H+(aq)]2
but now with new values for the various equilibrium concentrations.
Chemists sometimes say that adding acid makes the ‘position of equilibrium
Tip shift to the right’. The effect is visible because the yellow colour of the
Changing the concentrations does chromate(vi) ions turns to the orange colour of dichromate(vi) ions. This
not alter the value of the equilibrium is as Le Chatelier’s principle predicts. The advantage of using Kc is that it
constant so long as the temperature makes quantitative predictions possible.
stays constant. Remember that in dilute
solutions [H2O(l)] is constant, so it
does not appear in the equilibrium law Test yourself
expression. 21 Describe and explain the effect of adding alkali to a solution of
dichromate(vi) ions.
22 a)
Use the equilibrium law to predict and explain the effect of adding
pure ethanol to an equilibrium mixture of ethanoic acid, ethanol,
ethyl ethanoate and water:
CH3COOH(l) + C2H5OH(l) ⇋ CH3COOC2H5(l) + H2O(l)
b) Show that your prediction is consistent with Le Chatelier’s
principle.
310 11 Equilibrium II
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The effects of pressure changes on systems at
equilibrium
Changes of pressure are not generally significant for equilibria that involve
only solids or liquids, but they have a marked effect on gaseous equilibria.
Lowering the pressure on a system at equilibrium favours the direction of
change that produces more molecules, while increasing the pressure favours
the change that produces fewer molecules. This is what Le Chatelier’s
principle predicts.
low pressure
fewer molecules ⇋ more molecules
high pressure
The effects of increasing or decreasing the total pressure of a gas mixture at
equilibrium can be predicted quantitatively with the help of the equilibrium
law. Take the example of the reaction used to make ammonia in the Haber
process:
N2(g) + 3H2(g) ⇋ 2NH3(g)
The equilibrium law expression in terms of partial pressures is:
(pNH3)2
Kp =
pN × (pH )3
2 2
Suppose that the equilibrium partial pressures of nitrogen, hydrogen
and ammonia are a atm, b atm and c atm, respectively. Substituting in the
expression for Kp gives:
2
Kp = c 3 atm−2
ab
Now suppose that the total pressure is suddenly doubled. At that instant all the
partial pressures double so that pN2 = 2a atm, pH2 = 2b atm and pNH3 = 2c atm. Tip
Substituting these values in the ‘equilibrium constant ratio’ gives: Changing pressure does not alter the
(pNH3)2 (2c)2 2 value of the equilibrium constant Kp so
3 = 3 = 14 × c 3 atm−2 long as the temperature stays constant.
pN × (pH )
2 2
2a × (2b) ab
So at that moment the ‘equilibrium constant ratio’ is one-quarter of the value
of Kp. The system is not at equilibrium. In order to restore equilibrium, some
of the nitrogen and hydrogen must react to decrease their partial pressures
and form more ammonia to increase its partial pressure. This happens until
the values are such that the ‘equilibrium constant ratio’ again equals Kp. In
other words, increasing the pressure causes the equilibrium to shift to the
right. In this way the equilibrium law makes it possible to predict not only
the direction, but also the extent of the shift (Table 11.5).
Table 11.5
Total pressure/atm 10 50 100 200
Percentage by volume of ammonia
1.2 5.6 10.6 18.3
at equilibrium at 773 K
11.5 Factors affecting systems at equilibrium 311
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Test yourself
23 Predict the effect of increasing the pressure on these systems at
equilibrium:
a) 2SO2(g) + O2(g) ⇋ 2SO3(g)
b) CH4(g) + H2O(g) ⇋ CO(g) + 3H2(g)
c) N2(g) + O2(g) ⇋ 2NO(g)
24 For the reaction N2O4(g) ⇋ 2NO2(g), the value of Kp is 0.11 atm at
298 K. Is a mixture containing N2O4(g) with a partial pressure of
2.4 atm and NO2(g) with a partial pressure of 1.2 atm at equilibrium at
298 K? If not, which gas tends to increase its partial pressure?
25 Use the expression for Kp to predict and explain the effect of the
following changes on an equilibrium mixture of hydrogen, carbon
monoxide and methanol:
100 2H2(g) + CO(g) ⇋ CH3OH(g)
a) adding more hydrogen to the gas mixture at constant total pressure
b) compressing the mixture to increase the total pressure
Percentage conversion to SO3
80
c) adding an inert gas such as argon while keeping the total
60
pressure constant.
40 The effects of temperature changes on systems
at equilibrium
20 Le Chatelier’s principle predicts that raising the temperature makes the
equilibrium shift in the direction which is endothermic. For example, for
0
the reaction which produces sulfur trioxide during the manufacture of
600 700 800 900 sulfuric acid, raising the temperature lowers the percentage of sulfur trioxide
Temperature/K at equilibrium.
Figure 11.7 The effect of raising the 2SO2(g) + O2(g) ⇋ 2SO3(g) ΔH = −98 kJ mol–1
temperature on the equilibrium between
The equilibrium shifts to the left as the temperature rises because this is the
SO2, O2 and SO3.
direction in which the reaction is endothermic (Figure 11.7).
Table 11.6 Values of Kp at four Tip
temperatures for the equilibrium
If ΔH for the forward reaction is negative, then ΔH for the reverse reaction has the
2SO2(g) + O2(g) ⇋ 2SO3(g).
same magnitude but the opposite sign. So if the forward reaction is exothermic, then
Temperature/K Kp/atm –1 the reverse reaction is endothermic.
298 4.0 × 1024
500 2.5 × 1010
The reason that temperature changes cause a shift in the position of
700 3.0 × 104
equilibrium is that the value of the equilibrium constant changes. This is
1100 1.3 × 10–1 illustrated by the values in Table 11.6.
312 11 Equilibrium II
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Tip
The value of an equilibrium constant for an exothermic reaction becomes smaller
as the temperature rises. The value of the equilibrium constant for an endothermic
reaction rises as the temperature rises.
Test yourself
26 Show that graph in Figure 11.7 and the values in Table 11.6
are consistent with predictions for the equilibrium based on
Le Chatelier’s principle.
27 The value of Kp for the equilibrium N2O4(g) ⇋ 2NO2(g) is 4.79 atm at
400 K and 347 atm at 500 K.
a) What is the effect of raising the temperature on the position of
equilibrium?
b) How can your answer to (a) account for the appearance of the gas
mixtures in Figure 11.8?
c) What is the sign of ΔH for the reaction? Figure 11.8 Sealed tubes containing
28 For the reaction between hydrogen and iodine to form hydrogen equilibrium mixtures of NO2(g) which
iodide, the value of Kp is 794 at 298 K but 54 at 700 K. What can you is orange-brown and N2O4(g) which is
deduce from this information? colourless. The tube on the left is in hot
water and the tube on the right in ice.
The effects of catalysts on systems at equilibrium
Catalysts speed up reactions but are not used up as they do so (Figure 11.9).
It is important to note that while a catalyst speeds up the rate at which a
reaction gets to an equilibrium state, it has no effect on the final position of
equilibrium. In other words, a catalyst provides a faster route to the same
equilibrium state. The alternative route with a catalyst has a lower activation
energy but speeds up the forward and back reactions to the same extent, so
that the dynamic equilibrium is unchanged.
Figure 11.9 Crystals of palladium seen
under an electron microscope. Palladium is
a rare and precious metal which is used to
catalyse hydrogenation reactions.
11.5 Factors affecting systems at equilibrium 313
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Chapter summary
Chapter 11 Equilibrium II l Kp values for heterogeneous equilibria do not
include terms for any separate solid phases.
l Equilibrium constants make it possible to l Changes to the concentration, or pressure, of an
describe the behaviour of equilibrium mixtures equilibrium mixture do not change the value of Kc
quantitatively. or Kp so long as the temperature stays constant.
l For the reaction: aA + bB ⇋ cC + dD the
l Equilibrium constant expressions can be used to
equilibrium law expression is: predict the effects of changing the concentrations,
[C]c[D]d or partial pressures, of substances in equilibrium
Kc = mixtures.
[A]a[B]b l The reason that temperature changes cause a shift
l In dilute aqueous solutions, the concentration of in the position of equilibrium is that the value of
water is so large that it is effectively constant, so the the equilibrium constant changes.
term [H2O(l)] does not appear in the expression l The value of an equilibrium constant becomes
for Kc. smaller for an exothermic reaction as the
l In a mixture of gases, the total pressure is the sum temperature rises.
of the partial pressures of the gases. The partial l The value of an equilibrium constant becomes
pressure of a gas in a mixture is calculated by larger for an endothermic reaction as the
multiplying the total pressure by its mole fraction. temperature rises.
l K p is the symbol for the equilibrium constant for an l The value of an equilibrium constant is unchanged
equilibrium involving gases with partial pressures in the presence of a catalyst. A catalyst speeds up
replacing concentrations. For the homogeneous the rate at which a reaction mixture reaches an
equilibrium: equilibrium state, but it has no effect on the final
rR(g) + sS(g) ⇋ tT(g) + uU(g) the equilibrium law position of equilibrium.
expression is:
(pT )t × (pU)u
Kp =
(pR)r × (pS)s
314 11 Equilibrium II
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Exam practice questions
1 At 298 K the value of Kc for the following b) i) Calculate the mole fraction of each of
equilibrium is 1 × 1010: the three gases in the mixture at 10 atm
and 650 K. (3)
Sn2+(aq) + 2Fe3+(aq) ⇋ Sn4+(aq) + 2Fe2+(aq)
ii) Calculate the partial pressures of the
a) i) Write the expression for Kc. (2) three gases. (2)
ii) Give the units of Kc for this reaction. iii) Calculate a value for Kp and give the
Explain your answer. (2) units. (3)
b) Calculate the value of Kc for:
4 Evaluate each of the following statements by
i) Sn4+(aq) + 2Fe2+(aq) identifying what is wrong with them, and the
⇋ Sn2+(aq) + 2Fe3+(aq) (2) extent to which there is truth in them.
ii) 21 Sn2+(aq) + Fe3+(aq) a) Once a reaction mixture reaches
⇋ 21 Sn4+(aq) + Fe2+(aq). (2) equilibrium, there is no further reaction. (3)
b) Adding more of one of the reactants to
2 a) A flask contains an equilibrium mixture of
an equilibrium mixture increases the
hydrogen gas (0.010 mol dm–3), iodine gas
yield of products because the value of the
(0.010 mol dm–3) and hydrogen iodide gas
equilibrium constant increases. (3)
(0.070 mol dm–3) at a constant temperature.
c) Adding a catalyst to make a reaction go
Calculate Kc for the reaction of hydrogen
faster can increase the amount of product at
with iodine to form hydrogen iodide. (3)
equilibrium. (3)
b) Enough hydrogen is added to the
d) Raising the temperature to make a reaction
mixture in (a) to suddenly double the
go faster can increase the amount of
hydrogen concentration in the flask
product at equilibrium. (3)
to 0.020 mol dm−3. After a while the
e) Adding a catalyst can mean that a reaction
mixture settles down with a new iodine
that is only feasible at a high temperature
concentration of 0.0070 mol dm−3 at the
becomes feasible at a much lower
same temperature as before.
temperature. (3)
i) Calculate the new concentrations of
hydrogen and hydrogen iodide. (2) 5 A solution of ammonia in water was shaken
ii) Show that the new mixture is at with an equal volume of an organic solvent
equilibrium.(2) until the system reached equilibrium with the
c) Explain the effect of a sudden doubling of ammonia distributed between the two solvents.
the hydrogen concentration on the position In a series of titrations, 10 cm3 of the aqueous
of equilibrium. (2) layer was neutralised by an average of 17.0 cm3
of 0.50 mol dm–3 hydrochloric acid. In a second
3 Nitrogen and hydrogen react to form ammonia
series of titrations, 10 cm3 of the organic layer
when heated under pressure in the presence of
was neutralised by 6.0 cm3 of 0.010 mol dm–3
a catalyst.
hydrochloric acid.
N2(g) + 3H2(g) ⇋ 2NH3(g) a) Calculate the concentration of the ammonia
ΔH = −92 kJ mol–1 in the aqueous layer at equilibrium. (2)
b) Calculate the concentration of the ammonia
Analysis of an equilibrium mixture of the
in the organic solvent at equilibrium. (2)
gases at 10 atmospheres and 650 K found that
c) Determine the value of Kc for the
it contained 1.41 mol NH3, 6.75 mol H2 and
equilibrium:
2.15 mol N2.
a) Explain, in this context, the term ‘dynamic NH3(org) ⇋ NH3(aq). (2)
equilibrium’.(2)
315
Exam practice questions
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6 At 473 K, the value of Kc for the b) Predict, qualitatively, the conditions
decomposition of PCl5 is 8 × 10–3 mol dm–3. which favour a high yield of NO in the
equilibrium mixture. (3)
PCl5(g) ⇋ PCl3(g) + Cl2(g) c) The industrial process typically runs at 1175 K
ΔH = +124 kJ mol–1 and a pressure of about 7 atm with a mixture
a) Write the expression for Kc for the of 10% ammonia and 90% air. Comment
reaction.(1) on the similarities and differences between
b) Calculate the value of Kc for the reverse these conditions and the conditions that you
reaction at 473 K and give its units. (2) predicted in (b). (3)
c) A sample of pure PCl5 is heated to 473 K d) Explain the advantage of using a
in a vessel containing no other chemicals. At heterogeneous catalyst in this process. (1)
equilibrium the concentration of PCl5 is e) Explain why the gas mixture leaving
5 × 10–2 mol dm–3. Calculate the equilibrium the reactor does not contain as high a
concentrations of PCl3 and Cl2. (3) percentage of NO as predicted by the
d) Explain how the concentrations of PCl5, equilibrium law. (2)
PCl3 and Cl2 change in the equilibrium f) The hot gas mixture leaving the reactor has to
mixture if: be cooled before the next step. Describe how
i) more PCl5 is added (2) this might be done in a way that improves the
ii) the pressure is increased (2) overall energy efficiency of the process.(2)
iii) the temperature is increased. (2)
e) Explain the effect on the value of Kc if: At 488 K, for this equilibrium:
9
i) more PCl5 is added (1) COCl2(g) ⇋ CO(g) + Cl2(g)
ii) the pressure is increased (1)
iii) the temperature is increased. (2) Kp = 0.2 Pa. The fraction of COCl2 that splits
up in this way can be represented as the degree
7 Hydrogen is made from natural gas by partial of dissociation, α.
oxidation with steam. This involves the a) Determine the relationship between Kp, α,
following reaction: and the total pressure P. (6)
CH4 (g) + H2O(g) ⇋ CO(g) + 3H2(g) b) Assuming that the temperature remains
ΔH = +210 kJ mol–1 constant, calculate the degree of dissociation
a) Write an expression for Kp for this at these two pressures:
reaction. (2) i) 105 Pa (2)
b) State how the value of Kp is affected by: ii) 2 × 105 Pa. (2)
i) increasing the pressure (1) Make the assumption that α is small so that
ii) increasing the temperature (1) (1 + α) ≈ 1 and (1 − α) ≈ 1.
iii) using a catalyst. (1) 10 During the manufacture of sulfuric acid,
c) Describe how the composition of an sulfur dioxide and air pass through reactors
equilibrium mixture of gases changes when: at about 700 K, which convert the sulfur
i) the pressure rises (1) dioxide to sulfur trioxide. The volume ratio of
ii) the temperature rises (1) oxygen : sulfur dioxide is 1 : 1. The gas pressure
iii) a catalyst is added. (1) is 1–2 atmospheres.
Ammonia is converted to nitric acid on a large
8
scale. In the first step, ammonia is mixed with 2SO2(g) + O2(g) ⇋ 2SO3(g)
air and compressed before passing through ΔH = −192 kJ mol−1
a reactor containing catalyst gauzes. The
The gases pass through a series of four beds
catalyst is an alloy of platinum and rhodium.
of catalyst. The gas mixture is cooled in heat
The reversible, exothermic reaction produces
exchangers as it flows from one catalyst bed to
nitrogen monoxide (NO) and steam.
the next.
a) Write an equation for the reaction of
ammonia with oxygen to form nitrogen The catalyst is vanadium(v) oxide. The catalyst
monoxide.(2) is not effective if the temperature is lower than
316
11 Equilibrium II
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700 K. Between the third and fourth beds of to concentrations (or partial pressures) and
catalyst, the gases pass through an absorption substitute in the expression for the equilibrium
tower to remove the sulfur trioxide produced constant. Rearrange the expression to arrive at
in the first three stages. At the end of the a quadratic equation that you can solve using
process over 99.5% of the sulfur dioxide is the formula given.
converted to sulfur trioxide. a) 2.0 mol hydrogen and 1.0 mol iodine are
a) i) Show that the oxygen is in excess in mixed in a 1.0 dm3 container at 710 K.
this process. (2) The value of Kc for the decomposition of
ii) Explain why excess oxygen is used. (2) hydrogen iodide into hydrogen and iodine
iii) Suggest a reason why the process does is 0.020 at this temperature. Calculate the
not operate with an even larger excess amounts of hydrogen, iodine and hydrogen
of oxygen. (2) iodide present when the system reaches
b) Explain the choice of 700 K as the equilibrium. (6)
temperature for the process. (3) b) 1.0 mol ethanoic acid is mixed with 3.0 mol
c) Explain why the process operates at a ethanol and 3.0 mol water in the presence
pressure close to atmospheric pressure. (3) of an acid catalyst. The value of Kc for
d) Explain why the gas mixture is cooled the reaction to form the ester is 4.0 at
between the catalyst beds. (2) the temperature of the mixture. Calculate
e) Explain why the sulfur trioxide is removed the amount of ethyl ethanoate formed at
from the gas stream before the gases pass equilibrium. (6)
into the fourth catalyst bed.(2) c) 0.01 mol iodine is mixed with 0.01 mol
f ) Give reasons why it is important to convert of iodide ions in 1 dm3 of an aqueous
nearly 100% of the sulfur dioxide to sulfur solution at 298 K. Calculate the equilibrium
trioxide. (2) concentrations of I3− ions and iodide ions in
the solution. (6)
11 Note that each part of this question requires
you to use algebra and solve a quadratic I3−(aq) ⇋ I2(aq) + I−(aq)
equation. This is not expected for the Pearson Kc = 1.5 × 10−3 mol dm−3 at 298 K.
Edexcel specification. The general form of a
quadratic equation is: d) 0.20 mol of carbon monoxide and 0.10 mol
of chlorine are mixed in a 3.0 dm3 container
ax2 + bx + c = 0, where a, b and c are constants. and allowed to reach equilibrium. Calculate
The general solution to the equation is the equilibrium concentration of COCl2.
CO(g) + Cl2(g) ⇋ COCl2(g)
–b± b 2 – 4ac
x = Kc = 0.41 dm3 mol−1 at the temperature of
2a
the mixture. (6)
Solving for x with a calculator always gives two −1
e) Kp = 7.1 atm for this equilibrium at
possible values. In equilibrium calculations, 298 K:
only one of the values turns out to be a
2NO2(g) ⇋ N2O4(g)
possible solution to the problem.
Calculate the partial pressures of the two
In each question work out the amounts
gases at equilibrium when starting with
at equilibrium in terms of the unknown
pure N2O4(g) at 1 atmosphere pressure. (6)
quantity x. Where necessary then convert
317
Exam practice questions
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Acid–base equilibria
12
Acids and bases are very common, not only in laboratories but also in living
things, in the home and in the natural environment. Acid–base reactions are
reversible and governed by the equilibrium law. This means that chemists
are able to predict reliably and quantitatively how acids and bases behave.
This is important for the supply of safe drinking water, the care of patients in
hospital, the formulation of shampoos and cosmetics, as well as the processing
of food and many other aspects of life.
12.1 Theories to explain reactions
of acids and bases
Figure 12.1 Red ants attacking a insect
on leaf. Methanoic acid is an ingredient of
Jabir ibn-Hayyan and his discoveries
the sting of red ants. The traditional name The mineral acids that are now taken for granted were discovered by
for the acid was formic acid based on the Jabir ibn-Hayyan (c. 722–c. 815), who worked in the alchemical tradition
Latin name for ants: formica. but pioneered experimental chemistry (Figure 12.2). He developed the
techniques of crystallisation and distillation and used them to discover
sulfuric, hydrochloric and nitric acids. He also studied ethanoic acid in
Tip vinegar and tartaric acid in wine.
Section 12.1 traces the development
Acids were first recognised by their chemical properties. Acidic solutions have a
of ideas about acids and bases. This
sour taste; they tend to corrode metals and they change the colour of indicators.
section reminds you of Arrhenius’s
theory, which you have used until
now to explain the reactions of acids
and bases. Section 12.2 introduces
a new theory which can be applied to
reactions that you studied in the first
year of the course. The chapter goes
on to show how this theory uses the
equilibrium law to explain quantitatively
the behaviour of acids and bases.
Figure 12.2 Jabir ibn-Hayyan in a coloured engraving, published in 1883, which shows
him teaching at the school at Edessa in Mesopotamia (now Sanliurfa in Turkey). He
played a key part in turning chemistry from a mystical practice (alchemy) into a science.
He pioneered experimental techniques and invented much of the equipment that is still
commonly used in laboratories.
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Test yourself Tip
1 Jabir ibn-Hayyan discovered sulfuric acid and then studied its All the names for acids in this chapter
reactions on heating with salts such as sodium chloride and are the modern chemical names; they
potassium nitrate. How could these studies, and his improved are not the names generally used at the
methods of distillation, lead to the discovery of hydrochloric and time when these chemicals were first
nitric acids? discovered.
The theories of Antoine Lavoisier and
Humphry Davy
The scientist who laid the foundations of modern chemical theory was
the French nobleman Antoine Lavoisier (1743–1794). His experiments and
insights led to the oxygen-theory of burning and a systematic approach to
quantitative chemistry.
The name ‘oxygen’ means ‘acid former’. Lavoisier gave the gas this name
because he thought that all acids were compounds containing this element.
The generalisation that all acids contain oxygen was disproved by a series of
experiments carried out by the English scientist Humphry Davy between
1809 and 1810. He heated hydrogen chloride (then called muriatic acid) to
high temperatures with a range of metals and non-metals. He could find no
trace of oxygen in the compound. After further work, Davy proposed, in
1816, that what all acids have in common is that they contain hydrogen.
Test yourself
2 Why is it not surprising that Lavoisier thought that all acids contain
oxygen?
3 Identify three acids, other than hydrochloric acid, which do not contain
oxygen.
4 Give examples which show that:
a) acids contain hydrogen
b) not all compounds that contain hydrogen are acids.
Arrhenius’s theory
As a young man in his mid-20s, the Swedish chemist Svante Arrhenius wrote
a doctoral thesis which proposed that some compounds are ionised in solution
all the time. This was the start of the ionic theory of solutions that we now
take for granted. In 1884 it was highly controversial. At the time, Arrhenius
was bitterly disappointed to be awarded the bottom grade for his paper. Later
he was vindicated and awarded the Nobel prize for chemistry in 1903. Key term
Arrhenius used his ionic theory to come up with an explanation of why it
Acids dissociate when they dissolve in
is that all acids have similar properties when dissolved in water. His theory
water. This means that they form ions in
could also account for what happens when an acid is neutralised by an alkali
the solution. The extent of dissociation
and explain the difference between strong and weak acids. He realised that
into ions distinguishes strong and weak
acids dissociate when they dissolve in water to form ions in the solution.
acids.
The extent of dissociation into ions distinguishes strong and weak acids.
12.1 Theories to explain reactions of acids and bases 319
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In 1887, Arrhenius defined an acid as a compound that could produce hydrogen
Key terms ions when dissolved in water, and an alkali as a compound that could produce
hydroxide ions in water. According to Arrhenius’s theory, hydrochloric acid
A strong acid is an acid which is fully
is a strong acid which is fully ionised when dissolved in water.
ionised (dissociated) when it dissolves
in water. HCl(aq) → H+(aq) + Cl−(aq)
A weak acid is an acid which is only Ethanoic acid is a weak acid which is only slightly ionised.
ionised (dissociated) to a slight extent
when it dissolves in water. CH3COOH(aq) ⇋ CH3COO−(aq) + H+(aq)
A proton is the nucleus of a hydrogen Arrhenius’s theory was a big advance in its time. It could account for the
atom, so a hydrogen ion, H+, is just a similarities between acids. In this theory, the typical reactions of dilute acids
proton. in water are the reactions of aqueous hydrogen ions.
With metals: Mg(s) + 2H+(aq) → Mg 2+(aq) + H2(g)
With carbonates: CO32−(s) + 2H+(aq) → CO2(g) + H2O(l)
With bases: O2−(s) + 2H+(aq) → H2O(l)
The Arrhenius theory is still useful today and equations for the reactions of
acids and alkalis are often written in a form based on the theory. However,
the theory has a number of weaknesses, one of which is that it is limited to
aqueous solutions.
Test yourself
5 What, according to Arrhenius’s theory, happens when an acid
Tip neutralises an alkali?
6 Suggest a simple practical demonstration of the difference between
Weak acids are only very slightly
equimolar solutions of a strong acid and of a weak acid.
ionised. Do not describe weak
acids as ‘not completely ionised (or 7 Write ionic equations to show how Arrhenius’s theory describes the
dissociated)’. This could be taken to reactions of nitric acid with:
mean 95% ionised, which could be true a) zinc b) potassium carbonate
of a strong acid. c) calcium oxide d) lithium hydroxide.
The theory of Johannes Brønsted and
Thomas Lowry
The preferred theory for discussing acid–base equilibria today was put
forward independently in 1923 by the Danish chemist Johannes Brønsted and
the English chemist Thomas Lowry. This theory describes acids as proton
donors, and bases as proton acceptors, as explained in the next section.
12.2 Acids, bases and proton transfer
Acids as proton donors
According to the Brønsted–Lowry theory, hydrogen chloride molecules give
hydrogen ions (protons) to water molecules when they dissolve in water,
producing hydrated hydrogen ions called oxonium ions. The water acts
as a base.
320 12 Acid–base equilibria
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HCl(aq) + H2O(l) ⇋ H3O+(aq) + Cl−(aq)
Tip
The proton transfer between hydrogen chloride molecules and water
It is more correct to represent hydrogen
molecules is reversible. A proton from the oxonium ion can transfer back to
ions in aqueous solution as H3O+(aq);
the chloride ion to give hydrogen chloride and water.
however, chemists commonly use a
Hydrochloric acid is a strong acid. What this means is that it readily gives up shorter symbol for hydrated protons,
its protons to water molecules and the equilibrium in solution lies well over H+(aq).
to the right. Hydrochloric acid is effectively completely ionised in solution.
Other examples of strong acids are sulfuric acid and nitric acid.
Some acids can give away (donate) one proton per molecule. Examples are
hydrochloric acid, HCl; nitric acid, HNO3; and ethanoic acid, CH3COOH.
These acids are sometimes described as monobasic acids because one mole of
the acid neutralises one mole of hydroxide ions (a base). They are also called
monoprotic acids because one proton per molecule can ionise.
There are other acids that can give away (donate) two protons per molecule.
Examples are sulfuric acid, H 2SO4; and ethanedioic acid, HOOC−COOH.
These acids are sometimes described as dibasic acids because one mole of the
acid neutralises two moles of a base such as sodium hydroxide ions. They are
also called diprotic acids.
Key terms
According to the Brønsted–Lowry theory, acids are proton donors.
According to the Brønsted–Lowry theory, a base is a proton acceptor.
Test yourself
8 a) What type of bond links the water molecule to a proton in an
oxonium ion?
b) Draw a dot-and-cross diagram to show the bonding in an oxonium
ion. Use your diagram to explain why the ion has a positive charge.
c) Predict the shape of an oxonium ion.
9 Write a balanced ionic equation for the reaction of 1 mol ethanedioic
acid with 2 mol NaOH, showing the structural formulae for the acid
and for the ethanedioate ion formed.
Bases as proton acceptors
According to the Brønsted–Lowry theory, a base is a molecule or ion which can
accept a hydrogen ion (proton) from an acid. A base has a lone pair of electrons
which can form a dative covalent bond with a proton (Figures 12.3 and 12.4).
+
H H
2– H N + H+ H N H
O H+ O H–
H H
Figure 12.3 Oxide ions have lone pairs of Figure 12.4 The lone pair on the nitrogen
electrons which can form dative covalent atom of ammonia allows it to act as a base.
bonds with hydrogen ions.
12.2 Acids, bases and proton transfer 321
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NH2 An ionic oxide, such as calcium oxide, reacts completely with water to
form calcium hydroxide. The calcium ions do not change; but the oxide
C N ions, which are powerful proton acceptors, all take protons from water
N C
C H molecules. An oxide ion is a strong base. Common bases include the
C C oxide and hydroxide ions, ammonia, amines, as well as the carbonate and
H N N hydrogencarbonate ions.
H In biochemistry the term ‘base’ often refers to one of the five nitrogenous
Figure 12.5 The displayed formula of bases which make up nucleotides and the nucleic acids DNA and RNA.
adenine. This is one of the bases in DNA. These compounds (adenine, guanine, cytosine, uracil and thymine) are bases
in the chemical sense because they have lone pairs on nitrogen atoms which
can accept hydrogen ions (Figure 12.5).
Tip
Many compounds of the Group 1 and Group 2 metals form alkaline solutions. This is
because metals such as sodium, potassium, magnesium and calcium (unlike other
metals) form oxides, hydroxides and carbonates which are soluble (to a greater or
lesser extent) in water. It is important to realise that it is the oxide, hydroxide or
carbonate ions in these compounds that are bases, and not the metal ions.
Test yourself
10 a)
Identify the products of the reaction when concentrated sulfuric
acid reacts with sodium chloride.
b) Show that this is a proton transfer reaction and identify the base.
c)
Account for the fact that this reaction can give a good yield of
hydrogen chloride gas, despite the fact that concentrated sulfuric
acid and hydrogen chloride are strong acids.
11
Show that the reactions between these pairs of compounds are
acid–base reactions and identify as precisely as possible the
molecules or ions which are the acid and the base in each example.
a) MgO + HCl
b) H2SO4 + NH3
c) NH4NO3 + NaOH
d) HCl + Na2CO3
12 a) What type of bond links the ammonia molecule to a proton in an
ammonium ion?
b)
Draw a dot-and-cross diagram to show the bonding in an
ammonium ion.
c) Predict the shape of an ammonium ion.
322 12 Acid–base equilibria
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Conjugate acid–base pairs Key terms
Any acid–base reaction involves competition for protons. This is illustrated
by a solution of an ammonium salt, such as ammonium chloride, in water. An acid turns into its conjugate base
when it loses a proton. A base turns
NH4+(aq) + H2O(l) ⇋ NH3(aq) + H3O+(aq)
into its conjugate acid when it gains a
acid 1 base 2 base 1 acid 2 proton. Any pair of compounds made
up of an acid and a base that can be
In this example there is competition for protons between ammonia molecules
converted from one to the other by
and water molecules. On the left-hand side of the equation the protons are
proton transfer is a conjugate acid–
held by lone pairs on the ammonia molecules. On the right-hand side they
base pair.
are held by lone pairs on water molecules. The position of equilibrium shows
which of the two bases has the stronger hold on the protons. The definition of pH is:
Chemists use the term conjugate acid–base pair to describe a pair of pH = −log10 [H+(aq)]
molecules or ions which can be converted from one to the other by the gain
or loss of a proton. The equilibrium in a solution of the ammonium salt
above involves two examples of conjugate acid–base pairs:
● NH4
+ and NH3
● H 3O
+ and H2O.
Test yourself
13
Identify and name the conjugate bases of these acids: HNO3,
CH3COOH, H2SO4, HCO3−.
14 Identify and name the conjugate acids of these bases: O2−, OH−,
NH3, CO32−, HCO3−, SO42−.
15 Explain and illustrate these two statements:
a) The stronger the acid, the weaker its conjugate base.
b) The stronger the base, the weaker its conjugate acid.
12.3 The pH scale
The concentration of hydrogen ions in aqueous solutions commonly ranges
from about 2 mol dm−3 to about 1 × 10−14 mol dm−3. The concentration of
aqueous hydrogen ions in dilute hydrochloric acid is about 100 000 000 000 000
times greater than the concentration of hydrogen ions in dilute sodium
hydroxide solution.
Figure 12.6 A scientist measuring the pH
Given such a wide range of concentrations, scientists find it convenient to of glacier melt water during research into
use a logarithmic scale to measure the concentration of aqueous hydrogen air and water pollution.
ions in acidic or alkaline solutions (Figures 12.6 and 12.7). This is the pH
scale, where:
pH = −log10 [H+(aq)]
pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11 –12 –13 –14
[H+(aq)]/mol dm–3 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10
increasingly acidic neutral increasingly alkaline
Figure 12.7 The pH scale showing the colours of a full-range indicator at the different
pH values.
12.3 The pH scale 323
469983_12_Chem_Y1-2_318-[Link] 323 13/04/19 10:06 PM
Tip Examples
See Section 3 in ‘Mathematics in A Example 1
Level chemistry’, which you can access
online at [Link]. What is the pH of 0.020 mol dm−3 hydrochloric acid?
[Link]/EdexcelChemistry, to learn
about logarithms to base 10 and how Notes on the method
they are used to handle values which Hydrochloric acid is a strong acid so it is fully ionised. Note that 1 mol HCl
range over several orders of magnitude. gives 1 mol H+(aq). Use the log button on your calculator. Do not forget the
minus sign in the definition of pH. Give your answer to 2 decimal places.
In pH calculations, always use the ‘log’
button on your calculator, not the ‘ln’
Answer
button.
[H+(aq)] = [HCl(aq)] = 0.020 mol dm−3
pH = −log (0.020) = 1.70
Example 2
The pH of human blood is 7.40. What is the aqueous hydrogen ion
concentration in blood?
Notes on the method
pH = −log [H+(aq)]
From the definition of logarithms this rearranges to [H+(aq)] = 10−pH
x
Use the inverse log button (10 ) on your calculator. Do not forget the
minus sign in the definition of pH.
Give your answer to 2 decimal places.
Answer
pH = 7.40
[H+(aq)] = 10 −7.40 = 3.98 × 10 −8 mol dm−3
Test yourself
16
What is the pH of solutions of hydrochloric acid with these
concentrations?
a) 0.10 mol dm−3
b) 0.010 mol dm−3
c) 0.0010 mol dm−3
17 Calculate the pH of a 0.080 mol dm−3 solution of nitric acid. Give
your answer to 2 decimal places.
18
Calculate the concentration of hydrogen ions in each solution:
a) orange juice with a pH of 3.30
b) coffee with a pH of 5.40
c) saliva with a pH of 6.70
d) a suspension of an antacid in water with a pH of 10.50.
324 12 Acid–base equilibria
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The ionic product of water
There are hydrogen and hydroxide ions even in pure water because of a
transfer of hydrogen ions between water molecules. This only happens to
a very slight extent.
H2O(l) + H2O(l) ⇋ H3O+(aq) + OH−(aq)
This equilibrium can be written more simply as:
H2O(l) ⇋ H+(aq) + OH−(aq)
The equilibrium constant is:
[H+ (aq)][OH – (aq)]
Kc =
[H2 O(1)]
There is such a large excess of water that [H 2O(l)] is a constant, so the
relationship simplifies to: Key term
Kw = [H+(aq)][OH−(aq)] Kw is the ionic product of water. It
where Kw is the ionic product of water. is the equilibrium constant for the
ionisation of water. It is defined by the
The pH of pure water at 298 K is 7.0. So the hydrogen ion concentration at expression:
equilibrium is:
Kw = [H+(aq)][OH−(aq)]
[H+(aq)] = 1.0 × 10−7 mol dm−3
Also, in pure water [H+(aq)] = [OH−(aq)], so:
[OH−(aq)] = 1.0 × 10−7 mol dm−3
Tip
Refer to Section 2 in ‘Mathematics in A
Hence:
Level chemistry’, which you can access
Kw = 1.0 × 10−14 mol 2 dm−6 online at [Link].
[Link]/EdexcelChemistry, for help
Kw is a constant in all aqueous solutions at 298 K. This makes it possible to
with multiplying together numbers in
calculate the pH of alkaline solutions.
standard form.
Example
What is the pH of a 0.050 mol dm−3 solution of sodium hydroxide at 298 K?
Notes on method
Sodium hydroxide is fully ionised in solution. So in this solution:
[OH−(aq)] = 0.050 mol dm−3
pH = −log [H+(aq)]
Answer
For this solution:
Kw = [H+(aq)] × 0.050 mol dm−3 = 1.0 × 10 −14 mol2 dm−6
1.0 × 10 −14 mol2 dm−6
So [H+(aq)] = = 2.0 × 10 −13 mol dm−3
0.050 mol dm−3
Hence pH = −log (2.0 × 10 −13) = 12.7
12.3 The pH scale 325
469983_12_Chem_Y1-2_318-[Link] 325 13/04/19 10:06 PM
Test yourself
19 The value of Kw varies with temperature. At 273 K its value is
1.10 × 10 −15 mol2 dm−6, while at 303 K it is 1.50 × 10 −14 mol2 dm−6.
a) Is the ionisation of water an exothermic or an endothermic
process?
b) What happens to the hydrogen ion concentration in pure water,
and hence the pH, as the temperature rises?
c) Does pure water stop being neutral if its temperature is above or
below 298 K?
20 Calculate the pH of these solutions at 298 K:
a) 1.0 mol dm−3 NaOH
b) 0.020 mol dm−3 KOH
c) 0.0010 mol dm−3 Ba(OH)2.
Working in logarithms
The logarithmic form of equilibrium constants is particularly useful for pH
calculations. Taking logarithms produces a conveniently small range of values.
Kw = [H+(aq)][OH−(aq)] = 1.0 × 10−14 at 298 K
Taking logarithms, and applying the rule that log xy = log x + log y, gives:
log Kw = log [H+(aq)] + log [OH−(aq)] = log 10−14 = −14
Multiplying through by −1 reverses the signs:
−log Kw = −log [H+(aq)] − log [OH−(aq)] = 14
The term −log Kw is given the symbol pKw.
Key term
The term −log [OH−(aq)] is represented as pOH.
pKw is defined as −log Kw.
Hence: pKw = pH + pOH = 14
So: pH = 14 − pOH
This makes it easy to calculate the pH of alkaline solutions at 298 K.
Example
What is the pH of a 0.050 mol dm−3 solution of sodium hydroxide?
Tip
See Section 3 in ‘Mathematics in A Note on method
Level chemistry’, which you can access Sodium hydroxide, NaOH, is a strong base so it is fully ionised.
online at [Link].
Use the log button on your calculator to find the values of the logarithms.
[Link]/EdexcelChemistry, for help
with logarithms. You do not have to be
Answer
able to work in logarithms, but some
people find it easier. Do not try to [OH−(aq)] = 0.050 mol dm−3
remember the formula pH = 14 − pOH.
pOH = −log 0.050 = 1.3
Only use it if you can work it out quickly
for yourself from the definition of Kw. pH = 14 − pOH = 14 − 1.3 = 12.7
326 12 Acid–base equilibria
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12.4 Weak acids and bases
Most organic acids and bases ionise to only a slight extent in aqueous solution.
Carboxylic acids (see Section 17.3.3), such as ethanoic acid in vinegar,
citric acid in fruit juices and lactic acid in sour milk, are all weak acids
(Figure 12.8). Ammonia and amines (see Section 18.2.3) are weak bases.
Weak acids
In a 0.10 mol dm−3 solution of ethanoic acid, only about 1 in 100 molecules
ionise to produce hydrogen ions. In other words, they only dissociate into
ions to a very slight extent.
CH3COOH(aq) ⇋ CH3COO−(aq) + H+(aq)
This means that the pH of a 0.10 mol dm−3 solution of ethanoic acid is 2.9 and
not 1.0, as it would be if it were a strong acid.
There is a very important distinction between acid strength and concentration.
Strength is the extent of ionisation. Concentration is the amount in moles
of acid in a cubic decimetre. It takes just as much sodium hydroxide to
neutralise 25 cm3 of a 0.10 mol dm−3 solution of a weak acid such as ethanoic Figure 12.8 Bacteria added to milk
acid as it does to neutralise 25 cm3 of a 0.10 mol dm−3 solution of a strong acid ferment the lactose sugar and turn it into
such as hydrochloric acid. lactic acid. Lactic acid is a weak acid that
turns the milk into yogurt and also restricts
Test yourself the growth of food poisoning bacteria.
21 E xplain why measuring the pH of a solution of an acid does not
provide enough evidence to show whether or not the acid is strong Tip
or weak. In everyday life people use ‘weak’ to
22 E xplain why it takes the same amount of sodium hydroxide to mean dilute, as in, “I’ll have a cup of
neutralise 25.0 cm3 of 0.10 mol dm−3 ethanoic acid as it does weak tea, please.” In chemistry the
to neutralise 25 cm3 of 0.10 mol dm−3 hydrochloric acid. word has a technical meaning and
refers to the degree of ionisation and
Weak bases not to the concentration. In chemistry:
weak ≠ dilute.
Weak bases only react to form ions to a slight extent when they dissolve in
water. In a 0.1 mol dm−3 solution of ammonia, for example, 99 in every 100
molecules do not react but remain as dissolved molecules. Only 1 molecule
in 100 reacts to form ammonium ions.
NH3(aq) + H2O(l) ⇋ NH4+(aq) + OH−(aq)
As with weak acids, it is important to distinguish between strength and
concentration.
Acid dissociation constants
Chemists use the equilibrium constant for the reversible ionisation of a weak
acid as a measure of its strength. The equilibrium constant shows the extent
to which acids dissociate into ions in solution.
A weak acid can be represented by the general formula HA, where A− is the
ion produced when the acid ionises.
HA(aq) ⇋ H+(aq) + A−(aq)
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According to the equilibrium law, the equilibrium constant Kc, takes this
Key term form with the subscript ‘c’ for concentration replaced by subscript ‘a’ for acid:
Acid dissociation constant is the name [H+ (aq)][A – (aq)]
given to the equilibrium constant Kc for Ka =
[HA(aq)]
the ionisation of a weak acid. This is such
an important type of equilibrium constant In this context the equilibrium constant K a is called the acid dissociation
that it is given its own symbol: Ka. constant. Given the value for K a and the concentration, it is possible to
calculate the pH of a solution of a weak acid.
Test yourself
23 If a weak acid is shown as HA, what is A− in the particular case of:
a) hydrogen fluoride
b) methanoic acid
c) chloric(i) acid?
Example
Calculate the hydrogen ion concentration and the pH of a 0.010 mol dm−3
solution of propanoic acid. Ka for the acid is 1.3 × 10−5 mol dm−3.
Notes on the method
Two approximations simplify the calculation.
1 The first assumption is that at equilibrium [H+(aq)] = [A−(aq)]. In this
example A− is the propanoate ion CH3CH2COO−. This assumption
seems obvious from the equation for the ionisation of a weak acid,
but it ignores the hydrogen ions from the ionisation of water. Water
produces far fewer hydrogen ions than most weak acids, so its
ionisation can usually be ignored. This assumption is acceptable so
long as the pH of the acid is below 6.
2 The second assumption is that so little of the propanoic acid ionises in
water that at equilibrium [HA(aq)] ≈ 0.01 mol dm−3. Here HA represents
propanoic acid. This is a riskier assumption which has to be checked,
because in very dilute solutions the degree of ionisation may become
quite large relative to the amount of acid in the solution. Chemists
generally agree that this assumption is acceptable so long as less than
5% of the acid ionises.
Answer
CH3CH2COOH(aq) ⇋ H+(aq) + CH3CH2COO −(aq)
[H+ (aq)] [CH3CH2COO– (aq)] [H+ (aq)]2
Ka = =
[CH3CH2COOH(aq)] 0.010 mol dm–3
Ka = 1.3 × 10 −5 mol dm−3
Therefore
[H+(aq)]2 = 0.010 mol dm−3 × 1.3 × 10−5 mol dm−3 = 1.3 × 10−7 mol2 dm−6
328 12 Acid–base equilibria
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So [H+(aq)] = 3.61 × 10 −4 mol dm−3
pH = −log [H+(aq)]
= −log (3.61 × 10 −4)
= 3.4
Check the second assumption: in this case less than 0.0004 mol dm−3
of the 0.0100 mol dm−3 of acid (4%) has ionised. In this instance the
degree of dissociation is small enough to justify the assumption that
[HA(aq)] ≈ the concentration of un-ionised acid.
One method which can, in principle, be used to measure K a for a weak
acid is to measure the pH of a solution when the concentration of the
acid is accurately known. This is not a good method for determining the
size of K a because the pH values of dilute solutions are very susceptible to
contamination – for example by dissolved carbon dioxide from the air.
Example
Calculate the Ka of lactic acid given that pH = 2.43 for a 0.10 mol dm−3
solution of the acid.
Notes on the method
The same two approximations simplify the calculation.
1 Assume that [H+(aq)] = [A−(aq)], where A−(aq) here represents the
aqueous lactate ion. Since the pH is well below 6 this is certainly
justified.
2 Also assume that so little of the lactic acid ionises in water that at
equilibrium [HA(aq)] ≈ 0.1 mol dm−3. Here HA represents lactic acid.
This is a riskier assumption, which again can be checked during the
calculation.
Answer
pH = 2.43
[H+(aq)] = 10 −2.43 = 3.72 × 10 −3 mol dm−3
[H+(aq)] = [A−(aq)] = 3.72 × 10 −3 mol dm−3
In this example less than 5% of the acid is ionised (less than 0.004 out
of 0.100 mol in each litre).
So [HA(aq)] ≈ 0.1 mol dm−3
Substituting in the expression for Ka:
+
[H (aq)] [A – (aq)] (3.72 × 10–3 moldm–3 )2
Ka = =
[HA (aq)] 0.1moldm–3
Ka = 1.4 × 10 −4 mol dm−3
12.4 Weak acids and bases 329
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Test yourself
24 Calculate the pH of a 0.010 mol dm−3 solution of hydrogen cyanide
given that Ka = 4.9 × 10 −10 mol dm−3.
Calculate the pH of a 0.050 mol dm−3 solution of ethanoic acid given
25
that Ka = 1.7 × 10 −5 mol dm−3.
26 Calculate Ka for methanoic acid given that pH = 2.55 for a
0.050 mol dm−3 solution of the acid.
Calculate Ka for butanoic acid, C3H7COOH, given that pH = 3.42 for
27
a 0.010 mol dm−3 solution of the acid.
Working in logarithms
Chemists find it convenient to define a quantity pK a = −log K a when working
with weak acids. This definition means that hydrocyanic acid, HCN, with
a pK a value of 9.3, is a much weaker acid than nitrous acid, HNO2, with a
pK a value of 3.3.
Data tables show pK a values. The relationship between acid strength and pH
can be expressed simply because both are logarithmic quantities.
[H+ (aq)][A – (aq)]
Ka =
[HA(aq)]
The two common assumptions when using this expression in calculations
are that:
[H+(aq)] = [A−(aq)]
[HA(aq)] = cA, where cA = the concentration of the un-ionised acid.
Substituting in the expression for K a gives:
+
[H (aq)]2
Ka =
cA
Tip
Hence: K a × cA = [H+(aq)]2
See Section 3 in ‘Mathematics in A
Level chemistry’, which you can access Taking logarithms:
online at [Link]. log (K a × cA) = log [H+(aq)]2
[Link]/EdexcelChemistry, for help
with logarithms. You do not have to Applying the rules that log xy = log x + log y and that log xn = n log x, gives:
be able to work in logarithms, but log K a + log cA = 2 log [H+(aq)]
some people find it easier. Do not try
to remember this logarithmic form of which on multiplying by −1 becomes:
the equilibrium law. The relationship is −log K a − log cA = −2 log [H+(aq)]
easy to use, but only apply it if you can
derive it quickly from first principles Hence: pK a − log cA = 2 × pH
as shown here. Do not forget that This shows that, for a solution of a weak acid which is less than 5% ionised:
this form of the law has two built-in
assumptions, so it only applies when pH = 21 (pK a − log cA) which rearranges to pK a = 2 pH + log cA.
these assumptions are acceptable.
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Test yourself
28 What is the value of pKa for methanoic acid, given that
Ka = 1.6 × 10 −4 mol dm−3?
29 What is the value of Ka for benzoic acid, given that pKa = 4.2?
30 Show that the logarithmic relationship pKa = 2pH + log cA gives the
same answers from the data as the methods used in the worked
examples on pages 328 and 329.
Activity
The effect of dilution on the degree of dissociation of
a weak acid
Two students used a pH meter to investigate the effect of dilution on the dissociation
of ethanoic acid. They started by preparing the solutions shown in Table 12.1
by diluting a 0.10 mol dm−3 solution of the acid.
Next they calibrated a pH meter by dipping the probe into a solution of known pH
(a buffer solution, see Section 12.8). After rinsing the probe with distilled water,
they dipped it into the least concentrated of the solutions to measure the pH. They
continued to rinse the probe and then measure the pH of the next solution, until
they had recorded pH values for all four solutions.
Table 12.1
Concentration of Measured pH of Calculated pH of solutions
ethanoic acid/mol dm−3 ethanoic acid of hydrochloric acid with
the same concentration
0.00010 4.2
0.0010 3.5
0.010 3.0
0.10 2.7 1.0
1 Describe how the students could prepare a 0.010 mol dm−3 solution of ethanoic
acid from the 0.10 mol dm−3 solution.
2 The water that the students used for the dilutions had been boiled and then
allowed to cool to room temperature. Explain why this improved the accuracy of the
measurements.
3 Explain why it was necessary to calibrate the pH meter.
4 Why did the students measure the pH of the most dilute solution first and then work
up to the more concentrated solutions in the order they are listed in Table 12.1?
5 What are the calculated pH values for hydrochloric acid that are missing from
the table?
6 a) What does the table tell you about the degree of dissociation of ethanoic acid
compared to hydrochloric acid at any concentration?
b) Study the difference in the pH values for the two acids at each dilution. What do
the differences show about the effect of dilution on the degree of dissociation
of ethanoic acid?
c) Use the equilibrium law to explain the effect of dilution on the degree
of dissociation of ethanoic acid.
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12.5 Acid–base titrations
The equilibrium law helps to explain what happens during acid–base
titrations and it provides a rationale for the selection of the right indicator
for a titration.
During a titration, the pH changes as a solution of an alkali runs from a
burette and mixes with an acid in a flask (Figure 12.9). Plotting pH against
volume of alkali added gives a graph whose shape is determined by the
nature of the acid and the base. Usually there is a marked change in pH near
the equivalence point, and it is this which makes it possible to detect the
end-point of the titration with an indicator.
At the end-point, the colour change of the indicator shows that enough of
the solution in the burette has been added to react with the amount of the
chemical in the flask. In a well-planned titration, the colour change observed
at the end-point corresponds exactly with the equivalence point.
The equivalence point is the point during any titration when the amount in
moles of one reactant added from a burette is just enough to react exactly
Figure 12.9 A pH meter can be used to with all of the measured amount of chemical in the flask as shown by the
measure the pH of the solution in the flask balanced equation.
during an acid–base titration, and to detect
the end-point.
Key terms
The equivalence point during a titration is reached when the amount of reactant
added from a burette is just enough to react exactly with all the measured amount of
chemical in the flask according to the balanced equation.
The end-point in a titration the point at which a colour change or pH change indicates
that just enough of the solution in the burette has been added to react with the
chemical in the flask.
Activity
Titration of a strong acid with a strong base
Strong acids and bases are fully ionised in solution. Figure 12.10 shows the shape of
the pH curve for the titration of a strong acid, hydrochloric acid, with a strong base, 12
sodium hydroxide. 10
1 Show that pH = 1.0 for a solution of 0.10 mol dm−3 HCl(aq). 8
equivalence
2 Why does pH = 7 at the equivalence point of a titration of a strong acid with a strong
pH
point pH = 7
6
base?
3 Calculate the pH of 25 cm3 of a solution of sodium chloride after adding: 4
a) 0.05 cm3 (1 drop) of 0.10 mol dm−3 HCl(aq) 2
b) 0.05 cm3 (1 drop) of 0.10 mol dm−3 NaOH(aq).
0
(In both instances assume that the volume change on adding 1 drop is insignificant.) 0 5 10 15 20 25 30
4 Calculate the pH of the solution produced by adding 5.0 cm3 of 0.10 mol dm−3 Titre/cm3
NaOH(aq) to 25.0 cm3 of a solution of sodium chloride. Figure 12.10 The pH change on adding
5 Show that your answers to Questions 1, 2, 3 and 4 are consistent with Figure 12.10. 0.10 mol dm−3 NaOH(aq) from a burette
6 What features of the curve plotted in Figure 12.10 are important for the accuracy to 25.0 cm3 of a 0.10 mol dm−3 solution
of acid–base titrations of this kind? of HCl(aq).
332 12 Acid–base equilibria
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Titration of a weak acid with a strong base
If the acid in the titration flask is weak, then the equilibrium law applies and
the pH curve up to the equivalence point has to be calculated with the help
of the expression for K a.
Consider, for example, the reaction of ethanoic acid with sodium hydroxide
during a titration (Figure 12.11). At the start the flask contains the pure acid.
CH3COOH(aq) ⇋ H+(aq) + CH3COO−(aq)
Figure 12.11 The pH change on adding
12
0.10 mol dm−3 NaOH(aq) from a burette to
10 equivalence point 25.0 cm3 of a 0.10 mol dm−3 solution of
solution of sodium ethanoate
8 pH = 8.7 CH3COOH(aq).
acid
half-neutralised
pH
6 pH = 4.8
4
pure ethanoic acid
2 pH = 2.9
0
0 5 10 15 20 25 30
Titre/cm3
The pH of the pure acid can be calculated from K a, as shown in Section 12.4.
However, when some strong alkali runs in from the burette, some of the
ethanoic acid reacts to produce sodium ethanoate. Once this has happened,
[H+(aq)] ≠ [CH3COO−(aq)], and the method of calculating the pH has to
change to account for this. The following worked example shows how this is
done. (The reason why this method is necessary is explained in Section 12.8.)
Example
What is the pH of a mixture formed during a titration The total volume of the solution = 45.0 cm3.
after adding 20.0 cm3 of 0.10 mol dm−3 NaOH(aq) to 5.0 cm3 of the 0.10 mol dm−3 ethanoic acid remains
25.0 cm3 of a 0.10 mol dm−3 solution of CH3COOH(aq) not neutralised. This is now diluted to a total volume
if Ka = 1.7 × 10−5 mol dm−3? of 45 cm3 solution.
Concentration of ethanoic acid molecules
Notes on the method
5.0 cm3
The pH of the mixture can be estimated quite = × 0.10 mol dm−3
45.0 cm3
accurately using the equilibrium law by assuming that: Also the concentration of ethanoate ions
20.0 cm3
● the concentration of ethanoic acid molecules at = × 0.10 mol dm−3
45.0 cm3
equilibrium is determined by the amount of acid
which has yet to be neutralised [CH3COOH(aq)] 5.0
So the ratio
● the concentration of ethanoate ions is determined by [CH3COO −(aq)] = 20.0
the amount of acid converted to sodium ethanoate.
Substituting in the rearranged expression for the
equilibrium law gives:
Answer 5.0 5.0
[H+(aq)] = Ka × = 1.7 × 10 −5 mol dm−3 ×
[H+ (aq)][CH3COO– (aq)] 20.0 20.0
Ka = [H+(aq)] = 4.25 × 10 −6 mol dm−3
[CH3COOH(aq)]
This rearranges to give: pH = − log [H+(aq)] = −log (4.25 × 10 −6) = 5.4
Ka × [CH3COOH(aq)]
[H+ (aq)] =
[CH3COO- (aq)]
12.5 Acid–base titrations 333
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The pH change for this titration is shown in Figure 12.11. Note that halfway
to the equivalence point, the added alkali converts half of the weak acid to
its salt. In this example, at this point:
[CH3COOH(aq)] = [CH3COO−(aq)]
K a × [CH3COOH(aq)]
So: [H+(aq)] = = Ka
[CH COO-(aq)]
3
Hence, at this point: [H+(aq)] = K a
It follows that pH = pK a halfway to the equivalence point (see Core practical
9 in Section 12.8).
At the equivalence point, the solution contains sodium ethanoate. As
Figure 12.11 shows, the solution at this point is not neutral. A solution of a
salt of a weak acid and a strong base is alkaline (see Section 12.7). Sodium
ions have no effect on the pH of a solution, but ethanoate ions are basic. The
ethanoate ion is the conjugate base of a weak acid.
Beyond the equivalence point, the curve is determined by the excess of
strong base, and so the shape is very close to the shape of the curve after the
end-point in Figure 12.10.
Test yourself
31 Calculate the pH of a 0.10 mol dm−3 solution of CH3COOH(aq).
32 Calculate the pH of a mixture formed during a titration after adding
10.0 cm3 of 0.10 mol dm−3 NaOH(aq) to 25.0 cm3 of a 0.10 mol dm−3
solution of CH3COOH(aq).
33 Explain why a solution of sodium ethanoate is alkaline.
34 Why is the equivalence point reached at 25.0 cm3 in both the
titrations illustrated in Figures 12.10 and 12.11?
Titration of a strong acid with a weak base
12
During the titration of a strong acid with a weak base, the flask contains
10 a strong acid at the start and the titration curve follows the same line as in
8
Figure 12.10. In a titration of hydrochloric acid with ammonia solution, for
example, the salt formed at the equivalence point is ammonium chloride.
pH
6 equivalence
point Since ammonia is a weak base, the ammonium ion is an acid. So a solution of
4
ammonium chloride is acidic and the pH is below 7 at the equivalence point. As
2 shown in Figure 12.12, after the equivalence point the curve rises less far than
in Figure 12.10 because the excess alkali is a weak base and is not fully ionised.
0
0 5 10 15 20 25 30
Titre/cm3
Figure 12.12 The pH change on adding a
Test yourself
0.10 mol dm−3 solution of the weak base 35 Explain why a solution of ammonium chloride is acidic.
NH3(aq) from a burette to 25.0 cm3 of a
36 Write a balanced equation for the neutralisation of ethanoic acid
0.10 mol dm−3 solution of the strong acid
by ammonia solution.
HCl(aq).
334 12 Acid–base equilibria
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Titration of a weak acid with a weak base 12
In practice it is not usual to titrate a weak acid with a weak base. As shown 10
in Figure 12.13, the change of pH around the equivalence point is gradual
and not very marked if both the acid and base are weak. This means that it is 8
pH
hard to fix the end-point precisely. If the dissociation constants for the weak 6
acid and for the weak base are approximately equal (as is the case for ethanoic
4
acid and ammonia) then the salt formed at the equivalence point is neutral
and pH = 7 at this point. 2
0
Working with logarithms 0 5 10 15 20
Titre/cm3
25 30
There can be an advantage to working with a logarithmic form of the
Figure 12.13 The pH change on adding
equilibrium law when calculating the pH of a mixture of a weak acid and
0.10 mol dm−3 NH3(aq) from a burette
one of its salts during titrations.
to 25.0 cm3 of a 0.10 mol dm−3 solution
In general, for a weak acid HA: of CH3COOH(aq). Before the end-point
the curve is essentially the same as in
HA(aq) ⇋ H+(aq) + A−(aq)
Figure 12.11, while after the end-point it is
[H+ (aq)][A− (aq)] as in Figure 12.12. Note the resulting small
Ka = change of pH around the equivalence point.
[HA(aq)]
This rearranges to give:
[HA(aq)]
[H+(aq)] = K a ×
[A− (aq)]
Taking logs and substituting pH for −log [H+(aq)] and pK a for −log K a gives:
[A– (aq)]
pH = pK a + log
[HA(aq)]
Note the change of sign and the inversion of the log ratio. This follows because:
[A− (aq)] [HA(aq)]
– log = + log −
[HA(aq)] [A (aq)]
In a mixture of a weak acid and one of its salts, the weak acid is only slightly
ionised, while the salt is fully ionised, so it is often accurate enough to make Tip
the assumption that all the negative ions come from the salt present and
You are not required to use the
all the un-ionised molecules from the acid.
logarithmic form of the equilibrium
Hence: pH = pKa + log [salt] law. If you choose to do so, make sure
[acid] that you can derive it for yourself.
Also check that you understand the
This form of the equilibrium law cannot be used to calculate the pH of a solution
assumptions made when deriving this
of a weak acid on its own. However, it can help to explain the properties of
form of the law, so that you know when
acid–base indicators (Section 12.6) and to account for the behaviour of buffer
it applies.
solutions (Section 12.8).
12.5 Acid–base titrations 335
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12.6 Indicators
Acid–base indicators change colour when the pH changes. They signal
the end-point of a titration. No one indicator is right for all titrations and
the equilibrium law can help chemists to choose the indicator that gives
accurate results for a particular combination of acid and base.
The indicator chosen for a titration must change colour completely in the
pH range of the near vertical part of the pH curve (see Figures 12.10, 12.11
and 12.12). This is essential if the visible end-point is to correspond to the
equivalence point when exactly equal amounts of acid and base are mixed.
Table 12.2 gives some data for four common indicators. Note that each indicator
changes colour over a range of pH values, which differs from one indicator to
the next (Figures 12.14, 12.15 and 12.16). These indicators are themselves weak
acids or bases which change colour when they lose or gain protons.
Table 12.2 Properties of selected indicators (the Pearson Edexcel Data booklet includes
information about more indicators).
Indicator pKa Colour change HIn/In− pH range over which the
colour change occurs
Methyl orange 3.7 Red/yellow 3.2–4.4
Methyl red 5.1 Yellow/red 4.2–6.3
Bromothymol blue 7.0 Yellow/blue 6.0–7.6
Phenolphthalein 9.3 Colourless/pink 8.2–10.0
Tip
Methyl orange can be a difficult indicator to use because it is hard to spot the point
at which an orange colour marks the end-point. Sometimes a dye is mixed with the
indicator to produce ‘screened methyl orange’. This changes from purple to grey at the
end-point and then goes green with excess alkali. Some people find it much easier to
detect the end-point with the screened indicator.
Figure 12.14 The colours of screened methyl orange Figure 12.15 The colours of Figure 12.16 The colours of
indicator at pH 6 (left), pH 4 (middle) and pH 2 (right). phenolphthalein indicator at pH 7 bromothymol blue indicator at
This indicator includes a green dye to make the colour (left) and pH 11 (right). pH 5 (left) and pH 8 (right).
change easier to see.
336 12 Acid–base equilibria
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When added to a solution, an indicator gains or loses protons depending on
the pH of the solution. It is conventional to represent a weak acid indicator as
HIn, where In is a shorthand for the rest of molecule other than the ionisable
hydrogen atom. In water:
HIn(aq) ⇋ H+(aq) + In−(aq)
un-ionised indicator indicator after losing a proton
colour 1 colour 2
Note that an analyst only adds a drop or two of indicator during a titration. This
means that there is so little indicator that it cannot affect the pH of the mixture.
The pH is determined by the titration (as shown in Figures 12.10, 12.11 and
12.12). The position of the equilibrium for the ionisation of the indicator shifts
one way or the other as dictated by the changing pH of the solution in the
titration flask.
The pH range over which an indicator, HIn, changes colour is determined
by its strength as the acid (Figure 12.17). Typically the range is given roughly
by pK a ± 1. The logarithmic form of the equilibrium law derived at the end
of Section 12.5 shows why this is so. For an indicator it takes this form:
pH = pK a + log ( [In– (aq)]
[HIn (aq)] (
When pH = pK a, [HIn(aq)] = [In−(aq)] and the two different colours of the
indicator are present in equal amounts. The indicator is mid-way through its
colour change.
Add a few drops of acid and the pH falls. If the two colours of the indicator
are equally intense, it turns out that the human eye sees the characteristic acid
colour of the indicator clearly when [HIn] = 10 × [In−(aq)].
At this point pH = pK a + log 0.1 = pK a − 1 (since log 0.1 = −1).
Add a few drops of alkali and the pH rises. Similarly, the human eye
sees the characteristic alkaline colour of the indicator clearly when
[In−(aq)] = 10 × [HIn(aq)].
At this point: pH = pK a + log 10 = pK a + 1 (since log 10 = +1).
Figure 12.17 shows structures of methyl orange in acid and alkaline solutions.
In acid solution the added hydrogen ion (proton) localises two electrons to
form a covalent bond. In alkaline solution the removal of the hydrogen ion
allows the two electrons to join the other delocalised electrons (see Section
18.1.3). The change in the number of delocalised electrons causes a shift
in the peak of the wavelengths of light absorbed, so the colour changes and
the molecule acts as an indicator.
+ H+ +
– –
O3S N N N CH3 O3S N N N CH3
+
–H
CH3 H CH3
yellow red
Figure 12.17 The structures of methyl orange in acid (right) and alkaline solutions (left).
12.6 Indicators 337
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Test yourself
37 a) If HIn represents methyl red, what are the colours of HIn and In−?
b) Explain, qualitatively why methyl red is yellow when the pH = 3 but
is red at pH = 7.
38 a) Why is methyl orange an unsuitable indicator for the titration
illustrated by Figure 12.11?
b) Why is phenolphthalein an unsuitable indicator for the titration
illustrated by Figure 12.12?
c) From Table 12.2, choose the indicators that can be used to detect
the equivalence points of the titrations illustrated by Figures
12.10, 12.11 and 12.12.
d) Explain why it is not possible to use an indicator to give a sharp and
accurate end-point for the titration illustrated by Figure 12.13.
39 Suggest an explanation for the fact that the indicators shown in
Table 12.2 do not all change colour over a pH range of 2 units.
12.7 Neutralisation reactions
Chemists use the term ‘neutralisation’ to describe any reaction in which an
acid reacts with a base to form a salt, even when the pH does not equal 7 on
mixing equivalent amounts of the acid and the alkali.
The pH of salts
Mixing equal amounts (in moles) of hydrochloric acid with sodium
hydroxide produces a neutral solution of sodium chloride. Strong acids, such
as hydrochloric acid, and strong bases, such as sodium hydroxide, are fully
ionised in solution. The salt formed from the reaction of hydrochloric acid
and sodium hydroxide, sodium chloride, is also fully ionised. Writing ionic
equations for these examples shows that neutralisation is essentially a reaction
between aqueous hydrogen ions and hydroxide ions. This is supported by
the values for enthalpies of neutralisation – see the next part of this section.
H+(aq) + OH−(aq) ⇋ H2O(l)
The surprise is that ‘neutralisation reactions’ do not always produce neutral
solutions. ‘Neutralising’ a weak acid, such as ethanoic acid, with an equal
amount, in moles, of a strong base, such as sodium hydroxide, produces a
solution of sodium ethanoate, which is alkaline (see Figure 12.11).
‘Neutralising’ a weak base, such as ammonia, with an equal amount of the
strong acid hydrochloric acid produces a solution of ammonium chloride,
which is acidic (see Figure 12.12).
Where either the ‘parent acid’ or ‘parent base’ of a salt is weak, the salt
Figure 12.18 Raponzolo di roccia grows in
dissolves to give a solution which is not neutral (Figure 12.18). The ‘strong
the moist and shady crevices of limestone
parent’ in the partnership ‘wins’:
cliffs of the Italian Alps. Weathering of the
limestone keeps the pH high so that the ● weak acid/strong base – the salt is alkaline in solution
soil water is alkaline. ● strong acid/weak base – the salt is acidic in solution.
338 12 Acid–base equilibria
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Test yourself
Predict whether the salt formed on mixing equivalent amounts of
40
these acids and alkalis gives a solution with pH = 7, pH above 7
or pH below 7:
a) nitric acid and potassium hydroxide
b) chloric(i) acid and sodium hydroxide
c) hydrobromic acid and ammonia
d) propanoic acid and sodium hydroxide.
Enthalpy change of neutralisation
Strong acids and bases
The standard enthalpy change of neutralisation is the enthalpy change
for the reaction when an acid neutralises an alkali to form one mole of water. Key term
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) The standard enthalpy change of
∆n H 1 = −57.5 kJ mol−1 neutralisation is the enthalpy change
when the acid and alkali in the equation
The standard enthalpy change of neutralisation for dilute solutions of strong for the reaction neutralise each other
acid with strong base is always close to −57.5 kJ mol−1. The reason is that these under standard conditions to form one
acids and alkalis are fully ionised. So, in every instance, the reaction is the same: mole of water.
H+(aq) + OH−(aq) → H2O(l) ∆H 1 = −57.5 kJ mol−1
Enthalpy changes of neutralisation can be measured approximately by mixing
solutions of acids and alkalis in a calorimeter (Figure 12.19).
thermometer –10 to 50°C
Weak acids and bases
The standard enthalpy changes of neutralisation reactions involving
weak acids and weak bases are less negative than those for neutralisation polystyrene cup and lid
reactions between strong acids and bases. The standard enthalpy changes
for the neutralisation of ethanoic acid by sodium hydroxide, for example, is reaction mixture
−56.1 kJ mol−1.
This is partly because the weak acids and bases are not fully ionised at the start,
so that the neutralisations cannot be described simply as reactions between Figure 12.19 Apparatus for measuring the
aqueous hydrogen ions and aqueous hydroxide ions. Also, the solutions are enthalpy change of the neutralisation of an
not neutral at the equivalence point. acid by a base.
Example
50 cm3 of 1.0 mol dm−3 dilute nitric acid were mixed with 50 cm3 of
1.0 mol dm−3 dilute potassium hydroxide solution in an expanded
polystyrene cup. The temperature rise was 6.7 °C. Calculate the enthalpy
change of neutralisation for the reaction.
Notes on the method
Note that the total volume of solution on mixing is 100 cm3.
Assume that the density and specific heat capacity of the solutions is the
same as for pure water.
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The density of water is 1.0 g cm−3, so the mass of 100 cm3 water is 100 g.
The specific heat capacity of water is 4.18 J g−1 K−1.
A temperature change of 6.7 °C is the same as a change of 6.7 K on the
Kelvin scale.
The energy from the exothermic reaction is trapped in the system by
the expanded polystyrene, so it heats up the mixture.
Answer
The energy change = 4.18 J g−1 K−1 × 100 g × 6.7 K = 2800 J
50 dm3 × 1.0 mol dm−3 = 0.050 mol
Amount of acid neutralised =
1000
HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
− 2800 J = −56 000 J mol−1 = −56 kJ mol−1
Δ nH =
0.050 mol
Test yourself
41 Account for the discrepancy between the value calculated in the
worked example from experimental results and the expected value
of about −57.5 kJ mol−1.
42 Suggest an explanation for the difference in the values of ∆nH 1
for HCl/NaOH and CH3COOH/NaOH.
43 Here are three pairs of acids and bases which can react to form
salts: HBr/NaOH, HCl/NH3, CH3COOH/NH3.
Here are three values for the standard enthalpy change of
neutralisation:
● −50.4 kJ mol−1
● −53.4 kJ mol−1
● −57.6 kJ mol−1.
Write the equations for the three neutralisation reactions and match
them with the corresponding value of ∆nH 1.
12.8 Buffer solutions
Buffer solutions are mixtures of molecules and ions in solution which
help to keep the pH more or less constant when small quantities of acid
or alkali are added to a solution. Buffer solutions help to stabilise the pH
of blood, medicines, shampoos, water in swimming pools, and of many
other solutions in living things, domestic products and in the environment
(Figure 12.20).
Buffer mixtures are important in the food industry because many properties
of drinks and foods depend upon their being formulated to the correct pH.
Figure 12.20 Eye drops contain a buffer The quality of food can be affected significantly if the pH shifts too far from
solution to make sure that they do not the required value. Maintaining a low pH can help to prevent the growth of
irritate the sensitive surface of the eye. bacteria or fungi that spoil food or cause food poisoning.
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Buffers are also important in living organisms. The pH of blood, for example,
is closely controlled by buffers within the narrow range 7.35 to 7.45. Chemists
Key term
use buffers when they want to investigate chemical reactions at a fixed pH.
A buffer solution is a mixture of
Buffers are equilibrium systems which illustrate the practical importance of molecules and ions in solution which
the equilibrium law. A typical buffer mixture consists of a solution of a weak help to keep the pH more or less
acid and one of its salts; for example, a mixture of ethanoic acid and sodium constant. A buffer solution cannot
ethanoate (Figure 12.21). There must be plenty of both the acid and its salt. prevent pH changes, but it evens out
– +
the large changes in pH which can
CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq)
happen without a buffer when small
acid molecules base ions – with the
are a reservoir of capacity to accept
amounts of acid or alkali are added to
stays roughly an aqueous solution. A typical buffer
H+ ions H+ ions
constant so the pH
mixture consists of a solution of a weak
plenty of weak acid to plenty of the ions from hardly changes
+ acid and one of its salts.
supply more H ions if the salt able to
alkali is added combine with H+ ions if
acid is added
Figure 12.21 The action of a buffer solution.
Le Chatelier’s principle provides a qualitative interpretation of the buffering
action. Adding a little strong acid temporarily increases the concentration Test yourself
of H+(aq) so at that instant the system is not equilibrium. So the reaction
mixture shifts towards the left-hand side of the equation to reduce the 44 Explain why a weak acid
hydrogen ion concentration, thus counteracting the change and establishing on its own cannot make a
a new equilibrium. Conversely, the effect of adding a little strong alkali buffer solution, but a mixture
temporarily decreases the concentration of H+(aq) so the reaction mixture of a weak acid and one of its
shifts to the right of the equation to replace some, though not all, of the salts can.
hydrogen ions that have been neutralised.
The pH of buffer solutions
Tip
By choosing the right weak acid, it is possible to prepare buffers at any pH
value throughout the pH scale. If the concentrations of the weak acid and its The theory of acid–base indicators
salt are the same, then the pH of the buffer is equal to pK a for the acid. The and the theory of buffer solutions is
pH of a buffer mixture can be calculated with the help of the equilibrium law. essentially the same. The difference is
that a large amount of buffer mixture is
[H+(aq)[A – (aq)]
Ka = added to dictate the pH of a solution,
[HA(aq)]
whereas the drop or two of an indicator
[A – (aq)] in a titration flask is too little to affect
This rearranges to give: [HA(aq)] = K a ×
[H+ (aq)] the pH. An indicator follows the pH
So the equilibrium law makes it possible to calculate the pH of a buffer changes dictated by the mixture of acid
solution made from a mixture of a weak acid and its conjugate base. and alkali during the titration.
In a mixture of a weak acid and its salt, the weak acid is only slightly ionised while
the salt is fully ionised. This means that it is often accurate enough to assume that:
● all the molecules HA come from the added acid
● all the negative ions, A−(aq), come from the added salt.
So the calculation of the hydrogen ion concentration of a buffer solution
can be based on the formula:
[acid]
[H+(aq)] = Ka ×
[salt]
Alternatively, you can use the logarithmic form of this relationship
(see Section 12.5).
12.8 Buffer solutions 341
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Activity
Blood buffers
In a healthy person the pH of blood lies within a narrow range Two major organs help to control the total amounts of carbonic
(7.35—7.45). Chemical reactions in cells tend to upset the acid and hydrogencarbonate ions in the blood. The lungs
normal pH. Respiration, for example, continuously produces remove excess carbon dioxide from the blood (Figure 12.22)
carbon dioxide. The carbon dioxide diffuses into the blood and the kidneys remove excess hydrogencarbonate ions.
where it is mainly in the form of carbonic acid. However,
The brain responds to the level of carbon dioxide in the blood.
the blood pH stays constant because it is stabilised by
During exercise, for example, the brain speeds up the rate of
buffer solutions, in particular by the buffer system based
breathing.
on the equilibrium between carbon dioxide, water and
hydrogencarbonate ions. This is the carbonic The consequences can be fatal if the blood pH moves
acid–hydrogencarbonate buffer. outside the normal range. Patients who have been badly
burned or suffered other serious injuries are treated quickly
Proteins in blood, including haemoglobin, can also contribute
with a drip into a vein. One of the purposes of an intravenous
to the buffering action of blood pH. This is because the
drip is to help maintain the pH of the blood close to its
molecules contain both acidic and basic functional groups
normal value.
(see Section 18.2.7).
1 Write an equation to show aqueous carbon dioxide reacting
with water to form hydrogen ions and hydrogencarbonate ions.
2 Explain why breathing faster and more deeply tends to raise
the blood pH.
3 a) Suggest two reasons why the blood pH tends to fall during
strenuous exercise.
b) Why do people breathe faster and more deeply when
running?
4 a) Write the expression for Ka for the equilibrium
between carbon dioxide, water, hydrogen ions and
hydrogencarbonate ions
b) In a sample of blood, the concentration of hydrogencarbonate
ions = 2.50 × 10−2 mol dm−3. The concentration of aqueous
carbon dioxide = 1.25 × 10−3 mol dm−3. The value of
Ka = 4.5 × 10−7 mol dm−3. Use this information to calculate:
i) the hydrogen ion concentration in the blood
ii) the pH of the blood.
c) What can you conclude about the person who gave the
blood sample?
5 Explain why a mixture of carbon dioxide, water and
hydrogencarbonate ions can act as a buffer solution.
6 Why are blood buffers on their own unable to maintain the
correct blood pH for any length of time?
7 Suggest reasons why people may need treatment to adjust
Figure 12.22 The lungs have a vital part to play in maintaining their blood pH if they have been rescued after breathing
the pH of the blood. thick smoke during a fire.
Diluting a buffer solution with water does not change the ratio of the
concentrations of the salt and acid, so the pH does not change, unless
the dilution is so great that the assumptions used to arrive at this formula
break down.
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Buffer solutions form during the titration of a weak acid with a strong base.
This is illustrated by Figure 12.23. Once some of the base has been added
the pH does not change by much until the titration begins to approach the buffer
end-point. pH 7
range
pH = pKa
halfway to the
end-point
Example
0
0 25
What is the pH of a buffer solution containing 0.40 mol dm−3 methanoic
Added alkali/cm3
acid and 1.00 mol dm−3 sodium methanoate?
Figure 12.23 In the buffering range, the
Notes on the method pH changes little on adding substantial
Look up the value of Ka in a table of data. Ka of methanoic acid is volumes of strong alkali. Over this range the
1.6 × 10−4 mol dm−3. flask contains significant amounts of both
the acid and the salt formed from the acid.
Make the assumptions that all the molecules of methanoic acid come
from the added acid, and that all the methanoate ions come from the
added salt.
Answer
From the information in the question:
[acid] = 0.40 mol dm−3
[salt] = 1.00 mol dm−3
Substituting in the formula gives:
[acid]
[H+(aq)] = Ka ×
[salt]
0.40 mol dm−3
= 1.6 × 10 −4 mol dm−3 ×
1.00 mol dm−3
[H+(aq)] = 6.4 × 10 −5 mol dm−3
pH = −log [H+(aq)] = −log [6.4 × 10 −5] = 4.2
Test yourself
45 Calculate the pH of these buffer mixtures.
a) A solution containing equal amounts in moles of H2PO4−(aq)
and HPO42−(aq). Ka for the dihydrogenphosphate(v) ion is
6.3 × 10 −7 mol dm−3.
b) A solution containing 12.2 g benzoic acid (C6H5COOH) and 7.2 g
of sodium benzoate in 250 cm3 solution. Ka for benzoic acid is
6.3 × 10 −5 mol dm−3.
c) A solution containing 12.2 g benzoic acid (C6H5COOH) and 7.2 g
of sodium benzoate in 1000 cm3 solution.
46
What must be the ratio of the concentrations of the ethanoic acid
molecules and ethanoate ions in a buffer solution with pH = 5.4 if
Ka = 1.7 × 10 −5 mol dm−3 for ethanoic acid?
12.8 Buffer solutions 343
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Core practical 9
Finding the Ka value for a weak acid
A good method to determine Ka for a weak acid is to measure 14
the pH of the solution in the flask during a titration of the
acid with a dilute solution of a strong alkali such as sodium 12
hydroxide. The procedure can be related to the steps for
determining equilibrium constants described in Section 11.2. 10
Procedure
Step 1: Mix measured quantities of chemicals and allow the 8
mixture to come to equilibrium.
pH
6
Sodium hydroxide solution is added from a burette to a
measured volume of the weak acid solution in a flask. The
added alkali neutralises some of the acid and turns it into 4
its sodium salt. After each addition of alkali there is a new
equilibrium mixture in the flask. The reaction is fast, so after 2
each addition the mixture reaches an equilibrium state instantly.
0
Step 2: Analyse the mixture to find the equilibrium 0 5 10 15 20 25 30
concentration of one of the reactants or products. Volume of NaOH(aq)/cm3
In this case the hydrogen ion concentration in the solution can Figure 12.24 Plot of pH against titre for a titration of
be determined easily using a pH meter. chloroethanoic acid with sodium hydroxide.
Step 3: Use the equation for the reaction and the equilibrium 1 Why is it possible to determine the equilibrium concentrations
law to find the value of the equilibrium constant. in acid–base equilibria without upsetting the position of
In this example the value of Ka can be determined by taking equilibrium?
readings from the graph. 2 a) Show, with the help of values read from the graph in Figure
12.24, that the flask contained a series of buffer solutions
This procedure was used to determine the acid dissociation during the titration.
constant for chloroethanoic acid, CH2ClCOOH. b) Write the equation for the reversible reaction in the buffer
Results solutions.
Figure 12.24 shows the results of plotting pH against titre for 3 a) Take and note down the readings from the graph that you
a titration of 25.0 cm3 of a roughly 0.1 mol dm−3 solution of need to work out the value of Ka for chloroethanoic acid.
chloroethanoic acid, CH2ClCOOH, with 0.10 mol dm−3 sodium b) Calculate the value for Ka, showing your working. State any
hydroxide solution. assumptions that you make in the calculation.
4 Why is it not necessary to know the concentration of the acid or
the alkali precisely when this method is used to measure Ka?
Tip
Refer to Practical skills sheet 16, ‘Finding the Ka value
for a weak acid’, which you can access online at www.
[Link]/EdexcelChemistry.
344 12 Acid–base equilibria
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Chapter summary
l A pH meter can be used to monitor the changes
Chapter 12 Acid–base equilibria
to the hydrogen ion concentration during an
l Strong acids are fully dissociated in aqueous acid-base titration. This gives rise to characteristic
solution. Weak acids are only slightly dissociated. titration curves for strong acid–strong base, strong
l According to the Brønsted–Lowry theory, acids
acid–weak base, weak acid–strong base and weak
are proton donors while bases are molecules or ions acid–weak base titrations.
that can accept a hydrogen ion (proton) from an l Titration curves can be used to explain the choice
acid. Bases are proton acceptors. of indicator for particular titrations.
l The term conjugate acid–base pair is used to
l Acid–base indicators are weak acids that
describe a pair of molecules or ions that can be change colour over a pH range roughly given by
converted from one to the other by the gain or loss pK a ± 1. The indicator chosen for a titration must
of a proton. change colour completely in the pH range of the
l The pH scale is used to measure the concentration
near vertical part of the titration curve.
of hydrogen ions in aqueous solutions. l The standard enthalpy change of neutralisation
pH = −log10[H+(aq)]. This expression can be used is the enthalpy change when the acid and alkali
to calculate the pH of strong acids that are fully in the equation for the reaction neutralise each
ionised in solution. other under standard conditions to form one mole
l The hydrogen ion concentration of a solution
of water.
can be calculated from pH using the expression l The standard enthalpy changes of neutralisation
[H+(aq)] = 10−pH. for dilute solutions of strong acids with strong
l Kw is the ionic product for water. It is the
bases all have approximately the same value. The
equilibrium constant for the ionisation of water. standard enthalpy changes of neutralisation for
Kw = [H+(aq)][OH−(aq)]. reactions involving weak acids or weak bases are
l The expression for Kw can be used to calculate the
less negative.
pH of an aqueous solution of a strong base from its l A buffer solution helps to keep the pH of a solution
concentration. more or less constant. This can be explained in
l pKw is defined as −log Kw.
terms of the equilibrium law.
l The relative strength of weak acids can be
l The pH of a buffer solution made from a weak acid
compared using equilibrium constants for the and one of its salts can be calculated using the value
ionisation of the acids in aqueous solution. For a of K a for the acid. Simplifying assumptions make
weak acid HA, the acid dissociation constant takes the calculation easier.
[H+(aq)] [A−(aq)] l During the titration of a weak acid with a strong
the form: Ka = base, the pH = pK a when the titre is half that
[HA(aq)]
needed to reach the equivalence point. The flask
l pK a is defined as −log K a.
contains a series of buffer solutions for values of the
l K a for a weak acid can be calculated given the pH
titre on either side of this point.
and the concentration of the solution.
l Blood pH is kept within a narrow range thanks to
l K a can be used to calculate the hydrogen ion
buffer solutions in the blood. The buffer solution
concentration and pH of a weak acid.
based on carbonic acid and hydrogencarbonate ions
l The calculations using K a can often be simplified
is particularly important.
by making assumptions about the equilibrium
concentrations.
Chapter summary 345
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Exam practice questions
1 a) Write an equation for the reaction of 4 a) Give examples to explain the difference
ammonia with water and identify the two between a strong acid and a weak acid. (4)
conjugate acid–base pairs in the solution. (3) b) At 298 K, calculate the pH of:
b) Explain why a solution of sodium chloride i) 0.010 mol dm−3 HNO3(aq) (1)
in water is neutral but a solution of sodium ii) 0.010 mol dm−3 KOH(aq). (2)
ethanoate is alkaline. (4) c) Butanoic acid, C3H7COOH, has an
c) Explain why acid dissociation constants are acid dissociation constant, Ka, of
used to compare the strengths of weak acids 1.5 × 10−5 mol dm−3 at 298 K.
and not the pH of the acids in aqueous i) Calculate the pH of a 0.020 mol dm−3
solution. (3) solution of butanoic acid at this
temperature. (2)
2 a) Write an equation for the reaction which
ii) Draw a sketch graph to show the
occurs when a weak acid HX is added to
change of pH when 50 cm3 of
water. (1)
0.020 mol dm−3 NaOH(aq) is
b) Write an expression for the acid dissociation
added to 25 cm3 of 0.020 mol dm−3
constant of the weak acid HX. (1)
butanoic acid. (4)
c) The ionisation of HX in aqueous solution is
iii) Identify from the table the indicator that
endothermic. Predict the effect, if any, of:
would be most suitable for detecting
i) an increase in temperature on the
the end-point of a titration between
value of the acid dissociation constant
butanoic acid and sodium hydroxide.
for HX (1)
Give your reasons. (2)
ii) an increase in temperature on the pH
of an aqueous solution of the weak acid Indicator Colour change pH range over
HX (1) acid/alkaline which colour
iii) a decrease in concentration of the acid change occurs
HX on the value of its acid dissociation Thymol blue Red/yellow 1.2–2.8
constant. (1) Congo red Violet/red 3.0–5.0
3 a) Explain what is meant by the term ‘ionic Thymolphthalein Colourless/blue 8.3–10.6
product of water’. (2)
b) i) Show that the value of the ionic 5 a) Describe and explain the use of buffer
product of water is 1 × 10−14 based on solutions with the help of examples. (6)
the fact that for pure water pH = 7 at b) i) Calculate the pH of a buffer solution
298 K. (2) in which the concentration of ethanoic
ii) At 303 K the value of the ionic product acid is 0.080 mol dm−3 and the
of water is 1.47 × 10−14. Deduce from concentration of sodium ethanoate is
the difference in values between 298 K 0.040 mol dm−3.
and 303 K whether the ionisation of Ka for ethanoic acid is
water is exothermic or endothermic. (2) 1.7 × 10−5 mol dm−3. (2)
c) i) Calculate the pH of a 0.300 mol dm−3 ii) Calculate the new pH value if
solution of NaOH at 298 K. (2) 0.020 mol of NaOH is dissolved in
ii) Calculate the pH of a solution formed 1 dm3 of the buffer solution in (i). (2)
by mixing 25.0 cm3 of a 0.300 mol dm−3 iii) Calculate the pH of a solution of
solution of NaOH with 225 cm3 of 0.020 mol of NaOH in 1 dm3 of
water at 298 K. (3) water. (2)
iii) Calculate the pH of a solution formed iv) Comment on the effectiveness of the
by mixing 25.0 cm3 of 0.300 mol dm−3 buffer solution. (1)
NaOH with 75.0 cm3 of 0.200 mol dm−3
hydrochloric acid at 298 K. (4)
346
12 Acid–base equilibria
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6 For a solution containing 0.050 mol dm−3 8 Introducing halogen atoms into the structure
chloric(i) acid (HClO) and 0.050 mol dm−3 of carboxylic acids can have a marked effect
sodium chlorate(i), the pH = 7.43 at 298 K. on their acid strength. This is illustrated by
a) i) Write an equation for the ionisation the values in the table.
of chloric(i) acid. (1)
ii) Write an expression for the acid Acid Ka/mol dm−3 pKa
dissociation constant of chloric(i) Ethanoic acid 1.7 × 10 −5 4.8
acid. (1) Fluoroethanoic acid 2.2 × 10 −3 2.7
b) Calculate the value of Ka for chloric(i) Chloroethanoic acid 1.3 × 10 −3 2.9
acid, showing your working. (4) Iodoethanoic acid 7.6 × 10 −4 3.1
7 The graph shows the changes in pH during the Dichloroethanoic acid 5.0 × 10 −2 1.3
titration of 10 cm3 of a monobasic acid with a Trichloroethanoic acid 2.3 × 10 −1 0.7
0.010 mol dm−3 solution of sodium hydroxide. Butanoic acid 1.5 × 10 −5 4.8
2-Chlorobutanoic acid 1.4 × 10 −3 2.8
11 3-Chlorobutanoic acid 8.7 × 10 −5 4.0
10 4-Bromobutanoic acid 3.0 × 10 −5 4.5
9
a) Calculate the pH of a 0.10 mol dm−3
8 solution of:
7 i) butanoic acid (2)
ii) trichloroethanoic acid. (2)
pH
6
5 b) i) Determine the pattern in the acid
4 strength of fluoro-, chloro- and
3 iodo-ethanoic acids when compared
2 with the value for ethanoic acid.(1)
1
ii) Give a reason to explain the
pattern.(4)
0
0 5 10 15 c) i) Determine the pattern in the acid
Volume of alkali added/cm3 strength of chloro-, dichloro- and
trichloro-ethanoic acids when
a) Answer these questions, giving your reasons. compared with the value for
i) Calculate the concentration of the ethanoic acid. (1)
acid at the start.(2) ii) Assess whether or not the pattern
ii) Determine the pH of the acid before is consistent with your suggested
any alkali was added. (1) explanation in (b)(ii). (2)
iii) Show that the monobasic acid was a d) i) Determine the pattern in the acid
weak acid. (1) strength of the chlorinated butanoic
b) Use your answers to part (a) to calculate acids when compared with the
a value of Ka for the acid. (2) value for butanoic acid.(1)
c) i) State the range of titration readings ii) Suggest an explanation for the
over which there was an effective buffer pattern. (3)
solution in the flask. Explain your
answer. (3) 9 The graph on page 348 shows the results from
ii) Use a value read from the buffer an experiment in which measured volumes
region to determine a value for the of a 2.0 mol dm−3 solution of an acid,
Ka of the acid. (2) HnX, were mixed with measured volumes of
d) i) State the pH of the mixture in 2.0 mol dm−3 NaOH(aq). The temperatures
the flask at the equivalence point.(1) of the two solutions were the same before
ii) Explain the pH value given in your mixing. The temperature rise after mixing was
answer to (i). (2) measured and recorded.
347
Exam practice questions
469983_12_Chem_Y1-2_318-[Link] 347 13/04/19 10:07 PM
Temperature rise/K 20
15
10
0
Volume of 2.0 mol dm–3 acid 100 90 80 70 60 50 40 30 20 10 0 cm3
Volume of 2.0 mol dm–3 alkali 0 10 20 30 40 50 60 70 80 90 100 cm3
a) Give a suitable container for mixing d) Consider the mixture of acid and alkali which
the two solutions. (1) would react to exactly neutralise each other.
b)* Give reasons to account for the shape i) Use the graph to determine the
of the plot on the graph.(6) temperature rise on making this
c) i) Determine the volumes of the acid mixture.(1)
and the alkali which would react to ii) Assuming that the mixed solution
exactly neutralise each other.(1) has a specific heat capacity of
ii) Determine the value of n in the 4.18 J g−1 K−1 and a density of
formula of the acid using your answer 1.0 g cm−3, calculate the energy change
to (i).(2) from the reaction in this mixture.(2)
iii) Calculate the enthalpy change of
neutralisation per mole of the acid. (2)
iv) Use your answer to (iii) to assess
whether HnX is a strong or a
weak acid. (2)
348
12 Acid–base equilibria
469983_12_Chem_Y1-2_318-[Link] 348 13/04/19 10:07 PM
Lattice energy
13.1
13.1.1 Ionic bonding and structures
Compounds of metals with non-metals, such as sodium chloride and magnesium
oxide, are composed of ions. When such compounds form, the metal atoms
lose electrons and form positive ions. At the same time, the non-metal atoms
gain electrons and form negative ions. For example, when sodium reacts with
chlorine (Figure 13.1.1), each sodium atom loses its one outer electron forming
a sodium ion, Na+. Chlorine atoms gain these electrons and form chloride
ions, Cl− (Figure 13.1.2).
Tip
The first two sections of this chapter remind you of the model of ionic giant structures
that you learned about in Year 1 of your chemistry course. In these sections, the ‘Test
yourself’ questions help you to check your understanding of ionic compounds and
enthalpy changes. From Section 13.1.3, the chapter goes on to show that this model
can be tested quantitatively by studying the energy changes involved in the formation
of crystals held together by ionic bonding.
Figure 13.1.1 Hot sodium reacting with
chlorine. In Figure 13.1.2, the electrons of one element are shown as dots and those of
the other reactants are shown as crosses. Diagrams of this kind provide a useful
balance sheet for keeping track of the electrons when ionic compounds form.
Na Cl
sodium atom, Na chlorine atom, Cl
2,8,1 2,8,7
+ –
Na Cl
sodium ion, Na+ chloride ion, Cli–
2,8 2,8,8
Figure 13.1.2 The formation of ions in sodium chloride when sodium reacts with chlorine.
13.1.1 Ionic bonding and structures 349
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Very often it is sufficient to show simply the outer shell electrons in dot-
Key terms and-cross diagrams and two of these simplified dot-and-cross diagrams
are shown in Figure 13.1.3.
Dot-and-cross diagrams show the way
in which the outer electrons of atoms are –
transferred, or shared, when chemical Na + Cl Na+ + Cl
bonds form. Despite the use of dots and sodium atom chlorine atom sodium ion chloride ion
crosses for the electrons coming from (2, 8, 1) (2, 8, 7) (2, 8) (2, 8, 8)
different atoms, all the electrons are the
– –
same. Once bonds form it is impossible to Ca + F F Ca2+ + F F
say which electron came from which atom. calcium atom two fluorine atoms calcium ion two fluoride ions
(2, 8, 8, 2) (2, 7) (2, 8, 8) (2, 8)
A lattice is a regular three-dimensional
arrangement of atoms or ions in a crystal. Figure 13.1.3 Dot-and-cross diagrams for the formation of sodium chloride and
calcium fluoride showing only the electrons in the outer shells of the reactants
Ionic bonding refers to the strong
and products.
electrostatic forces between oppositely
charged ions in a lattice.
Ionic crystals
Ionic crystals consist of giant lattices containing billions of positive and
negative ions packed together in a regular pattern (Figure 13.1.4). In
the lattice, each Na+ ion is surrounded by Cl− ions, and each Cl− ion is
surrounded by Na+ ions. The oppositely charged ions attract each other.
At the same time, the chloride ions repel other chloride ions and sodium
ions repel other sodium ions, but overall there are strong net electrostatic
attractions between ions in all directions throughout the lattice. These
electrostatic attractions between oppositely charged ions are described as
ionic bonding.
Many other compounds have the same lattice structure as sodium chloride
including the chlorides, bromides and iodides of lithium, sodium and
potassium and the oxides and sulfides of magnesium, calcium and strontium.
Figure 13.1.4 A three-dimensional model
of the structure of sodium chloride. The
smaller red spheres represent Na+ ions. Test yourself
The larger green spheres represent Cl−.
1 Draw dot-and-cross diagrams, similar to those in Figure 13.1.3 for:
Tip a) potassium oxide
Electrostatic forces operate between b) magnesium sulfide.
the ions in a crystal. Oppositely charged 2 Why do metals form positive ions, whereas non-metals form negative
ions attract each other while ions with ions?
like charges repel each other. The size of 3 Why do you think the melting temperature of magnesium oxide
the electrostatic force, F, between two (2852 °C) is so much higher than that of sodium fluoride (993 °C)?
charges is given by the equation:
4 a) Why do ionic compounds conduct electricity when molten but not
Q ×Q
F∝ 1 2 2 when solid?
d
b) Write equations for the reactions at the cathode and anode during
● The larger the charges, Q1 and Q 2,
electrolysis of molten magnesium chloride.
the stronger the force.
● The greater the distance, d, between
5 Why are chloride ions larger than sodium ions as shown in Figure 13.1.4?
the charges, the smaller the force and 6 Why are the electrostatic attractions between ions with opposite
this has a big effect because it is the charges in an ionic lattice overall greater than the repulsive forces
square of the distance that matters. between ions with the same charge?
350 13.1 Lattice energy
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13.1.2 Energy changes and ionic
bonding
Standard enthalpy changes Reactants: Na(s) + 12 Cl2(g)
When sodium reacts with chlorine, a very exothermic reaction occurs and
energy is given out to the surroundings. As the product, sodium chloride, cools
(energy content)
down to room temperature, the system loses energy to its surroundings. This ∆ f H = –411 kJ mol–1
Enthalpy
can be represented by an energy level diagram for the reaction (Figure 13.1.5).
In order to compare energy changes fairly and consistently it is important
+ –
to make thermochemical measurements under the same conditions. The Product: Na Cl (s)
conditions chosen for comparing enthalpy changes and other thermochemical
measurements are called standard conditions. These standard conditions are:
● a temperature of 25 °C (298 K) Course of reaction
● a pressure of 1 × 105 Pa = 100 kPa (this is very close to standard atmospheric Figure 13.1.5 An energy level diagram for
pressure at sea level, which is 101.3 kPa) the formation of sodium chloride.
● all reactants and products in their standard (stable) states at 25 °C and
1 atmosphere pressure
● any solutions at a concentration of 1 mol dm .
−3
The symbol for these standard enthalpy changes is Δ H 1, and Δf H 1 for
standard enthalpy changes of formation.
The enthalpy change shown in Figure 13.1.5 relates to the formation of
one mole of sodium chloride from its elements sodium and chlorine. If the
measurements have been made at 25 °C (298 K) and 1 atmosphere pressure
the result is described at the standard enthalpy change of formation of Tip
sodium chloride. This can be written either as:
1
The superscript sign in ΔH1 shows
Na(s) + 2 Cl 2(g) → Na+Cl−(s) Δf H 1 = −411 kJ mol−1 that the value quoted is for standard
or as: Δf H 1[NaCl(s)] = −411 kJ mol−1 conditions. The symbol is pronounced
‘delta H standard’.
Key term
The standard enthalpy change of formation of a compound, Δ f H1, is the enthalpy
change when one mole of the compound forms from its elements under standard
conditions with the elements and the compound in their standard (stable) states.
Test yourself
7 Why does Δ f H1 = 0 kJ mol−1 for an element?
8 Why are values for the standard enthalpy changes of formation of
compounds containing carbon based on graphite and not diamond?
9 Write an equation for the reaction for which the enthalpy change is the
standard enthalpy change of formation of calcium iodide.
13.1.2 Energy changes and ionic bonding 351
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13.1.3 Enthalpy changes when
ions form
Figure 13.1.2 shows the formation of sodium chloride from its elements,
but it simplifies the process in many ways. As far as sodium is concerned,
Figure 13.1.2 ignores the following facts:
● Sodium starts as a giant lattice of metal atoms.
● Energy is required to separate the sodium atoms in the giant lattice to
produce gaseous atoms. The energy change for this process is the standard
enthalpy change of atomisation of sodium.
● Energy is also required to remove one electron from each gaseous sodium
atom to form positive sodium ions, Na+. This is the first ionisation
energy of sodium.
As far as chlorine is concerned, Figure 13.1.2 ignores the following facts:
● Chlorine consists of Cl 2 molecules.
● Energy is required to break the bonds between Cl atoms in the Cl 2
molecules and form separate gaseous Cl atoms. The energy change, for
each mole of chlorine atoms formed, is the standard enthalpy change of
atomisation of chlorine.
● An energy change also occurs when one electron is added to each gaseous
Cl atom forming chloride ions, Cl−. The energy change for this process is
called the first electron affinity of chlorine.
Finally, and equally importantly, Figure 13.1.2 ignores the fact that energy
is given out when gaseous Na+ and Cl− ions come together forming a giant
ionic lattice of sodium chloride, Na+Cl−(s). The energy change for this
process is called the lattice energy of sodium chloride.
Tip Key terms
Note that the standard enthalpy of
The standard enthalpy change of atomisation of an element is the energy change
atomisation of chlorine is per mole of
needed to produce one mole of gaseous atoms of the element.
atoms formed – not per mole of molecules
atomised. So the standard enthalpy For sodium this is:
change of atomisation of chlorine is half Na(s) → Na(g) ΔatH 1[Na(s)] = +107 kJ mol−1
the size of the Cl—Cl bond energy.
And for chlorine this is:
1 1
Cl (g)
2 2
→ Cl(g) ΔatH 1[ 2 Cl2(g)] = +122 kJ mol−1
The first ionisation energy of an element is the energy needed to remove one electron
from each atom in one mole of gaseous atoms of the element under standard
conditions.
For sodium this is:
Na(g) → Na+(g) + e− 1st IE[Na(g)] = +496 kJ mol−1
352 13.1 Lattice energy
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Successive ionisation energies for the same element measure the energy needed to
remove a second, third, fourth electron, and so on. For example, the third ionisation
energy of sodium relates to the process:
Na2+(g) → Na3+(g) + e−
The first electron affinity of an element is the energy change when each atom in one
mole of gaseous atoms gains one electron to form one mole of gaseous ions with a
single negative charge.
The following two equations define the first and second electron affinities for oxygen:
O(g) + e− → O−(g) 1st EA = −141 kJ mol−1
O−(g) + e− → O2−(g) 2nd EA = +798 kJ mol−1
The gain of the first electron is exothermic, but adding a second electron to a
negatively charged ion is endothermic.
Lattice energies
The lattice energy of a compound is defined as the energy change when
one mole of an ionic compound is formed from free gaseous ions. For sodium
chloride, this is summarised by the equation:
Na+(g) + Cl−(g) → Na+Cl−(s) ΔlattH 1[NaCl(s)] = −787 kJ mol−1
This is the lattice energy for the process shown diagrammatically in Figure
13.1.6.
Lattice energies are important because they can be used as a measure of the
strength of the ionic bonding in different compounds.
The strength of ionic bonds, measured as lattice energies in kJ mol−1, arises
+
from the energy given out as billions upon billions of positive and negative Na Cl–
ions come together to form a crystal lattice. Figure 13.1.6 Lattice energy is the energy
The overall force of attraction between the ions is stronger and this results in that would be given out to the surroundings
a more exothermic lattice energy if: (red arrows) if one mole of an ionic
compound could be formed directly from
● the charges on the ions are large free gaseous ions coming together (black
● the ionic radii are small, allowing the ions to get closer to each other. arrows) and arranging themselves into a
It is important to distinguish between the lattice energy of an ionic compound crystal lattice.
and its standard enthalpy change of formation. The lattice energy relates to the
formation of one mole of a compound from its free gaseous ions, whereas the
standard enthalpy change of formation relates to the formation of one mole of
the compound from its elements in their stable states under standard conditions.
During the early part of the twentieth century, scientists found ways in which
to measure enthalpy changes of formation and atomisation, ionisation energies Key term
and electron affinities of various elements. This led two German scientists,
The lattice energy of an ionic
Max Born (1882–1970) and Fritz Haber (1868–1934), to analyse the energy
compound is the energy change when
changes in the formation of different ionic compounds. Their work resulted
one mole of the compound forms from
in Born–Haber cycles, which are thermochemical cycles for calculating lattice
free gaseous ions.
energies and for investigating the stability and bonding in ionic compounds.
13.1.3 Enthalpy changes when ions form 353
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Test yourself
10 Write equations for the processes with the 13 Why is the first electron affinity of chlorine
following energy changes: exothermic, but the second electron affinity
a) the standard enthalpy change of formation of endothermic?
magnesium bromide 14 Why are ionisation energies always endothermic?
b) the lattice energy of magnesium bromide 15 Why is the second ionisation energy of any
c) the second ionisation energy of magnesium element more endothermic than the first
ionisation energy?
d) the enthalpy change of atomisation of
magnesium 16 Why does the lattice energy of lithium fluoride
indicate that the ionic bonding in lithium fluoride
e) the first electron affinity of bromine.
is stronger than that in sodium chloride?
11 Draw an energy level diagram for the formation of
ΔlattH1 [LiF(s)] = −1031 kJ mol−1;
aluminium oxide.
ΔfH1 [Al2O3(s)] = −1676 kJ mol−1 ΔlattH1 [NaCl(s)] = −780 kJ mol−1
12 Why should water in equations to represent
standard enthalpy changes be shown as H2O(l)
not H2O(g)?
13.1.4 Born–Haber cycles
Born–Haber cycles are an application of Hess’s law (Section 8.5). They make
it possible to calculate lattice energies from other quantities that can be
measured. They also enable chemists to test the ionic model of bonding in
different substances.
A Born–Haber cycle identifies all the energy changes which contribute to
the standard enthalpy change of formation of a compound.
These overall changes, shown in Figure 13.1.7, involve:
●
the energy required to create free gaseous ions by atomising and then
ionising the elements
● t he energy given out (the lattice energy) when the ions come together to
form a crystal lattice.
Figure 13.1.7 The overall structure of a gaseous ions
Born–Haber cycle.
total enthalpy
change for atomising
and ionising the
two elements
lattice
energy
metal element and
non-metal element
enthalpy change
of formation of
the compound compound
354 13.1 Lattice energy
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A Born–Haber cycle is often set out as an energy level diagram. All the
processes in the cycle can be determined from experimental data except the
lattice energy. So, by using Hess’s law, it is possible to calculate the lattice
energy. Figure 13.1.8 shows the Born–Haber cycle for sodium oxide, Na 2O.
Sum of the first and second 2Na+(g) + O2–(g) Figure 13.1.8 The Born–Haber cycle for
electron affinities of O(g) sodium oxide.
1st EA + 2nd EA = –141 + 798
Eaff.1 + Eaff.2 = +657 kJ mol–1 + –
2Na (g) + 2e + O(g)
Atomisation to form 1 mol O(g) + –
2Na (g) + 2e + 12 O2(g)
∆ atH = +249 kJ mol–1
Ionisation of 2 mol Na(g)
2 × 1st IE = +992 kJ mol–1 1
Lattice energy
2Na(g) + 2 O2(g) sodium oxide,
∆lattH [Na2O(s)] = ?
Atomisation to form 2 mol Na(g)
2Na(s) + 12 O2(g)
2 × ∆ at H = +214 kJ mol–1
Standard enthalpy change
of formation of Na2O(s)
∆ f H [Na2O(s)] = –414 kJ mol–1
Na2O(s)
Starting with the elements sodium and oxygen, the measured value for the
standard enthalpy change of formation of sodium oxide has been written
beside a downwards arrow on the cycle, showing that it is exothermic.
Above that, the terms and values for the atomisation and then ionisation
of sodium are written beside arrows that point upwards as endothermic
processes.
Notice also that the amount of sodium required is two moles because there
are two moles of sodium in one mole of sodium oxide.
The terms and values for sodium are followed by those required for the
1
conversion of half a mole of oxygen molecules, 2 O2(g), to one mole of oxide
ions, O2−(g). This involves the atomisation of oxygen, followed by its first
and second electron affinities. All these experimentally determined values
make it possible to calculate the lattice energy.
Example
Calculate the lattice energy of sodium oxide, ΔlattH1[Na2O(s)], using the
data in Figure 13.1.8.
Notes on the method
Apply Hess’s law to the cycle in Figure 13.1.8 and remember that an
endothermic change in one direction becomes an exothermic change with
the opposite sign in the reverse direction.
Answer
ΔlattH1 [Na2O(s)] = (−657 − 249 − 992 − 214 − 414) kJ mol−1
= −2526 kJ mol−1
13.1.4 Born–Haber cycles 355
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Tip
Lattice energies are negative. This means that the descriptions ‘larger’ and ‘smaller’
can be ambiguous when comparing the values of lattice energies. For this reason, it is
better to describe one lattice energy as more or less exothermic than another.
Test yourself
17 Why are lattice energies:
a) always negative
b) impossible to measure directly?
18 Explain why a Born–Haber cycle is an application of Hess’s law.
19 Look carefully at Figure 13.1.9, which is a Born–Haber cycle for
magnesium chloride.
Mg2+(g) + 2e– + 2Cl(g)
∆H6 = –698 kJ mol–1
∆H5 = +244 kJ mol–1 Mg2+(g) + 2e– + Cl (g)
2
2+ –
Mg (g) + 2Cl (g)
∆H4 = +1451 kJ mol–1
+ –
Mg (g) + e + Cl2(g)
∆H7
–1
∆H3 = +738 kJ mol
Mg(g) + Cl2(g)
∆H2 = +148 kJ mol–1 Mg(s) + Cl2(g)
∆H1 = –641 kJ mol–1
MgCl2(s)
Figure 13.1.9 A Born–Haber cycle for magnesium chloride.
a) Identify the energy changes ΔH11, ΔH12, ΔH13, ΔH14, ΔH15, ΔH16 and
ΔH17.
b) Calculate the lattice energy of magnesium chloride.
13.1.5 Testing the ionic model –
ionic or covalent?
One way in which scientists can test their theories and models is by comparing
the predictions from their theoretical models with the values obtained by
experiment.
Born–Haber cycles are very helpful in this respect because they enable chemists
to test the ionic model and check whether the bonding in a compound is truly
ionic. The experimental lattice energy obtained from a Born–Haber cycle
can be compared with a theoretical value calculated by applying the laws of
electrostatics and assuming that the only bonding in the crystal is ionic.
356 13.1 Lattice energy
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Using the laws of electrostatics, it is possible to calculate a theoretical value + – +
for the lattice energy of an ionic compound by summing up the effects
of all the attractions and repulsions between the ions in the crystal lattice
– + –
(Figure 13.1.10).
Table 13.1.1 shows both the experimentally determined lattice energies and
– + –
the theoretical lattice energies for a number of compounds.
Table 13.1.1 + – +
Compound Experimental lattice energy from Theoretical lattice energy Figure 13.1.10 Some of the many
a Born–Haber cycle/kJ mol−1 calculated assuming that the attractions (red) and repulsions (blue)
only bonding is ionic/kJ mol−1 which must be taken into account in
NaCl −780 −770 calculating a theoretical value for the lattice
NaBr −742 −735 energy of an ionic crystal.
NaI −705 −687
KCl −711 −702
KBr −679 −674
KI −651 −636
AgCl −905 −833
MgI2 −2327 −1944
Pure ionic bonding arises solely from the electrostatic forces between the
ions in a crystal. Notice in Table 13.1.1 that there is close agreement between
the experimental and theoretical values of the lattice energies for sodium and
potassium halides. In all these compounds, the difference between the actual
value found from experimental data and the theoretical value is less than 3%.
This shows that ionic bonding can account almost entirely for the bonding
in sodium and potassium halides.
But look at the experimental and theoretical lattice energies of silver chloride
and magnesium iodide in Table 13.1.1. In these two compounds, the
theoretical values based on the assumption that the bonding is purely ionic
are much less exothermic than the experimental values. The actual bonding
is clearly stronger than that predicted by a pure ionic model. This suggests
that there is covalent bonding as well as ionic bonding in these substances.
Polarisation of ions
In ionic compounds, positive metal ions attracts the outermost electrons of
negative ions. The attraction can pull these electrons into the space between
Key term
the ions. This distortion of the electron clouds around anions by positively
Polarisation is the distortion of the
charged cations is an example of polarisation. As a result of polarisation, in
electron cloud in a molecule or ion by a
some ionic compounds there is a significant degree of electron sharing, that
nearby positive charge.
is covalent bonding.
The contribution from covalent bonding makes the size of the lattice
energy greater numerically than that expected from the purely ionic
model. The values in Table 13.1.1 show clearly that both silver chloride
and magnesium iodide, although mainly ionic, have significant extents of
covalent bonding.
13.1.5 Testing the ionic model – ionic or covalent? 357
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Figure 13.1.11 shows three examples of ionic bonding with increasing degrees
of electron sharing as a positive cation polarises the neighbouring negative
ion. In general, results show that:
● the polarising power of a cation is greater if it has a larger charge and a
smaller radius
● the polarisability of an anion is greater if it has more electrons in shells
and so a larger radius.
Figure 13.1.11 Ionic bonding with size of the
increasing degrees of electron sharing as a unpolarised
ion
positive cation polarises the neighbouring
negative anion. (Dotted circles show the
size of unpolarised ions.)
ion pair
substantial covalent bonding
Key terms increasing polarisation of the negative ion by the positive ion
In a larger negative anion with more electrons, the outermost electrons are
The polarising power of a positive
further from the attraction of its positive nucleus. Consequently, its outermost
ion (cation) is its ability to distort
electrons are more readily attracted to a neighbouring positive ion and are
the electron cloud of a neighbouring
therefore more polarisable. Also, a negative ion with a 2− charge is more
negative ion (anion).
polarisable than an ion with a 1− charge.
Polarisability is an indication of the
This means that iodide ions are more polarisable than bromide ions, bromide
extent to which the electron cloud in a
ions are more polarisable than chloride ions, and fluoride ions are very difficult
molecule, or an ion, can be distorted by
to polarise. In fact, fluorine, with its small singly charged fluoride ion, forms
a nearby electric charge.
compounds that are more ionic than those of any other non-metal.
Test yourself
20 Table 13.1.2 shows the ionic radii of some ions. 21 The lattice energy of LiF is −1031 kJ mol−1 and that of
Table 13.1.2 LiI is −759 kJ mol−1.
a) Why is the lattice energy of LiI less exothermic
Ion Li+ Na+ K+ Mg2+ Al3+
than the lattice energy of LiF?
Ionic radius/nm 0.074 0.102 0.138 0.072 0.053
b) Which of these two compounds would you
Ion N3− O2− F−
expect to have the closer agreement between
Ionic radius/nm 0.171 0.140 0.133
the Born–Haber experimental value of its
a) Why do the ionic radii decrease from N3− through lattice energy and its theoretical value based
O2− to F−? on the ionic model?
b) Use the data in Table 13.1.2 to explain why: c) Explain your answer to part (b).
i) the polarising power of Mg2+ is greater than 22 Here are four values for lattice energy in kJ mol−1:
that of Li+ −3791, −3299, −3054 and −2725. The four ionic
ii) the polarising power of Li+ is greater than compounds to which these values relate are BaO,
that of K+ MgO, BaS and MgS. Match the formulae with the
values and justify your choice.
iii) the polarising power of Al3+ is much greater
than that of Na+
iv) the polarisability of N3− is greater than that
of F−.
358 13.1 Lattice energy
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Activity
The stability of ionic compounds
Almost all the compounds of metals with non-metals are regarded as ionic and these compounds
have standard enthalpy changes of formation which are exothermic. This means that the compounds
are at a lower energy level and therefore more stable than the elements from which they are formed.
The Born–Haber cycles in Figures 13.1.8 and 13.1.9, show that an ionic compound has an exothermic
standard enthalpy of formation if its negative lattice energy outweighs the total energy needed to
produce gaseous ions from the elements.
Using a Born–Haber cycle with a theoretically calculated value for the lattice energy, it is possible to
estimate the standard enthalpy change of formation for compounds which do not normally exist.
For example, consider the Born–Haber cycle for the hypothetical compound MgCl in Figure 13.1.12.
Mg+(g) + e– + Cl(g)
∆ at H [ 12Cl 2 (g) ] = +122 kJ mol–1 + –
Mg (g) + e + 12 Cl2(g)
EA[ C l] = –349 kJ mol–1
+ –
Mg (g) + Cl (g)
–1
1st IE[Mg ] = +738 kJ mol
Mg(g) + 12 Cl2(g)
Theoretical
–1 ∆la tt H [ M g C l( s ) ] = –753 kJ mol–1
∆ at H [Mg(g) ] = +148 kJ mol
Mg(s) + 12 Cl2(g)
∆ f H [ M gC l( s ) ]
MgCl(s)
Figure 13.1.12 A Born–Haber cycle for the hypothetical compound MgCl.
1 Use Figure 13.1.12 to calculate a value for the standard b) Suggest why the value of Δ f H1[MgCl3(s)] is so
enthalpy change of formation of MgCl(s). endothermic.
2 What does your answer to Question 1 suggest about the 6 The estimated lattice energy of MgCl3(s) is −5440 kJ mol−1.
stability of MgCl(s)? a) Write an equation to summarise the lattice energy of
3 Using the Hess cycle in Figure 13.1.13, calculate the MgCl3.
standard enthalpy change for the reaction: b) Why is the lattice energy of MgCl3 more exothermic than
2MgCl(s) → MgCl2(s) + Mg(s) that of MgCl2(s)?
given that Δ f H1[MgCl2(s)] = −641 kJ mol−1. 2MgCl(s) MgCl2(s) + Mg(s)
4 What does your result for Question 3 tell you about the
stability of MgCl(s)?
5 A Born–Haber cycle for the hypothetical compound MgCl3
suggests that Δ f H1[MgCl3(s)] = +3950 kJ mol−1. 2Mg(s) + Cl2(g)
a) What does the value of Δ f H1[MgCl 3(s)] tell you about the Figure 13.1.13 A Hess cycle for the reaction
stability of MgCl3(s)? 2MgCl(s) → MgCl2(s) + Mg(s).
13.1.5 Testing the ionic model – ionic or covalent? 359
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13.1.6 Enthalpy changes during
dissolving
Why do ionic crystals dissolve in water, even though ions in the lattice
are strongly attracted to each other? What, in general, are the factors that
determine the extent to which an ionic salt dissolves in water (Figure 13.1.14)?
Chemists look for answers to questions of this kind by analysing the energy
changes that take place as crystals dissolve.
An ionic compound, such as sodium chloride, does not dissolve in a non-
polar solvent like hexane, but it does dissolve in a polar solvent like water.
Figure 13.1.14 The concentration of When one mole of sodium chloride dissolves in a large volume of water to
sodium chloride in the Dead Sea is so high produce a very dilute solution, there is an enthalpy change of +3.8 kJ mol−1.
that salt crystallises out in some places. This enthalpy change is described as the enthalpy change of solution of
sodium chloride.
The process can be summarised by the equation:
NaCl(s) + aq → Na+(aq) + Cl−(aq) Δsol H 1 = +3.8 kJ mol−1
or simply as Δsol H 1[NaCl(s)] = +3.8 kJ mol−1.
In the equation above, ‘+ aq’ is short for the addition of water.
Sodium chloride readily dissolves in water despite the fact that the process is
slightly endothermic. As this example shows, the sign of ΔH is not a reliable
guide to whether or not a process happens. This is particularly the case when
the magnitude of ΔH is small (see Chapter 13.2).
Key terms
It is not immediately obvious why the charged ions in a crystal such as sodium
The enthalpy change of solution, chloride separate and go into solution in water. Where does the energy come
ΔsolH1, is the enthalpy change when from to overcome the attractive forces between oppositely-charged ions?
one mole of a compound dissolves to
When sodium chloride dissolves in water, the overall process can be pictured
form a solution of infinite dilution. An
in two stages; these are shown in Figure 13.1.15.
infinitely dilute solution is one where
there is so much water that there is no ● First of all, Na+ and Cl− ions must be separated from the solid NaCl crystals
further energy change if more water to form well-spaced ions in the gaseous state, Na+(g) and Cl−(g). This is
is added. the reverse of the lattice energy and labelled −ΔlattH 1 = +787 kJ mol−1 in
Figure 13.1.15.
The enthalpy change of hydration is + −
● In the second stage, gaseous Na (g) and Cl (g) ions are hydrated by polar
the enthalpy change when one mole of
water molecules forming a dilute solution of sodium chloride, Na+(aq) +
gaseous ions dissolve in water to give
Cl−(aq). Under standard conditions, this process is the sum of the standard
an infinitely dilute solution.
enthalpy changes of hydration of Na+(g) and Cl−(g). This is written as
For example, for sodium ions:
Δhyd H 1[Na+] + Δhyd H 1[Cl−] = −784 kJ mol−1 in Figure 13.1.15.
Na+(g) + aq → Na+(aq)
ΔhydH 1 = −444 kJ mol−1 It is now possible to see from Figure 13.1.15 why sodium chloride dissolves in
water. The explanation is that the ions, Na+ and Cl−, are so strongly hydrated
Enthalpy changes of hydration are
by the polar water molecules that the exothermic enthalpy changes of
sometimes just called hydration
hydration nearly balance the energy required to separate the ions (the reverse
enthalpies.
lattice energy).
360 13.1 Lattice energy
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+ –
Na (g) + Cl (g) + aq
–∆lattH = +787 kJ mol–1 ∆hydH [Na+] + ∆hydH [Cl– ] = – 784 kJ mol–1
+ –
Na (aq) + Cl aq
Na+Cl–(s) + aq ∆solH = +3 kJ mol–1
Figure 13.1.15 An energy level diagram for sodium chloride dissolving in water.
The enthalpy change of solution is the difference between the energy needed
to separate the ions from the crystal lattice (the reverse of the lattice energy)
and the energy given out as the ions are hydrated (the sum of the hydration
enthalpies).
Figure 13.1.16 shows the structure of hydrated sodium and chloride ions. In
water molecules, there is a δ+ charge in the region between the hydrogen
atoms and a δ− charge on the oxygen atoms. This means that the polar water
molecules are attracted to both positive cations and negative anions. The bond
between the ions and the water molecules is an electrostatic attraction.
With cations, the electrostatic attraction involves the positive charge
on the cations and the δ− charges on the oxygen atoms of the water
molecules. In contrast, with anions, the attraction involves the negative charge
on the anions and the δ+ charge between the hydrogen atoms in the water
molecules.
δ–
δ+
δ–
δ+ δ– δ+ δ+
δ+ δ– δ+
δ–
δ– Cl–
Na+ δ+ δ+
δ– δ–
δ– δ–
δ+ δ– δ+
δ+
δ+
δ–
Figure 13.1.16 Sodium and chloride ions are hydrated when they dissolve in water. Polar
water molecules are attracted to both cations and anions.
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Test yourself
23 Sodium chloride is soluble in water. Why is it not soluble in hexane?
24 Refer to the data sheet headed ‘Lattice energies and enthalpy
changes of hydration’, which you can access online at
[Link]/EdexcelChemistry.
a) How is the enthalpy change of hydration affected by increasing
ionic charge?
b) List an appropriate series of ions and their enthalpy changes of
hydration to illustrate your conclusion in part (a).
25 Refer to Table 13.1.2, which shows the radii of some ions, and to
the data sheet headed ‘Lattice energies and enthalpy changes of
hydration’, which you can online at [Link]/
EdexcelChemistry.
a) How is the enthalpy change of hydration affected by increasing
ionic radius?
b) List an appropriate series of ions, their radii and their enthalpy
changes of hydration to illustrate your conclusion in part (a).
The effect of ionic charge on enthalpy change of
hydration and lattice energy
Enthalpy changes of hydration and lattice energies both involve electrostatic
attractions between opposite charges. Because of this, both processes are
exothermic.
In addition, both processes become more exothermic as the charge on an ion
increases because the charge density of the ion increases, and therefore its
attraction for any opposite charge increases.
This is illustrated very well by the enthalpy changes of hydration for Na+ and
Mg2+ and the lattice energies of NaF and MgO in Table 13.1.3.
Table 13.1.3 Comparing some enthalpy Enthalpy change of hydration/kJ mol−1 Lattice energy/kJ mol−1
changes of hydration and lattice energies.
Na+ −444 NaF −918
Mg2+ −2003 MgO −3791
Li+ −559 LiF −1031
K+ −361 KF −817
The effect of ionic radius on enthalpy change of
hydration and lattice energy
Comparing ions with the same charge, the larger the radius of the ion the
smaller its charge density. This results in a weaker attraction for oppositely
charged ions and for the δ+ or δ− charges on polar molecules such as water.
So, an increase in ionic radius leads to less exothermic values for enthalpy
changes of hydration and lattice energies. This point is neatly illustrated by
the enthalpy changes of hydration for Li+ and K+ and the lattice energies of
LiF and KF in Table 13.1.3.
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Trends in solubility
It is difficult to use enthalpy cycles, like that in Figure 13.1.15, to account for
trends in the solubilities of ionic compounds for three reasons.
First, the enthalpy change of solution is generally a small difference between
two large enthalpy changes, neither of which can be measured directly. So,
even small errors in estimating trends in the values of lattice energies and
hydration enthalpies can lead to large percentage errors in the predicted
enthalpy changes of solution.
Second, both lattice energies and hydration enthalpies are affected in the
same way by changes in the size of ions and their charges. This tends to
reduce the likelihood of any clear trends in enthalpy changes of solution.
Finally, it is clear from the small endothermic value for sodium chloride that
the sign and magnitude of the enthalpy change of solution is not a reliable
guide as to whether or not a solid will dissolve. Other factors must be taken
into account; these are considered in Chapter 13.2.
Test yourself
26 a) Use the data sheet headed ‘Lattice energies and enthalpy
changes of hydration’, which you can access online at
[Link]/EdexcelChemistry, to calculate the
enthalpies of solution of lithium fluoride and lithium iodide.
b) Account for the relative values of the lattice enthalpies and
hydration enthalpies of the two compounds in terms of ionic radii.
c) To what extent, if at all, can your answers to part (a) explain the
differences in the solubilities of the two compounds?
(Solubilities: LiF = 5 × 10 −5 mol in 100 g water; LiI = 1.21 mol in
100 g water.)
27 Why do you think the lattice energy of magnesium oxide,
ΔlattH1 [MgO (s)] = −3791 kJ mol−1, is roughly four times more
exothermic than that of sodium fluoride,
ΔlattH1[NaF(s)] = −918 kJ mol−1?
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Chapter summary
Chapter 13.1 Lattice energy with theoretical values calculated on the
assumption that the bonding in the crystals is solely
l The lattice energy of an ionic compound is defined the result of electrostatic forces between ions. The
as the energy change when one mole of the comparison makes it possible to estimate the extent
compound forms from free gaseous ions. to which the ionic model can account for the
l Lattice energies provide a measure of the strength
bonding in compounds.
of ionic bonding in different compounds. l Polarisation is the distortion of the electron cloud
l The lattice energy of a compound is more
in an ion (or molecule) by a nearby positive charge.
exothermic if the charges on the ions are relatively l As a result of polarisation, in some ionic
large and the ionic radii are relatively small. compounds there is a significant degree of electron
l The lattice energy of a compound cannot be
sharing, that is, covalent bonding.
measured directly but can be determined with the l The polarising power of a cation is greater if it has
help of a Born–Haber cycle by the application of a larger charge and a smaller radius.
Hess’s law. l The polarisability of an anion is greater if it has
l The energy level diagram for a Born–Haber cycle
more electrons in shells and so has a larger radius.
shows: The larger the charge on a negative ion, the more
• the total energy required to create free gaseous polarisable it is.
ions by atomising and then ionising the elements l The enthalpy change of solution for an ionic
• the total energy given out when the ions come compound is the enthalpy change when one mole
together to form a crystal lattice (the lattice of the compound dissolves to form a solution of
energy) infinite dilution.
• the standard enthalpy change of formation of the l The enthalpy change of solution is the difference
compound. between the energy needed to separate the ions
l The definitions of the energy quantities in a Born–
from the crystal lattice (the reverse of the lattice
Haber cycle are important. They include: the energy) and the energy given out when the ions
standard enthalpy of atomisation of the elements, are hydrated (the sum of the hydration enthalpies).
the ionisation energies of the metal element, the This can be represented by an energy level
electron affinities of the non-metal element, the diagram.
lattice energy of the compound and the standard l The larger the charge on an ion, and the smaller
enthalpy change of formation of the compound. its size, the more negative its enthalpy change of
l Lattice energies determined from experimental
hydration.
quantities by Born–Haber cycles can be compared
364 13.1 Lattice energy
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Exam practice questions
1 a) i) Show that the formation of CaF2 from Lattice energy of silver(ii) fluoride, AgF2(s),
its elements is a redox reaction.(2) ΔlattH 1[AgF2(s)] = −2650 kJ mol−1
ii) Draw a dot-and-cross diagram to represent
Lattice energy of silver(i) fluoride, AgF(s),
the bonding in calcium fluoride.(2)
ΔlattH 1[AgF(s)] = −955 kJ mol−1
b) i) State two characteristic physical
properties of compounds with the a) Use the outline of the Born–Haber cycle
types of structure and bonding found for silver(ii) fluoride shown below to:
in calcium fluoride.(2) i) write the formulae and state symbols of
ii) Describe the model that chemists use the species that should appear in boxes
to describe the structure and bonding A, B and C (3)
of a compound such as calcium ii) Identify the enthalpy changes D, E
fluoride. Explain how this model and F.(3)
accounts for the physical properties of b) State what is meant by the term ‘lattice
such compounds. (3) energy’. (2)
2 a) State what is meant by the term ‘first F
electron affinity’.(2) Ag2+(g) + 2F–(g) AgF2(s)
b) The equation below represents the change
for which the energy change is the second D
electron affinity of nitrogen. E
B
N−(g) + e− → N2−(g)
i) Explain why the second electron affinity
for all elements is endothermic(2) C
ii) Write equations for the first and third
electron affinities of nitrogen.(2) A
c) In magnesium iodide, MgI2, the iodide ions
are polarised, but in sodium iodide they are
not polarised.
i) State what is meant by the term Ag(s) + F2(g)
‘polarisation’ of an ion.(2)
ii) State two factors which help to explain
c) Use the diagram and the data supplied to
the polarisation of iodide ions in
magnesium iodide.(2) calculate the enthalpy change of formation
of silver(ii) fluoride. Give a sign and units in
3 The following data can be used in Born–Haber your answer.(3)
cycles for silver fluorides AgF and AgF2 d) Draw a similar Born–Haber cycle for
Enthalpy change of atomisation of fluorine, silver(i) fluoride and use the data supplied to
ΔatH 1[ 12 F2(g)] = +79 kJ mol−1 calculate the enthalpy change of formation
of silver(i) fluoride. Give a sign and units in
Enthalpy change of atomisation of silver, your answer. (6)
ΔatH 1[Ag(s)] = +289 kJ mol−1 e) i) Use the values of enthalpy change of
First ionisation energy of silver, formation from parts (c) and (d) to
1st IE[Ag(g)] = +732 kJ mol−1 calculate an enthalpy change for the
Second ionisation energy of silver, reaction:
2nd IE[Ag+(g)] = +2070 kJ mol−1 AgF2(s) → AgF(s) + 12 F2(g) (2)
Electron affinity of fluorine, ii) Comment on the relative stability of
EA[F(g)] = −348 kJ mol−1 AgF2(s) compared to its elements and
also compared to AgF(s) (2)
365
Exam practice questions
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4 The lattice energy of rubidium iodide, RbI, can d) Explain why the hydration enthalpies of
be determined indirectly using a Born–Haber both anions and cations are negative. (2)
cycle. 6 The table shows the values of the first five
a) Use the data in the table below to draw a ionisation energies of boron.
Born–Haber cycle for rubidium iodide. (6)
Ionisation energies of boron Value/kJ mol−1
Enthalpy change Energy/kJ mol−1
First +800
Formation of rubidium iodide −334
Second +2 400
Atomisation of rubidium +81
Third +3 700
Atomisation of iodine +107
Fourth +25 000
First ionisation energy of rubidium +403
Fifth +32 800
First electron affinity of iodine −295
a) i) Give the electronic configuration of a
b) Determine a value for the lattice energy of boron atom.(1)
rubidium iodide. (2) ii) Use your answer to (i) to explain the
c) Explain why the lattice energy of lithium pattern of the five ionisation energies
iodide, LiI, is more exothermic than that of of boron. (3)
rubidium iodide. (3) b) Boron forms an oxide with the formula
d) A theoretical value for the lattice energies B2O3. Use the table of ionisation energies
of LiI and RbI can be calculated assuming and this data to calculate a value for the
the compounds are purely ionic. When this lattice energy of boron oxide.
is done, the experimentally determined Enthalpy change of atomisation of oxygen,
value of the lattice energy of lithium ΔatH 1[ 12 O2(g)] = +250 kJ mol−1
iodide is greater than the calculated value Enthalpy change of atomisation of boron,
by 21 kJ mol−1, whereas the experimental ΔatH 1[B(s)] = +590 kJ mol−1
value for rubidium iodide is greater than First electron affinity of oxygen,
the calculated value by only 11 kJ mol−1. 1st EA[O(g)] = −140 kJ mol−1
Explain the difference in these values. (6) Second electron affinity of oxygen,
5 When calcium chloride dissolves in water, the 2nd EA[O−(g)] = +790 kJ mol−1
process can be represented by the equation: Standard enthalpy change of formation
CaCl2(s) + aq → Ca2+(aq) + 2Cl−(aq) of boron oxide,
Δf H 1[boron oxide] = −1270 kJ mol−1 (6)
The enthalpy change for this process is called c) Predict whether or not the value for the lattice
the enthalpy change of solution. Its value can energy for boron oxide calculated in (b) is in
be calculated from a Born–Haber cycle using good agreement with a value calculated in
the following data. theory by applying the laws of electrostatics to
ΔlattH 1[CaCl2(s)] = −2258 kJ mol−1 boron oxide assuming that has an ionic lattice.
Give reasons to explain your prediction. (2)
Δhyd H 1[Ca2+(g)] = −1657 kJ mol−1 d) One mole of boron oxide slowly dissolves
Δhyd H 1[Cl−(g)] = −340 kJ mol−1 in water to given a product that dissolves
in water forming a solution that is slightly
a) Draw and label the Born–Haber cycle acidic. One mole of the oxide reacts with
linking the enthalpy change of solution three moles of water.
of calcium chloride with the enthalpy i) Write a balanced equation for the
changes in the data above. (4) reaction of boric oxide with water. (1)
b) Use your Born–Haber cycle to calculate ii) Comment on this property of boron
the enthalpy change of solution of calcium oxide in relation to your answer to (b). (2)
chloride. (3) e) Boron forms a chloride which melts at
c) Comment on the factors that affect the 166 K and boils at 286 K. Describe the likely
size of the enthalpy change of hydration of structure and bonding of this chloride and
Ca2+(g) compared with that of Li+(g). (2) predict how it reacts when added to water. (3)
366
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Entropy
13.2
13.2.1 Enthalpy changes and the
direction of change
Chemists have devised a range of ways for predicting the direction and
extent of change. They use equilibrium constants (Chapters 11 and 12) and
electrode potentials (Chapter 14) to explain why some reactions go while
others do not. These quantities are related and there is a more fundamental
concept which links them together; however this concept is not the enthalpy
change for reactions, but the entropy change. Many exothermic reactions
with a negative enthalpy change of reaction do tend to go naturally
(Figure 13.2.1), but change can also happen in directions that are endothermic
if there are other changes in the surroundings (Figure 13.2.2).
Figure 13.2.1 A cheetah hunting its prey Figure 13.2.2 Energy can drive change in the
in Kenya. The cheetah gets its energy from direction opposite to the natural direction of
respiration, taking advantage of the natural change. Photosynthesis effectively reverses
direction of change. Carbohydrates react the changes of respiration. Leaves harness
with oxygen in muscle cells to form carbon energy from the Sun to convert carbon dioxide
dioxide and water, releasing energy. and water into carbohydrates.
Spontaneous changes
A spontaneous reaction is a reaction which tends to go without being driven
by any external agency. Spontaneous reactions are the chemical equivalent of
water flowing downhill (Figure 13.2.3). Any reaction which naturally tends
to happen is spontaneous in this sense even if it is very slow, just as water
has a tendency to flow down a valley even when held up behind a dam. The
chemical equivalent of a dam is a high activation energy for a reaction.
Figure 13.2.3 Metals such as magnesium,
iron and aluminium react spontaneously In practice, chemists also use the word ‘spontaneous’ in its everyday sense
with oxygen. They are ingredients of to describe reactions which not only tend to go, but go fast on mixing the
fireworks. They burn, when heated, by the reactants at room temperature. Here is a typical example:
spontaneous reactions between sulfur, ‘The hydrides of silicon catch fire spontaneously in air, unlike methane
carbon and potassium nitrate in gunpowder. which has an ignition temperature of about 500 °C.’
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The reaction of methane with oxygen is also spontaneous in the thermodynamic
sense, even at room temperature. However, the activation energy for the
reaction is so high that nothing happens until the gas is heated with a flame.
There are many examples of reactions that theory shows are spontaneous, but
which in practice happen very slowly (Figure 13.2.4).
The possible ambiguity in the use of the term ‘spontaneous’ means that
chemists generally prefer an alternative term. They describe a reaction as
feasible if it tends to go naturally.
Figure 13.2.4 Theory shows that the change
of diamond to graphite is spontaneous. Test yourself
Graphite is more stable than diamond.
Fortunately for owners of valuable jewellery 1 Classify these changes as feasible/fast, feasible/slow or not feasible:
the change is very, very slow. a) ice melting at 5 °C
b) ammonia gas condensing to a liquid at room temperature
Key term c) diamond reacting with oxygen at room temperature
d) water splitting up into hydrogen and oxygen at room temperature
A feasible reaction is one that naturally
tends to happen, even if it is very slow e) sodium reacting with water at room temperature.
because the activation energy is high.
Enthalpy changes and feasible reactions
Most reactions that are feasible are also exothermic. They have a negative
standard enthalpy change of reaction. This is such a common pattern that
chemists often use the sign of ΔH to decide whether or not a reaction is
likely to go.
However, some endothermic processes are feasible too. This shows the
limitation of using the enthalpy change to decide the likely direction of
change. One example is the reaction of ethanoic acid with ammonium
carbonate. The mixture fizzes vigorously while getting colder and colder.
ΔH for the reaction is positive. There is a similar reaction between citric acid
Figure 13.2.5 Antacid tablets that contain
and sodium hydrogencarbonate (Figure 13.2.5).
citric acid and sodium hydrogencarbonate
fizz when added to water. The reaction of Another example is the reaction between solid barium hydroxide and
the acid and the hydrogencarbonate is a solid ammonium chloride. When these chemicals are mixed in a flask,
spontaneous endothermic reaction. the endothermic reaction means that the temperature inside the beaker
drops dramatically. If the reaction flask is sitting on a damp wooden
block, the temperature drop is enough to freeze the water, sticking the
flask and block together (Figure 13.2.6).
13.2.2 Feasible by chance alone
Diffusion
Open a bottle of perfume in the corner of a room, and it is not long
before everyone else in the room can smell it. The perfume molecules
spread out naturally and mix with the air molecules. This is diffusion
Figure 13.2.6 The endothermic reaction
(Figures 13.2.7 and 13.2.8).
between the solids barium hydroxide and
ammonium chloride cools the flask so The opposite never happens. If there is a bad smell in a room it is impossible
much that the water between the flask to get rid of it by persuading all the smelly molecules to collect together in a
and the wooden block freezes. small bottle before putting in a stopper.
368 13.2 Entropy
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Figure 13.2.7 The right-hand bottle contains
bromine vapour. The left-hand bottle contains
air. The two are separated by a barrier.
Figure 13.2.8 After removing the barrier,
the bromine diffuses from the right-hand
bottle until the bromine molecules are
evenly spread between the two bottles.
Once mixed, the air and bromine never
unmix. When the concentrations are the
same in both bottles, molecules continue
to diffuse between the bottles, but overall
there seems to be no change. This is an
example of dynamic equilibrium.
Diffusion happens by chance alone. This is shown by the very simple
example in Figure 13.2.9. This shows an imaginary situation with just
six molecules of bromine in the right-hand jar ( jar R). The left-hand jar
is empty and there is a barrier between the jars. This situation can be
represented as RRRRRR.
jar L jar R
Figure 13.2.9 Two gas jars separated by a barrier with six molecules of bromine in the
right-hand jar. The molecules are in rapid, random motion (RRRRRR).
The molecules in jar R are moving around randomly, bumping into each
other and the sides of the jar. Figure 13.2.10 shows what happens immediately
after removing the barrier. One molecule has moved into the left-hand jar.
This can be represented as RRRRRL.
jar L jar R
Figure 13.2.10 After removing the barrier one molecule has moved into the left-hand jar
(RRRRRL).
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As the molecules move around randomly in the two jars, there are many
possible arrangements. Each molecule can either be R or L. For six molecules
moving between the jars, there are 2 × 2 × 2 × 2 × 2 × 2 = 26 = 64 possible
arrangements.
Only one of these arrangements is RRRRRR, so there is a 1 in 64 chance
that all the molecules will end up back in jar R. There is a much bigger
chance that the molecules will be arranged in some other way.
In practice the number of molecules is always many more than six. In
Figure 13.2.7 there are about 1022 molecules of bromine in the right-hand
jar. The number of ways of distributing the molecules between the two jars
22
is 210 . This is a huge number – too big to be imagined. Only one of all this
number of arrangements has all the molecules back in the right-hand jar.
The chance of this happening is impossibly small.
It is overwhelmingly likely, by chance alone, that bromine molecules will
spread out and mix with air molecules unless there is a barrier to stop them.
This is a spontaneous change. In general, gas molecules naturally tend to mix
up and disperse themselves randomly.
Molecules and energy
Table 13.2.1 The number of ways that two It is not only the arrangement of molecules that matters. Even more important
molecules can share four energy quanta. is the way that energy is spread out between molecules. The molecules in
Number of quanta bromine gas are moving around, spinning and vibrating. The energies of the
tiny molecules are quantised, in a similar way to the energies of electrons in
Molecule 1 Molecule 2
atoms. All the time the molecules are bumping into each other; as they do so,
2 2
they lose and gain energy quanta to and from other molecules.
3 1
1 3 Taking a very simple situation, Table 13.2.1 shows the number of ways
0 4 that two molecules can share four energy quanta. In all there are five different
4 0
ways. The more molecules, and the more energy quanta, the more ways there
are of sharing the energy between the molecules, as shown in Table 13.2.2.
Number of molecules Number of quanta Number of ways of sharing the
molecules between the quanta
10 100 ≈ 1012
100 10 ≈ 1013
100 100 ≈ 1060
Table 13.2.2 The number of ways of
200 110 ≈ 1086
sharing energy between molecules.
Increasing the temperature of a substance increases the number of energy
quanta in the system and so increases the number of possible ways that the
energy quanta can be shared out between the molecules.
13.2.3 Entropy changes
Entropy
Random changes, which happen by chance, always tend to go in the direction
that increases the number of ways of distributing the molecules and energy
quanta. This is a fundamental principle.
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However, the number of ways, W, of distributing the energy quanta between
the molecules in a mole of gas at room temperature is huge. The numbers
involved are very hard to deal with. Fortunately, the Austrian physicist
Ludwig Boltzmann (1844–1906) showed how the established laws of
thermodynamics could be explained in terms of molecules and their energies.
Physicists had already developed the concept of entropy to account for the
way that steam engines work. What Boltzmann was able to show was that
there is a relationship between this quantity entropy and the number of
ways that any chemical system could distribute its molecules and energy. He
demonstrated the truth of this formula:
entropy, S = k ln W
Figure 13.2.11 The reaction mixture
where S is the entropy of the system, k is a constant named after Boltzmann in the flask is ‘the system’. The air
and ln W is the natural logarithm of the number of ways of arranging the and everything else around the flask
particles and energy in the system. makes up the ‘surroundings’. It is the
total entropy in the system and the
surroundings which determines whether
Tip or not the reaction is feasible.
Find out more about natural logarithms, ln, in Section 3 of ‘Mathematics in A Level
chemistry’, which you can access online at [Link]/
gas
EdexcelChemistry.
Now chemists can use this quantity called entropy, S, to decide whether or
not a reaction is feasible. The formula shows that as W increases, S increases.
Entropy
So change happens in the direction which leads to a total increase in entropy. liquid
Chemists sometimes describe entropy as a measure of disorder or randomness.
These descriptions have to be interpreted with care because the disorder
refers not only to the arrangement of the particles in space, but much more solid
significantly to the numbers of ways of distributing the energy of the system
across all the available energy levels.
When considering chemical reactions it is essential to calculate the total melting boiling
temperature temperature
entropy change in two parts: the entropy change of the system and the
Temperature
entropy change of the surroundings (Figure 13.2.11).
Figure 13.2.12 The entropy of a chemical
ΔS total = ΔS system + ΔS surroundings increases as its temperature rises. There
is a jump in the entropy values when the
Standard molar entropies chemical melts or boils.
In a perfectly ordered crystal at 0 K the entropy is zero. The entropy of a chemical
rises as the temperature rises. This is because increasing the temperature raises Key term
the number of energy quanta to share between the atoms and molecules. There
are also jumps in the entropy of the chemical wherever there is a change of state. Standard molar entropy, S, is the
This is because energy is added to change a solid to a liquid or a liquid to a gas entropy per mole for a substance under
(Figure 13.2.12). Also there is an increase in the number of ways of arranging the standard conditions. Chemists use
atoms or molecules as a solid changes to a liquid and then to a gas. values for standard molar entropies
to calculate entropy changes and so
Standard molar entropy values are quoted for pure chemicals under
to predict the direction and extent of
standard conditions (298 K and 1 atmosphere pressure). Gases generally have
chemical change.
higher entropies than comparable liquids, which have higher entropies than
similar solids (Figures 13.2.13–13.2.15).
13.2.3 Entropy changes 371
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Figure 13.2.13 Structure of a solid. Solids Figure 13.2.14 Structure of a liquid. In Figure 13.2.15 Atoms of a noble gas. Gases
have relatively low values for standard general, liquids have higher standard molar have even higher standard molar entropies
molar entropies. In diamond the carbon entropies than comparable solids because than comparable liquids because the atoms
atoms are held firmly in place by strong, the atoms or molecules are free to move. or molecules are not only free to move but
highly directional covalent bonds. The There are many more ways of distributing also widely spaced. There are even more
standard molar entropy of diamond is low. the particles and energy – there is more ways of distributing the particles and energy
Lead has a higher value for its standard disorder. The standard molar entropy – the disorder is even greater. The standard
molar entropy because metallic bonds are of mercury is higher than that of lead. molar entropy of argon is higher than that
not directional. The heavier, larger atoms Molecules with more atoms have higher of mercury. As with liquids, molecules with
can vibrate more freely and share out standard molar entropies because they can more atoms have even higher standard molar
their energy in more ways than can carbon vibrate, rotate and arrange themselves in entropies because they can vibrate, rotate
atoms in diamond. yet more ways. and arrange themselves in more ways.
Tip
The units for standard molar entropy are joules per kelvin per mole (J mol−1 K−1). Note
that the units are joules and not kilojoules.
Table 13.2.3 Standard molar entropies for selected solids, liquids and gases.
Solids S 1/ Liquids S 1/ Gases S 1/
J mol−1 K−1 J mol−1 K−1 J mol−1 K−1
Carbon 2.4 Mercury 76.0 Argon 155
(diamond)
Magnesium 26.9 Water 69.9 Ammonia 192
oxide
Copper 33.2 Ethanol 160 Carbon 214
dioxide
Lead 64.8 Benzene 173 Propane 270
Test yourself
2 Refer to Table 13.2.3. Why is the value of the standard molar entropy of:
a) mercury higher than the value for copper
b) ammonia higher than the value for water
c) propane higher than the value for argon?
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3 Which substance in each of the following pairs is expected to have the
higher standard molar entropy at 298 K:
a) Br2(l) or Br2(g)
b) H2O(s) or H2O(l)
c) HF(g) or NH3(g)
d) CH4(g) or C2H6(g)
e) NaCl(s) or NaCl(aq)
f) ethane or poly(ethene)
g) pentene gas or cyclopentane gas?
The entropy change of the system
Tables of standard molar entropies make it possible to calculate the entropy
change of the system of chemicals during a reaction.
∆S system
1
= the sum of the standard molar entropies of the products Tip
− the sum of the standard molar entropies of the reactants
The symbol Σ is the Greek capital letter
This can be shortened to: for sigma. It is used in science and
maths to mean ‘sum of’.
∆S system
1
= ΣS 1[products] − ΣS 1[reactants]
Example
Calculate the entropy change for the system, ∆S system
1 , for the synthesis of
ammonia from nitrogen and hydrogen. Comment on the value.
Notes on the method
Write the balanced equation for the reaction.
Look up the standard molar entropies on the data sheet headed
‘Thermodynamic properties of selected elements and compounds’,
which you can access online at [Link]/
EdexcelChemistry. Take careful note of the units.
Answer
N2(g) + 3H2(g) → 2NH3(g)
Sum of the standard molar entropies of the products
= 2S 1[NH3(g)]
= 2 × 192.4 J mol−1 K−1 = 384.8 J mol−1 K−1
Sum of the standard molar entropies of the reactants
= S 1[N2(g)] + 3S 1[H2(g)]
= 191.6 J mol−1 K−1 + (3 × 130.6 J mol−1 K−1) = 583.4 J mol−1 K−1
∆S system
1 = 384.8 J mol−1 K−1 − 583.4 J mol−1 K−1
= −198.6 J mol−1 K−1
This shows that the entropy of the system decreases when nitrogen and
hydrogen combine to form hydrogen. This is not surprising since the
change halves the number of molecules so the amount of gas decreases.
13.2.3 Entropy changes 373
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Test yourself
4 Without doing any calculations, predict whether the entropy of the
system increases or decreases as a result of these changes:
a) KCl(s) + aq → KCl(aq) b) H2O(l) → H2O(g)
c) Mg(s) + Cl2(g) → MgCl2(s) d) N2O4(g) → 2NO2(g)
e) NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
f) 2CH3COOH(l) + (NH4)2CO3(s) → 2CH3COONH4(s) + H2O(l) + CO2(g)
5 Refer to the data sheet headed ‘Thermodynamic properties of selected
elements and compounds’, which you can access online at www.
[Link]/EdexcelChemistry. Use values from the data
sheet to calculate the entropy change for the system, ∆S1, for the
catalytic reaction of ammonia with oxygen to form NO and steam.
Comment on the value.
The entropy change of the surroundings
It is not enough to consider only the entropy of the system. What matters
is the total entropy change, which is the sum of the entropy changes of the
system and the entropy change in the surroundings.
It turns out that the entropy change of the surroundings during a chemical
reaction is determined by the size of the enthalpy change, ∆H, and the
temperature, T. The relationship is:
∆H
∆S surroundings = −
T
The minus sign is included because the entropy change becomes more
positive, the more energy that is transferred to the surroundings. For an
exothermic reaction, which transfers energy to the surroundings, ∆H is
negative so −∆H is positive.
What this relationship shows is that the more energy transferred to the
surroundings by an exothermic process, the larger the increase in the entropy of
Figure 13.2.16 The thermite reaction
the surroundings (Figure 13.2.16). It also shows that, for a given quantity of energy,
between iron(iii) oxide and aluminium metal
the increase in entropy is greater when the surroundings are cool than when they
is highly exothermic. The molten iron formed
are hot. Adding energy to molecules in a cool system has a proportionately greater
can be used to weld railway lines. The
effect on the number of ways of distributing matter and energy than adding the
reaction gives out a great deal of energy to
same quantity of energy to a system that is already very hot.
its surroundings. The entropy change in the
surroundings is large and positive.
Total entropy changes
A reaction is only feasible if the total entropy change, ∆S total, is positive. This
means that:
● Most exothermic reactions tend to go because around room temperature
the value of −∆H/T is much larger and more positive than ∆S system, which
means that ∆S total is positive.
● An endothermic reaction can be feasible so long as the increase in the
entropy of the system is greater than the decrease in the entropy of the
surroundings.
● A reaction that does not tend to go at room temperature may become feasible
as the temperature rises, because ∆S surroundings decreases in magnitude as T
increases.
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Test yourself
6 Consider the reaction of magnesium with oxygen:
2Mg(s) + O2(g) → 2MgO(s) ΔH 1 = −602 kJ mol−1
ΔS system
1 = −217 J mol−1 K−1
a) Why does the entropy of the system decrease?
b) Show why the reaction of magnesium with oxygen is feasible at
298 K despite the decrease in the entropy.
7 Consider the reaction of ammonium chloride with barium hydroxide:
2NH4Cl(s) + Ba(OH)2(s) → BaCl2.2H2O(s) + 2NH3(g)
Refer to the data sheet headed ‘Thermodynamic properties of
selected elements and compounds’, which you can access online at
[Link]/EdexcelChemistry.
a) Use the data from the data sheet to calculate the entropy change
for the system and comment on the value you get.
b) Calculate the enthalpy change for the reaction.
c) Calculate the entropy change of the surroundings at 298 K.
d) Work out the total entropy change for the reaction and decide
whether or not it is feasible under standard conditions. Comment
on your answer.
Entropy changes during dissolving
Sodium chloride dissolves readily in water, despite the fact that the process
is slightly endothermic. This is another example that shows that the sign of
ΔH is not a reliable guide to whether or not a process will happen. This is
particularly the case when the magnitude of ΔH is small.
When sodium chloride dissolves, the disorder increases as the ions leave the
regular lattice and mix with the molecules of liquid water. This means that
the entropy of the system increases as a salt dissolves. The entropy change in
the surroundings is negative for this endothermic change, but the increase in
the entropy of the system is more than enough to compensate for this and so
the total entropy change is positive.
Dissolving depends on the balance between the change in entropy of the
solution and the change in entropy of the surroundings. This balance starts to
alter as soon as a salt starts to dissolve and the concentration of the solution rises.
In a more concentrated solution the ions are closer together and there are fewer
free water molecules to hydrate the ions. These changes modify the values of
the hydration energies and the entropies of the ions in solution. Overall, this
means that once a certain amount of a salt has dissolved, the processes ceases to
be feasible and no more solid dissolves. At this point the solution is saturated.
The salt crystals and the saturated solution are in equilibrium.
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Example
Table 13.2.4 Use the values in Table 13.2.4 to calculate the total entropy change when
sodium chloride dissolves in water under standard conditions.
S 1/J mol−1 K−1
NaCl(s) +72.1 NaCl(s) → Na+(aq) + Cl−(aq) ∆H system
1 = +3.8 kJ mol−1
Na+(aq) +321 Notes on the method
Cl−(aq) +56.5 Calculate the entropy change of the surroundings using the formula:
ΔS surroundings = − ΔH
1 1
T
Remember to convert the value of the enthalpy change to J mol−1.
Answer
3800 J mol −1
ΔS surroundings
1 =− = −12.8 J mol−1 K−1
298 K
ΔS system
1
= ΣS 1[products] − ΣS 1[reactants]
= 321 J mol−1 K−1 + 56.5 J mol−1 K−1 − 72.1 J mol−1 K−1
= +305 J mol−1 K−1
ΔS total
1 = ΔS system
1
+ ΔS surroundings
1
= +305 J mol−1 K−1 − 12.8 J mol−1 K−1 = +292 J mol−1 K−1
The total entropy change is positive. Sodium chloride dissolving in water
is a feasible process.
Test yourself
8 a) Use the values in Table 13.2.5 to calculate the total entropy
Table 13.2.5 change when ammonium nitrate dissolves in water under standard
S 1/J mol−1 K−1 conditions.
NH4NO3(s) 151 NH4NO3(s) → NH4+(aq) + NO3−(aq) ΔH 1 = +28.1 kJ mol−1
NH4+(aq) 113 b) Comment on the fact that ammonium nitrate dissolves in water
NO3−(aq) 146 even though the process is endothermic.
13.2.4 Free energy
A chemical change is feasible if the total entropy change is positive. There
is no doubt about this. The problem is that using entropy to decide on the
direction change involves three steps: working out the entropy change of
the system, working out the entropy change of the surroundings, and then
putting the two together to calculate the total entropy change. This can be
laborious and so chemists are grateful to the American physicist Willard
Gibbs, who discovered an easier way of unifying all that chemists know
about predicting the extent and direction of change.
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Free energy and entropy
Willard Gibbs (1839–1903) was the first to define the thermochemical
quantity free energy. The symbol for a free energy change is ∆G. If ∆G is
Key term
negative the reaction is feasible and tends to go.
The free energy change, ∆G, is the
The advantage of ∆G 1
values for chemists is that tables of standard free thermochemical quantity used by
energies of formation can be used to calculate the standard free energy chemists to decide whether a reaction
change for any reaction. The calculations follow exactly the same steps as tends to go and how far it will go.
the calculations to calculate standard enthalpy changes for reactions from ∆G is the test for the feasibility of a
standard enthalpies of formation. reaction. If ∆G is negative the reaction
is feasible.
The quantity ‘free energy’ is closely related to the idea of entropy and can be
thought of as the ‘total entropy change’ in disguise. Willard Gibbs defined
free energy as:
ΔG 1 = −TΔS total
1
Given that:
ΔS total
1
= ΔS system
1
+ ΔS surroundings
1
And that for a change at constant temperature and pressure:
−ΔH 1
ΔS surroundings
1
=
T
It follows that:
ΔS total
1
= ΔS system
1
+ ( −ΔHT )
1
Hence: −TΔS total
1
= −TΔS system
1
+ ΔH 1
From Gibbs’ definition this becomes: ΔG 1 = ΔH 1 − TΔS system
1
The great advantage of this equation is that all the terms refer to the system
and so it is no longer necessary to calculate changes in the surroundings.
Given that this is the case, the equation is usually written as here, with the
understanding that the entropy change is ΔS system
1
.
ΔG 1 = ΔH 1 − TΔS 1
Table 13.2.6 summarises the implications of this important relationship.
Table 13.2.6
Enthalpy change Entropy change of the system Is the reaction feasible?
Exothermic Increase (∆S positive) Yes, ∆G is negative
(∆H negative)
Exothermic Decrease (∆S negative) Yes, if the number value
(∆H negative) of ∆H is greater than the
magnitude of T∆S
Endothermic Increase (∆S positive) Yes, if the magnitude of T∆S
(∆H positive) is greater than the number
value of ∆H
Endothermic Decrease (∆S negative) No, ∆G is positive
(∆H positive)
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The possibilities listed in Table 13.2.6 show why chemists sometimes say that
the feasibility of a reaction depends on the balance between the enthalpy
changes and the entropy changes for the process.
Table 13.2.6 also shows that a change that is not feasible at a lower temperature
may become feasible if the temperature rises. Generally the values of ΔH 1
and ΔS 1 do not change markedly with temperature, so it is possible to
estimate the higher temperature at which a reaction that is not feasible at
room temperature becomes feasible.
Full data are not available for all reactions, so it is not always possible to calculate
the free energy change. At relatively low temperatures the TΔS 1 term is often
relatively small compared to the enthalpy change, so that ΔG 1 ≈ ΔH 1.
This means that chemists can often use the sign of ΔH 1 as a guide to feasibility.
This can be misleading if the magnitude of the entropy change for the reaction
system is large. Also, the approximation becomes less justified at higher
temperatures when T is bigger and so the magnitude of TΔS 1 is bigger.
Example
Calculate ΔS 1 and ΔH 1 for the synthesis of methanol from carbon
monoxide and hydrogen. Work out the temperature at which the synthesis
ceases to be feasible.
Notes on the method
Start by writing the equation for the reaction.
Look up the standard enthalpies changes of formation on the data
sheet headed ‘Thermodynamic properties of selected elements and
compounds’, which you can access online at [Link].
[Link]/EdexcelChemistry.
Remember that all temperatures are measured on the Kelvin scale. Note
too that the values for the enthalpy change and entropy change must be
converted to be in the same units.
From the equation ΔG 1 = ΔH 1 − TΔS 1, it follows that ΔG 1 becomes
positive and an exothermic reaction ceases to be feasible when the
temperature is high enough for −TΔS 1 to be positive and large enough to
balance the negative value for the enthalpy change.
Answer
CO(g) + 2H2(g) → CH3OH(l)
Δr H 1 = Σ Δf H 1[products] − Σ Δf H 1[reactants]
∆H 1 = −239 kJ mol−1 − (−110 kJ mol−1) = −129 kJ mol−1
∆S system
1
= ΣS 1[products] − ΣS 1[reactants]
∆S 1 = 240 J mol−1 K−1 − (260 J mol−1 K−1 + 198 J mol−1 K−1)
= −218 J mol−1 K−1 = −0.218 kJ mol−1 K−1
The reaction ceases to be feasible when −TΔS becomes more positive
than −129 kJ mol−1.
−T∆ S 1 = +129 kJ mol−1
T = 129 kJ mol−1 ÷ 0.218 kJ mol−1 K−1 = 592 K
The synthesis ceases to be feasible when the temperature is above 592 K.
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Test yourself
9 Can an exothermic reaction which is not feasible at room
temperature become feasible at a higher temperature if the entropy
change for the reaction is negative?
10 Consider the reduction of iron(iii) oxide by carbon:
2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)
∆H 1 = + 468 kJ mol−1
∆S 1 = +558 J mol−1 K−1
a) Why does the entropy of the system increase for this reaction?
b) Show that the reaction is not feasible at room temperature (298 K).
c) Assuming that ∆H 1 and ∆S 1 do not vary with temperature,
estimate the temperature at which the reaction becomes feasible.
Free energy and equilibrium constants
Chapter 11 shows that the direction and extent of change can be described in
terms of equilibrium constants. This chapter has introduced the idea that the
feasibility of a reaction can be determined by calculating its standard free energy
change. It turns out that these two methods for determining the direction and
extent of chemical change are directly related. The theory of thermochemistry
shows that, for reactions involving gases, they are related by this formula, where
R is a constant (the same constant as in the ideal gas equation):
∆G 1 = −RT ln K
Tip
This equation applies to reactions involving gases and gives the value of Kp
(see Section 11.4). For guidance about the meaning and
use of natural logarithms, ln, refer to
This relationship shows that the values for ∆G 1 and value for K offer Section 1 in ‘Mathematics for A Level
alternative ways of answering the same questions for a reaction: chemistry’, which you can access online
● Will the reaction go? at [Link]/
● How far will it go? EdexcelChemistry.
The question that cannot be answered by these thermodynamic quantities
is: ‘How fast will it go?’.
It is very important to keep in mind that even if a reaction is feasible, it may
be very slow. This is fortunate, otherwise it would not be possible to fill a
car’s petrol tank in the presence of air.
Tip
Chapter 14 introduces a third method of predicting the direction and extent of change
for redox reactions based on standard electrode potentials. In practice, chemists use
the quantity that it is more convenient to measure. Knowing the value of one of the two
quantities, it is possible to calculate the others.
13.2.4 Free energy 379
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Example
Nitrogen can react with hydrogen to make ammonia, ∆G 1 = −32.9 kJ mol−1.
N2(g) + 3H2(g) ⇋ 2NH3(g)
Calculate the equilibrium constant, K, for this reaction at 298 K. Comment
on the value of K.
Notes on the method
The value of the gas constant R = 8.31 J K−1 mol−1.
Work in consistent units by converting kJ to J.
Rearrange ∆G 1 = −RT ln K to find ln K.
Refer to Section 3 in ‘Mathematics in A Level chemistry’, which you can
access online at [Link]/EdexcelChemistry, to find
out how to calculate the value of K from ln K.
Logarithms do not have units, so the calculation based on this
thermodynamic equation gives a value for K that does not have units.
Answer
ln K = −∆G
1
RT
−(−32 900 J mol−1)
ln K= = 13.3
8.31 J K−1 mol−1 × 298 K
K = e13.3 ≈ 6 × 105
The value of K is very large. This shows that, at room temperature, the
equilibrium is well over to the product side. This is consistent with a
negative value of ∆G 1, which shows that the reaction is feasible under
these conditions. However, at room temperature, in the absence of a
catalyst, the reaction is very, very slow.
Test yourself
11 ∆G 1 = +163 kJ mol−1 for the conversion of oxygen to ozone at 298 K.
3
O (g) ⇋ O3(g)
2 2
a) Calculate a value for the equilibrium constant, K, of this reaction
at 298 K.
b) Comment on the values of ∆ G 1 and K.
Stable or inert?
The study of energetics (thermochemistry) and rates of reaction (kinetics)
Key term helps to explain why some chemicals are stable (Figure 13.2.17) while others
react rapidly.
A chemical or mixture of chemicals is Compounds are stable if they have no tendency to decompose into their
thermodynamically stable if there is no elements or into other compounds. Magnesium oxide, for example, has no
tendency for a reaction. A positive free tendency to split up into magnesium and oxygen. However, a compound
energy change, ΔG, indicates that the that is stable at room temperature and pressure may become more or less
reaction does not tend to occur. stable as conditions change.
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Chemists often use standard enthalpy changes as an indicator of stability.
Strictly they should use standard free energy, ΔfG 1 , values but in many cases
Key term
ΔfG 1 ≈ Δf H 1.
A chemical or mixture of chemicals
A chemical is inert if it has no tendency to react even when the reaction is is kinetically inert when a reaction
feasible (Figure 13.2.18). Nitrogen, for example, is a relatively unreactive gas does not go even though the reaction
that can be used to create an ‘inert atmosphere’ free of oxygen (which is much appears to be feasible. The reaction
more reactive). However, nitrogen is not inert in all circumstances. It reacts tends to go according to the free energy
with hydrogen in the Haber process to form ammonia and with oxygen at high change, yet nothing happens. There is
temperatures in motor engines and power stations to form nitrogen oxides. no change because the rate of reaction
is too slow to be noticeable. There is a
barrier preventing change – usually a
high activation energy. The compound
or mixture is inert.
Figure 13.2.17 The Giant Buddha in a temple at Bangkok Thailand. Gold is stable in air
and water. It has no tendency to react and tarnish.
A compound such as the gas N2O, for example, is thermodynamically unstable Figure 13.2.18 Distorted reflections
but it continues to exist at room temperature because it is kinetically inert of surrounding park and buildings in an
(∆fH 1 = +82 kJ mol−1 and ∆fG 1 = +104 kJ mol−1). The free energy change for aluminium street sculpture in Los Angeles.
the decomposition reaction is negative so the compound tends to decompose Aluminium is a reactive metal, but it is inert
into its elements, but the rate is very slow under normal conditions. in air and water because it is protected
Examples of kinetic inertness are: from corrosion by a thin layer of the metal
oxide on its surface.
● a fuel tank containing petrol and air
● mixture of hydrogen and oxygen at room temperature
● a solution of hydrogen peroxide in the absence of a catalyst
● aluminium metal in dilute hydrochloric acid.
‘Kinetic stability’ is a term sometimes used for ‘kinetic inertness’. It helps,
however, to make a sharp distinction between two quite different types of
explanation. For clarity, chemists refer to:
● systems with no tendency to react as ‘stable’ and
● systems that should react but do not do so for a rate (kinetic) reason as
‘kinetically inert’.
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Table 13.2.7 Sometimes there is no tendency for a reaction to go because the reactants
are stable. This is so if the free energy change for the reaction is positive. Sometimes
there is no reaction even though thermochemistry suggests that it should go. The free
energy change is negative, so the change is feasible, but a high activation energy means
that the rate of reaction is very, very slow.
ΔG 1 Activation Change Stability
(≈ ΔH 1) energy observed
Positive High No reaction Reactants stable relative to
products
Negative High No reaction Reactants unstable relative to
products but kinetically inert
Positive Low No reaction Reactants stable relative to
products
Negative Low Fast reaction Reactants unstable relative
to products
Test yourself
12 Suggest examples of reactions to illustrate each of the four
possibilities in Table 13.2.7.
13 Draw a reaction profile to show the energy changes from reactants to
products for reactants that are thermodynamically unstable relative
to the products, but kinetically inert.
Activity
The thermal stability of Group 2 carbonates
The decomposition of Group 2 metal carbonates is used on a large scale to make oxides
such as magnesium and calcium oxides.
MgCO3(s) → MgO(s) + CO2(g)
∆H 1 = +117 kJ mol−1, ∆S 1 = +175 J mol−1 K−1
BaCO3(s) → BaO(s) + CO2(g)
∆H 1 = +268 kJ mol−1, ∆S 1 = +172 J mol−1 K−1
The carbonates of Group 2 metals do not decompose at room temperature. They do
decompose on heating.
Figure 13.2.19 Crystals of chalcopyrite on
1 Why does the entropy increase when a Group 2 carbonate decomposes? dolomite with a large calcite crystal. Dolomite
is a calcium magnesium carbonate rock.
2 a) Calculate the free energy change for the decomposition of:
Calcite is pure calcium carbonate.
i) magnesium carbonate
ii) barium carbonate.
b) Are these two compounds stable or unstable relative to decomposition into their
oxide and carbon dioxide at 298 K?
3 Assuming that ∆H 1 and ∆S 1 for the reactions do not vary with temperature,
estimate the temperatures at which the two decomposition reactions become feasible.
4 Down Group 2, do the metal carbonates become more or less stable relative to
decomposition into the oxide and carbon dioxide?
382 13.2 Entropy
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One of the ways that chemists explain the trend in thermal stability of the Group 2
carbonates is to analyse the related energy changes.
enthalpy change
for breaking
M2+(g) + CO32–(g) M2+(g) + O2–(g) + CO2(g)
bonds in the
carbonate ion
lattice energy for lattice energy
the metal carbonate for the metal oxide
enthalpy change
2+ for the decomposition
M CO32–(s) M2+O2–(s) + CO2(g)
of the metal carbonate
to the metal oxide
Figure 13.2.20 An energy cycle for the decomposition of the carbonate of a Group 2 metal, M.
CO32–
O2–
CO32– M2+ CO32– O2– M2+ O2–
O2–
CO32–
Figure 13.2.21 Decomposition of the Group 2 carbonate into its oxide.
5 Which of the enthalpy changes in Figure 13.2.20 are exothermic and which are
endothermic?
6 Why is the lattice energy of magnesium oxide more negative than the lattice energy of
barium oxide?
7 Why is the lattice energy of magnesium oxide more negative than the lattice energy
of magnesium carbonate?
8 Why is the difference between the lattice energies of the metal carbonates and oxides
significant in explaining the trend in thermal stability of the Group 2 carbonates?
9 How does the trend in thermal stability of the metal carbonates down Group 2 relate
to the polarising power of the metal ions?
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Chapter summary
Chapter 13.2 Entropy l Standard molar entropies (in J mol−1 K−1) can be
used to calculate the entropy change of the system
l A reaction is feasible if it tends to go naturally. during a reaction.
l Most reactions that are feasible are also exothermic, ΔS system
1
= Σ ΣS 1[products] − ΣS 1[reactants].
but some endothermic reactions are feasible too. l The entropy change in the surroundings during a
This shows that the enthalpy change alone for a reaction is determined by the size of the enthalpy
reaction does not control whether or not it tends change. ΔS surrounding
1
= − ΔH 1/T.
to occur. l The tendency of an ionic lattice to dissolve in
l Examples of endothermic reactions that are
water is determined by the balance between the
feasible include the reaction of ethanoic acid with change in entropy of the solution and the change
ammonium carbonate and the reaction of solid, in entropy of the surroundings.
hydrated barium hydroxide with solid ammonium l The free energy change, ΔG, is the
chloride. thermochemical quantity (related to the total
l Entropy is a measure of the disorder of a system
entropy change) that chemists usually use to decide
that refers not only to the arrangement of the whether or not a reaction is feasible.
particles in space but also to the numbers of ways ΔG 1 = ΔH 1 − ΔS system
1
of distributing the energy of the system across all l A reaction is feasible if ΔG is negative.
available energy levels. l A reaction that is not feasible at one temperature
l Gases generally have higher standard molar
may become feasible at a different temperature.
entropies than comparable liquids, which have l The free energy change and the equilibrium
higher entropies than similar solids. constant for a reaction both indicate the extent and
l The entropy of a substance increases as the
direction of change. The two quantities are related:
temperature rises, with sudden jumps in the value ΔG 1 = −RT ln K.
of the entropy when there are changes of state l Reactions that are feasible in terms of ΔG have
(melting and boiling). large values of the equilibrium constant.
l The natural direction of change is the direction in
l Some reactions that are feasible according to their
which the total entropy increases. negative ΔG values do not occur because there are
l The total entropy change for a reaction is the
kinetic factors that mean that the change is very
sum of the entropy change of the system and slow.
the entropy change of the surroundings
ΔS total = ΔS system + ΔS surroundings.
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Exam practice questions
1 The graph below shows how the entropy of 3 Electrolysis is usually used to extract aluminium
water changes from 0 K to 450 K. metal. In principle, it should be possible to extract
the metal by heating its oxide with carbon. Use
the data in the table to answer the questions.
Substance Al2O3(s) C(s) Al(s) CO(g)
Entropy/J mol–1 K–1
∆ H 1/kJ mol−1
f −1669 0 0 −111
S 1J mol−1 K−1 50.9 5.7 28.3 198
a) Write the equation for the reduction of
aluminium oxide by carbon, assuming that the
carbon is converted to carbon monoxide.(1)
b) Calculate the standard enthalpy change for
0 the reduction reaction.(3)
0 T1 T2 c) Calculate the standard entropy change for
Temperature/K the reduction reaction.(3)
a) Explain why the molar entropy of water is d) i) Calculate the standard free energy
zero at 0 K.(2) change for the reduction at 298 K.(2)
b) Give reasons for the entropy changes at ii) Determine whether or not the reaction
temperatures T1 and T2. Explain why one is is feasible at 298 K using your
bigger than the other.(3) answer to (d)(i).(1)
c) i) Explain why ΔG = 0 kJ mol−1 for the e) Calculate the minimum temperature at
change of water to steam at 373 K and which the reduction of aluminium oxide by
1 atmosphere pressure.(2) carbon becomes feasible. (2)
ii) The enthalpy change of vaporisation of f) Give a reason why the reaction between
water is +41.1 kJ mol at 373 K. Calculate aluminium oxide and carbon does not
the value of ΔS for the conversion happen to a significant extent until the
of water into steam at 373 K and temperature is about 1000 degrees higher
1 atmosphere pressure. (4) than your answer to (e). (1)
2 Write equations, with state symbols, for these 4 The equation shows the reaction of carbon
changes. Without doing any calculations, state dioxide and water to form glucose:
whether the entropy change for the system 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g)
is positive or negative during each change.
Explain your answers. ΔH 1 = +2879 kJ mol−1
a) Ammonium nitrate decomposing to form ΔS 1 = −256 J K−1 mol−1
dinitrogen oxide and steam. (2) a) Use the equation and the data in the
b) Dissolving potassium chloride in water.(2) table to calculate the standard entropy
c) Reacting nitrogen and oxygen to form of glucose. (2)
nitrogen dioxide. (2) Molecule CO2(g) H2O(l) O2(g)
d) Adding a catalyst to an aqueous solution of
S 1/J K−1 mol−1 214 70 205
hydrogen peroxide.(2)
e) Mixing ammonia and hydrogen iodide to b) Calculate ∆G 1 for the reaction.(2)
form ammonium iodide. (2) c) i) Discuss whether or not this reaction is
feasible at any temperature.(3)
ii) Explain why plants are able to make
glucose from carbon dioxide and water
by photosynthesis. (1)
385
Exam practice questions
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5 Carbon monoxide is one of the products when 6 a) Explain why the standard entropy of
methane burns in a limited supply of air. bromine is greater than that of iodine.(2)
3 b) The enthalpy change of vaporisation of
CH4(g) + 2 O2(g) → CO(g) + 2H2O(g)
∆H 1 = −519 kJ mol−1 ethoxyethane is +26.0 kJ mol−1 at its boiling
∆S 1 = +82 J mol−1 K−1 temperature, which is 35 °C. Calculate the
value of ∆S system
1
for the change shown in
a) Calculate ∆G 1 for the reaction. (2)
this equation. (4)
b) i) Plot a graph to show how ∆G for this
reaction varies with temperature using C2H5OC2H5(l) → C2H5OC2H5(g)
your answer to (a) and the values c) Predict (without any calculations) the sign
below. (2) (positive or negative) of the values of ∆H 1,
Temperature/K 1500 3000 4000 ∆S 1 and ∆G 1 for each of these changes.
∆G/kJ mol−1 −640 −765 −845 i) The decomposition of water into
hydrogen and oxygen. (3)
ii) Determine how ∆S for this reaction ii) The combustion of a hydrocarbon such
varies with temperature based on the as octane in petrol.(3)
shape of the line on the graph.(2) d) The reaction of carbon with steam produces
c) i) Deduce whether or not the reaction of synthesis gas.
methane with oxygen to form carbon
C(s) + H2O(g) ⇋ CO(g) + H2(g)
dioxide is feasible in the temperature
∆H 1 = +132 kJ mol−1
range 200 K to 3000 K. (2)
∆S 1 = +233 J mol−1 K−1
ii) Explain why methane does not burn in
air at 298 K. (2) i) Calculate the value of ∆G 1 for the
d) Sooty carbon is another of the products when reaction.(2)
methane burns in a limited supply of air. ii) Calculate the value of the equilibrium
constant K for the reaction. (3)
CH4(g) + O2(g) → C(s) + 2H2O(g)
The gas constant R = 8.31 J K−1 mol−1.
∆S 1 = −8 J mol−1 K−1
iii) Comment on the values of ∆G 1
Explain the difference in the values for ∆S 1
and K. (2)
for the reaction producing carbon monoxide
iv) Determine the temperature at which
and the reaction that forms soot. (4)
the reaction becomes feasible.(2)
386
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Redox II
14
14.1 Redox reactions
Redox reactions are very important in the natural environment, in living
things and in modern technology. It is no surprise that the Earth, with its
oxygen atmosphere, has an extensive range of redox chemistry. Every year,
oxidation of ions such as Fe2+ in weathered rocks, and oxidation of molecules
such as hydrogen sulfide, carbon monoxide and methane in volcanic gases
(Figure 14.1), removes about one thousand billion (1012) moles of oxygen
from the atmosphere.
Redox reactions are also involved in the metabolic pathways of respiration.
These pathways produce the molecule adenosine triphosphate (ATP). ATP
transfers the energy released when food is oxidised to make possible the
Figure 14.1 Volcanoes release millions movement, growth and all the other activities in living things that need a
of tonnes of reducing gases into the source of energy.
atmosphere where they react with In addition, the voltages of chemical cells are obtained from the energy of
oxygen. The April 2010 eruption of the redox reactions, and redox is also involved in manufacturing processes that
Eyjafjallajokull volcano in Iceland created use electrolysis to make products such as chlorine and aluminium.
an ash cloud that grounded air traffic
around the world. Tip
This first section of this chapter revisits ideas that you met when studying Chapter 3.
The ‘Test yourself’ questions in this section are to help you revise your understanding
of the theory of redox reactions.
Definitions of redox
Descriptions and theories of oxidation and reduction have developed over
Key terms the years. Although there are now several definitions of redox, oxidation
and reduction always go together.
Redox stands for Reduction + Oxidation.
Oxidation originally meant addition of oxygen or removal of hydrogen,
Oxidation involves the loss of electrons
but the term now covers all reactions in which atoms or ions lose electrons.
or an increase in oxidation number.
Chemists have further extended the definition of oxidation to include
An oxidation number is a number molecules by defining oxidation as a change that makes the oxidation
assigned to an atom or ion to describe number of an element more positive, or less negative.
its relative state of oxidation or
Similarly, reduction originally meant removal of oxygen or addition of
reduction.
hydrogen, but the term now covers all reactions in which atoms, molecules
Reduction involves the gain of electrons or ions gain electrons. Defining reduction as a change in which the oxidation
or a decrease in oxidation number. number of an element decreases further extends the concept of reduction.
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Oxidation states and oxidation numbers
Elements in the s block of the Periodic Table have only one oxidation state in
Tip their compounds: this is +1 for Group 1 elements and +2 for Group 2 elements.
However, most elements in the p block and d block form compounds in
Remember the mnemonic OIL RIG –
which their atoms have different oxidation states. Displaying the compounds
Oxidation Is Loss; Reduction Is Gain of
of an element on an oxidation state diagram provides a ‘map’ of its chemistry
electrons.
and shows the different oxidation numbers that it can have (Figure 14.2).
There are strict rules for assigning oxidation numbers; these are shown in
Figure 14.3.
+6 SO3 H2SO4 SO42–
Oxidation number rules
1 The oxidation number of the atoms in uncombined elements is zero.
+4 SO2 H2SO3 SO32– 2 In simple ions, the oxidation number of the element is the charge on the ion.
3 In neutral molecules, the sum of the oxidation numbers of the constituent
elements is zero.
+2
4 In ions containing two or more elements, the charge on the ion is the sum of
the oxidation numbers.
0 S 5 In any compound, the more electronegative element has a negative oxidation
number and the less electronegative element has a positive oxidation number.
6 The oxidation number of hydrogen in all its compounds is +1, except in metal
–2 H2S S2–
hydrides in which it is −1.
Figure 14.2 The oxidation numbers of 7 The oxidation number of oxygen in all its compounds is −2, except in peroxides
sulfur in its different oxidation states. in which it is −1 and OF2 in which it is +2.
8 The oxidation number of chlorine in all its compounds is −1, except in
compounds with oxygen and fluorine in which it is positive.
Figure 14.3 Rules for assigning oxidation numbers.
Half-equations
Half-equations are ionic equations used to describe either the gain or the
loss of electrons during a redox process. Half-equations help to show what is
happening during a redox reaction. Two half-equations can be combined to
give the full equation for a redox reaction.
For example, zinc metal can reduce Cu2+ ions in copper(ii) sulfate solution,
forming copper metal and Zn 2+ ions in zinc sulfate solution. This can be
shown as two half-equations:
● electron loss (oxidation) Zn(s) → Zn 2+(aq) + 2e−
● electron gain (reduction) Cu2+(aq) + 2e− → Cu(s)
This leads to the full equation by balancing the number of electrons lost by
Zn with the number gained by Cu2+(aq):
Zn(s) + Cu2+(aq) → Zn 2+(aq) + Cu(s)
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Test yourself
1 Describe, in terms of gain or loss of electrons, the c) H2O2
redox reactions in these examples: d) SF6
a) the reaction of calcium with iodine to form e) NaH?
calcium iodide
5 Which sulfur molecules or ions in Figure 14.2 can:
b) the changes at the electrodes during the
a) act as an oxidising agent or as a reducing agent
manufacture of aluminium from molten (liquid)
depending on the conditions
aluminium oxide
b) only act as an oxidising agent
c) the reaction of iron with chlorine to form iron(iii)
chloride. c) only act as a reducing agent?
2 Explain why the oxidation number of oxygen is: 6 Draw charts similar to Figure 14.2 to show that:
a) +2 in OF2 a) nitrogen can exist in the −3, 0, +1, +2, +3, +4 and
+5 oxidation states
b) −1 in peroxides such as Na2O2.
b) chlorine can exist in the −1, 0, +1, +3, +5 and
3 State the changes in oxidation number when
+7 oxidation states.
concentrated sulfuric acid reacts with potassium
bromide. 7 Are the named elements below oxidised or reduced
in the following conversions?
2KBr(s) + 3H2SO4(l)
→ 2KHSO4(s) + Br2(l) + SO2(g) + 2H2O(l) a) magnesium to magnesium sulfate
4 What is the oxidation number of each element in: b) iodine to aluminium iodide
a) K IO3 c) hydrogen to lithium hydride
b) N2O5 d) iodine to iodine monochloride, ICl
14.2 Balancing equations for
redox reactions
Half-equations and stoichiometry
One way to arrive at the balanced equation for a redox reaction is to combine
two half-equations. In the reaction, the number of electrons given up in one Key term
half-equation must equal the number taken in the other. The amounts of
The stoichiometry of a reaction is the
different substances involved in a balanced chemical equation is sometimes
amounts in moles of the reactants and
described as the reaction stoichiometry.
products as shown in the balanced
For example, when hydrogen peroxide in acid solution reacts with iron(ii) equation for a reaction. A stoichiometric
ions, the half-equations are: reaction is one that uses up reactants and
produces products in amounts exactly as
H2O2(aq) + 2H+(aq) + 2e− → 2H2O(l)
predicted by the balanced equation.
Fe2+(aq) → Fe3+(aq) + e−
Multiplying the second equation by 2 balances the number of electrons given
up with those taken, and the overall equation can be obtained by adding the
Tip
two half-equations together. When balancing ionic equations for
redox reactions, always check that the
H2O2(aq) + 2H+(aq) + 2e− → 2H2O(l)
overall charges balance.
2Fe2+(aq) → 2Fe3+(aq) + 2e−
Overall: H2O2(aq) + 2H+(aq) + 2Fe2+(aq) → 2H2O(l) + 2Fe3+(aq)
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Oxidation numbers and stoichiometry
Key term It is also possible to use oxidation numbers to balance equations for redox
reactions. This is done by matching the decrease in oxidation number of the
An oxoanion is an ion with the general element that is reduced to the increase in oxidation number of the element
formula is X x Oyz−, where X represents that is oxidised.
any element (metal or non-metal) and O
The oxidation of iron(ii) ions by manganate(vii) ions in acid solution shows
represents an oxygen atom.
how this can be done to balance an equation when oxoanions are involved.
Example
What is the balanced equation for the reaction in which manganate(vii)
ions, MnO4−, in acid solution are reduced to manganese(ii) ions as they
oxidise iron(ii) ions to iron(iii) ions?
Notes on the method
When oxoanions such as MnO4− (and Cr2O72−) react in acid solution, the
oxygen atoms are converted to water molecules by H+ ions. The atoms of
the metal in the oxoanions become stable simple ions.
The initial and final states of the metal element must be known before it
is possible to write the balanced equation.
Answer
Step 1 Write the formulae of the atoms, ions and molecules involved in
the reaction.
MnO4− + H+ + Fe2+ → Mn2+ + H2O + Fe3+
Step 2 Identify the elements which change in oxidation number and the
extent of change.
change of −5 in Mn
MnO4− + H+ + Fe2+ → Mn2+ + H2O + Fe3+
change of +1 in Fe
Step 3 Balance the equation so that the decrease in oxidation number of
one element equals the increase in oxidation number of the other
element.
In this example, the decrease of −5 in the oxidation number of
manganese is balanced by five Fe2+ ions, each increasing their
oxidation number by +1.
MnO4− + H+ + 5Fe2+ → Mn2+ + H2O + 5Fe3+
Step 4 Balance for oxygen and hydrogen.
In this example, the four oxygen atoms of the MnO4− ion join with
eight hydrogen ions to form four water molecules.
MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
Step 5 Finally, check that the overall charges on each side of the equation
balance and then add state symbols.
The net charge on the left is 17+, which is the same as that on
the right. So, the equation for the reaction is:
MnO4−(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
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Oxidising agents and reducing agents
Oxidising agents (oxidants) are chemical reagents that can oxidise other
substances. They do this either by taking electrons away from these substances, Key term
or by increasing their oxidation number. Common oxidising agents include
Oxidising agents (oxidants) take
oxygen, chlorine, bromine, hydrogen peroxide, the manganate(vii) ion in
electrons and are reduced when they
potassium manganate(vii) and the dichromate(vi) ion in potassium or sodium
react.
dichromate(vi) (Figure 14.4).
2 × decrease in oxidation
number of –3
Cr2O72–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l)
6Fe2+(aq) 6Fe3+(aq) + 6e–
6 × increase
in oxidation
number of +1
Figure 14.4 Dichromate(vi) ions act as oxidising agents by taking electrons from iron(ii)
ions in acid solution. An oxidising agent is itself reduced when it reacts.
Some reagents change colour when they are oxidised which makes them
useful for detecting oxidising agents. In particular, a colourless solution of
iodide ions is oxidised to iodine, which turns the solution to a yellow-brown
colour, so long as excess iodide ions are present. Iodine is only very slightly
soluble in water, but it dissolves in a solution containing iodide ions to form
the tri-iodide ion, I3−(aq).
I2(s) + I−(aq) → I3−(aq)
A reagent labelled ‘iodine solution’ is normally I2(s) in KI(aq) which forms
KI3(aq). The I3−(aq) ion is yellow-brown, which explains the colour change
when iodine is produced from iodide ions.
2I−(aq) → I2(aq) + 2e −
electrons taken by oxidising agent
This can be a very sensitive test for oxidising agents if starch is also present
because starch forms an intense blue-black colour with iodine. Moistened
starch-iodide paper is a version of this test which can detect oxidising gases
such as chlorine and bromine vapour.
Reducing agents (reductants) are chemical reagents that can reduce other Key term
substances. They do this either by giving electrons to these substances or by
Reducing agents (reductants) give up
decreasing their oxidation number. Common reducing agents include metals
electrons and are oxidised when they
such as zinc and iron, often with acid, sulfite ions (SO32−), iron(ii) ions and
react.
iodide ions (Figure 14.5).
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2 × decrease in oxidation
number of –5
2MnO4–(aq) + 16H+(aq) + 10e– 2Mn2+(aq) + 8H2O(l)
5SO32–(aq) + 5H2O(l) 5SO42–(aq) + 10H+(aq) + 10e–
5 × increase
in oxidation
number of +2
Figure 14.5 Sulfite ions act as reducing agents by giving electrons to manganate(vii)
ions. A reducing agent is itself oxidised when it reacts.
Some reagents change colour when they are reduced, which makes them
useful for detecting reducing agents (Figure 14.6 and Figure 14.7).
Figure 14.6 A test for reducing agents. Figure 14.7 Another test for reducing agents.
The test: add a solution of purple The test: add orange dichromate(vi) solution
potassium manganate(vii) acidified with acidified with dilute sulfuric acid to the
dilute sulfuric acid to the reducing agent. reducing agent.
The result: the purple solution turns The result: the orange solution turns green as
colourless as purple MnO4− ions are orange Cr2O72− ions are usually reduced to
reduced to very pale pink Mn2+ ions. green Cr3+ ions but sometimes to blue Cr2+ ions.
Test yourself
8 Write half-equations to show what happens when the following act as
oxidising agents:
a) Fe3+(aq) b) Br2(aq)
c) H2O2(aq) in acid solution.
9 Write half-equations to show what happens when the following act as
reducing agents:
a) Zn(s) b) I −(aq)
c) Fe2+(aq).
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10 E xplain why moist starch-iodide paper can be used as a very
sensitive test for chlorine.
11 a) Write the half-equations involved in the reaction between Fe3+
ions in a solution of iron(iii) chloride and iodide ions in potassium
iodide solution.
b) Write a full redox equation for the reaction involved.
c) Describe the change you would see in the solution when the
reaction occurs.
d) How many moles of Fe3+ ions react with one mole of I− ions?
e) Why does iron(iii) iodide not exist?
12 Write a balanced equation for each of these redox reactions:
a) manganese(iv) oxide with hydrochloric acid to form manganese(ii)
ions and chlorine
b) copper metal with nitrate ions in nitric acid to form copper(ii) ions
and nitrogen dioxide gas.
14.3 Redox titrations
In a redox titration, an oxidising agent reacts with a reducing agent. During
the titration, the aim is to measure the volume of a standard solution of
an oxidising agent or a reducing agent that reacts exactly with a measured
volume of the other reagent.
Measuring reducing agents – potassium
manganate(vii) titrations
Potassium manganate(vii) is often chosen to measure reducing agents because
it can be obtained as a pure, stable solid which reacts in acid solution exactly
as in the following equation:
MnO4−(aq) + 8H+(aq) + 5e − → Mn 2+(aq) + 4H 2O(l)
Because of these properties, potassium manganate(vii) is a primary standard
used to make up standard solutions.
No indicator is required in potassium manganate(vii) titrations. On adding
the potassium manganate(vii) solution from a burette, the purple MnO4−
ions change rapidly to the very pale, pink Mn 2+ ions which look colourless
in dilute solution. At the end-point, one drop of excess MnO4− is sufficient
to produce a permanent pale pink colour.
Key terms
A primary standard is a chemical which can be weighed out accurately to make up a
standard solution. A primary standard must:
● be very pure
● not gain or lose mass when exposed to the air
● have a relatively high molar mass so weighing errors are minimised
● react exactly and rapidly as described by the chemical equation.
A standard solution is a solution with an accurately known concentration. The method
of preparing a standard solution is to dissolve a weighed sample of a primary standard
in water and then make the solution up to a definite volume in a graduated flask.
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Example
Two iron tablets (mass 1.30 g) containing iron(ii) 2 Amount of MnO4− reacting in the titration
sulfate were dissolved in dilute sulfuric acid and
made up to 100 cm3 (Figure 14.8). 10.0 cm3 of this 12.00
= dm3 × 0.00500 mol dm−3
solution required 12.00 cm3 of a standard solution 1000
of 0.00500 mol dm−3 KMnO4 to produce a faint red
Therefore the amount of Fe2+ reacting in the titration
colour. What is the percentage of iron in the iron
tablets? (Fe = 55.8) 12.00
= × 0.00500 × 5 mol
Notes on the method 1000
1 Write the half-equations and work out the amounts 3 Amount of Fe2+ in 100 cm3 of solution (2 tablets)
in moles of Fe2+ and MnO4− that react. 12.00
= × 0.00500 × 5 × 10.0 mol
2 Calculate the amount of MnO4− that reacts in the 1000
titration, and hence the amount of Fe2+ which reacts.
3.00
3 Work out the amount of Fe2+ in the whole solution, = mol
and hence in the tablets dissolved. 1000
4 Mass of Fe2+ in 2 tablets
4 Calculate the percentage of iron in the tablets.
3.00
= mol × 55.8 g mol−1
Answer 1000
1 The half-equations for the reaction are:
= 0.1674 g
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O and
Therefore the percentage of iron in the tablets
(Fe2+ → Fe3+ + e−) × 5
0.1674
= × 100 = 12.9%
Therefore 5 mol Fe2+ react with 1 mol MnO4 −.
1.30
safety filler
pipette
solution containing
1.30 g iron tablets KMnO4 solution,
3
in 100 cm concentration = 0.0050 mol dm–3
burette
100 cm3 conical flask
solution
average accurate titration
3
10.0 cm = 12.00 cm3 KMnO4 solution
Figure 14.8 Finding the percentage of iron in iron tablets.
394 14 Redox II
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Test yourself
13 Write half-equations and the full ionic equation for the reactions
between acidified potassium manganate(vii) and:
a) iron(ii) sulfate solution
b) hydrogen peroxide solution.
14 Use your answers to Question 13 to calculate the volume of
0.0200 mol dm−3 potassium manganate(vii) solution required to
oxidise 20.0 cm3 of:
a) 0.100 mol dm−3 iron(ii) sulfate solution
b) 0.200 mol dm−3 hydrogen peroxide solution.
Measuring oxidising agents – iodine/
thiosulfate titrations
Many oxidising agents rapidly convert iodide ions to iodine by taking away
electrons.
2I−(aq) → I2(aq) + 2e−
The iodine which forms can be titrated with thiosulfate ions and be reduced
back to iodide ions.
I2(aq) + 2e− → 2I−(aq)
2S2O32−(aq) → S4O62−(aq) + 2e−
thiosulfate ion tetrathionate ion
These two half-equations show that 2 mol of thiosulfate ions reduce 1 mol of
iodine molecules.
This system can be used to investigate quantitatively any oxidising agent that
can oxidise iodide ions to iodine. Oxidising agent that can be estimated by this
method include iron(iii) ions, copper(ii) ions, chlorine, and manganate(vii)
ions in acid.
The procedure is:
● Add excess potassium iodide to a measured quantity of the oxidising agent,
which then converts iodide ions to iodine.
● Titrate the iodine formed with a standard solution of sodium thiosulfate. Tip
The greater the amount of oxidising agent added, the more iodine is formed When using thiosulfate to titrate iodine
and the more thiosulfate is needed from the burette to react with it. formed by an oxidising agent, iodide
ions are first oxidised to iodine and
Iodine is not soluble in water. It dissolves in potassium iodide solution to
then reduced back to iodide. This
form a solution that is dark brown when concentrated but pale yellow when
means you do not need to include
dilute.
the iodine in the calculation; it is
At the end-point, the pale yellow iodine colour disappears to give a colourless present as a ‘go between’. You can
solution. Adding a few drops of starch solution just before the end-point relate the thiosulfate directly to the
makes the colour change much sharper. Starch gives a deep blue colour with oxidising agent with the help of the half-
iodine that disappears at the end-point. equations.
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The following practical illustrates the use of iodine/thiosulfate titrations to
determine the concentration of supermarket bleaches.
Tip
Refer to Practical skills sheet 17, ‘Measuring chemical amounts by titration’, which you
can access online at [Link]/EdexcelChemistry.
Core practical 11
A redox titration
The active reagent in household bleaches is sodium chlorate(i),
NaClO (Figure 14.9). To increase the cleaning power of these
bleaches manufacturers usually add detergents, and to improve
their smell they add perfumes. Sodium chlorate(i) is a strong
oxidising agent which bleaches by oxidising coloured materials
to colourless or white substances.
The half-equation when sodium chlorate(i) acts as an oxidising
agent is:
ClO −(aq) + 2H+(aq) + 2e− → Cl−(aq) + H2O(l)
A student is asked to determine the concentration of
sodium chlorate(i) in a supermarket bleach. The bleach is
too concentrated to be titrated directly, so it first has to be
diluted.
Using a measuring cylinder, 100 cm3 of the bleach is added
to a graduated flask and made up to a volume of 1000 cm3.
10.0 cm3 of the diluted solution is then pipetted into a conical Figure 14.9 Pouring concentrated bleach into a bucket
flask, followed by the addition of excess potassium iodide. before use for cleaning.
The iodine produced is finally titrated with 0.100 mol dm−3 7 Give examples of random and systematic errors that can
sodium thiosulfate solution, giving an average accurate titre of affect the results from this procedure and explain how can
26.60 cm3. they be minimised.
8 With the help of Practical skills sheet 15, ‘Identifying errors
1 Write a half-equation for the oxidation of iodide ions to iodine. and estimating uncertainties’, which you can access online
2 Write a balanced equation for the reaction of chlorate(i) ions at [Link]/EdexcelChemistry,
with iodide ions in acid solution to form iodine, chloride ions calculate:
and water. a) the uncertainty and percentage uncertainty in:
3 Write a balanced equation for the reaction of iodine with i) the volume of undiluted bleach taken
thiosulfate ions during the titration. ii) the volume of diluted bleach pipetted
4 Using the equations from Questions 2 and 3, work out the iii) the volume of thiosulfate titrated
number of moles of thiosulfate that react with the iodine iv) the concentration of the thiosulfate solution
produced by 1 mol of chlorate(i) ions. b) the total percentage uncertainty in the mass of sodium
5 Calculate the number of moles of thiosulfate in the average chlorate(i) in 100 cm3 of undiluted bleach.
accurate titration, and hence the amount in moles of sodium 9 Finally, write your result for the mass of sodium chlorate(i) in
chlorate(i) in 10.0 cm3 of the diluted bleach. undiluted bleach in the form x ± y g per 100 cm3.
6 Calculate the mass of sodium chlorate(i) in 100 cm3 of
undiluted bleach. (Na = 23.0, Cl = 35.5, O = 16.0)
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14.4 Electrode potentials
Electrochemical cells
Redox reactions, like all reactions, tend towards a state of dynamic
equilibrium. Redox reactions involve electron transfer and chemists have Key term
developed electrochemical cells based on redox changes. Some of these
cells have great practical and technological importance while others, Electrochemical cells produce
particularly fuel cells, are becoming a serious alternative to oil-based fuels an electric potential difference
for vehicles. Measurements of the voltages of cells help to assess the feasibility (voltage) from a redox reaction. In
and likelihood of redox reactions. an electrochemical cell, the two half
reactions happen in separate half
Redox reactions involve the transfer of electrons from a reducing agent to cells. The electrons flow from one cell
an oxidising agent. The electron transfer can be shown by writing half- to the other through a wire connecting
equations. So, for example, when zinc is added to copper(ii) sulfate solution, the electrodes. The electric circuit is
Zn atoms give up electrons to form Zn 2+ ions. At the same time, the electrons completed by a salt bridge connecting
are transferred to Cu2+ ions, which form Cu atoms. the two solutions.
The two half-equations for the reaction are:
Zn(s) → Zn 2+(aq) + 2e −
and Cu2+(aq) + 2e − → Cu(s)
The overall balanced equation is:
Zn(s) + Cu2+(aq) → Zn 2+(aq) + Cu(s)
Instead of mixing two reagents, it is possible to carry out a redox reaction
in an electrochemical cell so that electron transfer takes place along a wire
connecting the two electrodes. This harnesses the energy from the redox
reaction to produce an electrical potential difference (voltage).
Test yourself
15 Write two ionic half-equations and the overall balanced equation for
each of the following redox reactions. In each example, state which
atom, ion or molecule is oxidised and which is reduced:
a) magnesium metal with copper(ii) sulfate solution
b) aqueous chlorine with a solution of potassium bromide
c) a solution of silver nitrate with copper metal.
One of the first useable cells was based on the reaction of zinc metal with
aqueous copper(ii) ions (Figure 14.10). In the cell, zinc is oxidised to zinc(ii)
ions as copper(ii) ions are reduced to copper metal.
In electrochemical cells, the two half-reactions happen in separate half-cells.
The electrons flow from one cell to the other through a wire connecting
the electrodes. A salt bridge connecting the two solutions completes the
electrical circuit.
14.4 Electrode potentials 397
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high resistance voltmeter
V
zinc strip copper strip
salt bridge
solution of Zn2+(aq) solution of Cu2+(aq)
–3
(1 mol dm–3) (1 mol dm )
Figure 14.10 An electrochemical cell based on the reaction of zinc metal with aqueous
copper(ii) ions. In this cell, electrons tend to flow from the negative zinc electrode to the
positive copper electrode through the external circuit.
The salt bridge makes an electrical connection between the two halves of
Key term the cell by allowing ions to flow while preventing the two solutions from
mixing. At its simplest, a salt bridge consists of a strip of filter paper soaked in
The electromotive force (e.m.f.) of a
saturated potassium nitrate solution and folded over each of the two beakers.
cell measures the maximum ‘voltage’
produced by an electrochemical cell. All potassium salts and all nitrates are soluble so the salt bridge does not
The symbol for e.m.f. is E and its SI react to produce precipitates with any of the ions in the half-cells. In more
unit is the volt (V). The e.m.f. is the permanent cells, a salt bridge may consist of a porous solid such as sintered
energy transferred in joules per coulomb glass.
of charge flowing through the circuit
Chemists measure the tendency for the current to flow in the external circuit
connected to a cell. Cell e.m.f.s are at a
by using a high-resistance voltmeter to measure the maximum cell e.m.f.
maximum when no current flows because
when no current is flowing.
under these conditions no energy is lost
due to the internal resistance of the cell In Figure 14.10, electrons tend to flow out of the zinc electrode (negative)
as the current flows. through the external circuit to the copper electrode (positive). The maximum
voltage of the cell, usually called its electromotive force (e.m.f.), or cell
potential, is 1.10 V under standard conditions.
Test yourself
16 Identify the oxidation and reduction reactions that take place in the
cell in Figure 14.10 when a current flows and state where these
processes take place.
17 Why is the copper strip in Figure 14.10 the positive electrode and
the zinc strip the negative electrode?
Standard conditions
In order to compare the voltages (e.m.f.s) developed by different
electrochemical cells, scientists carry out the measurements under standard
conditions. These standard conditions for electrochemical measurements are
the same as those for thermochemical measurements. They are:
● temperature 298 K (25 °C)
● gases at a pressure of 100 kPa
−3
● solutions at a concentration of 1.0 mol dm .
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Chemists have also developed a convenient shorthand called a cell diagram
for describing cells. The cell diagram for the cell in Figure 14.10 is shown in
Tip
Figure 14.11 with an explanation below each entry. Under standard Use ‘ROOR’ as a reminder that cells
conditions, the symbol for the e.m.f. of the cell is E cell
1
and this is called the are always written with the reduced
standard e.m.f. of the cell. form (the reducing agent) of each
half-electrode on the outside and the
Zn(s) Zn2+(aq) Cu2+(aq) Cu(s) oxidised form (the oxidising agent) on
metal electrode metal ion in solution metal ion in solution metal electrode
the inside.
(reduced form (oxidised form (oxidised form (reduced form
of the electrode) of the electrode) of the electrode) of the electrode)
Figure 14.11 A cell diagram for the cell composed of the Zn2+(aq) ∣ Zn(s) and
Cu2+(s) ∣ Cu(s) half-cells. The solid vertical lines, ∣ , separate the different physical states
of each half-electrode. The double dotted line, , represents the salt bridge. Note that the
reduced form of each electrode appears towards the outside of the cell diagram. This is
the general rule.
If the cell e.m.f. is positive, the reaction in the cell tends to go according to
the cell diagram reading from left to right. As a current flows in the external
circuit connecting the two electrodes in Figure 14.11, zinc atoms turn into
zinc ions and go into solution, while copper ions turn into copper atoms and
deposit on the copper electrode (Figure 14.12).
reaction
– +
Zn(s) Zn2+(aq) Cu2+(aq) Cu(s) Ecell = +1.10 V
Zn(s) Zn2+(aq) + 2e– 2e– + Cu2+(aq) Cu(s)
elec
via
ext trons it
ernal circu
Figure 14.12 The direction of change in an electrochemical cell.
Test yourself
18 Consider a cell based on the redox reaction below that tends to go in
the direction shown.
Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s)
The potential difference between the electrodes is +1.61 V.
a) Write the half-equations for the electrode processes when the
cell supplies a current.
b) Write the conventional cell diagram for the cell, including the
value for E cell
1
.
Standard electrode potentials
The study of many cells has shown that a half-electrode, such as the
Cu2+(aq) ∣ Cu(s) electrode, makes the same contribution to the cell e.m.f. in
any cell, so long as the measurements are made under the same conditions.
But there is no way of measuring the e.m.f. of an isolated, single electrode
because it has only one terminal.
14.4 Electrode potentials 399
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Chemists have solved this problem by selecting a standard electrode system
Tip as a reference electrode against which they can compare all other electrode
It is only in a standard hydrogen systems. The chosen reference electrode is the standard hydrogen electrode.
electrode that the platinum metal is
By convention, the electrode potential of the standard hydrogen electrode is
covered with finely divided platinum
zero. This is represented as:
black. This helps to maintain an
equilibrium between hydrogen gas and Pt[H2(g)] ∣ 2H+(aq) E 1 = 0.00 V
hydrogen ions and ensure a reversible
The standard electrode potential for any half-cell is measured relative to
reaction between them.
a standard hydrogen electrode under standard conditions (Figure 14.13). A
standard hydrogen electrode sets up an equilibrium between hydrogen ions
in solution (1.00 mol dm−3) and hydrogen gas (100 kPa pressure) at 298 K on
the surface of a platinum electrode coated with platinum black.
Figure 14.13 The e.m.f. of this cell under V
standard conditions is, by definition, H2(g) at high resistance voltmeter
the standard electrode potential of the 298 K and copper strip
100 kPa pressure
Cu2+(aq) ∣ Cu(s) electrode for which
E 1 = +0.34 V. salt bridge
acid solution
containing
+ –3
H (aq) (1 mol dm )
solution of
platinum electrode Cu2+(aq)
coated with finely (1 mol dm–3)
divided platinum
black holes in glass
for bubbles of H2
gas to escape
By convention, when a standard hydrogen electrode is the left-hand
electrode in an electrochemical cell, the cell e.m.f. is the electrode potential
of the right-hand electrode.
Key terms So, the conventional cell diagram for the cell which defines the standard
electrode potential of the Cu2+(aq) ∣ Cu(s) electrode is:
A standard hydrogen electrode is
a half-cell in which a 1.00 mol dm−3 Pt[H2(g)] ∣ 2H+(aq) Cu2+(aq) ∣ Cu(s) E 1 = +0.34 V
solution of hydrogen ions is in
The electrode and its standard electrode potential are often represented more
equilibrium with hydrogen gas at
simply as:
100 kPa pressure on the surface of
a platinum electrode coated with Cu2+(aq) + 2e− ⇋ Cu(s) E 1 = +0.34 V
platinum black at 298 K. oxidised form reduced form
The standard electrode (reduction) This also serves to emphasise that standard electrode potentials represent
potential, E 1, of a standard half-cell reduction processes.
is the e.m.f. of that half-cell relative
A hydrogen electrode is difficult to set up and maintain, so it is much easier
to a standard hydrogen electrode
to use a secondary standard such as a silver/silver chloride electrode or a
under standard conditions. Standard
calomel electrode as a reference electrode. These electrodes are available
electrode potentials are sometimes
commercially and are reliable to use. They have been calibrated against
called standard redox potentials.
a standard hydrogen electrode. Calomel is an old-fashioned name for
A reference electrode is used to mercury(i) chloride. The cell reaction and electrode potential relative to a
measure electrode potentials in place hydrogen electrode for a calomel electrode are:
of the standard hydrogen electrode.
Hg2Cl 2(s) + 2e− ⇋ 2Hg(l) + 2Cl−(aq) E 1 = +0.27 V
400 14 Redox II
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When used as a reference electrode, on the left of the cell, the reverse
reaction, E 1 has the opposite sign. This gives:
2Hg(l) + 2Cl−(aq) ⇋ Hg2Cl 2(s) + 2e− E 1 = −0.27 V
Figure 14.13 shows how to measure the standard electrode potential of metals
in contact with their ions in aqueous solution. However, it is also possible
to measure the standard electrode potentials of electrode systems in which
both the oxidised and reduced forms are ions in solution, such as ions of the
same element in different oxidation states. In these cases, the electrode in the
system is platinum (Figures 14.14 and 14.15).
V
H2(g) at high resistance voltmeter
298 K and
solution of Fe3+(aq)
100 kPa pressure
and Fe2+(aq) both
salt bridge at 1 mol dm–3
platinum
electrode
coated with
finely divided shiny platinum
platinum black electrode
acid solution
+
containing H (aq)
(1 mol dm–3)
Figure 14.14 The apparatus in a cell for measuring the standard electrode potential of
the redox reaction Fe3+(aq) + e− ⇋ Fe2+(aq).
+ 3+ 2+
Pt[H2(g)] 2H (aq) Fe (aq) , Fe (aq) Pt(s)
hydrogen gas hydrogen ion metal ion metal ion shiny platinum
on Pt electrode in solution in solution in solution (inert electrode)
coated with (oxidised form (oxidised form (reduced form
Pt black of the electrode) of the electrode) of the electrode)
(reduced form
of the electrode)
Figure 14.15 The cell diagram for the cell in Figure 14.14. Here both the reduced and
oxidised forms of the chemicals in right-hand half-cell are in solution. As in Figure 14.11, the
reduced form of each electrode system appears towards the outside of the cell diagram.
Test yourself
19 Suggest why a hydrogen electrode is difficult to set up and maintain.
20 Why is it not possible to measure the electrode potential for the
Na+(aq) ∣ Na(s) system using the method illustrated in Figure 14.14?
21 Why do you think that platinum metal is used as the electrode for
systems in which both the oxidised and reduced forms are ions in
solution, such as Fe3+(aq) and Fe2+(aq)?
22 What are the half-equations and standard electrode potentials of the
right-hand electrode in each of the following cells?
a) Pt[H2(g)] ∣ 2H+(aq) Sn2+(aq) ∣ Sn(s)
E 1 = −0.14 V
b) Pt[H2(g)] ∣ 2H+(aq) Br2(aq), 2Br−(aq) ∣ Pt(s)
E 1 = +1.07 V
c) Pt ∣ [2Hg(l) + 2Cl−(aq)], Hg2Cl2(s) Cr3+(aq) ∣ Cr(s)
E 1 = −1.01 V
14.4 Electrode potentials 401
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Test yourself
23 The standard electrode potential for the Cu2+(aq) ∣ Cu(s) electrode is
+0.34 V. For the cell, Cu(s) ∣ Cu2+(aq) Pb2+(aq) ∣ Pb(s), the standard
cell e.m.f., E cell
1
, equals −0.47 V.
What is the standard electrode potential for the Pb2+(aq) ∣ Pb(s)
electrode?
24 a)
What is the e.m.f. when a standard calomel electrode is
connected to a standard Cu2+(aq) ∣ Cu(s) electrode?
b) Write half-equations for the reactions at the electrodes.
14.5 Cell e.m.f.s and the direction of
change
Chemists use standard electrode potentials:
● to calculate the e.m.f.s (standard potentials) of electrochemical cells
Key term ● to predict the direction (feasibility) of redox reactions.
Feasibility: a feasible reaction is one The data sheet entitled ‘Standard electrode potentials’, which you can access
that naturally tends to happen, even online at [Link]/EdexcelChemistry, lists redox half-
if it is very slow because it has a high reactions in order of their standard electrode (reduction) potentials from the
activation energy. most negative to the most positive.
The size and sign of a standard electrode potential shows how likely it is
that a half-reaction will occur. The more positive the standard electrode
potential, the more likely it is that the half-reaction will occur.
So, the half-reaction H2O2(aq) + 2H+(aq) + 2e− → 2H2O(l) with a standard
electrode potential of +1.77 V near the bottom of the table has a stronger
tendency to change in the direction shown by the equation than the half-
reaction Li+(aq) + e− → Li(s) with a standard electrode potential of −3.03 V
at the top of the list. Indeed, this second half-reaction is much more likely
to occur in the opposite direction, as it does when it contributes a voltage of
+3.03 V to any electrochemical cell (Figure 14.16).
Using a table of standard electrode potentials, it is possible to calculate cell
e.m.f.s (E cell
1
values) by combining the standard electrode potentials of the
two half-cells that make up the full cell. This provides a prediction of the
expected direction of chemical change for the redox reaction in the cell.
Look again at Figure 14.12 and the value of the cell e.m.f., E cell
1
. The value
of +1.10 V arises from the sum of the E values for two half-reactions:
Cu2+(aq) + 2e− → Cu(s) E 1 = +0.34 V
Figure 14.16 Lithium batteries come in a and the reverse of:
range of sizes.
Zn 2+(aq) + 2e− → Zn(s) E 1 = −0.76 V
Combining these two half-equations gives:
Cu2+(aq) + 2e− → Cu(s) E 1 = +0.34 V
Zn(s) → Zn 2+(aq) + 2e− E 1 = +0.76 V
Overall: Zn(s) + Cu2+(aq) → Zn 2+(aq) + Cu(s) E cell
1
= +1.10 V
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This shows that:
Tip
E cell
1
= −E (left-hand
1 1
electrode) + E (right-hand electrode)
Half-equations may have to be
= E (right-hand
1 1
electrode) − E (left-hand electrode) multiplied to balance the electrons
gained with the electrons lost when
Positive values of E cell
1
indicate that the sign of the right-hand electrode is positive
balancing equations. Note that this
and that a reaction is tends to go in the direction of the overall equation. The
does not apply to E 1 values.
more positive the value of E cell 1
, the greater the tendency for the reaction to
happen. On the other hand, negative values for E cell 1
indicate that reactions tends
to go in the opposite direction to that shown in the cell diagram.
Example
Write the cell diagram for a cell based on the two half-equations below.
Work out the e.m.f. of the cell and write the overall equation for the
reaction which is likely to occur (the feasible reaction).
Fe3+(aq) + e− ⇋ Fe2+(aq) E 1 = +0.77 V
Cu2+(aq) + 2e− ⇋ Cu(s) E 1 = +0.34 V
Notes on the method
Write the cell diagram with the more positive electrode on the right.
Then use the equation:
E cell = E (right-hand electrode) − E (left-hand electrode)
1 1 1
to calculate the cell e.m.f.
Answer
The Fe3+(aq), Fe2+(aq) electrode is the more positive so it should be on
the right-hand side of the cell diagram. Both the oxidised and reduced
forms are in solution so a shiny platinum electrode is needed.
Fe3+(aq), Fe2+(aq) ∣ Pt(s)
The left-hand electrode is Cu2+(aq) ∣ Cu(s) so the reduced from is copper metal
which can also be the conducting electrode. The cell diagram is therefore:
Cu(s) ∣ Cu2+(aq) Fe3+(aq), Fe2+(aq) ∣ Pt(s)
and E cell
1
= (+0.77 V) − (+0.34 V) = +0.43 V
E cell
1
is positive and so the reaction tends to go from left to right in the
direction of the cell diagram. Balancing the two half-equations in this
direction gives the overall equation:
Cu(s) + 2Fe3+(aq) → Cu2+(aq) + 2Fe2+(aq)
Test yourself
25
Write the cell diagram for a cell based on each of the a) V3+(aq) + e− ⇋ V2+(aq)
following pairs of half-equations. For each example, Zn2+(aq) + 2e− ⇋ Zn(s)
find the standard electrode potentials from the data b) Br2(aq) + 2e− ⇋ 2Br −(aq)
sheet for this chapter accessed online at www. I2(aq) + 2e− ⇋ 2I−(aq)
[Link]/EdexcelChemistry. Work out the
c) Cl2(aq) + 2e− ⇋ 2Cl−(aq)
e.m.f. of the cell and write the overall equation for the
PbO2(s) + 4H+(aq) + 2e− ⇋ Pb2+(aq) + 2H2O(l)
reaction that tends to happen (the feasible reaction):
14.5 Cell e.m.f.s and the direction of change 403
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Core practical 10
Investigating some electrochemical cells
The unlabelled diagram in Figure 14.17 can be used to investigate cells made
using strips of metals dipping into solutions of their own ions.
V
Figure 14.17 Outline of an apparatus for setting up chemical cells. Figure 14.18 The strip of zinc in this test tube
was dipped into a solution of copper(ii) sulfate
1 Copy Figure 14.17 and label it to show how to investigate a cell combining solution that was an even darker blue at the start.
an Ag+(aq) ∣ Ag(s) electrode and a Zn2+(aq) ∣ Zn(s) electrode under standard
conditions.
2 What is used to make the part of the apparatus that links the solutions in the
two beakers? Explain its purpose.
Table 14.1 shows the results of measuring the cell e.m.f.s of three cells. The
concentration of the silver nitrate solution used was 0.10 mol dm−3 because of the
high cost of the silver salt.
Table 14.1
Cell Negative electrode Positive electrode Cell e.m.f./V
A Cu2+(aq) ∣ Cu(s) Ag+(aq) ∣ Ag(s) 0.40
B Zn2+(aq) ∣ Zn(s) Cu2+(aq) ∣ Cu(s) 1.06
C Zn2+(aq) ∣ Zn(s) Ag+(aq) ∣ Ag(s) 1.48
3 For cell A:
a) write the half-equation for the reaction taking place at the copper electrode
and explain why this electrode is negative
b) write the half-equation for the reaction taking place at the silver electrode
and explain why this electrode is positive.
Figure 14.19 The copper wire in this test tube
4 Write half-equations for the reactions at each of the electrodes in cells B and C. was originally added to a colourless solution of
5 Write the overall cell reaction for each of the three cells. silver nitrate solution.
6 Give the conventional cell diagram for each cell in Table 14.1. Use Table 14.2
on page 405 to work out the expected e.m.f. of each cell under standard
conditions and comment on any differences with the experimental values.
7 Explain the observations in Figures 14.18 and 14.19 and show that they are
consistent with the results shown in Table 14.1.
404 14 Redox II
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The electrochemical series Table 14.2 The standard electrode
potentials of some common metals.
A list of electrode systems set out in order of their electrode potentials (as on
the data sheet ‘Standard electrode potentials’, which you can access online Metal ion/metal Standard electrode
electrode potential, E 1/V
at [Link]/EdexcelChemistry) is a useful guide to the
behaviour of oxidising and reducing agents. It is an electrochemical series. Li+(aq) ∣ Li(s) −3.03
K+(aq) ∣ K(s) −2.92
The metal ion ∣ metal electrodes with highly negative electrode potentials
involve half-reactions for Group 1 metal ions and metals (Table 14.2). Na+(aq) ∣ Na(s) −2.71
Lithium is the most reactive of these metals when it reacts as a reducing Al3+(aq) ∣ Al(s) −1.66
agent forming metal ions. Consequently, the reverse reaction of Li+ ions −0.76
Zn (aq) ∣ Zn(s)
2+
forming Li metal is least likely and this results in the most negative standard
Fe (aq) ∣ Fe(s)
2+ −0.44
electrode potential.
Pb (aq) ∣ Pb(s)
2+ −0.13
Cu (aq) ∣ Cu(s)
2+ +0.34
Tip
Ag (aq) ∣ Ag(s)
+ +0.80
The data in tables such as Table 14.2 show reduction potentials for changes that can
be represented as:
oxidised form (oxidising agent) + electron(s) → reduced form (reducing agent).
The metal ion ∣ metal electrodes with positive electrode potentials involve
half-reactions for d-block metal ions and metals low in the reactivity series,
such as copper and silver. These metals are relatively unreactive as reducing Tip
agents and they do not react with dilute acids to form hydrogen gas. However, In whatever order electrode (reduction)
their ions are readily reduced to the metal, which results in positive standard potentials are tabulated, it is always
electrode potentials. true that:
The order of metal ion/metal systems in Table 14.2 closely corresponds to ● the half-cell with the most positive
the reactivity series for metals and the reactions shown by metal/metal ion electrode potential has the greatest
displacement reactions (Figures 14.18 and 14.19). tendency to gain electrons, so the
The electrode potentials of the half-equations involving halogen molecules species on the left-hand side of the
and halide ions are positive. The F2(aq) ∣ 2F−(aq) system is the most positive, half-equation is the most powerfully
showing that fluorine is the most reactive of the halogens as an oxidising oxidising
agent. The next most reactive halogen is chlorine, then bromine and finally ● the half-cell with the most negative
iodine is the least reactive. This corresponds to the order of reactivity of the electrode potential has the greatest
halogens and the results of their displacement reactions. tendency to give up electrons, so the
species on the right-hand side of the
An electrochemical series based on electrode potentials can be used to predict half-equation is the most powerfully
the direction of change in redox reactions. This is an alternative approach to reducing.
the use of cell diagrams to make predictions.
Test yourself
For Questions 26 and 27, refer to the standard electrode potential values
from the data sheet for this chapter, which you can access online at www.
[Link]/EdexcelChemistry.
26 Using the standard electrode potential values, arrange the following
sets of metals in order of decreasing strength as reducing agents:
a) Ca, K, Li, Mg, Na
b) Cu, Fe, Pb, Sn, Zn.
14.5 Cell e.m.f.s and the direction of change 405
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Test yourself
27 Using the standard electrode potential values, arrange the following
sets of molecules or ions in order of decreasing strength as
oxidising agents in acid solution:
a) Cr2O72−, Fe3+, H2O2, MnO4−
b) Br2, Cl2, ClO −, H2O2, O2.
Example
Use electrode potentials to predict what happens when chlorine is added
to a solution of iodide ions.
Notes on the method
The first step is to identify the two half-equations. Write them down one
above the other.
The more positive half-reaction tends to go from left to right, taking in
electrons, while the more negative half-reaction goes from right to left.
Answer
Figure 14.20 shows how to predict the direction of change for the two
half-equations involved when chlorine reacts with iodide ions.
More
negative I2(aq) + 2e– 2I–(aq) E = +0.54 V
electrode
More
positive CI2(aq) + 2e– 2CI–(aq) E = +1.51 V
electrode
Figure 14.20 Chlorine is a stronger oxidising agent than iodine. Iodide ions are stronger
reducing agents than chloride ions.
As expected, the electrode potentials predict that chlorine displaces
iodine from a solution of aqueous iodide ions.
Disproportionation reactions
Electrode potentials to predict whether or not disproportionation reactions
are likely to occur. During a disproportionation reaction, the same element
both increases and decreases its oxidation number. Figure 14.21 on the next
page shows that copper(i) ions do tend to disproportionate in aqueous solution
while iron(ii) ions do not.
Key term
A disproportionation reaction is a change in which the same element both increases
and decreases its oxidation number. Some of the element is oxidised while the rest of
it is reduced.
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Figure 14.21 Copper(i) ions
More disproportionate in aqueous solution, but
negative Cu2+(aq) + e– Cu+(aq) E = +0.15 V iron(ii) ions do not.
electrode
copper(I)
disproportionates
More
positive Cu+(aq) + e– Cu(s) E = +0.52 V
electrode
More
negative Fe2+(aq) + 2e– Fe(s) E = –0.44 V
electrode
iron(III) reacts with iron
metal to form iron(II)
More
positive Fe3+(aq) + e– Fe2+(aq) E = +0.77 V
electrode
Test yourself Tip
28
Using the standard electrode potential values from the data Note that strictly the term
sheet for this chapter, which you can access online at www. ‘disproportionation’ refers to a
[Link]/EdexcelChemistry, show that: particular element in a compound not
to the compound as a whole. However,
a) hydrogen peroxide tends to disproportionate to give oxygen and
as in Question 28, it is common to
water under acid conditions
refer to the disproportionation of
b) nitrous acid, HNO2, tends to disproportionate to give nitrate(v) compounds.
ions and nitrogen monoxide under acid conditions.
The limitations of predictions from E 1 data
Although the electrode potentials for a redox reaction suggest that a reaction
should take place, in practice the reaction may be too slow for any change to
be observed. In other words, the E 1 values show whether or not a reaction
is thermodynamically feasible, but they do not give any indication about
the rate of the reaction. There may be something to inhibit the reaction
kinetically, so that the reaction mixture is inert.
For example, E 1 values predict that Cu2+(aq) should oxidise H2(g) to
H+ ions:
E cell
Cu2+(aq) + H2(g) → Cu(s) + 2H+(aq) 1
= +0.34 V
However, nothing happens when hydrogen is bubbled into copper(ii) sulfate
solution because the activation energy is so high that the reaction rate is
effectively zero.
A second important point about E 1 values is that they relate only to
standard conditions. Changes in concentration, temperature and pressure
affect electrode potentials. In particular, all electrode (reduction) potentials
become more positive if the concentration of reactant ions is increased and
less positive if their concentration is reduced. This means that some reactions
which are not possible under standard conditions occur under non-standard
conditions, and vice versa.
14.5 Cell e.m.f.s and the direction of change 407
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For example, under standard conditions, MnO2 does not oxidise 1.0 mol dm−3
HCl(aq) to Cl 2.
MnO2(s) + 4H+(aq) + 2e− ⇋ Mn 2+(aq) + 2H2O E 1 = +1.23 V
Cl 2(aq) + 2e− ⇋ 2Cl−(aq)
E 1 = +1.36 V
Chlorine is the stronger oxidising agent under standard conditions.
But if MnO2 is heated with concentrated HCl, the electrode potentials
of both half-equations change in such a way that chlorine is produced, as
predicted by equilibrium theory. This is because, in hydrochloric acid, the
hydrogen ion concentration is about 12 mol dm−3. Under these non-standard
conditions manganese(iv) oxide is a stronger oxidising agent with a more
positive electrode potential. The higher chloride ion concentration in the
concentrated acid means that chlorine is less powerfully oxidising, and so its
electrode potential is less positive. These changes are sufficient to reverse the
predicted direction of change.
So, predictions from cell e.m.f.s about the feasibility of redox reactions may
not occur in practice due to kinetic effects (slow reaction rates) or non-
standard conditions of concentration and temperature.
Test yourself
For these questions, refer to the standard electrode potential values from
the data sheet for this chapter, which you can access online at www.
[Link]/EdexcelChemistry.
29 a) Using the standard electrode potential values, show that
aluminium is expected to react with dilute hydrochloric acid.
b) Suggest a reason why there is very little change at first when a
piece of aluminium foil is added to 1.0 mol dm−3 hydrochloric acid
at room temperature.
30 Using the standard electrode potential values, explain why when
copper reacts with dilute nitric acid the product is nitrogen dioxide
and not hydrogen.
14.6 How far and in which direction?
Chapter 11 shows that the direction and extent of a chemical reaction can
be described by its equilibrium constant, Kc. The larger the value of the
equilibrium constant, the greater is the proportion of products to reactants
at equilibrium.
Chapter 13.2 introduces the connection between the total entropy change of
a reaction, ΔS total
1
, and its free energy change, ΔG 1. The two are related by
this formula: ΔG 1 = −TΔS total 1
. The connection between the free energy
change and the equilibrium constant for a reaction involving gases is
summarised in this equation:
ΔG 1 = −RT ln K
where R is the gas constant = 8.31 J K−1 mol−1.
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This chapter has now shown that for redox reactions, electrode potentials
offer yet another way of deciding the direction and extent of a reaction.
A positive value for the e.m.f. of a cell means that the reaction that it is based
on is feasible. The more positive E cell 1
, the greater the tendency for the
reaction to go. So it is not surprising that scientists have found that the total
entropy change of a redox reaction, its equilibrium constant and E cell 1
are
also closely related.
Thus:
E cell
1
∝ ΔS total
1
E cell
1
∝ ln K
What this all shows is that ΔS total, ΔG 1, Kc and E cell
1
are different ways of
presenting what is essentially the same information. Given the value for one
of these quantities it is possible, in principle, to calculate any of the others.
They are quantities that can all answer two key questions for any reaction:
● Will the reaction go? and
● How far will it go?
In practice, chemists use the quantity that is most easily determined by direct
experiment or by calculation from experimental data. For example, they
use equilibrium constants to explain the behaviour of weak acids; they use
standard electrode potentials to explain what happens during redox reactions;
they use free energy changes to determine the conditions needed to extract
metals from oxide ores.
Table 14.3 shows how the values of these predictors are related to the extent
of a reaction.
Table 14.3 Predicting the direction and
ΔG 1� ΔS total/ E cell/V Kc (units depend Extent of reaction
1 1
kJ mol−1 J mol−1 K−1 on the reaction) extent of chemical reactions from the
values of ΔG 1, ΔS total
1
, E cell
1
and Kc.
More More positive More Greater than 1010 Goes to completion
negative than +200 positive
than −60 than +0.6
≈ −10 ≈ +40 ≈ +0.1 ≈ 102 Equilibrium with more
products than reactants
≈0 ≈0 ≈0 ≈1 Roughly equal amounts
of reactants and
products
≈ +10 ≈ −40 ≈ −0.1 ≈ 10−2 Equilibrium with more
reactants than products
More More More Less than 10−10 No reaction
positive negative than negative
than +60 −200 than −0.6
The fact that ΔS total, ΔG 1, Kc and Ecell are all related to each other has
had a profound impact on scientific thinking and on chemists in particular.
It has brought together concepts of entropy, equilibrium and electrochemistry,
showing that ideas developed in different areas and different contexts of
chemistry are all related to the over-riding concept of the thermodynamic
feasibility of chemical reactions.
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14.7 Modern storage cells
Mobile phones, tablets and laptop computers depend on the existence of
chemical cells that can be recharged. These are storage cells because they
store the electricity - usually from the mains power supply. Often a single
cell does not have a voltage that is high enough, and so cells are linked
together in series to make a battery of cells.
When a storage cell is recharged, an electric current passes through it in the
opposite direction to the current that the cell produces. Recharging is an
example of electrolysis as chemical reactions occur to reform the chemicals
that make up the electrodes.
Key terms
A storage cell is an electrochemical cell that is based on reversible chemical changes
so that it can be recharged by an external electricity supply.
A battery is two or more electrochemical cells connected in series.
Lead–acid cells
Vehicles that run on petrol or diesel need storage batteries for the starter
motor and for the lights when the engine is not running. Most of batteries
in these vehicles are composed of six lead–acid cells in series, giving a total
battery potential of 12 volts.
Lead–acid cells are also used to provide power for the motor in battery-
operated vehicles such as electric wheelchairs, bicycles and scooters
(Figure 14.22). Another important use of lead–acid batteries is to provide
back-up power to computer systems, emergency lighting and hospital
equipment in case power cuts interrupt the mains supply.
The negative terminal in a lead–acid cell is lead. The electrolyte is fairly
concentrated sulfuric acid (about 6 mol dm−3). The lead gives up electrons,
forming lead(ii) ions when the cell is working normally (discharging). In the
presence of sulfate ions, the lead(ii) ions precipitate to form lead(ii) sulfate.
Pb(s) + SO42−(aq) → PbSO4(s) + 2e−
The positive terminal is lead coated with lead(iv) oxide. During discharge,
the lead(iv) oxide reacts with H+ ions in the sulfuric acid electrolyte and
takes electrons. Here, too, the lead(ii) ions precipitate as lead(ii) sulfate.
PbO2(s) + SO42−(aq) + 4H+(aq) + 2e− → PbSO4(s) + 2H2O(l)
The formation of insoluble lead(ii) sulfate creates a problem for lead–acid
cells. If the cells are discharged for long periods, the precipitate of lead(ii)
sulfate becomes coarser and thicker and the process cannot be reversed when
the cells are recharged.
When a lead–acid cell is recharged, the current is reversed and the reactions
at each terminal are reversed. This turns Pb2+ ions back to lead metal at one
Figure 14.22 Most battery-operated terminal and back to PbO2 at the other, with sulfate ions going back into
wheelchairs are powered by lead–acid cells. the electrolyte.
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Test yourself
31 a) Use the data sheet headed ‘Standard electrode potentials’, which
you can access online at [Link]/
EdexcelChemistry, to write down the value of E 1 for a Pb2+(aq) ∣ Pb(s)
half-cell. Suggest a reason why this only gives a very approximate
estimate of the potential of the negative electrode of a lead–acid cell.
b) What is the approximate cell potential for one lead–acid cell?
c) What is the approximate electrode potential of the
[PbO2(s) + 4H+(aq)], [Pb2+(aq) + 2H2O(l)] ∣ Pb(s) half-cell in a
lead–acid cell?
d) Write the half-equations for the processes at the two terminals
when the cell is being recharged.
Lithium cells
Modern mobile phones and laptop computers use lithium batteries. One
advantage of electrodes based on lithium is that the metal has a low density,
so that cells based on lithium electrodes can be relatively light. Also, lithium
is very reactive, which means that the electrode potential of a lithium half-
cell is relatively high and each cell has a large e.m.f.
The difficulty to overcome is that lithium is so reactive that it readily combines
with oxygen in the air, forming a layer of non-conducting oxide on the
surface of the metal. The metal also reacts rapidly with water. Research
workers have solved these technical problems by developing electrodes with
lithium atoms and ions inserted into the crystal lattices of other materials.
In addition, the electrolyte is a polymeric material rather than an aqueous
solution (Figure 14.23).
device powered electron flow in Figure 14.23 A schematic diagram of a
by the battery the external circuit lithium battery discharging. The electrode
processes are reversible, so the battery
e– e–
can be recharged.
Li+
Li+ Li+
Li+ Li+
Positive electrode Polymer electrolyte Negative electrode
+
Layer lattice of MnO2 with Li ions Carbon (graphite) with a
into which Li+ ions layer lattice containing
can move lithium atoms
2MnO2(s) + 2e– + 2Li+ +
2Li 2Li + 2e–
from from in
electrolyte electrode electrolyte
Mn2O3(s) + Li2O(s)
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14.8 Fuel cells
Fuel cells are electrochemical cells in which the chemical energy of a fuel
Key term is converted directly into electrical energy. Fuel cells differ from typical
electrochemical cells such as lithium cells and lead–acid cells in having
A fuel cell is an electrochemical cell
a continuous supply of reactants from which to produce a steady electric
which is continuously supplied with
current. Fuel cells use a variety of fuels including hydrogen, hydrocarbons
fuel and oxidising agent. A fuel cell
(such as methane) and alcohols. Inside a fuel cell, energy from the redox
produces electric power from a fuel,
reaction between a fuel and oxygen is used to create a potential difference
directly, without having to burn it.
(voltage). Hydrogen and alcohol fuel cells have been used in the development
of electric cars and in space exploration.
One of the most important fuel cells is the hydrogen–oxygen fuel cell
(Figure 14.24).
external
circuit
electron electron
flow flow
2e– 2e–
1
hydrogen O
2 2
oxygen
2H+
H2
excess
hydrogen H2O water
porous electrode porous electrode
solid polymer
impregnated impregnated
electrolyte
with a catalyst with a catalyst
– +
Figure 14.24 A hydrogen–oxygen fuel cell operating with a polymer electrolyte that is
permeable to protons.
Figure 14.24 shows a hydrogen–oxygen fuel cell working in acid conditions.
Fuel cells of this kind are used in buses and to provide back-up power for
telecommunications and in data centres.
Hydrogen gas flows into the negative terminal where the H 2 molecules split
into single H atoms in the presence of the catalyst. The H atoms then lose
electrons and form H+ ions.
Negative terminal: H2(g) → 2H+ + 2e− Equation 1
The electrons flow into the external circuit as an electric current while the
hydrogen ions migrate through the electrolyte.
Oxygen flows onto the positive terminal, where the nickel/nickel(ii) oxide
catalyses the splitting into single oxygen atoms. These oxygen atoms then
combine with hydrogen ions from the electrolyte and electrons to form water.
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Positive terminal: 1 O (g) + 2H+ + 2e− → H2O(l) Equation 2
2 2
The overall reaction in the hydrogen–oxygen fuel cell (obtained by adding
Equations 1 and 2) is:
1
H2(g) + 2 O2(g) → H2O(l)
NASA has used hydrogen–oxygen fuel cells in space missions since the 1960s
to provide both electricity and drinking water. Typically, these fuel cells
used aqueous potassium hydroxide as the electrolyte so they operated under
alkaline conditions.
The two reduction half-equations with electrode potentials for an alkaline
fuel cell using hydrogen are:
2H2O(l) + 2e− → H2(g) + 2OH−(aq)
E 1 = −0.83 V
1
O (g)
2 2
+ H2O(l) + 2e− → 2OH−(aq)
E 1 = +0.40 V
When an alcohol such as methanol is used as the energy source in place of
hydrogen in a cell such as that shown in Figure 14.24, the following half-
reactions occur at the terminals:
Negative terminal: CH3OH(l) + H2O(l) → CO2(g) + 6H+(aq) + 6e−
Positive terminal: 32 O2(g) + 6H+(aq) + 6e− → 3H 2O(l)
Fuel cells are no different in principle from more familiar electrochemical
cells. The innovation is that new reactants (such as H 2, or CH3OH, and O2)
are constantly fed into the cell and the products (H 2O and sometimes CO2
or other products) are drawn off. This continuous flow of materials allows
the cell potential to remain constant and the power output is uninterrupted.
The great advantage of all fuel cells is that they convert energy from chemical
changes directly into electricity and in doing so achieve a remarkable
efficiency of about 70%. In comparison, modern power plants and petrol
engines using fossil fuels have a conversion efficiency for the energy from
chemical reactions to electrical energy or kinetic energy of only about 40%.
Test yourself
32 a) W
hat are the changes at the negative and positive terminals of a
hydrogen fuel cell of the type used by NASA in spacecraft?
b) What is the overall equation for the reactions taking place in this
type of fuel cell?
c) What is the cell e.m.f.?
33 a) What is the overall reaction for a fuel cell based on methanol?
b) Suggest one advantage and one disadvantages of using methanol
rather than hydrogen.
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Chapter summary
Chapter 14 Redox II oxidised and reduced forms in solution). For
example: Br2(aq), 2Br−(aq)|Pt(s).
l Elements in the s-block of the Periodic Table have l The standard e.m.f. of a cell can be calculated by
only one oxidation state. Most elements in the combining the two standard electrode potentials.
p- and d-blocks form compounds in which their E 1cell = E (right-hand electrode) − E (left-hand electrode).
1 1
atoms can have different oxidation states. l Positive values of E cell indicate that the sign of
1
l Titrations are used to investigate redox reactions
the right-hand electrode is positive and that
quantitatively. Potassium manganate(vii) is used to the reaction is feasible in the direction shown
estimate the concentrations of aqueous reducing by reading the conventional cell diagram from
agents. No indicator is needed. Oxidising agents left to right.
in solution can be estimated by adding an excess of l An electrochemical series based on electrode
potassium iodide, then titrating the iodine formed potentials can also be used to predict the feasible
with a standard solution of sodium thiosulfate. direction of change in redox reactions, including
Starch solution is added as an indicator near the disproportionation reactions.
end-point. l Although electrode potentials for a redox reaction
l An electrochemical cell produces an electric
may suggest that a reaction is feasible, in practice
potential difference from a redox reaction. The the reaction may be too slow for any change to
two half reactions happen in separate half cells be observed.
connected by a salt bridge. l Standard electrode potential values refer to
l The standard conditions for electrochemical
standard conditions. Some reactions that are not
measurements are a temperature of 298 K, any feasible under standard conditions may become
gases at a pressure of 100 kPa and solutions at a feasible if the conditions are changed (and vice
concentration of 1.00 mol dm−3. versa).
l There is no way of measuring the e.m.f. of an
l E cell for a reaction is directly proportional to the
1
isolated, single electrode, so electrode potentials total entropy change for the reaction and to ln K
are measured using a standard reference electrode. for the reaction. This shows that these related
l The standard electrode potential, E°, of a half-cell
quantities all give essentially the same information
is the e.m.f. of that half-cell relative to a standard and can be used to decide whether or not a
hydrogen electrode. reaction will go and how far it will go. In practice,
l A standard hydrogen electrode is a half-cell in
for a given reaction, chemists use the quantity that
which a 1.00 mol dm−3 solution of hydrogen is most easily measured or calculated.
ions is in equilibrium with hydrogen gas at l Rechargeable batteries for storing electricity
100 kPa pressure on the surface of a platinum are electrochemical cells. Standard electrode
electrode coated with platinum black at 298 K. potentials can be used to predict the cell e.m.f. of a
Pt[H2(g)]|2H+(aq) . storage cell.
l Standard electrode potentials can be measured for
l A fuel cell is an electrochemical cell that is
two types of half-cell: continuously supplied with a fuel (such as hydrogen
• a metal electrode (the reduced form) dipping or methanol) and an oxidising agent, usually
into a solution of its own ions (the oxidised oxygen.
form) Zn 2+(aq)|Zn(s). l Hydrogen–oxygen fuel cells can be made with
• a shiny (inert) platinum electrode dipping into both acid and with alkaline electrolytes.
a solution containing two ions of the same
element in different oxidation states (both the
414 14 Redox II
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Exam practice questions
1 a) For each of these reactions, identify the An electrochemical cell was set up with
changes of oxidation number and state the Fe3+(aq) ∣ Fe2+(aq) as the right-hand electrode
element that is oxidised and the element a) Write half-equations for the reactions that
that is reduced. occur in each half-cell when a current flows.
i) 2NH3(g) + 3Cl2(g) State which half-equation involves oxidation
→ N2(g) + 6HCl(g)(2) and which involves reduction.(3)
ii) Cu2O(s) + H2SO4(aq) b) Calculate the change in oxidation
→ CuSO4(aq) + Cu(s) + H2O(l) (2) number of the oxidised and reduced
iii) 2KNO3(s) → 2KNO2(s) + O2(g) (2) elements in each half-cell.(2)
b) i) Write the half-equations involved c) Write the conventional cell diagram
when dichromate(vi) ions in acid and give the e.m.f. of the cell.(3)
solution react with sulfite ions,
4 A student set up the electrochemical cell shown
SO32−, to form chromium(iii)
in the diagram.
ions and sulfate ions. (2)
ii) Write a full, balanced redox high resistance voltmeter
equation for the reaction. (1) V
iii) Describe the change you would
see in the solution when the copper foil silver foil
reaction occurs. (2)
iv) Determine how many moles of
sulfite ions react with one mole of
salt bridge
dichromate(vi) ions. (1)
2 A 20.0 cm3 sample of water was taken from 1.0 mol dm–3
copper(II) sulfate
a swimming pool which had recently been
–3
disinfected with chlorine. The sample was 1.0 mol dm
silver nitrate
added to excess potassium iodide solution.
The iodine formed was titrated with The standard electrode potentials are:
0.00500 mol dm−3 sodium thiosulfate solution. Cu2+ ∣ Cu E 1 = +0.34 V
The volume of sodium thiosulfate solution
needed to reach the end-point was 19.4 cm3. Ag+ ∣ Ag E 1 = +0.80 V
a) Write ionic equations for: a) State how the student made the
i) the reaction of chlorine with salt bridge.(1)
iodide ions (1) b) Write half-equations to show the
ii) the reaction of iodine molecules reactions that occurred in:
with thiosulfate ions.(2) i) the Cu2+ ∣ Cu half-cell(1)
b) Name the indicator used to detect ii) the Ag+ ∣ Ag half-cell.(1)
the end-point of the titration.(1) c) Write an equation for the overall cell
c) Describe the colour changes reaction.(1)
observed at each stage of the analysis.(3) d) Calculate the e.m.f. (potential difference)
d) Calculate the concentration of for this cell.(2)
chlorine in the swimming pool water e) Identify the electrode at which reduction
in mol dm−3.(4) occurs. Explain your answer.(2)
f) The student found that the cell e.m.f.
3 Use the standard electrode potentials below to was less than the calculated value.
answer the questions that follow. Give two reasons for this.(2)
Fe3+(aq) ∣ Fe2+(aq) E 1 = +0.77 V
Cu2+(aq) ∣ Cu(s) E 1 = +0.34 V
415
Exam practice questions
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5 a) Fuel cells can be made using these A second 25.00 cm3 sample of the original
electrode systems. solution was reduced with zinc and acid.
2H+(aq) + 2e− → H2(g) E 1 = 0.00 V After filtering off the zinc, the solution
O2(g) + 4H+(aq) + 4e− → 2H2O(l) was titrated with the same solution of
E 1 = +0.40 V KMnO4(aq). This time the mean titre
i) Write an overall equation for the cell was 22.50 cm3. Calculate the
reaction. Show your working.(2) concentrations of the iron(ii) and
ii) Identify the half-cell from which iron(iii) ions in the original solution. (5)
electrons flow into the external Calculate the specified quantities for each of
7
circuit.(1) the following redox titrations. In each case give
b) State the principal differences the equation for any reactions, show the steps
between a hydrogen–oxygen fuel cell and of your working in full and give an answer to
a storage cell, such as a lead–acid cell.(3) an appropriate number of significant figures.
c) Describe and explain two advantages a) Copper(ii) ions oxidise iodide ions to iodine.
that are gained by generating electricity A pale, off-white precipitate of a copper
using fuel cells rather than in thermal compound forms at the same time. 3.405 g
power stations.(4) of CuSO4.5H2O was dissolved in water and
d)* Discuss the advantages and disadvantages made up to 250 cm3. Excess potassium iodide
of using hydrogen fuel cells as the power was added to 25.0 cm3 of the copper(ii) sulfate
supply for motor vehicles rather than solution. In a titration, 18.00 cm3 of a solution
using engines that burn fuels. (6) of 0.0760 mol dm−3 sodium thiosulfate was
6 Calculate the specified quantities for each of required to react with the iodine formed.
the following redox titrations. In each case give Determine the oxidation number of the
the equation for any reactions, show the steps copper in the precipitated copper compound
of your working in full and give an answer to formed during the reaction of copper(ii) ions
an appropriate number of significant figures. with iodide ions.(5)
a) Sodium ethanedioate, Na2C2O4, is b) 0.275 g of an alloy containing copper was
a primary standard that can be used dissolved in nitric acid and then diluted
to standardise solutions of potassium with water, producing a solution of
manganate(vii). Under acidic conditions, copper(ii) nitrate. An excess of potassium
KMnO4 oxidises ethanedioate ions on iodide was then added. The copper(ii) ions
heating to carbon dioxide. In a titration it reacted with the iodide ions to form a
was found that 28.85 cm3 of a solution of precipitate of a copper iodide and iodine.
KMnO4 oxidised 25.00 cm3 of a solution In a titration, the iodine reacted with
containing 7.445 g dm−3 of sodium 22.50 cm3 of 0.140 mol dm−3 sodium
ethanedioate. Calculate the concentration of thiosulfate solution. (Cu = 63.5)
the potassium manganate(vii) solution. Calculate the percentage by mass of
(Na = 23.0, C = 12.0, O = 16.0)(5) copper in the alloy.(5)
b) Iron in 1.34 g of iron ore was dissolved 8 The list below shows three standard electrode
in acid and reduced to iron(ii) ions. potentials.
The solution was then titrated with
0.0200 mol dm−3 potassium manganate(vii) I I2(aq) + 2e− ⇋ 2I−(aq) E 1 = +0.54 V
solution. The titre was 26.75 cm3. Calculate II Cd2+(aq) + 2e− ⇋ Cd(s) E 1 = −0.40 V
the percentage by mass of
iron in the ore. (Fe = 55.8)(5) III Fe2+(aq) + 2e− ⇋ Fe(s) E 1 = −0.44 V
c) A 25.00 cm3 sample of a solution containing a) Using the standard electrode potentials I, II
iron(ii) and iron(iii) ions was titrated with and III above, write:
potassium manganate(vii) in the presence i) one equation for a reaction which
of acid. The titre was 18.00 cm3 with a would go to completion, and explain
0.0200 mol dm−3 solution of KMnO4(aq). your choice (2)
416 14 Redox II
14 Redox II
469983_14_Chem_Y1-2_387-[Link] 416 13/04/19 10:04 PM
ii) one equation for a reaction which c) The electrode potentials, E, of a number
might occur, but only to an of copper/copper(ii) sulfate half-cells with
equilibrium position, and explain different Cu2+(aq) concentrations were
your choice. (2) measured against a standard hydrogen
b) i) Draw a labelled diagram of the cell electrode at 298 K. The results are shown in
formed by connecting half-cells I the graph below.
and III.(5) i) From the graph, determine the
ii) Label your diagram to show the value of E when log [Cu2+(aq)]
direction of electron flow in the is zero.(1)
external circuit.(1) ii) Explain the significance of the
iii) Calculate the e.m.f. (standard cell value of E when log [Cu2+(aq)]
potential) of this cell.(2) is zero.(2)
iv) Predict the effect on the cell e.m.f. iii) Give a formula to describe the
of decreasing the concentration of shape of the graph and explain
Fe2+(aq). Explain your answer. (3) your answer.(3)
0.36
0.35
Electrode potential, E/V
0.34
0.33
0.32
0.31
0.30
–1.0 –0.5 0
log [Cu2+]
417
Exam practice questions
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Transition metals
15
15.1 The atoms and ions of
transition elements
The transition metals are vital to life and bring colour to our lives. They
are also metals of great engineering and industrial importance. Chemically,
these elements, which occupy the d block of the Periodic Table, are more
alike than might be expected. Across the ten d-block metals from scandium
to zinc in Period 4, the similarities are as striking as the differences. Chemists
explain the characteristics of transition metals in terms of the electronic
configurations of their atoms. Transition metal chemistry is colourful
because of the range of oxidation states and complex ions. Transition metals
matter because their properties are fundamental, not only to life, but also to
modern technology (Figure 15.1).
Figure 15.1 Specimens of some d-block Tip
elements. The chemistry of an element is Chapter 1 introduces the description of atomic energy levels in terms of s, p and d
determined to a large extent by its outer energy levels. Then Chapter 4 shows how an understanding of atomic structure can
shell electrons because they are the first explain the arrangement of elements in the Periodic Table with particular reference
to get involved in reactions. All the d-block to elements in the s and p blocks of the table. This chapter builds on these ideas to
elements have their outer electrons in the explain the chemistry of the metals in the d block.
4s sub-shell.
4p
Orbitals in
the 4th shell
Electronic configurations
3d As the shells of electrons around the nuclei of atoms get further from the
4s
nucleus, they become closer in energy. Therefore, the difference in energy
between the second and third shells is less than that between the first and
3p
second. When the fourth shell is reached, there is an overlap between the
Orbitals in
the 3rd shell
orbitals of highest energy in the third shell (the 3d orbitals) and that of lowest
energy in the fourth shell (the 4s orbital) (Figure 15.2).
3s
Figure 15.2 The relative energy levels of
The 3d sub-shell is on average nearer the nucleus than the 4s sub-shell, but
orbitals in the third and fourth shells.
at a higher energy level. So, once the 3s and 3p sub-shells are filled, the next
418 15 Transition metals
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electrons go into the 4s sub-shell because it occupies a lower energy level
than the 3d sub-shell.
This means that potassium and calcium have the electron structure [Ar]4s1
and [Ar]4s2 respectively (Table 15.1).
Table 15.1 Electron configurations from potassium to zinc in Period 4 of the Periodic
Table. ([Ar] represents the electronic configuration of argon.) Note the way that the
electron configurations for chromium and copper atoms do not fit the general pattern.
Element Symbol Electronic structure
s,p,d,f notation Electrons-in-boxes
notation
Potassium K [Ar]4s1 [Ar] ↑
Calcium Ca [Ar]4s2 ↑
[Ar] ↑
↑
Scandium Sc [Ar]3d14s2 [Ar] ↑ ↑
↑
Titanium Ti [Ar]3d24s2 [Ar] ↑ ↑ ↑
↑
Vanadium V [Ar]3d34s2 [Ar] ↑ ↑ ↑ ↑
Chromium Cr [Ar]3d54s1 [Ar] ↑ ↑ ↑ ↑ ↑ ↑
↑
Manganese Mn [Ar]3d54s2 [Ar] ↑ ↑ ↑ ↑ ↑ ↑
↑ ↑
Iron Fe [Ar]3d64s2 [Ar] ↑ ↑ ↑ ↑ ↑ ↑
↑ ↑ ↑
Cobalt Co [Ar]3d74s2 [Ar] ↑ ↑ ↑ ↑ ↑ ↑
↑ ↑ ↑ ↑
Nickel Ni [Ar]3d84s2 [Ar] ↑ ↑ ↑ ↑ ↑ ↑
↑ ↑ ↑ ↑ ↑
Copper Cu [Ar]3d104s1 [Ar] ↑ ↑ ↑ ↑ ↑ ↑
↑ ↑ ↑ ↑ ↑ ↑
Zinc Zn [Ar]3d104s2 [Ar] ↑ ↑ ↑ ↑ ↑ ↑
Look carefully at Table 15.1. In Period 4, the d-block elements run from
scandium (1s22s22p63s23p63d14s2) to zinc (1s22s22p63s23p63d104s2). But, notice
that the electronic configurations of chromium and copper do not fit the
general pattern. The explanation of these irregularities lies in the stability
associated with half-filled and filled sub-shells. So, the electronic structure
of chromium, [Ar]3d54s1, with half-filled sub-shells and an equal distribution
of charge around the nucleus, is more stable than the electronic structure
[Ar]3d4 4s2.
Similarly, the electronic structure of copper, [Ar]3d10 4s1, with a filled 3d
sub-shell and a half-filled 4s sub-shell is more stable than [Ar]3d94s2.
Along the series of d-block elements from scandium to zinc, the number
of protons in the nucleus increases by one from one element to the next.
However, the added electrons go into an inner d sub-shell, but the outer
electrons are always in the 4s sub-shell. This means that there are clear
similarities amongst the transition elements. Changes in their chemical
properties across the series are much less marked than the big changes across
a series of p-block elements such as aluminium to argon.
15.1 The atoms and ions of transition elements 419
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In this way, the energy-level model for electronic structure can help to
account for the similarities in properties of transition metals. Later in this
chapter, Section 15.5 shows that there are limitations to this energy-level
model so that more sophisticated explanations are needed.
Ions of the transition metals
When transition metals form their ions, electrons are lost initially from the
4s sub-shell and not the 3d sub-shell. This may seem somewhat illogical
because, prior to holding any electrons, the 4s level is more stable than the 3d
level. But once the 3d sub-shell is occupied by electrons, these 3d electrons,
being closer to the nucleus, repel the 4s electrons to a higher energy level.
The 4s electrons are, in fact, repelled to an energy level higher than those
occupying the 3d sub-shell. So, when transition metals form ions, they lose
electrons from the 4s before the 3d level. This further emphasises the fact that
transition metals have similar chemical properties dictated by the behaviour
of the 4s electrons in their outer shells.
Test yourself
1 Write the full s,p,d electronic configuration of:
a) a scandium atom
b) a scandium(iii) ion
c) a manganese atom
d) a manganese(ii) ion.
2 Look at the electronic structures of iron and copper in Table 15.1.
a) Write the electronic structure of an iron(ii) ion.
b) Write the electronic structure of an iron(iii) ion.
c) Which ion, Fe2+ or Fe3+, would you expect to be the more stable?
Explain your choice.
d) Write the formula for the ion of copper that you would expect to be
the more stable. Explain your choice.
15.2 Defining the transition metals
The simplest and neatest way to define the transition metals would be to say
that they are the elements in the d block of the Periodic Table. But this simple
definition leads to the inclusion of scandium and zinc as transition metals and
ignores the fact that these two metals have some clear differences from the
metals between them in the Periodic Table from titanium to copper. For
instance:
● Scandium and zinc have only one oxidation state in their compounds
(scandium +3, zinc +2), whereas the elements from titanium to copper
have two or more.
● The compounds of scandium and zinc are usually white, unlike those of
transition metals which are generally coloured.
● Scandium, zinc and their compounds show little catalytic activity.
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Activity
Studying the ionisation energies of transition
metals
Experimental evidence for the electronic configurations of Look carefully at Figure 15.3, which shows graphs of the first,
transition metals atoms and ions can be obtained from the second and third ionisation energies of the elements from
ionisation energies of the elements concerned. scandium to zinc.
4000
Zn third
Cu
3500 Ni
Mn Co
Cr
3000
V
Ionisation energy/kJ mol –1
Ti Fe
2500
Sc
Cu
2000
Ni
Cr Co second
Fe
V Mn Zn
1500
Ti
Sc
1000 Mn Fe Co Ni Cu Zn
Sc Ti V Cr first
Figure 15.3 Graphs of the first, second
and third ionisation energies of the
20 21 22 23 24 25 26 27 28 29 30 31 elements from scandium to zinc in the
Atomic number Periodic Table.
1 Write the electronic structures of the following atoms and ions b) How does this relate to the electronic configuration of
using [Ar] for the electronic structure of argon. copper atoms and ions?
a) Zn b) Cu+ c) Zn+ d) Cr+ e) Mn+ 6 a) What does the high second ionisation energy of chromium,
2 Write an equation for: relative to its neighbours in the Periodic Table, tell you
a) the second ionisation energy of chromium about chromium?
b) the third ionisation energy of iron. b) How does this relate to the electronic configuration of
3 Explain the general trend in ionisation energies as atomic chromium atoms and ions?
number increases. 7 a) Which elements have relatively high third ionisation
4 a) How does the first ionisation energy of zinc compare with energies compared with their neighbours in the d-block
those of the other d-block elements in Period 4? elements of Period 4?
b) What does this tell you about zinc relative to the other b) How do these relatively high third ionisation energies
elements? provide further evidence for the proposed electronic
c) How does this relate to the electronic configuration of zinc configurations of the elements concerned?
atoms?
5 a) What does the high second ionisation energy of copper,
relative to its neighbours in the Periodic Table, tell you
about copper?
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As scandium and zinc do not show the typical properties of a transition
Key term metal, chemists looked for a more satisfactory definition. This definition
A transition metal is an element
should exclude scandium and zinc, but include all the elements from titanium
that has one or more stable ions with
to copper. In order to achieve this, chemists describe transition metals as
incompletely filled d orbitals.
those elements that form one or more stable ions with incompletely filled d
orbitals.
Characteristics of the transition metals
In general, transition elements share a number of common properties (see the
Data sheet ‘Properties of selected elements – d-block metals’, which you can
access online at [Link]/EdexcelChemistry).
● They are hard metals with useful mechanical properties, high melting and
high boiling temperatures.
● They show variable oxidation numbers in their compounds.
● They form coloured ions in solution.
● They can act as catalysts both as the elements and as their compounds.
● They form complex ions involving monodentate, bidentate and polydentate
ligands (Section 15.8).
15.3 The transition elements as
metals
Most of the transition elements have a close-packed structure in which
each atom has 12 nearest neighbours (Figure 15.4). In addition, transition
elements have relatively low atomic radii because an increasingly large
nuclear charge is attracting electrons that are being added to an inner sub-
shell. The dual effect of close packing and small atomic radii results in strong
metallic bonding. So, transition metals have higher melting temperatures,
Figure 15.4 Close packing of atoms in one
higher boiling temperatures, higher densities and higher tensile strengths
layer of a metal crystal. Each atom is in
than s-block metals such as calcium and p-block metals such as aluminium
contact with six atoms in the same layer,
and lead. A plot of physical properties against atomic number often has two
three atoms in the layer above and three
peaks or two troughs associated with a half-filled and then a filled d sub-shell
atoms in the layer below, so 12 in all.
(Figure 15.5).
2500
2163
2130
1933
1808
1814
1768
2000
1728
Melting temperature/K
1517
1356
1500
1112
1000
693
500
0
Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn
Figure 15.5 A plot of melting temperature against atomic number for the elements
calcium to zinc in the Periodic Table.
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The transition metals are much less reactive than the s-block metals.
However, the electrode potentials listed in Table 15.2 suggest that all of
them, except copper, should react with dilute strong acids such as 1 mol dm−3
hydrochloric acid. In practice, many of the metals react very slowly with
dilute acids because the metal is protected by a thin, unreactive layer of
oxide. Chromium provides a very good example of this. Despite the
predictions from its standard electrode potential, it is used as a protective,
non-rusting metal owing to the presence of an unreactive, non-porous layer
of chromium(iii) oxide, Cr2O3.
Table 15.2 Standard electrode potentials of the transition metals from V to Cu.
Element Standard electrode potential,
E 1 for M2+(aq) | M(s)/V
Vanadium −1.20
Chromium −0.91
Manganese −1.19
Iron −0.44
Cobalt −0.28
Nickel −0.25
Copper +0.34
Copper is the least reactive of the transition metals in Period 4. It does
not react with dilute non-oxidising acids, such as dilute HCl and dilute
H2SO4, and it oxidises only very slowly in moist air. Copper is also a good Figure 15.6 Saxophones are made of
conductor of electricity, which leads to its use in electricity cables, and for brass. Brass is an alloy of 60–80% copper
domestic water pipes. Copper’s mechanical properties are enhanced by and 20–40% zinc. It is easily worked, has an
making alloys such as brass and bronze (Figure 15.6). attractive gold colour and does not corrode.
Test yourself
3 Why can scandium and zinc be described as d-block elements, but not
as transition metals?
4 Suggest a reason why zinc only forms compounds in the +2 oxidation state.
5 a) What is the general trend in standard electrode potentials of the
M2+(aq) | M(s) systems for the transition metals in Table 15.2?
b) What does this suggest about the reactivity of transition metals
across Period 4 in the Periodic Table?
6 Explain why the atomic radius falls from 0.15 nm in titanium to
0.14 nm in vanadium and then 0.13 nm in chromium.
15.4 Variable oxidation numbers
Most of the d-block elements in Period 4 form a range of compounds in which
they are present in different oxidation states. The main reason for this is that the
transition metals from titanium to copper have electrons of similar energy in
both the 3d and 4s levels. This means that each of these elements can form ions
15.4 Variable oxidation numbers 423
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of roughly the same stability in aqueous solution or in crystalline solids by losing
different numbers of electrons. This contrasts with the metals in Groups 1 and
2 of the Periodic Table for which there is a large jump in ionisation energy after
the electrons in the outer shell have been removed.
The formulae of the common oxides of the elements from scandium to
zinc are shown in Figure 15.7 along with the stable oxidation states of each
element in its compounds. The main oxidation states of the elements are
shown in bold blue print.
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Common Sc2O3 Ti2O3 V2O3 Cr2O3 MnO FeO CoO NiO Cu2O ZnO
oxides TiO2 V2O5 CrO3 MnO2 Fe2O3 Co2O3 CuO
Mn2O7
+7
+6 +6
+5
+4 +4 +4
+3 +3 +3 +3 +3 +3 +3
+2 +2 +2 +2 +2 +2 +2 +2 +2
+1
Figure 15.7 Oxidation states and common oxides of the elements scandium to zinc with
the main oxidation states in bold blue print.
The elements at each end of the series in Figure 15.7 give rise to only one
oxidation state. The elements near the middle of the series have the greatest
range of oxidation states. Most of the elements form compounds in the +2
state corresponding to the use of both 4s electrons in bonding.
The +2 state is a main oxidation state for all elements in the second half of the
series, whereas +3 is a main oxidation state for all elements in the first part.
Across the series, the +2 state becomes more stable relative to the +3 state.
From scandium to manganese, the highest oxidation state corresponds to
the total number of electrons in the 3d and 4s energy levels. However, these
higher oxidation states never exist as simple ions. Typically, they occur in
compounds in which the metal is covalently bonded to an electronegative
atom, usually oxygen, as in the dichromate(vi) ion, Cr2O72−, and the
manganate(vii) ion, MnO4−.
One of the most attractive and effective demonstrations of the range of
oxidation states in a transition element can be shown by shaking a solution
of ammonium vanadate(v), NH4VO3, in dilute sulfuric acid with zinc.
Before adding zinc, H+ ions in the sulfuric acid react with VO3− ions to
form dioxovanadium(v) ions and the solution is yellow.
VO3−(aq) + 2H+(aq) → VO2+(aq) + H2O(l)
When the yellow solution, containing dioxovanadium(v) ions, is shaken
with zinc, it is reduced first to blue oxovanadium(iv) ions, VO2+(aq), then
to green vanadium(iii) ions, V3+(aq), and finally to violet vanadium(ii) ions,
V2+(aq) (Figure 15.8).
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Figure 15.8 The oxidation states of
vanadium showing the colours of its ions in
the +5, +4, +3 and +2 oxidation states.
+ 2+ 3+ 2+
VO2 (aq) VO (aq) V (aq) V (aq)
Test yourself
For Questions 10 and 11, refer to the data sheet for Chapter 15 headed
‘Standard electrode potentials’, which you can access online at www.
[Link]/EdexcelChemistry.
7 Write down four generalisations about the oxidation states of
transition metals based on Figure 15.7 and the text in Section 15.4.
8 Give examples of compounds other than oxides of:
a) chromium in the +3 and +6 states
b) manganese in the +2 and +7 states
c) iron in the +2 and +3 states
d) copper in the +1 and +2 states.
9 a) Show that the oxidation state of vanadium in VO2+ ions is +5.
b) Refer to Figure 15.8. Predict the colour of the solution when a
solution containing VO2+ ions is half reduced to VO2+ ions.
10 a) Write half-equations for:
i) the reduction of dioxovanadium(v) ions, VO2+, to oxovanadium(iv)
ions in acid solution
ii) the oxidation of zinc to zinc(ii) ions.
b) Use the data sheet ‘Standard electrode potentials’ to show that:
i) iodide ions reduce VO2+ ions to VO2+ ions in acid solution
ii) tin reduces VO2+ ions to V3+ ions in acid solution
iii) zinc reduces VO2+ ions to V2+ ions in acid solution.
11 a) Write a half-equation involving electrons for:
i) the oxidation of Cu+(aq) to Cu2+(aq)
ii) the reduction of Cu+(aq) to Cu(s).
b) Use the data sheet ‘Standard electrode potentials’ to find the
standard electrode potentials for the two half-equations in part (a).
c) Using your data in part (b), explain why Cu+(aq) ions
disproportionate in aqueous solution.
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Chromium forms compounds in three oxidation states, +2, +3 and +6. In
the +3 state, chromium exists as Cr3+ ions, which can be both oxidised and
reduced (Figure 15.9).
Under alkaline conditions, hydrogen peroxide oxidises green chromium(iii)
ions, Cr3+(aq), to yellow chromium(vi) in chromate ions, CrO42−(aq).
H2O2(aq) + 2e− → 2OH−(aq)
Cr3+(aq) + 8OH−(aq) → Cr2O42−(aq) + 4H2O(l) + 3e−
In contrast to this, zinc reduces green Cr3+(aq) to blue-violet Cr2+(aq) ions.
Figure 15.9 Solutions containing ions in
Zn(s) → Zn 2+(aq) + 2e−
two of the oxidation states of chromium.
On the left, a solution of dichromate(vi) Cr3+(aq) + e− → Cr2+(aq)
ions, Cr2O72–; on the right, a solution of
Chromium(ii) ions are powerful reducing agents which are rapidly converted
chromium(iii) ions, Cr 3+.
to chromium(iii) by oxygen in the air. This means that air has to be excluded
when zinc and acid are used to reduce chromium(iii) to the +2 state.
Note that chromium in the +6 state can be either orange or yellow, depending
on the pH, as a result of this equilibrium:
2CrO42−(aq) + 2H+(aq) ⇋ Cr2O72−(aq) + H2O(l)
yellow orange
Test yourself
12 Use the data sheet ‘Standard electrode potentials’, which you can
access online at [Link]/EdexcelChemistry, to
show that zinc reduces Cr3+(aq) to Cr2+(aq) ions.
13 a) E xplain why an orange solution of dichromate(vi) ions turns yellow
on adding alkali, and then orange again if the solution is acidified.
b) Is the change of CrO42−(aq) to Cr2O72−(aq) a redox reaction?
15.5 Coloured ions
Most coloured compounds get their colour by absorbing some of the
radiation in the visible region of the electromagnetic spectrum with
wavelengths between 400 nm and 700 nm. When light hits a substance, part
is absorbed, part is transmitted (if the substance is transparent) and part is
usually reflected. If all the light is absorbed, the substance looks black. If all
the light is reflected, the substance looks white. If very little light is absorbed,
and all the radiations in the visible region of the electromagnetic spectrum
are transmitted equally, the substance is colourless like water.
However, many compounds, and particularly those of transition metals,
absorb radiations in only certain areas of the visible spectrum. This means
that the substances take on the colour of the light that they transmit or
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reflect. For example, if a material absorbs all radiations in the green-
blue-violet region of the spectrum, it appears red-orange in white light
(Figure 15.10).
Colour of compound Wavelength absorbed/nm Colour of light absorbed
greenish yellow 400–430 violet
yellow to orange 430–490 blue
red 490–510 blue-green
purple 510–530 green
violet 530–560 yellow-green
blue 560–590 yellow
greenish blue 590–610 orange
blue-green to green 610–700 red
Figure 15.10 A chart showing complementary colours in the left and right-hand columns.
The colour of a compound is the colour complementary to the light it absorbs.
It is the electrons in coloured compounds that absorb radiation and jump
from their normal state to a higher excited state. According to the quantum
theory, there is a fixed relationship between the size of the energy ‘jump’ and
the wavelength of the radiation absorbed. In many compounds, the electron
‘jumps’ between one sub-level and the next are so large that the radiation
absorbed is in the ultraviolet region of the spectrum. These compounds
are therefore white or colourless because they are not absorbing any of the
radiation in the visible region of the electromagnetic spectrum.
Tip
The quantum theory states that radiation is emitted or absorbed in tiny, discrete
amounts called energy quanta. Quanta have energy, E = hν where h is Planck’s
constant and ν is the frequency of the radiation.
However, the colour of transition metal ions arises from the possibility of
transitions between the orbitals within the d sub-shell.
In a free gaseous atom or ion, the five 3d orbitals are all at the same energy
level even though they do not all have the same shape. But when the ion of
a d-block element is surrounded by other ions in a crystalline solid, or by
molecules such as water in aqueous solutions, the differences in shape cause
the five orbitals to split into two groups. When there are six molecules or
ions around the central metal atom, two of the 3d orbitals move to a slightly
higher energy level than the other three. As a result, ions such as Cu2+(aq)
appear coloured because light of a particular frequency can be absorbed
from visible light as electrons jump from a lower to a higher 3d orbital
(Figure 15.11). If all the d orbitals are full, or empty, there is no possibility of
electronic transitions between them.
15.5 Coloured ions 427
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Figure 15.11 The energy between the the d orbitals split into two groups with
separated d orbitals in an aqueous transition different energies in an aqueous ion
metal ion, like Cu2+(aq), allows electron
transitions from a lower orbital to a higher five d orbitals with the same
energy in a free gaseous ion
orbital. The ion absorbs light with a particular absorption of light at a
frequency in the visible region of the particular frequency can
promote an electron to a
electromagnetic spectrum. 2+
Cu (g) higher level
2+
Cu (aq)
Tip
The colour of transition metal ions results from the absorption of part of the visible
radiation in white light as electrons move from a lower to a higher level. This contrasts
with flame colours, which arise from the emission of radiation as electrons fall from a
higher to a lower level.
The explanation of the colour of transition metal ions, illustrates the limitations
of the simple energy-level model of the electronic structures of atoms. The
need for more sophisticated explanations is clear, bearing in mind the existence
of sub-shells and the different shapes of orbitals within d sub-shells.
Test yourself
xplain why Zn2+, Cu+ and Sc3+ ions are usually colourless
14 a) E
in solution and white in solids by writing out their electronic
configurations.
b) What colours of light are absorbed most effectively by a Cu2+ ion?
Tip 15.6 Formation of complex ions
The terms ‘hydronium ion’ and The symbol H+(aq) does not represent a simple ion in acid solutions. In
‘hydroxonium ion’ are sometimes used aqueous solution, H+ ions are strongly attached to water molecules by dative
for the oxonium ion, H3O+. covalent bonds forming oxonium ions, H3O+ (Figure 15.12).
O
H
H+
H
Figure 15.12 An oxygen atom in a water molecule forming a dative covalent bond
with a hydrogen ion to form an aqueous H3O+ ion. The oxygen atom donates both
electrons of a lone pair to form the bond.
In the same way as H+, other cations can also exist in aqueous solution as
hydrated ions. So Cr3+(aq), Cu2+(aq) and Ag+(aq) can be represented more
completely as [Cr(H2O)6]3+(aq), [Cu(H2O)6]2+(aq) and [Ag(H2O)2]+(aq) in
aqueous solution. The larger size of these other cations relative to H+ enables
them to associate with up to six water molecules (Figure 15.13).
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H H
H O
H
H O
O
H
Cu2+
H
O H
O
H
H
O
H H
Figure 15.13 Dative covalent bonding in an aqueous Cu2+ ion. Each water molecule uses a
lone pair of electrons to form a dative covalent bond with the central metal ion.
Tip
In aqueous solution, the copper(ii) ion is surrounded by six water molecules to form the
complex ion [Cu(H2O)6]2+. In solid hydrated copper(ii) sulfate (CuSO4.5H2O), however,
there are only four water molecules co-ordinated with each copper(ii) ion. The fifth
water molecule in the solid copper(ii) sulfate is associated with a sulfate ion, SO42−.
Other polar molecules, besides water, can form dative covalent bonds
with metal ions. For example, in excess ammonia solution, Cr3+ ions form
[Cr(NH3)6]3+, Cu2+ ions form [Cu(NH3)4(H2O)2]2+ and Ag+ ions form
[Ag(NH3)2]+. In addition to polar molecules, anions can also associate with
cations using dative covalent bonds. For example, when anhydrous copper(ii)
sulfate is added to concentrated hydrochloric acid, the solution contains
yellow [CuCl4]2− ions.
Ions such as [Cu(H2O)6]2+, [Cu(NH3)4(H2O)2]2+ and [CuCl4]2− in which
a metal ion is associated with a number of molecules or anions are called Tip
complex ions, and the anions and molecules attached to the central metal When anions act as ligands, the overall
ion are called ligands. Each ligand must have at least one lone pair of charge on the complex ion does not
electrons which it uses to form a dative covalent bond with the metal ion. equal the oxidation number of the
The number of ligands in a complex ion is typically two, four or six. central metal ion.
Chemists have an alternative name for dative covalent bonds which they
often prefer when describing complex ions. The alternative name is ‘co- Key terms
ordinate bond’, which also gives rise to the terms ‘co-ordination compound’
and ‘co-ordination number’. A co-ordination compound is one that contains A complex ion is an ion in which a
a complex ion, and the co-ordination number of a complex ion is the number of molecules or anions are
number of co-ordinate bonds from the ligands to the central metal ion. bound to a central metal cation by co-
ordinate bonds.
Co-ordination compounds contain complexes which may be cations, anions
A ligand is a molecule or anion bound
or neutral molecules (Figure 15.14). Examples of co-ordination compounds
to the central metal ion in a complex
include:
ion by co-ordinate bonding.
● K 3[Fe(CN)6] containing the negatively charged complex ion [Fe(CN)6]3−
The co-ordination number of a metal
● Fe(NO3)3.6H 2O containing the positively charged complex ion
ion in a complex is the number of co-
[Fe(H2O)6]3+
ordinate bonds to the metal ion from
● Ni(CO)4 containing a neutral complex between nickel atoms and carbon
the surrounding ligands.
monoxide molecules.
15.6 Formation of complex ions 429
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Figure 15.14 Crystals of co-ordination compounds. From left to right these are:
NiSO4.7H2O, FeSO4.7H2O, CoCl2.6H2O, CuSO4.5H2O, Cr2(SO4)3.18H2O and K3[Fe(CN)6].
There are two common visible signs that a reaction has occurred during the
formation of a new complex ion:
● a colour change
● an insoluble solid dissolving.
A familiar example of a colour change occurs when excess ammonia solution
is added to copper(ii) sulfate solution. Ammonia molecules displace water
molecules from hydrated copper(ii) ions forming [Cu(NH3)4(H2O)2]2+(aq)
ions and the colour changes from pale blue to deep blue.
[Cu(H2O)6]2+(aq) + 4NH3(aq) → [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
The test for chloride ions, using aqueous silver nitrate followed by ammonia
solution, is an example of an insoluble solid dissolving as a complex ion
forms. Adding silver nitrate to a solution of chloride ions produces a white
precipitate of silver chloride, AgCl. This precipitate dissolves on adding
ammonia solution as silver ions form the complex ion, [Ag(NH3)2]+(aq),
with ammonia molecules.
AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Cl−(aq)
Co-ordination compounds and complex ions are not only important in the
inorganic chemistry of transition metals. They are also very important in the
natural world. Chlorophyll in the leaves of plants, myoglobin in muscles and
haemoglobin in red blood cells are all examples of complexes between metal
ions and organic molecules. Zinc is an essential trace element in the human
diet because zinc ions are an essential part of important enzymes. Amino
acids in the protein chain form dative covalent bonds with the zinc.
15.7 Naming complex ions
There are four rules to follow when naming a complex ion.
1 Identify the number of ligands around the central cation using Greek
prefixes: mono-, di-, tri-, tetra-, and so on.
2 Name the ligand using names ending in -o for anions, e.g. chloro- for Cl−,
fluoro- for F−, cyano- for CN−, hydroxo- for OH−. Use aqua for H2O and
ammine for NH3.
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3 Name the central metal ion using the normal name of the metal for
positive and neutral complex ions and the Latinised name ending in
-ate for negative complex ions, e.g. ferrate for iron, cuprate for copper,
argentate for silver.
4 Finally, add the oxidation number of the central metal ion.
The examples in Table 15.3 illustrate how you should use the rules.
Table 15.3 Writing the systematic names of complex ions.
Formula of 1 Identify 2 Name the 3 Name the 4 Add the oxidation
complex ion the number ligand central metal number of the
of ligands ion central metal ion
[Ag(NH3)2]+ di ammine silver ( i)
[Cu(H2O)6]2+ hexa aqua copper (ii)
[CuCl4]2− tetra chloro cuprate (ii)
[Fe(CN)6]3− hexa cyano ferrate (iii)
Test yourself Tip
15 What is the co-ordination number of the named ions in the given Notice that ammonia, NH3, in
complex ion? complexes is described as ‘ammine’,
whereas the –NH2 group in organic
a) Cu2+ ions in [CuCl4]2−
compounds such as CH3NH2 is
b) Cu2+ ions in [Cu(H2O)6]2+ described as ‘amine’.
c) Fe3+ ions in [Fe(CN)6]3−
16 Write the systematic name of each complex ion.
a) [Co(NH3)6]3+
b) [Zn(OH)4]2−
c) [AlH4]−
d) [Ni(H2O)6]2+
17 What is the oxidation state of the metal ion in the following complex
ions?
a) [NiCl4]2−
b) [Ag(NH3)2]+
c) [Fe(H2O)6]3+
d) [Fe(CN)6]4−
18 A solution of thiosulfate ions, S2O32−, can dissolve a precipitate
of silver bromide. Each silver ion forms a complex ion with two
thiosulfate ions as the silver bromide dissolves. Write an equation
for this reaction.
15.8 The shapes of complex ions
The shapes of complex ions depend on the number of ligands around the
central metal ion. There is no simple, definitive rule for predicting the shapes
of complexes from their formulae, but:
● in complexes with a co-ordination number of six, the ligands usually
occupy octahedral positions so that the six electron pairs around the
central atom are repelled as far as possible (Figure 15.15) .
15.8 The shapes of complex ions 431
469983_15_Chem_Y1-2_418-[Link] 431 13/04/19 10:03 PM
● in complexes with a co-ordination number of four, the ligands usually
occupy tetrahedral positions although there are a few complexes with
four-fold co-ordination, such as [Pt(NH3)2Cl 2], that have a square planar
structure (Figure 15.15)
● in complexes with a co-ordination number of two, the ligands usually
Tip
form a linear structure with the central metal ion (Figure 15.15).
The shapes of complex ions cannot be
Two common tetrahedral complexes are [CuCl4]2− and [CoCl4]2−. The
predicted by the electron pair repulsion
relatively large size of the chloride ions, compared to oxygen atoms in water
theory that applies to molecules and
molecules, means that it is not possible for more than four ligands to fit round
the more simple ions.
the central metal ion.
NH3 Cl–
NH3
Pt2+
Cl–
H3N
A few complexes with NH3
NH3
a co-ordination number
Cr3+ Cl– of 4 are planar
H3N
NH3
Complexes with a Cr3+
co-ordination number of Cl–
6 are usually octahedral NH3 Cl– Cu+ Cl–
Complexes with a
Cl– co-ordination number
of 2 are usually linear
Complexes with a Cl–
co-ordination number of
4 are usually tetrahedral
Figure 15.15 The shapes of complex ions.
Types of ligand
Most ligands use only one lone pair of electrons to form a co-ordinate bond
with the central metal ion. These ligands are described as monodentate
Key terms because they have only ‘one tooth’ to hold onto the central cation (dens is
Latin for tooth). Examples of monodentate ligands include H2O, NH3,
Monodentate ligands form one dative
Cl−, OH− and CN−.
covalent (co-ordinate) bond with a
central metal ion in a complex. Some ligands have more than one lone pair of electrons that can form co-
ordinate bonds with the same metal ion. Bidentate (‘two-toothed’) ligands,
Bidentate ligands form two dative
for example, form two dative covalent bonds with metal ions in complexes.
covalent (co-ordinate) bonds with a
Bidentate ligands include 1,2-diaminoethane, H2NCH2CH2NH2, the
central metal ion in a complex.
ethanedioate ion, C2O42−, and amino acids (Figure 15.16).
432 15 Transition metals
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2+ 2+
H2N CH2
en
CH2
H2
N NH2
H2C
Ni2+ en Ni2+
H2C
N NH2
H2
CH2 en
H2N CH2
Figure 15.16 Representations of a complex formed by the bidentate ligand
1,2-diaminoethane with nickel(ii) ions. Note the use of ‘en’ as an abbreviation for the ligand.
The hexadentate ligand EDTA4− is particularly impressive because it
O
can form six co-ordinate bonds with the central metal ion in complexes.
– 2–
EDTA4− is the common abbreviation for this ion, which binds so firmly with O C
metal ions that it holds them in solution and makes them chemically inactive. O CH2
Figure 15.17 shows how the hexadentate ligand can fold itself around metal C CH2
–O N
ions, such as Pb2+, so that four oxygen atoms and two nitrogen atoms form
CH2
co-ordinate bonds to the metal ion. This is the ion formed when EDTA4− is
used to treat lead poisoning. The EDTA4− ion forms such a stable complex Pb2+
with Pb2+ ions that they can be excreted through the kidneys. CH2
The disodium salt of EDTA4−
is added to commercially produced salad –
O N
dressings to extend their shelf life. The EDTA4− ion traps traces of metal ions C H2C
that would otherwise catalyse the oxidation of vegetable oils. The disodium O –
CH2
O C
salt of EDTA4− is also an ingredient of bathroom cleaners to help remove
scale by dissolving Ca 2+ ions from the calcium carbonate left by hard water. O
Figure 15.17 The complex ion formed by
Tip the EDTA4− ion with a Pb2+ ion.
EDTA crystals consist of the disodium salt of ethylenediaminetetraacetic acid. Chemists
sometimes use the abbreviation Na2H2Y for the salt, where Y represents the 4− ion.
Ligands like those in Figures 15.16 and 15.17, which form more than one
co-ordinate bond with metal ions, are sometimes called multidentate
ligands, and the complexes which these ligands form are called chelates
(pronounced ‘keelates’). The term ‘chelate’ comes from a Greek word for Tip
a crab’s claw, reflecting the claw-like way in which chelating ligands grip
Multidentate ligands are sometimes
metal ions. Powerful chelating agents trap metal ions and effectively isolate
called polydentate ligands.
them in solution.
Key terms
Multidentate ligands form more than one co-ordinate bond with the same metal ion.
Chelates are complex ions involving multidentate ligands.
15.8 The shapes of complex ions 433
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Activity
Cis-platin – an important chemotherapy drug
The neutral complex, PtCl2(NH3)2, in which Cl− ions and NH3 Unfortunately, cis-platin is not a miracle cure without risks or
molecules act as ligands, has two isomers. These isomers drawbacks. It is toxic, resulting in unpleasant side-effects, and
have different melting temperatures and different chemical can cause kidney failure. Clinical trials have, however, led to
properties. One isomer called cis-platin is used as a the discovery of other platinum complexes that cause fewer
chemotherapy drug in the treatment of certain cancers, whereas problems and are already used as anti-cancer drugs.
the other isomer is ineffective against cancer. The trans isomer
1 Why is it possible to conclude that cis-platin has a square
is more toxic and not effective as a cancer drug.
planar rather than a tetrahedral structure?
Patients are given an intravenous injection of cis-platin, which 2 What type of isomerism do cis-platin and its isomer show?
circulates all around the body, including the cancerous area. 3 a) What is the oxidation number of platinum in cis-platin?
Cis-platin diffuses relatively easily through the tumour cell b) Write the systematic name of cis-platin.
membrane because it has no overall charge, like the cell c) Draw the structure of cis-platin.
membrane. 4 Why does cis-platin diffuse easily through the membrane
of cells?
Once inside the cell, cis-platin exchanges one of its chloride
5 What is meant by the term ‘active principle’ applied to
ions for a molecule of water to form [Pt(NH3)2(Cl)(H2O)]+, which
[Pt(NH3)2(Cl)(H2O)]+?
is the ‘active principle’ (Figure 15.18). This positively charged
6 When [Pt(NH3)2(Cl)(H2O)]+ has formed inside the cell, it
ion then enters the cell nucleus where it readily bonds with two
cannot diffuse out through the cell membrane. Why is this?
sites on the DNA. Binding involves co-ordinate bonding from
7 Why is a cell with cis-platin binding to its DNA unable to
the nitrogen or oxygen atoms in the bases of DNA to the
replicate?
platinum ion.
8 Why is the binding to cis-platin from nitrogen and oxygen
The cis-platin binding changes the overall structure of the DNA atoms rather than from carbon and hydrogen atoms in the
helix, pulling it out of shape and shortening the helical turn. bases of DNA?
The badly shaped DNA can no longer replicate and divide to 9 Why is cis-platin more likely to affect cancerous cells than
form new cells, although the affected cells continue to grow. normal cells?
Eventually, the cells die and, if enough of the cancerous cells 10 Why is any anti-cancer chemotherapy drug that acts like
absorb cis-platin, the tumour is destroyed. cis-platin likely to have undesirable side-effects?
11 Why is it important to test samples of cis-platin to make
sure that they are free of the trans isomer?
H3N Cl H3N Cl
Pt Pt
H3N Cl H3N Cl
cis-platin is neutral – Cl–
and so diffuses + H2O
through the cell +
membrane H3N Cl cell
nucleus
Pt DNA
H3N H2O
the active principle enters
the nucleus and binds
with its DNA
Figure 15.18 The action of cis-platin.
434 15 Transition metals
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Test yourself
19 Predict the likely shape of the following complex ions:
a) [Ag(CN)2]− b) [Fe(CN)6]3−
c) [NiCl4]2− d) [CrCl2(H2O)4]+
20 Explain how the amino acid glycine (H2NCH2COOH) can act as a
bidentate ligand.
21 a) Why is EDTA4− described as a hexadentate ligand?
b) What is the overall shape of the EDTA4− complex in Figure 15.17?
22 a) Draw a diagram to represent the complex ion formed between a
Cr3+ ion and three ethanedioate ions − O2C–CO2−.
b) What is the overall shape of this complex ion?
23 a) Predict an order of stability for the complex ions [Ni(NH3)6]2+,
[Ni(en)3]2+ and [Ni(EDTA)]2−. Figure 15.19 The blue pigment in this rare
Mauritius two pence stamp from 1847 is
b) Explain your prediction.
called Prussian blue. Its correct chemical
24 Refer to Figure 15.19. Write the formula of: name is iron(iii) hexacyanoferrate(ii).
a) the hexacyanoferrate(ii) ion b) iron(iii) hexacyanoferrate(ii).
15.9 Ligand exchange reactions
Complex ions often react by exchanging one ligand for another. These ligand
exchange reactions are often reversible and the changes of ligand are sometimes
accompanied by colour changes. For example, when excess concentrated
ammonia solution is added to pale blue copper(ii) sulfate solution, ammonia
molecules are exchanged for water molecules around the central Cu2+ ion
and the colour changes to a deep blue. The reaction takes place in two stages.
At first the alkaline solution of ammonia removes protons from the hydrated
copper(ii) ions to give a pale blue precipitate of the hydrated hydroxide.
[Cu(H2O)6]2+(aq) + 2OH−(aq) ⇋ [Cu(H2O)4(OH)2](s) + 2H2O(l)
pale blue solution pale blue precipitate
Then the ligand exchange takes place as the precipitate redissolves to give
the deep blue solution.
[Cu(H2O)4(OH)2](s) + 4NH3(aq)
pale blue precipitate
⇋ [Cu(NH3)4(H2O)2]2+(aq) + 2H2O(l) + 2OH−(aq)
deep blue solution
The ligands NH3 and H2O are both uncharged and similar in size. This
allows exchange reactions between these ligands without a change in co-
ordination number of the metal ion.
A ligand exchange reaction also occurs when concentrated hydrochloric acid
is added to copper(ii) sulfate solution. This time, the colour changes from
pale blue to yellow as Cl− ions replace water molecules around the Cu2+ ion. Tip
[Cu(H2O)6]2+(aq) + 4Cl−(aq) ⇋ [CuCl4]2−(aq) + 6H2O(l) A solution of a copper(ii) salt may turn
pale blue yellow green and not yellow if the concentration
of added chloride ions is not high
In this case, however, the ligand exchange involves a change in co-ordination
enough to convert all the blue hydrated
number. Chloride ions are larger than water molecules, so fewer chloride
ions to the yellow chloro complex.
ions can fit round the central Cu2+ ion.
15.9 Ligand exchange reactions 435
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Another example of a colour change associated with a change of ligand
and a change of co-ordination number is observed on adding concentrated
hydrochloric acid to an aqueous solution of a cobalt(ii) salt. The octahedral
hexaaquocobalt(ii) ion is pink. The solutions turns to a deep blue colour as
the tetrahedral tetrachlorocobaltate(ii) ion forms.
[Co(H2O)6]2+(aq) + 4Cl−(aq) ⇋ [CoCl4]2−(aq) + 6H2O(l)
N
Ligand exchange reactions also take place in living organisms. Haemoglobin
N Fe2+ N
O is the red protein in blood that carries oxygen from the lungs to the cells in
C body tissues. A haemoglobin molecule consists of four polypeptide chains,
N
OH each with a nitrogen atom forming a dative bond to an Fe2+ ion in a haem
O
C group (Figure 15.20). So there are four haem groups in each haemoglobin
molecule.
OH
Nitrogen atoms in the porphyrin ring form four more dative bonds to the
Figure 15.20 A haem group with its Fe2+
Fe2+ ion. So the porphyrin ring acts as a multidentate ligand. This leaves one
ion. There are four co-ordinate bonds
remaining site on the metal ion which can accept a pair of electrons from
between N atoms in the haem group and
an oxygen molecule (in oxyhaemoglobin which is bright red) or a water
the metal ion. In haemoglobin there is a
molecule (in deoxyhaemoglobin which is dull red).
fifth co-ordinate bond between an N atom
in one of the polypeptide chains and the The reactions between haem groups and oxygen or water are reversible
Fe2+ ion. This leaves one site on the Fe2+ ligand substitution reactions, allowing haemoglobin to pick up and release
ion that can accept a pair of electrons from oxygen. The reaction with carbon monoxide is irreversible, which explains
an oxygen molecule. why the gas is dangerously toxic.
Test yourself
25 Explain why breathing in carbon monoxide leads to death.
26 Write equations for the ligand exchange reactions that occur when:
a) hexaaquacobalt(ii) ions react with ammonia molecules to form
hexaamminecobalt(ii) ions
b) hexaamminecobalt(ii) ions react with chloride ions to from
tetrachlorocobaltate(ii) ions
c) hexaaquairon(ii) ions react with cyanide ions to form
hexacyanoferrate(ii) ions.
27 A dilute solution of cobalt(ii) chloride is pink because it
contains hydrated cobalt(ii) ions. The solution turns blue on
adding concentrated hydrochloric acid with the formation of
tetrachlorocobaltate(ii) ions.
Figure 15.21 Filter paper soaked in pink a) Write an equation for the reaction that occurs when concentrated
cobalt(ii) chloride solution and dried in an HCl is added to dilute cobalt(ii) chloride solution and indicate the
oven until it is blue, can be used to test for colour of all species.
the presence of water. b) Explain the chemical basis for the test illustrated in Figure 15.21.
Relative stability of complex ions
In aqueous solution, the simple compounds of most transition metals contain
complex ions with formulae such as [Cu(H2O)6]2+, [Cr(H2O)6]3+ and
[Co(H2O)6]2+.
When solutions containing other ligands, such as Cl−, are added to aqueous
solutions of these hydrated cations, the mixture comes to an equilibrium in
436 15 Transition metals
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which the water molecules of some complexes have been replaced by the
added ligands. For example, the equilibrium which results when concentrated
sodium chloride solution is added to aqueous copper(ii) ions is:
[Cu(H2O)6]2+(aq) + 4Cl−(aq) ⇋ [CuCl4]2−(aq) + 6H2O(l)
The equilibrium constant, Kc, for this reaction is:
[CuCl42–(aq)]
Kc =
[Cu(H2O)62+(aq)][Cl–(aq)]4
[H2O(l)] is constant and therefore it is not included in the equation for Kc.
Equilibrium constants like this for the formation of complex ions in aqueous
solution are called stability constants and the symbol K stab is sometimes used
in place of Kc.
Stability constants enable chemists to compare the stabilities of the complex
ions of a cation with different ligands. The larger the stability constant, the
more stable is the complex ion compared with that containing water.
Table 15.4 shows the stability constants of three complexes of the copper(ii) ion.
These show that the relative stabilities of the three copper(ii) complexes are:
[Cu(EDTA)]2− > [Cu(NH3)4(H2O)2]2+ > [CuCl4]2−
Ligand Complex ion K Table 15.4 The stability constants of three
Cl− [CuCl4]2− 4.0 × 105 copper(ii) complexes.
NH3 [Cu(NH3)4(H2O)2]2+ 1.3 × 1013
EDTA4− [Cu(EDTA)]2− 6.3 × 1018
Complex ions and entropy
When a bidentate ligand, such as 1,2-diaminoethane, replaces a monodentate
ligand, such as water, there is an increase in entropy of the system. One
molecule of the bidentate ligand replaces two molecules of the monodentate
ligand and this results in an increase in the number of product particles. For
example, in the reaction:
[Cu(H2O)6]2+(aq) + 3H2NCH2CH2NH2(aq)
→ [Cu(H2NCH2CH2NH2)3]2+(aq) + 6H2O(l)
there are seven product particles but only four reactant particles. In this
reaction the enthalpy change is very small and so the entropy change in the
surroundings is close to zero.
As the entropy of any system depends on the number of particles present, the
entropy of the system increases when this reaction occurs. In other words, Tip
∆S system is positive.
For a spontaneous change to occur,
When a polydentate ligand, such as EDTA4−, replaces a monodentate ligand, ∆Stotal must be positive.
an even larger increase occurs in the entropy of the system.
∆Stotal = ∆Ssystem + ∆Ssurroundings
[Cu(H2O)6]2+(aq) + EDTA4−(aq) → [Cu(EDTA)]2−(aq) + 6H2O(l)
If ∆Stotal is positive, the products of the
Because of this increase in the entropy of the system, complexes with
system are more thermodynamically
polydentate ligands are usually more stable than those with bidentate ligands,
stable than the reactants.
which in turn are more stable than complexes with monodentate ligands.
15.9 Ligand exchange reactions 437
469983_15_Chem_Y1-2_418-[Link] 437 13/04/19 10:03 PM
Test yourself
Table 15.5 28 The stability constants and colour of some cobalt(ii) complexes are
shown in Table 15.5.
Complex Stability Colour
constant What would you expect to see when:
[Co(H2O)6]2+ 1.0 Pink a) ammonia solution is added to an aqueous solution of cobalt(ii)
[Co(NH3)6]2+ 3 × 104 Green chloride
[Co(EDTA)]2− 2 × 1016 Pink b) EDTA solution is added to a solution of cobalt(ii) chloride in
aqueous ammonia?
29 Suggest two reasons why the stability constant of [Co(EDTA)]2− is so
much larger than those of [Co(H2O)6]2+ and [Co(NH3)6]2+.
Core practical 12
The preparation of a transition metal complex
This is a summary of the procedure for preparing and purifying 3 Suggest a reason for adding ammonium chloride to the
the complex salt called hexamminecobalt(iii) chloride. reaction mixture as well as concentrated ammonia.
4 Step C involves the oxidation of the cobalt(ii) complex to the
Preparation of the salt
required cobalt(iii) complex.
A Add 8 g ammonium chloride and 12 g hydrated cobalt(ii)
a) Use these standard electrode potentials to explain why
chloride, [Co(H2O)6]Cl3, to a measured volume of water in a
the oxidation is possible with the ammine complex but
flask. Then add a measured amount of powdered charcoal
not the aqua complex.
and bring to the boil.
B Cool the mixture from A and then add 25 cm3 of [Co(NH3)6]3+(aq) + e− ⇋ [Co(NH3)6]2+(aq)
concentrated ammonia solution, and cool again. E 1 = +0.10 V
C Add a total of 25 cm3 of 20-volume hydrogen peroxide a
H2O2(aq) + 2H+(aq) + 2e− ⇋ 2H2O(l)
small amount at a time, shaking after each addition. Then
E 1 = +1.77 V
heat the mixture to about 60 °C and keep the solution at
this temperature for 30 minutes. [Co(H2O)6]3+(aq) + e− ⇋ [Co(H2O)6]2+(aq)
D Cool the flask in iced water to precipitate the impure product.
E 1 = +1.82 V
Purification of the salt
b) What do the electrode potentials show about the relative
E Filter off the impure crystals with the charcoal catalyst.
stability of the cobalt(ii) and cobalt(iii) states when
F Add the solid from the filter paper to boiling water acidified
complexed with water or with ammonia?
with concentrated hydrochloric acid. Stir and the filter again,
5 Write the overall balanced ionic equation for the oxidation
retaining the filtrate.
reaction.
G Cool the filtrate in iced water, then filter off the crystals that
6 The charcoal is added as a catalyst for the formation of
separate.
the required product. What is the evidence that this is a
H Rinse the crystals on the filter paper first with a very little
heterogeneous catalyst?
cold water and then with ethanol.
7 What is the name of the procedure in steps F and G and
I Leave the crystals to dry, then measure the mass of the
why does it purify the product?
product.
8 Suggest a reason for adding hydrochloric acid in step F to
Questions increase the yield of the product.
1 Identify the particularly hazardous chemicals used in the 9 Explain the purpose of the small amount of cold water and
preparation and state the special precautions needed when then the ethanol used in step H.
handling them. 10 Calculate the theoretical yield of product and the mass of
2 There is a ligand exchange reaction in step B to change the the golden-brown crystals obtained if the procedure gives
hexaqua complex into a hexamminecobalt(ii) ion. Write an a 70% yield.
ionic equation for the reaction.
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15.10 Reactions of transition metal
ions with aqueous alkalis
The acid–base and ligand-exchange reactions of aqueous metal ions can Key term
be used in qualitative analysis to identify the positive ions in salts. Adding
An amphoteric hydroxide is one that
aqueous sodium hydroxide produces a precipitate if the metal hydroxide is
can dissolve in either aqueous acid or
insoluble. The precipitate dissolves in excess of the alkali if the hydroxide is
aqueous alkali.
amphoteric.
Adding ammonia solution also precipitates insoluble hydroxides. These
redissolve in excess if the metal ion forms stable complex ions with ammonia
molecules (Table 15.6).
Table 15.6 Results of adding aqueous sodium hydroxide and aqueous ammonia solutions
to samples of transition metal ions.
Positive ion in Observations on adding Observations on adding
solution sodium hydroxide solution drop ammonia solution drop by
by drop and then in excess drop and then in excess
Chromium(iii), Green precipitate which Grey-green precipitate slightly
Cr3+ dissolves in excess reagent to soluble in excess reagent to
green form a dark green solution give a purple solution
Iron(ii), Fe2+ Dirty green precipitate insoluble Green precipitate insoluble in
pale green in excess reagent excess reagent
Iron(iii), Fe3+ Browny-red precipitate insoluble Browny-red precipitate
yellow in excess reagent insoluble in excess reagent
Cobalt(ii), Co2+ Blue precipitate from a pink Blue precipitate from a
pink solution, insoluble in excess pink solution dissolving in
reagent excess reagent to give a pale
brown solution that gradually
darkens.
Copper(ii), Cu2+ Pale blue precipitate insoluble Pale blue precipitate dissolving
blue in excess reagent in excess reagent to form a
dark blue solution
The behaviour of aqueous chromium(iii) ions shown in Table 15.6 is
typical of hydrated metals ions in the 3+ state. The behaviour of aqueous
iron(iii) is the exception. The chromium(iii) ions are hydrated. In the
complex, the electrons in the water molecules are pulled towards the
highly polarising Cr3+ ion, making it easier for the water molecules
linked to the chromium ion to give away protons. The hydrated ion is
partially ionised in solution. It is a strong enough acid to form carbon
dioxide when added to a solution of sodium carbonate.
[Cr(H2O)6]3+ ⇋ [Cr(H2O)5OH]2+(aq) + H+(aq)
[Cr(H2O)5OH]2+(aq) ⇋ [Cr(H2O)4(OH)2]+(aq) + H+(aq)
Adding hydroxide ions, a base, to a solution of chromium(iii) ions removes a
third proton, producing an uncharged complex. The uncharged complex is
much less soluble in water and precipitates as a white jelly-like precipitate of
hydrated chromium(iii) hydroxide.
[Cr(H2O)4(OH)2]+(aq) + OH−(aq) ⇋ [Cr(H2O)3(OH)3](s) + H2O(l)
15.10 Reactions of transition metal ions with aqueous alkalis 439
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Core practical 15 (part 1)
Analysis of an inorganic unknown
A student carried out a series of tests on a compound X as described in the first column
of Table 15.7. The results are shown in column 2 of the table. Column 3 shows some, but
not all, of the student’s interpretation of the observations.
Table 15.7
Test Observations Inferences
A Describe the appearance of X. Pale green crystals X might be a salt of a
transition metal.
B Heat a sample of X in a test At first condensation appears on the cooler parts of the X is a hydrated salt. It
tube, first gently and then more test tube. On strong heating gas is given off and the decomposes on strong
strongly. solid turns reddish-brown. The gas turns blue litmus red heating. The gas is a
and reacts with a solution potassium dichromate(vi) reducing agent.
turning it from orange to green.
C Add a few drops of NaOH(aq) to a A green precipitate forms. The precipitate does not X could be an iron(ii)
solution of X, then add an excess dissolve in excess alkali. salt. It cannot be a
of the alkali. chromium(iii) salt.
D Add a few drops of NH3(aq) to a A green precipitate forms. The precipitate does not This is consistent with
solution of X, then add an excess dissolve in excess ammonia solution. the result for test C.
of the ammonia solution.
E Add aqueous chlorine to a solution The very pale green solution turns yellow on adding
of X, then add a few drops of chlorine. Then adding alkali produces a browny-red
NaOH(aq) to the solution followed precipitate insoluble in excess reagent.
by an excess of the alkali.
F Acidify a solution of X with dilute No precipitate forms.
nitric acid, then add a few drops of
AgNO3(aq) to the solution.
G Acidify a solution of X with dilute A white precipitate forms.
nitric acid, then add a few drops
of Ba(NO3)2(aq) to the solution.
1 Test A: Which of the first row of transition metals, in which 5 Tests F and G:
oxidation states, form salts that are green? ● What can be inferred from the results of tests F and G?
2 Test B: ● Write an ionic equation for the reaction in test G.
● Why is the conclusion that the gas is a reducing agent 6 What is the evidence that a redox reaction takes place
justified? on heating crystals of X? Suggest an equation for the
● Given an example of an acidic gas that is a reducing agent. decomposition reaction.
3 Test C: Why do the results of the test show that X cannot be a 7 The traditional name for iron(ii) sulfate is green vitriol. This
chromium(iii) salt? is because it produces oil of vitriol (sulfuric acid) when the
4 Test E: gases given off on heating the salt are condensed. Use
● What can be inferred from the results of this test? your answer to Question 6 to suggest an explanation for the
● Write an ionic equation for the reaction between X and formation of sulfuric acid in this way.
chlorine.
Tip
Analysis of an organic unknown is covered in Core practical 15 (part 2), in Section 18.3.3.
Refer to Practical skills sheet 19, ‘Analysing inorganic unknowns’, which you can
access online at [Link]/EdexcelChemistry.
440 15 Transition metals
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Adding excess alkali removes yet another proton, now producing a negatively
charged ion which is soluble in water so that the precipitate redissolves. This
explains the amphoteric properties of chromium(iii) hydroxide.
[Cr(H2O)3(OH)3](s) + OH−(aq) ⇋ [Cr(H2O)2(OH)4]−(aq) + H2O(l)
All these changes are reversed by adding a solution of a strong acid such as
hydrochloric acid, which reacts with the hydroxide ions in the complex to
form water molecules.
Test yourself
30 E xplain the following changes with the help of 31 E xplain the following observations with the help of
ionic equations. ionic equations.
a) Adding a small amount of ammonia solution a) A solution of iron(iii) chloride is acidic. A
to a pale blue solution of hydrated copper(ii) browny-red precipitate forms on adding
ions produces a pale blue precipitate of the aqueous sodium hydroxide but the precipitate
hydrated hydroxide. is not soluble in excess alkali.
b) On adding more ammonia solution, the b) Adding aqueous sodium hydroxide to a solution
precipitate dissolves to give a deep blue of cobalt(ii) chloride produces a blue precipitate
solution. which does not dissolve in excess alkali.
15.11 Catalysts based on transition
elements and their compounds
Transition elements and their compounds play a crucial role as catalysts in
industry. Table 15.8 lists some important examples of transition metals and
their compounds as catalysts.
Catalysts can be divided into two types – heterogeneous and homogeneous.
Heterogeneous catalysis
Heterogeneous catalysis involves a catalyst in a different state from the
reactants it is catalysing. It is used in almost every large-scale manufacturing
process, for example the manufacture of ammonia in the Haber process
(Table 15.8), in which nitrogen and hydrogen gas flow through a reactor
containing lumps of iron or iron(iii) oxide.
Transition element/ Reaction catalysed Table 15.8 Some important examples of
compound used as catalyst transition metals and their compounds as
Vanadium(v) oxide, V2O5 Contact process in the manufacture of sulfuric acid: heterogeneous catalysts.
2SO2(g) + O2(g) ⇋ 2SO3(g)
Iron or iron(iii) oxide Haber process to manufacture ammonia:
N2(g) + 3H2(g) ⇋ 2NH3(g)
Nickel, platinum or Hydrogenation of unsaturated vegetable oils:
palladium RCH=CH2(l) + H2(g) → RCH2CH3(l)
Platinum or platinum– Conversion of NO and CO to CO2 and N2 in catalytic
rhodium alloys converters in vehicles:
2CO(g) + 2NO(g) → 2CO2(g) + N2(g)
Platinum Reforming straight-chain alkanes as cyclic alkanes
and arenes:
CH3(CH2)5CH3 → CH3̶ C6H5 + 4H2
heptane methylbenzene
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Heterogeneous catalysts work by adsorbing reactants at active sites
Key term on their surface. Nickel, for example, acts as a catalyst for the addition of
hydrogen to unsaturated compounds with carbon–carbon double bonds
A heterogeneous catalyst is a catalyst
(Table 15.8). Hydrogen molecules are adsorbed on the catalyst surface,
that is in a different state from the
where they are thought to split into single atoms (free radicals). These
reactants. Generally, a heterogeneous
highly reactive hydrogen atoms undergo addition with molecules of
catalyst is a solid, while the reactants
unsaturated compounds, like ethene, when the unsaturated compounds
are gases or in solution. The advantage
approach the catalyst surface (Figure 15.22).
of heterogeneous catalysts is that they
can be separated from the reaction
products easily.
C
C
C
C C C
H H H H step 1 H H H step 2 H H
Ethene approaches the catalyst Ethene adds one hydrogen After adding a second hydrogen
surface where hydrogen gas is atom and the CH3CH2• radical atom the hydrocarbon, now
adsorbed as single atoms is attached to the surface ethane, escapes from the surface
Figure 15.22 A possible mechanism for the hydrogenation of an alkene using a nickel
catalyst. The reaction takes place on the surface of the catalyst, which adsorbs hydrogen
molecules and then splits them into atoms.
If a metal is to be a good catalyst for the addition of hydrogen, it must not
adsorb the hydrogen so strongly that the hydrogen atoms become unreactive.
This happens with tungsten. Equally, if adsorption is too weak there are
insufficient adsorbed atoms for the reaction to occur at a useful rate, and
this is the case with silver. The strength of adsorption must have a suitable
Key terms
intermediate value, which is the case with nickel, platinum and palladium.
Adsorption is a process in which atoms, The Contact process for making sulfuric acid gets its names from the
molecules or ions are held onto the ‘contact’ between the reacting gases and the surface of the heterogeneous
surface of a solid. catalyst. The vanadium(v) oxide catalyst is effective because the metal can
Desorption is the opposite of change its oxidation state reversibly. First the vanadium(v) oxide is reduced
adsorption when atoms, molecules or to vanadium(iv) oxide as it oxidises sulfur dioxide to sulfur trioxide. Then
ions are released from a solid surface. the vanadium(iv) oxide is reoxidised to vanadium(v) oxide by oxygen in the
mixture of reacting gases.
Test yourself
32 a) Write equations to describe the catalytic action of vanadium(v)
oxide in the Contact process.
b) W
hy is the vanadium(v) oxide in the Contact process described
as a catalyst given that it reacts with sulfur dioxide?
442 15 Transition metals
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Activity
Catalytic converters
Catalytic converters have done a great deal to improve air quality in our towns Gases from the engine
containing unburned
and cities. The catalyst speeds up reactions that remove pollutants from motor hydrocarbons, carbon
vehicle exhausts. The reactions convert oxides of nitrogen to nitrogen and monoxide and oxides
oxygen. They also convert carbon monoxide to carbon dioxide, and unburnt of nitrogen.
hydrocarbons to carbon dioxide and steam (Figure 15.23).
The catalyst in a catalytic converter is made from a combinations of platinum,
palladium and rhodium. The pollutants are adsorbed onto the surface of the
catalyst, where they react (Figure 15.24). Then the products are desorbed Ceramic block with a
structure like a honey- Exhaust gas
into the stream of exhaust gases. containing
comb. The channels
have a very large surface carbon dioxide,
The catalyst must not adsorb molecules so strongly that the reactive sites on area which is coated nitrogen
the surface of the metal are inactivated. However, the interaction between with the catalyst. and steam.
pollutant molecules and the metal surface has to be strong enough to weaken Figure 15.23 Function of a catalytic converter.
bonds and provide a reaction mechanism that is fast enough under the
conditions in the exhaust system. Then the reaction products have to be so
weakly attracted that they are quickly released into the gas stream.
Figure 15.24 The surface of the metal catalyst in
a catalytic converter adsorbs the pollutants NO
and CO, where they react to form N2 and CO2. In
this computer graphic, oxygen atoms are coloured
red, nitrogen atoms are coloured blue and carbon
atoms are coloured green.
1 Why do you think the catalyst in a catalytic converter is 5 Why would the catalyst be ineffective if the bonding between
present as a very thin layer on the surface of many fine holes the catalyst surface:
running through a block of inert ceramic? a) and the reactants is too weak
2 Suggest reasons why the catalyst in a catalytic converter is b) and the products is too strong?
only fully effective: 6 Suggest a reason why using petrol containing lead additives
a) after the engine has been running for some time in a car engine rapidly stops the catalytic converter being
b) if the engine is properly maintained so that it runs with the effective.
right mixture of air and fuel. 7 Identify two ways, other than fitting catalytic converters, to
3 Write equations for the reactions catalysed by a catalytic reduce air pollution from motor vehicles in cities.
converter that remove carbon monoxide and nitrogen 8 What contribution, if any, do catalytic converters make to
monoxide from exhaust gases. solving the problem of climate change?
4 How does the catalyst speed up the reactions that destroy
pollutants?
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Key term Homogeneous catalysis
Homogeneous catalysis involves a catalyst in the same state as the reactants
A homogeneous catalyst is a catalyst it is catalysing. Homogeneous catalysts are very important in biological
that is in the same state as the systems because enzymes (proteins) act as catalysts in the metabolic processes
reactants. Usually, the reactant and of all organisms. Very often transition metal ions act as co-enzymes in
the catalyst are dissolved in the same these processes, enhancing the catalytic activity of the associated enzyme.
solution. Cytochrome oxidase is an important enzyme containing copper. This
enzyme is involved when energy is released from the oxidation of food.
In the absence of copper, cytochrome oxidase is totally ineffective and the
animal or plant is unable to metabolise successfully.
Transition metal ions can also be effective as homogeneous catalysts because
they can gain and lose electrons, changing from one oxidation state to
another. The oxidation of iodide ions by peroxodisulfate(vi) ions using
iron(iii) ions as a catalyst is a good example of this.
2I−(aq) + S2O82−(aq) → I2(aq) + 2SO42−(aq)
In the absence of Fe3+ ions the reaction is very slow, but with Fe3+ ions in
Tip
the mixture the reaction is many times faster. A possible mechanism is that
Do not confuse the reaction of iodide Fe3+ ions are reduced to Fe2+ as they oxidise iodide ions to iodine. Then
ions with peroxodisulfate(vi) ions with the S2O82− ions oxidise Fe2+ ions back to Fe3+, ready to oxidise more of the
the reaction of iodine molecules with iodide ions, and so on.
thiosulfate ions.
Sometimes one of the products of a reaction can act as a catalyst for the
process. This is called autocatalysis. An autocatalytic reaction starts slowly,
but then speeds up as the catalytic product is formed. Mn 2+ ions act as an
autocatalyst in the oxidation of ethanedioate ions, C2O42−, by manganate(vii)
ions in acid solution. This is also an example of homogeneous catalysis.
2MnO4−(aq) + 16H+(aq) + 5C2O42−(aq)
→ 2Mn 2+(aq) + 8H2O(l) + 10CO2(g)
Test yourself
33 What is the advantage of using a solid heterogeneous catalyst in:
a) a continuous industrial process
b) an industrial batch process?
34 a) S
uggest a reason why the reaction of between iodide ions and
peroxodisulfate ions is slow in the absence of a catalyst.
b) W
rite half-equations to explain the mechanism by which iron(iii) ions
catalyse the reaction between iodide ions and peroxodisulfate ions.
o you think Fe2+ ions can also catalyse this reaction? Explain
c) D
your answer.
35 a) S
uggest two methods of speeding up the reaction between
MnO4−(aq) and C2O42−(aq) from the start of the reaction.
b) W
hat would you expect to see when a solution of potassium
manganate(vii) is added to an acidified solution of potassium
ethanedioate:
i) at the start of the reaction
ii) as the reaction gets underway?
444 15 Transition metals
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The development of new catalysts
One of the real challenges and priority areas in chemical research today is
the development of new and improved catalysts. Catalysts are particularly
important in the drive for a ‘greener world’ and a ‘greener future’.
● Catalysts speed up reactions and products are obtained faster.
● Catalysts allow processes to operate at lower temperatures with savings on
energy and fuel.
● Catalysts make possible processes that have high atom economies and
produce less waste.
● Catalysts can be highly selective so that only the desired product is formed
without the obvious waste from side-reactions.
The manufacture of ethanoic acid illustrates the advantages of developing
new catalysts. Until the 1970s, the main method of manufacturing ethanoic
acid was to oxidise hydrocarbons from crude oil in the presence of a cobalt(ii)
ethanoate catalyst. The process operated at 180–200 °C and 40–50 times
atmospheric pressure. Only 35% by mass of the products was ethanoic acid.
Today, ethanoic acid is manufactured as the only product in the direct
combination of methanol and carbon monoxide. The catalyst of iridium
metal is mixed with ruthenium compounds that act as catalyst promoters
and triple the rate of reaction. In addition, the process operates at lower
temperature and lower pressure, producing only ethanoic acid.
As scientists develop new techniques, such as the improved catalytic process
for ethanoic acid, and propose new ideas, it is important that their work is
reported, checked and validated. This reporting and validating is carried out
in three ways.
● Through reports, journals and conferences at which scientists discuss their work
with others.
● By peer review in which others working in the same area look closely at
the validity of the experimental methods used, the accuracy of the results
obtained and the appropriateness of the conclusions drawn.
● By replicating experiments, which is perhaps the ultimate test of reliability
and validity for any innovation. If other scientists repeat the work and
achieve the same results, then its integrity is not questioned.
Test yourself
36 a) Write an equation for the production of ethanoic acid from
methanol and carbon monoxide.
b) What is the atom economy of the process?
37 You hear, from one of the popular media, about a newly discovered
catalyst that promises great economic and environmental benefits.
Suggest some of the questions that you should ask before deciding
whether or not to take the claims seriously.
15.11 Catalysts based on transition elements and their compounds 445
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Chapter summary
Chapter 15 Transition metals examples of complexes with four-fold coordination
that are square planar.
l The electron configurations of d-block elements l The square, planar complex called cisplatin disrupts
run from scandium, [Ar]3d14s2, to zinc, DNA replication and so can be used to treat
[Ar]3d104s2. The configurations of chromium and cancer, unlike its trans isomer, which is inactive.
copper do not fit the general pattern. l A bidentate ligand forms two coordinate bonds
l The elements in the d-block are transition
with a central metal ion in a complex. Multidentate
elements if they form one or more stable ions ligands form more coordinate bonds with the
with incompletely filled d orbitals. This definition central atom (for example EDTA4−, which has six
excludes scandium and zinc. electron pairs that can form dative bonds).
l Transition metals form compounds with more
l Complex ions can react by exchanging one ligand
than one oxidation number; this can be explained for another. There are often colour changes
in terms of the relative energies of the 3d and 4s associated with these ligand exchange reactions.
orbitals. l Ligand exchange can lead to a change of
l A complex ion is a central metal ion surrounded
coordination number when smaller uncharged
by ligands. The ligands have lone pairs of electrons ligands are replaced by larger charged ligands.
that form dative (coordinate) bonds with the l Haemoglobin is an iron(ii) complex containing a
metal ion. multidentate ligand. A ligand exchange reaction
l The number of coordinate bonds to the metal ion
takes place when oxygen bound to haemoglobin is
from ligands is the coordination number. replaced by carbon monoxide.
l Colour in transition metal ions can arise from
l Substitution of a monodentate ligand by a bidentate
electronic transitions between the orbitals within or multidentate ligand leads to a more stable
the 3d sub-shell, which do not all have the complex because of the large, positive increase in
same energy when the metal ion is hydrated or ΔS system.
combined with other ligands in a complex. l Many hydrated transition metal ions are acidic –
l Ions are colourless if the d orbitals are empty (as in
they form a metal hydroxide precipitate on adding
Sc3+) or full as in (Cu+ and Zn 2+). alkali. This can be reversed by adding acid. The
l The colour of a transition metal ion may change if
hydroxide precipitate also dissolves in excess alkali
there is a change in oxidation number of the metal, if it is amphoteric.
a change in the type of ligand, or a change of l Transition metals and their compounds can act as
coordination number. catalysts.
l The oxidation states of vanadium have
l Examples of heterogeneous catalysts are V2O5 in
characteristic colours and the interconversion of the contact process and the platinum alloy in a
these states can be explained in terms of E 1 values. catalytic converter. The reactions take place on the
l E values can be used to account for the reduction
1
surface of the solid catalyst.
of Cr(iii) to Cr(ii) by zinc and the oxidation of l Transition metal ions can act as homogeneous
Cr(iii) to Cr(vi) by hydrogen peroxide under catalysts because they can gain or lose electrons as
alkaline conditions. they change oxidation state. An example is the use
l Ligands with one lone-pair of electrons are
of Fe2+ as a catalyst for the reaction of iodide ions
monodentate (such as H2O, OH− and NH3). with peroxodisulfate(vi) ions.
l Complexes with a coordination number of six
l The product of a reaction can act as a catalyst for
usually have an octahedral shape. the process. This is autocatalysis, illustrated by the
l If the coordination number is four, the ligands
oxidation of ethanedioate ions by manganate(vii)
usually occupy tetrahedral positions, especially ions.
if they are relatively large; but there are a few
446 15 Transition metals
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Exam practice questions
1 a) Write the electronic structure of: d) A 0.012 mol sample of an oxochloride of
i) a scandium atom(1) vanadium,VOClx, required 20.0 cm3 of
ii) a copper atom(1) 0.100 mol dm−3 potassium dichromate(vi)
iii) a Cu2+ ion.(1) solution for oxidation of the vanadium to its
b) Both scandium and copper are d-block +5 oxidation state.
elements, but only copper is a transition i) Complete the half-equation below
element. Explain the meaning of the for the action of dichromate(vi) as an
terms in italics.(3) oxidising agent.
c) Aqueous Cu2+ ions react with Cr2O72−(aq) + …..H+(aq) + …..e−
excess ammonia solution to form → …..Cr3+(aq) + …H2O(l) (1)
[Cu(NH3)4(H2O)2]2+ ions. ii) Calculate the amount, in moles, of
i) Write the name of the dichromate(vi) that reacted with the
[Cu(NH3)4(H2O)2]2+ ion.(1) oxochloride of vanadium.(1)
ii) Name the overall shape of the iii) Calculate the amount, in moles, of
[Cu(NH3)4(H2O)2]2+ ion.(1) electrons that were gained by the
d) Explain why the complexes of copper(ii) Cr2O72− ions.(1)
ions are usually coloured.(4) iv) Calculate the change in oxidation
2 This question concerns the chemistry of state of the vanadium during the
transition metals. reaction.(1)
a) State and explain what is meant by the v) Write the formula of the oxochloride
terms giving examples: of vanadium showing the correct
i) transition metal(3) value of x.(1)
ii) oxidation number(3) 4 a) State whether these catalysts are
iii) complex ion.(3) homogeneous or heterogeneous:
b)* Discuss, with examples, equations and i) the platinum alloy in a catalytic
observations, the typical reactions of the converter (1)
ions of transition metals.(6) ii) the sodium hydroxide used to
3 Chromium shows its highest oxidation hydrolyse a halogenoalkane (1)
state in the oxoanion, CrO42−. iii) the zeolite use to crack oil fractions (1)
a) State the oxidation number of iv) the nickel used to hydrogenate
chromium in CrO42−.(1) unsaturated fats.(1)
b) The mixture changes colour when b) The activation energy for the
dilute acid is added to a solution of decomposition of ammonia into
CrO42− ions. State the changes in nitrogen and hydrogen is 335 kJ mol−1
colour and write an equation for the in the absence of a catalyst but
reaction that occurs.(3) 162 kJ mol−1 in the presence of a
c) When sulfur dioxide is bubbled into tungsten catalyst. Explain the
a solution of CrO42− ions, the colour significance of these values in terms
changes and chromium is reduced to of transition state theory. (4)
a simple ion. State the new colour and
the formula of the new simple ion.(2)
447
Exam practice questions
469983_15_Chem_Y1-2_418-[Link] 447 13/04/19 10:04 PM
H2O(I) dilute NH3(aq)
CuSO4.5H2O(s) blue solution blue precipitate
A B
excess excess
concentrated concentrated
HCl(aq) NH3(aq)
yellow-green deep blue
solution solution
C D
5 The reaction scheme above involves various tends to be poor at the ends of the transition
compounds of copper. series, but high in the middle of the series.
a) Write the formulae of the species i) Give two reasons why the catalytic
responsible for the colour in each of efficiency of metals is poor at the ends
the products A to D.(4) of the transition series.(2)
b) Describe and explain, with an equation, ii) Give a possible reason why the catalytic
what you would see when solution C efficiency is high in the middle of the
is diluted with excess water.(6) transition series.(2)
c) When aqueous sodium hydroxide is added to d) In catalytic converters used to ‘clean’ the
copper(ii) sulfate solution, a blue precipitate exhaust gases from petrol engines, a catalyst
is formed. If, however, excess EDTA4− reduces nitrogen oxides using another
solution is first added to the copper(ii) pollutant gas as the reducing agent. State
sulfate solution before the aqueous sodium a suitable catalyst for catalytic converters,
hydroxide, no precipitate forms. identify the reducing agent and write
an equation for a possible reaction that
Write an equation for the formation of results.(3)
the blue precipitate with aqueous sodium
hydroxide, and explain why no precipitate 7 Hydrazine, H2NNH2, is a powerful reducing
forms if excess EDTA4− is added to the agent in alkaline solution. It is oxidised to
copper(ii) sulfate solution before the nitrogen gas and water.
sodium hydroxide.(5) Vanadium exists in several oxidation states and
There are two main types of catalyst:
6 two of its electrode (reduction) potentials are
homogeneous and heterogeneous. shown below.
a) Explain the term ‘homogeneous catalysis’ VO2+(aq) + H2O(l) + e−
and state the most important feature of ⇋ V3+(aq) + 2OH−(aq)
transition metal ions that allows them to E 1 = −1.32 V
act as homogeneous catalysts.(2)
b) In aqueous solution, I− ions slowly reduce VO2+(aq) + H2O(l) + e−
S2O82− ions to SO42− ions. ⇋ VO2+(aq) + 2OH−(aq)
i) Write an equation (or two half- E 1 = −0.66 V
equations) for the reaction.(2) a) Deduce the ionic half-equation for the
ii) Give a possible reason why the oxidation of hydrazine in alkaline
activation energy of the reaction is solution.(2)
high, resulting in a slow reaction in the b) Hydrazine reduces vanadium(v) but not
absence of a catalyst.(1) vanadium(iv) in alkaline solution. Explain
iii) Write two equations (or two pairs of what this shows about the value of the
half-equations) to show the role of electrode potential for the reduction half-
iron salts in catalysing the reaction.(2) equation that is the reverse of your
c) In Periods 5 and 6, the catalytic efficiency of answer to part (a).(2)
transition metals as heterogeneous catalysts
448
15 Transition metals
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c) Write the overall ionic equation for the 8 The complexes [Ni(NH3)2Cl2] and
reduction of vanadium(v) by hydrazine in [Co(NH3)4Cl2]+ both form cis and trans
alkaline solution.(2) stereoisomers.
d) i) Explain, with an example, what is a) State what is meant by the terms ‘complex’
meant by the term and ‘stereoisomer’.(4)
‘disproportionation’.(2) b) Give the oxidation number of:
ii) Explain why an element must have at i) Ni in [Ni(NH3)2Cl2](1)
least three oxidation states if it is to ii) Co in [Co(NH3)4Cl2]+.(1)
undergo disproportionation.(1) c) Give the co-ordination number of:
iii) Write an equation for the i) Ni in [Ni(NH3)2Cl2](1)
disproportionation of vanadium(iv) ii) Co in [Co(NH3)4Cl2]+.(1)
into vanadium(iii) and vanadium(v) d) Write the systematic name for:
in alkaline solution.(1) i) [Ni(NH3)2Cl2](1)
iv) State whether this disproportionation ii) [Co(NH3)4Cl2]+.(1)
will occur and explain your answer.(3) e) Draw and describe the shape of the cis
and trans isomers of [Ni(NH3)2Cl2].(4)
f) Draw and describe the shape of the cis
and trans isomers of [Co(NH3)4Cl2]+. (4)
449
Exam practice questions
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Kinetics II
16
16.1 Factors that affect
reaction rates
Reaction kinetics is the study of the rates of chemical reactions. Several
factors influence the rate of chemical change including the concentration
of the reactants, the surface area of solids, the temperature of the reaction
mixture and the presence of a catalyst. Chemists have found that they can
learn much more about reactions by studying these effects quantitatively.
Tip
The first section of this chapter is a
very brief summary of ideas introduced
in Chapter 9. The rest of this chapter
shows that there is much that chemists
can learn about reactions from the
quantitative study of kinetics. In
particular, the chapter ends by showing
how chemists’ understanding of the
mechanisms of organic reactions
depends on key evidence from kinetics
experiments.
Figure 16.1 Understanding the factors which determine the rate of chemical change is
essential in the design of processes to manufacture drugs for the pharmaceutical industry.
Chemists use a collision model to explain the effects of the factors that alter
reaction rates. This model is based on kinetic theory and the Maxwell–
Key term Boltzmann distribution of energies in a collection of molecules. The idea
The activation energy of a reaction is that a chemical reaction happens when the molecules or ions of reactant
is the minimum energy needed in a collide, making some bonds break and allowing new bonds to form.
collision between molecules if they are However, it is not enough for the molecules to collide. In soft collisions
to react. The activation energy is the the molecules simply bounce off each other. Molecules are in rapid random
height of the energy barrier separating motion and if every collision led to reaction all reactions would be explosively
reactants and products during the fast. Only pairs of molecules that collide with enough energy to stretch and
progress of a reaction. break chemical bonds can lead to new products. Reactant molecules have to
overcome the activation energy.
Chemists apply the theory of chemical kinetics to drug design and to the
formulation of medicines to make sure that patients receive treatments
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that are effective for some time without causing harmful side-effects. The
theory can also help to account for the damage arising from pollutants in the
Key term
atmosphere and to explore ways for reducing or preventing the problems.
The rate of reaction measures the rate
of formation of a product or the rate of
Test yourself removal of a reactant.
change in measured property
rate =
1 Each of the following factors can change the rate of a reaction. Give time
an example of a reaction to illustrate each one:
a) the concentration of reactants in solution
b) the pressure of gaseous reactants
A C
c) the surface area of a solid
d) the temperature
e) the presence of a catalyst.
–3
[Product]/mol dm
2 How does collision theory account for the effects of altering each of
the factors (a) to (e) in Question 1?
rate at time t
16.2 Measuring reaction rates B
=
AB
AC
–3 –1
mol dm s
Balanced chemical equations say nothing about how quickly the reactions
occur. In order to get this information, chemists have to do experiments to
measure the rates of reactions under various conditions. O t
Time/s
The amounts of the reactants and products change during any chemical Figure 16.2 A concentration–time graph
reaction. Products form as reactants disappear. The rates at which these for the formation of a product. The rate of
changes happen give a measure of the rate of reaction. formation of product at time t is the gradient
Chemists define the rate of reaction as the change in concentration of a (or slope) of the curve at this point.
product, or a reactant, divided by the time for the change. Usually the rate is
not constant but varies as the reaction proceeds. Normally the rate decreases
with time as the concentrations of reactants fall. However, a reaction may get
faster and faster if it is exothermic and the temperature rises (Section 16.5),
or if the reaction gives a product that can act as a catalyst for the reaction
(Section 15.11).
[Reactant]/mol dm–3
The first step in analysing the results of an experiment is to plot a
concentration–time graph. The gradient (or slope) of the graph at any point P
rate at time t
gives a measure of the rate of reaction at that time (Figures 16.2 and 16.3).
PQ –3 –1
=– mol dm s
QR
Test yourself
3 In a study of the hydrolysis of an ester, the concentration of the ester
fell from 0.55 mol dm−3 to 0.42 mol dm−3 in 15 seconds. What was Q R
the average rate of reaction in that period? O
Time/s
4 The gaseous oxide N2O5 decomposes to NO2 gas and oxygen. Figure 16.3 A concentration–time graph
a) Write a balanced equation for the reaction. for the disappearance of a reactant. Here
b) If the rate of disappearance of N2O5 is 3.5 × 10 −4 mol dm−3 s−1, the gradient is negative and so the rate of
what is the rate of formation of NO2? reaction at time t is minus the gradient (or
slope) of the curve at this point.
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Tip Practical methods
Ideally, chemists look for methods for measuring reaction rates that do
The units for rate of reaction are
not interfere with the reaction mixture, as shown in Figures 16.4–16.7.
always mol dm−3 s−1. Do not confuse
Sometimes, however, it is necessary to withdraw samples of the reaction
this with the units for rate constants,
mixture at regular intervals and analyse the concentration of a reactant or
k, which vary according to the order
product by titration, as illustrated in Figure 16.8.
of the reaction. See Section 4 in
‘Mathematics in A Level chemistry’,
which you can access online at
[Link]/
EdexcelChemistry, for guidance
on how to work out and interpret the gas syringe
gradient of a tangent to a graph.
dilute hydrochloric acid
magnesium turnings
Figure 16.4 Following the course of a reaction with time by collecting and measuring the
volume of a gas formed.
light source filter reaction light sensitive meter
mixture cell
Figure 16.5 Using a colorimeter to follow the formation of a coloured product or the
removal of a coloured reactant.
conductivity
platinum meter
electrode
Figure 16.6 Using a conductivity cell and meter to measure the changes in electrical
conductivity of the reaction mixture as the number or nature of the ions changes.
452 16 Kinetics II
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cotton wool plug
reaction mixture
giving off a gas
reactant in solid reactant folded
aqueous in folded paper paper
solution
top pan
balance
Figure 16.7 Following the course of a reaction by measuring the change of mass as the
reaction gives off a dense gas that is lost from the system.
graduated
pipette
standard
solution of alkali
reaction mixture
containing an acid
ice cold water
sample after
stopping reaction
Figure 16.8 Following the course of a reaction during which there is a change in concentration
of acids present by removing measured samples of the mixture at intervals, stopping the
reaction by running the sample into an alkali, and then determining the concentration of one
reactant or product by titration. Further samples are taken at regular intervals.
Test yourself
5 Suggest a suitable method for measuring the rate of each of these
reactions:
a) Br2(aq) + HCOOH(aq) → 2HBr(aq) + CO2(g)
b) CH3COOCH3(l) + H2O(l) → CH3COOH(aq) + CH3OH(aq)
c) C4H9Br(l) + H2O(l) → C4H9OH(l) + H+(aq) + Br−(aq)
d) MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l)
16.2 Measuring reaction rates 453
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Activity
Investigating the effect of concentration on the rate of a reaction
Bromine oxidises methanoic acid in aqueous solution to carbon 4 Plot a graph of bromine concentration against time using the
dioxide. The reaction is catalysed by hydrogen ions. results in Table 16.1.
5 Draw tangents to the graph and measure the gradient
Br 2(aq) + HCOOH(aq) → 2Br−(aq) + 2H+(aq) + CO2(g)
to obtain values for the rate of reaction at two points
The reaction can be followed using a colorimeter. during the experiment. Take values at 100 s and 500 s.
(Remember when calculating the gradients that the bromine
Table 16.1 shows some typical results. The concentration of
concentrations are 1000 times less than the numbers in the
methanoic acid was kept constant throughout the experiment
table.)
by having it present in large excess.
6 Table 16.2 shows values for the reaction rate obtained by
Table 16.1 Results of an experiment to investigate the rate of drawing gradients at other times on the concentration–time
reaction of bromine with methanoic acid. Note that the bromine graph. Plot a graph of rate against concentration using the
concentrations are multiplied by 1000. The actual bromine
concentration at 90 seconds, for example, was 7.3 × 10 −3 mol dm−3
two values you have found in answering Question 5 and the
= 0.0073 mol dm−3. values in Table 16.2.
Time/s Concentration of Table 16.2 Rate values obtained by finding gradients of tangents to
bromine/10−3 mol dm−3 the concentration–time graph. Note that the rate of reaction values
0 10.0 are multiplied by 100 000. The actual rate at 300 seconds, for
example, was 1.2 × 10−5 mol dm−3 s−1.
10 9.0
30 8.1 Time/s Concentration Rate of reaction
of bromine/ from gradients to
90 7.3 10−3 mol dm−3 the concentration–
120 6.6 time graph/
180 5.3 10−5 mol dm−3 s−1
240 4.4 50 8.3 2.9
360 2.8 200 5.0 1.7
480 2.0 300 3.5 1.2
600 1.3 400 2.5 0.8
1 Explain why it is possible to follow the rate of this reaction 7 How does the bromine concentration change with time?
using a colorimeter. 8 How does the rate of reaction change with time?
2 Suggest a suitable chemical to use as the catalyst for the 9 How does the rate of reaction depend on the bromine
reaction. concentration?
3 Explain the purpose of adding a large excess of methanoic
acid.
Key term 16.3 Rate equations
Chemists have found that they can summarise the results of investigating the
In a rate equation such as
rate of reaction in the form of a rate equation. A rate equation shows how
rate = k[A]m[B]n, the constant k is the
changes in the concentrations of reactants affect the rate of a reaction.
rate constant.
Take the example of a general reaction for which x mol A react with y mol B
to form products:
xA + yB → products
The equation that describes how the rate varies with the concentrations of
the reactants takes this form:
rate = k[A]m[B]n
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where [A] and [B] represent the concentrations of the reactants in moles per
cubic decimetre. Key terms
The powers m and n are the reaction orders. The reaction above is order m The powers m and n are the reaction
with respect to A and order n with respect to B. The overall order of the orders with respect to the reactants
reaction is (m + n). A and B that appear in this equation.
The rate constant, k, is only constant for a particular temperature. In other The overall order of the reaction is
words, the value of k varies with temperature (see Section 16.5). The units sum of the orders for all the substances
of the rate constant depend on the overall order of the reaction (Table 16.3). that appear in the rate equation; here
(m + n).
Example Table 16.3 The units of the rate constant
The decomposition of ethanal to methane and carbon monoxide is for different reaction orders. These units
second order with respect to ethanal. When the concentration of can be worked out from the rate equations.
ethanal in the gas phase is 0.20 mol dm−3, the rate of reaction is Overall order Units of the rate
0.080 mol dm−3 s−1 at a certain temperature. What is the value of the rate constant
constant at this temperature? (Give the units with your answer.) Zero mol dm−3 s−1
First s−1
Notes on the method
Second dm3 mol−1 s−1
Start by writing out the rate equation based on the information given.
There is no need to write the equation for the reaction because the rate
equation cannot be deduced from the balanced chemical equation.
Substitute values in the rate equation, including the units as well as the
values. Then rearrange the equation to find the value of k. Check that the Tip
units are as expected for a second order reaction.
A rate equation cannot be deduced
Answer from the balanced equation: it has to
The rate equation: rate = k[ethanal]2 be found by experiment. In the general
Substituting: 0.080 mol dm−3 s−1 = k × (0.20 mol dm−3)2 example, rate = k[A]m[B]n, the values
−3 −1 of m and n in the rate equation may or
Rearranging: k = 0.080 mol dm −3s2 may not be the same as the values of x
(0.20 mol dm )
and y in the balanced equation for the
Hence: k = 2.0 dm3 mol−1 s−1
reaction.
Test yourself
6 The rate of decomposition of an organic peroxide is first order with
respect to the peroxide. Calculate the rate constant for the reaction at
107 °C if the rate of decomposition of the peroxide at this temperature
is 7.4 × 10 −6 mol dm−3 s−1 when the concentration of peroxide is
0.02 mol dm−3. Show that the unit of the rate constant is s−1.
7 The hydrolysis of the ester methyl ethanoate in alkali is first order
with respect to both the ester and hydroxide ions. The rate of reaction
is 0.00069 mol dm−3 s−1, at a given temperature, when the ester
concentration is 0.05 mol dm−3 and the hydroxide ion concentration
is 0.10 mol dm−3. Write the rate equation for the reaction and
calculate the rate constant. Show that the unit of the rate constant is
dm3 mol−1 s−1.
16.3 Rate equations 455
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First order reactions
A reaction is first order with respect to a reactant if the rate of reaction is
Rate of reaction
proportional to the concentration of that reactant. The concentration term
for this reactant is raised to the power one in the rate equation.
rate = k[X]1 = k[X]
This means that doubling the concentration of the chemical X leads to a
Concentration of reactant doubling of the rate of reaction.
Figure 16.9 The variation of reaction rate The fact that the rate of reaction is directly proportional to the concentration
with concentration for a first order reaction. of the reactant means that a plot of rate against concentration gives a straight
line passing through the origin (Figure 16.9).
One of the easier ways to spot a first order reaction is to plot a concentration–
Key term time graph and then study the time taken for the concentration to fall by half
(Figure 16.10). This is the half-life, t ½. It can be shown mathematically from
The half-life of a reaction is the time
the rate equation that for a first order reaction:
for the concentration of one of the
reactants to fall by half. ln 2 0.69
t½ = =
k k
where k is the rate constant.
This shows that, at a constant temperature, the half-life of a first order
reaction is the same wherever it is measured on a concentration–time graph.
Concentration of a reactant
It is independent of the initial concentration.
Test yourself
8 Refer to your answers to the activity in Section 16.2.
a) From your rate–concentration graph, what is the order of the
reaction of the reaction of bromine with methanoic acid with
Time respect to bromine?
equal half-lives
b) i) Determine three values for half-lives for the reaction from your
Figure 16.10 The variation of concentration–time graph.
concentration of a reactant plotted against
ii) Are your values consistent with your answer to part (a)?
time for a first order reaction. The half-life
for a first order reaction is a constant, 9 Explain how the rate constant can be found from a rate–concentration
so it is the same wherever it is read off graph such as in Figure 16.9.
the curve. It is independent of the initial
concentration.
Second order reactions
Tip A reaction is second order with respect to a reactant if the rate of reaction is
proportional to the concentration of that reactant squared. This means that
Logarithms, including natural logarithms
the concentration term for this reactant is raised to the power two in the rate
(ln), are explained in Section 3 of
equation. At its simplest, the rate equation for a second order reaction takes
‘Mathematics in A Level chemistry’,
this form:
which you can access online at
[Link]/ rate = k[reactant]2
EdexcelChemistry. There is also more
This means that doubling the concentration of X increases the rate by a
information about half-lives in Section 4.
factor of four.
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A concentration–time graph of a second order reaction has unequal half-lives
(Figure 16.11). The time for the concentration to fall from its initial value c
c c
to c is half the time for the concentration to fall from to . The half-life is
2 2 4
inversely proportional to the starting concentration.
The variation of rate with concentration for a second order reaction can be
found, as before, by drawing tangents to the curve of the concentration–time
graph. However, a rate–concentration graph is not a straight line for a second
order reaction; instead it is a curve, as shown in Figure 16.12.
Concentration of a reactant
Rate of reaction
Time
unequal half-lives Concentration of reactant
Figure 16.11 The variation of Figure 16.12 The variation of reaction
concentration of a reactant plotted against rate with concentration for a second order
time for a second order reaction. reaction.
Test yourself
10 The rate of reaction of 1-bromopropane with hydroxide ions is first
order with respect to the halogenoalkane and first order with respect
to hydroxide ions.
a) Write the rate equation for the reaction.
b) What is the overall order of reaction?
c) What are the units of the rate constant?
Zero order reactions
At first sight is seems odd that there can be zero order reactions. However,
there are examples of reactions for which the gradient of a concentration–
time graph does not change with time. The constant gradient in
Figure 16.13 shows that the rate of the reaction stays the same, even though
the concentration of the reactant is falling.
A reaction of this kind is zero order with respect to a reactant because the Tip
rate of reaction is unaffected by changes in the concentration of that reactant
Any term raised to the power zero equals
(Figure 16.14). Chemists have found a way to account for zero order reactions
1 (see Section 1 of 'Mathematics in A
in terms of the mechanisms of these reactions (see Section 16.4).
Level chemistry', which you can access
In a rate equation for a zero order reaction, the concentration term for the online at [Link].
reactant is raised to the power zero and so the rate is equal to the rate constant. [Link]/EdexcelChemistry. So
[reactant]0 = 1.
rate = k[reactant]0 = k
16.3 Rate equations 457
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Concentration of a reactant
Rate of reaction
Time Concentration of reactant
Figure 16.13 The variation of Figure 16.14 The variation of reaction
concentration of a reactant plotted against rate with concentration for a zero order
time for a zero order reaction. reaction.
Test yourself
11 Ammonia gas decomposes to nitrogen and hydrogen in the presence
of a hot platinum wire. Experiments show that the reaction continues
at a constant rate until all the ammonia has disappeared.
a) Sketch a concentration–time graph for the reaction.
b) Write both the balanced chemical equation and the rate equation
for this reaction.
The nitration of methylbenzene is an example where the conditions can be
such that the reaction is zero order with respect to the aromatic compound.
The methylbenzene reacts with nitronium ions (NO2+) formed from nitric
acid (see Section 18.1.6). The reaction creating the NO2+ ions is relatively
slow, but as soon as the ions form they react with methylbenzene. As a result
the rate is not affected by the methylbenzene concentration.
The initial-rate method
The most general method for determining reaction orders is the initial-
rate method. The method is based on finding the rate immediately after
the start of a reaction. This is the one point when all the concentrations
are known.
The investigator makes up a series of mixtures in which all the initial
concentrations are the same except one. A suitable method is used to
measure the change of concentration with time for each mixture (see
Section 16.2). The results are used to plot concentration–time graphs. The
initial rate for each mixture is then found by the drawing tangent to the
curve at the start and calculating the gradient.
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Example
The initial-rate method was used to study the reaction:
BrO3−(aq) + 5Br−(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l)
The initial rate was calculated from four graphs plotted to show
how the concentration of BrO3−(aq) varied with time for different
initial concentrations of reactants. The results are shown in
Table 16.4.
What is:
a) the rate equation for the reaction
b) the value of the rate constant?
Table 16.4
Experiment Initial Initial Initial Initial rate
concentration concentration concentration of reaction/
of BrO3−/ of Br−/ of H+/ mol dm−3 s−1
mol dm−3 mol dm−3 mol dm−3
1 0.10 0.10 0.10 1.2 × 10−3
2 0.20 0.10 0.10 2.4 × 10−3
3 0.10 0.30 0.10 3.6 × 10−3
4 0.20 0.10 0.20 9.6 × 10−3
Notes on the method
Recall that the rate equation cannot be worked out from the balanced
equation for the reaction.
First, study the experiments in which the concentration of BrO3− varies
but the concentration of the other two reactants stays the same. How
does doubling the concentration of BrO3− affect the rate?
Then study, in turn, the experiments in which the concentrations of first
Br− and then H+ vary, while the concentrations of the other two reactants
stay the same. How does doubling or tripling the concentration of a
reactant affect the rate?
Substitute values for any one experiment in the rate equation to find the
value of the rate constant, k. Take care with the units.
Answer
From experiments 1 and 2:
doubling [BrO3−]initial increases the rate by a factor of 2. So rate ∝ [BrO3−]1.
From experiments 1 and 3:
tripling [Br−]initial triples the rate. So rate ∝ [Br−]1
From experiments 2 and 4:
doubling [H+]initial increases the rate by a factor of 4 (22). So rate ∝ [H+]2.
16.3 Rate equations 459
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The reaction is first order with respect to BrO3− and Br− but second order
with respect to H+.
The rate equation is: rate = k [BrO3−][Br−][H+]2
Rearranging this equation, and substituting values from experiment 4:
rate
k =
[BrO 3 ][Br – ][H+ ]2
–
9.6 × 10 –3 mol dm–3 s –1
=
0.2 mol dm–3 × 0.1 mol dm–3 × (0.2 mol dm–3 )2
k = 12.0 dm9 mol−3 s−1
Test yourself
12 This data refers to the reaction of the halogenoalkane
1-bromobutane (here represented as RBr) with hydroxide ions.
The results are shown in Table 16.5.
a) Deduce the rate equation for the reaction.
b) Calculate the value of the rate constant.
Table 16.5
Experiment [RBr]/mol dm−3 [OH−]/mol dm−3 Rate of reaction/
mol dm−3 s−1
1 0.020 0.020 1.36
2 0.010 0.020 0.68
3 0.010 0.005 0.17
13 This data refers to the reaction of halogenoalkane
2-bromo-2-methylbutane (here represented as R′Br) with
hydroxide ions. The results are shown in Table 16.6.
a) Deduce the rate equation for the reaction.
b) Calculate the value of the rate constant.
Table 16.6
Experiment [R'Br]/mol dm−3 [OH−]/mol dm−3 Rate of reaction/
mol dm−3 s−1
1 0.020 0.020 40.40
2 0.010 0.020 20.19
3 0.010 0.005 20.20
14 Hydrogen gas reacts with nitrogen monoxide gas to form steam and
nitrogen. Doubling the concentration of hydrogen doubles the rate of
reaction. Tripling the concentration of NO gas increases the rate by a
factor of nine.
a) Write the balanced equation for the reaction.
b) Write the rate equation for the reaction.
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Clock reactions
A variant on the initial-rate method is to use a ‘clock reaction’, so called
because the reaction is set up to produce a sudden colour change after a
certain time when it has produced a fixed amount of one reactant.
The reaction of hydrogen peroxide with iodide ions in acid solution can be
set up as a clock reaction:
H2O2(aq) + 2H+(aq) + 2I−(aq) → I2(aq) + 2H2O(l)
A small, known amount of sodium thiosulfate ions is added to the reaction
mixture, which also contains starch indicator. At first the thiosulfate ions
react with any iodine, I2, as soon as it is formed, turning it back to iodide
ions, so there is no colour change. At the instant when all the thiosulfate ions
have been used up, free iodine is produced and this immediately gives a deep
blue-black colour with the starch.
A
Amount of product
0
0 tA tB Time
Figure 16.15 Two plots showing the formation of a product with time under different
conditions. In a clock reaction the reaction mixture includes an indicator that gives a
sudden colour change when the amount x of product has formed.
The experiment can be repeated with different conditions but each time
with the same amount of sodium thiosulfate added. This means that the
sudden colour change always happens when the same amount of iodine has
been formed (represented by amount x in Figure 16.15).
Line A shows the formation of a product under one set of conditions. An
amount of product x forms in time tA. Line B shows the formation of the
same product under a different set of conditions. The same amount of product
x forms in the longer time t B.
x
The average rate of formation of product on line A =
tA
x
The average rate of formation of product on line B = t Test yourself
B
If x is kept the same, it follows that the average rate near the start ∝ 1t . 15 Explain why the estimate of
1 the initial rate of a reaction
This means that it is possible to use t as a measure of the initial rate of determined by a clock
a reaction by determining how long the reaction takes to produce the
reaction is close to, but not
small, fixed amount of product needed for the colour change in the clock
equal to, the true initial rate.
reaction.
16.3 Rate equations 461
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Core practical 13b
Investigating the reaction of hydrogen peroxide
with iodide ions by a clock reaction
The equation for the reaction of hydrogen peroxide with iodide Questions
ions in acid solution is: 1 Explain why the instructions require that the total volume of
potassium iodide solution and water is 25 cm3 for each run
H2O2(aq) + 2H+(aq) + 2I−(aq) → I2(aq) + 2H2O(l)
of the experiment.
The rate equation for the reaction takes the form: 2 What is the relationship between the concentration of iodide
ions in each run and the volume of K I(aq)?
Rate = k[H2O2]m[I−]n[H+]p
3 Give the equation for the reaction of sodium thiosulfate with
1
A student followed these instructions to determine the order of iodine and explain why t is a measure of the initial rate of
the reaction with respect to iodide ions. reaction where t is the time for the blue colour to appear.
4 Show that under the conditions of this experiment the rate
A Use a pipette to add 10 cm3 of 0.10 mol dm−3 hydrogen
equation takes the form: rate = constant × [I−]n
peroxide to a clean beaker. Then use a measuring cylinder to
5 Explain why a graph of log (rate) against log [I−] gives a
add 25 cm3 of 0.25 mol dm−3 sulfuric acid to the same beaker.
straight line with gradient n (see Section 4 in ‘Mathematics
B Use burettes to add to a second clean beaker: 5.0 cm3
in A Level chemistry’, which you can access online at
of 0.10 mol dm−3 potassium iodide solution; 20.0 cm3 of
[Link]/EdexcelChemistry).
distilled water, 2.0 cm3 of 0.050 mol dm−3 sodium thiosulfate
6 Draw up a table with the headings shown and enter the
solution and 1 cm3 of starch solution.
values for each run.
C Pour the contents of the first beaker into the second beaker,
start timing and swirl the contents to mix thoroughly. Record Volume of
0.01 mol dm−3
log (volume of
aqueous K I)
1 −1
t
/s log ( 1t )
the time (to an appropriate accuracy) taken for the blue
KI(aq)/cm3
colour of the starch–iodine complex to appear.
D Repeat steps A and C using the same quantities in step A
1
but with different volumes of the potassium iodide solution 7 Plot a graph of log ( t ) against log (volume of aqueous K I)
and water in step B (making sure that the total volume of and hence determine the order of the reaction with respect
potassium iodide solution and water is 25 cm3 each time). to iodide ions.
The student’s results are shown in Table 16.7. 8 What changes to the procedure would be needed to determine
the order of the reaction with respect to hydrogen peroxide?
Table 16.7
Run Volume of Volume Time, t, taken for Tip
0.01 mol dm−3 K I(aq)/ of water/ the blue colour to
cm3 cm3 appear/s For practical guidance, refer to Practical skills sheet
1 5.0 20.0 258 18, ‘Investigating reaction orders and activation
2 10.0 15.0 136 energies’, which you can access online at www.
3 15.0 10.0 98 [Link]/EdexcelChemistry.
4 20.0 5.0 72
5 25.0 0.0 58
16.4 Rate equations and reaction
mechanisms
Rate equations were some of the first pieces of evidence to set chemists
thinking about the mechanism of reactions. They wanted to understand
why a rate equation cannot be predicted from the balanced equation for
the reaction. They were puzzled that similar reactions turned out to have
different rate equations.
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Multi-step reactions Diffusion to
the catalyst (fast)
Diffusion away from
the catalyst (fast)
The key to understanding the mechanism of a reaction was the realisation
that most reactions do not take place in one step, as suggested by the balanced
equation, but in a series of steps.
It is unexpected that the decomposition of ammonia gas in the presence
of a hot platinum catalyst is a zero order reaction. How can it be that
the concentration of the only reactant does not affect the rate? A possible
explanation is illustrated in Figure 16.16.
Ammonia rapidly diffuses to the surface of the metal and is adsorbed onto
the surface. This happens fast. Bonds break and atoms rearrange to make new
molecules on the surface of the metal. This is the slowest process. Once formed,
the nitrogen and hydrogen rapidly break away from the metal into the gas phase.
So there is a rate-determining step which can only happen on the surface
of the platinum. The rate of reaction is determined by the surface area of Bonds breaking and new bonds forming.
Rate determined by the surface area
the platinum, which is a constant. This means that the rate of reaction is of the catalyst, which is a constant
a constant as long as there is enough ammonia to be adsorbed all over the (rate-determining)
metal surface. The rate is independent of the ammonia concentration. Figure 16.16 Three steps in the
decomposition of ammonia gas in the
Test yourself presence of a platinum catalyst.
16
Give an analogy from the everyday world to explain the idea of a
rate-determining step. You could base your example on people
getting their meals in a busy self-service canteen, or heavy traffic on Key terms
a motorway affected by lane closures.
The mechanism of a reaction describes
Hydrolysis of halogenoalkanes how the reaction takes, place showing
step by step the bonds that break and
Another puzzle for chemists was the discovery that there are different rate the new bonds that form.
equations for the reactions between hydroxide ions and two isomers with the
formula C4H9Br (see ‘Test yourself ’ Questions 12 and 13 in Section 16.3). Adsorption is a process in which
atoms, molecules or ions are held on
Hydrolysis of a primary halogenoalkane, such as 1-bromobutane, is overall the surface of a solid.
second order. The rate equation has the form: rate = k[C4H9Br][OH−].
The rate-determining step in a multi-
To account for this chemists have suggested a mechanism showing the C– Br step reaction is the slowest step: the
bond breaking at the same time as the nucleophile, OH−, forms a C–OH one with the highest activation energy.
bond. In this mechanism both reactants are involved in the single, rate-
determining step (Figure 16.17).
H H H – H Figure 16.17 A one-step mechanism for
H H the hydrolysis of 1-bromobutane.
–
HO C Br HO C Br HO C + Br–
C3H7 C3H7 C3H7
1-bromobutane transition state butan-1-ol
In this example of a substitution reaction the nucleophile is the hydroxide Key term
ion. Chemists label this mechanism SN2, where the ‘2’ shows that there are
two molecules or ions involved in the rate-determining step. An SN2 reaction is a nucleophilic
substitution reaction with a mechanism
Hydrolysis of tertiary halogenoalkanes such as 2-bromo-2-methylpropane,
that involves two molecules or ions in
however, is overall first order. The rate equation has the form:
the rate-determining step.
rate = k[C4H9Br]
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The suggested mechanism shows the C–Br bond breaking first in a slow step
to form an ionic intermediate. This is the rate-determining step that forms a
tertiary carbocation. Then the nucleophile, OH−, rapidly forms a new bond
with the positively charge carbon atom (Figure 16.18).
CH3 CH3 CH3
CH3
slow step
C Br C+ + Br–
rate-determining
CH3 CH3
carbocation intermediate (planar)
CH3 CH3 CH3
CH3
– fast
C+ OH C OH
step
CH3 CH3
Figure 16.18 A two-step mechanism for the hydrolysis of 2-bromo-2-methylpropane.
In this example of a substitution reaction the nucleophile is also the hydroxide
Key term ion. Chemists label the mechanism SN1, where the ‘1’ shows that there is just
one molecule or ion involved in the rate-determining step. The concentration
An SN1 reaction is a nucleophilic
of the hydroxide ions does not affect the rate of reaction because hydroxide
substitution reaction with a mechanism
ions are not involved in the rate-determining step.
that involves only one molecule or ion in
the rate-determining step. What these examples show is that it is generally the molecules or ions
involved (directly or indirectly) in the rate-determining step that appear in
the rate equation for the reaction.
Tip Test yourself
More evidence to suggest that the SN1 17 Explain, in terms of bonding, why the first step in the SN1
and SN2 mechanisms provide a correct mechanism is slow while the second step is fast.
description of the reactions comes from
18
In the proposed two-step mechanism for the reaction of nitrogen
the study of the shapes of molecules
dioxide gas with carbon monoxide gas, the first step is slow while
(see Section 17.1.3).
the second step is fast:
2NO2(g) → NO3(g) + NO(g) slow
NO3(g) + CO(g) → NO2(g) + CO2(g) fast
a) What is the overall equation for the reaction?
b) Suggest a rate equation that is consistent with this mechanism.
c) What, according to your suggested rate equation, is the order of
reaction with respect to carbon monoxide?
The reaction of iodine with propanone
The reaction of iodine with propanone is another example which shows that
it is not possible to deduce the rate equation from the balanced equation for
the reaction.
The results of experiments, such as those outlined in Core practical 13a,
suggest that the rate-determining step, for the reaction between iodine and
propanone, involves propanone and hydrogen ions but not iodine.
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Tip
Refer to Practical skills sheet 18, ‘Investigating reaction orders and activation
energies’, which you can access online at [Link]/
EdexcelChemistry.
Contrast the reaction of propanone with iodine in acid conditions in Core practical 13a
with the reaction forming triiodomethane under alkaline conditions (Section 17.2.6).
Core practical 13a
Investigating the rate equation for the reaction
of iodine with propanone
Iodine reacts with propanone in the presence of an acid catalyst. The student carried out a second experiment, with a series of
runs using the initial rate method, to check the result of the
I2(aq) + CH3COCH3(aq) → CH2ICOCH3(aq) + H+(aq) + I−(aq)
first experiment and to find the order with respect to propanone
A student first investigated the order of reaction with respect to and hydrogen ions. She made up a series of mixtures of acid,
iodine by using a titration method. propanone and water as shown in Table 16.8. Then she added
a measured volume of iodine solution to each mixture and
She mixed 50 cm3 of 0.020 mol dm−3 aqueous iodine with
recorded the time taken for the iodine colour to disappear. Her
50 cm3 of an acidified 0.25 mol dm−3 solution of propanone in
results are shown in Table 16.8.
a flask. Every five minutes after the start, she used a pipette
to remove 10.0 cm3 of the reaction mixture which she ran into Table 16.8
an excess of sodium hydrogencarbonate solution. She then Run 1 Run 2 Run 3 Run 4
titrated the iodine remaining against a standard solution of Volume of 2.0 mol dm−3 20.0 10.0 20.0 20.0
sodium thiosulfate solution. She plotted a graph of her results, HCl(aq)/cm3
as shown in Figure 16.19. Volume of 2.0 mol dm−3 8.0 8.0 4.0 8.0
CH3COCH3(aq)/cm3
20 Volume of water/cm3 0 10.0 4.0 2.0
Volume of thiosulfate solution, V/cm3
Volume of 0.001 mol 4.0 4.0 4.0 2.0
19 dm−3 I2(aq)/cm3
Time, t, for iodine 115 264 243 58
18 colour to disappear/s
Rate of reaction/cm3 0.035 0.015 0.016 0.034
iodine solution per
17
second
16 4 Explain why the shape of the line in Figure 16.19 justifies
the method used in the second experiment.
15
5 Why was it important to measure the volumes of solutions
0 5 10 15 20 25 30 with pipettes or burettes?
Time, t/min 6 Show how the student arrived at the values of the rate of
Figure 16.19 reaction in runs 3 and 4.
7 Show that her results confirm the value for the order of
1 Why did the student add the 10 cm3 samples of the reaction reaction with respect to iodine found in the first experiment.
mixture to excess sodium hydrogencarbonate immediately 8 What are the orders of reaction with respect to propanone
after removing them from the flask? and hydrogen ions?
2 How did the rate of change of iodine concentration change 9 Give the rate equation for the reaction.
during the experiment? 10 What does the rate equation suggest about the
3 What is the order of the reaction with respect to iodine? mechanism of the reaction of propanone with iodine?
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The suggested mechanism for the acid-catalysed reaction is that propanone
molecules react relatively slowly to form an intermediate molecule with a
double bond and an –OH group. This is followed by fast steps in which
the intermediate reacts with iodine, as shown in Figure 16.20. Only the
concentrations of chemical species involved the rate-determining step appear
in the rate equation.
O OH O
slow reaction fast reaction
CH3 C CH3 + CH3 C CH2 CH3 C CH2 I + HI
with H ions + I2
Figure 16.20 Outline of the suggested mechanism to account for the rate equation for
the reaction of propanone with iodine.
This reaction shows that the form of the rate equation, the order of reaction
and the value of the rate constant are all likely to be different when a catalyst
is added to speed up a reaction. The reaction of iodine with propanone is an
example of the way that a catalyst can change the mechanism of a reaction by
combining with one of the reactants to form an intermediate. The intermediate
then reacts to give the products and the catalyst is released so that it is freed up
to interact with further reactant molecules and continue the reaction.
Test yourself
19 Why do chemists use the term ‘enol’ to describe the intermediate
formed during the reaction of iodine with propanone?
20 Why does the formula for hydrogen ions appear in the rate equation
for the iodination of propanone but not as a reactant in the balanced
equation for the reaction?
21 Bromine reacts with propanone in a similar way to iodine. The
mechanism for the reaction is the same. Explain why bromine reacts
with propanone at the same rate as iodine under similar conditions.
16.5 The effect of temperature
on reaction rates
Raising the temperature often has a dramatic effect on the rate of a reaction,
especially reactions that involve the breaking of strong covalent bonds. This
explains why the practical procedure for most organic reactions involves
heating the reaction mixtures. With the help of collision theory it is possible
to make predictions about the effect of temperature changes on rates.
Key term The constant k in a rate equation is only a constant at a specified temperature.
Generally, the value of the rate constant increases as the temperature rises
A transition state is the state of the and this means that the rate of reaction increases.
reacting atoms, molecules or ions when
Collision theory accounts for the effect of temperature on reaction rates
they are at the top of the activation
by supposing that chemical changes pass through a transition state. The
energy barrier for a reaction step.
transition state is at a higher energy than the reactants so there is an energy
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barrier or activation energy as shown by the reaction profile in Figure 16.21.
Reactant molecules must collide with enough energy to overcome the
Key term
activation energy barrier. This means that the only collisions which lead to
A reaction profile is a plot that shows
reaction are those with enough energy to break existing bonds and allow the
how the total energy of the atoms,
atoms to rearrange to form new bonds in the product molecules.
molecules or ions changes during the
progress of a change from reactants to
products.
transition state
Energy
activation energy
reactants
products
Progress of reaction
Figure 16.21 Reaction profile showing the activation energy for a reaction.
Activation energies account for the fact that reactions go much more slowly
than would be expected if every collision in a mixture of chemicals led to
reaction. Only a very small proportion of collisions bring about chemical
change because molecules can only react if they collide with enough
energy to overcome the energy barrier. For many reactions, at around room
temperature only about 1 in 1010 molecules have enough energy to react.
The Maxwell–Boltzmann curve shows the distribution of the kinetic energies
of molecules. As Figure 16.22 shows, the proportion of molecules which can
collide with energies greater than the activation energy is small at around 300 K.
300 K number of molecules able to Figure 16.22 The Maxwell–Boltzmann
collide with energies greater distribution of molecular kinetic energies
Number of molecules with
310 K than the activation energy
at 300 K in a gas at two temperatures. The modal
kinetic energy E
energy gets higher as the temperature
number of molecules able to
collide with energies greater
rises. The area under the curve gives the
than the activation energy total number of molecules. This does not
at 310 K change as the temperature rises so the
activation energy peak height falls as the curve widens.
Kinetic energy E
The shaded areas in Figure 16.22 are a measure of the proportions of molecules
able to collide with enough energy for a reaction at two temperatures. The Test yourself
area is bigger at a higher temperature. So at a higher temperature there are 22
Why is it that many reactions
more molecules with enough energy to react when they collide, and the have activation energies
reaction goes faster. that range between about
50 kJ mol−1 and 250 kJ mol−1?
The effect of temperature changes on rate
constants
The Swedish physical chemist Svante Arrhenius (1859–1927) found that he
obtained a straight line if he plotted the natural logarithm of the rate constant
for a reaction against 1/T (the inverse of the absolute temperature). His equation
to describe the relationship between rate constant and temperature is:
k = A e−Ea/RT
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Key terms where k is the rate constant, R is the gas constant, T the absolute
temperature, E a is the activation energy for the reaction and A is another
The gas constant is the constant R in constant.
the ideal gas equation pV = nRT. The After taking natural logarithms of both sides, the Arrhenius equation takes
value of the constant depends on the this form:
units used for pressure and volume.
If all quantities are in SI units, then
–E 1
ln k = R a × T + constant
R = 8.314 J K−1 mol−1.
The absolute temperature is the A useful rough guide, based on the Arrhenius equation, is that at about room
temperature on the Kelvin scale. The temperature the value of the rate constant doubles for each 10 degree rise
absolute zero of temperature is at 0 K, in temperature if the activation energy for the reaction is about 50 kJ mol−1.
which is approximately −273 °C.
4
Tip 2
Exponential functions and natural 0
–Ea
logarithms are explained in Section 3 gradient =
–2 R
of ‘Mathematics in A Level chemistry’,
ln k
which you can access online at –4
[Link]/
–6
EdexcelChemistry. See also Section
4 for guidance on how to work out the –8
gradient of a straight-line graph based
on the general formula y = mx + c. –10
7 8 9 10 11 12
1 × 104/K–1
T
Figure 16.23 A plot of ln k against 1/T for a reaction. The activation energy can be
calculated from the gradient. The general equation for a straight line is y = mx + c,
where m is the gradient and c is the intercept on the y-axis. Here c is the constant in the
Arrhenius equation and the gradient m is −E a/R.
Test yourself
23
Table 16.9 shows the value of the rate constant for c) What is the effect of a 10 degree rise in
the reaction of a diazonium salt with water at temperature on the rate of the reaction?
four temperatures.
24 Show that the Arrhenius equation signifies that:
Table 16.9
a) the higher the temperature, the greater the
Temperature/K Rate constant/10 −5 s−1 value of k and hence the faster the reaction
278 0.15 b) a reaction with a relatively high activation
298 4.1 energy has a relatively small rate constant.
308 20 25 The rate constant for the decomposition of
323 140 hydrogen peroxide is 4.93 × 10 −4 s−1 at 295 K.
a) What do the units of the rate constant tell It increases to 1.40 × 10 −3 s−1 at 305 K.
you about the form of the rate equation? Estimate the activation energy for the reaction.
b) What do the values tell you about the effect
of temperature on the rate of the reaction?
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Core practical 14
Finding the activation energy of a reaction
A student used the clock method for determining initial In a series of experiments he kept the concentrations of the
rates to find the activation for the oxidation of iodide ions by reactants constant while varying temperature of the reaction
peroxodisulfate(vi) ions: mixture over a range of values.
S2O82−(aq) + 2I−2(aq) → 2SO42−(aq) + I2(aq) The reaction mixture included a small, measured amount of
sodium thiosulfate and some starch solution, as shown in
Figure 16.24.
thermometer thermometer
hot water bath
3 held at a constant
5 cm 0.5 mol dm−3 KI(aq) 10 cm30.02 mol dm−3 contents of
temperature
+ 5 cm3 0.01 mol dm−3 Na2S2O3(aq) K2S2O8(aq) tubes mixed
+ 2.5 cm3 starch solution
Figure 16.24 Outline of the experimental procedure.
Table 16.10 shows the student’s results from a series of runs 1 Why did the tube containing potassium iodide solution also
with temperatures in the range 30–51 °C. include sodium thiosulfate and starch?
Table 16.10 2 Why was the solution of potassium peroxodisulfate(vi)
measured into a separate tube at the start, and when should
Temperature/°C 30 36 39 45 51 the solutions be mixed and timing started?
Temperature/K 303 309 3 The table includes some calculated values for ln (1/t) and
Time, t, for the 204 138 115 75 55 1000/T K−1 corresponding to the variables in the Arrhenius
blue colour equation. Copy and complete the table by calculating the
to appear/s
values to include in the empty cells.
ln (1/t) −5.32 −4.93 4 What is the advantage of calculating 1000/T rather than
1000 −1 3.30 3.24 1/T?
/K
T 5 Plot a graph of ln (1/t) against 1000/T.
6 Use the graph to calculate the activation energy for the
reaction.
Tip
Refer to Practical skills sheet 18, ‘Investigating reaction orders and activation
energies’, which you can access online at [Link]/
EdexcelChemistry.
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Chapter summary
Chapter 16 Kinetics II based on finding the rate immediately after the
start of the reaction, when all the concentrations
l The rates at which products form, or reactants are known.
disappear, give a measure of the rates of reactions. l A clock reaction is a variant of the initial rate
l Experimental methods for obtaining rate data
method. The reaction mixture is set up to produce
include titration, colorimetry, changes in mass and a sudden colour change shortly after the reaction
changes in the volume of a gaseous product. has started.
l The rate of reaction at any time can be found by
l The mechanism of a reaction describes how the
drawing tangents to the concentration time graph reaction takes place, showing step by step the
and calculating the gradients. bonds that break and the new bonds that form. In
l The results of investigating the rates of reaction can
a multi-step reaction, the slowest step is called the
be summarised in the form of a rate equation. For rate determining step.
a reaction xA + yB → products, l The rate equation for a reaction indicates the
the rate equation takes the form: species that are likely to be involved in the rate
rate = k[A]m[B]n determining step.
In this equation k is the rate constant. The powers l The rate equation for the hydrolysis of a primary
m and n are the reaction orders with respect to halogenoalkane provides evidence for the SN2
reactants A and B. The overall order of reaction is mechanism.
(m + n). l The rate equation for the hydrolysis of a tertiary
l The rate constant, k, is only constant at a particular halogenoalkane provides evidence for the SN1
temperature. mechanism.
l The units of the rate constant depend on the l The initial rate method can be used to find the
overall order of the reaction. rate equation for the acid-catalysed iodination
l The reaction orders cannot be deduced from the of propanone. The results make it possible
balanced chemical equation but have to be found to determine the species involved in the rate
by experiment. determining step, and to propose a possible
l For a zero order reaction, the concentration–time mechanism for the reaction.
graph is a straight line with constant gradient, and l The presence of a catalyst can change the
the rate–concentration graph is a horizontal line. mechanism of a reaction. This is true of both
l For a first order reaction, the concentration–time homogeneous and heterogeneous catalysts.
graph is a curve with a half-life that is the same l Generally, the value of the rate constant increases
whatever the starting concentration, and the rate– when the temperature rises.
concentration graph is a straight line with gradient l The Arrhenius equation describes the relationship
equal to the rate constant. between the rate constant and temperature. The
l For a second order reaction, the concentration– logarithmic form of the equation is:
time graph is a curve with a half-life that is ln k = −E a/RT + constant
inversely proportional to the starting concentration, where E a is the activation energy for the reaction.
and the rate–concentration graph is a curved line l The activation energy can be determined by
with a rising gradient. measuring the rate constant over a range of
l The initial rate method is the most general temperatures. Plotting ln k against 1/T gives a
method for determining reaction orders. It is straight line with gradient −E a/R.
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Exam practice questions
1 Hydrogen peroxide oxidises iodide ions to Time/103 s [I 2]/10 −3 mol dm−3
iodine in the presence of hydrogen ions. The 0 20.0
other product is water. The reaction is first 1 15.6
order with respect to hydrogen peroxide, first 2 12.8
order with respect to iodide ions, but zero
3 11.0
order with respect to hydrogen ions.
4 9.4
a) Write a balanced equation for the
5 8.3
reaction.(1)
b) Write a rate equation for the reaction.(2) 6 7.5
c) State the overall order of the reaction.(1) 7 6.8
d) A proposed mechanism for the reaction 8 6.2
involves three steps:
H2O2 + I− → H2O + IO− a) Plot a concentration–time graph and show
H+ + IO− → HIO that the half-life is not constant.(5)
HIO + H+ + I− → I2 + H2O b) From your graph, determine the rate of
Identify the step that is likely to be the rate- reaction at four concentrations. (3)
determining step, giving your reasons.(2) c) Use your results from part (b) to plot a
2 The data in the table refers to the graph of log (rate) against log (concentration)
decomposition of hydrogen peroxide, H2O2. to determine the order of the reaction with
respect to iodine.(6)
Time/103 s [H2O2]/10 −3 mol dm−3 Two gases X and Y react according to this
4
0 20.0 equation:
12 16.0 X(g) + 2Y(g) → XY2(g)
24 13.1
36 10.6
This reaction was studied at 400 K, giving the
results shown in the table below.
48 8.6
60 6.9 Experiment Initial Initial Initial rate
72 5.6 number concentration concentration of formation
96 3.7 of X/mol dm−3 of Y/mol dm−3 of XY2/
mol dm−3 s−1
120 2.4
1 0.10 0.10 0.0001
2 0.10 0.20 0.0004
a) Plot a concentration–time graph for the
3 0.10 0.30 0.0009
decomposition reaction. (3)
4 0.20 0.10 0.0001
b) Determine three half-lives from the graph
5 0.30 0.10 0.0001
and show that this is a first-order reaction.(3)
c) Draw tangents to the curve in your graph at
a) Determine the order of the reaction with
four different concentrations and determine
respect to:
the gradient of the curve at each point.(4)
i) X
d) Plot a graph of rate against concentration
ii) Y.
using your results from (c) and hence
Explain your answers.(4)
determine the value for the rate constant at
b) Write a rate equation for the reaction
the temperature of the experiment.(4)
of X with Y.(2)
3 The results in the table come from a study c) Use the results of the first experiment to
of the rate of reaction of iodine with a large calculate a value of the rate constant and
excess of hex-1-ene dissolved in ethanoic acid. give its units.(2)
471
Exam practice questions
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d) Give a possible mechanism for the reaction. a) Explain why each reaction mixture
(3) included a small amount sodium
e) Explain why chemists are interested in thiosulfate solution and starch solution.(3)
determining rate equations and measuring b) Use the results to plot a graph to find
rate constants.(3) a value for the activation energy in the
presence of iron(iii) ions. (The gas
5 The table below shows data obtained for the
constant R = 8.31 J K−1 mol−1.)(8)
reaction between nitrogen monoxide and
c) In the absence of iron(iii) ions the
hydrogen at 750 °C.
activation energy for the reaction is
2NO(g) + 2H2(g) → N2(g) + 2H2O(g) 52.9 kJ mol−1. Give a possible explanation
for the effect of adding iron(iii) ions.(4)
Experiment 1 2 3
Initial concentration of 0.012 0.012 0.024 Chlorate(i) ions disproportionate on
7
hydrogen/mol dm−3 heating an aqueous solution of sodium
Initial concentration of NO/ 0.002 0.004 0.002 chlorate(i). The rate equation is:
mol dm−3 Rate = k[ClO−]2
Initial rate/mol dm−3 s−1 1.20 2.40 4.80 a) Write an equation for the disproportionation
of chlorate(i) ions and show changes in the
a) Explain the advantage of investigating oxidation states of chlorine.(3)
reaction kinetics by measuring initial rates.(2) b) Predict the effect on the rate of reaction of
b) Determine the order of the reaction halving the concentration of chlorate(i) ions
with respect to: if all other conditions are kept the same?(1)
i) hydrogen(1) c) Give a mechanism for the reaction
ii) nitrogen monoxide.(1) that is consistent with the rate equation.
c) Write the rate equation for the reaction.(1) Give your reasoning.(5)
d) Calculate the value of the rate constant d) Give possible evidence that you might look
and give the units. (2) for to test whether your proposed
e) Explain why is it not possible to deduce the mechanism is correct.(2)
reaction orders from the balanced chemical
equation.(2) 8 a) Explain, with the help of sketch graphs,
f) i) State what happens to the value of the how the Maxwell–Boltzmann distribution
rate constant for this reaction as the can be used to explain the effect of
temperature rises above 750 °C.(1) temperature on reaction rates.(4)
ii) Explain why the rate constant changes b) Explain why calculations based on theory
as it does when the temperature show that the rates of chemical reactions
increases.(4) increase with a rise in temperature far more
than would be predicted just from the
6 The table below shows the results of a series increase in the rate of collisions between
of experiments to determine the activation molecules. (3)
energy for the oxidation of iodide ions by c) The rate constants (k) for the
peroxodisulfate(vi) ions in the presence of decomposition of hydrogen iodide at
iron(iii) ions. different temperatures are given in the table.
S2O82−(aq) + 2I−(aq) → 2SO42−(aq) + I2(aq) Plot a graph and use it to determine the
activation energy for the reaction. (The gas
The results were obtained by the ‘clock’ constant R = 8.31 J K−1 mol−1.)(7)
method. Each reaction mixture included a
small, measured amount of aqueous sodium Rate constant, k/dm−3 mol−1 s−1 Temperature, T/K
thiosulfate and a few drops of starch solution. 3.74 × 10 −9 500
Temperature, T/K 288 292 299 308 315 6.65 × 10 −6 600
Time, t, for the blue 10.0 7.0 5.0 3.5 2.5 1.15 × 10 −3 700
colour to appear/s 7.75 × 10 −2 800
472
16 Kinetics II
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Chirality
17.1
17.1.1 Isomerism
Tip
The first section of this chapter revisits ideas about isomerism first introduced in
Chapters 6.1 and 6.2. Structural isomerism and the E/Z form of stereoisomerism were
introduced in the first year of the A Level course. This chapter expands on what you
already know about stereoisomerism and introduces chirality and optical isomerism.
Some of the ‘Test yourself’ questions are designed to help you revise ideas from the
first year of the A Level course.
Key terms
Structural isomerism occurs where
compounds have the same molecular
formula but different structural formulas.
Stereoisomerism occurs where
molecules have the same structural
formula but the atoms are arranged
differently in space.
Figure 17.1.1 Caraway seeds (left) and spearmint leaves (right). The compound carvone
is largely responsible for the different tastes and smells of caraway and spearmint. There
are two forms of carvone molecules (see Figure 17.1.15). These forms have the same
H H H
formula and structure but subtly different shapes and so different smells and tastes.
H C C C Cl Chemists describe them as optical isomers (see Section 17.1.3).
H H H
If two molecules have the same molecular formula, but a different arrangement
1-chloropropane
of their atoms, they are isomers. These isomers are distinct compounds with
different physical properties and, in most cases, different chemical properties.
H H H Isomers and isomerism occur most commonly with carbon compounds
because of the way in which carbon atoms can form chains and rings.
H C C C H
H Cl H
There are two ways in which the atoms can be rearranged to give isomers.
2-chloropropane ● The atoms are joined together in a different order forming different
Figure 17.1.2 The two structural isomers structures. This is called structural isomerism (Figure 17.1.2).
of C3H7Cl. ● The atoms are joined together in the same order, but they occupy different
positions in space. This is called stereoisomerism.
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Figure 17.1.3 shows how the two different types of isomerism are further
Key terms divided. Structural isomerism can be divided into three different types: chain
isomerism, position isomerism and functional group isomerism.
Chain isomerism occurs where
structural isomers have different
arrangements of the carbon skeleton. Isomerism
Position isomerism occurs where
Structural isomerism Stereoisomerism
structural isomers have the same
functional group attached to the same
carbon chain, but in a different position. Chain Position Functional group E/Z (cis–trans) Optical
isomerism isomerism isomerism isomerism isomerism
Functional group isomerism occurs
where structural isomers have different Figure 17.1.3 A ‘family tree’ showing the relationship between different forms of isomerism.
functional groups. There are two different types of stereoisomerism: E/Z (or cis–trans)
isomerism and optical isomerism. In both these forms of stereoisomerism,
the stereoisomers have the same molecular formula and the same structural
formula, but different three-dimensional shapes in which their atoms occupy
different positions in space.
Test yourself
1 Draw the structures and name the chain isomers of
2,2-dimethylbutane.
2 Draw the structures and name the position isomers of
1-bromopentane.
3 Draw the structures and name two functional group isomers with the
molecular formula C4H10O.
4 There are three isomers with the formula C5H12. Their boiling
temperatures are: 10 °C, 28 °C and 36 °C. Draw the structures of the
three compounds. Match the structures with the boiling temperatures
and justify your answer.
17.1.2 E/Z isomerism
The traditional system for naming the isomers of alkenes, in which the same
groups are arranged differently, is to name them as cis or trans. This is illustrated
by the sex attractant, bombycol, secreted by female silk moths (Figures 17.1.4
and 17.1.5). This messenger molecule strongly attracts male moths of the
same species. Analysis shows that two double bonds help to determine the
shape of the compound. Chemists are interested in pheromones because they
offer an alternative to pesticides for controlling damaging insects. By baiting
insect traps with sex attractants it is possible to capture large numbers of
Figure 17.1.4 Silk moths and cocoons. insects before they mate.
1 2 3 9
HOCH2CH2CH2CH2CH2CH2CH2CH2CH2 H
10 11 15 16
C C CH2CH2CH3
12 13
H C C
H H
Figure 17.1.5 The sex attractant bombycol, which is hexadeca-trans,10-cis,12-dien-1-ol.
474 17.1 Chirality
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However, there are many examples where the cis–trans system is not easily
applied and the E/Z system was developed to name these more complex
Tip
molecules. The advantage of the E/Z system is that it always works, whereas Z comes from the German word
the cis–trans system can break down in some cases. zusammen, meaning together.
E comes from the German word
Follow these steps in applying the E/Z naming system:
entgegen, meaning opposite.
1 Look at the atoms bonded to the first carbon atom in the C=C bond. The
atom with the higher atomic number has the higher priority.
2 If two atoms with the same atomic number, but in different groups, are
attached to the first carbon atom, then the next bonded atom is taken into
account. Thus, CH3CH2– has precedence over CH3 –.
3 Similarly, identify the group with the higher priority of the two attached
to the second carbon atom in the C=C bond.
4 If the two groups of higher priority are on the same side of the double bond,
the isomer is designated Z-, but if the two groups of highest priority are on
opposite sides of the double bond, the isomer is designated E- (Figure 17.1.6).
Br Br H Br
C C C C
H H Br H
Z-1,2-dibromoethene E-1,2-dibromoethene
melting temperature – 53 °C melting temperature – 9 °C
boiling temperature 110 °C boiling temperature 108 °C
Figure 17.1.6 The E and Z isomers of 1,2-dibromoethene are distinct compounds with
different melting temperatures and different boiling temperatures. The relative atomic
masses of bromine atoms are higher than the relative atomic masses of hydrogen atoms
so they have the higher priority.
Tip
In many cases a compound classed as trans in one system is E in the other and a
compound classed as cis is Z in the other, but there are examples where this is not
the case.
Test yourself
5 Draw the E/Z isomers of the following compounds and name them
using the E/Z system. State also whether each structure is a cis or a
trans isomer.
a) pent-2-ene
b) 2-bromobut-2-ene
c) 1-chloro-2-methylbut-1-ene
6 Draw the skeletal formula of:
a) (1E,4Z)-1,5-dichlorohexa-1,4-diene
b) (E)-3-methyl-4-propyloct-3-ene.
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17.1.3 Chirality and optical
isomerism
Mirror-image molecules
Every molecule has a mirror image. Generally, the mirror image of a
molecule can be turned around to show that it is identical to the original
molecule. Sometimes, however, a molecule and its mirror image are not
quite the same. The molecule and its mirror image cannot be superimposed.
Figure 17.1.7 shows left and right hands. If a mirror is placed vertically
between the two hands, the reflection of the left hand in the mirror is the
same as the right hand. Each hand is the mirror image of the other. But
the right hand does not match the left when placed on top of it (without
turning either over) so the right hand (the mirror image of the left) cannot
be superimposed on the left. The hands are non-superimposable.
Key term A chiral molecule, like a hand, cannot be superimposed on its mirror image.
The word ‘chiral’ (pronounced ‘kiral’) comes from the Greek for ‘hand’.
A chiral molecule is one that cannot be
Many everyday objects are chiral. Figure 17.1.8 shows some objects which
superimposed on its mirror image.
are chiral and some which are not.
Figure 17.1.7 Left and right hands are
mirror images of each other.
Figure 17.1.8 Everyday objects – chiral
or not?
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The most common chiral compounds are organic molecules in which there
is a single chiral centre due to a carbon atom attached to four different atoms
or groups. These carbon atoms are said to be asymmetric.
Figure 17.1.9 shows the two structures of lactic acid, which form when milk
turns sour. The two molecules each have the same four atoms or groups
attached to their central carbon atom: a CH3 – group, an –OH group, a
–COOH group and an H– atom. However, it is impossible to superimpose
the mirror images of lactic acid. No matter how the molecules are rotated, it
is not possible to get the two to look identical with groups and atoms in the
same positions in space.
Tip
Chemists have conventions for drawing
CH3 CH3
three-dimensional molecules on paper.
C C bond behind
HO COOH HOOC OH bonds in the the plane of
H H plane of the paper
the paper C
bond in front of the
plane of the paper
mirror
Figure 17.1.9 Molecules of lactic acid (2-hydroxypropanoic acid) are chiral. It is not
possible to superimpose the two mirror-image molecules.
The two forms of lactic acid behave identically in all their chemical reactions
and all their physical properties except for their effect on polarised light. This
optical property is the only way of telling the two forms of lactic acid apart.
So, chemists call them optical isomers. The word ‘enantiomers’ is also
used to describe mirror-image molecules that are optical isomers. The word
‘enantiomer’ comes from a Greek word meaning ‘opposite’.
Key terms
An asymmetric carbon atom is joined to four different groups or atoms.
Asymmetric molecules are molecules with no centre, axis or plane of symmetry.
Asymmetric molecules are chiral and exist in mirror-image forms. Any carbon atom
with four different groups or atoms attached to it is asymmetric and chiral.
Optical isomers or enantiomers occur in pairs made up of a chiral molecule and its
non-superimposable mirror image. One mirror-image form rotates the plane of plane-
polarised light clockwise. The other form rotates the plane of plane-polarised light
anticlockwise.
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Key term Optical isomerism and polarised light
A light beam becomes polarised after passing through a sheet of Polaroid®,
A racemic mixture is a mixture of equal the material used to make some sunglasses. The Polaroid prevents vibrations
amounts of the two mirror-image forms of of the light waves in all but one plane. So, in polarised light, all the waves are
a chiral compound. The mixture does not vibrating in the same plane (Figure 17.1.10).
rotate polarised light because the two
optical isomers have equal and opposite
effects so they cancel each other out.
sheet of Polaroid
ordinary
light plane
polariser polarised
light
Figure 17.1.10 Ordinary light and polarised light.
Light is said to be plane polarised after passing through a sheet of Polaroid.
If the polarised light is then directed at a second sheet of Polaroid, all the
polarised beam passes through if the second sheet of Polaroid is aligned in the
same way as the first (Figure 17.1.11a). However, no light gets through if the
second sheet is rotated through 90° relative to the first sheet (Figure 17.1.11b).
Figure 17.1.11 The effect of a second
sheet of Polaroid on polarised light. a)
b)
Tip
Because enantiomers have identical
properties, it is very difficult to separate
individual isomers from a racemic
mixture. Louis Pasteur laboriously used
a magnifying glass and tweezers to pick
individual left- and right-handed crystals
out of a mixture. Methods used today When polarised light passes through a solution of just one of a pair of optical
include the use of chromatography isomers, it rotates the plane of polarisation. One isomer rotates the plane
(see Chapter 19). If a column is packed of plane-polarised light clockwise. This is named the + isomer. The other
with a chiral stationary phase, the (+) isomer rotates the plane of plane-polarised light anticlockwise and this is
and (−) forms in a racemic mixture will the − isomer.
interact with it differently, and so one
optical isomer will leave the column
When polarised light passes through a solution containing equal amounts of
before the other.
both optical isomers, the effects cancel so this solution is optically inactive.
Solutions of this type are described as containing a racemic mixture.
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For accurate results, chemists measure the rotations with monochromatic
light (light of one colour or frequency) in an instrument called a polarimeter
(Figure 17.1.12).
Tip
Several systems of naming chiral compounds are in use including the (+)/(−) system
described in this section and the D/L system. The (+)/(−) system depends on the
effect that optical isomers have on plane polarised light. The D/L system, on the other
hand, is based on the actual stereochemical structure at the chiral centre.
tube containing solution
of one enantiomer rotates second polaroid must be rotated
the plane-polarised light to allow the maximum amount of
plane-polarised light through
angle of
rotation
first polaroid
produces
plane-polarised plane-polarised light
light has been rotated
anticlockwise
Figure 17.1.12 The effect of passing plane-polarised light through a solution of a
chiral compound.
Test yourself
7 Identify the chiral objects in Figure 17.1.8.
8 With the help of molecular models, decide which of the following
molecules are chiral:
NH3, CH2Cl2, CH2ClBr, CH3CHClBr, CH3CH(NH2)COOH.
9 Which of these alcohols can exist as optical isomers?
butan-1-ol, butan-2-ol, pentan-1-ol, pentan-2-ol, pentan-3-ol
10 Look at the upper representations of lactic acid in Figure 17.1.9.
Using the same convention, draw three-dimensional representations
of the two enantiomers of 2-chlorobutane, showing that they are
mirror images.
17.1.4 Optical isomerism and
reaction mechanisms
The optical activity of the reactants and products of organic reactions can
help chemists to determine the mechanisms of reactions. This is illustrated
by the outcomes of the SN1 and SN2 mechanisms in nucleophilic substitution
reactions (see Section 16.4). This information provides additional evidence
for the mechanism.
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During the one-step SN2 mechanism, the three groups that remain attached
to the central carbon atom are turned inside out. The molecule is inverted
like an umbrella in a high wind (Figure 17.1.13).
Figure 17.1.13 Nucleophilic substitution by H H3C H – H
H3C CH3
the SN2 mechanism with a molecule that
–
is chiral. HO C Br HO C Br HO C + Br–
C3H7 C3H7 C3H7
2-bromopentane transition state pentan-2-ol
This means that an optically active halogenoalkane gives rise to an optically
active alcohol if substitution takes place by the SN2 mechanism.
Tip
There no simple relationship between the three-dimensional shape of a chiral
compound and the direction that it rotates polarised light. What this means is that
molecular inversion of the + isomer of a halogenoalkane during an SN2 reaction does
give an optical isomer of the alcohol, but it is not possible to predict whether it will be
the + or the − isomer.
In the two-step SN1 mechanism, a planar intermediate is formed after the first
step. However, attack by the nucleophile during the second step can happen
with equal probability from either above or below the planar intermediate.
The result is that starting with one optical isomer of a halogenoalkane leads
to a product that is a racemic mixture of the two forms of the chiral alcohol.
This means that the product is optically inactive (Figure 17.1.14).
Figure 17.1.14 Hydrolysis of an optically OH
active halogenoalkane by the SN1
– C
mechanism. The product is a mixture of OH C3H7 CH3
equal amounts of the two optical isomers Br
C2H5
of the alcohol. This is a racemic mixture. + CH3
C C3H7 C
C3H7 CH3
C2H5
C2H5 –
OH C2H 5
C3 H 7
C CH3
OH
Test yourself
11 Suggest explanations to account for the fact that the reaction of
iodide ions with an optically active isomer of 3-chloro-3-methylhexane
gives:
a) a mixture of the two optical isomers of 3-iodo-3-methylhexane
b) a product mixture which is slightly optically active.
12 Account for the fact that the reaction of 2-bromooctane with sodium
hydroxide in an aqueous solvent is stereospecific: (+) 2-bromooctane
reacts to form (−) octan-2-ol while (−) 2-bromooctane gives (+) octan-2-ol.
480 17.1 Chirality
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Activity
Chirality and living things
Human senses are sensitive to molecular shape. The optical isomers of some molecules
have different tastes and smells (Figure 17.1.15).
O O HOOC COOH
CH3 CH3
(–) carvone (+) carvone
C C
H2N CHC2H5 H5C2CH NH2
H H
spearmint caraway L-isoleucine D-isoleucine
bitter sweet
Figure 17.1.15 Optical isomers with differing tastes and smells.
What is true of the sensitive cells in the nose and on the tongue is also true of most of
the rest of the human body. Living cells are full of messenger and carrier molecules that
interact selectively with the active sites and receptors in other molecules such as
enzymes. These messenger and carrier molecules are all chiral and the body works with
only one of the mirror-image forms. This is particularly true of amino acids and proteins
(see Chapter 18.2).
The chemists who synthesise and test new drugs have to pay close attention to chirality,
as molecular shape can also subtly alter the physiological effects of drugs. Perhaps the
most tragic example of this is thalidomide (Figure 17.1.16). This drug was first used in
1957 as a mild sedative that reduced morning sickness in early pregnancy. It was banned
in 1961 after children were born with stunted and distorted limbs and evidence showed
that thalidomide was responsible. The drug had been produced and sold as a racemic
mixture. It is now known that one of the optical isomers was beneficial and harmless,
while the other enantiomer was toxic. This tragedy caused many countries to introduce
much stricter rules for the testing of new drugs before licences are granted.
1 Identify the functional groups in:
O O
a) carvone
b) isoleucine. NH
2 Identify the chiral centres in: N O
a) isoleucine
b) carvone O
c) thalidomide.
3 Explain why the amino acid alanine, CH3CHNH2COOH, has optical isomers while the O O
amino acid glycine, CH2NH2COOH, does not. NH
4 Suggest reasons why chemists working on new drugs need to develop effective N O
methods to separate optical isomers of new compounds or methods to synthesise
selectively each of the pairs of isomers.
O
5 In the last few years thalidomide has been discovered to be effective in the treatment
of leprosy, some AIDS/HIV-related conditions and also of certain cancers. Should the Figure 17.1.16 The two enantiomers of
thalidomide.
use of this drug be allowed despite the risk?
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Chapter summary
Chapter 17.1 Chirality molecules are a pair of optical isomers, also called
enantiomers.
l Isomerism occurs in two main types: structural l One optical isomer rotates the plane of plane-
isomerism and stereoisomerism. polarised light clockwise. Its mirror-image
l Structural isomers are compounds with the same
form rotates the plane of plane-polarised light
molecular formula but different structural formulae. anticlockwise.
This occurs in three ways: different arrangements l A racemic mixture is a mixture of equal amounts
of the carbon atoms in a chain; different functional of enantiomers. The mixture does not rotate
groups; and different positions of the functional plane-polarised light because the two optical
groups on the carbon chain (see Section 6.1). isomers have equal and opposite effects so they
l Stereoisomerism occurs when molecules with the
cancel each other out.
same molecular formula and the same structural l Nucleophilic substitution by hydroxide ions with
formula have a different spatial arrangement of a halogenoalkane can occur either by SN1 or SN2
bonds. This occurs in two ways: E/Z isomerism mechanisms.
and optical isomerism. l In the two-step SN1 mechanism, a planar
l E/Z isomerism describes the arrangement of
carbocation is first formed. Attack by the
groups around a C=C bond in alkenes. E is for nucleophile can then happen with equal
entgegen, a German word meaning opposite, and probability from either above or below this planar
Z is for zusammen, meaning together. Groups intermediate. So, starting with one optical isomer
are assigned a priority depending on the atomic of the halogenoalkane, an optically inactive
number of the atoms attached to the double bond. racemic mixture containing both forms of the
l Optical isomerism results from a chiral centre in
chiral alcohol is formed.
a molecule, i.e. an asymmetric carbon atom. An l During the one-step SN2 mechanism, the three
asymmetric carbon atom is joined to four different groups that remain attached to the central carbon
groups or atoms. atom are turned inside out. The molecule is
l A chiral molecule is one that cannot be
inverted, so an optically active halogenoalkane
superimposed on its mirror image; these two forms an optically active alcohol.
482 17.1 Chirality
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Exam practice questions
1 Draw the displayed formulae of all the isomers At 25°C, an optically active product is
of the following, stating which types of formed. At 80°C, the product shows little
isomerism are involved: optical activity.
a) C3H7Cl (2) Explain why the products are different at the
b) C2HFClBr (6) two temperatures. (6)
c) C2H4ClBr. (3) c) If the sodium hydroxide reacts in ethanolic
2 a) State what is meant by a ‘chiral centre’ in a solution with 2-bromobutane, a mixture of
molecule. (1) three isomeric alkenes is formed. Draw and
b) Explain why a chiral centre in a molecule name these three isomers. (3)
gives rise to optical isomers. (2) 5 a) Name and draw a mechanism for the
c) Identify which of the following compounds reaction of pent-1-ene with HBr to form
can exist as optical isomers. (1) the major product. (5)
CH2(NH2)COOH CH2OHCH2COOH b) Explain why major and minor products are
CH3CHOHCOOH formed. (3)
d) Explain why a racemic mixture of optical c) Explain why the major product obtained in
isomers has no effect on the plane of plane- the reaction shows no optical activity. (4)
polarised light. (2)
e) Salbutamol is a drug used to relieve the 6 a) Draw four non-cyclic functional group
symptoms of asthma. isomers of C3H6O. (4)
b) Describe how you could distinguish between
H them using:
OH
N i) test-tube reactions (give the reagents
HO used and the observations you would
make) (9)
HO ii) infrared spectroscopy, excluding the
i) Deduce the molecular formula of fingerprint region (refer to the infrared
salbutamol. (1) spectroscopy data in the Pearson
ii) Identify any chiral centres in the Edexcel Data booklet). (4)
salbutamol molecule. (1) c) Draw two cyclic isomers of C3H6O and
comment on their relative stabilities. (4)
3 Explain the term ‘stereoisomerism’ and describe
the different types of stereoisomerism. Give 7 P is a neutral compound with the formula
examples to illustrate your answer. (15) C10H18O2. P slowly dissolves on boiling
with aqueous sodium hydroxide to produce
4 The halogenoalkane 2-bromobutane has ethanol and a solution of the sodium salt of
optical isomers. It reacts with aqueous sodium Q. Q is obtained by acidifying this solution.
hydroxide. The formula of Q is C6H10O4. The solid Q
a) State why 2-bromobutane has optical decomposes at its melting temperature to give
isomers. (1) a racemic mixture of isomers of R with the
b)* The reaction of an optical isomer of formula C5H10O2.
2-bromobutane with aqueous sodium
hydroxide is studied at different Explain these observations and give structures
temperatures. for P, Q and R. (8)
483
Exam practice questions
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Carbonyl compounds
17.2
17.2.1 The carbonyl group
The carbonyl group consists of the C=O bond. In aldehydes and ketones,
which are known as the carbonyl compounds, only carbon or hydrogen atoms
are attached to the carbon atom of the carbonyl group. However, the carbonyl
group is also present in carboxylic acids (RCOOH) and their derivatives,
acyl chlorides (RCOCl), esters (RCOOR) and amides (RCONH2). In these
compounds the carbon of the carbonyl group is also attached to another
electronegative atom. These electronegative atoms modify the properties of
the carbonyl group and for this reason carboxylic acids and their derivatives
are treated as separate functional groups (see Chapter 17.3 and Chapter 18.2).
As expected for compounds containing a double bond, the characteristic
reactions of carbonyl compounds are addition reactions. The C=O bond is
polar because oxygen is highly electronegative. As a result, the mechanism
of addition to carbonyl compounds (Section 17.2.5) is different from the
mechanism of addition to alkenes.
Tip 17.2.2 Aldehydes
Sections 17.2.2–17.2.4 of this chapter
revisit and build on ideas about Names and structures
alcohols and their oxidation first Aldehydes are carbonyl compounds in which a carbonyl group (C=O) is
introduced in Chapter 6.3. Some of attached to a hydrogen atom and a hydrocarbon group, or in the case of the
the ‘Test yourself’ questions are also first aldehyde, methanal, to two hydrogen atoms. So the carbonyl group in
designed to help revise ideas from the aldehydes is at the end of a carbon chain. The names are based on the alkane
first year of the A Level course. with the same carbon skeleton with the ending changed from -ane to -anal.
O O O O
H C CH3 C CH3CH2 C C
H H H H
methanal ethanal propanal benzenecarbaldehyde
(benzaldehyde)
Figure 17.2.1 Structures and names of aldehydes. The —CHO group is the functional
group that gives aldehydes their characteristic reactions.
Tip
Because the aldehyde functional group is always on the end carbon in a chain, no
number is needed in the name of an aldehyde to show where this functional group is –
it always includes carbon number 1. Always write the aldehyde group as –CHO. Writing
–COH is unconventional and easily leads to confusion with alcohols.
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Occurrence and uses
Biologists traditionally used a solution of methanal to preserve specimens.
Such use is now restricted because of the toxicity of methanal. It has also
been the main ingredient of the fluids used by embalmers. Methanal is
still an important industrial chemical because it is a raw material for the
manufacture for a range of thermosetting plastics.
O
Retinal is a naturally occurring aldehyde (Figure 17.2.2). Combined with Figure 17.2.2 The skeletal formula of retinal.
a protein, it forms the light-sensitive part of the visual pigment in the rod
cells of the retina (Figure 17.2.3). When light falls on a rod cell, the retinal
molecule changes shape. This sets off a series of changes that lead to a signal
being sent to the brain.
Figure 17.2.3 A coloured scanning
electron micrograph of rod cells in the
retina of an eye. The cells are magnified
about 3000 times. Rod cells contain a
visual pigment that can respond to dim
light but cannot distinguish colours.
Formation
Aldehydes are formed by the oxidation of primary alcohols using a mixture
of potassium or sodium dichromate(vi) and dilute sulfuric acid. An excess of
the alcohol is heated with the oxidising agent and the aldehyde distilled off as
it forms (see Figure 17.2.4). Unlike ketones, aldehydes can easily be oxidised
further to carboxylic acids by longer heating with an excess of the oxidising
agent and using a reflux condenser to prevent escape of the aldehyde (see
Section 17.2.5).
Figure 17.2.4 Apparatus used to oxidise
a primary alcohol to an aldehyde. The
aldehyde distils off as it forms.
to fume
cupboard
or sink
excess propan-1-ol
heat + sodium dichromate(VI)
+ dilute sulfuric acid
propanal
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Oxidation of a primary alcohol can be represented by a simplified equation,
where [O] represents the oxidising agent (see also Section 6.3.8):
CH3CH2CH2OH + [O] → CH3CH2CHO + H2O
propan-1-ol propanal
Test yourself
1 a) Draw the displayed formula of ethanal.
b) Draw the structural formula of 2-methylbutanal.
c) Draw the skeletal formula of 3-methylpentanedial.
n aldehyde can be made by adding butan-1-ol a few drops at a time
2 A
to a mixture of sodium dichromate(vi) and dilute sulfuric acid. The
product is then distilled from the reaction mixture.
a) Write an equation for the reaction. (Represent the oxygen from the
oxidising agent as [O].)
b) Suggest a reason for adding the butan-1-ol a few drops at a time.
c) E xplain why the reaction mixture is not heated in a flask fitted with
a reflux condenser before distilling off the product.
17.2.3 Ketones
Tip
Names and structures
In ketones, the carbonyl group is in
In ketones the carbonyl group is attached to two hydrocarbon groups.
the middle of a chain of carbon atoms.
Chemists name ketones after the alkane with the same carbon skeleton by
This means that the simplest ketone
changing the ending –ane to –anone. Where necessary a number in the
is propanone, which has the minimum
name shows the position of the carbonyl group.
number of three carbon atoms.
O CH3 O
6 5 4 3 2 1
CH3 C CH3 CH3 CH CH2 C CH2 CH3
propanone 5-methylhexan-3-one
Figure 17.2.5 Structures and names of two ketones.
Tip
The number of the principal functional group in a name is always as low as possible.
So the correct name of the second ketone in Figure 17.2.5 uses numbering from
the right to give 5-methylhexan-3-one, rather than the alternative name of
2-methylhexan-4-one found by numbering from the left.
Occurrence and uses
The most widely used ketone is propanone, which is a common solvent
and is familiar as nail polish remover. It has a low boiling temperature
and evaporates quickly, making it suitable for cleaning and drying parts of
precision equipment. Propanone is also the starting point for producing the
monomer of the glass-like addition polymer in display signs, plastic baths
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and the cover of car lights. Propanone, and other ketones, form during
normal metabolism, especially at night and during fasting when the levels of
propanone and other ketones in the blood rise. People with diabetes produce
larger amounts of propanone than normal. Other ketones, such as menthone
(Figures 17.2.6 and 17.2.7), are used in oils and perfumes.
O
Figure 17.2.7 The skeletal formula of the naturally occurring ketone called menthone,
which is found in the oils extracted from some plants.
Formation
Oxidation of secondary alcohols with hot, acidified potassium dichromate(vi)
produces ketones which, unlike aldehydes, are not easily oxidised further.
Oxidation of a secondary alcohol can be represented as a simplified equation, Figure 17.2.6 A bottle of peppermint oil
where [O] represents the oxidising agent (see also Section 6.3.8): and leaves of the peppermint plant. The
CH3CH(OH)CH3 + [O] → CH3COCH3 + H2O oil is used in aromatherapy. Peppermint oil
contains menthone, together with a range
propan-2-ol propanone
of chemicals which include menthol, methyl
ethanoate and volatile oils.
Test yourself
3 a) Draw the displayed formula of propanone.
b) Draw the structural formula of 4,4-dimethylpentan-2-one.
c) Draw the skeletal formula of 2,4-dimethylcyclohexanone.
4 What is the molecular formula of menthone?
5 Show that propanone and propanal are functional group isomers.
6 Write an equation for the oxidation of butan-2-ol to butanone.
(Represent the oxygen from the oxidising agent as [O].)
7 The simplest ketone that contains a benzene ring is C6H5COCH3,
which is called phenylethanone. Why might the ethanone part of the
name for this ketone be considered unusual?
δ+ δ–
C O
17.2.4 Physical properties of
carbonyl compounds δ+ δ–
C O
C
C O
The C=O bond in carbonyl compounds is polar (Figure 17.2.8). As a
result, the intermolecular forces include dipole–dipole attractions as well as
London forces. There are no hydrogen atoms bonded to oxygen in carbonyl Figure 17.2.8 Representations of the
compounds so hydrogen bonding cannot occur between the molecules carbonyl group in aldehydes and ketones.
C O
C
of aldehydes or ketones. However, hydrogen bonding is possible between The bond is polar because the oxygen atom
oxygen atoms in carbonyl groups and the –OH group in water. is more electronegative than carbon.
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Methanal is a gas at room temperature. Ethanal boils at 21 °C, so it may be
a liquid or gas at room temperature, depending on the conditions. Other
common aldehydes are also liquids. Similarly, the common ketones are liquids
with boiling temperatures similar to those of the corresponding aldehydes.
The simpler aldehydes such, as methanal and ethanal, are freely soluble in
water. The simplest ketone, propanone, mixes freely with water.
Test yourself
8 Refer to the data sheet headed ‘Properties of alkanes, alcohols,
aldehydes and ketones’, which you can access online at
[Link]/EdexcelChemistry.
a) Show that the boiling temperatures of aldehydes are higher than
those of alkanes with similar relative molecular masses, but lower
than those of the corresponding alcohols.
b) Account for the values of the boiling temperatures of aldehydes
relative to those of alkanes and alcohols in terms of intermolecular
forces.
9 Explain why propanone is freely soluble in water.
10 Why does an aldehyde such as ethanal mix freely with water whereas
hexanal is much less soluble?
17.2.5 Reactions of aldehydes and
ketones
Oxidation
Oxidising agents easily convert aldehydes to carboxylic acids (Figure 17.2.9).
It is much harder to oxidise ketones. Oxidation of ketones is only possible
with powerful oxidising agents which break up the molecules. Chemical
tests to distinguish aldehydes and ketones are based on the difference in the
ease of oxidation (Section 17.2.6).
O O
Cr2O72–(aq)/H+(aq)
CH3 CH2 C + [O] CH3 CH2 C
heat
H OH
Figure 17.2.9 Oxidation of propanal to propanoic acid.
Acidified potassium dichromate(vi) is orange and contains Cr2O72− ions.
After oxidising an aldehyde to a carboxylic acid, a green solution is formed
containing green Cr3+ ions.
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Activity
Chemicals in perfumes
The perfume ‘Chanel No 5’ was innovative when produced for the first time in 1921. As well as
natural extracts from flowers, the scent includes a high proportion of synthetic aldehydes,
such as dodecanal. This produces a highly original perfume.
O
undecanal OH
OH
O
citral O geraniol citronellol
O
hexamethyl O
tetralin musk methyl dihydrojasmonate
Figure 17.2.10 Skeletal formulae of some perfume chemicals.
The people who devise new perfumes think of the mixture as a sequence of ‘notes’. You first
smell the ‘top notes’, but the main effect depends on the ‘middle notes’, while the more
lasting elements of the perfume are the ‘end notes’. The overall balance of the three is critical.
This means that the volatility of perfume chemicals is of great importance to the perfumer.
Table 17.2.1 Natural and synthetic chemicals used to make perfumes.
Note Natural chemicals Synthetic chemicals Boiling temperature or
melting temperature
Top Citrus oils Octanal (citrus) (boils) 168 °C
Lavender Undecanal (green) (boils) 117 °C
Middle Rose Geraniol (floral) (boils) 146 °C
Violet Citronellol (rosy) (boils) 224 °C
End Balsam Indane (musk) (melts) 53 °C
Musk Hexamethyl tetralin (musk) (melts) 55 °C
1 Draw the skeletal formula of octanal. a) Draw the structure of 2-methylbuta-1,3-diene.
2 Suggest two advantages for the perfumer of using synthetic b) How many 2-methylbuta-1,3-diene units are needed to
chemicals instead of chemicals extracted from living things. make up the hydrocarbon skeleton of geraniol?
3 Identify the carbonyl compounds among the compounds shown 5 Use your knowledge of intermolecular forces to explain why:
in Figure 17.2.10. In each case state whether the compound a) aldehydes are useful as top notes while alcohols are more
includes the functional group of an aldehyde or of a ketone. commonly used as middle notes
4 Like many perfume chemicals, geraniol is a terpene. b) the musks used as ‘end notes’ also help to ‘fix’, or retain,
Terpene molecules are built from units derived from the more volatile components of a perfume
2-methylbuta-1,3-diene. c) geraniol is more soluble in water than undecanal.
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Test yourself
11 Identify the oxidation states of chromium before and after the
oxidation of an aldehyde by a solution of acidified potassium
dichromate(vi).
12 a) Outline the reagents and conditions for converting butanal to
butanoic acid.
b) What apparatus is used i) to carry out the reaction and ii) to
separate the product from the reaction mixture?
c) Write a half-equation for the oxidation of butanal to butanoic acid.
13 Write equations for the oxidation of:
a) ethanediol to ethanedial and
b) ethanedial to ethanedioic acid.
(In both cases, represent the oxygen from the oxidising agent as [O].)
Tip Reduction
LiAlH4 is also known as lithium
Metal hydrides can reduce carbonyl compounds to alcohols. Lithium
aluminium hydride, which is sometimes
tetrahydridoaluminate(iii), LiAlH4, is a powerful reducing agent that converts
abbreviated to lithal.
aldehydes to primary alcohols, and ketones to secondary alcohols. LiAlH4 is
easily hydrolysed so the reagent is dissolved in dry ether (ethoxyethane).
The reaction involves two steps: reduction involving LiAlH4 followed by
addition of dilute acid to complete the reaction. Simplified equations, as in
Figure 17.2.11, are written for the overall reaction; these use [H] to represent
the reducing agent.
O
LiAlH4
CH3 CH2 C + 2[H] CH3 CH2 CH2OH
propanal propan-1-ol
H
CH3
Tip C O + 2[H]
LiAlH4
CH3 CHOH CH3
The reduction step in the reaction of propan-2-ol
CH3
carbonyl compounds with LiAlH4 can be
propanone
represented by a nucleophilic addition
mechanism (see next section). Figure 17.2.11 Reduction of propanal and propanone. The 2[H] comes from the reducing
agent. This is a shorthand way of balancing a complex equation involving reduction.
Test yourself
14 Name the products of reducing butanal and butanone and state
which is a primary alcohol and which a secondary alcohol.
15 Show that reduction of an aldehyde or ketone with LiAlH4 has the
effect of adding hydrogen to the double bond.
16 Write equations for the reduction, using LiAlH4, of:
a) 2-methylbutanal
b) cyclohexanone.
Use [H] to represent the reducing agent.
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Reaction with hydrogen cyanide – a nucleophilic
addition
Hydrogen cyanide rapidly adds to carbonyl compounds at room temperature
(Figure 17.2.12). Hydrogen cyanide is a highly toxic gas that is formed in
the reaction mixture by adding potassium cyanide and dilute sulfuric acid.
The potassium cyanide must be in excess to ensure that there are free
cyanide ions ready to start the reaction. The product of the addition is a Tip
hydroxynitrile. The nitrile group, –CN, can be hydrolysed to a carboxylic
Note that the carbon atom in the —CN
acid (see Section 17.3.2). The reaction with the cyanide ion adds a carbon
group counts as part of the carbon
atom to the carbon skeleton of the original molecule and is therefore a useful
chain when naming nitriles.
step in synthetic routes to valuable compounds.
O OH
CH3 C + HCN CH3 C CN
H
H
ethanal 2-hydroxypropanenitrile
Figure 17.2.12 Addition of hydrogen cyanide to ethanal.
Nucleophilic addition
The reaction of a carbonyl compound with hydrogen cyanide and also
its reduction with LiAlH4 are both examples of nucleophilic addition.
The carbon atom in a carbonyl group is electron deficient because the
electronegative oxygen draws electrons away from it. This leaves it open to
attack by a nucleophile.
The incoming nucleophile uses its lone pair to form a new bond with the
δ+ carbon atom. This displaces one pair of electrons from the double bond Tip
onto oxygen. Oxygen has gained one electron from carbon and now has a It is the lone pair on carbon, not
single negative charge (Figure 17.2.13). nitrogen, that is used for nucleophilic
attack.
CH3
H3C
– –
NC C O NC C O
H H
Key terms
Figure 17.2.13 The first step of the nucleophilic addition of hydrogen cyanide to ethanal.
A nitrile is a compound with the
To complete the reaction, the negatively charged oxygen acts as a base and functional group —C≡N.
gains a proton from a hydrogen cyanide molecule (Figure 17.2.14) or from a
water molecule. Nucleophilic addition to a carbonyl
compound occurs when an electron-
rich species, a nucleophile, attacks the
CH3 CH3
electron-deficient carbon atom of
–
NC C O H CN NC C OH + CN– the unsaturated C=O bond, leading
to the formation of a carbon–oxygen
H H single bond.
Figure 17.2.14 The second step of the nucleophilic addition of hydrogen cyanide to ethanal.
A nucleophile is an electron–pair donor.
Note that taking a proton from HCN produces another cyanide ion.
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This mechanism helps to account for the fact that if the product of addition is
chiral, the outcome is a racemic mixture of the two optical isomers. This is
illustrated by Figure 17.2.15, which shows the addition of hydrogen cyanide
to ethanal in three dimensions. The atoms around the carbon atom of a
carbonyl group lie in a plane. The attacking nucleophile has an equal chance
of bonding to the carbon atom from either side of this plane.
CH3 CH3
H3C CH3
–
NC C C O C –
C O NC CN CN
OH HO
H H+ H H H+ H
Figure 17.2.15 Formation of two optical isomers by addition of HCN to ethanal.
Nucleophiles attack with equal probability from either side of the ethanal molecules,
giving rise to equal numbers of the two isomer molecules.
Test yourself
17 These questions are about the nucleophilic addition of hydrogen
cyanide to a carbonyl compound.
a) What features of the cyanide ion means that it is a nucleophile?
b) What type of bond breaking takes place in each step?
c) E
xplain why there is a negative charge on the oxygen atom at the
end of step 1.
d) Which molecule acts as an acid in step 2?
18 a) Show that the hydroxynitrile formed from ethanal and HCN is
chiral, but that formed from propanone is not.
b) Name the product of each of these reactions.
19 a) Write equations to show a nucleophilic addition mechanism for
the reduction of ethanal by LiAlH4. You may assume that the
nucleophile is the hydride ion, H−, and that water is involved in the
second step of the process.
b) E
xplain why LiAlH4 reduces the double bond in carbonyl
compounds but not the double bond in alkenes.
17.2.6 Tests for aldehydes and
Key term ketones
Chemists used to use a derivative to
identify an unknown organic compound.
Recognising carbonyl compounds
Converting a compound to a crystalline Today chemists can identify aldehydes and ketones with the help of instrumental
derivative produces a compound that techniques such as mass spectrometry and infrared spectroscopy (Chapter 19).
can be purified by recrystallisation and Traditionally chemists characterised these compounds by combining them
then identified by measuring its melting with a reagent that could convert them to a solid product. The solid, a so-called
temperature. crystalline derivative, is a chemical which can be purified by recrystallisation
and then identified by measuring its melting temperature.
492 17.2 Carbonyl compounds
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The reagent 2,4-dinitrophenylhydrazine reacts with carbonyl compounds
to form 2,4-dinitrophenylhydrazone derivatives, which are solid at room
temperature and bright yellow or orange (Figure 17.2.16). Figure 17.2.17 shows
the equation for the formation of the derivative formed with ethanal. The solid
derivative can be filtered off, recrystallised and identified by measuring its
melting temperature. This value is then compared to data values, such as those on
the data sheet ‘2,4-Dinitrophenylhydrazine derivatives of carbonyl compounds’,
which you can access online at [Link]/EdexcelChemistry.
Together with the boiling temperature of the original aldehyde or ketone, this
information makes it possible to identify the carbonyl compound.
O2N
H3C
C O + H2N NH NO2
H
2,4-dinitrophenylhydrazine
Figure 17.2.16 A bright orange
O2N 2,4-dinitrophenylhydrazone derivative.
H3C
C N NH NO2 + H2O
H Tips
ethanal-2,4-dinitrophenylhydrazone 2,4-Dinitrophenylhydrazine is
sometimes abbreviated to 2,4-DNPH.
Figure 17.2.17 Equation showing the formation of ethanal-2,4-dinitrophenylhydrazone.
Be careful to name the test reagent
as 2,4-dinitrophenylhydrazIne but
Tip the crystalline derivative it forms as a
For some analysis exercises, recrystallisation of the derivative is not necessary. The 2,4-dinitrophenylhydrazOne.
formation of an orange precipitate in a test-tube reaction can be used to show the Their polarity and high Mr guarantees
presence of a carbonyl compound. Only aldehydes and ketones form these derivatives. that all derivatives are solids at room
temperature, however small the
Distinguishing aldehydes and ketones carbonyl compound.
Aldehydes are easily oxidised. It is more difficult to oxidise ketones, but they
can be oxidised by stronger oxidising agents. In order to be absolutely sure
that ketones are not affected, three very mild oxidising agents are used to
distinguish aldehydes from ketones. These are Fehling’s solution, Benedict’s
solution and Tollens’ reagent.
Fehling’s reagent does not keep, so it is made when required by mixing
two solutions. One solution is copper(ii) sulfate in water. The other is a
solution of 2,3-dihydroxybutanedioate (tartrate) ions in strong alkali. The
2,3-dihydroxybutanedioate salt forms a complex with copper(ii) ions so that
they do not precipitate as copper(ii) hydroxide with the alkali.
Benedict’s solution is similar to Fehling’s solution but is more stable. It is less
strongly alkaline and does not react so reliably with all aldehydes.
Aldehydes reduce the blue copper(ii) ions in Fehling’s, or Benedict’s, solution
to copper(i), which then precipitates in the alkaline conditions to give an
orange-brown precipitate of copper(i) oxide, Cu2O (Figure 17.2.18).
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Tollens’ reagent (ammoniacal silver nitrate) similarly distinguishes
aldehydes from ketones. Tollens’ reagent consists of an alkaline solution of
diamminesilver(i) ions, [Ag(NH3)2]+. Forming a complex ion with ammonia
keeps the silver(i) ions in solution under alkaline conditions (see Section 15.6).
Aldehydes reduce the silver ions to metallic silver (Figure 17.2.19).
Key term The mild oxidising agents Fehling’s, Benedict’s and Tollens’ oxidise the
aldehyde to a carboxylic acid, which is present as a carboxylate ion in the
A carboxylate ion is formed when a alkaline conditions. Ketones do not react with these mild oxidising agents
carboxylic acid molecule loses a proton. and so the colour remains unchanged.
Figure 17.2.18 The test tube in the middle Figure 17.2.19 Warming Tollens’ reagent
contains Fehling’s reagent that has been with an aldehyde produces a precipitate of
reduced by an aldehyde to form an orange- silver, which coats clean glass with a shiny
brown precipitate of copper(i) oxide. The layer of silver so that it acts like a mirror
test tubes on the left and right contain (left). There is no reaction with a ketone
Fehling’s reagent and ketones. (right).
Test yourself
20 Write an ionic equation for the reaction of 23 Hydrolysis of A, C4H9Cl, with hot, aqueous sodium
copper(ii) ions with an alkali in the absence of hydroxide produces B, C4H10O.
2,3-dihydroxybutanedioate ions. Heating B with an acidic solution of potassium
21 a) Write an equation for the reaction of Tollens’ dichromate(vi) and distilling off the product as it
reagent with propanal using the symbol [O] to forms gives C, C4H8O.
represent the reagent. C gives a yellow precipitate with
b) Use the oxidation numbers of the metal ions 2,4-dinitrophenylhydrazine and forms a silver
and atoms to show that propanal reduces mirror when warmed with Tollens’ reagent.
Tollens’ reagent. Identify compounds A, B and C.
22 Write half-equations for the reduction of:
a) copper(ii) ions in alkaline conditions to
copper(i) oxide as in Fehling’s test
b) [Ag(NH3)2]+ ions in Tollens’ reagent to silver.
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Activity
Identifying an unknown carbonyl compound
Figure 17.2.20 shows stages in making, purifying and identifying a carbonyl compound.
stir impure minimum volume of
impure
solid ethanol needed to
derivative
dissolve the solid
2,4-dinitrophenyl- Hirsch funnel
hydrazine and a carbonyl
compound
to pump
hot water
hot solution
orange of derivative
filtrate plus impurities
precipitate of the
derivative
rinse with ice-cold
solvent
solvent evaporates purified
to leave dry crystals solid
to pump
ice and
water
impurities in
capillary tube
solution in
ethanol
measure the melting
pure temperature of a filtrate with
crystals sample of crystals impurities recrystallised 2,4-dinitrophenyl-
hydrazone derivative
Figure 17.2.20 Making a pure crystalline derivative of a carbonyl compound.
1 Why is it necessary to purify the derivative before measuring its melting temperature?
2 Explain how the procedure illustrated in Figure 17.2.20 removes soluble impurities
from the derivative.
3 In this instance, ethanol is the solvent used for recrystallising the derivative.
What determines the choice of solvent?
4 When measuring the melting temperature, what are the signs that the derivative is pure?
5 Identify the carbonyl compound that forms a 2,4-dinitrophenylhydrazone that melts
at 115 °C. The carbonyl compound boils at 80 °C and does not give an orange
precipitate with Fehling’s solution.
6 Suggest how the procedure outlined in Figure 17.2.20 could be modified so that
insoluble impurities were also removed. Discuss how any changes you suggest might
affect the yield of crystals obtained.
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Tip The triiodomethane reaction
A compound containing the CH3CO– group produces a yellow precipitate
An alternative reagent for the
of triiodomethane (iodoform) when warmed with a mixture of iodine and
triiodomethane reaction is a mixture
sodium hydroxide. The test shows the presence of a methyl group next to a
of potassium iodide and sodium
carbonyl group in an organic molecule.
chlorate(i).
Iodine in sodium hydroxide reacts to form iodate(i) ions (see Section 4.11).
Tip These ions oxidise secondary alcohols to ketones. So alcohols containing the
CH3CHOH– group will also produce a yellow precipitate of triiodomethane.
The triiodomethane reaction involves These include all methyl secondary alcohols and one primary alcohol,
breaking a carbon–carbon bond. It can ethanol.
be used to shorten a carbon chain at
room temperature by removing a methyl The reaction takes place in two main steps, substitution of iodine for
group. hydrogen then hydrolysis (see Figure 17.2.21). An overall equation for the
reaction is given in Figure 17.2.22.
H3C I3C
I2 OH–
C O C O CHI3(s) + R COOH
substitution hydrolysis
R R yellow
precipitate
Figure 17.2.21 The two steps in the triiodomethane reaction.
H3C
C O + 3I 2 + 4OH– CHI3 + RCOO– + 3I– + 3H2O
R
Figure 17.2.22 An overall equation for the triiodomethane reaction.
Test yourself
24 a) E xplain why alcohols with the group CH3CHOH – also give a
positive result with the triiodomethane reaction.
b) E xplain why a mixture of iodine with sodium hydroxide solution
is chemically equivalent to a mixture of potassium iodide and
sodium chlorate(i) solutions.
25 Name and write the displayed formulae of the two isomers of C5H10O
that form a yellow precipitate when they react with iodine in the
presence of alkali.
26 Which is the only aldehyde to undergo the triiodomethane reaction?
27 a) Write equations for:
i) t he reaction of iodine with hydroxide ions to form I− ions,
IO − ions and water
ii) the reaction of IO − ions with the ketone RCOCH3 to form
RCOCI3 and hydroxide ions
iii) the reaction of RCOCI3 with hydroxide ions to form CHI3 and
RCOO − ions.
b) Show that combining these three equations produces the overall
equation given in Figure 17.2.22.
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Chapter summary
Chapter 17.2 Carbonyl the copper(ii) ions in Benedict’s or Fehling’s
solutions to form a red–brown precipitate of
compounds copper(i) oxide. Aldehydes also reduce the silver
l The carbonyl group consists of the C=O bond. ions in Tollens’ reagent (ammoniacal silver nitrate)
Carbonyl compounds, aldehydes and ketones, have to metallic silver. Ketones have no effect on
only carbon or hydrogen atoms attached to the Benedict’s or Fehling’s solutions or on Tollens’
carbon atom of the carbonyl group. reagent.
l Aldehydes are carbonyl compounds in which a l Metal hydrides reduce carbonyl compounds to
carbonyl group is attached to a hydrogen atom, so alcohols. Lithium tetrahydridoaluminate(iii),
the carbonyl group in aldehydes is at the end of a LiAlH4, converts aldehydes to primary alcohols
carbon chain. In ketones, the carbonyl group is in and ketones to secondary alcohols. LiAlH4 is easily
the middle of a carbon chain. hydrolysed so the reagent is dissolved in dry ether
l The C=O bond in carbonyl compounds is polar. (ethoxyethane). The reaction involves two steps:
Their strongest intermolecular forces are dipole– reduction involving LiAlH4 followed by addition
dipole attractions, therefore, apart from methanal, of dilute acid.
the simplest aldehydes and ketones are liquids at l Carbonyl compounds react with hydrogen
room temperature. cyanide in excess potassium cyanide to form
l There are no hydrogen atoms bonded to oxygen hydroxynitriles by a nucleophilic addition
in carbonyl compounds so hydrogen bonding mechanism. The reaction with cyanide adds a
cannot occur between these molecules. However, carbon to the carbon skeleton of the original
hydrogen bonding is possible between oxygen molecule and is useful in synthesis. With
atoms in carbonyl groups and the –OH group in aldehydes and unsymmetrical ketones a racemic
water. Therefore the lower aldehydes and ketones mixture of two enantiomers is formed.
are water soluble. l 2,4-dinitrophenylhydrazine (2,4-DNPH) reacts
l Aldehydes are formed by the oxidation of primary with carbonyl compounds to form an orange
alcohols using acidified potassium dichromate(vi). precipitate. This reaction can be used as a
The aldehyde is distilled off as it is formed to qualitative test for the presence of an aldehyde or
prevent further oxidation to a carboxylic acid. If ketone. Alternatively, the melting temperature of
further oxidation is required, the reaction mixture a recrystallised derivative can be used to identify
is heated using a reflux condenser to prevent loss a particular carbonyl compound using data for the
of the aldehyde. melting temperatures of derivatives.
l Ketones are formed by the oxidation of secondary l Compounds containing the CH 3CO– group
alcohols with acidified potassium dichromate(vi). produce a yellow precipitate of triiodomethane
Unlike aldehydes, ketones are not easily oxidised. (iodoform) when warmed with iodine and sodium
Chemical tests to distinguish between aldehydes hydroxide. The reaction involves breaking a
and ketones are based on this difference. carbon–carbon bond and can be used to shorten a
l Aldehydes can be distinguished from ketones carbon chain at room temperature.
using mild oxidising agents. Aldehydes reduce
Chapter summary 497
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Exam practice questions
1 Copy and complete the table to show three 3* An optically active compound X with the
different reactions of propanal. (6) molecular formula C4H8O3 reacts with sodium
carbonate giving off a colourless gas. Oxidation
Reactant Reagent Organic product
of X forms Y which gives a yellow precipitate
Name Displayed
with 2,4-dinitrophenylhydrazine. Y does not
formula
react with Fehling’s solution or with iodine in
CH3CH2CHO Tollens’
sodium hydroxide.
reagent
CH3CH2CHO Propanoic acid Explain these observations and give the
CH3CH2CHO LiAlH4 structures of X and Y. (6)
4 The molar mass of a hydrocarbon W is
2 Citral and β-ionone are perfume chemicals.
56 g mol–1 and it contains 85.7% carbon.
β-ionone is one of the chemicals in the oil
W reacts with hydrogen bromide to form X.
extracted from violets.
Heating X under reflux with aqueous sodium
O hydroxide produces Y. Heating Y with an
acidified solution of potassium dichromate(vi)
converts it to Z. Z gives a yellow precipitate
with 2,4-dinitrophenylhydrazine but it does
not give a precipitate with Benedict’s reagent.
β-ionone
a) Identify W, X,Y and Z and explain your
answers. (8)
H O
b) Write equations for the reactions mentioned
citral
of W, X and Y. (3)
a) Deduce the molecular formula of citral. (1) c) Name the mechanisms for the reactions of
b) i) Name the two functional groups in W and X. (2)
citral. (2)
ii) Name the functional group which is 5 Consider the following pairs of compounds:
present in β-ionone but not in citral.(1) a) pentanal and pentan-3-one
c) Describe the observations you would expect b) pentan-3-one and pentan-3-ol
with each of the two compounds on: c) pentan-3-ol and pentan-2-ol.
i) warming them with Fehling’s i) Describe how you could distinguish
solution (2) between the compounds in each pair,
ii) mixing them with a solution of using a test-tube reaction. In each case,
2,4-dinitrophenylhydrazine (2) state a reagent and describe what you
iii) warming them with a mixture of would observe when it is added to each
iodine and sodium hydroxide. (2) compound. (9)
d) i) Draw the skeletal formula of the ii) Describe how you could distinguish
product of treating β-ionone with between the compounds in each pair
LiAlH4 in dry ether. (1) using a physical or a spectroscopic
ii) Draw the skeletal formula of the method other than the using the
compound formed when citral is infrared fingerprint region. In your
warmed with an acidic solution of answer, you should use data from the
potassium dichromate(vi). (1) Pearson Edexcel Data booklet. (6)
e) i) State whether the two compounds have 6 a) Write a half-equation for the oxidation of
E/Z isomers. (1) propanal to propanoic acid. (1)
ii) Draw an E/Z isomer of one of the two b) Write a half-equation for the oxidation of
compounds. (2) propanal to the propanoate ion by Tollens’
iii) State whether the two compounds have reagent in alkaline conditions. (1)
optical isomers. (1)
498 17.2
17.2 Carbonyl compounds
Carbonyl compounds
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c) Write a half-equation for the oxidation f ) Explain how the open chain molecule reacts
of cyclohexanone to hexanedioic acid. (1) to form the ring structure. (5)
Describe the mechanisms for the reactions of
9
7 a) Draw the mechanism for the reaction of propene with bromine, and propanone with
propanal with HCN. Include curly arrows, HCN. Identify similarities and differences
and any relevant dipoles and lone pairs. (4) between the two mechanisms with reference to
b) The reaction is usually performed in the the nature of the bonds and bond breaking, the
presence of KCN at a pH of about 5. reagents involved, the formation of intermediates
Explain why the rate of this reaction is slow and the overall effects of the change. (10)
if the pH is either too high or too low. (4)
10 a) Write an equation for the reaction of
c) Explain why the product obtained is
butanone with LiAlH4 in dry ether, using
optically inactive. (2)
[H] to represent the reducing agent. (1)
Glucose is a carbohydrate. Glucose molecules
8 b) Explain why:
usually exist in a ring form. In solution, about i) LiAlH4 reacts with carbonyl compounds
1% of the molecules exist in an open chain form. but not with alkenes although both
H
contain a double bond (3)
O
ii) dry ether rather than water is used as a
CH2OH C
solvent with LiAlH4. (2)
O H C OH c) An alternative to LiAlH4 for the reduction
H OH of aldehydes and ketones is NaBH4, sodium
H HO C H
tetrahydridoborate(iii). State which of
OH H H C OH LiAlH4 and NaBH4 is the stronger reducing
OH H agent and justify your answer (3)
H C OH
11 a) Compound P has the formula C4H6O2 and
H OH CH2OH an unbranched carbon chain. P reacts with
a) i) Deduce the molecular formula of HCN to form a compound Q, C6H8O2N2. P
glucose. (1) is easily oxidised by acidified potassium
ii) By inspection of the formula explain dichromate(vi) to an acidic compound R,
why glucose and related compounds are C4H6O4. When 1.0 g R is dissolved in water
called carbohydrates. (1) and titrated with 1.00 mol dm−3 sodium
b) Explain why glucose is a solid at room hydroxide, the mean titre is 16.90 cm3.
temperature and is also very soluble in Deduce structures for P, Q and R and
water. (4) explain the reactions. (9)
c) Sugars such as glucose are classified as b) A compound Z contains 64.3% carbon,
aldoses or ketoses based on the type of 7.1% hydrogen and 28.6% oxygen by mass.
carbonyl group present in the open chain The mass-to-charge ratio of the molecular
forms of the molecules. Deduce whether ion in its mass spectrum is 56. Z reduces
glucose is an aldose or a ketose. (1) Fehling’s solution to give an orange
d) i) State what you would expect to observe precipitate. When Z reacts with hydrogen
on mixing a solution of glucose with a in the presence of a nickel catalyst, 0.1 g
solution of 2,4-dinitrophenylhydrazine.(1) Z reacts with 85.4 cm3 hydrogen under
ii) Explain why this reaction is slower with conditions in which 1 mol gas occupies
glucose than with compounds such as 24 dm3. Deduce a structure for Z and
propanal. (2) explain your answer. (8)
e) i) State what you would expect to observe
on warming a mixture of a solution of 12 Propanoic acid can be prepared in a four-step
glucose and Tollens’ reagent. (1) synthesis starting from 2-bromobutane.
ii) State why glucose is classified as a Identify the three intermediate compounds and
reducing sugar. (1) give the reagents needed for each step. (7)
499
Exam practice questions
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Carboxylic acids and their
17.3 derivatives
17.3.1 Carboxylic acids
Occurrence
Carboxylic acids are compounds with the formula R–COOH where R
represents an alkyl group, aryl group or a hydrogen atom. The carboxylic
acid group –COOH is the functional group which gives the acids their
characteristic properties. Some carboxylic acids are found naturally in insects
(Figure 17.3.1) and in plants (Figure 17.3.2). Many organic acids are instantly
recognisable by their odours. Ethanoic acid, for example, gives vinegar its
taste and smell. Butanoic acid is responsible for the foul smell of rancid butter,
while the body odour of goats is a blend of the three unbranched organic
acids with 6, 8 and 10 carbon atoms.
Tip
The first two sections of this chapter revisit work on the oxidation of primary alcohols
covered in Section 6.3.8. Some of the ‘Test yourself’ questions are also designed to
help revise ideas from the first year of the A Level course.
Figure 17.3.1 The traditional names for organic acids were based Figure 17.3.2 Many vegetables contain ethanedioic acid, which
on their natural origins. The original name for methanoic acid is commonly called oxalic acid. The level of the acid in rhubarb
was formic acid because it was first obtained from red ants and leaves is high enough for it to be dangerous to eat the leaves. The
the Latin name for ‘ant’ is formica. This red wood ant can spray acid kills by lowering the concentration of calcium ions in blood to
attackers with methanoic acid (magnification ×5). a dangerously low level.
Names and structures
The carboxylic acid group can be regarded as a carbonyl group, C=O,
attached to an –OH group (see Figures 17.3.3 and 17.3.4), but is better seen
as a single functional group with distinctive properties.
Chemists name carboxylic acids by changing the ending of the corresponding
alkane to -oic acid. So ethane becomes ethanoic acid.
500 17.3 Carboxylic acids and their derivatives
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O O O
H C CH3 C CH3CH2 C
OH OH OH
methanoic acid ethanoic acid propanoic acid
O O O
C C C
HO OH OH
ethanedioic acid benzoic acid
Figure 17.3.3 Names and structures of carboxylic acids.
Carboxylic acids form a wide range of derivatives, each with their own
characteristics. This is illustrated by the derivatives of ethanoic acid in Figure
17.3.5. All of these compounds contain the acyl group, CH3CO–.
Figure 17.3.4 A ball-and-stick model of a
O O O carboxylic acid.
CH3 C CH3 C CH3 C
– +
OH O Na CI
acid sodium salt acyl chloride
ethanoic acid sodium ethanoate ethanoyl chloride
O
CH3 C O O
O CH3 C CH3 C
CH3 C O CH3 NH2
O
anhydride ester amide
ethanoic anhydride methyl ethanoate ethanamide
Figure 17.3.5 Compounds related to carboxylic acids.
Test yourself Key term
1 Write out the structural formulae and give the IUPAC names of the The acyl group consists of all the
three carboxylic acids which were traditionally derived from the Latin parts of a carboxylic acid except the
word ‘caper’, meaning goat: caproic acid (6C), caprylic acid (8C) and –OH group. The ethanoyl group is an
capric acid (10C). example of an acyl group.
2 Give the molecular formula, skeletal formula and name of the acid
shown in Figure 17.3.4.
3 Carboxylic acids and esters are structural isomers.
a) Name the type of structural isomerism they show
b) Draw the structure of
i) the ester which is an isomer of ethanoic acid
ii) the carboxylic acid which is an isomer of methyl ethanoate.
17.3.1 Carboxylic acids 501
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Physical properties
Even the simplest acids such as methanoic acid and ethanoic acid are liquids
at room temperature because of hydrogen bonding between the carboxylic
Tip acid groups. Carboxylic acids with more than eight carbon atoms in the
Benzoic acid is strictly called chain are solids. Benzoic acid is also a solid at room temperature.
benzenecarboxylic acid. However, the
Carbon–oxygen bonds are polar. There is also the possibility of hydrogen
simpler name is also commonly used.
bonding between water molecules and the –OH groups and oxygen atoms
in carboxylic acid molecules. This means that the simplest acids are soluble
in water. However, the solubility falls as the non-polar carbon chain length
increases and both hexanoic and benzoic acids are only very slightly soluble
at room temperature.
Test yourself
Tip 4 Draw a diagram to show hydrogen bonding between ethanoic acid
molecules and water molecules.
Benzoic acid is used as a food
5 In a non-polar solvent, ethanoic acid molecules dimerise through
preservative (E210) as it inhibits the
hydrogen bonding.
growth of mould and bacteria. Water-
soluble salts of benzoic acid, such a) Suggest a reason why the acid dimerises in a non-polar solvent but
as sodium benzoate (E211), are also not in water.
used as food preservatives, as these b) Draw a diagram to show an ethanoic acid dimer with two hydrogen
are converted to benzoic acid in acidic bonds between the molecules.
conditions. 6 a) Explain why sodium ethanoate is a solid at room temperature while
ethanoic acid is a liquid.
b) Explain why benzoic acid is insoluble in cold water whereas sodium
benzoate is soluble.
17.3.2 Preparation of carboxylic
acids
In the laboratory, carboxylic acids are normally made by oxidising primary
alcohols or aldehydes (Section 17.2.5). The usual oxidising agent is an acidic
solution of potassium dichromate(vi).
Carboxylic acids can also be made by hydrolysing nitriles. The reagent for
speeding up the hydrolysis can either be a solution of a strong acid or a
solution of a strong base.
RCN + 2H2O + HCl → RCOOH + NH4Cl
RCN + H2O + NaOH → RCOONa + NH3
Figure 17.3.6 Hydrolysis of nitriles heat with R COOH + NH4Cl
produces carboxylic acids or their salts. dilute HCl(aq)
R C N
heat with
dilute NaOH(aq)
R CO2– Na+ + NH3
502 17.3 Carboxylic acids and their derivatives
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Test yourself
7 Give the name and structure of the carboxylic acid formed by:
a) heating pentan-1-ol under reflux with an acidic solution of
potassium dichromate(vi) and then distilling off the product
b) heating butanal under reflux with an acidic solution of potassium
dichromate(vi) and then distilling off the product
c) heating propanenitrile under reflux with aqueous sodium hydroxide,
acidifying the mixture and then distilling off the product.
8 Write the equation for the hydrolysis of ethanenitrile with an excess of:
a) dilute hydrochloric acid
b) aqueous sodium hydroxide.
9 After hydrolysis of a nitrile with aqueous alkali, why must the solution
be acidified before distilling off a carboxylic acid from the mixture of
products?
17.3.3 Reactions of carboxylic acids
Reactions as acids
Carboxylic acids are weak acids (see Section 12.4). They are only slightly
ionised when they dissolve in water.
CH3COOH(aq) ⇋ CH3COO−(aq) + H+(aq)
The aqueous hydrogen ions in the solutions of these compounds mean
that they show the characteristic reactions of acids with metals, bases and
carbonates.
Carboxylic acids are sufficiently acidic to produce carbon dioxide when
added to a solution of sodium carbonate or sodium hydrogencarbonate
(Figure 17.3.7). This reaction distinguishes carboxylic acids from weaker
acids such as phenols (see Section 18.1.8).
O O
–
CH3 C (aq) + HCO3 (aq) CH3 C (aq) + H2O(l) + CO2(g)
OH O–
Figure 17.3.7 The reaction of ethanoic acid with hydrogencarbonate ions.
Citric acid is the weak acid found in the juice of all citrus fruits (Figure
17.3.8). Citric acid contains three carboxylic acid functional groups and has
the molecular formula of C6H8O7.
Tip
Carboxylic acids are not readily oxidised (except by combustion) as they are the end
products of the oxidation of primary alcohols and aldehydes. However, methanoic acid
can be oxidised by acidified potassium manganate(vii) to carbonic acid, H2CO3,
Figure 17.3.8 Citrus fruits including
which decomposes to give carbon dioxide and water. Investigation of the displayed
lemons, grapefruits, limes, clementines
formula of methanoic acid should indicate why it can be oxidised to carbonic acid.
and oranges.
17.3.3 Reactions of carboxylic acids 503
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Test yourself
10 For each of these pairs of chemicals in aqueous solution, describe
what you would observe when they react, write equations for the
reactions and name the organic products:
a) ethanoic acid and potassium hydroxide
b) propanoic acid and sodium carbonate
c) butanoic acid and ammonia.
lternative names for citric acid include 3-carboxy-3-
11 a) A
hydroxypentanedioic acid and 2-hydroxypropane-1,2,3-tricarboxylic
acid. Draw the structure of citric acid and justify the first as the
name that follows the IUPAC rules.
b) Write an equation for the complete neutralisation of citric acid by
sodium hydroxide.
c) Describe in outline how the reaction of sodium hydroxide with
citric acid could be used to estimate the concentration of citric
acid in fruit juices, indicating reasons why the result might not be
accurate.
Reduction
Carboxylic acids are much harder to reduce than carbonyl compounds.
However, they can be reduced to primary alcohols by the powerful reducing
agent lithium tetrahydridoaluminate(iii), LiAlH4 (Figure 17.3.9). The
reagent is suspended in dry ether (ethoxyethane). Adding dilute acid after
the reaction is complete destroys any excess reducing agent.
Figure 17.3.9 The reduction of ethanoic O
acid with LiAlH4. LiAH4
CH3 C (l) + 4[H] CH3CH2OH (l) + H2O(l)
in ether
OH
Reaction with phosphorus(v) chloride
Phosphorus(v) chloride or phosphorus pentachloride, PCl5, reacts vigorously
with carboxylic acids at room temperature. The reaction replaces the –OH
group with a chlorine atom, forming an acyl chloride (Figure 17.3.10). The
other product is hydrogen chloride gas, which fumes in moist air.
Figure 17.3.10 The reaction of PCl5 with O O
ethanoic acid.
CH3 C (l) + PCl5(s) CH3 C (l) + POCl3(l) + HCl(g)
OH Cl
ethanoyl chloride
Tip
When treated with PCl5, the O –H group in an alcohol and the O –H group in a
carboxylic acid behave identically. In most other reactions of carboxylic acids, the
C=O group modifies the properties of the O –H group.
504 17.3 Carboxylic acids and their derivatives
469983_17.3_Chem_Y1-2_500-[Link] 504 13/04/19 10:01 PM
Test yourself
12 LiAlH4 contains the [AlH4]− ion.
a) Draw a dot-and-cross diagram to show the bonding in the ion.
b) Deduce the shape of the ion and justify your answer.
13 Write the structural formula and give the name of the product of the
reaction of:
a) butanoic acid with LiAlH4
b) propanoic acid with PCl5.
14 a) How can the reaction of carboxylic acids with PCl5 be used as a
test-tube test and what does the test show?
b) Alcohols and carboxylic acids both give a positive result when
tested with PCl5. How can they be distinguished?
Esterification
Carboxylic acids react with alcohols to form esters (see Section 17.3.5). The
two organic compounds are mixed and heated under reflux in the presence
of a small amount of a strong acid catalyst such as concentrated sulfuric acid
(Figure 17.3.11).
O O
H+(aq)
CH3 C (l) + CH3CH2CH2OH(l) CH3 C (l) + H2O(l)
heat
OH OCH2CH2CH3
Figure 17.3.11 The formation of the ester propyl ethanoate from ethanoic acid and
propan-1-ol.
This reaction is reversible. The conditions for reaction have to be arranged
to increase the yield of the ester. One possibility is to use an excess of either
the acid or the alcohol, depending on which is the more available or cheaper.
Using more concentrated sulfuric acid than needed for its catalytic effect
can also help because the acid reacts with the water formed. However, the
amount of catalyst added is small so this effect is limited. In some esterification
reactions it is possible to distil off either the ester or the water as they form,
which encourages the reaction to go to completion.
Test yourself
15 Use Le Chatelier’s principle to discuss methods which could be used
to increase the yield of an ester formed from an acid and an alcohol.
16 Figure 17.3.11 shows that the H2O molecule formed in the
esterification reaction is made up of the –OH group from the acid
and the H atom from the alcohol (all in black type). How did chemists
confirm that this happened rather than the alternative use of H from
the acid and OH from the alcohol? (Hint: see Section 6.1.8.)
17.3.3 Reactions of carboxylic acids 505
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Activity
Preparation of an ester
The sequence of diagrams in Figure 17.3.12 shows the procedure for preparing a
small sample of an ester.
A B
reaction mixture
after refluxing
heat
ethanol and pure
ethanoic acid with
concentrated impure
sulfuric acid product
heat
D C
E
ester
aqueous
reagent
anti-bumping
granule
organic layer granules of
heat from separating calcium chloride Shake with sodium carbonate
funnel (a drying agent) solution. Run off aqueous layer,
ethyl ethanoate then shake the ester with calcium
(fraction boiling between chloride solution to remove
74 °C and 79 °C) unchanged ethanol
Figure 17.3.12 Stages in the preparation of an ester.
1 Identify what is happening at each of the stages A, B, C, D and E.
2 Write an equation for the reaction which forms the ester, and name the product.
3 What is the purpose of the concentrated sulfuric acid?
4 What are the visible signs of reaction during stage C and what practical precautions
are necessary during this stage? Tip
5 A volatile by-product distils off in the boiling range 35–40 °C before the ester in Refer to Practical skills sheet 13,
stage E. Suggest a structure for this by-product, which has the molecular ‘Assessing hazards and risks’,
formula C4H10O. which you can access online at
6 Calculate the percentage yield if the actual yield is 50 g from 40 g ethanol [Link]/
and 52 g ethanoic acid. EdexcelChemistry.
506 17.3 Carboxylic acids and their derivatives
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17.3.4 Acyl chlorides
Chemists value acyl chlorides as reactive compounds for synthesis, both on
a small laboratory scale and on a large scale in industry. These compounds
cause acylation and often provide the easiest way to make important
products such as esters.
Key term
Acylation is a reaction which substitutes an acyl group for a hydrogen atom. The H
atom may be part of an –OH group, an –NH2 group or a benzene ring.
Reaction with water
Ethanoyl chloride is a colourless liquid that fumes as it reacts with moisture
in the air. The reaction between ethanoyl chloride and water is violent at
room temperature and forms both ethanoic acid and hydrogen chloride. Figure 17.3.13 Ethanoyl chloride reacts
vigorously with water, releasing hydrogen
CH3COCl(l) + H2O(l) → CH3COOH(l) + HCl(g) chloride. A smoke of ammonium chloride
Although hydrogen chloride is a colourless gas, it reacts with moisture in the forms when an ammonia-soaked glass rod
air to form a mist of droplets of hydrochloric acid. However, the formation of is held in these fumes.
hydrogen chloride can be shown more clearly by its reaction with ammonia
vapour to form a white smoke of ammonium chloride (Figure 17.3.13).
NH3(g) + HCl(g) → NH4Cl(s) Tip
The rapid hydrolysis of acyl chlorides in
Reaction with alcohols moist air makes them difficult to store
Ethanoyl chloride and other acyl chlorides also react rapidly with alcohols and use on a large scale. Because of
at room temperature to form esters (Figure 17.3.14). The reaction of acyl this, acid anhydrides are sometimes
chlorides with alcohols is fast and not reversible and is much preferred as used in industry instead of acyl
a way to prepare esters to the slow and reversible reaction of acids with chlorides. Anhydrides are less reactive
alcohols (Section 17.3.3). so are easier to store, and they also
O O produce carboxylic acids rather than the
more corrosive fumes of hydrochloric
CH3 C + CH3CH2CH2OH CH3 C + HCl acid when they react. Anhydrides are
Cl OCH2CH2CH3 also cheaper than acyl chlorides. (See
Core practical 16: The preparation of
Figure 17.3.14 The formation of an ester from ethanoyl chloride and propan-1-ol.
aspirin, in Section 17.3.5.)
Reaction with ammonia
Acyl chlorides react rapidly with concentrated ammonia to form amides
(see Section 18.2.5). For example, when ethanoyl chloride is carefully added
to a concentrated aqueous solution of ammonia, a vigorous reaction takes
place producing fumes of hydrogen chloride and ammonium chloride plus a
residue of ethanamide.
CH3COCl(l) + NH3(aq) → CH3CONH2(s) + HCl(g)
ethanamide
HCl(g) + NH3(g) → NH4Cl(s)
ammonium chloride
17.3.4 Acyl chlorides 507
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Reaction with amines
Amines react with acyl chlorides to form N-substituted amides (see
also Section 18.2.3). This, and other reactions of ethanoyl chloride, are
summarised in Figure 17.3.15.
CH3COCl(l) + CH3CH2NH2(aq) → CH3CONHCH2CH3(s) + HCl(g)
N-ethyl ethanamide
The pain reliever paracetamol is an N-substituted amide (see Activity:
Paracetamol – an alternative to aspirin, in Section 18.2.3).
Figure 17.3.15 A summary of the reactions O O
of ethanoyl chloride. All the reactions
CH3 C CH3 C
happen quickly at room temperature. The NH (l)
3 (aq) H 5OH
C2
ethanoyl group in the products is shown NH2 O C2H5
in red. ethanamide O
ethyl ethanoate
amide ester
CH3 C
)
O (aq Cl H
2O
NH
2 (l) O
H5
CH3 C C2
CH3 C
N C2H5
OH
H ethanoic acid
acid
N-ethyl ethanamide
N-substituted amide
Test yourself
17 Write an equation for the reaction between ethanoyl chloride and
water. Show that this is an example of hydrolysis.
18 Write an equation for the formation of ethanamide from ethanoyl
chloride to show why two moles of ammonia are required for the
reaction with one mole of the acyl chloride.
19 Draw the structure and name the product of the reaction of propanoyl
chloride and butylamine.
20 The ester, propyl propanoate, can be prepared by reacting propan-1-ol
with either propanoic acid or propanoyl chloride. Write an equation
for each method and discuss any advantages and disadvantages of
using propanoyl chloride.
17.3.5 Esters
Occurrence and uses
Many of the sweet-smelling compounds found in perfumes and fruit
flavours are esters. Some drugs used in medicine are esters, including aspirin,
paracetamol and the local anaesthetics novocaine and benzocaine. The
insecticides malathion and pyrethrin are also esters. Compounds with more
than one ester link include fats and oils, as well as polyester fibres. Other
esters are important as solvents and plasticisers.
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Some esters have odours that resemble particular fruit flavours. Examples
are propyl ethanoate (pear), ethyl butanoate (pineapple), octyl ethanoate
(orange), 2-methylpropyl ethanoate (apple). Natural fruit flavours are
complex mixtures of esters and other compounds, including carboxylic
acids. Artificial flavourings using only a few esters are therefore unlikely to
replicate natural flavours exactly.
Names and structures
The general formula for an ester is RCOOR′, where R and R′ are alkyl or
aryl groups.
The name of an ester is in two parts derived from the acid and the alcohol
used to prepare the ester.
The parent carboxylic acid gives the ending of the name. For instance, esters
of ethanoic acid all contain the CH3COO group, so all have names which
end in ethanoate.
The rest of the ester molecule, the R′, comes from the alcohol and is an alkyl
or aryl group with a name such as methyl or ethyl (Figure 17.3.16).
O
O O
C
CH3 C H C
O CH2CH3
O CH3 O CH2CH3
methyl ethanoate ethyl methanoate ethyl benzoate
Figure 17.3.16 The names and structures of some esters.
Physical properties
Esters such as ethyl ethanoate are volatile liquids and only slightly soluble
in water. All esters contain polar C=O and C–O bonds, but they do not
contain O–H bonds and therefore are unable to form hydrogen bonds to
each other. This makes esters much more volatile than acids or alcohols with
similar Mr values.
Esters with short carbon chains are slightly soluble in water. However, as the
non-polar carbon chain length increases, the attractions between the polar
bonds in esters and water molecules become insufficient to cause overall
solubility.
Test yourself
21 Give the name and displayed formulae of the b) ethyl propanoate
esters formed when: c) 2-methylpropyl ethanoate.
a) butanoic acid reacts with propan-1-ol 23 Explain, in terms of intermolecular forces, why:
b) ethanoic acid reacts with methanol a) the boiling temperature of ethyl ethanoate is
c) ethanoic acid reacts with butan-1-ol. similar to that of ethanol but lower than that of
22 Draw each ester: ethanoic acid
a) propyl ethanoate b) ethyl ethanoate is less soluble in water than
either ethanol or ethanoic acid.
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Hydrolysis reactions
Hydrolysis splits an ester into an alcohol and an acid (or the salt of an acid).
Acids or bases can catalyse the hydrolysis.
Hydrolysis catalysed by an acid is a reversible reaction. (Figure 17.3.17). It is
the reverse of the reaction used to synthesise esters from carboxylic acids (see
Section 17.3.3).
Figure 17.3.17 Hydrolysis of an ester. O O
H+(aq)
These are the products when an ester + H2O + CH3CH2OH
CH3 C CH3 C
is heated with an excess of dilute acid,
such as hydrochloric acid. This reaction is O CH2CH3 OH
reversible. ester acid alcohol
Base catalysis is generally more efficient because it is not reversible
(Figure 17.3.18). This is because the acid formed loses its proton by reacting
with excess alkali. This turns it into a negative ion which does not react with
the alcohol.
Figure 17.3.18 The result of hydrolysing O O
ethyl ethanoate by heating it with an –
CH3 C + OH CH3 C + CH3CH2OH
aqueous alkali such as sodium hydroxide.
The salt and alcohol produced do not react O CH2CH3 O–
with each other, so this reaction is not ester salt alcohol
reversible.
Test yourself
24 Identify the products of heating:
a) propyl butanoate with dilute hydrochloric acid
b) ethyl methanoate with aqueous sodium hydroxide.
25 Under acid conditions the reaction of ethyl ethanoate with water is
reversible.
a) What conditions favour the hydrolysis of the ester?
b) How do these conditions compare with those for the synthesis of
the ester?
Tip
In Core practical 16 the purity of the aspirin formed is checked using melting
temperature data. The purity of the aspirin can also be checked by chromatography
(see Section 19.5).
For practical guidance, refer to Practical skills sheet 21, ‘Synthesis of an organic
solid’, which you can access online at [Link]/EdexcelChemistry.
510 17.3 Carboxylic acids and their derivatives
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Core practical 16
The preparation of aspirin
From ancient times, the bark of willow trees and Recrystallisation
meadowsweet flowers had been used to relieve pain and F Transfer the impure solid to a flask and insert a reflux
reduce fevers. In the nineteenth century, chemists identified condenser.
that material in the willow bark was converted in the body to G Add about 5 cm3 of ethanol down the condenser and heat
salicylic acid (2-hydroxybenzoic acid), which they named after the flask in a water bath until the crystals dissolve. If the
Salix, the willow tree genus. solid does not all dissolve, add a further 5 cm3 of ethanol
For several years, salicylic acid was given to patients to and continue to warm the solution.
H When the solid has all dissolved, add about 20 cm3 of warm
relieve pain. Although it was effective, the compound tasted
water. Allow the solution to cool slowly and crystals will
bitter and caused irritation of the mouth and stomach, so
form. Then cool the flask in ice-water.
alternatives were investigated. The sodium salt of salicylic I Filter off the crystals using a Buchner funnel. Wash the crystals
acid was found to relieve pain but tasted unpleasant and with about 10 cm3 of ice-cold ethanol then suck air through the
frequently made patients vomit. solid for about 15 minutes to dry it as much as possible.
In 1897, Felix Hoffman, working for the Bayer Company, J Weigh the pure, dry aspirin.
synthesised the ester acetylsalicylic acid, which was effective, K Measure the melting temperature of the aspirin produced.
tasted less unpleasant and was less irritating. He called his Questions
product aspirin from A (for acetyl) and spirin (from the name 1 Identify the particularly hazardous chemicals used in the
Spirea ulmaria for the meadowsweet flower). preparation and purification and state the precautions
Aspirin sold extremely well and its production is thought to needed when using them.
mark the start of the modern pharmaceutical industry. 2 Confirm that the ethanoic anhydride is in excess and
calculate the theoretical yield of aspirin.
Preparation of aspirin
3 If a student produced 1.8 g aspirin, calculate the percentage
To prepare aspirin, salicylic acid (2-hydroxybenzoic acid) is reacted
yield.
with ethanoic anhydride. A small amount of phosphoric acid is
4 Suggest why a reflux condenser is used in steps B and G.
used as a catalyst to speed up the reaction (Figure 17.3.19).
5 Why can crystallisation be sudden during stirring in step D?
The following is a summary of the procedure to prepare and
6 State why the volume of ethanol used in the recrystallisation
purify aspirin.
is as low as possible.
Preparation 7 Why is the flask cooled in ice in step H?
A Add 2.0 g of 2-hydroxybenzoic acid to a flask and, in a fume 8 Why was the suction turned off when the ice-cold ethanol
cupboard, add 4.0 cm3 of ethanoic anhydride (density = was added in step I?
1.08 g cm−3) and 5 drops of 85% phosphoric acid.
9 Give practical details in step J to describe how the mass of
B Attach a reflux condenser to the flask and, using a water
the aspirin is found.
bath, heat the mixture under reflux for 10 minutes.
C Remove the flask for the water bath, allow to cool for a few 10 The melting point of pure aspirin is 138–140 °C. Give a
minutes then carefully add 10 cm3 water down the condenser. reason in each case why a measured melting point might be
D Cool the flask in ice-water and leave until crystallisation is lower than 138 °C or higher than 140 °C.
complete. It may be necessary to stir with a glass rod to 11 Suggest why old aspirin tablets that have been exposed to
start crystallisation. moisture often smell of vinegar.
E Filter off the impure solid using a Buchner funnel.
O OH O O OH
C C
CH3 C O
OH + O O
O + CH3 C
C
CH3 C OH
O CH3
2-hydroxybenzoic ethanoic anhydride 2-ethanoyloxybenzenecarboxylic ethanoic acid
acid (salicylic acid) acid (aspirin)
Figure 17.3.19 Formation of aspirin from salicylic acid.
17.3.5 Esters 511
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Activity
Triglycerides – fats, oils and fatty acids
Fats and vegetable oils are esters of naturally occurring, long- It is accepted that saturated fats contribute to heart
chain carboxylic acids, often called fatty acids, with the alcohol disease, so there is emphasis on eating unsaturated and
propane-1,2,3-triol, better known as glycerol. The three – OH polyunsaturated fats rather than saturated animal fats.
groups in a glycerol molecule can form three ester links with
Alkaline hydrolysis of triglycerides produces soaps and glycerol.
fatty acids, giving rise to triglycerides (Figure 17.3.20). The
Soaps are the sodium or potassium salts of fatty acids and help
fatty acids may be saturated or unsaturated, see Table 17.3.1.
to remove greasy dirt because they have an ionic (water-loving)
O head and a long hydrocarbon (water-hating) tail. Most toilet
CH2 O C R1 soaps are made from a mixture of animal fat and coconut palm
O oil. Soaps from animal fat are less soluble and longer lasting.
Soaps from palm oils are more soluble so that they lather quickly
CH O C R2 but also wash away more quickly. Bars of soap also contain a
O
dye and perfume, together with an antioxidant to stop the soap
CH2 O C R3 and air combining to make irritant chemicals.
from from In a similar reaction to hydrolysis, if a triglyceride is heated
glycerol fatty acids
with methanol in the presence of a base catalyst, a trans-
three ester links esterification reaction takes place. Methyl esters of the fatty
Figure 17.3.20 The general structure of a triglyceride. In natural fats acids are produced together with glycerol. These esters are
and vegetable oils the hydrocarbon chains, R1, R2 and R3, may all be used as the renewable fuel, biodiesel.
the same or they may be different.
1 Write definitions for the terms ‘triglyceride’ and ‘fatty acid’.
Table 17.3.1 Examples of fatty acids. The cis–trans system for naming 2 Classify the acids in Table 17.3.1 as saturated or unsaturated.
geometric isomers is still generally used for fatty acids.
3 Give the name for the relevant acids in Table 17.3.1 using
Fatty Chemical name Formula E-, Z- terminology.
acid 4 Draw the skeletal formulae of palmitic acid and of linoleic acid.
Palmitic Hexadecanoic CH3(CH2)14COOH 5 Use your answers to Question 4 to explain why triglycerides
Stearic Octadecanoic CH3(CH2)16COOH of palmitic acid are solids at room temperature whereas
Oleic cis-Octadec-9-enoic CH3(CH2)7CH=CH(CH2)7COOH triglycerides of linoleic acid are liquid oils.
Linoleic cis, cis-Octadec- CH3(CH2)4CH=CHCH2CH=CH 6 Write an equation for the alkaline hydrolysis of the
9,12-dienoic (CH2)7COOH triglyceride in Figure 17.3.20 to form glycerol and soap.
Fats are solid at around room temperature (below 20 °C) and 7 Write an equation for the reaction of the triglyceride in Figure
contain triglycerides with a high proportion of saturated fatty 17.3.20 with methanol to form glycerol and three methyl
acids. Solid triglycerides are generally found in animals. In esters. Use your equation to explain why the process is
lard, for example, the main fatty acids are palmitic acid (28%), called ‘trans-esterification’.
stearic acid (8%) and about 56% oleic acid. 8 Suggest a reason why the product of reacting a vegetable oil with
methanol produces a much better fuel than the original oil itself.
Triglycerides with unsaturated fatty acids have lower melting 9 Suggest a reason why the alcohol and the triglyceride must
temperatures and have to be cooler before they solidify. be very dry to ensure a good yield of biodiesel.
Triglycerides of this kind occur in plants and are liquids at around 10 Why is it increasingly important to develop new raw
room temperature. In a vegetable oil such as olive oil, the main materials for making biofuels in order to avoid using
fatty acids are oleic acid (80%) and linoleic acid (10%). vegetable oils or corn starch?
512 17.3 Carboxylic acids and their derivatives
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17.3.6 Polyesters Key terms
Polyesters are polymers in which the monomers are linked together by
A condensation reaction is a reaction in
ester groups. The formation of polyesters involves a series of condensation
which molecules join together by splitting
reactions where water molecules are lost between the monomer molecules
off a small molecule such as water.
as they react to create ester links (see Section 17.3.5). So polyesters are a type
of condensation polymer (see also Section 18.2.9). A condensation polymer is a polymer
formed by a series of condensation
Polyesters are formed by condensation reactions:
reactions.
● between acids with two carboxylic acid groups and alcohols with at least
two –OH groups
● or between monomers that have both a carboxylic acid group and an
Tip
–OH group. Diacyl chlorides can also be used
instead of diacids.
The most common polyester is formed from ethane-1,2-diol (ethylene
glycol) and benzene-1,4-dicarboxylic acid (terephthalic acid).
This is used to make the plastic containers for fizzy drinks (Figure 17.3.21),
where the polymer is called PET. The initials are short for the traditional
name for the polymer, which is polyethylene terephthalate.
PET is also used as fibres to make clothing (Figure 17.3.22), where it is
often called simply ‘polyester’ or by the commercial name Terylene® (from
terephthalic acid). Fabrics made from polyester are hard wearing, washable
and relatively cheap. In a third form, called Mylar ®, the polyester is used as
plastic sheets. Because of the polymer’s strength, but low density, these sheets
can be used to make aviation balloons or sails for hang-gliders.
Figure 17.3.21 Bottles for sparkling drinks
are made of the polyester, PET. After heat
O O O O treatment, this polymer is impermeable to
C C C C gases.
HO OH HO OH
HO CH2 CH2 OH HO CH2 CH2 OH
O O O O
C C O CH2 CH2 O C C O CH2 CH2 OH
HO
+ H2O + H 2O + H2O
Figure 17.3.23 Condensation polymerisation to produce the polyester Terylene®.
The condensation reactions shown in Figure 17.3.23 can be repeated again
and again to produce a polymer with the repeat unit shown in Figure 17.3.24.
O O
C C O CH2 CH2 O
repeat unit
Figure 17.3.24 The repeat unit of Terylene®. Figure 17.3.22 The blazer, tie, shirt and
trousers that this schoolboy is wearing may
all contain polyester (Terylene®).
17.3.6 Polyesters 513
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Perhaps the most important development in polyester chemistry in recent
years concerns poly(2-hydroxypropanoic acid), commonly called poly(lactic
acid) or PLA. Poly(lactic acid) is possibly the most useful and most versatile
of the new biodegradable plastics. It is already used in such diverse goods as
plant pots, disposable nappies and absorbable surgical sutures (stitches).
Poly(lactic acid) is manufactured by the condensation polymerisation of
lactic acid, a single monomer that contains both a carboxylic acid group and
an alcohol group (Figure 17.3.25).
CH3 O CH3 O
lactic acid
HO C C + HO C C monomers
H OH H OH
CH3 O CH3 O
HO C C O C C + H2O
H H OH
CH3 O
many more reactions at
each end of the molecule H O C C OH
H
n
Figure 17.3.25 The synthesis of poly(lactic acid) by condensation polymerisation.
Test yourself
26 Draw the repeat unit of the polyester made from propane-1,3-diol and
pentanedioic acid.
CH3 27 Name the monomer used to make the polymer represented by the
repeat unit in Figure 17.3.26.
O C CH2 CH2 CH
28 a) Identify the types of intermolecular force that act:
O
i) between the chains in polyesters
Figure 17.3.26 The repeat unit of a
polyester. ii) between the chains in polyalkenes.
b) Explain why polyesters are generally biodegradeable whereas
polyalkenes are not.
514 17.3 Carboxylic acids and their derivatives
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Chapter summary
Chapter 17.3 Carboxylic acids moist air forms misty fumes containing droplets of
hydrochloric acid. Although laboratory synthesis
l Carboxylic acids contain the functional group often uses acyl chlorides, industrialists may prefer
–COOH. The carboxylic acid group could be to use acid anhydrides which are safer to store and
regarded as a carbonyl group, C=O, attached to an do not produce corrosive fumes when they react.
–OH group, but is better seen as a single functional (See Core practical 16 – the preparation of aspirin.)
group with its own distinctive properties. l Acyl chlorides react rapidly with alcohols at room
l Hydrogen bonding between carboxylic acid
temperature to form esters. The reaction is fast and
groups means that even the simplest acids, such as not reversible, so is preferred as a way to prepare
methanoic acid and ethanoic acid, are liquids at esters to the reaction of acids with alcohols.
room temperature. The lower acids are also soluble l Acyl chlorides react rapidly with concentrated
in water, but the solubility falls as the non-polar ammonia to form amides (RCONH2) and with
carbon chain length increases. amines to form N-substituted amides. For instance,
l Carboxylic acids can be prepared by the oxidation
ethanoyl chloride reacts with ethylamine to form
of primary alcohols or aldehydes using acidified N-ethyl ethanamide, CH3CONHCH2CH3.
potassium dichromate(vi) or by the hydrolysis of l Esters are sweet-smelling compounds found in
nitriles in acid or alkaline conditions. perfumes and fruit flavours. Hydrolysis splits an
l They are weak acids and react with bases to form
ester into an alcohol and an acid (or the salt of an
carboxylate salts. acid). Acids or bases can catalyse the hydrolysis, but
l Carboxylic acids are less easy to reduce
base catalysis is generally more efficient because it
than carbonyl compounds, but can be is not reversible.
reduced to primary alcohols by lithium l Polyesters are polymers in which the monomers
tetrahydridoaluminate(iii), LiAlH4, in dry ether are linked by ester groups formed by condensation
(ethoxyethane). reactions in which water molecules are eliminated.
l Phosphorus(v) chloride (phosphorus pentachloride),
Polyesters such as polyethylene terephthalate (PET)
PCl5, reacts vigorously with a carboxylic acid at are formed in reactions between acids with two
room temperature forming an acyl chloride. carboxylic acid groups and alcohols with at least
l In the presence of a small amount of strong acid
two –OH groups. Alternatively polyesters such
catalyst, carboxylic acids react with alcohols to as poly(lactic acid) (PLA) are formed from single
form esters. The reaction is reversible and an monomers, e.g. 2-hydroxypropanoic acid, that
equilibrium mixture is formed. have both a carboxylic acid group and an –OH
l Acyl chlorides contain the functional group
group.
–COCl. They react vigorously with water to form
a carboxylic acid and hydrogen chloride, which in
Chapter summary 515
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Exam practice questions
1 Although butane, propan-1-ol, ethanoic acid 3 The structure of an ester is shown:
and methyl methanoate all have similar relative CH3CH=CHCH2CH2COOCH3
molecular masses, their boiling temperatures
are 273 K, 371 K, 391 K and 305 K, respectively. Give the structures of the organic products
Explain why their boiling temperatures vary. (4) formed when this ester reacts with:
a) a solution of bromine in hexane at
Copy and complete the diagram below, using
2
room temperature (1)
propanoic acid as the example, to summarise
b) hydrogen in the presence of a nickel
reactions that form or use carboxylic acids and
catalyst at 140 °C (1)
their derivatives.
c) aqueous sodium hydroxide when
a) Each box should contain the type of
heated under reflux. (1)
compound with the name and formula
of the example. (6) 4 a) Describe two examples of test tube reactions
b) Each arrow should be labelled with the you could use to show the similarities and
type of reaction together with the differences between ethanoic acid and
reagents and conditions. (8) hydrochloric acid. (3)
b) Explain, with the help of equations, the
observations you have described in (a). (6)
Oxidation propanal
CH3CH2CH2OH
Treat with LiAlH4 in dry Heat with K2Cr2O7(aq) in
ethoxyethane dilute sulfuric acid
Carboxylic acid
propanoic acid
Esterification
Reaction with PCl5 at room
temperature
Hydrolysis
ethyl propanoate
Acyl chloride Reaction with aqueous
sodium hydroxide or
aqueous
sodium carbonate at room
temperature Hydrolysis
Salt of carboxylic acid
CH3CH2COO−Na+
516
17.3 Carboxylic acids and their derivatives
469983_17.3_Chem_Y1-2_500-[Link] 516 13/04/19 10:01 PM
5 Give the names and structures of the main 8 a) Explain the meaning of the term
organic products of each of these reactions: ‘condensation polymerisation’. (2)
a) propanoic acid and phosphorus(v) b) A section of a polyester is shown.
chloride (1)
b) butanoic acid and LiAlH4 in C C O CH2 CH2 CH2 O
ethoxyethane (1)
c) pentyl ethanoate and hot, aqueous O O
sodium hydroxide (2) Give the structures of a pair of compounds
d) ethanoic acid and aqueous calcium that can react to form this polyester and
hydroxide (1) name the functional groups in the
e) ethanoyl chloride and propan-2-ol. (1) compounds. (4)
c) State one important use of polyester
6 Ibuprofen is a painkiller with this skeletal polymers and state the properties of the
formula: polymer on which the use depends. (2)
OH 9* Three isomeric acids W, X and Y have the
molecular formula C8H6O4. They all contain
O a benzene ring. One mole of each acid reacts
a) Deduce the molecular formula of with two moles of sodium hydroxide.
ibuprofen. (1) When each of the acids is heated separately,
b) Deduce whether or not ibuprofen is soluble W and X melt without decomposing. Acid
in water. Explain your answer in terms Y decomposes at about 250 °C to form Z,
of intermolecular forces. (3) C8H4O3.
c) Draw the displayed formula of the
organic products formed when ibuprofen Compound X is used to make a polyester
reacts with: with ethane-1,2-diol in which the polymer
i) dilute sodium hydroxide solution (1) molecules are linear. Deduce structures for
ii) ethanol and a little sulfuric acid on W, X,Y and Z and justify your answers. (6)
warming. (1) 10 A, B and C are three isomers of C2H4O2.
1 cm3
7 of ethanol and 1 cm3 of ethanoic acid A reacts with sodium carbonate to produce an
were added to a test tube followed by 3 drops effervescence of a colourless gas.
of concentrated sulfuric acid. The test tube
was then warmed in a hot water bath for few B reacts with Fehling’s solution to give a red
minutes, after which the contents of the tube precipitate.
were poured into a beaker containing about C is almost insoluble in water, but reacts with
50 cm3 of cold water. An immiscible liquid with dilute sodium hydroxide to form soluble
a glue-like smell floated on top of the water in products.
the beaker.
a) Write an equation for the reaction, showing Deduce the structure of each isomer and
the structures of the organic compounds give its name. Explain your deductions. (6)
involved. (2) 11 Consider this reaction sequence:
b) State the role of the sulfuric acid. (1)
step1 step2
c) State why the test tube was not heated C3H7CN C3H7COOH X
directly using a Bunsen burner. (2) step3
CH3CH2CH2CONH2
NH3
d) Name the substance with the glue-like
a) Give the reagents and conditions for
smell. (1)
steps 1 and 2. (2)
e) Explain why the mixture was poured into
b) Give the name and structure of X. (2)
cold water before testing the smell. (2)
c) Give the displayed formula and name
of the product of step 3. (2)
517
Exam practice questions
469983_17.3_Chem_Y1-2_500-[Link] 517 13/04/19 10:01 PM
12 Ester B is used as a solvent for paint strippers. d) Name ester B and give a precaution
It is formed from acid A as shown in the necessary when using it as a paint
following equation. stripper. (2)
COOH COOC2H5 13 The four isomeric esters W, X,Y and Z with
+ 2C2H5OH + 2H2O molecular formula C4H8O2 were separately
COOH COOC2H5 heated under reflux with sodium hydroxide.
acid A ester B The mixtures formed were then distilled.
∆H = –20 kJ mol–1
The distillates from W and from X gave a
In an experiment, 0.50 mol of acid A was yellow precipitate when treated with iodine
mixed with 0.80 mol of ethanol and a small and sodium hydroxide, but those from
amount of concentrated sulfuric acid, and the Y and Z did not.
mixture left to reach equilibrium at a given After cooling, the solutions remaining in the
temperature. The equilibrium mixture formed distillation flasks were acidified with dilute
contained 0.27 mol of ester B. The total volume sulfuric acid.
of the mixture was V dm3.
a) Calculate the amounts of each of the other Solutions in the flasks from W and Z
three substances present in the equilibrium decolourised potassium manganate(vii),
mixture with ester B. (3) but those from X and Y did not.
b) Hence calculate a value for Kc for the a) Draw the structures of the four
equilibrium at this temperature and give its isomeric esters. (4)
units. State why the volume V need not be b) Identify W, X,Y and Z and explain
known. (4) your deductions. (6)
c) State the effect, if any, on the value of
Kc of increasing the temperature of the
equilibrium mixture. Justify your answer. (3)
518
17.3 Carboxylic acids and their derivatives
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Arenes – benzene compounds
18.1
18.1.1 Arenes
Key term Arenes are hydrocarbons – such as benzene, methylbenzene and naphthalene.
They are ring compounds in which there are delocalised electrons. The
Arenes are hydrocarbons with a ring or simplest arene is benzene. Traditionally chemists have called the arenes
rings of carbon atoms in which there are ‘aromatic’ ever since the German chemist Friedrich Kekulé was struck by
delocalised electrons. the fragrant smell of oils such as benzene. In their modern name ‘arene’, the
‘ar-’ comes from aromatic and the ending ‘-ene’ points to the fact that they
Delocalised electrons are bonding are unsaturated hydrocarbons, like the alkenes. However, the chemistry of
electrons that are not fixed between two arenes is different from that of alkenes in many ways.
atoms in a bond but shared between
three or more atoms. Benzene is an important and useful chemical. It was first isolated in 1825 by
the fractional distillation of whale oil, which was commonly used for lighting
homes. Later, it was obtained by the fractional distillation of coal tar. Today,
it is obtained by the catalytic reforming of fractions from crude oil.
Many important compounds, including painkillers such as aspirin, paracetamol
and ibuprofen, antiseptics such as Dettol® and TCP® (Figures 18.1.1 and 18.1.2)
and polymers such as Terylene® and polystyrene, contain the remarkably stable
ring of six carbon atoms, the benzene ring, in their structures.
OH OH
Cl Cl
H3C CH3
Cl Cl
4-chloro-3,5-dimethylphenol 2,4,6-trichlorophenol
(Dettol) (TCP)
Figure 18.1.2 The antiseptics Dettol and TCP both contain a benzene ring in their
Figure 18.1.1 The antiseptics used in structure.
some throat sprays are similar in structure
to Dettol and TCP.
18.1.2 The structure of benzene
Friedrich Kekulé played a crucial part in our understanding of the structure
of benzene as the result of a dream. The dream helped Kekulé to propose a
possible structure for benzene which had an empirical formula of CH and a
molecular formula of C6H6. Kekulé had been working on the problem of the
structure of benzene for some time. Then, one day in 1865, while dozing in
front of the fire, he dreamed of a snake biting its own tail. This inspired him
to think of a ring structure for benzene (Figure 18.1.3).
18.1.2 The structure of benzene 519
469983_18.1_Chem_Y1-2_519-[Link] 519 13/04/19 10:00 PM
H
H C H
C C
C C
H C H
H
Figure 18.1.3 Kekulé’s snake and his structural and skeletal formulae for the structure of
benzene. Kekulé’s formula would have the systematic name cyclohexa-1,3,5-triene.
Kekulé’s structure explained many of the properties of benzene. It was
accepted for many years, but still left some problems.
Cl Cl The absence of isomers of 1,2-dichlorobenzene
Cl Cl Kekulé’s structure suggests that there should be two isomers of
1,2-dichlorobenzene, one in which the chlorine atoms are attached to
carbon atoms linked by a single carbon–carbon bond, the other in which
the chlorine atoms are attached to carbon atoms linked by a double carbon–
Figure 18.1.4 Possible isomers of carbon bond (Figure 18.1.4).
1,2-dichlorobenzene.
In practice, it has never been possible to separate two isomers of
1,2-dichlorobenzene or any other 1,2-disubstituted compound of benzene.
To get round this problem, Kekulé suggested that benzene molecules might
somehow alternate rapidly between the two possible structures, but this
failed to satisfy his critics.
The bond lengths in benzene: X-ray
diffraction data
The Kekulé structure shows a molecule with alternating single and double
bonds. This would imply that three of the bonds are similar in length to
the carbon–carbon single bond in alkanes while the other three are similar
in length to the carbon–carbon double bond in alkenes. X-ray diffraction
studies show that the carbon atoms in a benzene molecule are at the corners
of a regular hexagon (Figure 18.1.5). All the bonds are the same length,
shorter than single bonds but longer than double bonds (Figure 18.1.6).
0 0.1 nm
Figure 18.1.5 Electron density map of C C
C C
benzene.
0.154 nm 0.134 nm 0.139 nm
Figure 18.1.6 Carbon–carbon bond lengths in ethane, ethene and benzene.
The resistance to reaction of benzene
An inexperienced chemist looking at the Kekulé structure might expect
benzene to behave chemically like a very reactive alkene and to take part in
addition reactions with bromine, hydrogen bromide and similar reagents.
Benzene does not do this. The compound is much less reactive than alkenes
and its characteristic reactions are substitutions, not additions.
520 18.1 Arenes – benzene compounds
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The stability of benzene: thermochemical data Key term
A study of enthalpy (energy content) changes show that benzene is more stable
Hydrogenation is the reaction
than expected for a compound with the Kekulé formula. This conclusion
of hydrogen gas with a molecule
is based on a comparison of the enthalpy changes of hydrogenation of
containing a double bond. Hydrogen
benzene and cyclohexene.
adds across the double bond which
Cyclohexene is a cyclic hydrocarbon with one double bond. Like other becomes saturated.
alkenes, it adds hydrogen in the presence of a nickel catalyst at 140 °C to
form cyclohexane.
The enthalpy change of the reaction, ΔH 1, is −120 kJ mol−1 (Figure 18.1.7).
Figure 18.1.7 Hydrogenation of
+ H2 ∆H = –120 kJ mol–1
cyclohexene.
cyclohexene cyclohexane
So, if benzene has three carbon–carbon double bonds, as in Kekulé’s structure,
we might reasonably predict that ΔH 1 for the hydrogenation of benzene
should be −360 kJ mol−1. But, when the hydrogenation is carried out, the
measured enthalpy change is only −208 kJ mol−1.
The measured enthalpy change is much less exothermic than the estimated
value. This suggests that the addition of hydrogen to benzene does not
involve normal double bonds and that benzene is actually lower in energy
and much more stable than expected (Figure 18.1.8).
Figure 18.1.8 Comparing the measured
Kekulé’s enthalpy change of hydrogenation of
benzene + 3H2
benzene with the estimated enthalpy
change of hydrogenation for Kekulé’s
structure.
benzene + 3H2
Enthalpy
(energy
content) Estimated
∆H = –360 kJ mol–1
Measured
∆H = –208 kJ mol–1 cyclohexane
Bonding in benzene: infrared data
The fact that benzene does not have carbon–carbon bonds like ethane or ethene
is also reflected in its infrared absorption spectrum. IR spectra provide unique
fingerprints of compounds in which each bond shows characteristic absorptions at
specific frequencies in the infrared region of the electromagnetic spectrum. In IR
spectra, the frequencies are normally shown as wavenumbers in units of cm−1, i.e.
the number of waves in a centimetre rather than the number of waves per second.
Look closely at the infrared spectrum of benzene in Figure 18.1.9 and that of
oct-1-ene, CH3(CH2)5CH=CH2 in Figure 18.1.10 on the next page.
18.1.2 The structure of benzene 521
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Figure 18.1.9 The infrared absorption 100
spectrum of benzene.
80
Transmittance/%
60
40
20
0
4000 3000 2000 1500 1000 600
Wavenumber/cm–1
Figure 18.1.10 The infrared absorption 100
spectrum of oct-1-ene.
80
Transmittance/%
60
40
20
0
4000 3000 2000 1500 1000 600
Wavenumber/cm–1
Notice that benzene does not have the typical strong absorptions of C–H
bonds in –CH2 and –CH3 groups in the wavenumber range 2962–2853 cm−1,
nor the C=C absorption of an alkene, like oct-1-ene, just below 1700 cm−1.
Instead, and unlike alkanes and alkenes, benzene has strong absorptions
at about 3050 cm−1 and 750 cm−1. All this provides further evidence that
benzene does not have normal C–C or C=C bonds in its structure.
Test yourself
1 Assume that the empirical formula of benzene is CH. What further
information is needed to show that its molecular formula is C6H6?
What methods do chemists use to obtain this information?
2 Draw one possible structure for C6H6 that is not a ring. Why does this
structure not fit with Kekulé’s structure for benzene?
3 An arene consists of 91.3% carbon.
a) What is the empirical formula of the arene?
b) What is the molecular formula of the arene if its molar mass is
92 g mol−1?
c) Draw the structure of the arene.
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18.1.3 Delocalisation in benzene
The accumulation of the evidence discussed in Section 18.1.2 led to
increased activity in the search for a more accurate model for the structure
of benzene. One model was to treat the carbon–carbon bonds in benzene as
halfway between single and double bonds and draw them with a full line
and a dashed line side-by-side, as in Figure 18.1.6. This model explains the
absence of isomers of 1,2-dichlorobenzene, the equal carbon–carbon bond
lengths in benzene and also its resistance to reaction. In recent years, the
bonding between carbon atoms in benzene has been simplified to a circle
inside a hexagon:
Despite this evidence, benzene may be represented in equations or mechanisms
either by the Kekulé structure or by the simplified structure above.
Although the simplified structure allows an improved understanding of the
properties of benzene, a better insight comes from considering its electronic
and orbital structure.
Figure 18.1.11 shows benzene with normal covalent sigma bonds (σ bonds)
between its carbon and hydrogen atoms. Each carbon atom uses three of its
electrons to form three σ bonds with its three neighbours. This leaves each
carbon atom with one electron in an atomic p orbital.
Figure 18.1.11 Sigma bonds in benzene,
H H p orbital
with one electron per carbon atom
remaining in a p orbital.
H H
H H
σ bond
These six p electrons do not pair up to form three carbon–carbon double
bonds (consisting of a σ bond plus a π bond) as in the Kekulé structure. delocalised electrons
Instead, they are shared evenly between all six carbon atoms, giving rise to
H
circular clouds of negative charge above and below the ring of carbon atoms
(Figure 18.1.12). This is an example of a delocalised π electron system, which C C
occurs in any molecule where the conventional structure shows alternating H C C H
double and single bonds. Within the π electron system, the electrons are free
C C
to move anywhere.
H H
Molecules and ions with delocalised electrons, in which the charge is spread benzene
over a larger region than usual, are more stable than might otherwise be
Figure 18.1.12 Representation of the
expected. In benzene, this accounts for the compound being 152 kJ mol−1
delocalised π bonding in benzene. The
more stable than expected for the Kekulé structure.
circle in a benzene ring represents six
The development of ideas concerning the structure of benzene illustrates the delocalised electrons. This way of showing
way in which theories develop and get modified as new knowledge becomes the structure explains the shape and
available. stability of benzene.
18.1.3 Delocalisation in benzene 523
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Test yourself
4 a) Look carefully at Figure 18.1.8. How much more stable is real
benzene than Kekulé’s structure for benzene?
b) Predict the enthalpy changes for the complete hydrogenation
of cyclohexa-1,4-diene and of cyclohexa-1,3-diene. Justify your
predictions.
5 How does the model of benzene molecules with delocalised π
electrons account for the following?
a) The benzene ring is a regular hexagon.
b) There are no isomers of 1,2-dichlorobenzene.
c) Benzene is less reactive than cycloalkenes.
6 When chemists thought that benzene had alternate single and double
bonds, it was sometimes named cyclohexa-1,3,5-triene. Why is this
now an unsatisfactory systematic name for benzene?
18.1.4 Naming arenes
The name ‘benzene’ comes from gum benzoin, a natural product containing
benzene derivatives. These derivatives of benzene are named either as
substituted products of benzene or as compounds containing the phenyl
group, C6H5 –. The names and structures of some derivatives of benzene are
shown in Table 18.1.1.
Table 18.1.1 The names and structures of some derivatives of benzene.
Systematic name Substituent group Structure
Chlorobenzene Chloro, – Cl C6H5– Cl
Nitrobenzene Nitro, – NO2 C6H5– NO2
Methylbenzene Methyl, – CH3 C6H5– CH3
Phenol Hydroxy, – OH C6H5– OH
Phenylamine Amine, – NH2 C6H5– NH2
The names used for compounds with a benzene ring can be confusing. The
phenyl group C6H5 – is used to name many compounds in which one of the
hydrogen atoms in benzene has been replaced by another atom or group.
The use of phenyl in this way dates back to the first studies of benzene. At
this time ‘phene’ was suggested as an alternative name for benzene, based
Tip on a Greek word for ‘giving light’. The name ‘phene’ was suggested because
The name ‘benzyl’ has been used to benzene had been discovered in the tar formed on heating coal to produce
represent the group C6H5CH2– as in gas for lighting.
benzyl chloride, C6H5CH2Cl. Good
When more than one hydrogen atom is substituted, numbers are used to
practice now avoids this by using the
indicate the positions of substituents on the benzene ring (Figure 18.1.13). The
systematic name for this compound,
ring is numbered to get the lowest possible numbers. In phenyl compounds,
(chloromethyl)benzene, to remove any
such as phenol and phenylamine, the –OH and –NH2 groups are assumed to
confusion with phenyl.
occupy the 1 position.
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Cl Cl Cl NH2 OH
Cl Cl
CH3 Br NO2
Cl
1,2-dichlorobenzene 1-chloro-3-methylbenzene 1-bromo-3-chlorobenzene 3-nitrophenylamine 2,4-dichlorophenol
Figure 18.1.13 Naming disubstituted
products of benzene, phenylamine and
Test yourself phenol.
7 Why is the middle compound in Figure 18.1.13 named
1-bromo-3-chlorobenzene and not 1-chloro-3-bromobenzene?
8 An old bottle of chemical was found with the label m-dinitrobenzene. Tip
Draw the structure and give the systematic name of this compound. The prefixes ortho-, meta- and para-, or
9 Name each of these disubstituted arenes. their abbreviations o-, m- and p-, were
traditionally used to name, respectively,
a) b) c)
d) 1,2- 1,3- and 1,4-disubstituted
Br OH benzene compounds. So the first
CH3 COOH OH compound in Figure 18.1.13 used to
be called ortho-dichlorobenzene or
o-dichlorobenzene. The use of these
COOH prefixes is still seen in the name of
NO2 OH compounds such as paracetamol (see
the Activity in Section 18.2.3).
10
Draw and name the isomers of C6H4Cl2, C6H3Cl3, C6H2Cl4 and
C6HCl5. (Beware of duplicates!)
Tip
Take care with the number of hydrogen
18.1.5 The properties and reactions atoms when writing the formulae of
substituted benzenes. Chlorobenzene
of benzene and arenes is C6H5Cl, dichlorobenzene is C6H4Cl2
Arenes are non-polar compounds with weak London forces between their and trichlorobenzene is C6H3Cl3.
molecules. The boiling temperatures of arenes depend on the size of the
molecules. The bigger the molecules, the higher the boiling temperatures.
Benzene and methylbenzene are liquids at room temperature, while
naphthalene is a solid (Figure 18.1.14).
CH3 Figure 18.1.14 Three arenes: benzene,
methylbenzene and naphthalene.
benzene methylbenzene naphthalene
Tip
Benzene is toxic. It is also a carcinogen. Because of this, benzene is banned from
teaching laboratories. Arene reactions may be studied using other compounds such as
methylbenzene or methoxybenzene, C6H5OCH3.
18.1.5 The properties and reactions of benzene and arenes 525
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Arenes, like other hydrocarbons, do not mix with water, but they do mix
freely with non-polar solvents such as cyclohexane.
The reactions of arenes
Arenes burn in air. Unlike straight-chain alkanes and alkenes of similar
molar mass, they burn with a very smoky flame because of the high ratio of
carbon to hydrogen in their molecules (Figure 18.1.15).
Benzene and other arenes are similar to alkenes in having a prominent
electron-dense region. In arenes this is a delocalised ring of π electrons,
while in alkenes it is a single localised π bond. Because of this similarity,
both arenes and alkenes react with electrophiles in many of their reactions.
However, the similarity ends there, because the overall reactions of arenes
involve substitution, unlike those of alkenes, which involve addition.
It is quite easy to see why addition reactions are difficult for arenes. If, for
example, benzene reacted with bromine in an addition reaction, the ring of
Figure 18.1.15 A sample of delocalised π electrons would be broken (Figure 18.1.16). Given the stability
methylbenzene burning in air showing the associated with the delocalised π system, this would require much more
yellow flame and very smoky fumes. energy than that needed to break the one double bond in ethene. So, arenes
largely undergo substitution rather than addition reactions in order to retain
their delocalised π electrons.
Br
+ Br2
Test yourself
Br
Figure 18.1.16 Benzene does not combine 11 The circle in a benzene ring represents six delocalised electrons.
with bromine in an addition reaction Some structures for naphthalene show circles in both rings. Suggest
because its ring of delocalised π electrons why many chemists prefer to see naphthalene drawn as in Figure
would be broken. 18.1.14 and not with two circles.
12 E xplain the existence of weak attractive forces between benzene
molecules, which are uncharged and non-polar.
Tip
13 a) E xplain why benzene does not mix with water.
Some addition reactions to arenes do
b) Name a solvent, other than cyclohexane, with which you would
occur, but if sufficient energy is provided
expect benzene to mix freely.
to start these reactions, further addition
then occurs until saturated compounds 14 Write equations for:
are formed. See hydrogenation a) the complete combustion of benzene
reactions in Section 18.1.7. b) the incomplete combustion of methylbenzene to form carbon.
18.1.6 Electrophilic substitution
reactions of benzene
Halogenation – bromination and chlorination
Addition of bromine to an alkene can occur when bromine water alone
is added with no need for a catalyst. The π electrons induce a dipole in
the bromine molecule and the δ+ bromine atom is a sufficiently strong
electrophile for reaction to take place. However, by contrast, the greater
stability of the benzene ring means that the induced dipoles and the δ+
bromine atoms in Br2 are not sufficiently electrophilic to react with benzene.
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Electrophilic substitution by bromine can occur (Figure 18.1.17) but, as
in all electrophilic substitution reactions of benzene, the first step of the
Key terms
reaction involves use of a catalyst to produce a stronger electrophile. Electrophilic substitution reactions
in arenes involve the replacement of a
H Br
hydrogen atom following attack by an
H C H H C H electrophile.
C C C C
+ Br2 warm with + HBr Chemists sometimes use the term
C C Fe or FeBr3 C C halogen carrier to describe substances
H C H H C H such as iron(iii) bromide and aluminium
chloride which catalyse the reaction of
H H benzene with chlorine or bromine.
bromobenzene
Figure 18.1.17 The reaction of benzene with bromine.
When benzene is warmed with bromine in the presence of iron filings, the
bromine first reacts with the iron to form iron(iii) bromide.
2Fe(s) + 3Br2(l) → 2FeBr3(s)
The iron(iii) bromide then acts as a catalyst for the reaction of bromine with
benzene by polarising further bromine molecules until a positive Br+ ion is
formed. This ion acts as the electrophile.
δ+ δ−
Br−Br + FeBr3 → Br−Br........FeBr3 → Br+ + FeBr4−
The Br+ ion is a reactive electrophile, which is strongly attracted to the H
delocalised electrons in benzene. As it approaches the benzene ring, the Br+ + + Br
ion forms a covalent bond to one of the carbon atoms using two electrons Br
from the π system (Figure 18.1.18). This step produces an intermediate cation.
Figure 18.1.18 Electrophilic Br+ ions use
two of the delocalised electrons in benzene
Tip to form an intermediate cation.
In these intermediate cations, the positive charge is delocalised over five carbons. The
other carbon, the one at which substitution occurs, is attached to four atoms, so it is
saturated and therefore not part of the electron delocalisation.
This intermediate then breaks down to form bromobenzene as two electrons
are returned from the C–H bond to the π system and the stable, delocalised
ring is restored (Figure 18.1.19). At the same time, an H+ ion is released
from the intermediate cation. This H+ ion immediately combines with the
Br− ion released in stage 1 to form hydrogen bromide.
H Br
+ Br + H+
Figure 18.1.19 The intermediate cation breaks down to form bromobenzene.
A similar reaction occurs when benzene is warmed with chlorine in the
presence of iron, iron(iii) chloride or aluminium chloride. The catalysts
are often referred to as halogen carriers.
18.1.6 Electrophilic substitution reactions of benzene 527
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Test yourself
15 a) Give the reagents and conditions for the preparation of
bromobenzene from benzene.
b) Why does benzene not react with bromine unless a halogen
carrier is present?
16
Iodine does not react with hot benzene even in the presence of
iron, but a good yield of iodobenzene can be produced by reacting
benzene with iodine(i) chloride.
a) Why do you think iodine does not react with hot benzene, even in
the presence of iron?
b) How is iodine(i) chloride polarised?
c) Using your answer to part (b), explain why iodine(i) chloride reacts
with benzene to give good yields of iodobenzene.
Nitration
Nitration of benzene, and other arenes, is important because it produces
a range of important products including dyes and powerful explosives
such as TNT (trinitrotoluene, now called 1-methyl-2,4,6-trinitrobenzene)
(Figures 18.1.20 and 18.1.21).
CH3
O2N NO2
Figure 18.1.20 Nitrated organic
compounds, like TNT (trinitrotoluene) and
nitroglycerine, are useful explosives in
demolition, mining, tunnelling and road NO2
building. Figure 18.1.21 The structure of TNT.
When benzene is warmed to about 55 °C with concentrated nitric acid in
the presence of concentrated sulfuric acid, the major product is yellow, oily
nitrobenzene (Figure 18.1.22).
H H
H C H H C NO2
C C C C
conc. H2SO4
+ HNO3 + H2O
C C at 55 °C C C
H C H H C H
H H
benzene nitrobenzene
Figure 18.1.22 The nitration of reaction of benzene.
An electrophilic substitution reaction occurs in which hydrogen is replaced
Key term by a nitro group, −NO2. If the reaction mixture is heated above 55 °C,
further nitration occurs forming dinitrobenzene.
A nitration reaction of an arene is
At 55 °C, concentrated nitric acid on its own reacts very slowly with benzene,
an electrophilic substitution where a
and concentrated sulfuric acid by itself has practically no effect. However,
hydrogen atom is replaced by a nitro
in a mixture of the two, sulfuric acid reacts with nitric acid to produce the
group, –NO2.
nitronium ion, NO2+, which is a very reactive electrophile.
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Sulfuric acid is a stronger acid than nitric so initial protonation of nitric
acid occurs.
HNO3 + H2SO4 → H2NO3+ + HSO4− Tip
The protonated acid then loses water to form the nitronium ion. In this reaction, concentrated HNO3
acts as a base and accepts a proton
H2NO3+ → NO2+ + H2O
from the stronger acid, H2SO4.
An overall equation is:
HNO3 + H2SO4 → NO2+ + HSO4− + H2O
The NO2+ ion is a reactive electrophile. It replaces a hydrogen in the benzene Tip
ring in a two-step electrophilic substitution mechanism (Figure 18.1.23)
similar to that which occurs in the bromination of benzene. Make sure that the correct bonding is
shown. It is the nitrogen atom that is
H NO2 bonded to a carbon atom in the ring,
+ not an oxygen atom, so the bond to the
NO2 + NO2 + H+ ring from the nitro group must always
start at N, not O.
Figure 18.1.23 Electrophilic substitution mechanism for the nitration of benzene.
Test yourself
17 Explain why dilute nitric acid does not react with benzene.
18
Write an overall equation for the formation of the nitronium ion in
which the water produced is also protonated.
19
Three possible isomers of dinitrobenzene can be produced. One of
these isomers is called 1,2-dinitrobenzene.
a) Draw and name the structures of the other two dinitrobenzenes.
b) Why is there no isomer called 1,6-dinitrobenzene?
20
Why is TNT mixed with a compound containing a high proportion of
oxygen, such as potassium nitrate, when it is used as an explosive?
21
Write a) an overall equation and b) a mechanism for the formation of
1-methyl-4-nitrobenzene from methylbenzene.
Alkylation and acylation (Friedel–Crafts
reactions)
The Friedel–Crafts reaction is an important method for substituting
an alkyl group or an acyl group for a hydrogen atom in an arene. This
Key term
reaction was discovered and developed jointly by the French organic chemist The Friedel–Crafts reaction is an
Charles Friedel (1832–1899) and the American, James Crafts (1839–1917). electrophilic substitution where a
The reaction is used on both a laboratory and an industrial scale. hydrogen atom in an arene is replaced
In a Friedel–Crafts reaction, a halogenoalkane or an acyl chloride is refluxed by an alkyl or an acyl group following
with an arene in the presence of aluminium chloride as catalyst. For example, attack by a carbocation or an acylium
if benzene is refluxed with chloromethane and aluminium chloride, a ion electrophile.
substitution reaction occurs forming methylbenzene (Figure 18.1.24).
18.1.6 Electrophilic substitution reactions of benzene 529
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CH3
AlCl3 catalyst
+ CH3Cl + HCl
heat
methylbenzene
Figure 18.1.24 Friedel–Crafts alkylation of benzene with chloromethane forming
methylbenzene.
A similar reaction occurs when benzene is refluxed with the acyl chloride,
ethanoyl chloride, plus aluminium chloride as a catalyst. This time, the product
is phenylethanone, also known as methylphenylketone (Figure 18.1.25).
CH3
O C
AlCl3 catalyst O
+ CH3C + HCl
heat
Cl
phenylethanone
Figure 18.1.25 Friedel–Crafts acylation of benzene with ethanoyl chloride forming
phenylethanone.
In Friedel–Crafts reactions, the aluminium chloride plays the important
catalytic role in creating the electrophiles which attack benzene. So, when
chloromethane is mixed with aluminium chloride, AlCl3 molecules remove
Cl− ions from polar δ+CH3 –Clδ− molecules, allowing reactive carbocations,
CH3+ ions, to act as electrophiles.
δ+CH δ− + AlCl → CH3+ + AlCl4−
3 –Cl 3
electrophile
These reactive CH3+ electrophiles then attack the delocalised π system of
benzene molecules to form an intermediate cation, which breaks down
producing methylbenzene and H+ ions (Figure 18.1.26).
CH3
CH3
+ H
+ CH3 + + H+
intermediate methylbenzene
cation
Figure 18.1.26 The reaction of CH3+ electrophiles with benzene in the Friedel–Crafts
alkylation reaction to produce methylbenzene.
Finally, the aluminium chloride catalyst is regenerated as H+ ions released in
the electrophilic substitution react with AlCl4− ions.
H+ + AlCl4− → HCl + AlCl3
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In a Friedel–Crafts acylation reaction, an acyl chloride, RCOCl, reacts
with aluminium chloride to from an acylium ion (Figure 18.1.27). This ion Tip
acts as the electrophile in the two-step electrophilic substitution reaction Acylium ions are formed in mass
(Figure 18.1.28). spectrometers as fragment ions of
compounds such as ketones or acid
+ – derivatives (Chapter 19).
R C CI + AICI3 R C + AICI4
O O
an acylium ion
Figure 18.1.27 Formation of an acylium ion.
H C
+
C R + C R O + H+
O O
Figure 18.1.28 Electrophilic substitution mechanism for Friedel–Crafts acylation of
benzene.
Tip
In Friedel–Crafts alkylation, the initial product contains an alkyl group attached to a
benzene ring. Alkyl groups are electron releasing, so the electron density on the ring is
greater than on benzene, which makes further substitution likely.
In Friedel–Crafts acylation, the initial product contains a carbonyl group attached to a
benzene ring. The carbonyl group withdraws electron density from the benzene ring, so
further substitution is unlikely.
Test yourself
22
This question is about the Friedel–Crafts reaction between benzene
and ethanoyl chloride, CH3COCl, in the presence of aluminium
chloride, AlCl3, as catalyst.
a)
Write an equation to show how AlCl3 molecules react with
polarised CH3COCl molecules to produce reactive acylium ions.
b)
Write an equation for the electrophilic substitution of benzene
by acylium ions to produce phenylethanone.
c)
Write an equation to show how molecules of the aluminium
chloride catalyst are regenerated.
23 a)
Why must the reaction mixture be completely dry during
a Friedel–Crafts reaction?
b)
Draw the structures of the products of a Friedel–Crafts reaction
of benzene with:
i) 2-iodo-2-methylpropane
ii) propanoyl chloride.
c)
Suggest a reason for using an iodoalkane instead of
a chloroalkane in a Friedel–Crafts reaction.
18.1.6 Electrophilic substitution reactions of benzene 531
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18.1.7 Addition reaction of benzene
with hydrogen
The characteristic reactions of benzene involve substitution because this
type of reaction retains the delocalised π electron system with its associated
stability. However, addition reactions involving disruption of the π electron
system do occur. Benzene, like alkenes, will undergo addition with hydrogen
in the presence of a nickel catalyst, but at considerably higher temperatures
(Figure 18.1.29).
Figure 18.1.29 The addition of hydrogen Raney nickel
to benzene forming cyclohexane. + 3H2
200 °C
cyclohexane
A higher temperature is needed with benzene in order to break up the stable
π electron system and allow addition to occur. A special finely divided form
of nickel, called Raney nickel, is also used because this has an extremely high
surface area. The addition of hydrogen to benzene is thought to occur in
three stages. First, the formation of cyclohexa-1,3-diene, then cyclohexene
and finally cyclohexane.
The catalytic hydrogenation of benzene is important industrially in the
manufacture of cyclohexane, which is used to make nylon.
Tip
In the presence of ultraviolet light, chlorine will add to benzene in a free-
radical reaction to form a mixture of chlorinated cyclohexanes including
1,2,3,4,5,6-hexachlorocyclohexane, shown below. There are several isomers of
C6H6Cl6, one of which has been used as commercial insecticide, but such use is now
restricted because of concerns about its toxicity to humans.
CI
CI CI
+ 3CI2
CI CI
CI
Test yourself
24 a) Why is Raney nickel used in the manufacture of cyclohexane from
benzene?
b) Write equations to show the three stages in the hydrogenation
of benzene via cyclohexa-1,3-diene and cyclohexene to form
cyclohexane.
c) Why do cyclohexa-1,3-diene and cyclohexene react more readily
with hydrogen than benzene?
25 E xplain why 1,2,3,4,5,6-hexachlorocyclohexane shows geometric
isomerism.
532 18.1 Arenes – benzene compounds
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Activity
Studying the reaction of benzene with chlorine
Figure 18.1.30 shows the apparatus that might once have 4 Write a mechanism for the reaction of the electrophile Cl+ with
been used to prepare chlorobenzene by heating benzene with benzene to form an intermediate cation in a first step, and
chlorine gas in the presence of iron filings. This preparation is then the formation of chlorobenzene and H+ in a second step.
now banned in teaching laboratories in schools and colleges. 5 The reaction shown in Figure 18.1.30 is just as effective if
aluminium chloride or iron(iii) chloride are used in place of
1 Why is this reaction banned in the teaching laboratories
iron. The substances FeCl3 and AlCl3 are often described as
of schools and universities?
catalysts and halogen carriers.
2 a) Why should the reaction be carried out in a fume
a) Why are these substances described as halogen carriers
cupboard?
for the reaction?
b) Why is a hotplate used?
b) Why is it correct to describe aluminium chloride and
c) Why is the oil bath at 70 °C?
iron(iii) chloride as catalysts?
3 During the reaction, iron reacts with chlorine to form iron(iii)
c) Why is it incorrect to describe iron as a catalyst for the
chloride, which then acts as an electron-pair acceptor, first
reaction?
polarising the Cl2 molecules as Clδ+–Clδ− and then forming
6 Aluminium chloride acts as a catalyst for the chlorination
the electrophile Cl+.
of benzene by polarising Cl2 molecules in the same way as
Write equations to show:
iron(iii) chloride.
a) the formation of iron(iii) chloride
a) Do you think aluminium chloride is likely to be more or
b) the polarisation of Cl2 as Clδ+–Clδ− by iron(iii) chloride
less effective than iron(iii) chloride?
and the formation of Cl+.
b) Explain your answer to part (a).
chlorine supply
oil bath
at 70 °C
benzene and
excess
iron filings
chlorine
magnetic
anti-bumping
stirrer
granules
hotplate
Figure 18.1.30 Preparing chlorobenzene.
18.1.8 Phenol
Phenol is an example of a compound with a functional group directly attached
to a benzene ring. In phenol, the functional group is –OH. Experiments
show that the –OH group affects the behaviour of the benzene ring while
the benzene ring modifies the properties of the –OH group. As a result of
this, phenol has some distinctive and useful properties.
As expected, the –OH group gives rise to hydrogen bonding in phenol and
therefore much stronger intermolecular forces than in benzene. This results
in phenol being a solid at room temperature (Figure 18.1.31). Figure 18.1.31 Crystals of phenol.
18.1.8 Phenol 533
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The −OH group in phenol also allows it to hydrogen bond with water. As
a result of this, phenol dissolves slightly in water and much more readily in
alkalis, such as sodium hydroxide solution (Figure 18.1.32), with which it
forms a soluble ionic compound. In this reaction, phenol is behaving as an
acid. However, phenol does not ionise significantly in water and it does not
react with carbonates to produce carbon dioxide.
Figure 18.1.32 Phenol reacts with
aqueous sodium hydroxide to form a OH(s) + NaOH(aq) O–Na+(aq) + H2O(l)
colourless solution of sodium phenoxide.
The derivatives of phenol are named in a similar fashion to those of benzene,
by numbering the carbon atoms in the benzene ring starting from the −OH
group (Figure 18.1.33).
Figure 18.1.33 The structure of phenol OH OH OH OH
and other substituted phenols.
O2N 6
1 NO2 CH3 2
1
2 6
5 3
3 5
4 4 NO2
NO2 Br
phenol 2,4,6-trinitrophenol 2-methyl-5-nitrophenol 4-bromophenol
(not 6-methyl-3-nitrophenol)
Test yourself
26 Explain, in terms of intermolecular forces, why:
a) phenol is a solid while benzene is a liquid at room temperature
b) phenol, unlike benzene, is slightly soluble in water
c) phenol does not mix with water as freely as ethanol.
27 What would you expect to observe on heating phenol until it burns?
28 Identify one way in which the –OH group behaves similarly in phenol
and ethanol, and one way in which it behaves differently.
29 a) What would you expect to observe if you added enough dilute
hydrochloric acid to a solution of phenol in sodium hydroxide to
make the mixture acidic?
b) Explain the reaction that occurs.
18.1.9 Reactions of the benzene ring
in phenol
Comparing the reactivity of benzene and phenol
The −OH group in phenol activates its benzene ring and makes it more
reactive than benzene itself. A lone pair of electrons on the −OH group
interacts with the delocalised electrons in the benzene ring, releasing
electrons into the ring and increasing the electron density in the ring. This
makes electrophilic attack easier so, as a result, electrophilic substitution
takes place under much milder conditions with phenol than with benzene.
534 18.1 Arenes – benzene compounds
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Reaction with bromine
An aqueous solution of phenol reacts readily with bromine water to produce
an immediate white precipitate of 2,4,6-tribromophenol as the orange/
yellow bromine colour fades. The reaction is rapid at room temperature
with bromine water. There is no need to heat the mixture or use a catalyst
(Figure 18.1.34).
OH OH Figure 18.1.34 The reaction of phenol
with bromine.
Br Br
(aq) + 3Br2(aq) (s) + 3HBr(aq)
Br
2,4,6-tribromophenol
Reaction with nitric acid
Dilute nitric acid reacts rapidly with phenol at room temperature to form a
brown mixture. The main products of the reaction are 2-nitrophenol and
4-nitrophenol (Figure 18.1.35). Notice how the conditions for nitrating
phenol are so much milder than those needed to nitrate benzene.
OH OH OH Figure 18.1.35 Nitrating phenol with dilute
nitric acid.
NO2
2 + 2HNO3 + + 2H2O
NO2
2-nitrophenol 4-nitrophenol
Tip
The compound 4-nitrophenol is an important intermediate in the production
of paracetamol (see the Activity: Paracetamol – an alternative to aspirin, in
Section 18.2.3).
Test yourself
30 a) Write an equation for the reaction of chlorine with phenol and
name the organic product.
b) Explain why the reaction of phenol with chlorine does not require
a catalyst whereas the chlorination of benzene does.
18.1.9 Reactions of the benzene ring in phenol 535
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Activity
Manufacturing phenol
Phenol is manufactured from benzene, propene and oxygen in two stages. The process
is known as the cumene process (Figure 18.1.36).
The first stage of the process involves the acid-catalysed electrophilic substitution of
benzene with propene to form cumene.
The second stage involves the air oxidation of cumene. This produces equimolar
amounts of phenol and propanone, a valuable co-product. About 100 000 tonnes of
phenol are manufactured each year in the UK using the cumene process.
CH3 CH3 Figure 18.1.36 The manufacture of phenol by
the cumene process.
CH OH
CH3 CH3
H+ O2
+ CH3 CH CH2 + C
O
benzene propene cumene phenol propanone
(1-methylethylbenzene)
1 Benzene and propene are obtained for the cumene process from crude oil. What
processes, starting with crude oil, are used to produce:
a) benzene
b) propene?
2 In the first stage of the cumene process, H+ ions react with propene to produce
electrophiles.
a) Write the formulae of two possible electrophiles produced when H+ ions react
with propene.
b) Explain why one of these electrophiles is more stable than the other.
c) Name and draw the structure of a second possible product of this first stage
besides cumene.
3 a) Write a mechanism for the reaction of the more stable electrophile, identified in
Question 2(b), with benzene to produce cumene.
b) Why is this reaction described as acid-catalysed?
4 Write an equation for the second stage of the process in which cumene is oxidised to
phenol and propanone.
5 The actual yield in the cumene process is 85%. Calculate the mass of benzene
required to manufacture 1 tonne of phenol and the mass of propanone formed at
the same time.
536 18.1 Arenes – benzene compounds
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Chapter summary
l When benzene is warmed with bromine and iron
Chapter 18.1 Arenes
filings, the catalyst iron(iii) bromide is formed.
l Arenes are hydrocarbons with a ring of carbon This reacts with bromine to form the electrophile
atoms in which there are delocalised electrons; Br+ and FeBr4 –. Electrophilic substitution of Br+
these are bonding electrons that are not fixed with benzene forms bromobenzene and releases
between two atoms in a bond but shared between H+ which reacts with FeBr4 – to reform iron(iii)
three or more atoms. bromide.
l The ring structure of benzene was first suggested
l In the nitration of benzene, a mixture of
by Kekulé, but the hexagon he proposed with concentrated nitric and sulfuric acids reacts
alternative single and double bonds could not to form the nitronium ion, NO2+, a very
explain why benzene is a regular hexagon with reactive electrophile. With benzene, this forms
equal bond lengths and is unreactive. The structure nitrobenzene. Nitration of benzene and other
was eventually explained in terms of the overlap of arenes leads to a range of important products
a p orbital on each carbon to form π bonds which including dyes and explosives such as TNT.
contain six delocalised electrons above and below l Friedel–Crafts reactions are electrophilic
the plane of the molecule. substitutions where a hydrogen atom in an
l Data from enthalpy changes of hydrogenation show
arene is replaced by an alkyl or an acyl group.
that benzene is 152 kJ mol–1 more stable than the When benzene is refluxed with chloromethane
Kekulé structure because of the delocalised ring of and aluminium chloride as catalyst, attack by a
electrons. carbocation on benzene occurs and methylbenzene
l The delocalisation energy would be lost if bromine
is formed. With the acyl chloride, ethanoyl
added to benzene in the same way as it adds across chloride and aluminium chloride, an acylium ion
the double bond in an alkene. Rather than addition electrophile is produced, and with benzene this
reactions, benzene and other arenes undergo forms phenylethanone.
substitution in order to retain their delocalised π l In phenol an –OH group is directly attached to
electrons. a benzene ring. This group activates the ring
l Arenes burn in air with a smoky flame because
and makes phenol more reactive than benzene
of the high ratio of carbon to hydrogen in their itself. A lone pair of electrons on the −OH group
molecules. interacts with the delocalised electrons in the
l Benzene reacts by electrophilic substitution.
benzene ring and increases the electron density in
In each reaction an electrophile is formed in a the ring. This makes electrophilic attack easier so
preliminary step. The electrophile reacts with electrophilic substitution takes place under much
benzene to form an intermediate cation. This milder conditions with phenol than with benzene.
intermediate breaks down to release an H+ ion. Phenol reacts immediately with bromine water to
Two electrons are returned from the C–H bond form a white precipitate of 2,4,6-tribromophenol.
to the π system so the stable, delocalised ring is
restored.
Chapter summary 537
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Exam practice questions
1 a)* Describe the structure and bonding in c) i) State the conditions used to
benzene.(6) produce A from benzene.(1)
b) Explain, in terms of structure and bonding, ii) Give one possible use for A. (1)
why benzene and ethene react differently
with electrophiles.(4) d) Substance D can be dehydrogenated to
form phenylethene (styrene) which is
2 Benzene reacts with bromomethane, CH3Br, used to manufacture poly(phenylethene).
to form C6H5CH3. The reaction involves i) Write an equation for the formation
electrophilic substitution. of poly(phenylethene) from
a) Explain the term ‘electrophilic substitution’.(2) phenylethene. (1)
b) Name the product and identify a catalyst ii) State one use of poly(phenylethene). (1)
that could be used for the reaction. (2)
c) Draw a mechanism for the reaction, 4 a) Chlorobenzene can be produced from
including curly arrows. Write an equation benzene and chlorine with a suitable
for the step that forms the attacking catalyst.
electrophile. (3) i) Name the catalyst. (1)
d) Give the names of the two chemists ii) Describe briefly how chlorobenzene
who are associated with this type of could be prepared. (3)
reaction of benzene. (1) b) Under suitable conditions methyl
benzene can be used to make the
3 Benzene is one of the most important aromatic
halogen-containing compounds I and II
compounds in industry. The flow chart below
shown below.
shows the formation of three useful products
from benzene. CH3 CH2CI
CH2CH3
B CH 3 CH 2 Cl
Ni +C
A benzene D CI
E +F I II
i) Name compounds I and II. (2)
nitrobenzene
ii) State the type of reaction that has
a) Name substances A to F. (6) occurred in each case. (2)
b) Reactions to form D and nitrobenzene c) Only one of the compounds reacts on
involve electrophilic substitution. warming with aqueous sodium hydroxide.
i) Write the formula of the electrophile i) Identify which compound reacts and
in each case. (2) justify your answer. (3)
ii) Give one condition needed for ii) Draw the structure of the organic
the nitration of benzene to form product formed with excess sodium
nitrobenzene, and outline the hydroxide solution and state the type
mechanism of the reaction using of reaction that has occurred. (2)
curly arrows where appropriate. (4)
iii) Give one use for the compounds
formed by the nitration of arenes. (1)
538
18.1 Arenes – benzene compounds
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5 A student prepared a sample of methyl a) Methyl 3-nitrobenzoate can be nitrated
3-nitrobenzoate by nitration of methyl further to form dinitro- and trinitro-
benzoate according to the following derivatives. Give two ways in which the
instructions. procedure tried to prevent this. (2)
b) Name two nitro-compounds that
O OCH3 O OCH3 may contaminate the impure methyl
C C
3-nitrobenzoate. (2)
c) State why the impure methyl
+ HNO3 + H2O 3-nitrobenzoate was washed with water
before recrystallisation. (1)
NO 2
d) Explain why impurities that are only
present in small amounts are removed
A Put 5 cm3 of methyl benzoate into a 100 cm3 during recrystallisation. (2)
conical flask and carefully add 1 cm3 of e) i) State why some methyl 3-nitrobenzoate
concentrated sulfuric acid. When the methyl is lost during recrystallisation.(1)
benzoate has dissolved, cool the mixture ii) State how the procedure tried to
in ice. minimise this loss. (2)
B Prepare the nitrating mixture in a separate f) State the effect in step H of cooling the
boiling tube by carefully adding 4 cm3 solution in ice rather than allowing it to
of concentrated nitric acid to 4 cm3 of cool slowly. (1)
concentrated sulfuric acid and cool this g) Assuming that excess nitric acid is used,
mixture in the ice as well. calculate the theoretical yield of methyl
C Add the nitrating mixture dropwise to the 3-nitrobenzoate. (Density of methyl
methyl benzoate solution in sulfuric acid. benzoate = 1.1 g cm−3) (4)
Stir the mixture using a thermometer and h) If 3.5 g methyl 3-nitrobenzoate are
use the ice bath to keep the temperature formed, calculate the percentage yield. (1)
between 5 and 15 °C. When the addition is Three reactions of phenol are summarised in
6
complete, remove the flask from the ice and the flow diagram below.
let it stand at room temperature for
15 minutes. OH
D After 15 minutes, pour the reaction mixture
over about 50 g of crushed ice and stir until 2-nitrophenol Reaction 1 Reaction 2
+ B
all the ice has melted and crystals of methyl 4-nitrophenol A Br2
3-nitrobenzoate have formed.
E Filter the solid by suction using a Buchner Reaction 3 NaOH(aq)
funnel, wash the solid with cold water and
then transfer it to a small conical flask. C
F Add 25 cm3 of ethanol to the solid and a) Name reagent A. (1)
warm the contents of the conical flask to b) Draw the structural formulae of B and
about 50 °C by immersing it in a beaker of of C. (2)
hot water. c) Under suitable conditions, nitrophenols
G Once the impure solid has dissolved, pure can be converted to dinitrophenols.
methyl 3-nitrobenzoate can be recovered by i) Give the reaction conditions
cooling the solution in an ice bath and then for converting nitrophenols to
collecting the crystals that form by suction dinitrophenols. (2)
filtration. ii) One possible dinitrophenol is
H Dry the crystals using absorbent paper and 2,3-dinitrophenol. Write the names of
weigh them. all the other possible dinitrophenols. (3)
539
Exam practice questions
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7 A synthesis of 1-phenylpropene from benzene c) The benzene ring in L-dopa is more
is shown below. reactive with electrophiles than that in
benzene.
COCH2CH3 CH(OH)CH2CH3 Explain why the benzene ring in L-dopa is
more susceptible to attack by electrophiles
step 1 step 2 than benzene itself. (2)
d) In the human body, an enzyme called
‘L-dopa decarboxylase’ eliminates the
step 3
carboxylic acid group from L-dopa to
produce the primary amine, dopamine.
CH=CHCH3 i) State what is meant by the term
‘primary amine’. (1)
ii) Draw the structural formula of
dopamine. (1)
The following data are provided to answer
9
a) i) Identify a reagent and a catalyst for
this question.
step 1. (2)
ii) Write an equation for the reaction Mean bond energy of C=C = 612 kJ mol−1
between them to form an electrophile Mean bond energy of C–C = 347 kJ mol−1
and draw a mechanism for the reaction Mean bond energy of C–H = 413 kJ mol−1
of this electrophile with benzene. (3)
b) Identify a reagent for step 2. Name the Enthalpy change of atomisation of
type of mechanism involved. (2) H2(g) = 218 kJ mol−1 (Hint: per mol of atoms
c) Identify a reagent for step 3. Name the formed.)
type of reaction involved. (2)
Enthalpy change of atomisation of
d) Discuss the stereochemistry of the
C(s) = 715 kJ mol−1
compounds formed in steps 2 and 3. (6)
a) Use the data provided to calculate the
The drug L-dopa is used in the treatment of
8 enthalpy change of formation of gaseous
Parkinson’s disease. Its structural formula is: Kekulé benzene. (7)
HO CH2
b) The experimental value for the enthalpy
COOH
CH
change of formation of gaseous benzene is
+82 kJ mol−1.
HO NH2 Calculate the difference between this figure
and the value you calculated in part (a) and
a) Write the structural formulae of the sodium give a reason for the difference. (3)
salts formed when L-dopa reacts with: 10 a)* The pKa values for ethanol and phenol are
i) excess sodium hydroxide solution (1) 16.0 and 10.0, respectively.
ii) excess sodium carbonate solution. (1) Compare the acid strengths of the two
b) Predict the structural formula of the compounds and explain why there is
organic ion produced when L-dopa such a large difference. (6)
reacts with excess dilute hydrochloric b) Both ethanol and phenol react with
acid. (1) ethanoyl chloride to form esters.
Draw and name the two esters. (2)
540
18.1 Arenes – benzene compounds
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Amines, amides, amino acids and
18.2 proteins
18.2.1 The structures and names
of amines
Amines are nitrogen compounds in which one or more of the hydrogen
atoms in ammonia, NH3, has been replaced by an alkyl or an aryl group.
The number of these groups determines whether the compound is a primary
amine, a secondary amine or a tertiary amine. If one H atom in ammonia
is replaced by an alkyl or aryl group, the compound is a primary amine. If
two H atoms in ammonia are replaced, the compound is a secondary amine,
and if all three H atoms in ammonia are replaced, the compound is a tertiary
amine.
The amine group is present in amino acids. As a result of this, the amine
group plays an important part in metabolism and is part of the structure
of many medicinal drugs (Figure 18.2.1).
Chemists have two systems for naming amines.
Figure 18.2.1 The active constituent
of asthma inhalers is salbutamol, which
contains the amine functional group.
Tip
Note that the terms ‘primary’, ‘secondary’ and ‘tertiary’ do not have the same meaning
with amines as they do with halogenoalkanes or alcohols. For halogenoalkanes or
alcohols, the term depends on the number of alkyl or aryl groups attached to the
carbon bearing the halogen or OH group.
Simple amines
Simple amines are treated as a combination of the alkyl or aryl group
followed by the ending -amine. So, CH3CH2CH2CH2NH2 is butylamine,
C6H5NH2 is phenylamine and CH3CH2NHCH3 is ethylmethylamine. The
prefixes di- and tri- are used when there are two or three of the same alkyl
or aryl group (Figure 18.2.2).
H CH3 CH3
CH3 N CH3 N CH3 N
H H CH3
methylamine dimethylamine trimethylamine
(a primary amine) (a secondary amine) (a tertiary amine)
Figure 18.2.2 The structures and names of primary, secondary and tertiary amines
containing the methyl group.
18.2.1 The structures and names of amines 541
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More complex amines
The prefix ‘amino’ is used in compounds that have a second functional
group, such as amino acids. So the systematic name for H2NCH2COOH is
aminoethanoic acid. The prefix ‘diamino’ is used for compounds containing
two amino groups, such as 1,2-diaminoethane, H2NCH2CH2NH2.
Test yourself
1 Draw the structures of:
a) diethylamine
b) ethylmethylpropylamine
c) 1,6-diaminohexane
d) 1,2-diaminopentane, which contributes to the smell of rotting flesh
and has the common name cadaverine
e) 1-phenyl-2-aminopropane, an amphetamine that is an addictive
stimulant.
2 Salbutamol is the active ingredient in asthma inhalers. Its structure is
shown below.
H
HOCH2 CHOH N
CH2 C(CH3)3
HO
a) Is its amine group primary, secondary or tertiary?
b) What other functional groups does salbutamol contain?
3 a) Draw the structures and name all the primary amine isomers of
C4H9NH2.
b) Draw the structures of all the secondary and tertiary amines that
are isomers of C4H9NH2.
4 Classify the halogenoalkanes, alcohols and amines below as primary,
secondary or tertiary.
a) a) b) b)
CH3 CH
CH3 CH3
CH CH3 CH3 CH
CH3 CH3
CH CH3
OH OH NH2 NH2
c) c) d) d)
CH3 CH3
CH3 CH3
CH3 C3
CH CH
C3 CH3
CH3 N3
CH CH
N3 CH3
CI CI
e) e) f) f)
CH3 CH3 CH3 CH3
CH3 C3
CH NHCH
C 3NHCH3 CH3 C3
CH CH
C 2OHCH2OH
CH3 CH3 CH3 CH3
542 18.2 Amines, amides, amino acids and proteins
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18.2.2 The structures and names of Tip
amides Amides were once called acid amides
to reflect their relationship to acids.
Amides are nitrogen compounds derived from carboxylic acids in which an Take care not to confuse the amine
–NH2 group replaces the –OH group. The chemistry of amides is important functional group, which has no C=O
because amide groups link up the monomers in proteins and in synthetic bond, with the amide functional group,
polymers such as nylon and Kevlar ®. The link is also found in many medicinal which does contain C=O.
drugs (Figure 18.2.3).
Amides have the general structure R C NH2
O
and the amide functional group is C N
O H
Amides are named using the suffix -amide after a stem that indicates
the number of carbon atoms in the molecule including that in the C=O
group (Figure 18.2.4). Figure 18.2.3 Paracetamol molecules
O O contain the amide group.
CH3 C CH3 CH2 C
NH2 N CH2CH3
H
ethanamide N-ethylpropanamide
Figure 18.2.4 The structures and names of amides. Note that N-ethylpropanamide has
an ethyl group substituted for one of the hydrogen atoms of the −NH2 group. The prefix N
indicates this and should be included in the name.
Test yourself
5 Draw the structures of:
a) butanamide b) N-methylpentanamide
c) hexanediamide d) N-phenylethanamide.
18.2.3 The properties and reactions
of amines
The physical and chemical properties of the simplest amines are similar to those
of ammonia. So, methylamine and ethylamine are gases at room temperature
and they smell like ammonia, though with a fishy character (Figure 18.2.5).
Propylamine and butylamine are liquids at room temperature.
Like ammonia, alkyl amines with short hydrocarbon chains dissolve readily
in water because they can hydrogen bond with it. Phenylamine, with its large Figure 18.2.5 The smell of fish is partly
non-polar benzene ring, is only slightly soluble in water. due to ethylamine.
18.2.3 The properties and reactions of amines 543
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Amines as bases
Primary amines, like ammonia, can act as Brønsted–Lowry bases. The lone
pair of electrons on the nitrogen atom of ammonia and amines is a proton
(H+ ion) acceptor.
Reaction with water
Ammonia acts as a Brønsted–Lowry base and a small proportion of the
dissolved molecules react with water to form a weakly alkaline solution
containing hydroxide ions.
NH3(aq) + H2O(l) ⇋ NH4+(aq) + OH−(aq)
Butylamine and other amines act as Brønsted–Lowry bases in a similar
manner by removing H+ ions from water molecules to form an alkaline
solution containing hydroxide ions (Figure 18.2.6).
Figure 18.2.6 The reaction of butylamine H H
with water. + –
C4H9 N + H O H C4H9 N H + O H
H H
Tip C4H9NH2 (aq) + H2O(I) C4H 9NH3+(aq) + OH– (aq)
butylamine butylammonium ion
Note that the shorthand C4H9 used
here to represent the carbon chain in
As with ammonia, the reaction of amines with water is reversible so alkyl
butylamine can also represent several
amines are also weak bases, although stronger than ammonia. This is because
other arrangements of the carbon
the alkyl group is electron releasing and increases the electron density on the
chain.
lone pair on the nitrogen. This effect makes the lone pair more attractive to
protons than the lone pair on the nitrogen in ammonia. The equilibrium in
Figure 18.2.6 lies further to the right than the equilibrium involving ammonia.
By contrast, phenylamine is a much weaker base than ammonia because
the lone pair in phenylamine is delocalised into the π cloud of the benzene
H
ring (Figure 18.2.7) and is less attractive to protons than the lone pair in
N
ammonia. Therefore, the equilibrium for the reaction of phenylamine with
H water lies further to the left than that for ammonia.
C6H5NH2(l) + H2O(l) ⇋ C6H5NH3+(aq) + OH−(aq)
Figure 18.2.7 Delocalisation of the lone
pair into the π cloud in phenylamine.
Reaction with acids – formation of salts
Amines react even more readily with acids than they do with water. The
lone pair on the nitrogen atom rapidly accepts an H+ ion from the acid to
form a substituted ammonium salt.
Tip C4H9NH2(g) + HCl(g) → C4H9NH3+Cl−(s)
butylamine butylammonium chloride
The pKa values of their conjugate acids
(see Section 12.2) give a measure When the vapour of gaseous amines such as ethylamine reacts with hydrogen
of the strength of ammonia (pKa = chloride gas, the product, ethylammonium chloride, forms as a white smoke.
9.25), butylamine (pKa = 10.61) and The smoke settles as a white solid (Figure 18.2.8).
phenylamine (pKa = 4.62) as bases.
CH3CH2NH2(g) + HCl(g) → CH3CH2NH3+Cl−(s)
A higher pKa value corresponds to a
ethylamine ethylammonium chloride
stronger base.
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This reaction is very similar to that of ammonia with hydrogen chloride to
form ammonium chloride. glass rod dipped
in conc. HCI
NH3(g) + HCl(g) → NH4+Cl−(s)
Phenylamine, C6H5NH2, is only slightly soluble in water, but it dissolves
in concentrated hydrochloric acid very easily. This is because it reacts with
H+ ions in the acid to form phenylammonium ions which are soluble in the
aqueous mixture.
C6H5NH2(l) + H+(aq) → C6H5NH3+(aq)
If a strong base, such as sodium hydroxide, is added to the aqueous
phenylammonium ions, H+ ions are removed from the phenylammonium white smoke
ions and yellow, oily phenylamine reforms.
C6H5NH3+(aq) + OH−(aq) → C6H5NH2(l) + H2O(l)
phenylamine
Test yourself
concentrated solution
6 Methylamine, like ammonia, mixes with and dissolves in water of ethylamine
whatever proportions of the two are mixed together. Why is this? Figure 18.2.8 The vapours from
7 a) Ethane (boiling temperature −89 °C) and methylamine (boiling ethylamine solution and concentrated
temperature −6 °C) have very similar molar masses, but very hydrochloric acid react to form a white
different boiling temperatures. Why is this? smoke of ethylammonium chloride.
b) Consider the boiling temperatures of methylamine (−6 °C),
dimethylamine (7 °C) and trimethylamine (4 °C). Why do you think
the boiling temperature of trimethylamine, (CH3)3N, is lower than
that of dimethylamine?
8 a) Write equations for the reactions of cyclohexylamine, C6H11NH3,
and phenylamine with water.
b) Explain why cyclohexylamine is a stronger base than phenylamine.
9 a) Write an equation to show the formation of a salt when
propylamine vapour reacts with hydrogen bromide gas.
b) E xplain why the reactants are both gases, but the product is
a solid.
10
Write equations for the following reactions and name the products:
a) methylamine with concentrated sulfuric acid
b) dimethylamine with concentrated sulfuric acid.
Amines as ligands
When ammonia and amine act as bases and accept a proton, they do so by
Figure 18.2.9 The result of adding
donating a lone pair of electrons to the proton. Ammonia and amines can
butylamine to a solution of copper ions.
also donate a lone pair of electrons to transition metal ions and act as ligands
The hydrated copper(ii) ions give the light
(Section 15.6).
blue colour, while the dark blue colour is
When butylamine is added to aqueous copper(ii) sulfate solution, a deep blue due to the formation of a complex between
solution is formed (Figure 18.2.9). Four butylamine molecules replace four the copper and the butylamine.
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water ligands and a deep blue complex is formed, similar to the complex
Key term formed by ammonia.
An ammine is a complex in which one 4C4H9NH2(aq) + [Cu(H2O)6]2+(aq)
or more ammonia molecules are co- → [Cu(C4H9NH2)4(H2O)2]2+(aq)+ 4H2O(l)
ordinately bonded to a metal ion.
Diamines such as 1,2-diaminoethane, H2NCH2CH2NH2, donate two lone
pairs and are called bidentate ligands (see Section 15.8).
Tip
Note that ammonia, NH3, in complexes is described as ‘ammine’, whereas the – NH2
group in organic compounds such as C4H9NH2 is described as ‘amine’.
Amines as nucleophiles
Reaction with halogenoalkanes
Amines are nucleophiles as well as bases and ligands, just like ammonia.
As nucleophiles, their lone pair of electrons is attracted to any positive
ion or positive centre in a molecule.
So, amines react with the δ+ carbon atoms in the C–Hal bond of
halogenoalkanes in a nucleophilic substitution reaction. The protonated
amine formed in the first step then loses a proton to form the secondary
amine, butylmethylamine (Figure 18.2.10).
H H
c+ c– +
H C Br C4H9 N CH3 + Br–
Tip H H
In these reactions, the loss of H+ or C4H9NH2 butylmethylammonium bromide
the formation of HBr is shown in the
equations. In the presence of amines H H
that are bases, these acids form + +
salts, but to avoid complication in the C4H9 N CH3 C4H9 N CH3 + H
equations the salts are not shown here.
H butylmethylamine
Figure 18.2.10 The reaction of butylamine with bromomethane to form the secondary
amine butylmethylamine.
As with the reaction of ammonia with halogenoalkanes (Section 18.2.4),
further reaction is possible.
The lone pair on the nitrogen atom of the secondary amine product is more
reactive than the lone pair on the primary amine reagent because of the
inductive effect of the extra alkyl group. So the secondary amine can also
react with the halogenoalkane in a reaction which forms a tertiary amine
(Figure 18.2.11).
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H CH3
+
C4H9 N CH3 + CH3Br C4H9 N CH3 + Br–
CH3 CH3
+
C4H9 N CH3 + Br– C4 H9 N CH3 + HBr
H butyldimethylamine
Figure 18.2.11 Formation of the tertiary amine butyldimethylamine.
The tertiary amine similarly can react further to form a quaternary
ammonium salt (Figure 18.2.12). Key term
CH3 CH3 Quaternary ammonium salt is
+ an ammonium salt where all four
C4H 9 N CH3 + CH3 Br C4H9 N CH3 + Br–
hydrogens are replaced by alkyl or aryl
CH3 groups.
butyltrimethylammonium bromide
Figure 18.2.12 Formation of the quaternary ammonium salt butyltrimethylammonium
bromide.
It is possible to limit further reaction by using an excess of the primary amine
so that there is a much greater chance of the primary amine rather than the
Tip
secondary amine acting as nucleophile with the halogenoalkane molecules. Quaternary ammonium salts where two
of the alkyl groups are long chains,
If an excess of the halogenoalkane is used, the quaternary ammonium salt is
such as [(CH3(CH2)17]2N(CH3)2+Cl−, are
the main product.
used in fabric softeners.
Reaction with acyl chlorides O
Amines also react as nucleophiles with the δ+ carbon atoms in the C Cl
group of acyl chlorides such as ethanoyl chloride (Figure 18.2.13). The
reaction forms an N-substituted amide (see also Section 17.3.4).
c– Figure 18.2.13 The reaction of butylamine
O
c+ with ethanoyl chloride.
CH3 C + CH3 CH2CH2CH2NH2
CI O
CH3 C
NCH2CH2CH2CH3 + HCI
H
N-butyl ethanamide
A reaction of this type is involved in the manufacture of paracetamol; this is
discussed in the activity that follows.
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Activity
Paracetamol – an alternative to aspirin
Aspirin and paracetamol (Figure 18.2.14) are by far the largest selling pain relievers HO O
available without a doctor’s prescription. However, aspirin is not without hazards. C
Every year, about 200 people die of aspirin poisoning due to deliberate or accidental O CH3
overdose. A large proportion of these deaths are of young children who die accidentally C
after eating the tablets. Aspirin is also recognised as a cause of internal bleeding and
gastric ulcers. O
Paracetamol has been produced and marketed as a safer alternative to aspirin. It is a aspirin
good analgesic (pain reliever) without the harmful side-effects of aspirin, as long as the
OH
recommended doses are not exceeded.
Compared to other non-prescription pain relievers, paracetamol is much more toxic
when an overdose is taken because it can cause potentially fatal liver damage.
The compound from which paracetamol is produced and the physiologically active
compound that paracetamol produces in the body is 4-aminophenol. Unfortunately, this N CH3
is toxic, so its harmful effect is reduced by conversion to its ethanoyl derivative. H C
Paracetamol can be produced by reacting 4-aminophenol with ethanoyl chloride. In O
industry, however, ethanoic anhydride, (CH3CO)2O, is used in preference to ethanoyl
paracetamol
chloride because it is cheaper and less vigorous in its reactions.
Figure 18.2.14 The structural formulae of
aspirin and paracetamol.
1 Draw the structure of 4-aminophenol.
2 a) Write an equation for the reaction of 4-aminophenol with ethanoyl chloride to
produce paracetamol.
b) Why does ethanoyl chloride react in this way with 4-aminophenol?
3 When ethanoyl chloride reacts with 4-aminophenol, the –OH group in 4-aminophenol
is susceptible to attack as well as the –NH2 group.
a) Why is the –OH group in 4-aminophenol also susceptible to reaction with
ethanoyl chloride?
b) Write an equation for the reaction of ethanoyl chloride with the – OH group in
4-aminophenol.
4 Fortunately, the –NH2 group in 4-aminophenol is more reactive than the –OH group.
So, in industry the reaction conditions can be carefully chosen so that only the –NH2
group is ethanoylated using ethanoic anhydride. Write an equation for the reaction
of 4-aminophenol with ethanoic anhydride to produce paracetamol.
5 Aspirin, like paracetamol, is manufactured by ethanoylation using ethanoic anhydride.
a) What do you understand by the term ‘ethanoylation’?
b) Draw the structure of the compound that is ethanoylated to produce aspirin.
6 a) What is the main benefit of aspirin tablets?
b) Summarise the risks posed by aspirin tablets.
c) What are the advantages of paracetamol over aspirin?
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Test yourself
11 Write the formula of the cobalt-containing ions formed by the
reaction of Co(H2O)62+ ions with an excess of 1,2-diaminoethane.
12 Write equations for the following reactions:
a) chloromethane with propylamine
b) propanoyl chloride with ethylamine.
13 a) Write an equation and draw a mechanism for the reaction
of chloromethane with an excess of ethylamine.
b) Write equations for any further reactions that occur if an
excess of chloromethane reacts with ethylamine.
18.2.4 The preparation of amines
Preparing aliphatic amines
From halogenoalkanes
Aliphatic amines can be prepared by heating the corresponding
halogenoalkanes in a sealed flask with excess ammonia in ethanol.
Tip
Use of excess ammonia limits the
During the first step of the reaction ammonia acts as a nucleophile and then
chance of further substitution, so a
in the second step it acts as a base to remove a proton from the salt initially
primary amine is the major product.
formed (Figure 18.2.15).
H H H H H H H H Br– H H H H
heat H H
δ+ δ– with excess + excess
H C C C C Br H C C C C N H H C C C C N + NH4Br
conc. NH3 conc. NH3
H H H H in ethanol H H H H H H H H H H
N
H NH3
H
H
Figure 18.2.15 The preparation of butylamine from 1-bromobutane and excess
concentrated ammonia in ethanol.
From nitriles
Reduction of nitriles produces primary amines. Unlike the nucleophilic
substitution reaction of ammonia with halogenoalkanes considered above,
this reaction produces a pure product as no further reaction can occur.
Reduction can be achieved in two ways:
a) Hydrogenation using hydrogen gas in the presence of a nickel catalyst.
CH3CH2CH2C≡N + 2H2 → CH3CH2CH2CH2NH2
Tip
butanenitrile butylamine
The overall equation is complex, so
b) Reduction using LiAlH4 in ethoxyethane, followed by dilute acid.
simplified equations of this sort, using
CH3CH2CH2C≡N + 4[H] → CH3CH2CH2CH2NH2 [H] to represent the reducing agent, are
accepted in A Level examinations.
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Preparing aromatic amines
The usual laboratory method for introducing an amine group into an
aromatic compound is a two-step process – first nitration to make a nitro
compound and then reduction (Figure 18.2.16). The reduction of the
aromatic nitro-compound is achieved by boiling under reflux with tin
and concentrated hydrochloric acid (Figure 18.2.17). The aromatic amine
dissolves in excess concentrated hydrochloric acid, forming a salt. The
free amine can be liberated from the solution by adding sodium hydroxide
solution. It is then separated from the mixture by steam distillation.
Figure 18.2.16 The two-step preparation NO2 NH2
conc. HNO3 Sn metal
of phenylamine from benzene.
conc. H2SO4 + conc. HCl
50–60 °C heat
benzene nitrobenzene phenylamine
Equations for the reduction of nitrocompounds are usually simplified by
using [H] to represent the reducing agent.
C6H5NO2 + 6[H] → C6H5NH2 + 2H2O
Figure 18.2.17 Reducing nitrobenzene to
phenylamine by refluxing with tin and hot
concentrated hydrochloric acid.
water out
concentrated
hydrochloric acid
water in
Tip
Reduction of nitrobenzene to
phenylamine is an important reaction
in industry notably in the preparation cold water while adding
of dyes. Tin is an expensive metal so in nitrobenzene the acid, then boiling to
tin complete the reaction
industry the cheaper metal iron is used.
18.2.5 The preparation of amides
Amides form rapidly at room temperature when acyl chlorides, such as
Tip
ethanoyl chloride, react with ammonia or with amines. For example, when
The reaction of a carboxylic acid with ethanoyl chloride is carefully added to a concentrated aqueous solution of
ammonia or an amine forms a salt, ammonia, a vigorous reaction takes place producing fumes of hydrogen
so in the laboratory, reactions using chloride and ammonium chloride plus a residue of ethanamide.
acyl chlorides are preferred for making
amides. However, industrial methods CH3COCl(l) + NH3(aq) → CH3CONH2(s) + HCl(g)
to make polyamides use acids (see ethanamide
Section 18.2.9) rather than acyl HCl(g) + NH3(g) → NH4Cl(s)
chlorides because of the difficulty of ammonium chloride
storing acyl chlorides and using them
on a large scale (see Section 17.3.4). The preparation of paracetamol discussed in the Activity in Section 18.2.3
involves the synthesis of an N-substituted amide.
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Test yourself
14 1-Chloropropane reacts with an excess of concentrated ammonia
dissolved in ethanol.
a) Name and describe the mechanism, using appropriate curly
arrows.
b) Explain why the ammonia is dissolved in ethanol rather than in
water. (Hint: ammonia reacts with water to produce an alkaline
solution containing OH− ions.)
15 Describe two methods to prepare ethylamine from a
halogenoalkane. Give one disadvantage of each method.
16 Give the laboratory reagents for the reduction of 1,3-dinitrobenzene
to benzene-1,3-diamine. Write an equation for the reaction using [H]
to represent the reducing agent.
18.2.6 Amino acids and proteins
hair is made
of protein
the surface of the skin is the enzyme
protein amylase, found in
saliva, is a protein
the hormone
insulin is a protein the red haemoglobin
made by the in blood cells is a
pancreas protein
the fibres of nerve
cells are surrounded
by protein
muscle fibres are
made of protein
bones consist of minerals
embedded in collagen, the tendons which join
which is a protein muscle to bone
contain protein
toenails and
fingernails are made
of protein
Figure 18.2.18 Proteins in the human body.
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Amino acids are compounds that contain two functional groups – the amino
Key terms group, –NH2, and the carboxylic acid group, –COOH. Amino acids are the
Amino acids are compounds that
monomers that make up proteins, the naturally occurring polymers which
contain both an amino group and a
comprise 15% of the human body. There are many different protein molecules
carboxylic acid group.
in our bodies (Figure 18.2.18). Muscles, hair, enzymes and hormones all
consist of proteins. Some proteins contain thousands of amino acid units.
2-amino acids are amino acids with
the formula RCH(NH2)COOH. The amino
group is attached to the carbon next to
Names and formulae
the carboxylic acid group. About 20 different amino acids are found widely in naturally-occurring
proteins. The names and formulae of six of these are shown in Figure 18.2.19.
The simplest amino acid is glycine, H2N–CH2–COOH.
H H H
H2N C COOH H2N C COOH H2N C COOH
H CH3 CH2 SH
glycine (gly) alanine (ala) cysteine (cys)
H H H
H2N C COOH H2N C COOH H2N C COOH
CH2 CH2OH CH2
serine (ser)
CH2COOH
glutamic acid (glu)
phenylalanine (phe)
Figure 18.2.19 Six of the amino acids that occur in proteins.
Notice in Figure 18.2.19 that all six formulae have the amino group attached
to the carbon atom next to the carboxylic acid group, carbon number 2 in
the chain. This is the case with all the amino acids that occur naturally. This
carbon number 2 is sometimes described as the alpha (α) carbon atom. So all
the amino acids in proteins are 2-amino acids (or α-amino acids) and their
general formula can be written as RCH(NH 2)COOH.
R stands for the side groups in different amino acids (Table 18.2.1). The
common names and R side groups of several other amino acids are shown
on a data sheet headed ‘The common names and R side groups of some
amino acids’, which you can access online at [Link]/
EdexcelChemistry.
Table 18.2.1 The R side groups in some amino acids.
Common name Abbreviated name R side group
Glycine gly H–
Alanine ala CH3–
Cysteine cys HS–CH2–
Phenylalanine phe C6H5–CH2–
Aspartic acid asp HOOC–CH2–
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Many of the natural amino acids have complex structures, so it is simpler
and more convenient to use their common names rather than their
systematic names. These common names are sometimes abbreviated to a
‘three-letter code’, which is usually the first three letters in the name. So,
H2NCH2COOH is normally called ‘glycine’ rather than 2-aminoethanoic
acid and its abbreviated name is ‘gly’.
Amino acid structures Tip
All the amino acids that occur in proteins, except glycine, have a central
carbon atom attached to four different groups. This is shown clearly in Several systems of naming chiral
Figure 18.2.19. So, except for glycine, all these amino acids have chiral compounds are in use, including the
molecules that can exist as mirror images (Section 17.1.3). The mirror-image (+)/(−) system and the D/L system. The
forms of the amino acid alanine are shown in Figure 18.2.20. All amino (+)/(−) system depends on the effect
acids found in proteins occur in the L-configuration. that optical isomers have on plane-
polarised light. The D/L system, on
the other hand, is based on the actual
CH3 CH3 stereochemical structure at the chiral
C C
centre. Based on the D/L system, all
COOH HOOC naturally occurring amino acids are the
H2N NH2
H H L isomers, but some of these L amino
acids are (+)isomers and others are
(−)isomers. None exist naturally as
racemic mixtures.
mirror
Figure 18.2.20 The mirror-image forms of the amino acid alanine. The mirror images are
chiral and cannot be superimposed.
As a result of their chirality, the separate (+) and (−) isomers of all naturally
occurring amino acids, except glycine, can rotate the plane of plane-
polarised light.
Test yourself
17 State the systematic name for each amino acid:
a) alanine
b) phenylalanine
c) serine.
18 A dipeptide contains two amino acids linked together. How many
different dipeptides can be formed from the 20 naturally occurring
amino acids?
19 Explain why the amino acid glycine is not chiral.
20 How could you distinguish between samples of the two mirror-image
forms of an amino acid by experiment?
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18.2.7 The acid–base properties of
amino acids
As amino acids carry an amino group, –NH2, and a carboxylic acid group,
–COOH, they show both the basic properties of primary amines and the
acidic properties of carboxylic acids.
The formation of zwitterions
In aqueous solution, carboxylic acid groups ionise producing hydrogen ions,
H+(aq), whereas amino groups are basic and attract hydrogen ions. As a
result of this, amino acids form ions in aqueous solution because the amino
groups accept hydrogen ions (protons) and the acid groups give them away
(Figure 18.2.21). The ions formed are, however, unusual in that they have
both positive and negative charges. Chemists call them zwitterions, from a
German word ‘zwitter’ meaning ‘hybrid or hermaphrodite’.
Figure 18.2.21 Glycine forming a H H O H H O
zwitterion. + +
H N C C H N C C
proton from another
carboxylic acid group H H O H H H O–
the proton is taken
up by another zwitterion
amine group
An amino acid can only form zwitterions at a particular pH. If the pH is too
Key terms high, the solution is too alkaline and in these conditions OH− ions remove
H+ ions from the zwitterions, forming negative ions (Figure 18.2.22 right).
A zwitterion is an ion with both a
On the other hand, if the pH is too low, the solution is too acidic. In this
positive and a negative charge.
case, H+ ions react with the zwitterions, producing positive ions (Figure
The isoelectric point of an amino acid 18.2.22 left). Amino acids can therefore exist in three forms depending on
is the pH value at which it exists as a the pH: a cation form, a zwitterion and an anion form. However, at one
zwitterion. particular pH, molecules of the amino acid will be in the zwitterion form
(Figure 18.2.22 centre) and this pH value is called the isoelectric point.
Notice from Figure 18.2.22 that the net charge on an amino acid molecule
varies with the pH. The net charge is positive in acid solutions and negative
in alkaline solutions. At the isoelectric point, the positive and negative
charges balance and the net charge on the zwitterion is zero.
Figure 18.2.22 The ions formed by an R R R
amino acid at different pH values. + OH– + OH–
H3N C COOH H3N C CO2– H2N C CO2–
H+ H+
H H H
(aq) (aq) (aq)
At a lower, more acidic At the isoelectric point, At a higher, more alkaline
pH, a positive ion forms the zwitterion forms pH, a negative ion forms
All amino acids form zwitterions along the lines described above, but their
isoelectric points may differ because of the different character of their R
groups. In fact, some amino acids, like glutamic acid and aspartic acid, have
two –COOH groups and others have two –NH 2 groups, which influences
their isoelectric point significantly.
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The movement of H+ ions from the –COOH group of an amino acid to its
–NH2 group occurs in solution before the solid amino acid crystallises out.
Tip
This means that amino acids also exist as zwitterions in the solid state. This Note that, although an amino acid
ionic character of amino acids accounts for their high solubility in water and is usually written as a molecule
their high melting temperatures. RCH(NH2)COOH, it exists naturally as
an ionic solid containing the zwitterion
Test yourself RCH(NH3+)COO−.
21 The relative molecular masses of butylamine, CH3(CH2)3NH2,
propanoic acid, CH3CH2COOH, and glycine, H2NCH2COOH are very
similar. But glycine (melting temperature 262 °C) is a solid at room
temperature, whereas butylamine (melting temperature −49 °C) and
propanoic acid (melting temperature −21 °C) are liquids. Why is this?
22 a) Write equations to show the reactions of alanine with:
i) dilute hydrochloric acid
ii) aqueous sodium hydroxide.
b) How do the products from alanine of these two reactions differ
from the zwitterions of alanine at its isoelectric point?
23 Why do zwitterions of amino acids exist just as readily in the solid
state as they do in aqueous solution?
18.2.8 From amino acids to peptides
and proteins
Peptides are compounds made by linking amino acids together in chains.
The simplest example is a dipeptide with just two amino acids linked together
by a peptide bond. Figure 18.2.23 shows the formation of a peptide bond
between alanine and glycine to form the dipeptide, ‘ala–gly’.
CH3 O H H O Figure 18.2.23 The formation of a peptide
bond between two amino acids.
H2N C C + N C C
H OH H H OH
ala gly
CH3 O H O
H2N C C N C C + H2O
Key terms
H H H OH
Peptides are chains of amino acids
peptide bond
linked by peptide bonds.
ala–gly
O H
A peptide bond is an amide link formed
when the −NH2 group of one amino
For chemists, the peptide bond, C N , is simply an example of the amide
acid reacts with the − COOH group of
bond (Section 18.2.2). However, the tradition in biochemistry is to call it a
another.
‘peptide bond’ or a peptide link.
A condensation reaction is a reaction in
Notice in Figure 18.2.23 that when a peptide bond forms between two
which molecules join together by splitting
amino acid molecules, a molecule of water is eliminated at the same time.
off a small molecule such as water.
This is an example of a condensation reaction.
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Further condensation reactions can occur between the dipeptide and other
Key term amino acid molecules to produce polypeptides and eventually proteins. This
is what happens when proteins are synthesised from amino acids in our
In condensation polymerisation,
bodies. The overall process is an example of condensation polymerisation
a polymer is formed by a series of
(Section 17.3.6).
condensation reactions.
Polypeptides are long-chain peptides. There is no clear dividing line
between peptides and polypeptides or between polypeptides and proteins.
CH3 O H O Some chemists do, however, make a distinction between polypeptides and
H 2N C C N C C the longer amino acid chains in proteins. They restrict the definition of
polypeptides to chains with 10 to 50 or so amino acids.
H H H OH
+ H2O + 2H+ The hydrolysis of peptides and proteins
reflux with Digestive enzymes in the stomach and small intestine catalyse the hydrolysis
conc. HCl
of peptide bonds, splitting proteins into polypeptides and then polypeptides
CH3 O H O into amino acids. Chemists can achieve the same result and hydrolyse the
+ + peptide bond by treating proteins and peptides with suitable enzymes, or
H 3N C C + H3N C C
simply by heating in acidic or alkaline solution.
H OH H OH
When proteins and peptides are hydrolysed by refluxing with concentrated
add water hydrochloric acid, the product contains the cation forms of the α-amino
+ 2H2O
acids. These are converted to the α-amino acids on dilution with water
CH3 O H O (Figure 18.2.24).
H2N C C + H 2N C C + 2H3O+ After hydrolysis, the mixture of amino acids produced can be separated and
identified by chromatography (Chapter 19).
H OH H OH
alanine glycine
Figure 18.2.24 Hydrolysing a peptide with
Test yourself
acid to produce α-amino acids. 24 Draw the structures of the two dipeptides that can be produced
from serine and phenylalanine.
25 Show that splitting a dipeptide into two amino acids is an example
of hydrolysis.
26 a) Identify the functional groups in the sweetener, aspartame.
O H H O CH2 O
C C C C N C C
HO H NH2 H H OCH3
b) How does aspartame differ from a dipeptide?
c) S
uggest a reason why aspartame cannot be used to sweeten
food that will be cooked.
d) W
hy do you think that soft drinks sweetened with aspartame
carry a warning for people with the genetic disorder that means
that they must not eat phenylalanine?
556 18.2 Amines, amides, amino acids and proteins
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18.2.9 Condensation polymerisation Key terms
Peptides are condensation polymers. When amino acids react, water molecules
Polyesters are polymers with ester links
are eliminated and a peptide link is formed. There are two other important
between monomer units.
classes of condensation polymers: polyesters formed from dicarboxylic acids
and diols (see Section 17.3.6) and polyamides formed from dicarboxylic Polyamides are polymers with amide
acids and diamines. links between monomer units.
Polyamides
Polyamides are polymers in which the monomers are linked by an amide bond.
This is exactly the same as the amide bond in proteins, in which it is usually
called the peptide bond (Figure 18.2.23). So, proteins and polypeptides are
naturally occurring polyamides.
From your studies earlier in this topic, you will know that polypeptides and
proteins are synthesised in living things by condensation reactions between
amino acids. In these reactions, the amino group, –NH 2, of one amino acid
reacts with the carboxylic acid group, –COOH, of another amino acid to
split out water and form an amide link (Figure 18.2.25). This process is then
repeated time after time to produce a polymer (protein) with tens, hundreds
or, in some cases, thousands of units.
The first synthetic and commercially important polyamides were various
forms of nylon. These were not, however, produced from amino acids.
Instead, they were formed by condensation polymerisation between diamines
and dicarboxylic acids. One of the commonest forms of nylon is nylon-6,6.
This is made by a condensation reaction between 1,6-diaminohexane and
hexanedioic acid (Figure 18.2.25). The product is named nylon-6,6 because
both monomers contain six carbon atoms.
O O H H O O H
C (CH2)4 C N (CH2)6 N C (CH2)4 C N (CH2)6 NH2
HO OH H H HO OH H
O O O O
C (CH2)4 C N (CH2)6 N C (CH2)4 C N (CH2)6 NH2
HO H H H
+ H2O + H2O + H2O
Figure 18.2.25 Condensation polymerisation to make nylon-6,6.
Tip
Early polymer chemists found it easier to synthesise polyamides using separate
dicarboxylic acid and diamine molecules rather than have the carboxylic acid
functional group and the amine functional group on the same molecule, as nature
does in an amino acid.
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Nylon-6,6 can be produced more readily in the laboratory using hexanedioyl
dichloride in place of the less reactive hexanedioic acid. A solution of
hexanedioyl dichloride in cyclohexane is floated on an aqueous solution of
1,6-diaminohexane. Nylon-6,6 forms as a skin at the interface and can be
pulled out as fast as it is produced forming a long thread – the ‘nylon rope’
(Figure 18.2.26).
In this reaction hydrogen chloride molecules are eliminated in the
condensation reaction (Figure 18.2.27).
O O H H
n Cl C(CH2)4C Cl + nH N(CH2)6N H
Figure 18.2.26 The nylon rope trick. O O H H
C(CH2)4C N(CH2)6N + 2nHCl
n
Figure 18.2.27 The reaction used to make nylon-6,6 in the laboratory.
Although nylon is similar in structure to wool and silk, it does not have the
softness of the natural fibres. It is, however, much harder wearing and one
of its earliest uses was as a substitute for silk in the manufacture of ladies’
stockings (Figure 18.2.28).
Apart from their obvious use in stockings and tights, nylon fibres are used in
various forms of clothing. In fact, about 75% of the UK nylon consumption
goes on clothing, but its uses are many and varied. Nylon is used to make
nylon ropes that don’t rot, machine bearings that don’t wear out, and it is
mixed with wool to make durable carpets.
Nylon is the collective name for polymers with aliphatic hydrocarbon
sections linked by amide bonds. They are aliphatic polyamides in which the
polar amide bonds are fixed and inflexible, but the non-polar hydrocarbon
sections are free to flex, rotate and twist. So, as the hydrocarbon sections
become longer, we would expect the nylon polymers to become more
flexible with weaker bonding between the molecules and therefore also a
lower melting temperature.
This suggests that the properties of polyamides can be modified by changing
the length and nature of the hydrocarbon sections. Chemists have followed
Figure 18.2.28 The American firm Du Pont up these ideas to develop polyamides in which the hydrocarbon sections
patented nylon in February 1938. The first are aromatic rather than aliphatic. These polymeric aromatic amides
nylon stockings went on sale in the USA are described as aramids. Aramids, such as Kevlar ® (Figure 18.2.29), are
on 15 May 1940. In New York alone, four extremely strong, rigid, fire-resistant and lightweight. Much of the strength
million pairs were sold in a few hours. is due to the extensive hydrogen bonding between the chains.
558 18.2 Amines, amides, amino acids and proteins
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H
O N N
H H
O N N O
H H
O O N N
H H
O N N O
H
O
Figure 18.2.29 Polymer chains in Kevlar.
Tip
The longer the carbon chains in nylons the less the material absorbs water. Early
versions of nylon shirts were very popular as they did not need ironing, but they left
the wearer damp and sweaty as the material did not absorb perspiration as well as
natural fibres. Modern shirts using mixtures of different polymers and natural fibres
have generally solved this problem.
Test yourself
27 a) What type of polymerisation would produce the polymer with a
repeat unit like that below?
O O
C CH2 CH2 C
O O
repeat unit
b) Draw the structure of the monomer or monomers that would be
used to prepare the polymer.
28 The compound below can form a polymer.
O
C NH2
HO
a) Identify the functional groups involved in forming the polymer.
b) What type of polymerisation will the monomer undergo?
c) What other product forms during polymerisation?
d) Draw a short length of the polymer chain showing two repeat units.
29 State two similarities and two differences between the structure of
nylon-6,6 and the structure of a protein.
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Activity
Modelling and synthesising polyamides
Experiments show that nylon polymers with longer hydrocarbon sections to their
chains are more flexible than those with shorter sections. Kevlar is similar to
nylon-6,6 but with benzene rings rather than aliphatic chains linked by the amide
group. The repetition of benzene rings in its structure makes Kevlar exceptionally
strong and very inflexible compared with nylon-6,6. Because of this, it is used
extensively in tyres, brakes and clutch fittings, in ropes and cables and in protective
clothing (Figure 18.2.30).
1 Look closely at the structure of one chain of Kevlar in Figure 18.2.29.
a) Explain how Kevlar is a condensation polymer of benzene-1,4-dicarboxylic acid
and benzene-1,4-diamine.
b) Weight for weight, Kevlar is five times stronger than steel. This exceptional
strength of Kevlar is due to hydrogen bonding between the separate chains.
Use Figure 18.2.29 to explain why interchain hydrogen bonding is so strong in
Kevlar.
c) Suggest a reason why Kevlar is made from monomers with functional groups in
the 1,4 positions and not from isomers with functional groups in the 1,2 or 1,3
positions.
2 Using a molecular model kit, make one repeat unit for the structure of Kevlar Figure 18.2.30 This policeman is wearing a
and explore the flexibility of the structure. (Hint: use the Kekulé structure with bulletproof jacket made from Kevlar.
alternating double and single bonds for the benzene ring.)
Repeat the model-making and flexibility testing with one repeat unit for the
structure of nylon-6,6. Why is nylon-6,6 flexible whereas Kevlar is inflexible?
3 A condensation polymer can be prepared by mixing equal amounts of the monomers
below at room temperature.
O O
H2N(CH2)3CHCH2NH2 C(CH2)2CHCH2C
Cl Cl
CH2OH COOH
a) Draw the structure of one repeat unit of the polymer formed from the
two monomers.
b) The polymer forms even more rapidly if the reaction mixture contains sodium
carbonate. Why is this?
c) The polymer molecules obtained at room temperature can be linked to one
another (cross-linked) by a second reaction. Explain how this cross-linking can be
achieved and state the conditions needed for it to happen.
d) Explain how the choice of reaction conditions can control the extent of
polymerisation and the extent of cross-linking.
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18.2.10 Comparing addition and
condensation polymers
Formation of the polymers
Although both addition polymerisation (Section 6.2.12) and condensation
polymerisation result in the formation of long-chain organic molecules, known
as polymers, from relatively small organic molecules, known as monomers,
there are some clear differences between the two processes.
● The type of reaction involved.
As its name suggests, addition polymerisation involves only addition reactions,
whereas condensation polymerisation involves addition plus elimination. As
monomer units join together, a small molecule, usually water or hydrogen
chloride, is eliminated and splits off.
● The type of links along the polymer chain.
In addition polymers, the central chain consists of carbon atoms linked by
carbon–carbon single bonds. In condensation polymers, the central chain
consists of short aliphatic or aryl sections linked by ester groups or amide
groups.
● The type of monomer involved.
In addition polymerisation, the monomers have molecules with carbon–
carbon double bonds. In condensation polymerisation, the monomers have
molecules with at least two functional groups which may be the same or
different.
● The conditions for preparation of the polymers.
In general, addition polymerisations require an initiator together with high
temperature and high pressure, unless a catalyst is involved. In contrast,
condensation polymerisations do not require initiators and usually occur at a
much lower temperature and atmospheric pressure. Tip
Most addition polymers are not
Polymer properties biodegradeable, but a recent
These differences between addition and condensation polymerisation lead development is the use of poly(ethenol),
to considerable variations in the properties of polymers. Polymeric materials sometimes called polyvinyl alcohol.
include plastics, fibres and elastomers. As polymer science has grown, Poly(ethenol) is used to make plastic
chemists and materials scientists have learned how to develop new materials bags that dissolve in water and the
with particular properties. soluble capsules containing liquid
detergent, which slowly dissolve and
Some of the ways of modifying the properties of polymers include: release detergent as the washing cycle
● altering the average length of polymer chains progresses.
● changing the structure of the monomer to one with different side groups The repeat unit of poly(ethenol) is:
and different intermolecular forces
H H
● varying the extent of cross-linking between chains
● selecting a monomer that produces a biodegradable polymer C C
● producing a co-polymer
● adding fillers and pigments H OH
● making composites.
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Tip
Plastics are materials made of long-chain molecules that can be moulded into shapes
which are retained.
Elastomers are materials made of long-chain molecules that can be moulded into new
shapes but which spring back to their original shape when the pressure is removed.
Co-polymers are polymers made from two or more monomers, each of which could
produce a polymer.
Composites are materials made up of two or more recognisable constituents, each of
which contributes to the properties of the composite (Figure 18.2.31).
Figure 18.2.31 An electron micrograph of a glass fibre composite showing rods of
glass fibre embedded in a polyester matrix. Magnification is × 660.
562 18.2 Amines, amides, amino acids and proteins
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Chapter summary
Chapter 18.2 Amines, amides, l Amines can act as nucleophiles with the δ+ carbon
atoms in acyl chlorides: butylamine reacts with
amino acids and proteins ethanoyl chloride to form N-butyl ethanamide.
l Amines are nitrogen compounds in which the l Amines can act as ligands and donate a lone pair
hydrogen atoms in ammonia are replaced by alkyl of electrons to a transition metal ion. Butylamine
or aryl groups. If one H atom is replaced, the reacts with aqueous copper(ii) ions to form a
compound is a primary amine. If two are replaced, deep blue solution containing the complex ion
the compound is a secondary amine. If all three are [Cu(C4H9NH2)4(H2O)2]2+(aq).
replaced, the compound is a tertiary amine. l Primary aliphatic amines can be prepared by
l Amides are nitrogen compounds derived from heating halogenoalkanes with ammonia: excess
carboxylic acids in which an –NH2 group replaces ammonia limits further substitution.
the –OH group to give –CONH2. The amide l Reduction of nitriles, by catalytic hydrogenation
functional group –CONH– occurs in proteins and or using LiAlH4, produces primary amines. This
in synthetic polymers. product is purer as no further reaction can occur.
l Amines, like ammonia, are bases. The lone pair of l Aromatic amines are formed by reducing aromatic
electrons on the nitrogen atom is a proton acceptor. nitro-compounds using tin and concentrated
Ammonia and amines react with water to form hydrochloric acid.
a weakly alkaline solution containing hydroxide l Amides form when acyl chlorides react with
ions. ammonia or with amines.
l Primary aliphatic amines such as butylamine l Amino acids with the general formula
contain an electron releasing alkyl group. The RCH(NH2)COOH are called 2-amino acids.
electron density on the nitrogen lone pair is greater Except for glycine, they are chiral molecules and
than in ammonia so primary aliphatic amines are exist as mirror images. The separate isomers rotate
stronger bases than ammonia. the plane of plane-polarised monochromatic light.
l The lone pair on nitrogen in phenylamine is l Amino acids show the basic properties of primary
delocalised into the π cloud of the benzene ring amines and the acidic properties of carboxylic
so is less attractive to protons than the lone pair in acids. In aqueous solution, the amine group gains a
ammonia. Phenylamine is therefore a weaker base proton and the acid group loses a proton to form a
than ammonia. zwitterion.
l Amines react readily with acids. The lone pair l A peptide bond is an amide link, formed when
on the nitrogen accepts a proton to form a the −NH2 group of one amino acid reacts with
substituted ammonium salt such as C4H9NH3+Cl−, the −COOH group of another in a condensation
butylammonium chloride. reaction. Many amino acids combine by
l Phenylamine is only slightly soluble in water, but condensation polymerisation to form polypeptides
dissolves in concentrated hydrochloric acid to form and proteins.
phenylammonium ions C6H5NH3+ which are l The peptide bond in proteins and peptides can be
water soluble. hydrolysed using enzyme catalysts or by heating
l Amines can act as nucleophiles in nucleophilic in acidic or alkaline solution. After hydrolysis, the
substitution reactions: the lone pair on nitrogen mixture of amino acids produced can be separated
is attracted to the δ+ carbon atom in a and identified by chromatography.
halogenoalkane. The first product is a secondary l Condensation polymerisation occurs in the
amine which can react further with excess formation of polyamides from dicarboxylic acids
halogenoalkane to form a tertiary amine and then and diamines and in the formation of polyesters
a quaternary ammonium salt. from dicarboxylic acids and diols. These polymers
can be represented by repeat units.
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Exam practice questions
1 Consider the six compounds below. a) i) Draw the structure of the dipeptide,
a) CH3CH2CONH2 d) inserting the missing peptide link.(2)
NH ii) On your structure, draw a circle around
each chiral centre. (3)
b) What does the presence of a chiral centre
tell you about a compound?(3)
b) (CH3CH2)3 N e) c) Draw the structures of the products obtained
CH2 NH2
when the dipeptide is refluxed with excess
concentrated hydrochloric acid.(2)
4 Look closely at the structures of glycine and
c) H f) NH2
glutamic acid in Figure 18.2.19.
H2 N C COOH O C a) Using glycine as an example, explain the
meaning of the term ‘the isoelectric point’
NH2
CH3 of an amino acid.(3)
b) The isoelectric point of glycine is at
Identify the compounds as:
pH = 6.0. Explain why the isoelectric point
A primary amines
of glycine is not at pH = 7.(5)
B secondary amines
c) Predict how the isoelectric point of
C tertiary amines
glutamic acid compares with that of glycine.
D amides
Explain your answer.(3)
E amino acids. (6)
5 a) Ethylamine can be prepared by reaction
2 Amines such as butylamine and phenylamine
between bromoethane and ammonia.
both behave as bases.
i) Write an equation for the reaction
a) Give the meaning of the term ‘base’ and the
involved.(1)
feature of an amine molecule that causes it
ii) Name the type of reaction taking
to act as a base.(2)
place.(1)
b) Give the formula of the salt formed when
iii) Draw the structures of two other
butylamine reacts with sulfuric acid.(1)
molecular organic products that may
c) Explain why butylamine is a stronger base
be produced when bromoethane
than phenylamine.(3)
reacts with ammonia. (2)
d) Describe what you would see when an
iv) Explain why these two other organic
excess of butylamine is added with shaking
products may be formed.(3)
to an aqueous solution of copper(ii) sulfate.
b) Describe and explain what happens when
State the role of butylamine in the reaction
the apparatus and materials shown in the
and give the formula of the product.(3)
diagram are set up. (5)
3 An incomplete structure of the dipeptide
threonylisoleucine is shown in the diagram.
long glass cylindrical tube bung
H H
H2N C C COOH
H C OH H C CH3
CH3 CH2 cotton wool soaked cotton wool soaked
in ethylamine in conc. HCl
CH3
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18.2 Amines, amides, amino acids and proteins
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6 The diagram shows a series of reactions ii) State the class of amine shown by
beginning with the amine cadaverine. cadaverine and the class shown by
Cadaverine is formed when proteins decompose. piperidine. (2)
c) State with a reason whether the infrared
CH2 CH2 NH2
spectra of cadaverine and piperidine are
CH2 similar or not.(2)
CH2 CH2 NH2 7 Short sections of the molecular structures
cadaverine of two polymers, A and B, are shown in the
2HCl(aq)
diagram at the bottom of the page.
a) Draw the simplest repeat unit for each
Compound W polymer.(2)
b) Draw and name the structural formula
heat
of the monomer used to prepare
CH2 CH2 polymer A.(2)
c) i) Draw the structural formulae of the
CH2 N+H2 Cl– + Compound X two monomers that could be used to
CH2 CH2 prepare polymer B.(2)
ii) Name one of the two monomers.(1)
NaOH(aq) d) During the last decade, degradable polymers
have been developed to reduce the quantity
CH2 CH2
of plastic waste that is dumped in landfill
CH2 NH + Compound Y + Compound Z sites. State and explain why polymer B,
which is a polyester, is more likely to be
CH2 CH2
degradable than polymer A.(4)
piperidine
8 a) A compound containing carbon, hydrogen
a) i) State one characteristic physical and nitrogen contains 61.0% carbon and
property of cadaverine you would 15.3% hydrogen by mass.
expect to notice if you were provided i) Calculate its empirical formula(3)
with a sample of it.(1) ii) State what other piece of data is
ii) Give the systematic name of required to deduce its molecular
cadaverine.(1) formula.(1)
iii) Draw the structural formula of iii) If the molecular formula of the
compound W.(1) compound is the same as its empirical
iv) Write the name and formula of formula, draw and name all possible
compound X.(2) structures for the compound.(4)
v) Write the formulae of compounds b) Each of the structures you have drawn in part
Y and Z.(2) (a)(iii) can act as a base. Predict and explain
b) Amines are classed as primary, secondary the relative basic strength of the structures.(2)
and tertiary. c) In aqueous solution, tertiary amines are
i) Explain the difference in structure weaker bases than secondary amines.
between the three types of amine.(3) Explain why this is so.(3)
H CH3 H CH3 H CH3 O H H H O O H H H O
C C C C C C O O C C C C C O O C C C C C
CH3 H CH3 H CH3 H H H H H H H
polymer A polymer B
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Exam practice questions
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9 a) Poly(ethenol) is made from ethenyl 10 This question is concerned with the molecular
ethanoate, interactions between the polymer chains in
CH2 CH O C CH3
various plastics.
a) Tables of data show that the melting
O temperatures of low density polythene and
Ethenyl ethanoate is first polymerised to high density polythene differ by 23 °C.
produce poly(ethenyl ethanoate). This is
i) State the type of intermolecular
then converted to poly(ethenol) by replacing
interactions that occur between
about 90% of the CH3COO– groups in
polythene molecules.(1)
poly(ethenyl ethanoate) with –OH groups.
ii) Explain why the densities of the two
i) Draw the structure of ethenol and the
polythenes are different.(2)
repeat unit in pure poly(ethenol).(2)
iii) State which type of polythene has the
ii) Draw the repeat unit in poly(ethenyl
higher melting temperature and explain
ethanoate).(1)
your answer.(3)
iii) Calculate the relative molecular mass
b) The polymer Kevlar is a polyamide with
of a sample of poly(ethenol) containing
a much higher melting temperature than
2000 monomer units in which
polythene.
10% of the monomer units contain
CH3COO– groups that have not i) Explain the term ‘polyamide’.(1)
been replaced by –OH groups.(4) ii) Kevlar is manufactured from the
iv) Explain why poly(ethenol) is soluble in monomers benzene 1,4-dicarboxylic
water.(2) acid and benzene 1,4-diamine. Draw
b) The solubility in water of poly(ethenol) the structure of one repeat unit in a
depends on the percentage of CH3COO– molecule of Kevlar.(1)
groups replaced. Maximum solubility iii) Give the strongest type of
in water occurs when about 88% of the intermolecular force in Kevlar. (1)
CH3COO– groups are replaced by –OH iv) Explain which atoms and groups
groups. If fewer groups are replaced, the are involved in the strongest type of
poly(ethenol) becomes less soluble. intermolecular force in Kevlar. (2)
When more CH3COO– groups are c) Kevlar is a very strong polymer as it
replaced, the solubility decreases and when contains crystalline regions.
all the groups are replaced, the poly(ethenol) i) State how the arrangement of Kevlar
is insoluble in water. molecules differs in crystalline regions
i) Explain why poly(ethenol) becomes and non-crystalline regions.(1)
less soluble if fewer CH3COO– groups ii) State why the crystalline regions help
are replaced by –OH groups.(1) to make Kevlar strong. (1)
ii) Explain why the solubility decreases
11 The following describes a method for the
when more than 88% of the
laboratory preparation of phenylamine.
CH3COO– groups are replaced by
–OH groups. (1) Read the method and answer the questions
iii) When all the CH3COO– groups which follow.
are replaced by –OH groups, the 1 Place 9.0 g of tin and 4.2 cm3 nitrobenzene in a
poly(ethenol) crystallises and becomes round-bottomed flask and attach a condenser.
insoluble in water. Explain what 2 Slowly pour 4 cm3 of concentrated
may have happened to make the hydrochloric acid down the condenser and
poly(ethenol) insoluble.(2) shake the flask.
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3 Add further 4 cm3 portions of acid up to a total c) State why the flask was heated in step 4
of 20 cm3 with shaking. If the reaction becomes without the condenser.(1)
too vigorous, cool the flask in cold water but do d) In step 6, give two reasons why an excess
not allow the flask to get too cold. of sodium hydroxide is needed.(2)
4 When all the acid has been added and the e) State why sodium chloride was added
vigour of the reaction has subsided, remove to the aqueous layer.(1)
the condenser and heat the flask for f) Explain whether phenylamine is the upper
30 minutes in a boiling water bath. or lower layer in the separating funnel.(2)
5 If there is a residue of tin, add further g) State why the sodium hydroxide pellets
hydrochloric acid dropwise until it has were added in step 9.(1)
dissolved. h) Calculate the expected yield of phenylamine
6 Cool the flask to room temperature and starting from 4.2 cm3 of nitrobenzene.(3)
carefully with shaking add an excess of i) Write half-equations for the reduction
concentrated sodium hydroxide solution of nitrobenzene to phenylamine in acid
until there is a clear solution. conditions and the oxidation of tin to
7 Set up the flask for steam distillation and SnCl62− ions and combine them to give an
distil until the distillate is clear. overall equation for the reaction.(3)
8 Transfer the distillate to a separating funnel
12* Propylamine and ethanamide have similar
and add 6 g of sodium chloride to saturate
relative molecular masses, but otherwise they
the aqueous layer.
have different properties.
9 Run off the phenylamine into a boiling −1
● Ethanamide has an absorption at 1681 cm
tube, add sodium hydroxide pellets and
in its infrared absorption spectrum whereas
stopper the tube.
propylamine has no absorption in this
10 Leave to stand until the liquid is clear.
region of its spectrum.
The density of nitrobenzene is 1.20 g cm−3 and ● Ethanamide is not basic whereas
the density of phenylamine is 1.0 g cm−3. propylamine is basic.
a) State why a condenser is used in steps ● Ethanamide is a solid at room temperature
2 and 3.(1) whereas propylamine is a liquid.
b) State why the flask should not be too cold Explain these differences.(6)
in step 3.(1)
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Exam practice questions
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Organic synthesis
18.3
18.3.1 Organic synthesis
A lot of the purpose and pleasure of chemistry comes from making new
materials such as polymers, perfumes, drugs and dyes. This making of new
materials is called synthesis. The synthesis of organic compounds is very
important in the research and production of new and useful products. Many
features of modern life depend on the skills of chemists and their ability
to synthesise new and complex materials. New colours, dyes and fabrics
for the fashion industry are synthetic organic molecules. So also are the
liquid crystals used in the flat screens of laptops or tablets (Figures 18.3.1 and
18.3.2). These organic compounds in the computer screen have been tailor-
made by chemists to respond to an electric field and affect light.
Figure 18.3.1 Liquid crystals photographed through a microscope with polarised light.
Figure 18.3.2 The structure of a molecule
which makes up a liquid crystal.
O
O
O O
568 18.3 Organic synthesis
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CH2OH
Tip OH CH3
The synthetic routes discussed in this chapter use reactions of organic compounds HO CH CH2 NH C CH3
studied in earlier chapters of this book. Successful and efficient synthesis depends CH3
on a good knowledge and understanding of the reactions of all the functional groups salbutamol
studied. Some of the ‘Test yourself’ questions are designed to help revise ideas from
O
earlier in the A Level course.
HO CH2 CH C
NH2 OH
One of the major areas of chemical research today involves the synthesis HO
of drugs and medicines. Every day, large numbers of compounds are levodopa
synthesised for testing in pharmaceutical laboratories as potential drugs to
cure or alleviate a particular disease. Medicines that have been synthesised
by chemists include aspirin and paracetamol to relieve pain (Section 18.2.3), CH2OH O
salbutamol to prevent asthma, chloramphenicol to treat typhoid and levodopa O2N CH CH NH C CHCl2
to alleviate Parkinson’s disease (Figure 18.3.3).
OH
Three other important areas of synthetic chemical research involve catalysts chloramphenicol
(Section 15.11), antiseptics and polymers (Sections 17.3.6 and 18.2.9). Figure 18.3.3 Three important drugs that
The essential job of synthetic organic chemists is to consider the proposed have been synthesised by chemists –
structure for a target molecule and then devise a way of making it from salbutamol, levodopa and chloramphenicol.
simpler, readily available starting materials. The scale of work involved and
the difficulties encountered in a complex organic synthesis are illustrated by
the painstaking and ingenious first synthesis of the anti-cancer drug Taxol®
by a team of chemists led by Robert Halton at Florida State University in
1994.
The synthesis of Taxol
In 1962, it was discovered that an extract of the bark of the Pacific yew tree
(Figure 18.3.4) was effective as an anti-cancer drug, particularly against
ovarian cancer. The structure of the active compound (Figure 18.3.5) was
determined in 1971 and its method of action also discovered. But the Pacific
yew tree was very slow growing and several trees had to be felled to produce
even a small amount of the drug so, in order to supply the demand, chemists
had to find a way to make the drug from simpler substances.
O
O O OH Figure 18.3.4 The Pacific yew tree (Taxus
brevifolia) – about 2000 trees were used to
produce 1 kg of Taxol.
O NH O
H O
O O
OH O
OH
O
O
Figure 18.3.5 The structure of paclitaxel, better known under its trademark name Taxol®.
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It took many years before a total synthesis was achieved. From the start, the
project was planned in great detail. The chemists drew on their understanding
of the mechanisms of organic reactions to predict the likely products at each
stage and suggest routes to their target molecule.
The synthesis of Taxol would have been impossible without the newer
methods of separation, purification and identification that had become
available. The variety of spectroscopy techniques was also crucial to success.
Halton’s team published their paper in 1994 describing the successful 37-
step total synthesis and just beat several other teams in the race to discover
a synthesis. Halton also used a method starting from a related compound
found in the needles of the European yew which involved only four steps
and was therefore commercially more viable. Also, as yew needles were used
rather than bark, the trees did not need to be felled. Needles of the European
yew are also a source in the production of Taxotere®, which is potentially
an even better anti-cancer agent than Taxol. In 2002, a further development
came with the discovery of a biochemical route using a fermentation process
from yew-derived plant cells.
Organic analysis
When complex molecules such as Taxol have been synthesised, chemists
must use a variety of methods to analyse them and identify their precise
composition and structure.
Traditionally, chemical tests were used to identify functional groups in
organic molecules, together with combustion and quantitative analysis.
Nowadays, however, modern laboratories rely on a range of highly sensitive,
automated and instrumental techniques to identify the products of synthesis.
These include chromatography (Section 19.5), mass spectrometry (Section
19.2) and various kinds of spectroscopy (Sections 19.3 and 19.4).
Sensitive methods of analysis are very important in monitoring organic
syntheses for several reasons.
● Sensitive methods of analysis determine the degree of purity of a synthetic
product.
● Sensitive methods of analysis also identify any impurities, some of which
may be toxic and in very small concentration.
● If analysis reveals an impurity in the product, it may be possible to limit its
formation by changing the operating conditions for the reaction. Changes
in the temperature, the pressure, the solvent used or the choice of catalyst
may promote the formation of a desired product while reducing the
formation of impurities.
● Many pharmaceutical laboratories that specialise in the development of
new drugs produce thousands of compounds every year for further testing.
Some of their products are obtained in very small concentrations and
particularly sensitive techniques are needed to analyse and identify them.
The food and drugs industries operate very high standards of purity in
their products. Impurities, depending on their toxicity, may interfere with
the health and well-being of consumers. For example, traces of sodium
chloride in a medicine would probably not be considered a problem, but
the slightest trace of sodium cyanide would be cause for alarm.
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18.3.2 The formulae of organic
molecules
Empirical formulae
From previous study you should know that the empirical formula of a
compound shows the simplest ratio of the number of atoms of each element
in it (Section 5.2).
Modern analysis usually begins with mass spectrometry to determine the M r
of an organic compound (Section 7.1), but finding its empirical formula is an
important step in understanding its chemistry. The usual way of doing this is
to oxidise a weighed sample of the organic compound completely by burning
it in pure dry oxygen, and then determine the masses of carbon dioxide and
water produced. This process of combustion analysis then enables you to
calculate the masses of carbon and hydrogen in a sample of the compound
and hence its empirical formula.
Figure 18.3.6 shows a simplified diagram of the method and apparatus used.
Modern methods of combustion analysis include refinements to ensure that
the organic compound is completely oxidised and that all the carbon dioxide
and water are absorbed and weighed.
organic compound careful suction from
under test a water pump
pure dry
oxygen
heat
anhydrous soda lime
calcium chloride (sodium hydroxide
+ calcium oxide)
Figure 18.3.6 Using careful suction, draw pure dry oxygen over a heated sample of the
solid organic compound. (Liquid organic compounds can be burnt from a wick.) Pass the
product gases through anhydrous calcium chloride (or anhydrous copper(ii) sulfate) to
absorb any water produced, and then through anhydrous soda lime (sodium hydroxide
and calcium oxide) to absorb the carbon dioxide produced.
Molecular formulae
Molecular formulae are more helpful than empirical formulae because
they show the actual number of atoms of each element in one molecule
of a compound. All that is needed to find the molecular formula from the
empirical formula is the molar mass of the compound. This can be determined
from the mass spectrum of the compound.
A molecular formula is always a simple multiple of the empirical formula.
Methane, for example, has the empirical formula CH4 and the molecular
formula CH4, benzene has the empirical formula CH and the molecular
formula C6H6, and ethanoic acid has the empirical formula CH 2O and the
molecular formula C2H4O2.
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Structural formulae
An empirical formula can represent many different compounds and a
Tip molecular formula can represent many different isomers, but the structural
Don’t forget that pairs of stereoisomers formula of a compound shows how the atoms link together in a molecule
have the same structural formula but of a particular compound. Figure 18.3.7 shows empirical, molecular and
the atoms are arranged differently in structural formulae for cyclohexene.
space (see Chapter 17.1). Given the molecular formula of an organic compound, and knowing
something about its characteristic reactions, it is often possible to predict its
C3H5 C6H10 structure by assuming that:
empirical formula molecular formula
● carbon atoms form four covalent bonds
● nitrogen atoms form three covalent bonds
CH2 ● oxygen atoms form two covalent bonds
H2C CH ● hydrogen and halogen atoms form one covalent bond.
The structural formulae of organic compounds can often be determined
H2C CH by combustion analysis to find their percentage composition, followed
CH2 by a study of their chemical reactions. However, structural formulae can
structural formula be obtained more definitively, and more precisely, by spectrometry and
Figure 18.3.7 The empirical, molecular spectroscopy.
and structural formulae of cyclohexene. Using mass spectrometry it is possible to identify the fragments of an organic
molecule and then piece the whole molecule together.
Spectroscopic methods such as infrared spectroscopy (see Section 7.2 and
Section 19.3) and, particularly, nuclear magnetic resonance spectroscopy
(Section 19.4), provide information about the various bonds and functional
groups in organic molecules in order to confirm their structural formulae.
Sometimes it is enough to show structural formulae in a condensed form,
such as CH3CH2CH2CH=CHCH3 for hex-2-ene. At other times it is more
helpful to write a full structural formula showing all the atoms and all the
bonds. This type of formula is called a displayed formula.
Chemists have also devised a useful shorthand for showing the formulae of
more complex molecules as skeletal structures. These skeletal formulae need
careful study because they represent the hydrocarbon part of the molecule
simply as lines for the bonds between carbon atoms, leaving out the symbols
for carbon and hydrogen atoms (Figure 18.3.8).
H H H Br H H Br
H C C C C C C H
H H H H
Figure 18.3.8 The displayed and skeletal formulae of 3-bromohex-2-ene.
Although the structural, displayed and skeletal formulae of an organic
compound show how its atoms link together, they do not show its true shape
in three dimensions. Sometimes, it is important to know and understand
what the three-dimensional shape of a molecule is like and chemists use
Figure 18.3.9 A ball-and-stick model and various models to do this. These include ball-and-stick models, space-filling
a space-filling model of 2-methylpropane. models (Figure 18.3.9) and various types of computer models.
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Test yourself
1 A compound of carbon, hydrogen and oxygen with Mr = 46.0 contains
52.2% carbon and 13.0% hydrogen. Calculate its molecular formula and
draw and name two possible structural formulae for the compound.
2 Write out the empirical, molecular, structural, displayed and skeletal
formulae of 2-methylpropane in Figure 18.3.9.
3 Describe briefly how you would determine the percentages of carbon
and hydrogen in a solid organic compound.
4 A compound containing only carbon, hydrogen and fluorine was burnt
in excess oxygen. 0.32 g of the compound produced 0.44 g of carbon
dioxide and 0.09 g of water.
a) What is the empirical formula of the compound?
b) The relative molecular mass of the compound is 64.0. What is its
molecular formula?
c) Write all the possible displayed formulae of the compound and give
the systematic name for each formula.
5 A compound, containing only carbon, hydrogen and oxygen has
prominent peaks in its mass spectrum at m/z values of 60, 43, 31,
29 and 17.
a) What is the relative molecular mass of the compound?
b) Suggest possible fragment ions for the peaks at m/z values of 17,
29 and 31.
c) Suggest a possible structure for the compound.
18.3.3 Functional groups – the keys
to organic molecules
Functional groups provide the key to organic molecules. A knowledge of the
properties and reactions of a limited number of functional groups has opened
up our understanding of most organic compounds.
tertiary
alcohol primary Figure 18.3.10 The structure of the steroid
alcohol cortisone, labelled to show the reactive
carbonyl CH2OH
group HO functional groups and the hydrocarbon
CH3 C O skeleton.
O CH2 C carbonyl
group
C C CH2
CH3
CH2 CH CH CH2
carbonyl H2C C CH
group
C C CH2
O C CH2
H double bond
as in an alkene
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A functional group is the atom or atoms that give a series of organic compounds
Key term their characteristic properties and reactions. Chemists often think of an
organic molecule as a relatively unreactive hydrocarbon skeleton with one or
A functional group is the atom or
more functional groups in place of hydrogen atoms. The functional group
group of atoms that give an organic
in a molecule is responsible for most of its reactions. In contrast, the carbon–
compound its characteristic properties.
carbon single bonds and carbon–hydrogen bonds are relatively unreactive,
partly because they are both strong and non-polar.
Table 18.3.1 shows the major functional groups that you have met during
your A Level studies, together with an example of one compound containing
each group.
Table 18.3.1 The major functional groups.
Functional group Example Functional group Example
Alcohol OH propan-1-ol Ester O C methyl ethanoate
CH3CH2CH2OH CH3 O C CH3
O O
Alkene propene
C C CH3CH=CH2 Acyl chloride C Cl ethanoyl chloride
(acid chloride) CH3 C Cl
Halogenoalkane Hal 1-chloropropane O
O
Hal = F, Cl, Br, I CH3CH2CH2Cl
Amine NH2 propylamine
Ether methoxyethane CH3CH2CH2NH2
C O C CH3OCH2CH3
Amide C N propanamide
Aldehyde O propanal CH3CH2 C NH2
CH3CH2CHO O H
C O
H Nitrile C N ethanenitrile
CH3 C N
Ketone propanone
C O CH3COCH3 Phenyl (C6H5 —) benzene
H H
Carboxylic acid O propanoic acid C C
CH3CH2COOH H C C H
C
C C
OH H H
The characteristic properties and tests for most of these functional groups
are shown on the data sheets headed ‘Tests and observations on organic
compounds’ and ‘Tests for gases’, accessed online at [Link].
[Link]/EdexcelChemistry. These tests are used in Core practical 15 (part 2):
Analysis of some organic unknowns.
Functional groups can also be identified by spectroscopic methods (Chapter
19). An absorption in the infrared spectrum of a compound or a peak in its
NMR spectrum can identify a particular functional group in the compound
(see the Pearson Edexcel Data booklet).
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Core practical 15 (part 2)
Analysis of some organic unknowns
A student was provided with three organic liquids, A, B and C, 5 Deduce possible structures for compounds A, B and C.
each of which has molecules that contain four carbon atoms. A fourth compound, D, with formula CH3CH2COCH2OH, was
He carried out a series of tests using (a) 2,4-dinitrophenylhydrazine, also tested with the same three reagents.
(b) iodine and sodium hydroxide and (c) acidified potassium 6 Give the results of the three tests and draw the structure
dichromate(vi). Table 18.3.2 shows which reactions gave of the organic product, if any, of the reaction of D with (b)
positive results. iodine and sodium hydroxide and (c) acidified potassium
1 Describe what he saw when each test gave a positive result. dichromate(vi).
2 What deductions about the functional group present can be 7 Draw a structure for compound E, an isomer of D, which
made using the test with 2,4-dinitrophenylhydrazine? gave a negative test with each of the reagents in Table
3 What deductions about the functional group present can be 18.3.2 but gave an effervescence of a colourless gas when
made using the test with iodine and sodium hydroxide? added to aqueous sodium hydrogen carbonate.
4 What deductions about the functional group present 8 Draw a structure for compound F, also an isomer of D, which
can be made using the test with acidified potassium gave a negative test with each of the reagents in Table 18.3.2
dichromate(vi)? and did not react with aqueous sodium hydrogen carbonate.
Table 18.3.2
Liquid 2,4-Dinitrophenylhydrazine Iodine with Acidified potassium
sodium hydroxide dichromate(vi)
A ✓ ✗ ✓
B ✓ ✓ ✗
C ✗ ✓ ✓
Tip Tip
Test-tube reactions carried out to identify functional groups have the major Analysis of an inorganic unknown is
disadvantage that they use up some of the compound. A major advantage of covered in Core practical 15 (part 1)
spectroscopic analysis is that these techniques are either non-destructive (infrared in Chapter 15.
and NMR) or use up tiny amounts of compound (mass spectrometry). For practical guidance, refer to Practical
skills sheet 20, ‘Analysing organic
unknowns’, which you can access online
Test yourself at [Link]/
EdexcelChemistry.
6 Anaerobic respiration in muscle cells breaks down glucose to simpler
compounds including the following two molecules. Identify the
functional groups in these molecules:
a) HOCH2−CH(OH)−CHO
b) CH3−CO−COOH.
7 Pheromones are messenger molecules produced by insects to attract
mates or to give an alarm signal. Identify the functional groups in the
pheromone below, which is produced by queen bees.
O
CH3CCH2CH2CH2CH2CH2 H
C C
H COOH
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Test yourself
8 Use the data sheets headed ‘Tests and observations on organic
compounds’, accessed online at [Link]/
EdexcelChemistry, to predict six important properties or reactions of
the painkiller dextropropoxyphene and its mirror image, which is an
ingredient of cough mixtures.
CH3 H H CH3
N(CH3)2 N(CH3)2
O
O
COCH2CH3
CO
painkiller CH2CH3 cough suppressant
9 For each of parts (a) to (f) below only one of the compounds labelled A
to D is correct.
A CH3CH2NH2 B C6H5NO2 C C6H5NH2 D (CH3)4NCl
a) Which is a strong electrolyte?
b) Which dissolves in dilute hydrochloric acid but not in water?
c) Which is insoluble in water, dilute acid and dilute alkali?
d) Which best forms a blue complex with aqueous copper(ii) sulfate?
e) Which has the highest vapour pressure at room temperature?
f) Which combines most readily with H+ ions?
18.3.4 Organic routes
Organic chemists synthesise new molecules using their knowledge of
functional groups, reaction mechanisms and molecular shapes, as well as the
factors that control the rate and extent of chemical change.
A synthetic pathway leads from the reactants to the required product in one
step or several steps. Organic chemists often start by examining the ‘target
molecule’. Then, they work backwards through a series of steps to find
suitable starting chemicals that are cheap enough and available. This is called
‘retrosynthetic analysis’ and involves planning a synthesis by transforming
a target molecule into simpler precursors without necessarily making any
assumptions about starting materials.
Figure 18.3.11 shows an example of the systematic way in which working
back can be used in synthesising one ‘target molecule’ from a ‘starting
molecule’. In this case, the ‘target molecule’ is butanoic acid and the ‘starting
molecule’ is 1-bromobutane.
CH3CH2CH2C N
nitrile
CH3CH2CH2CH2OH
alcohol CH3CH2CH2CHO
halogenoalkane O
CH3CH2CH CH2 aldehyde
CH3CH2CH2CH2Br O CH3CH2CH2C
alkene CH3CH2CH2C OH
starting target
molecule CH3CH2CH2CH2NH2 ester OCH3 molecule
amine CH3CH2CH2CH2OH
alcohol
Figure 18.3.11 Working back from the target molecule to find a two-step synthesis of
butanoic acid from 1-bromobutane.
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1 Begin by writing down the formulae of those compounds that could be
readily converted to butanoic acid, the target molecule. These include the
nitrile butanenitrile, the aldehyde butanal, the ester methyl butanoate and
the alcohol butan-1-ol.
2 Then look at your starting molecule, 1-bromobutane, to see whether it
could be converted to one of the compounds that would readily form
butanoic acid. If necessary, write down the formulae of compounds that
might be produced from 1-bromobutane. These include the alcohol
butan-1-ol, the alkene but-1-ene, and the amine butylamine.
3 With any luck, you should now see a possible two-step synthetic route
from your starting molecule to the target molecule. In this case, the route
can go via butan-1-ol.
4 If a two-step route is not clear at this point, then you might need to
consider a three-step route involving the conversion of one of the products
from the starting material to one of the reactants that will readily form the
target material.
Chemists normally seek a synthetic route that has the least number of steps
and produces a high yield of the product. The larger the scale of production,
the more important it is to keep the yield high so as to avoid producing large
quantities of wasteful by-products.
Changing the functional groups
All the reactions in organic chemistry convert one compound to another, but
there are some reactions that are particularly useful for developing synthetic
routes. These useful reactions, which do not change the number of carbon
atoms, include:
● the addition of hydrogen halides to alkenes
● substitution reactions that replace halogen atoms with other functional
groups such as −OH or −NH2
● substitution of a chlorine atom for the −OH group in an alcohol or a
carboxylic acid
● elimination of a hydrogen halide from a halogenoalkane to introduce a
carbon–carbon double bond
● oxidation of primary alcohols to aldehydes and then carboxylic acids
● reduction of carbonyl compounds to alcohols
● hydrolysis of a nitrile to form a carboxylic acid group.
Test yourself
10 Draw the structural formula of the main organic product in each
of the following reactions. Classify each reaction as addition,
substitution or elimination and classify the reagent on the arrow as a
free radical, nucleophile, electrophile or base.
HBr(g)
a) CH2=CH2(g)
KOH(aq)
b) CH3CH2CH2Br(l)
conc. HNO3
c) C6H6 + conc. H2SO4
benzene
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Test yourself
11 Identify substances A to E in the flow diagrams below.
A in ether PCI5(s) conc. NH3
A in ether B PCI5(s) CH CH CI conc. NH 3 C
a) CH
CH3COOH
3COOH then water B 3 CH
CH 3
2 CI
2 in ethanol C
then water in ethanol
b)
OO
HO
HO CH
CH2CH 3
2CH3
CC (i) E in CC
(i) E in
D CH 3 ethoxyethane CH
CH3
D CH3 ethoxyethane 3
AICI3 (ii) H2O
AICI3 (ii) H2O
12 Give the reagents and conditions for converting:
a) but-1-ene to but-2-ene in two steps
b) propanone to propene in three steps
c) bromoethane to ethane-1,2-diol in three steps.
Changing the carbon chain
Although it is easy to change the reactive part of a molecule, the functional
group, chemists sometimes need to lengthen or shorten the carbon skeleton
of a molecule. Planning the synthesis of a new drug may involve computer-
aided drug design to predict whether replacing methyl groups with ethyl
groups, for instance, would increase the effectiveness of the drug.
Increasing the chain length
Use of cyanide ions
Tip The use of cyanide ions to add a single carbon to the chain is described in
Section 6.3.4. When a halogenoalkane is heated with a solution of potassium
The nitrile group can be reduced to form cyanide in ethanol, the halogen atom is replaced by the CN group and a
a primary amine using hydrogen with a nitrile is formed with a longer carbon chain.
nickel catalyst or LiAlH4 in ethoxyethane
(Section 18.2.4). The nitrile group can For example, the two-carbon chain in bromoethane becomes three in
also be hydrolysed to form a carboxylic propanenitrile:
acid by reacting with hydrochloric acid CH3CH2Br + CN− ⎯→ CH3CH2CN + Br−
(Section 17.3.2). bromoethane propanenitrile
Use of Grignard reagents
An alternative reaction that can add more than one carbon uses a Grignard
Key term reagent.
In most organic compounds, carbon is bonded to a more electronegative
A Grignard reagent is an
atom, such as halogen, oxygen or nitrogen, so that carbon is usually polarised
organometallic compound with the
δ+. In a Grignard reagent, carbon is bonded to the much less electronegative
general formula RMgX formed from a
element magnesium, so carbon becomes polarised δ−. The δ− carbon in a
halogenoalkane RX and magnesium.
Grignard reagent then bonds to the δ+ carbon of a carbonyl compound and
a reaction takes place in which a new C−C bond is formed.
Preparation of Grignard reagents
Grignard reagents are unstable and must be prepared immediately before use.
Magnesium metal is added to a solution of a halogenoalkane, usually a bromide
or an iodide, dissolved in dry ethoxyethane. The magnesium reacts vigorously
and dissolves to form a compound that can be represented as RMgX.
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For example, iodoethane reacts with magnesium in ethoxyethane to form
ethylmagnesium iodide.
Tip
New carbon–carbon bonds are formed
CH3CH2I + Mg ⎯→ CH3CH2MgI
in Friedel–Crafts reactions of arenes.
ethylmagnesium iodide
These involve the replacement of a
hydrogen atom by a carbon chain
Reaction of Grignard reagents
(see Section 18.1.6).
Carbonyl compounds or carbon dioxide react with Grignard reagents. The
reaction occurs in two steps. An initial product is formed in an addition reaction
and this is then hydrolysed in the second step by the addition of dilute acid.
The following examples use the Grignard reagent ethylmagnesium iodide
made from iodoethane.
a) Reaction with methanal to form a primary alcohol (Figure 18.3.12).
H Figure 18.3.12 Reaction of
H
H2O ethylmagnesium iodide with methanal to
CH3CH2MgI + C O CH 3CH2 C OH + Mg(OH)I
form propan-1-ol.
H
H
b) Reaction with other aldehydes to form a secondary alcohol (Figure 18.3.13).
CH3 CH3 Figure 18.3.13 Reaction of
H2O ethylmagnesium iodide with ethanal to
CH3CH2MgI + C O CH3CH2 C OH + Mg(OH)I
form butan-2-ol.
H H
c) Reaction with a ketone to form a tertiary alcohol (Figure 18.3.14).
CH3 CH3 Figure 18.3.14 Reaction of
H2O
+ Mg(OH)I
ethylmagnesium iodide with propanone to
CH3CH2 MgI + C O CH 3CH2 C OH
form 2-methylbutan-2-ol.
CH3 CH3
d) Reaction with carbon dioxide to form a carboxylic acid (Figure 18.3.15).
O Figure 18.3.15 Reaction of
H2O ethylmagnesium iodide with carbon dioxide
CH3CH2 MgI + O C O CH 3CH 2 C + Mg(OH)I
to form propanoic acid.
OH
Decreasing the chain length
Triiodomethane reaction
The reaction of methyl ketones or methyl secondary alcohols with iodine and
sodium hydroxide forms a yellow precipitate of triiodomethane (iodoform)
and a carboxylate salt with one fewer carbon (Section 17.2.6).
CH3CH2COCH3 + 3I2 + 4OH− ⎯→ C
H3CH2COO− + CHI3 + 3I− + 3H2O
butanone propanoate ions
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Test yourself
13 Draw the structural formula of the organic iii) cyclohexanone
product(s) in each of the following reactions: iv) butanal
a) 1-bromopropane with potassium cyanide in The intermediates formed are then hydrolysed.
ethanol Draw the structural formula and name the final
b) butan-2-ol with iodine in sodium hydroxide. product in each case.
14 a) Outline how a Grignard reagent is made from 15 Give reagents and conditions for the preparation
1-bromopropane. of pentylamine in two-steps starting from
b) The Grignard reagent formed in part (a) reacts 1-bromobutane.
separately with: 16 Give reagents and conditions for the preparation
i) methanal of 2-phenylbut-2-ene in three steps starting
from benzene and show the structure of the
ii) carbon dioxide
intermediates in the synthesis.
Activity
Converting one functional group to another
Make a copy of the flow chart in Figure 18.3.16. For each numbered 2 Using your completed copy of Figure 18.3.16, suggest three-
arrow, write the reagents and conditions needed for the conversion. step syntheses, showing the reagents and conditions for
each of the following conversions:
1 Using your completed copy of Figure 18.3.16, suggest two-
a) ethene to ethanoic acid
step syntheses, showing the reagents and conditions for
b) propan-2-ol to propane.
each of the following conversions:
a) ethene to ethylamine
b) ethanol to ethyl ethanoate (using ethanol as the only
carbon compound)
c) propanoic acid to propanamide.
Amine
Alkane 5 7
2
6
1 Halogenoalkane Nitrile
3 Ester
4 25 18 19
8 9
Alkene 20
26 27 Carboxylic
28 Alcohol Grignard 23
10 acid
12 16
21
11 13 17
Dihalogenoalkane Ketone 22
Aldehyde Acyl chloride
14 15
Hydroxynitrile 24
Amide
Figure 18.3.16 A flow diagram summarising the methods for converting one functional group to another.
580 18.3 Organic synthesis
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18.3.5 Synthetic techniques
Chemists have developed a range of practical techniques and procedures for
the synthesis of solid and liquid organic compounds. These methods allow water out
for the fact that reactions involving molecules with covalent bonds are often
slow and that it is difficult to avoid side reactions that produce by-products.
There are five key stages in the preparation of an organic compound.
vapour escaping
from the flask
Stage 1: Planning condenses here
The starting point of any synthesis is to choose an appropriate reaction or water in
series of reactions as described in the previous section. The next thing to do condensed liquid flows
is to work out suitable reacting quantities from the equation and decide on back to the flask
the conditions for reaction. reaction mixture
with volatile liquids
An important part of the planning stage is a risk assessment (see Practical
skills sheet 13, ‘Assessing hazards and risks’, which you can access online anti-bumping granules
at [Link]/EdexcelChemistry). This should ensure that heat
hazards have been identified and that appropriate safety precautions and
Figure 18.3.17 Heating in a flask with a
control measures are used in order to reduce the risk during any synthesis.
reflux condenser prevents vapours escaping
while the reaction is happening. Vapours
Stage 2: Carrying out the reaction from the reaction mixture condense and
During this stage, the reactants are measured out and mixed in suitable flow back (reflux) into the flask.
apparatus. Most organic reactions are slow at room temperature so it is usually
necessary to heat the reactants using a flame, heating mantle or hotplate.
One of the commonest techniques is to heat the reaction mixture in a flask
fitted with a reflux condenser (Figure 18.3.17).
Organic reagents do not usually mix with aqueous reagents. So another
common technique is to shake the immiscible reactants in a stoppered
container.
moistened
Stage 3: Separating the crude product from the filter paper impure crystals
porous plate
reaction mixture Buchner
funnel
Filtration suction from
water pump
If the crude product is a solid, it can be separated by filtration using a Buchner
or Hirsch funnel with suction from a water pump. This is illustrated in
Figure 18.3.18. filtrate
Distillation Figure 18.3.18 Filtering a solid using a
Liquids can often be separated by simple distillation, fractional distillation Buchner funnel by suction from a water
or steam distillation. Distillation with steam at 100 °C allows the pump.
separation of compounds that decompose if heated at their boiling
temperatures. The technique works only with compounds that do not mix
with water. When used to separate the products of organic preparations,
steam distillation leaves behind those reagents and products that are soluble
in water (Figure 18.3.19).
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Key term
Steam distillation is used to distil a
liquid which decomposes when heated
at its boiling temperature and does not water
mix with water. The distillate consists of water in
a layer of water with separate layer of
immiscible liquid.
water out
starting material
oil floats on water
anti-bumping water
granules
heat
Figure 18.3.19 Steam distillation.
Tip
Some methods of steam distillation generate the steam in a separate flask and pass
steam into the impure mixture. The amounts of immiscible liquid and water that distil
together depend on the relative vapour pressures of the two liquids at the boiling
temperature.
Solvent extraction
Solid or liquid products can be separated from an aqueous reaction mixture
using the technique of solvent extraction. The aqueous mixture is shaken
in a separating funnel with a solvent which is immiscible with water
(Figure 18.3.20). The organic product dissolves preferentially in the organic
layer, which in most cases is the upper layer. The lower aqueous layer can
then be drained off and the organic product can be obtained from the upper
layer by evaporation of the solvent.
Test yourself
17 Pairs of liquids can be separated by (i) simple distillation, (ii)
fractional distillation or (iii) steam distillation. For each of the
following pairs, suggest the best distillation method to obtain the
first liquid from a mixture of it with the second. Boiling temperatures
are given in brackets.
a) methanol (65 °C) from a mixture with ethyl ethanoate (54 °C)
b) nitrobenzene (211 °C) from a mixture with water (100 °C)
Figure 18.3.20 Separating funnels used in
pesticide research. c) pentan-1-ol (138 °C) from a mixture with pentane (36 °C)
18 A separating funnel is shown in stage C in Figure 17.3.12 on
page 506. Explain why the liquid does not drain out if the
apparatus is used exactly as shown in the diagram.
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Stage 4: Purifying the product
The ‘crude’ product separated from the reaction mixture is usually
contaminated with by-products and unused reactants. The methods of
purifying this ‘crude’ product depend on whether it is a solid or a liquid.
Purifying organic solids
The usual technique for purifying solids is recrystallisation, which is illustrated
in part of Figure 17.2.20 on page 495. The procedure for recrystallisation
is based on using a solvent that dissolves the product when hot, but not
when cold. The choice of solvent is usually made by trial and error. Use of
a Buchner or Hirsch funnel and suction filtration speeds up filtering and
facilitates recovery of the purified solid from the filter paper. The procedure
is as follows. Key term
1 Dissolve the impure solid in the minimum volume of hot solvent.
A desiccator is a container that is
2 If the solution is not clear, filter the hot mixture through a heated funnel
used to dry a solid and also to store
to remove insoluble impurities.
materials in a dry atmosphere.
3 Cool the filtrate so that the product recrystallises, leaving the smaller
amounts of soluble impurities in solution.
4 Filter to recover the purified product.
5 Wash the purified solid with small amounts of cold, pure solvent to wash
away any solution containing impurities.
6 Allow the solvent to evaporate from the purified solid in the air or place
the solid in a desiccator (Figure 18.3.21).
Tip
A common drying agent used in a desiccator to remove water is silica gel. This is often
used in a form containing a little cobalt (ii) chloride. Cobalt chloride is blue when
anhydrous, but turns pink in the presence of water. So if the silica gel turns pink, this
indicates that it cannot absorb any more water and needs to be heated to return it to
the anhydrous state. Figure 18.3.21 A desiccator used to store
bottles in a dry atmosphere created by the
desiccant silica gel.
Test yourself
19 Explain why connecting a desiccator to a vacuum pump increases
the rate of evaporation of the solvent.
20 Give the formula of the cobalt-containing complex ion formed when
silica gel drying agent containing cobalt(ii) chloride turns from blue to
pink in a desiccator.
Purifying organic liquids
Washing and drying
Chemists often begin to purify organic liquids that are insoluble in water by
shaking with aqueous reagents in a separating funnel to extract impurities. For
instance, in the Activity: Preparation of an ester (Section 17.3.3), the impure
ester is shaken with aqueous sodium carbonate to remove any acidic impurities.
This first washing is usually followed by a second washing with pure water
to remove any inorganic impurities.
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The organic liquid is then dried using a solid drying agent such as pieces of
anhydrous calcium chloride or anhydrous magnesium sulfate. The organic
liquid, left to stand in contact with the solid, goes clear when any water present
has been removed. Finally, distillation is used to obtain the pure liquid.
Fractional distillation
Fractional distillation separates mixtures of liquids with different boiling
temperatures. On a laboratory scale, the process takes place in the distillation
apparatus that has been fitted with a fractionating column between the flask
and the still-head (Figure 18.3.22). Separation is improved if the column
is packed with inert glass beads or rings to increase the surface area where
rising vapour can mix with condensed liquid running back to the flask. The
column is hotter at the bottom and cooler at the top. The thermometer reads
the boiling temperature of the compound passing over into the condenser.
thermometer
still head
condenser
fractionating
column
receiver
starting material
anti-bumping
granules
heat
Key terms
Figure 18.3.22 The apparatus for fractional distillation of a mixture of liquids.
A vapour is a gas formed by evaporation If the flask contains a mixture of liquids, the boiling liquid in the flask
of a substance that is usually liquid or produces a vapour that is richer in the most volatile of the liquids present
solid at room temperature. (the one with the lowest boiling temperature).
Chemists talk about ‘hydrogen gas’ Most of the vapour condenses in the column and runs back. As it does so, it
but ‘water vapour’. Vapours are easily meets more of the rising vapour. Some of the vapour condenses. Some of the
condensed by cooling or increasing liquid evaporates. In this way, the mixture evaporates and condenses repeatedly
the pressure because of their relatively as it rises up the column. But, every time it does so, the vapour becomes
strong intermolecular forces. richer in the most volatile liquid present. At the top of the column, the vapour
A volatile liquid evaporates easily,
contains 100% of the most volatile liquid. So, during fractional distillation, the
turning to a vapour.
most volatile liquid with the lowest boiling temperature distils over first, then
the liquid with the next lowest boiling temperature, and so on.
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Activity
Preparation and purification of an organic liquid
A pure sample of the ester methyl benzoate can be prepared
by the reaction of methanol with benzoic acid in the presence
of concentrated sulfuric acid as follows.
A To a 50 cm3 pear-shaped flask add 8 g of benzoic acid, 15 cm3
conc. sulfuric
conc.
conc.
sulfuric
conc. acid
sulfuric
acid
sulfuric methanol
acid acid methanol
methanol
methanol
methyl
methyl methyl benzoate-
benzoatemethanol
methyl
benzoatemethanol
benzoatemethanol
of methanol and 2 cm3 of concentrated sulfuric acid. Fit the benzoic
benzoic
acid
benzoic
acid acid
methanolbenzoic acid
flask with a reflux condenser and boil the mixture for about 45
minutes. Questions
B Cool the mixture to room temperature and pour it into a 1 Why is a reflux condenser necessary in step A?
separating funnel that contains 30 cm3 of cold water. Add 2 Suggest why the cold water is added in step B.
15 cm3 of hydrocarbon solvent to the pear-shaped flask and 3 Why is hydrocarbon solvent added to the pear-shaped flask
then pour this into the separating funnel. in step B?
C Mix the contents of the separating funnel by vigorous 4 Describe how the pressure in the separating funnel is
shaking, releasing the pressure carefully from time to time. released in step C.
Allow the contents of the flask to settle, then run the lower 5 Describe how the hydrocarbon solvent layer is washed with
aqueous layer into a conical flask. water in step D and suggest what this washing removes.
D Wash the hydrocarbon solvent layer in the separating funnel 6 Suggest what is removed in step D by washing the
first with 15 cm3 of water and secondly with 15 cm3 of hydrocarbon solvent layer with sodium carbonate.
0.5 mol dm−3 aqueous sodium carbonate. 7 Describe the appearance of the dry organic layer in step E.
E Dry the hydrocarbon solvent extract over anhydrous sodium 8 Confirm that benzoic acid is the limiting reagent. The density
sulfate, then filter off the solid. of methanol is 0.79 g cm−3.
F Remove the hydrocarbon solvent by careful distillation. 9 Calculate the percentage yield if the preparation produces
Collect the distillate boiling above 190 °C and weigh your 4.8 g of methyl benzoate.
product. 10 What are the hazards posed by the preparation and how
might their risk be reduced?
Stage 5: Measuring the yield, identifying the
product and checking its purity
Measuring the yield
Comparing the actual yield with the yield expected from the chemical
equation is a good measure of the efficiency of a process.
The yield expected from the equation, assuming that the reaction is 100%
efficient, is called the theoretical yield.
The efficiency of a synthesis, like that of other reactions, is calculated as a
percentage yield. This is given by the relationship:
actual yield of product
percentage yield = × 100%
theoretical yield of product
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Example
a) What is the theoretical yield of glycine (2-aminoethanoic acid) from
15.5 g of chloroethanoic acid?
b) What is the percentage yield if the actual yield of glycine is 7.9 g?
Notes on the method
Start by writing an equation for the reaction. This need not be a full
balanced equation so long as it includes the limiting reactant, the product
and the molar amounts of reactant and product in the equation.
In this case we must assume that any other reactants are in excess, the
limiting reactant is chloroethanoic acid and the mole ratio is 1 : 1.
Answer
a) Equation extract: ClCH2COOH → H2NCH2COOH
The molar mass of chloroethanoic acid, ClCH2COOH = 94.5 g mol–1
The molar mass of glycine, H2NCH2COOH = 75.0 g mol–1
According to the equation:
1 mol of chloroethanoic acid produces 1 mol of glycine
∴ 94.5 g of chloroethanoic acid produce 75.0 g of glycine
So, 15.5 g of chloroethanoic acid produce 15.5 × 75.0 g of glycine
94.5
= 12.3 g of glycine
Theoretical yield of glycine = 12.3 g
actual yield of product
b) percentage yield = × 100%
theoretical yieldof product
7.9
= × 100%
12.3
= 64%
Identifying the product and checking its purity
Qualitative tests
Simple chemical tests for functional groups can help to confirm the identity
of the product. These tests for functional groups are shown in the data sheets
headed ‘Tests and observations on organic compounds’, which you can access
online at [Link]/EdexcelChemistry.
Measuring melting temperatures and boiling temperatures
Pure solids have sharp melting temperatures, but impure solids soften and
melt over a range of temperatures. So watching a solid melt can often show
whether or not it is pure. As databases now include the melting temperatures
of all known compounds, it is possible to check the identity and purity of
a product by checking that it melts sharply at the expected temperature
(Figure 18.3.23).
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stirrer
capillary tube
containing the
sample
thermometer
oil with a
high boiling
temperature
capillary tube
containing
the sample
very gentle heat
Figure 18.3.23 Two methods of measuring the melting temperature of a solid.
Like melting temperatures for solids, boiling temperatures can be used to check
the purity and identity of liquids. If a liquid is pure it will all distil at the expected
boiling temperature or in a narrow range including it. The boiling temperature
can be measured as the liquid distils over during fractional distillation.
Chromatography and spectroscopy
The use of chromatography and spectroscopy in identifying compounds and
checking their purity is covered in Chapter 19. These are the most important
modern-day analytical tools.
Test yourself
21 In the diagram for stage E of Figure 17.3.12 (page 506), the
thermometer bulb is placed near the top of the apparatus.
a) Explain why the bulb is placed there.
b) How and why would the reading differ if atmospheric pressure
were higher than normal on the day of the experiment?
c) How would the reading differ if the thermometer bulb were placed
in the liquid in the flask?
22 A possible two-step synthesis of 1,2-diaminoethane first converts an
alkene to a dihalogenoalkane and then reacts this with ammonia.
a) Write out a reaction scheme for the synthesis, giving reagents
and conditions.
b) Calculate the mass of the alkene needed to make 2.0 g of the
1,2-diaminoethane, assuming a 60% yield in step 1 and a 40%
yield in step 2.
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Test yourself
23 Give three reasons why the actual yield in an organic synthesis is
always less than the theoretical yield.
24 A two-step synthesis converts 18 g of benzene first to 22 g of
nitrobenzene and then to 12 g of phenylamine.
a) State the reagents and conditions for each step.
b) Calculate the theoretical yield and the percentage yield for each step.
c) What is the overall percentage yield?
25 Read again the sub-section headed ‘Purifying organic solids’ on
page 583 and then answer the following questions.
a) Why should the impure solid be dissolved in the minimum volume
of hot solvent?
b) Why is the solution sometimes cooled in ice when the pure
product is being recrystallised?
c) How could you improve the evaporation of excess solvent from
the purified solid in the final stage?
Activity
Preparation and purification of N-phenylethanamide
Amines react very rapidly with acyl chlorides to form
N-substituted amides. The reaction with acid anhydrides
is more easily controlled so the following preparation uses
ethanoic anhydride rather than ethanoyl chloride.
phenylamine
phenylamine ethanoic acid
ethanoic acid ethanoic acid ethanoic
A Mix 5.0 cm3 of ethanoic anhydride with 5.0 cm3 of glacial ethanoic anhydride
ethanoic acid
anhydride Penylalanine
ethanoic ethanoic anhydride
(pure) ethanoic acid in a round-bottomed flask.
anhydride phenylamine
B Cool the flask in a beaker of cold water and add 5.0 cm3 of
phenylamine, dropwise, with gentle shaking. Questions
C Add anti-bumping granules, fit a reflux condenser and reflux 1 Write an equation for the reaction between phenylamine and
the mixture for 30 minutes. ethanoic anhydride to produce N-phenylethanamide.
D Pour the liquid from the flask into a beaker containing 2 Apart from difficulty in controlling the reaction rate, give
100 cm3 of cold water. Stir, then allow the mixture to stand another disadvantage of using ethanoyl chloride rather than
until no more crystals are formed. Filter off the crystals ethanoic anhydride in this preparation.
under reduced pressure. 3 Suggest why phenylamine is added dropwise to cold
E Wash the crystals with cold water and recrystallise from the ethanoic anhydride.
minimum volume of boiling water. 4 Explain the function of the anti-bumping granules in step C.
F Dry the crystals between filter papers and then by storing in 5 Give three practical details in step D which help to keep the
a desiccator. loss of product to a minimum.
G Weigh the dry crystals and measure their melting temperature. 6 Describe how the crystals are washed with cold water in step E.
7 Name a possible drying agent to use in the desiccator.
Tip 8 The yield of N-phenylethanamide is 70.0% of the
theoretical yield. Calculate the actual mass obtained given
For practical guidance, refer to Practical skills sheet 13,
that ethanoic anhydride is in excess and the density of
‘Assessing hazards and risks’, and also to Practical skills
phenylamine is 1.02 g cm−3.
sheet 21, ‘Synthesis of an organic solid’, both of which you
9 What are the hazards posed by this preparation and how is
can access online at [Link]/
their risk reduced?
EdexcelChemistry.
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Chapter summary
l Decreasing the carbon chain can be achieved
Chapter 18.3 Organic synthesis
by reacting methyl ketones or methyl secondary
l Combustion analysis of a weighed sample of an alcohols with iodine and sodium hydroxide. A
organic compound gives the masses of carbon yellow precipitate of triiodomethane is formed
dioxide and water formed. From these, the masses together with a carboxylate salt with one fewer
of carbon and hydrogen in the sample can be found carbon.
and hence the empirical formula of the compound. l Chemists have developed a range of practical
l The molar mass of the compound is needed to find
techniques and procedures for the synthesis of solid
the molecular formula from the empirical formula. and liquid organic compounds.
The mass spectrum of the compound can be used l The key stages in the preparation of an organic
to find the molar mass. compound are planning, carrying out the reaction,
l A molecular formula can represent many different
separating the crude product from the reaction
isomers. The structural formula of a particular mixture, purifying the product and measuring
compound shows the functional groups and the yield, identifying the product and checking its
how the atoms link together in a molecule. purity.
Traditionally, chemical tests were used to identify l Planning: suitable reacting quantities are worked
functional groups in organic molecules. Nowadays, out from the equation with the conditions for
modern laboratories rely on highly sensitive, the reaction. Appropriate control measures are
automated and instrumental techniques to identify identified to reduce risk, based on data about
compounds. These include chromatography, mass hazards.
spectrometry and infrared and particularly nuclear l Carrying out the reaction may involve heating the
magnetic resonance spectroscopy. reactants: this may be in a flask fitted with a reflux
l The characteristic properties and tests for
condenser to prevent loss of reagents.
functional groups are shown on the data sheets l The crude product may be separated from the
headed ‘Tests and observations on organic reaction mixture by filtration, distillation (simple,
compounds’ and ‘Tests for gases’. These tests are fractional or steam distillation) or by solvent
used in Core practical 15 (part 2): Analysis of some extraction.
organic unknowns. l Purifying the ‘crude’ product: the usual technique
l Organic chemists synthesise new molecules using
for purifying solids is recrystallisation using the
their knowledge of functional groups. A synthetic minimum amount of hot solvent. The solid can
pathway leads from the reactants to the required then be dried in a desiccator. Liquids may be
product in one step or several steps. Chemists purified by fractional distillation or by washing
normally seek a synthetic route that has the least using a separating funnel and then drying.
number of steps and produces a high yield of the l Comparing the actual yield with the yield expected
product. from the chemical equation is a good measure of
l Chemists sometimes need to change the carbon
the efficiency of a process.
skeleton of a molecule. Increasing the chain length l The purity and identity of a solid can be checked
by one can be done by reacting cyanide ions with by showing that it melts sharply at the expected
halogenoalkanes. One or more carbon atoms temperature. If a liquid is pure it will all distil at
can be added using Grignard reagents which are the expected boiling temperature or in a narrow
formed from halogenoalkanes and magnesium range including it. The boiling temperature can be
in dry ether. Grignards react with carbonyl measured as the liquid distils over during fractional
compounds to form alcohols and with carbon distillation.
dioxide to form carboxylic acids.
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Exam practice questions
1 A series of tests was carried out on three D Add two spatula measures of anhydrous
organic compounds, A, B and C. The results of sodium sulfate or anhydrous calcium
the tests are described below. chloride to the organic product.
E Finally, redistill the organic product,
State the deductions which you could make
collecting the liquid that boils between
from the tests on each of A, B and C. You are
75 and 79 °C.
not expected to identify compounds A, B and C.
a) Draw a diagram of the apparatus for
a) i) A is a colourless liquid which does
heating under reflux.(3)
not mix with water.(1)
b) State the reasons for each of the procedures
ii) After warming a few drops of A
in steps A to E.(8)
with aqueous sodium hydroxide, the
resulting solution was acidified with Salicylic acid has been used as a painkiller. Its
3
nitric acid. Silver nitrate solution was displayed formula is shown in the diagram.
then added and a cream-coloured
O
precipitate formed.(2)
b) i) B is a white solid that chars on heating C
and gives off a vapour which condenses OH
to a liquid that turns cobalt(ii) chloride OH
paper from blue to pink.(2)
ii) A solution of B turns universal
a) Identify the functional groups in salicylic
indicator red. (1)
acid.(2)
iii) A solution of B reacts with aqueous
b) Write the molecular formula of salicylic
sodium carbonate to produce a
acid.(1)
colourless gas that turns limewater
c) Draw the displayed formula of the organic
milky.(2)
product that forms when salicylic acid:
iv) When a little of B is warmed with
i) is heated under reflux with ethanol and
ethanol and one drop of concentrated
concentrated sulfuric acid(1)
sulfuric acid, a sweet-smelling product
ii) reacts with bromine water(1)
can be detected on pouring the
iii) is warmed with aqueous sodium
reaction mixture into cold water.(2)
hydroxide.(2)
c) i) C is a liquid that burns with a very
smoky, yellow flame.(1) 4 A sample of the hydrocarbon limonene was
ii) C does not react with sodium obtained from the peel of oranges. Small
carbonate solution.(1) pieces of chopped orange peel were placed in
iii) C fizzes with sodium and gives off a gas a 250 cm3 round-bottomed flask with 100 cm3
that produces a ‘pop’ with a burning of water. The flask was then heated on a wire
splint.(2) gauze as shown in the diagram. About 50 cm3
of liquid was collected.
2 The following steps were used in one method
of synthesising ethyl ethanoate (boiling
H2
temperature 77 °C). C CH3
A Heat ethanol and ethanoic acid under CH2 C
reflux for about 45 minutes with a little
CH3
concentrated sulfuric acid. CH CH
B Then, distil the reaction mixture, collecting C C
H2
all the liquid that distils below 84 °C.
CH2
C Shake the distillate with aqueous sodium
carbonate solution. limonene
590
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c) Compound D is a stereoisomer.
thermometer (0 –100 °C)
i) Draw the structure of D and identify
each stereochemical component with
an asterisk.(2)
condenser
ii) For each component, state the type of
stereoisomerism involved.(2)
water in iii) Deduce how many stereoisomers there
are with the skeletal formula shown.
water out
Explain your answer.(3)
chopped zest
from two fresh 6 a) Two isomeric compounds, F and G, with
oranges the molecular formula C3H8O can be
oxidised to H and J respectively.
gauze heat
H reacts with Fehling’s solution to produce
limonene a red-brown precipitate of copper(i) oxide.
water J has no reaction with Fehling’s solution,
but gives a yellow crystalline product with
a) State why the round-bottomed flask was 2,4-dinitrophenylhydrazine.
heated on a wire gauze.(1) Give the structural formulae and names
b) State why it was necessary to chop the of F, G, H and J.(8)
orange rind into small pieces.(1) b) i) Describe what you would observe
c) Name the process used to obtain the and name all the products formed
limonene and outline how limonene is when H is warmed with sodium
obtained from the liquid collected.(2) dichromate(vi) solution acidified
d) The sample of limonene that you have with dilute sulfuric acid.(4)
collected is probably contaminated with a ii) Write an equation or equations for
little water. State how could you dry your the reactions taking place.(3)
sample of limonene.(1)
e) State why limonene is not obtained by 7 An organic compound, X, containing carbon,
heating the orange rind alone without hydrogen and oxygen only, was found to have
adding water.(1) a relative molecular mass of about 70. When
f) Name the functional group present in 0.36 g of the compound was burned in excess
limonene and suggest two simple tests to oxygen, 0.88 g of carbon dioxide and 0.36 g of
show the presence of this functional group water were formed.
in limonene.(3) a) Calculate the empirical formula and the
molecular formula of X. (5)
5 The skeletal formula of compound D is shown b) X reacted with 2,4-dinitrophenylhydrazine
below. D is a constituent of jasmine oil and it is to produce an orange solid. State what you
partly responsible for the taste and smell of black tea. can deduce from this. (1)
c) Write all the possible non-cyclic structural
O
formulae for X and give the systematic
name for each formula. (3)
13
d) The C NMR spectrum of compound X
shows that there are carbon atoms in three
different environments in the molecule of
O O X. Identify which is the correct structure for
Compound D X and describe how you could confirm this
using the product from the reaction with
a) Deduce the molecular formula of D.(1) 2,4-dinitrophenylhydrazine.(3)
b) Name the functional groups in D.(3)
591
Exam practice questions
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8 Explain how you would distinguish between b) State what you can deduce from the
the members in each of the following pairs of observations about the functional groups
compounds using one simple chemical test for present in Y. (4)
each pair. c) Draw a possible structural formula for Y.(1)
In each case, state the reagents and conditions 10 The solid halogenoalkane D is known to
used and describe what happens with each contain bromine, but no other halogen.
compound during the test. a)* Describe an experiment to determine the
percentage by mass of bromine in D.(6)
a) CH3CH2CH2Cl and CH3CH2CH2I (4)
b) Explain the chemistry of your method
b) CH3COCH2CH2CH3 and and show how you would calculate the
CH3CH2COCH2CH3 (3) percentage of bromine in D from your
c) (3)
CH3 C CH2 O CH3 measurements.(5)
CH3 C CH2 O CH3
11 Give synthetic routes to prepare:
O
O a) N-butyl ethanamide in three steps
and starting from 1-bromopropane(6)
CH3 C O CH2CH3 b) but-2-ene using ethene as the only
CH3 C O CH2CH3
organic reagent. (13)
O
O Your answers should use as few steps as possible
and should give the reagents, conditions and
d) CH3 OH (3) product of each step. Balanced equations are
CH3 OH
not required.
and 12 Gaseous chloromethane reacts with an alloy of
aluminium and sodium to form the liquid A.
CH2OH
CH2OH The composition by mass of compound A is
50.0% carbon, 12.5% hydrogen and 37.5%
e) CH3CH2CH2CONH2 and aluminium.
CH3CH2CH2CH2NH2 (3)
0.24 g of A reacts with excess water to produce
9 Substance Y is an organic compound. The 0.24 dm3 of a gas, B, and a white gelatinous
mass spectrum of Y showed that its molecular precipitate, C. The 0.24 dm3 of B has a mass of
ion has a mass-to-charge ratio of 132. When 0.16 g. C dissolves in hydrochloric acid and in
2.64 g of Y were completely combusted in pure sodium hydroxide solution.
dry oxygen, the only products were 7.92 g of
(The volume of gas B was measured at room
carbon dioxide and 1.44 g of water.
temperature and pressure, at which the molar
Tests were carried out on Y and the following volume of a gas = 24 dm3 mol−1. H = 1.0,
observations recorded. C = 12.0, Al = 27.0)
Observations a) Calculate the empirical formula of A.(2)
Burns with a smoky flame b) Give a structural formula for A. (1)
The yellow/orange colour of bromine water is c) Calculate the molecular mass of B.(1)
decolourised d) Identify the compounds B and C.(2)
A yellow/orange precipitate is produced with e) Write equations for:
2,4-dinitrophenylhydrazine i) the reaction of chloromethane with
A silver mirror is formed with Tollens’ reagent. the Al/Na alloy(1)
ii) the reaction of A with excess water(3)
a) Calculate the mass of carbon in 7.92 g of
iii) the reaction of C with hydrochloric
carbon dioxide and the mass of hydrogen
acid (1)
in 1.44 g of water. Hence, calculate the
iv) the reaction of C with sodium
molecular formula of Y.(6)
hydroxide solution.(1)
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Modern analytical techniques II
19
19.1 Analytical techniques
Key terms Analytical chemists use a combination of techniques using spectroscopes and
spectrometers to identify organic compounds and determine their structures.
The instruments used for analysis ● Mass spectrometry gives the relative molecular mass of a compound and
are variously called spectroscopes can suggest a likely structure for a compound from fragmentation peaks.
(emphasising the use of the ● Infrared spectroscopy shows the presence of particular functional groups
techniques for making observations) by detecting their characteristic vibration frequencies.
or spectrometers (emphasising the ● Nuclear magnetic resonance (NMR) techniques help to detect groups
importance of measurements). with carbon atoms and hydrogen atoms in particular environments in
molecules and is the most useful tool for determining structure.
Tip 19.2 Mass spectrometry
Sections 19.2 and 19.3 of this Mass spectrometry is used to determine the relative molecular masses and
chapter revisit ideas first introduced in molecular structures of organic compounds. In this way it can be used to
Chapter 7. Some of the ‘Test yourself’ identify unknown compounds.
questions are also designed to help you
The combination of gas chromatography (GC) with mass spectrometry is
revise ideas from the first year of the A
of great importance in modern chemical analysis (Section 19.5). First, gas
Level course.
chromatography separates the chemicals in an unknown mixture, such
as a sample of urine; then mass spectrometry detects and identifies the
components (Figure 19.1).
Figure 19.1 Using a mass spectrometer in a forensic laboratory to detect drugs in a
urine sample.
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Ionisation of the Mass analyser Ion detector giving an
Gaseous sample sample by separating ions by electrical signal which is
bombardment mass-to-charge converted to a digital
from inlet system ratio, e.g. by magnetic
with electrons or response that is stored in
other methods field or time of flight a computer
Figure 19.2 A schematic diagram to show All mass spectrometers have the components shown in Figure 19.2.
the key features of a mass spectrometer.
In a mass spectrometer, a beam of high-energy electrons bombards the
molecules of the sample. This turns them into ions by knocking out one or
Tip more electrons.
Inside a mass spectrometer there is a Bombarding molecules with high-energy electrons not only ionises them
high vacuum. This allows ionised atoms but usually splits them into fragments. As a result, the mass spectrum consists
and molecules from the chemical of a ‘fragmentation pattern’.
being tested to be studied without
interference from atoms and molecules Molecules break up more readily at weak bonds or at bonds which give rise
in the air. to stable fragments. The highest peaks correspond to positive ions which are
relatively more stable, such as tertiary carbocations or ions such as RCO+
(the acylium ion) or the fragment C6H5+ from aromatic compounds related
to benzene, C6H6.
After ionisation and fragmentation, the charged species are separated to
produce the mass spectrum that distinguishes the fragments on the basis of
their ratio of mass to charge (m/z).
Chemists study mass spectra with these ideas in mind and, as a result, can
gain insight into the structure of new molecules. They identify the fragments
from their masses and then piece together likely structures with the help of
evidence from other methods of analysis, such as infrared spectroscopy and
NMR spectroscopy.
Chemists have also built up a very large database of mass spectra of known
compounds for use in analysis. They regard the spectra in databases as ‘fingerprints’
for identifying chemicals during analysis. The computer of a mass spectrometer
is programmed to search its database to find a good match between the spectrum
of a compound being analysed and a spectrum in the database (Figure 19.3).
Figure 19.3 The mass spectrum of methyl 100 105
benzoate, C6H5COOCH3. The pattern
of fragments is characteristic of this
80
compound.
Relative intensity/%
77
60
40 136
51
20
0
25 50 75 100 125
m/z
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Analysing mass spectra 64
100
When analysing molecular compounds, the peak of the ion with the largest
m/z value is usually the whole molecule ionised. So the mass of this ‘parent Tip
ion’, M+, gives the relative molecular mass of the compound. Take care with when describing the
28
peak for the molecular ion. Do not
Relative intensity/%
The presence of isotopes shows up in spectra of organic compounds containing
simply use the word highest to describe
29
chlorine or bromine atoms (Figure 19.4). Chlorine has two isotopes, 35Cl
the peak because ‘highest’ could be
and 37Cl. Chlorine-35 is three times more abundant than chlorine-37.
50 If a 27
confused with tallest, which applies to
molecule contains one chlorine atom, its two molecular ions appear as two
the most abundant ion.
peaks separated by two mass units. The peak with the lower value of m/z is 49 66
three times higher than the peak with the higher value of m/z. 26
51
Bromine consists largely of two isotopes, 79Br and 81Br, in roughly equal
proportions. If a molecule contains one bromine atom, the molecular
0 ion
0 20 40 60 80 100
shows up as two peaks of roughly equal intensity separated by two mass units. m/z
64 15
100 100
94
28
Relative intensity/%
29 Relative intensity/%
50 27 50
49 66
26
79
51
28 47
0 0
0 20 40 60 80 100 0 20 40 60 80 100
m/z m/z
Figure 19.4 The mass spectra of two compounds containing halogen atoms.
15
100
High-resolution mass spectrometry
94
Modern mass spectrometers can measure a relative isotopic mass to four
or five decimal places. Using these values, the relative atomic mass of
Relative intensity/%
an element can be found to four decimal places and therefore the accurate Table 19.1 Relative atomic masses to four
relative molecular mass of a compound can be calculated, given these accurate decimal places.
50
relative atomic masses. This makes it possible to identify one compound Element Relative atomic mass
from several with the same integral mass. H 1.0079
For example, both C4H10 and C3H6O have relative molecular mass equal to C 12.0107
58 to the nearest whole number. 79 N 14.0067
Table 19.1 shows relative
28 47
atomic masses to four decimal places. O 15.9994
0
0 20 40 60 80 100
m/z
Key terms
Tip
The relative isotopic mass of an isotope is the mass of the isotope on a scale on
‘Relative isotopic mass’ refers to an
which a 12C atom has a mass of exactly 12.000 units.
individual isotope. ‘Relative atomic
The relative atomic mass of an element is the weighted average of the masses of its mass’ refers to the mixture of isotopes
isotopes on a scale on which a 12C atom has a mass of exactly 12.000 units. of an element.
19.2 Mass spectrometry 595
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Using the figures from Table 19.1, the relative molecular mass of C4H10 is
Tip 58.1218 and that of C3H6O is 58.0789, both to four decimal places. It is
Note that the distinction using accurate therefore possible to distinguish between the two compounds.
Mr values is between different
molecular formulae. Since a molecular Test yourself
formula may represent several isomers,
1 An organic molecule M can be represented as a combination of two
the structure of the particular isomer
parts: m1.m2. Draw a diagram to represent the ionisation and then
cannot be determined using the Mr
fragmentation of the molecule and explain why only one of the two
value. However, an isomer may be
fragments shows up in the mass spectrum.
identified using the fragmentation
pattern of its molecular ion or, 2 Suggest the identity of the peaks labelled in the mass spectrum
alternatively, by studying its reactions. shown in Figure 19.3.
3 Account for these facts about the mass spectrum of dichloroethene:
a) It includes three peaks at m/z values of 96, 98 and 100 with
intensities in the ratio 9 : 6 : 1.
b) It includes two peaks at m/z values 61 and 63 with intensities in
the ratio of 3 : 1.
4 One of the mass spectra in Figure 19.4 is bromomethane and the
other is chloroethane. Match the spectra to the compounds and
Tip identify as many fragments in the spectra as you can.
Section 19.3 revisits the content of 5 An organic compound is found to have the accurate Mr = 46.0682.
Section 7.2 and is included here as Use the data in Table 19.1 to decide which of the following molecular
revision. formulae is correct: CH2O2, CH6N2 or C2H6O.
Tip 19.3 Infrared spectroscopy
Infrared absorptions are measured in Spectroscopists have found that it is possible to correlate absorptions in the
wavenumbers with the unit cm−1. The region 4000–1500 cm−1 of an absorption spectrum with the stretching or
wavenumber is the number of waves in bending vibrations of particular bonds. As a result, the infrared spectrum
1 cm. gives valuable clues to the presence of functional groups in organic molecules.
radiation
sample detector computer printer
source
Key term
Figure 19.5 The essential features of a modern single-beam IR spectrometer.
An absorption spectrum is a plot
showing how strongly a sample absorbs The important correlations between different bonds and observed absorptions
radiation over a range of frequencies. are shown in Figure 19.6. Hydrogen bonding broadens the absorption peaks
Absorption spectra from infrared of –OH groups in alcohols and even more so in carboxylic acids.
spectroscopy give chemists valuable
Wavenumber ranges
information about the bonding and
4000 cm–1 2500 cm–1 1900 cm–1 1500 cm–1 650 cm–1
structure of chemicals.
C H C C C C
O H C N C O
N H fingerprint
Tip single bond triple bond double bond
region
stretching stretching stretching
Most organic molecules contain vibrations
vibrations vibrations
C–H bonds. As a result most organic
compounds have a peak at around Figure 19.6 A chart to show the main regions of the infrared spectrum and important
3000 cm−1 in their IR spectrum. correlations between bonds and observed absorptions.
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Tra
A B
0
4000 3000 2000 1500 1000 500
–1
Wavenumber/cm
Molecules with several atoms can vibrate in many ways because the vibrations
of one bond affect others close to it. The complex pattern of vibrations,
particularly in the region 1500–650 cm−1, can be used as a ‘fingerprint’ to be
matched against the recorded IR spectrum in a database.
100 100
Transmittance/%
Transmittance/%
50 50
A B
C
0 0
4000 3000 2000 1500 1000 500 4000 3000 2000 1500 1000 500
Wavenumber/cm–1 Wavenumber/cm–1
Figure 19.7 Two IR spectra.
Test yourself
6 Figure 19.7 shows the infrared spectra of ethyl ethanoate and
100
ethanamide. Use the Pearson Edexcel Data booklet to work out:
a) which bonds give rise to the peaks marked with letters
b) which spectrum belongs to which compound.
Transmittance/%
7 A compound P is a liquid which does not mix with water; its molecular
formula is C7H6O and it has an infrared spectrum with strong, sharp
50
peaks at 2800 cm−1, 2720 cm−1 and 1700 cm−1, with a weaker
absorption peak between 3000 and 3100 cm−1. Oxidation of P gives
a white crystalline solid Q with a strong broad IR absorption band
in the region 2500–3300 cm−1 and another strong absorption at
C
1680–1750 cm−1.
0
a)
4000Suggest possible2000
3000 structures for P and
1500 Q.
1000 500
–1
Wavenumber/cm
b) What chemical tests could you use to check your suggestions?
19.4 Nuclear magnetic resonance
spectroscopy (NMR)
Nuclear magnetic resonance spectroscopy (NMR) is a powerful analytical
technique for finding the structures of carbon compounds. The technique is
used to identify unknown compounds, to check for impurities and to study
the shapes of molecules.
This type of spectroscopy studies the behaviour of the nuclei of atoms in
magnetic fields. It is limited to those nuclei which behave like tiny magnets
because they have a property called spin. In common organic compounds
the only nuclei to do so are those of carbon-13 atoms, 13C, and of hydrogen
atoms, 1H. The nuclei of the much more common carbon-12, oxygen-16
and nitrogen-14 atoms do not show up in NMR spectra.
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When placed in a very strong magnetic field, magnetic nuclei line up either in
Tip the same direction as the field or in the opposite direction to it. Those aligned
The theory of NMR is not required for in the same direction are slightly lower in energy and the difference between
your examinations. Questions will only the two energy levels is known as the energy gap (Figure 19.8).
expect you to interpret the spectra
obtained and deduce from them
information about the structure of radio
applied energy
molecules. field frequency gap
Figure 19.8 Energy levels and the energy gap for protons in an applied magnetic field.
This energy gap corresponds to a particular frequency in the radio-frequency
range. When exactly the right frequency is supplied, a proton in the lower
energy level can flip into the higher level and the protons are said to be in
resonance. The radio-frequency needed for this is recorded.
radio- radio-
frequency frequency computer recorder
oscillator detector
probe at room temperature
containing the sample which is
spun at 20 times per second to
average any inhomogeneity
N S in the sample or the glass tube
powerful
magnet
Figure 19.9 A schematic diagram to show the key features of NMR spectroscopy.
The tube with the sample is supported in a strong magnetic field in the
spectrometer (Figure 19.9). The operator turns on a source of radiation at
radio-frequencies. The radio-frequency detector records the intensity of the
signal from the sample as the oscillator emits pulses of radiation across a
range of wavelengths.
The sample is dissolved in a solvent. Also in the solution is some
Key term tetramethylsilane (TMS), which is a standard reference compound that
produces a single, sharp absorption peak well away from the peaks produced
A standard reference compound is by samples for analysis.
added to the solution of a substance
being tested by NMR. The standard Each peak corresponds to one or more magnetic atoms in a particular
produces as single sharp absorption chemical environment. Nuclei in different parts of a molecule experience
peak well away from other peaks and slightly different magnetic fields in an NMR machine. This is because
the position of other peaks is compared they are shielded to a greater or lesser extent from the field applied by the
to this peak. spectrometer by the tiny magnetic fields associated with the electrons of
neighbouring bonds and atoms.
The recorder prints out a spectrum that has been analysed by computer to
show peaks wherever the sample absorbs radiation strongly. The zero on the
scale is fixed by the absorption of magnetic atoms in the reference chemical.
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The distances of the sample peaks from this zero are called their ‘chemical
shifts’. The symbol used for chemical shift is δ. Tables of chemical shifts are
Key term
included in the Pearson Edexcel Data booklet.
The horizontal scale of an NMR
spectrum shows the chemical shifts
Carbon-13 NMR of the peaks measured in parts per
Carbon-13 NMR relies on the magnetic properties of the 13C isotope. This million (ppm). The symbol δ stands for
isotope makes up only about 1% of all naturally occurring carbon atoms, but the ‘chemical shift’ relative to the zero
this is enough for a signal to be detected in an NMR machine. on the scale that is given by the signal
from tetramethylsilane.
Figure 19.10 shows the 13C NMR spectrum for ethanol. Spectra of this
kind are available from the Spectral Data Base System (SDBS) for Organic
Compounds at the National Institute of Advanced Industrial Science and
Technology (AIST), Japan.
Figure 19.10 13C NMR spectrum for
ethanol dissolved in CDCl3.
Signal strength
200 180 160 140 120 100 80 60 40 20 0
Chemical shift, δ/ppm
There are two peaks in the 13C NMR spectrum for ethanol. This reflects
the fact that there are two carbon atoms in an ethanol molecule and they
are in different environments. The carbon in the CH3 group is attached to
three hydrogen atoms and a carbon atom. The carbon in the CH2 group is
attached to two hydrogen atoms, a carbon atom and an oxygen atom.
Spectra of the type shown in Figure 19.10 are usually recorded with the
sample in solution. The chosen solvent is commonly CDCl3. The molecules
of CDCl3 contain one carbon atom and so produce a single line in 13C
spectra that is easy to recognise. This line is usually removed from the spectra
in databases such as SDBS to avoid any confusion. The line produced by
the solvent is not shown in any of the 13C spectra in this book.
Also omitted from Figure 19.10 is the peak at zero produced by the reference
chemical tetramethylsilane (TMS). The chemical shifts are measured relative
to the TMS peak at the zero mark. This peak is usually removed from the
spectra for clarity.
Test yourself
8 a) What is the difference in the structure of the nuclei of 12C and 13C
atoms?
b) What is the difference between the formula of CDCl3 and the
formula of trichloromethane?
9 Other than providing a suitable peak, what other requirements must
there be for a standard added to a solution in an NMR test?
10 TMS is related to silane, SiH4, but has the four hydrogen atoms
replaced by methyl groups. Why does this reference chemical
produce just one peak in 13C NMR spectra?
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Figure 19.11 illustrates the principles of 13C NMR in a more complex
example. Here, too, every carbon atom, or group of carbon atoms, in a
chemically distinct environment gives a separate peak in the NMR spectrum.
Signal strength
200 180 160 140 120 100 80 60 40 20 0
Chemical shift, δ/ppm
Figure 19.11 13C NMR spectrum for 4-methylpentan-2-one dissolved in CDCl3.
There are six carbon atoms in 4-methylpentan-2-one but only five peaks in
the 13C NMR spectrum. The reason for this is shown in Figure 19.12. The
carbon atoms of two of the methyl groups in the molecule are in exactly the
same environment so they give rise to only one peak. The other four carbon
atoms, including the carbon atom in the third methyl group, are in slightly
different environments and give rise to separate peaks.
Figure 19.12 The structure of E
4-methylpentan-2-one labelled to show
A
the five different environments for the O CH3
six carbon atoms. The two carbon atoms
H3C C CH2 C CH3
labelled E are in exactly the same chemical
environment. H
B C D
Test yourself
11 Predict the number of peaks in the 13C NMR 13 Deduce the number of peaks in the 13C spectrum
spectrum of: of ibuprofen (Figure 19.13).
a) pentane
b) propyl ethanoate OH
c) 2-methylbutanal
O
d) benzene
e) methyl benzene.
12 E
xplain how you could distinguish between
propanal and propanone by inspection of the 13C Figure 19.13 A skeletal formula of ibuprofen.
NMR spectra of the two compounds.
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Generally carbon atoms attached to electronegative atoms such as oxygen
show the largest chemical shifts. Chemists have found that it is possible to
draw up tables to show the likely range of values for the chemical shifts of
13C atoms in different environments. For example, 13C atoms in alkanes
typically give chemical shifts in the range 5–50 ppm, while carbon atoms
in the carbonyl groups of aldehydes and ketones give chemical shifts in the
range 190–220 ppm (see the Pearson Edexcel Data booklet).
Tip
You are only expected to interpret 13C NMR spectra in which each chemical
environment for carbon atoms is represented by a single peak. Also note that you
cannot draw any conclusions by looking at the size of the peaks in 13C NMR. In these
two ways 13C NMR differs from proton NMR, where the peaks may be split and the
area of each peak is significant.
Test yourself
14 Refer to the data sheet from the Pearson Edexcel Data booklet
showing chemical shifts for 13C NMR. To what extent do the data
show that the presence of electronegative atoms increases the
chemical shift values?
se the 13C NMR chemical shift values from the data sheet to
15 U
suggest which carbon atoms give rise to which peaks in the
spectrum of:
a) ethanol (Figure 19.10)
b) 4-methylpentan-2-one (Figure 19.11).
16 Sketch the 13C NMR spectrum you would expect to observe for:
a) ethyl ethanoate
b) cyclohexene.
Proton NMR spectroscopy Tip
Information obtained from proton NMR spectra The word ‘proton’ is used in NMR
Proton NMR relies on the magnetic properties of the 1H isotope. As spectroscopy, but there are no H+
with 13C NMR, the number of main peaks in the spectrum shows how ions present in these molecules. The
many different chemical environments there are for hydrogen atoms. The term strictly refers to the proton in the
values of the chemical shifts are a useful indication of the types of chemical nucleus of the 1H atoms which are
environment for the hydrogen atoms corresponding to each peak. affected by the magnetic field.
Even more information can be deduced from a proton NMR spectrum
than from a 13C spectrum because it is possible to work out the number of
hydrogen atoms in each environment. In a proton NMR spectrum the area
under a peak is proportional to the number of nuclei.
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An NMR instrument (Figure 19.14) can work out the ratios of the areas
Tip below the peaks. Sometimes the results of the calculation are shown by an
The range of chemical shifts in integration trace, such as the blue line shown in the spectrum for methyl
1H spectroscopy is 0 to 12 ppm ethanoate in Figure 19.15. Alternatively, the instrument’s computer prints
compared with 0 to 220 ppm for 13C a number below each peak that is a measure of the relative area under the
spectroscopy. Because of this much curve, as shown for the spectrum in Figure 19.16.
smaller range, there is much greater
overlap of peaks.
Signal strength
integration trace
10 9 8 7 6 5 4 3 2 1 0
Chemical shift, δ/ppm
Figure 19.15 Proton NMR spectrum of methyl ethanoate in CDCl3.
Signal strength
Figure 19.14 Researcher adding a sample
to an NMR spectrometer.
17.3 integration value 155.4
Key term 10 9 8 7 6 5 4 3 2 1 0
Chemical shift, δ/ppm
Integration is a mathematical technique Figure 19.16 Proton NMR spectrum of a compound C5H10O in CDCl3.
that works out the area under a curve.
The ratios of the results of integrating the Low-resolution proton NMR spectra, such as those in Figures 19.15 and
peaks in a proton NMR spectrum show 19.16, show the main peaks but no fine detail.
the relative numbers of hydrogen atoms
Proton NMR spectra, like carbon-13 NMR spectra, are usually recorded
in each chemical environment.
with the sample in solution. It is important that the solvent does not
contain hydrogen atoms that would give peaks with chemical shifts
similar to those in the sample. One possibility is to use a solvent that
contains no hydrogen atoms, such a tetrachloromethane, CCl4. The other
is to use a solvent in which atoms of the 1H isotope have been replaced by
Tip deuterium atoms. A compound that is often used is CDCl 3. Deuterium
The integration curve gives a ratio atoms produce peaks in regions of the spectrum well away from the
of numbers of hydrogens and not chemical shifts for proton NMR.
necessarily the actual number. In
The reference chemical for proton NMR is again tetramethylsilane (TMS).
Figure 19.15 the two jumps are equal,
There are 12 hydrogen atoms in a molecule of TMS and they all have the
so all we can deduce is that the two
same chemical environment. TMS provides a single strong peak that marks
types of proton are in a 1 : 1 ratio.
the zero on the scale of chemical shifts.
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CH3 C CH3
Test yourself CH3
CI
C
CH2
CH3
C CH3
CH3 C CH3
O
17 Refer to the proton NMR spectrum in Figure 19.15 and
O the Pearson
O
Edexcel Data booklet for chemical shifts.
a) Explain why there are two peaks in the spectrum.
CI CH2 C CH3
b) Use the chemical shift values to decide
H3C which hydrogen
CH2 atoms
CH
CH
CI 3 CH C 2 CH C3 CH3 3
give rise to each peak. CH3 C C C CHO3
O theOintegration trace is as expected for
c) Show that CH2the molecule.
CH3
H C 3 O
18 Deduce the number of peaks in the low-resolution proton NMR
spectrum of each compound. H 3C CH2 CH3
3 HCIC CH2 C
CH2 CHCH
3 3 C OHC
a) CH 3 C CH3 CI2C CH
b) H 2 C CH3
C C O H3C CH2 CH3
H3C O CH2 CH3 H2C OH O
c) H3C CH2 CH3 d) H2C OH
CI CH2 C CH3 H3C CH2 CH3
H2C C OHC
H2 C COH C
O
HH3CC
2 OH CH2 CH3
H3C CH2 CH3
e) HH32CC OH CH2 CH3 f)
C C H2C OH
H2C OH
H3C CH2 CH3 H2C OH
19 R
efer to the proton NMR spectrum in Figure 19.16 and the chemical
C from
shiftH2data OH the Pearson Edexcel Data booklet.
a) How OH different chemical environments are there for
H2C many
hydrogen atoms in the molecule?
b) Use the integration values under each peak to work out the
ratios of hydrogen atoms in each environment.
c) Use your answers to (a) and (b) and chemical shift values to
suggest a structure for the compound.
d) Describe two chemical tests that could be used to confirm the
presence of the main functional group in the molecule. State
what you would do and what you would expect to observe.
20 With the help of the chemical shift data from the Pearson Edexcel
Data booklet, sketch the low-resolution proton NMR spectrum you
would expect for:
a) butanone
b) 2-methylpropan-2-ol.
Key term
Tip Equivalent protons are those in the
Do not confuse the solvent with the standard. The solvent, such as CDCl3, does not same chemical environment with the
contain protons so does not give rise to any peaks in the 1H NMR spectrum. The standard, same chemical shift. They do not couple
TMS, has 12 equivalent protons and a peak for it may be seen in spectra at δ = 0. with each other.
19.4 Nuclear magnetic resonance spectroscopy (NMR) 603
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Key term Coupling
At high resolution it is possible to produce proton NMR spectra with
Non-equivalent protons are those in more detail that provide even more information about molecular structures
different chemical environments with (Figure 19.17). The spins of non-equivalent protons connected to
different chemical shifts. Protons on neighbouring carbon atoms interact with each other. Chemists call this
adjacent, that is neighbouring, carbon interaction ‘spin–spin coupling’ and they find that the effect is to split the
atoms couple with each other. peaks into a number of lines.
Signal strength 2
TMS
reference
10 9 8 7 6 5 4 3 2 1 0
Chemical shift, δ/ppm
Figure 19.17 A high-resolution proton NMR spectrum for a hydrocarbon with a benzene
ring. Note the extra peaks compared with a low-resolution spectrum.
A peak from protons bonded to an atom that is next to an atom with two
protons splits into three lines, with the central line being twice as large as the
other two. This happens because there are three energy states available to the
two protons, depending on whether each proton is aligned with or against
the magnetic field:
● both aligned with the field
● one aligned with the field and one against the field (with two possible
combinations)
● both aligned against the field.
For similar reasons, a peak from protons bonded to an atom that is next to
an atom with three protons splits into four lines. In general the ‘n + 1’ rule
makes it possible to work out the splitting, where n is the number of protons
on the adjacent atom. The number of lines and their intensities can also be
worked out using Pascal’s triangle (Figure 19.18).
Figure 19.18 Pascal’s triangle predicts number of splitting pattern and
equivalent protons relative intensity of
the pattern of peaks and the relative peak causing splitting the peaks
heights.
1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
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Example
Explain the splitting pattern in the spectrum of ethanol, shown in Figure
19.19.
(3H)
Signal strength
(2H)
(1H)
TMS
reference
6 5 4 3 2 1 0
Chemical shift, δ/ppm
Figure 19.19 The high-resolution spectrum of ethanol.
Answer
Consider the peak at δ = 1.2 (integration value 3); this corresponds to
the CH3 group.
The group adjacent to the CH3 in the molecule is a CH2 group, so the
number of adjacent protons is 2, therefore n = 2.
Using the n + 1 rule gives n + 1 = 3 and so the peak is a triplet.
Therefore the CH3 peak is a triplet because it is split by the two protons
in CH2.
Consider the peak at δ = 3.8 (integration value 2); this corresponds to
the CH2 group.
The group adjacent to the CH2 in the molecule is a CH3 group, so the
number of adjacent protons is 3, therefore n = 3.
Using the n + 1 rule gives n + 1 = 4 and so the peak is a quartet.
Therefore the CH2 peak is a quartet because it is split by the three
protons in CH3.
The peak at δ = 2.5 due to the proton in the OH group is not split. This is
usual for hydrogens in alcohol groups and is explained in the section on
labile protons on page 606.
Tip
Ethyl groups, CH3CH2 –, are often present in organic molecules. If a proton NMR
spectrum contains a triplet (relative integration 3) and a quartet (relative integration 2)
this means that the molecule contains an ethyl group. The ethyl group is not adjacent
to other protons.
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Test yourself
21 Show that when a proton can couple with three protons in a methyl
group, there are four alignments of the methyl protons that can give
rise to peaks with relative intensities 1 : 3 : 3 : 1.
22 Predict the splitting pattern if a molecule contains
a) X– CH2CH2 –Y or
b) X– CH2CH2 –X, where X and Y contain no hydrogen atoms.
23 Use the chart of chemical shifts in the Pearson Edexcel Data
booklet to suggest a structure for the hydrocarbon with a benzene
ring that has the proton NMR spectrum shown in Figure 19.17.
24 Sketch the high-resolution proton NMR spectrum you would expect
to observe with:
a) propane
b) ethoxyethane.
Labile protons
Key term
Hydrogen bonding affects the properties of compounds with hydrogen
An atom is labile if it quickly and easily atoms attached to highly electronegative atoms such as oxygen or nitrogen
reacts or moves from one molecule to (Figure 19.20). These molecules can rapidly exchange protons as they move
another. from one electronegative atom to another. Chemists describe these protons
as labile.
δ–
O
δ+
C2H5 H C2H5
O δ–
H δ+
Oδ–
H C2H5
Figure 19.20 Hydrogen bonding in ethanol.
Labile protons do not couple with the protons linked to neighbouring atoms.
This means that the NMR peak for a proton in an –OH group appears as a
single peak in a high-resolution spectrum.
Note that in the high-resolution spectrum of ethanol (Figure 19.19) there is
no coupling between the proton in the –OH group and the protons in the
next-door –CH2 group.
A useful technique for detecting labile protons is to measure the NMR
spectrum in the presence of deuterium oxide (heavy water), D2O. Deuterium
nuclei can exchange rapidly with labile protons. Deuterium nuclei do not
show up in the proton NMR region of the spectrum and so the peaks of any
labile protons disappear.
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Test yourself
25 W
rite an equation to show the reversible b) How would the spectrum in Figure 19.21
exchange of deuterium and hydrogen nuclei change if 3-chloropropanoic acid were used in
between ethanol and deuterium oxide. the presence of D2O?
26 a) Account for the splitting pattern shown for c) Describe the high-resolution proton NMR
the peaks in the proton NMR spectrum of spectrum you would expect to observe with
3-chloropropanoic acid, Cl– CH2 – CH2 – COOH, in 2-chloropropanoic acid.
Figure 19.21. 27 T able 19.2 shows the main features of the
high-resolution NMR spectrum of a compound
containing carbon, hydrogen and oxygen. The
peak with a chemical shift of 11.7 disappears in
the presence of D2O. Deduce the structure of the
Signal strength
compound.
Table 19.2
Chemical Number of lines Integration ratio
shift/ppm
1.2 Triplet 3
12 10 8 6 4 2 0 2.4 Quartet 2
Chemical shift, δ/ppm 11.7 Singlet 1
Figure 19.21 Proton NMR spectrum of 3-chloropropanoic acid.
Medical benefits from NMR
In medicine, magnetic resonance imaging uses NMR to detect the
hydrogen nuclei in the human body, especially in water and lipids. A
computer translates the information from a body scan into images of the
soft tissue and internal organs that are normally transparent to X-rays
(Figure 19.22).
Figure 19.22 A brain scan in progress in
an advanced magnetic resonance imaging
(MRI) machine.
Test yourself
28 Suggest a reason why
doctors and radiographers
refer to MRI scanning rather
than to NMR imaging, even
though the technologies
used in medicine and
chemical research are
essentially the same.
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Activity
Analysing a perfume chemical
Jasmine blossom is a source of chemicals used in perfumes (Figure 19.23).
Analysis of an extract from the blossom by gas chromatography shows that it can
contain over 200 compounds. Two of the compounds in the mixture are mainly
responsible for the smell of the blossom. One of these two chemicals is jasmone,
which has the empirical formula C11H16O.
Figure 19.23 Harvesters gathering jasmine flowers for the French perfume industry.
The structure of jasmone has been studied by mass spectrometry and by NMR and
IR spectroscopy, with the results shown in Figures 19.24–19.26.
100 Figure 19.24 The mass spectrum of the perfume
chemical jasmone.
80
Relative intensity/%
60
40
20
0
25 50 75 100 125 150
m/z
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Figure 19.25 The 13C NMR spectrum of the perfume
chemical jasmone.
Signal strength
200 180 160 140 120 100 80 60 40 20 0
Chemical shift, δ/ppm
100 Figure 19.26 The IR spectrum of the perfume
chemical jasmone.
Transmittance/%
50
0
4000 3000 2000 1500 1000 500
–1
Wavenumber/cm
1 a) Use the mass spectrum of jasmone to determine its relative molecular mass.
b) What is the molecular formula of jasmone?
c) How many double bonds and/or rings are there in the molecule?
2 Refer to the 13C NMR spectrum in Figure 19.25.
a) How many different environments for carbon are there in the molecule?
b) Use the table of 13C NMR chemical shifts in the Pearson Edexcel Data booklet
to suggest which chemical environments for carbon are in the molecule.
c) What can you conclude about the structure of the molecule from the
spectrum?
3 Refer to the IR spectrum in Figure 19.26 and the IR correlation tables in
the Pearson Edexcel Data booklet. What functional groups are present in the
molecule?
4 Suggest a possible structure for jasmone that is consistent both with the
information from the spectra and with the fact that the full name of the compound
is cis-jasmone.
5 Describe what you would expect to observe if you tested a sample of
cis-jasmone with:
a) a solution of bromine in an organic solvent
b) Tollens’ reagent
c) 2,4-dinitrophenylhydrazine.
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19.5 Chromatography
solvent
(mobile
phase) Principles of chromatography
In 1903, the Russian botanist Michel Tswett developed the technique of
adsorbent components column chromatography to study plant pigments. ‘Chroma’ means colour.
solid of a mixture
(stationary separating on The name chromatography was chosen because the technique was first used
phase) the column to separate coloured chemicals from mixtures. The general set up for column
chromatography is shown in Figure 19.27.
glass wool
There are now a range of chromatography techniques that can be used to:
● separate and identify the components of a mixture of chemicals
● check the purity of a chemical
● identify the impurities in a chemical preparation
● purify a chemical product.
Every type of chromatography has a stationary phase and a mobile phase
that flows through it. Chemicals in a mixture separate because they differ in
the extent to which they mix with the mobile phase or stick to the stationary
phase.
Figure 19.27 Column chromatography. A
solution of the mixture to be analysed is Powdered solids now used in column chromatography include silicon oxide
added to the top of the column. Then a (silica) and aluminium oxide (alumina). Both these solids can adsorb chemicals
solvent is added slowly and continuously onto their surfaces (Figure 19.28). The greater the tendency for molecules to be
to run through the column. The substances adsorbed by the stationary phase, the slower they move during chromatography.
in the mixture separate and emerge at
different times from the column.
Figure 19.28 In column chromatography
tiny granule of the
there is an equilibrium between molecules stationary phase
adsorbed onto the surface of tiny particles
of the silica or alumina stationary phase
and molecules dissolved in the solvent.
molecule adsorbed
onto the surface of molecule dissolved
the stationary phase in the mobile phase
Key terms
The stationary phase in chromatography may be a solid or a liquid held by a solid
support. The mobile phase moves through the stationary phase and may be a liquid
or a gas.
In column chromatography the liquid flowing through the column is the eluent. It washes
the components of the mixture through the column. This is the process of elution.
Tip Solids can adsorb very thin films of liquids or gases onto their external surfaces.
Note the difference between absorption By contrast, a sponge absorbs water internally into its pores as it soaks up the liquid.
inside a material and adsorption only Paper is absorbent and soaks up the moving solvent during paper chromatography as
onto its surface. does silica or alumina on a TLC plate.
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Test yourself
29 In Michel Tswett’s column chromatography identify:
a) the stationary phase
b) the mobile phase.
30 In Figure 19.27 state the colour of the component of the mixture
being separated that has the greater tendency to:
a) dissolve in the eluting solvent
b) stick to the stationary phase.
31 Name two other techniques, other than chromatography, that can
be used to determine whether or not a product of chemical synthesis
is pure.
Liquid chromatography
The column chromatography developed by Michel Tswett was an example of
liquid chromatography. Modern versions of the column technique continue
to be widely used. Other variants include thin-layer chromatography and
high-performance liquid chromatography.
Thin-layer chromatography, TLC
Thin-layer chromatography is another type of liquid chromatography in
which the stationary phase is a thin layer of a solid, such as silica or alumina, Tip
supported on a glass or plastic plate. As in column chromatography, the rate TLC can be used quickly to check that a
at which a sample moves up a TLC plate also depends on the equilibrium chemical reaction is going as expected
between adsorption on the solid and solution in the solvent. The position of and making the required product. After
equilibrium varies from one compound to another so the components of a attempts to purify a chemical, TLC can
mixture separate. show whether or not all the impurities
TLC is quick, cheap and only needs a very small sample for analysis. The have been removed.
technique is widely used both in research laboratories and in industry.
The amino acids in peptides and proteins can be investigated by hydrolysing
the peptides and proteins with concentrated hydrochloric acid, then
separating and identifying the amino acids produced using either thin-layer
or paper chromatography (Figure 19.29).
solvent front – liquid
lid Figure 19.29 Thin-layer chromatography
rises up the plate or solvent tank – atmosphere or paper chromatography can be used to
paper by capillary saturated with solvent vapour
by lining it with paper soaked
separate and identify amino acids after a
action
in the solvent protein has been hydrolysed. The sample
under investigation is spotted at Q and
thin-layer chromatography
compounds in the plate or cylinder of known amino acids are spotted at P, R
sample under chromatography paper and S.
investigation (e.g.
amino acids in a spots of the solution under
hydrolysed protein) investigation and known
separating out reference compounds
(e.g. amino acids)
P Q R S
solvent
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Identifying amino acids – ninhydrin and R f values
Key term After chromatography, it is easy to see coloured compounds on the TLC plate
or chromatography paper. With colourless compounds such as amino acids,
The Rf value is the ratio of the distance
it is necessary to use a locating agent. Locating agents can be sprayed onto
moved by the chemical in a mixture to
the plate or paper, where they react with the separated colourless compounds
the distance moved by the solvent front.
to form coloured products. Alternatively, the plate can be placed in a covered
Rf values are used in thin-layer and
beaker with a locating agent such as iodine crystals, and the iodine vapour
paper chromatography.
stains the spots.
The locating agent for amino acids is ninhydrin (Figure 19.30).
O
OH
OH
O
Figure 19.30 The structure of ninhydrin.
When ninhydrin is sprayed on the chromatography plate or paper and then
Tip heated in an oven at about 100 °C, it reacts with any amino acids to form
An alternative method is to use a TLC purple spots which fade and turn brown with time.
plate impregnated with a fluorescent
R f values are used to record the distances moved by chemicals in a mixture
chemical. Under a UV lamp the whole
relative to the distance moved by the solvent. R f stands for ‘relative to the
plate glows except in the areas where
solvent front’.
organic compounds absorb radiation,
so that they show up as dark spots. The values are ratios calculated using this formula, where x and y are as
shown in Figure 19.31:
distance moved by chemical x
Rf = =
distance moved by solvent front y
Using R f values it is possible to identify the different separated spots from the
sample under investigation by comparison with known reference compounds.
R f values can help to identify components of mixtures so long as the
conditions are carefully controlled. The values vary with the type of TLC
plate (or paper) and the nature of the solvent.
solvent front
start line
Figure 19.31 The distances on a TLC plate used to calculate Rf values.
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Test yourself
32 Why is it important to handle the TLC plate only on the edges and
not to touch the surface with fingers?
33 Why is the air inside the container of the TLC solvent saturated with
vapour of the solvent before adding the TLC plate?
34 Which of the four samples P, Q, R and S in Figure 19.29:
a) is the mixture being analysed
b) are reference compounds also present in the mixture
c) is a reference compound not present in the mixture?
35 Estimate the Rf value for the yellow component in Figure 19.29.
36 Suggest two factors that will change the Rf value of a particular
amino acid.
Activity
Using TLC to investigate the aspirin produced in Core practical 16
Core practical 16 (Section 17.3.5) describes the preparation of E Place a lid on the tank and allow it to stand in a fume
aspirin and how the purity of the product can be assessed by cupboard until the solvent front has risen to about 1 cm from
measuring its melting temperature. the top of the plate. Remove the plate from the tank, mark
the position of the solvent front and allow the plate to dry.
A student followed the instructions below to investigate the
F Observe the plate under a UV lamp and mark any spots
purity of samples of the crude product, the recrystallised
observed carefully with a pencil. In a fume cupboard, place
aspirin and a commercial sample of aspirin by using thin-layer
the plate in a beaker containing two or three iodine crystals.
chromatography (TLC).
Cover the beaker and warm gently on a steam bath until
A Make sure that that you handle the TLC plate only by the spots begin to appear.
edges and do not touch the surface. Using a pencil, draw
Questions
a line across the plate about 1 cm from the bottom and
1 Why is the start line for spotting TLC samples on a plate
mark three evenly spaced points on this line. Place a small
drawn in pencil and not with ink?
amount (about one-third of a spatula measure) of the crude
2 The recrystallised aspirin and the commercial sample both
product, the recrystallised aspirin and the commercial
produced only one spot on the TLC plate at an Rf value of
sample of aspirin in three separate test tubes and label
0.65. The crude product also produced this same spot but
the tubes.
in addition gave spots at Rf values 0.22 and 0.48. Draw a
B In a fume cupboard or a well-ventilated room, place 2 cm3 of
diagram of the plate, similar to that in Figure 19.31, to show
ethanol and 2 cm3 of dichloromethane in a test tube. Ensure
these results.
the liquids are mixed, then add 1 cm3 of this solvent mixture
3 What conclusions can you draw about the nature of the
to each of the labelled test tubes to dissolve the samples.
three samples tested?
C Using separate capillary tubes, place a small spot of each
4 In a different experiment, the developing solvent used was
of the three sample solutions onto the TLC plate. Allow the
a mixture of ethyl ethanoate and hexane. The Rf value for
spots to dry and then add more sample to each spot, three
aspirin in this experiment was 0.15. Suggest a reason why
times in all. Do not let the spots become larger than about
the Rf value with this solvent mixture was lower that the Rf
2 mm across.
value with pure ethyl ethanoate.
D After the spots are dry, place the TLC plate in a developing
5 Why should the spot size not be larger than 0.2 mm in
tank containing 0.5 cm depth of ethyl ethanoate. Make sure
step C?
that the original pencil line and spots are above the level of
the developing solvent.
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High-performance liquid chromatography
High-performance liquid chromatography (HPLC) is a sophisticated
version of column chromatography. The technique can be used to separate
components in a mixture which are very similar to each other (Figure 19.32).
High performance is achieved by packing very small particles of a solid such
a silica into a steel column. A typical column is about 10–30 cm long and has
an internal diameter of about 4 mm.
The use of fine particles increases the surface area of the stationary phase.
This makes the separation efficient but it means that a high-pressure pump is
necessary to force the solvent through the tightly packed column. As a result,
the technique is sometimes called high-pressure liquid chromatography
An advantage of HPLC is that it is carried out at room temperature and so can
analyse mixtures that would decompose on heating, such as many biological
molecules. Organic molecules that break down on heating cannot be studied
using gas chromatography. HPLC is also suitable for separating biological
molecules such as proteins. Another important application of HPLC is to
study urine or blood samples to investigate what happens to drugs as they are
Figure 19.32 A scientist checking a metabolised in the body.
sample tube in front of a set of high-
performance liquid chromatography (HPLC)
columns. Gas chromatography
Gas chromatography (GC) is a sensitive technique for analysing complex
mixtures. The technique is used for compounds that vaporise on heating
without decomposing. This type of chromatography not only separates the
chemicals in a sample, but also gives a measure of how much of each is present.
In gas chromatography the mobile phase is a gas, commonly helium, argon
or nitrogen, which carries the mixture of volatile chemicals through a long
tube containing the stationary phase. The column is coiled inside an oven.
Heating the column makes it possible to analyse any chemicals that turn to
vapour at the temperature of the oven (Figure 19.33).
Figure 19.33 The main features of gas dried
chromatography. The carrier gas takes the carrier
gas in
mixture of chemicals through the column,
replaceable
where they separate. As the compounds silicon rubber
leave the column they are detected and septum detector recorder
measured. injector exit
port
detector oven
injector
oven column
column oven
The analyst injects a small sample into the column where it enters the oven.
Volatile solids are dissolved in a solvent before injection. The chemicals in
the sample turn to gases and mix with the carrier gas. The gases then pass
through the column.
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The components in the mixture separate as they pass through the column.
After a time the chemicals emerge one by one. They pass into a detector
which sends a signal to a recorder as each compound appears. A series of
peaks, one for each compound in the mixture, make up the chromatogram
(Figure 19.34).
10
8
Recorded response
0
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210
Time/s
Figure 19.34 The printout from a gas chromatography instrument.
The position of a peak on a GC printout is a record of how long it takes
for a compound to pass through the column. This is called the compound’s
Key term
retention time.
The retention time in gas
The area under each peak gives an indication of the relative amounts of the chromatography is the time it takes
compounds in the mixture. If the peaks are narrow it is sufficient to measure for a compound in a mixture to pass
the peak heights to get an indication of the relative amounts. through a chromatographic column and
reach the detector.
Some GC instruments have capillary columns. These are 20–60 metres
long with a very small internal diameter. Capillary columns are often made
of silica with an outer polymer coating. The stationary phase is the inner
surface of the column, which adsorbs chemicals to a greater or lesser extent.
The inner surface may be coated with a solid adsorbent or a thin film of a
liquid.
Other GC columns are steel or glass tubes packed with a powder. The powder
is an inert solid coated with a thin film of a liquid that has a high boiling
temperature. In these columns the stationary phase is the liquid coating.
Chemicals in the carrier gas separate in these columns because they differ
in their solubility in the liquid of the stationary phase. When this type of
column is used the technique is sometimes called gas–liquid chromatography.
Applications of gas chromatography include:
● tracking down the source of oil pollution from the pattern of peaks, which
acts like a fingerprint for any batch of oil
● monitoring the presence of chemicals in industrial processes
● measuring the level of alcohol in blood samples from drivers
● detecting pesticides in river water.
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Test yourself
37 Suggest a reason for choosing helium, argon or nitrogen as the
carrier gas in gas chromatography.
38 Refer to Figure 19.34.
a) How many chemicals were there in the mixture?
b) What was the retention time of the least abundant chemical in
the mixture?
39 Why must the liquid for the stationary phase in gas–liquid
chromatography have a high boiling temperature?
40 What are the implications of the fact that similar compounds may
have very similar retention times in gas chromatography?
41 E xplain why it is not possible to identify a previously unknown
chemical by gas chromatography.
Activity
Forensic investigations of arson
Arsonists sometimes use flammable liquids such as petrol or paraffin to accelerate fires.
Firefighters collect samples from the burnt remains which forensic scientists can analyse
in the search for clues as to how the fire started (Figure 19.35).
Suitable samples for analysis come from areas where furniture and fittings have not been
completely destroyed. Useful samples include carpet underlays, soil from pot plants,
bedding, clothing and material collected from underneath floorboards.
1 Suggest a reason why firefighters collect only partially burned materials for analysis.
2 Suggest a reason why soil from pot plants can provide good evidence that there have
been flammable liquids present.
Figure 19.35 Firefighters searching through
the wreckage of a burnt-out house. They are
looking for evidence of how the fire started
to determine whether it was an accident or
arson.
Analysts use solvents to extract chemicals from the samples and then investigate the
solutions by gas chromatography. They compare the chromatograms with those from
standard samples of common flammable substances (Figure 19.36).
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Fresh paraffin Fresh petrol
Detector current
Detector current
Time/min Time/min
Partially evaporated paraffin Partially evaporated petrol
Detector current
Detector current
Time/min Time/min
Figure 19.36 Gas chromatograms of standard samples of fuels.
3 The gas chromatogram for fresh petrol is very different from the chromatogram for
fresh paraffin. Describe and explain the differences.
4 Suggest why the chromatogram for partially evaporated petrol differs from the
chromatogram for fresh petrol.
The two chromatograms in Figure 19.37 show the results of analysing the chemicals from
samples collected from a burnt-out house. Sample A was collected from the charred
floorboards just inside the front door. Sample B came from the partly burned carpet
under a table near the window of the front room.
Sample A Sample B
Detector current
Detector current
Time/min Time/min
Figure 19.37 Gas chromatograms of chemical extracts from two samples collected from a
burnt-out house.
5 What can you conclude from the chromatogram for sample A?
6 What can you conclude from the chromatogram for sample B?
7 Suggest the further evidence that the analysts would need to collect before deciding
whether the house fire was an accident or the result of arson.
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Chromatography combined with mass
spectrometry
Coupling gas chromatography (GC) with mass spectrometry (MS) gives a
very powerful system for separating, identifying and measuring complex
mixtures of chemicals. The combined technique (GC–MS) is widely used in
drug detection in sport and elsewhere, in the forensics investigation of fires,
in environmental monitoring and in airport security (Figure 19.38). Some
space probes carry tiny GC–MS systems for analysing samples collected in
space or on the surface of planets (Figure 19.39).
Gas chromatography The chemicals from Mass analyser Ion detector gives an electrical
column in which the the GC column are separates ions by signal which is converted to a
sample chemicals in the separately ionised by mass-to-charge digital response that is stored in a
injected mixture separate and bombardment with ratio, e.g. by computer. There is a separate
leave the column one electrons or other magnetic field or mass spectrum for each chemical
by one methods time of flight in the mixture
Figure 19.38 A schematic diagram
showing the key features of a GC–MS GC–MS overcomes some of the limitations of gas chromatography. Similar
system. compounds often have similar retention times in GC, which means that they
cannot be identified by chromatography alone, even if the conditions are
carefully standardised. Also, GC alone cannot identify any new chemicals
because there are no standards that can be used to determine retention times
under given conditions.
Figure 19.39 The Philae lander (left) from GC–MS produces a mass spectrum for each of the chemicals separated on
the Rosetta mission in November 2014 the GC column. These spectra can be used like fingerprints to identify the
analysed samples of the comet 67P/ compounds because every chemical has a unique mass spectrum.
Churyumov-Gerasimenko using a GC–MS
A computer receives the data from the GC–MS system. This computer
system the size of a shoe box (right) and
can be linked to a library of spectra of known compounds. The computer
detected organic molecules just above the
compares the mass spectrum of each chemical in a mixture to mass spectra in
comet’s surface.
the library. It automatically reports a list of likely identifications along with
the probability that the matches are correct.
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Activity
Solving a pollution problem with GC–MS
With complex mixtures, computers can help an analyst to look for a ‘needle in a haystack’.
Figure 19.40 shows the chromatogram from an investigation of the air in a home where the
family was feeling very sick. During the analysis by GC–MS the computer stored 700 mass
spectra as the mixture of chemicals emerged from the chromatography column.
100 231
152
Relative abundance/%
80
131
60
51 101
40
356 382
584
20 519
345 646
0
100 200 300 400 500 600 700
1:50 3:40 5:30 7:19 9:09 10:59 12:49
Sample number for mass spectra
Retention time/min
Figure 19.40 The gas chromatogram of chemicals sampled from the air in a house. The numbers
1–700 on the time axis indicate the points at which mass spectra were recorded and stored.
Underneath these numbers are the retention times.
The analysts suspected that the chemicals causing the family’s sickness might have come
from petrol, so they asked the computer to plot a chromatogram showing only those
chemicals producing a peak with mass-to-charge ratio of 91 in their spectra. The result is
shown in Figure 19.41, which indicates that methylbenzene, ethylbenzene and three
dimethylbenzenes were present in the mixture. These are all chemicals that are distinctive
for the mixture of hydrocarbons found in petrol. With this evidence the investigators carried
out further searches and tracked down the source of the petrol vapour.
100 Figure 19.41 The GC–MS printout for the
231
same sample as in Figure 19.40 but showing
only the chemicals with a prominent peak with
Relative abundance/%
80
1,3-and 1,4-dimethylbenzene a mass-to-charge ratio of 91 in their mass
60 spectra.
methylbenzene 356 382
1,2-dimethylbenzene
40
345
ethylbenzene
20
0
100 200 300 400 500
Sample number for mass spectra
1 Why, in a mass spectrometer, does each chemical: 3 How does the computer identify a chemical with a mass
a) have to be ionised spectrum recorded at a particular retention time?
b) pass through a region with electric and/or magnetic fields 4 Suggest two reasons why forensic scientists find GC–MS
c) produce a spectrum with several peaks? particularly valuable.
2 Suggest the identity of the ion fragment with a mass-to- 5 HPLC can also be combined with MS. Give an example of a
charge ratio of 91 in the mass spectra of methylbenzene, sample that could be analysed by HPLC–MS but not by
C6H5–CH3, and related compounds. GC–MS and explain your choice.
19.5 Chromatography 619
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Chapter summary
Chapter 19 Modern analytical the spectrum can be predicted because of carbon
atoms in different environments in a molecule.
techniques II 1
l Data from high resolution H NMR spectroscopy
l In mass spectrometry, molecules are ionised and can be used to predict the different types of proton
split into fragments by bombardment with high- present in a molecule, given values of chemical
energy electrons. The mass spectrum contains shift, δ. The relative peak areas are used to predict
a ‘fragmentation pattern’ which, together with the relative numbers of 1H atoms in different
the relative molecular mass, can suggest a likely environments. Using the ‘n + 1’ rule, splitting
structure for a compound. patterns of adjacent, non-equivalent protons can
l High-resolution mass spectrometers give the be explained and possible structures for a molecule
relative atomic mass of elements to four decimal suggested. Chemical shifts and splitting patterns of
places so the accurate relative molecular mass of 1H atoms in a given molecule can be predicted.
compounds can be calculated, so one compound l Chromatography is used to separate and identify
can be distinguished from others with the same the components of a mixture, to identify
integral mass. impurities in a preparation and to purify a
l An infrared absorption spectrum shows the product.
absorption of radiation over a range of frequencies. l Chromatography uses a stationary phase and a
The complex pattern of vibrations, particularly mobile phase that flows through it. Substances in a
in the region 1500–650 cm−1, are used as a mixture separate because they differ by how much
‘fingerprint’ to be matched against IR spectra in they mix with the mobile phase or adsorb onto the
databases. stationary phase.
l Nuclear magnetic resonance (NMR) studies the l The R f value is the ratio of the distance moved by
behaviour of the nuclei in magnetic fields and the chemical in a mixture to the distance moved
provides information about 13C atoms by the solvent front.
(13C NMR spectroscopy) and of 1H atoms (high l In column chromatography, a solution of the
resolution proton NMR) in molecules. mixture to be analysed is added to the top of
l The horizontal scale of an NMR spectrum shows the column. Then solvent is added slowly and
the chemical shifts of the peaks measured in parts continuously to run through the column. The
per million (ppm). The symbol δ stands for the substances in the mixture separate and emerge at
‘chemical shift’ relative to zero on the scale that different times from the column.
is given by the signal from tetramethylsilane, l High-performance liquid chromatography (HPLC)
a standard reference compound, added to the is a version of column chromatography that
solution of a substance being tested. TMS produces uses fine particles to increase the surface area of
a single sharp peak away from other peaks and the the stationary phase. This makes the separation
position of other peaks is compared to this. efficient at room temperature although high-
l The solvent commonly used is CDCl 3. This pressure is necessary to force the solvent through
produces no absorption in 1H NMR in the region the tightly packed column.
for protons and a single line in 13C spectra that is l The retention time in gas chromatography (GC) is
easy to recognise. the time taken for a compound to pass through a
l Data from
13C NMR spectra can be used to column and reach the detector.
predict the different environments for carbon l HPLC and GC may be used in conjunction with
atoms present in a molecule, given values of mass spectrometry, in applications such as forensics
chemical shift, δ. The number of peaks present in or drugs testing in sport.
620 19 Modern analytical techniques II
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Exam practice questions
1 a) The relative molecular mass of a compound a) Identify the peaks at m/z = 77 and 105
is 74.1212. Use the data in Table 19.1 to in the top spectrum and m/z = 93 in the
confirm that the molecular formula bottom spectrum. (3)
of this compound is C4H10O rather b) State which isomer gives the top spectrum
than C3H6O2 or C2H6N2O. (2) and explain your answer. (2)
b) Draw the four isomeric alcohols with c) Describe two test-tube reactions that
molecular formula C4H10O and predict will confirm your choice. (6)
the number of peaks in the 13C NMR
spectra of each compound. (4) 4 The proton NMR spectrum of a compound of
c) The compound in part (a) is unaffected carbon, oxygen and hydrogen is shown below.
by acidified potassium dichromate(vi).
Identify this compound and justify
your choice. (2)
2 Benzene was heated with a mixture of
Signal strength
concentrated nitric and sulfuric acids and the
major product obtained was found to
be 1,3-dinitrobenzene.
a) Write an equation for the formation of
1,3-dinitrobenzene from benzene. (1)
b) Draw the structures of the three possible
11 10 9 8 7 6 5 4 3 2 1 0
isomers of dinitrobenzene. (3)
δ/ppm
c) Explain how 13C NMR could be used
to confirm that the major product was An integration trace gives the ratios for the
1,3-dinitrobenzene. (3) peaks, as shown in the table.
3 The diagram shows the mass spectra of two Chemical Relative values from the
shift/ppm integration trace
isomers: benzoic acid (benzenecarboxylic acid,
C6H5COOH) and 3-hydroxybenzaldehyde 1.0 0.9
(3-hydroxybenzenecarbaldehyde, 2.1 0.9
HOC6H4CHO). 2.5 0.6
100
105 a) i) State how many chemical environments
Relative intensity/%
122
77
for hydrogen atoms there are in the
molecule. (1)
50 51
ii) Give the ratio of the numbers of each
type of hydrogen atom. (1)
28 39 57 65 94 b) Give likely chemical environments
0 of the protons in the molecule with the
20 40 60 80 100 120
m/z help of the chart of chemical shift values in
the Pearson Edexcel Data booklet. (2)
122 c) State what you can deduce from the
100
Relative intensity/%
splitting patterns of the peaks at chemical
shifts 1.0 and 2.5. (2)
93
50
65 d) The compound gives an orange precipitate
39 with 2,4-dinitrophenylhydrazine but does not
53 74 react with Fehling’s solution or phosphorus(v)
0
20 40 60 80 100 120 chloride. State what this tells you about the
m/z functional groups in the molecule. (3)
e) Give a structure for the compound. (1)
621
Exam practice questions
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5 Gas chromatography can be used to analyse d) The mixture was analysed by GC–MS. The
the chemicals formed during the production mass spectrum of the fifth peak to emerge
of beer. Below is the GC chromatogram from the GC column is shown. State
for a sample taken during beer making. The whether this peak is butan-2-ol or propanal
instrument was fitted with a capillary column and explain your answer. (4)
with solid stationary phase. 100
Relative abundance
80
Relative abundance/%
60
19 20 21 22 23 24 25 26 27 40
Retention time/min
20
The table gives the retention times for GC under
the conditions used to analyse the beer sample.
0.0
20 30 40 50 60
Compound Retention time/min m/z
Methanol 19.5
Ethanol 20.5 6 A naturally occurring dipeptide, A, has the
Propan-1-ol 20.9 molecular formula C7H14O3N2. The dipeptide
Propan-2-ol 22.5
is hydrolysed forming two amino acids, B and
C, on heating with concentrated hydrochloric
2-Methylpropan-1-ol 22.7
acid. The two amino acids can be separated by
Butan-2-ol 24.6
paper chromatography using a solvent in which
Ethanal 20.2
B has an Rf value of 0.60 and C has an Rf value
Propanal 24.5 of 0.26.
Butanal 25.5 a) Draw a labelled diagram, to scale, showing
Propanone 23.8 the original and final spots and the solvent
Ethanoic acid 24.2 front on the chromatogram. (4)
Butanoic acid 26.2 b) Amino acid B is chiral, but C is non-chiral.
i) Draw the displayed formula of C. (1)
a) Explain the trend in the retention times of ii) Deduce the molecular formula of
the four primary alcohols. (2) amino acid B. (1)
b) Explain why 2-methylpropan-1-ol has a iii) Draw a possible structural formula
shorter retention time than butan-2-ol. (2) for B. (2)
c) i) Use the gas chromatogram and the c) Draw a possible structural formula for
table to identify the chemicals in the the dipeptide A. (2)
beer sample.(3) d) Describe the procedure you would use to
ii) State which peaks were hard to identify. make the ‘spots’ of amino acids visible on
Explain your answer. (2) the chromatogram. (2)
e) Explain how you would show that a sample
of amino acid B was chiral. (2)
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19 Modern analytical techniques II
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7 Two isomeric ketones, X and Y, with the 8 Identify the compound containing carbon,
formula C5H10O and unbranched carbon hydrogen and oxygen only that gives rise to
chains have the mass spectra shown below. the spectra below. Draw the displayed formula
100 Ketone X 43
of the compound and name it. Give your
reasoning and show how you can account for
80 the key features in the three spectra. (10)
Relative intensity/%
100 27 55
60 72
40 80
Relative intensity/%
86
20
71 60
0
10 20 30 40 50 60 70 80 90 40
m/z 45
20
100 Ketone Y
57
0
25 50 75
80
Relative intensity/%
m/z
29
60
40
Signal strength
86
20
0
10 20 30 40 50 60 70 80 90
m/z
a) Draw the skeletal formulae of the two
ketones and name them. (2) 200 180 160 140 120 100 80 60 40 20 0
b) Explain why both spectra have peaks at Chemical shift, δ/ppm
m/z values of 86. (2)
c) i) Give possible identities for the four 100
fragments in the two spectra with
m/z values of 29, 43, 57 and 71. (4)
Transmittance/%
ii) Deduce which spectrum
belongs to which compound. (2)
50
d) Use the chart of chemical shifts in the
Pearson Edexcel Data booklet to predict:
i) the carbon-13 NMR spectrum
of ketone X (3)
ii) the proton NMR spectrum of 0
ketone Y. (6) 4000 3000 2000 1500 1000 500
Wavenumber/cm–1
An organic compound Z contains carbon,
9
hydrogen, oxygen and one other element.
The carbon-13 NMR spectrum of Z has four
peaks, showing that there are carbon atoms
in a Z molecule in four distinct chemical
environments.
623
Exam practice questions
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The mass spectrum and the proton NMR 10 The proton NMR spectrum of a compound
spectrum of the compound are shown below. with the formula C6H12O3 is shown below.
100
80
Relative intensity/%
60
Signal strength
40
107 109
20
166 168
0
25 50 75 100 125 150 175
m/z
Mass spectrum of Z.
7 6 5 4 3 2 1 0
Chemical shift, δ/ppm
The integration trace gives the ratios for the
peaks as shown in the table.
Chemical shift Ratios of values from
4.0 3.5 3.0 2.5 integration trace
Chemical shift, δ/ppm 2.2 1.8
Proton NMR spectrum of Z. 2.7 1.2
3.4 3.6
a) State what you can deduce from the pattern 4.8 0.6
of the mass spectrum and, in particular, from
the two peaks at 166 and 168 in the mass a) i) State how many chemical environments
spectrum. (2) for hydrogen atoms there are in the
b) The carbon-13 NMR spectrum suggests molecule. (1)
that there are four carbon atoms in the ii) Give the ratio of the numbers of each
molecule. Use this information, your answer type of hydrogen atom. (1)
to (a) and the proton NMR spectrum to b) The compound gives an orange precipitate
give a possible molecular formula for the with 2,4-dinitrophenylhydrazine but
compound. (2) does not react with Fehling’s solution or
c) Measure the step heights on the integration phosphorus(v) chloride. Explain what this
trace of the proton NMR spectrum and tells you about the functional groups in the
explain what the values show. (2) molecule. (2)
d) State what you can deduce from the c) Give likely chemical environments of the
chemical shift values and splitting patterns in protons in the molecule with the help of a
the proton NMR spectrum. (4) table of chemical shift values. (3)
e) Draw a structure for the compound. (1) d) i) State what you can deduce from
f) State the fragment ions that produce the the splitting patterns of the peaks at
peaks at m/z = 107 and 109 in the mass chemical shifts 2.7 and 4.8.(2)
spectrum. (1) ii) State what you can deduce from the
absence of splitting of the peaks at
chemical shifts at 2.2 and 3.4. (1)
e) Give a structure for the compound. (1)
624
19 Modern analytical techniques II
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11 a)* C
ompound A has the molecular formula b) Compound B has the molecular formula
C4H8O2. The infrared spectrum of [Link] infrared spectrum of
compound A is shown below. compound B is shown below.
Compound A Compound B
100 100
Transmittance/%
Transmittance/%
50 50
0 0
4000 3000 2000 1500 1000 500 4000 3000 2000 1500 1000 500
–1
Wavenumber/cm –1 Wavenumber/cm
Infrared spectrum of compound A. Infrared spectrum of compound B.
The proton NMR spectrum of A contains The proton NMR spectrum of B contains
only three peaks. The table contains data only four peaks. The table contains data
about these peaks. about these peaks.
δ/ppm Integration Splitting δ/ppm Integration Splitting
4.12 2 Quartet 3.84 1.2 Triplet
2.04 3 Singlet 3.40 0.6 Singlet
1.26 3 Triplet 2.70 1.2 Triplet
2.20 1.8 Singlet
Peak data for compound A.
Peak data for compound B.
Identify compound A and justify your
answer. (6) Identify compound B and justify your
answer. (9)
625
Exam practice questions
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Mathematics in chemistry Year 1
A1
In order to be able to develop your skills, knowledge and understanding in
chemistry, you need to be competent in certain areas of mathematics. At least
20% of the marks in your examinations will require the use of mathematical
skills at the standard of higher tier GCSE mathematics, but applied in the
context of A Level chemistry.
A1.1 Mathematical operations
Mathematics uses the term operations to describe processes such as addition,
subtraction, multiplication, division and squaring. Certain rules apply to the
order in which these operations are carried out, irrespective of the order
in which they are written on the page. The most important rule is that
multiplication is always done before addition. For example, the correct answer
to the calculation 7 + 2 × 3 is 13 because the multiplication 2 × 3 is done
first and then the product, 6, is added to 7 to give 13.
This calculation could perhaps have been written more clearly using brackets
as 7 + (2 × 3) = 13. This shows that the calculation inside a bracket is done
first, before any multiplication or addition outside the brackets.
The order of operations is important, for instance, in the calculation of
relative molar masses.
Example
Calculate the Mr of aluminium sulfate, Al2(SO4)3.
Notes on the method
The Mr is the sum of the Ar for the atoms in the formula, with the
multiplications done before the additions.
Answer
Mr[Al2(SO4)3] = 2 × Ar(Al) + 3 × [Ar(S) + 4 × Ar(O)]
Tip The Mr of SO4 inside the bracket in the chemical formula is found first.
The relative atomic masses in the Mr(SO4) = Ar(S) + 4 × Ar(O)
Periodic Table in the Pearson Edexcel = 32.1 + 4 × 16.0
Data booklet are given to one decimal = 32.1 + 64.0 (multiplication before division)
place. You should use the figures for Ar
= 96.1
to this precision in your calculations.
626 A1 Mathematics in chemistry Year 1
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Then multiplication of the Mr of SO4 by the number 3 outside the bracket
in the chemical formula gives:
Mr[(SO4)3] = 3 × Mr(SO4)
= 3 × 96.1
= 288.3
Ar(Al) = 27.0 so 2 × Ar(Al) = 54.0
Final addition gives:
Mr[Al2(SO4)3] = 2 × Ar(Al) + Mr[(SO4)3]
= 54.0 + 288.3 = 342.3
A1.2 Positive and negative numbers Tip
Chemists use oxidation numbers (Section 3.3) to represent the numbers of Take care when using negative numbers,
electrons gained or lost by an atom. Oxidation numbers can be positive or for instance in calculations using Hess’s
negative. The oxidation number of chlorine in the chloride ion is –1 and of Law (Section 8.5) or bond energies
oxygen in the oxide ion is –2. (Section 8.7). Use brackets around
negative numbers and remember that
Beware of describing the oxidation number of oxygen as ‘bigger’ than that of
x − (−y) = x + y.
chlorine in these two ions. The comparisons bigger/smaller and higher/lower
are unclear when discussing negative numbers. You should always make it
totally clear what you mean, so you should say that the oxidation number of
oxygen in the oxide ion is more negative than that of chlorine in chloride.
Key terms
Similar care is needed when discussing exothermic reactions where heat
energy is evolved and the enthalpy change, ΔH is negative (Section 8.2). For Standard form writes a number in two
methane, the standard enthalpy of combustion, ΔH 1 = –890 kJ mol–1 and for parts multiplied together. The first part
ethane, the standard enthalpy of combustion, ΔH 1 = –1560 kJ mol–1. Take is a number greater than 1 and less
care to say that the value for ethane is more negative than that for methane, not than 10; the second part is the number
just bigger. 10 raised to a power which is a whole
number.
A1.3 Standard form and ordinary form In ordinary form a number is written
with no powers of ten included.
Numbers in chemistry vary from the extremely large, such as the Avogadro
constant (Section 5.1), to the extremely small, such as the mass of a proton
in kilograms. A convenient way to write both numbers is called standard
form. This avoids the long strings of zeros which would be needed if the Tip
number were written in ordinary form.
The statement ‘a is greater than 1
In standard form, a number is written in two parts which are multiplied and less than 10’ can be represented
together, a × 10b. mathematically as 1 < a < 10, where the
symbol < means ‘is less than’. Similarly
The number a is greater than 1 and less than 10, and b is a whole number.
the symbol > means ‘is greater than’
If the overall number < 1, then b is a negative number. and the symbol >> means ‘is very much
greater than’.
If the overall number > 10, then b is a positive number.
A1.3 Standard form and ordinary form 627
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Tip Examples of standard form
The number 0.002 46 in ordinary form becomes 2.46 × 10 –3 when expressed
The value of b gives the number of
in standard form. The decimal point is moved three places to the right.
places that the decimal point must be
moved when changing between ordinary The number 135 000.0 in ordinary form becomes 1.35 × 105 when expressed
form and standard form. in standard form. The decimal point is moved five places to the left.
The Avogadro constant (Section 5.1) in standard form is 6.02 × 1023 mol–1.
This is much more convenient to use than the ordinary form, which is
602 000 000 000 000 000 000 000 mol–1.
The mass of a proton expressed in standard form is 1.67 × 10 –27 kg.
This is much easier to use than the ordinary form, which is
Tip 0.000 000 000 000 000 000 000 000 001 67 kg.
Always use standard form if possible;
When adding or subtracting numbers written in standard form, make sure
it reduces the chance of error when
that the addition or subtraction is done on equivalent numbers in which the
counting numbers of zeros.
powers of ten are the same.
Example
A solution containing 1.60 × 10–2 mol HCl is added to a solution
containing 4.50 × 10–3 mol HCl.
Calculate the total amount in moles of HCl present.
Answer
In order to convert both numbers to the same powers of ten we can
rewrite 4.50 × 10–3 as 0.450 × 10–2
The addition is then (1.60 + 0.450) × 10–2 = 2.05 × 10–2
When multiplying numbers in standard form, the powers are added together,
as is usual with indices, so 102 × 103 = 105.
Example
Calculate the amount in moles of solute in 250 cm3 of a 0.0130 mol dm–3
solution.
Answer
The amount of solute (in moles) in a solution is given by multiplying the
volume in dm3 by the concentration in mol dm–3 (Section 5.5).
Volume = 250 cm3 = 2.50 × 10 –1 dm3
Concentration = 0.0130 mol dm –3 = 1.30 × 10 –2 mol dm –3
Amount of solute = (2.50 × 10 –1) × (1.30 × 10 –2)
= (2.50 × 1.30) × 10 –(1+2)
= 3.25 × 10 –3 mol
628 A1 Mathematics in chemistry Year 1
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A1.4 Handling data
Significant figures
When writing numbers, it is important to give a suitable number of digits.
The number of written digits is called the number of significant figures, often
abbreviated to sig. figs or s.f. For example, 6.8 g has 2 significant figures,
whereas 6.84 g has 3 significant figures.
Zeros to the left of a number are ignored, and the number of significant
figures is determined starting with the leftmost non-zero digit. So 0.0506
has 3 significant figures. This is more obvious if the number is written in
standard form as 5.06 × 10 –2.
Zeros to the right of a number are significant, so 3.740 has 4 significant figures.
When a measurement is made, it is important to know how precise that
measurement is (Section 5.9) and, therefore, what is an appropriate number
of significant figures to give. A titre might be given as 25.65 cm3, that is to
4 significant figures, but a time might only be given as 23 seconds (to 2 s.f.).
A value can be rounded to a smaller number of significant figures but should
never be written to a greater number of significant figures than could be
measured with the equipment used.
So a volume of 25.65 cm3 (4 s.f.) measured using a burette might be rounded to
25.7 cm3 (3 s.f.) but a volume of 26 cm3 measured using a measuring cylinder
with 2 cm3 divisions should never be given to more than 2 significant figures.
Tip
Rounding means replacing a number containing lots of digits with an approximate, but
shorter and more convenient, number.
When rounding to 3 significant figures, look at the fourth figure. If is it 5 or more, you
should round the third figure up. If the fourth figure is 4 or less, do not round up. So
you should round 123.4 to 123 (3 s.f.), but 234.5 rounds to 235 (3 s.f.). Therefore, the
number 123 (to 3 s.f.) is the approximation which represents values with 4 significant
figures between 122.5 and 123.4.
In a calculation, the number with the fewest significant figures determines
the number of significant figures in the answer. So, if a number with
4 significant figures is multiplied by one with only 2 significant figures, the
result should only be quoted to 2 significant figures.
Some experiments contain many sources of error, for instance the energy
losses in combustion experiments. Results from such experiments should not
be quoted to more than 2 or 3 significant figures (Section 8.3).
Tip
Sometimes rounding numbers allows you to estimate what the answer to a calculation
should be. Then, if your calculator produces a very different result, you know you have
made a mistake. For instance, you know that the answer of 4.95 × 2.12 should be
approximately 5 × 2 = 10.
A1.4 Handling data 629
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In calculations with several steps it is better not to use your calculator at each
Tip stage. The danger is that you round the answer of each step and introduce
If you do need to use your calculator ‘rounding errors’. You will get a more accurate answer if you use your
at stages during a problem and the calculator once at the end to work out the answer, and only round to the
final answer is required to 3 significant correct number of significant figures in this final answer.
figures, work in 4 or 5 significant figures
during the earlier stages and only round Units and decimal places
your answer to 3 significant figures at
Unlike pure numbers, a physical quantity includes both the number value
the very end.
and its units, so a mass may be 5.0 g, a volume 25.0 cm3 and a molar mass
98.1 g mol–1. The number value depends on the units and it is often sensible to
change, for instance, from one volume unit to another to simplify a number.
So a volume of 0.0035 dm3 of solution with several decimal places may
be more conveniently written as 3.5 cm3 (or of course in standard form as
3.5 × 10 –3 cm3). (See also the combustion example in Section 8.3, where an
energy change in joules is changed into kJ to simplify the numbers.)
The number of decimal places given in a measurement also depends on the
equipment used. A three decimal place balance will allow measurements
Tip such as 3.456 g to be made.
In calculations, using numbers with their
However, the number of decimal places is much less important than the
units leads directly to an answer with its
number of significant figures. You may think that a figure of 0.005 g looks
correct units. This is good practice as
impressively precise as it is given to 3 decimal places (precision is discussed
not only does this produce the correct
in Section 5.9). But 0.005 g is only given to 1 significant figure (and can
units easily but also acts as a check on
mean anything between 0.0045 and 0.0054 g), so it is less precise than a
the calculation.
figure such as 2.5 g which, although only given to 1 decimal place, is to
2 significant figures.
Key terms Arithmetic means
One way a series of numbers can be compared is by taking their average.
The arithmetic mean of a list of This usually means adding the series of numbers together and dividing by
numbers is the total value of the list size of the series. This ‘average’ is strictly called an arithmetic mean.
divided by the size of the list.
Outliers are experimental results which Tip
lie well away from the others. In statistics, average can also mean the median or the mode. These terms are not
needed in chemistry at this level of study.
Most experiments are repeated to check for anomalous results, that is for
results that lie away from the others. These results are sometimes called
outliers and may be due to experimental error. Once these anomalous
results are identified, the arithmetic mean can be calculated from the rest of
the data without these outliers.
Every volumetric analysis involves an initial rough titration which is used
to get an idea of the volume needed. Thereafter, the end-point can be
approached slowly, so that further titrations, if done carefully, should lead
to concordant results. Any result which is not concordant is ignored and the
average volume added, the arithmetic mean, is calculated by adding the titres
together and dividing this total by the number of results.
630 A1 Mathematics in chemistry Year 1
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Example
Table A1.1 shows a set of titration results. Use the concordant titres from
the table to calculate a mean titre.
Answer
Titre 2 is not concordant. It is not within 0.10 cm3 of the others and is
ignored. It is an outlier.
The other three titres are concordant and their total
= 23.40 + 23.30 + 23.35 = 70.05 cm3
70.05 cm3
The arithmetic mean of these three results = = 23.35 cm3
3
Table A1.1 Titration results.
Rough Accurate 1 Accurate 2 Accurate 3 Accurate 4
Final burette 26.5 27.00 25.75 25.90 26.45
reading
Initial burette 2.0 3.60 2.15 2.60 3.10
reading
Titre/cm3 24.5 23.40 23.60 23.30 23.35
Used in ✓ ✗ ✓ ✓
calculation
Weighted means Key terms
Another example of the need to find an arithmetic mean occurs in the
calculation of the relative atomic mass of an element using isotopic abundances A weighted mean is an arithmetic mean
(see the example in Section 1.4). of a set of numbers in which some of
the numbers carry more importance or
Examples of this type, however, do not use a simple arithmetic mean but a
weight than others.
weighted mean.
Percentage (symbol %) means ‘out of
A simple arithmetic mean of the relative atomic masses of the two chlorine 100’. A fraction can be converted into
isotopes, chlorine-35 and chlorine-37 would be 36. But, on average, there are a percentage simply by multiplying by
3 atoms of chlorine-35 to every 1 atom of chlorine-37, so these proportions 100, so 12 as a percentage is
must be taken into account. 1
( 2 × 100) = 50%.
The relative atomic mass of chorine is, therefore, a weighted mean given by
total mass of 3 atoms of 35Cl and 1 atom of 37Cl (3 × 35) + (1 × 37)
= Tip
the total number of atoms 4
= 35.5 Remember that relative atomic masses
do not have units because they are
Weighted means can also be calculated using percentage abundances for ratios (see Section 1.4).
the isotopes (see the example in Section 1.4 using the percentage abundance
of the magnesium isotopes).
Key terms
Ratios
In the calculation of the relative atomic mass of chlorine above, the ratio of A ratio is the comparison of two numbers.
35Cl atoms to 37Cl is 3 : 1.
Proportion is the ratio of a part
The proportion of 35Cl is 3 or 75% of the total number of chlorine atoms. compared to the whole amount.
4
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The formula used to describe a molecular substance, the molecular formula,
shows the actual number of atoms of each element in one molecule. The
formula of an ionic substance, however, is an empirical formula and shows
only the ratio of each of the ions present in the giant lattice. The empirical
formula is calculated from experiments which measure the mass of each
element in a compound. These masses are converted to amounts in moles
and then the simplest ratio of these amounts found (see the examples in
Sections 5.2 and 6.1.3).
A1.5 Mathematical equations and
expressions
Chemists use mathematical equations and expressions in calculations.
For instance, the amount in moles can be expressed in three ways:
● using masses
mass/g
amount/mol =
molar mass/g mol–1
● using gas volumes
volume/cm3
amount of gas/mol =
molar volume/cm 3 mol–1
● using solutions
volume of solution/dm 3 × concentration
amount of solute/mol =
mol dm–3
It is important to note that these quantities are combined with their
units.
The left-hand side of each expression gives the amount in moles.
Therefore, the units on the right-hand side of each expression should cancel
to give mol.
For the mass expression, the ‘g’ cancels and 1 = mol
mol–1
For the volume expression, the ‘cm3’ cancels and 1 = mol
mol–1
For the solution expression, dm 3 multiplied by dm–3 cancels leaving mol.
Tip
10–1 means 1 or 0.1 and 1−1 means 10.
10 10
–1 1 1
(Check: 10 = 0.1, so −1 = = 10; correct.)
10 0.1
Similarly, mol–1 means 1 and dm–3 means 1 .
mol dm3
Therefore 1 –1 means mol and 1–3 means dm3.
mol dm
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One commonly required mathematical skill is to change the subject of
an equation. This involves rearrangement of the terms in a mathematical
equation to leave the required term alone on one side. Values are then
substituted into the rearranged expression.
Example
0.258 mol of a compound has a mass of 46.3 g. Calculate the molar
mass of the compound.
Notes on the method
In order to calculate the molar mass of a substance, we start with the
expression which includes it:
mass/g
amount/mol =
molar mass/g mol –1
The molar mass, which is the denominator (the underneath) of the fraction,
needs to become the subject of a new expression. (It needs to be on the
left-hand side of the equation on its own.)
This can be done in stages, as shown below, and then the values given in
the question can be substituted into the rearranged expression.
Answer
Start by multiplying both sides of the original equation by
molar mass/g mol–1
amount/mol × molar mass/g mol−1
mass/g
= × molar mass/g mol –1
molar mass/g mol –1
The right-hand side cancels to give mass/g, so the overall equation
becomes:
amount/mol × molar mass/g mol−1 = mass/g
Then both sides of the equation are divided by amount/mol giving:
amount/mol mass/g
× molar mass/g mol –1 =
amount/mol amount/mol
Which simplifies to give the required equation:
mass/g
molar mass/g mol–1 =
amount/mol
Substituting the numerical values into the equation gives:
46.3 g
molar mass/g mol–1 =
0.258 mol
= 179.5 g mol –1
= 180 g mol –1 (3 s.f.) Tip
So the molar mass of the compound is 180 g mol–1 . Include the units at every stage.
(See also Section 5.5 and the rearrangement of the formula in the example
in Section 5.6.)
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A1.6 Chemical equations
Chemical equations are often called the language of chemistry. They translate
information about types and amounts of substances into a shorthand using
symbols and formulae.
A chemical equation shows the substances present at the start of a reaction
(the reactants, which are always on the left-hand side), the substances
present at the end of a reaction (the products, which are always on
the right-hand side) and the mole ratio of all the substances involved in
the reaction.
Chemical equations are always balanced, so that the number and type of
each element is equal on both sides of the equation and, therefore, the
total mass on each side of the equation is equal. The mass is, however,
present in different species, as elements or compounds, on the two sides
of the equation.
The formulae of the species and the mole ratios allow chemists to calculate
the amounts of substances used and formed in reactions.
Calculations involving masses
The four key steps for solving problems using equations were set out in
Section 5.4.
Tip Step 1: Write the balanced equation for the reaction.
Make sure you can write correct
If the formulae of the substances in the equation are wrong, the equation
formulae!
cannot be balanced and the calculation will be wrong.
Step 2: Write down the amounts in moles of the relevant reactants and
products in the equation.
If the equation isn’t balanced, the mole ratio will be wrong and the calculation
will be wrong.
Step 3: Convert these amounts in moles of the relevant reactants and
products to masses.
Work out the molar mass first, then multiply your answer by the ratio from
the equation.
Step 4: Scale the masses to the quantities required.
Example
This example uses the data in the first example in Section 5.4, which
showed how to calculate the amount of iron that can be obtained from
1.0 kg of iron ore.
What mass of carbon dioxide is also formed?
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Answer
Step 1: From the example in Section 5.4, the equation for the reaction is:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
Step 2: The mol ratio of Fe2O3 to CO2 = 1 : 3
Step 3: Work out the Mr of CO2 (44.0 g mol–1).
Then multiply the answer by the ratio in the equation (1 : 3).
So, starting from 1 mol of Fe2O3,
the mass of CO2 formed = 3 × 44.0 = 132.0 g
(Don’t work out the mass of 3C then 3O2.)
Step 4: Scale the masses to the quantities required.
Mr of Fe2O3 is 159.6 g mol–1
So, starting from 1.0 g Fe2O3,
the mass of CO2 formed = 1.0 × 132.0
159.6
= 0.827 g = 0.83 g (2 s.f.)
So for 1 kg of iron ore, the mass of carbon dioxide formed is 830 g.
Calculations using gas volumes
Gas volume calculations are straightforward when all the relevant substances
are gases, because the ratio of the gas volumes in the reaction is the same as
the ratio of the numbers of moles in the equation (Section 5.4).
If a question involves both masses and gas volumes, then use is made of the
first two expressions in Section A1.4.
The amount in moles of a solid can be found using the expression:
mass/g
amount/mol =
molar mass/g mol–1
The amount in moles of a gas can be found using the expression:
amount of gas/mol = volume/cm3
molar volume/cm 3 mol–1
The molar volume of a gas has the value 24 000 cm3 at room temperature and
pressure for all gases.
The two are then compared using the ratio in the equation (see the example
of the reaction of magnesium with acid in Section 5.4).
Calculations using solutions
Titration calculations are very common and there are several examples in
Sections 5.7 and 5.8. These calculations can be answered using equations such as:
cA × VA nA
c B × V B = nB
In these calculations, the concentration and volume of solutions A and B
and the mole ratio nA /nB from the equation are used. In any titration, all
but one of the values in this relationship are known and the one unknown
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is calculated using the results. This formula can be used to analyse titration
results, but it is generally better to work out the results step by step, as shown
in the worked example in Section 5.8.
If a question involves not only solutions, but also masses or gas volumes, then
a step-by-step method using the relationships at the start of Section A1.5
must be used.
Example
What volume of carbon dioxide (at room temperature and pressure) is
produced when 50.0 cm3 of 0.150 mol dm–3 hydrochloric acid reacts with
excess calcium carbonate?
The molar volume is 24 000 cm3 at room temperature and pressure.
Notes on the method
Start by writing the equation for the reaction.
Convert the volume and concentration of hydrochloric acid to an amount
in moles.
Use the equation to determine the amount of carbon dioxide formed, in
moles.
Multiply the amount of gas in moles by the molar volume at room
temperature and pressure.
Answer
The equation for the reaction is:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
50.0
amount of hydrochloric acid = dm3 × 0.150 mol dm –3
1000
= 7.50 × 10 –3 mol
From the equation, 2 mol hydrochloric acid produce 1 mol carbon dioxide.
∴ amount/mol of carbon dioxide formed = 12 (7.50 × 10 –3)
= 3.75 × 10 –3 mol
volume of carbon dioxide = 3.75 × 10 –3 mol × 24 000 cm3 mol–1
= 90.0 cm3 (3 s.f.)
Percentage yields and atom economy
The mass or volume of the product calculated in the examples above assumes
everything works perfectly. In reality and for several reasons, few reactions
produce exactly the mass predicted from the equation, the theoretical
Key terms yield, and instead only a proportion of this mass is formed, the actual yield
(Section 5.10).
The theoretical yield is the mass of
product obtained if the reaction goes
according to the equation. Tip
The actual yield is the mass of product Don’t forget to multiply the Mr of each substance by the number of moles of that
obtained from a reaction. substance in the equation when adding all the molar masses together.
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Key terms
actual yield
Percentage yield = × 100
theoretical yield
molar mass of the desired product
Atom economy = × 100
sum of the molar masses of all the products
The ratio of the two is called the percentage yield.
Manufacturers continually try to improve the efficiency of their processes
to increase the percentage yield of the desired product. However, there
is increasing concern about the amount of waste also produced in some
processes, even if the percentage yield of the desired product is high.
The overall efficiency of a chemical process is now referred to by the term
atom economy, which is the molar mass of the desired product expressed
as a percentage of the sum of the molar masses of all the products in the
equation for the reaction (see the example in Section 5.10).
Tip
When plotting a graph, choose a
A1.7 Graphs scale for each axis that allows you
to fill as much of the available space
Experimental data as possible, but make sure the scale
is simple to use. You do not have to
Drawing a graph can turn a list of numbers, or numerical data, into a picture
include zero or the origin in all cases.
which shows if there is a pattern in the results of an experiment. The pattern
Label each axis of the graph with a
may be a straight line or a curve, or there may be no pattern. Plotting a
quantity divided by its unit, for instance
graph by drawing a best-fit line is a way of averaging results and checking
Mass of red copper oxide/g.
for anomalous results (see the Activity: Finding the formula of red copper
oxide, in Section 5.2).
It is also possible to use a mathematical equation to represent the pattern
of the graph and so draw mathematical conclusions about the experiment
which produced the data (see the Activity: Investigation of the effect of Key terms
concentration on the rate of a reaction, in Section 9.3). If the graph is a
straight line through the origin, then it is possible to say that one variable is A best-fit line is a smooth line through
directly proportional to another variable. the middle of the points on a graph.
The use of graphs is a very important part of the study of kinetics. For A variable in an experiment is an
instance, in the study of the reaction between magnesium and hydrochloric item, factor, or condition that can be
acid (Section 9.2), the volume of hydrogen evolved is plotted against time. controlled, changed or measured.
Time (x-axis) is the independent variable and the volume of hydrogen The independent variable is the
(y-axis) is the dependent variable. one condition that is changed in an
experiment.
The rate of a reaction at a particular point is given by the gradient of a graph
at that point. The gradient is found by drawing a tangent to the graph; the The dependent variable is the variable
steeper the gradient, the faster the reaction. that is measured; its value depends on
the changes made to the independent
The rate of reaction at the start of the experiment, as soon as the reagents are
variable.
mixed, provides a useful way of studying the effect of changing one variable.
Measuring this initial rate is described in Section 9.2. Extrapolation of a graph extends the
line beyond the experimental range of
Sometimes it is useful to extend a graph beyond the range of values measured
values.
in the experiment in a process called extrapolation. Useful examples of
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extrapolation occur in experiments where temperature changes are measured
(see the Activity: Measuring and evaluating the enthalpy change for the
reaction of zinc with copper(ii) sulfate solution, in Section 8.4).
Because of energy losses, the maximum temperature measured is less
than it should have been. When the exothermic reaction has ended, the
temperature falls back to room temperature at a steady rate. By extending
this cooling curve back to the time of mixing, an estimate of what the
maximum temperature would have been, had there been no loss of energy to
the surroundings, can be obtained for use in the enthalpy change calculation.
Maxwell–Boltzmann distributions and
reaction profiles
One of the fundamental models of chemistry is the collision theory. This
states that molecules must collide before a reaction can occur and that the
molecules which collide must have sufficient activation energy to react.
A Maxwell–Boltzmann distribution shows the spread of molecular kinetic
energies for a gas at a particular temperature (Section 9.4). The curve shows
that there is a wide spread of molecular energies, meaning that in a sample of
gas at a given temperature the molecules do not all have the same energy and
so do not all move at the same speed. This variation occurs because molecules
collide, and when they do so there is a transfer of energy between them. The
distribution plots energy on the x-axis against the fraction of molecules with
a particular energy on the y-axis. Strictly, there is no maximum energy for
a molecule, so the x-axis should continue to infinity. The area under the
curve sums all the fractions of molecules and so represents the total number
of molecules.
You should note that these curves are not symmetrical. The peak of a curve,
showing the percentage of molecules with the most probable energy, lies
to the left of the average energy. The curve starts at the origin – there are
no molecules with zero energy. The curve approaches the x-axis but never
reaches it. There are very few molecules at high energy.
These curves help to explain the effect on the rate of a gaseous reaction of
changes in temperature, pressure, concentration and the addition of a catalyst.
Reaction profiles, such as that shown in Figure 9.17 in Section 9.4, are linked to
these distributions. These show the progress of a reaction from reactant to product,
overcoming activation energy barriers. The y-axis shows the energy level at each
stage of a reaction, including any intermediates formed. These diagrams give a
measure of the activation energy and the overall enthalpy change for a reaction.
The x-axis is not quantified and is usually simply labelled ‘progress of reaction’
with no scale.
Mass and infrared spectra
Mass spectra
The information produced in a mass spectrometer can be displayed in two
ways, either as a list of the relative abundance of each ion detected, including
the molecular ion and each fragment ion, or alternatively as a graphical trace
called a mass spectrum (Section 7.1).
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A mass spectrum shows all the information from the list but in an easily
understood graphical form. The x-axis is labelled mass to charge ratio (m/z)
but, since most ions are 1+, z = 1 and so the x-axis is effectively the relative
mass of the ions. The y-axis represents the relative abundance of the ions.
The tallest peak, the peak for the most abundant ion, is given the relative
abundance value of 100% and the abundance of each other ion is measured
relative to this peak. The most abundant ion is often not the molecular ion,
if fragmentation easily occurs.
Tip
There may be peaks with (m/z) values one or two units greater than the molecular ion.
This occurs when isotopes are present. The M+1 peak occurs in organic molecules
because about 1% of carbon atoms exist as 13C. If a molecule contains several 13C
atoms, then the relative abundance of this ion will become greater. M+2 peaks occur if
chlorine or bromine atoms are present.
Infrared spectra
An infrared spectrum (Section 7.2) is obtained by passing infrared radiation
through a sample and observing where radiation is absorbed. Absorption
corresponds to the natural frequencies at which vibrating bonds in the
molecule bend or stretch. The spectrum shows the absorption of energy
across a given range of frequencies.
Tip
The frequencies of vibrations lie in the infrared region between 1.20 × 1013
Scanning the range between 4000 cm–1
and 1.20 × 1014 Hz.
to 400 cm–1 in sequence takes a long
These values correspond to wavelengths between 2.5 × 10 –5 and 2.5 × 10–6 m. time. To speed up the process, modern
Rather than use wavelengths with these very small numbers, spectroscopists prefer infrared spectrometers pass infrared
to work in wavenumbers. The wavenumber is the number of waves in 1 cm, so radiation of several wavenumbers
the range of wavenumbers used is from 400 to 4000 cm–1. The spectra are drawn through the sample at the same time.
with the wavenumber values on the x-axis increasing from right (400 cm–1) to This produces extremely complicated
left (4000 cm–1) as this shows increasing wavelength from left to right. results which are analysed by a
mathematical technique called Fourier
The y-axis is labelled Transmittance/%. If there is no absorption, there
analysis. This is why modern infrared
is 100% transmittance but, when absorption occurs, there is a dip in
spectrometers are called ‘Fourier
transmittance. These dips are still called ‘peaks’ because they indicate high
transform infrared spectrometers’.
levels of absorption.
A1.8 Geometry
All giant structures are three-dimensional lattices. Apart from some small
molecules, molecular compounds are also 3D structures. Chemists need to
be able to think in 3D. They also need to be able to represent substances
using 3D models or in 2D on flat surfaces such as paper. They should realise,
for instance, that the nine structures shown on page 640 all represent the
same molecule.
Dichloromethane, CH2Cl2, is a tetrahedral molecule and can be represented either
in 3D or by ‘flat’ drawings. In all cases, the structure shown has a central carbon
atom surrounded by four bonds located 109.5° apart from each other.
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In each of the first three 3D representations, the two normal lines represent
covalent bonds in the plane of the paper. The solid wedge represents a bond
coming out of the paper, while the hashed bond represents a bond going into
the paper.
H H Cl
C C C
Cl Cl H
H Cl Cl H H Cl
Rotating the structures should demonstrate that they are all the same
molecule.
But drawing 3D structures is fairly difficult, so molecules are sometimes
represented with normal line or dotted line bonds, but still with an attempt
at 3D structures as shown here.
H H Cl
C C C
Cl Cl H
H Cl Cl H H Cl
More commonly, molecules are shown as flat structures (as below). In this
case, chemists have to remember that the bond angles shown as 90° or 180°
are all the tetrahedral angle, 109.5°. Although these look like flat crosses, the
four bonds are arranged tetrahedrally around the central carbon atom in
exactly the same way as in the six structures above.
H H Cl
H C Cl Cl C Cl H C H
Cl H Cl
The shapes of molecules with a central atom surrounded by between 2 and 6
pairs of electrons are discussed in Section 2.4, together with the effect of the
extra repulsion if some of the pairs are lone pairs.
You should learn the five common molecule shapes and try to recognise
them whenever they appear.
So, the trigonal planar arrangement of BF3, with bond angles of 120°, is
also seen in the bonding around the carbons in ethene or in a carbocation
(Section 6.2.10) or the carbonate ion, CO32–.
The tetrahedral arrangement is seen in every alkane but also in the ammonium
ion, NH4+, and the sulfate ion, SO42–.
The trigonal bipyramid structure is seen in the transition state when
nucleophilic substitution occurs at a primary halogenoalkane (SN2)
(Chapter 6.3). The bond which is forming, the bond which is breaking, plus
the three unchanged bonds are as far apart from each other as possible – the
same arrangement as in a molecule of PF5 – so there are bond angles of 120°
and 90° in the structure.
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Preparing for the exam
A2
A2.1 Revision
Understanding what you need to know and do
Your revision should be systematic and based on a copy of the Pearson
Edexcel A Level specification. The specification tells you what you have to
know, understand and be able to do. However, the language is very concise
and mainly written for teachers. This textbook has been written to cover
the specification. The chapters are in the same order as the specification, so,
if you are puzzled by a statement in the content, look for guidance in the
related chapter in this book.
The specification includes a table showing the assessment objectives for the
course. This table may seem rather technical and unimportant, but it will
help you to understand what you have to be able to do when answering
questions in examinations. There are three assessment objectives: AO1, AO2
and AO3.
In the AS exams at the end of the first part of the Pearson Edexcel course,
about 35−37% of the marks test your knowledge and understanding of
the content. This is AO1. There will be questions asking you to show that
you can recall facts, patterns and principles. There will also be questions
asking you to translate information from one form to another, carry out
simple calculations of a kind you have seen before, and to solve problems in
familiar contexts.
About 41−43% of the marks test AO2, which covers your ability to apply
your knowledge and understanding of scientific ideas, processes, techniques
and procedures in a range of contexts, which could be familiar or unfamiliar:
● in a theoretical context
● in a practical context
● when handling qualitative data
● when handling quantitative data.
Then about 20−23% of the marks are allocated to AO3, which covers your
ability to analyse, interpret and evaluate scientific information, ideas
and evidence which you have not seen before. This requires you to be
able to:
● make judgements and reach conclusions
● develop and refine practical design and procedures.
So, you can see that it is very important that you have the confidence to apply
your chemical understanding to unfamiliar situations in which you may
have to interpret new sets of data and information, including that presented
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in tables, charts or graphs. This confidence comes with practice. You need
to be able to link together ideas from different fields of chemistry to offer
explanations or construct arguments.
Chemistry is a practical subject and high-scoring students not only develop
good technical skills in the laboratory, but also develop a strategic sense of
Tip the ways that experiments are planned and carried out to give meaningful
results. Examination questions will ask you to describe how experiments
See the Practical skills sheets
were carried out and what happened; they will also ask you to explain the
which you can access online at
rationale for the procedures. You will also be asked to interpret the results of
[Link]/
experiments qualitatively and quantitatively, and to draw valid conclusions
EdexcelChemistry for more guidance
from data, taking into account measurement uncertainty.
on practical procedures, measurements
and the evaluation of experimental data.
Understanding the course as a whole
Some questions in the examinations expect you to bring together your
knowledge and understanding of different areas of chemistry, applying
them in contexts that may be new to you. This is sometimes called synoptic
assessment. The term ‘synoptic’ implies that you have an understanding of
the course as a whole, and an appreciation of how the different topics in the
course hang together and relate to each other.
So, any synoptic questions require you to work across different parts of the
specification and to show that you can draw on ideas and information from
different topics to answer questions or solve problems.
Revision notes
Check that you have your own notes on all sections of the specification.
Organise your notes with clear titles and subheadings. Highlight key points
in colour. Include mnemonics if you find them helpful, such as:
● OIL RIG (oxidation is loss, reduction is gain
● MEPrB (methane, ethane, propane, butane)
● ALSUB (axes, labels, scales, units, points).
Flow diagrams can be helpful in giving you an overview of organic chemistry.
Alkenes
Ketones
Alkanes Alcohols
Halogenoalkanes Aldehydes
Nitriles Amines Carboxylic
acids
Figure A2.1 Relationships between series of organic compounds. Make a copy of
the diagram, then add examples with formulae and label the arrows to show how one
functional group can be converted to another.
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Active revision
You need to make your revision active, so that you maintain concentration
and really test that you understand the key points, while developing the
necessary skills. For each revision topic, try writing the title of a topic in the
centre of a sheet of paper. Then use the definitions and explanations in the
main part of this book to help you build up a mind map or concept map to
show how the ideas in the topic link together.
INTERM
OL DS
EC
SING UN
PO
ULA
M
LE
DO
CO
UBLE ES
R
CO
VALENT EN KAN
BO AL
G
NS A
EMPIR ARBO LKENES
HALO
IC URAL O C
N
A AT DR
DI
MOLEC N Y
TIC HOLS
NG
ULA
L
G C HA INS S Y N T H E CO L CO
H
ST
RUCTURAL
R FORMU
L LON IES OXYGE
N MPOUNDS A
R
SE
AE
AR
C
(cis – trans) M US BONYL CO
E–Z
CARB
ISOMERISM OL GO MPOUNDS
AL LO
STRUCT
UR ECULE O O
XY
S M LIC ACI S
NIC
HO D
A
G
OR ING FUELS
REFIN
ET
ADDITION CHANGING MOLE ROCHEMICALS
P
CU
C H E M I ST R Y
LE
IO N S
IO N
S
SUBSTITUT N REACT
S PR
D–BASE T Y P E
IO
C I ACTICA
A
T
I M I NA X B L ANALYSIS
EL O OND
RED BREAKIN
G HOMOLYT
RADICALS YN
IS
IC
FREE
S
THESIS
YS
L
HE
RO TE
HYD R O LY T I C ELECTROPHIL
ES
N
CL
U
EOPHILE
S
Figure A2.2 The start of a mind map for introductory organic chemistry.
Suppose you are revising the chemistry of the halogens. Have a pile of scrap
paper to hand and a pencil. Now, as you read, make jottings, small lists,
summary phrases, write equations, sketch diagrams and practise labelling
them. Now tear up the paper, close the textbooks and notes, and write out
those lists, equations, diagrams and so on. Then check to see whether you
have remembered correctly.
Tip
When it comes to learning the reactions of a family of organic compounds
Learn key definitions thoroughly. Even
such as the alcohols, consider using a set of cards. Write the equation for each
top candidates frequently lose marks
alcohol reaction you have to know on one side of the card. Write the names
by missing out key words when asked to
of the reactants and the conditions for the reaction on the other side. Now
state what is meant by chemical terms.
you can use the cards for revision. Look at one side of the card and try to
Use the online glossary to help you.
recall what is on the other side. Similarly, test yourself using the expanded
glossary available online with this textbook.
Practise calculations with the help of worked examples. Even if you have
answered all the ‘Test yourself ’ questions in this book, it is worth working
through them again, including the calculations, checking your responses
with the answers provided online.
One of the key characteristics of a high-scoring candidate is the ability to
carry out complex calculations, setting out the working step by step and
including the correct symbols and units. The worked examples in this
textbook show you how to do this. Among the important calculations are
those to work out titration results and thermochemistry calculations.
A2.1 Revision 643
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A2.2 Exam technique
Past papers, mark schemes and examiners’
reports
Look at past papers and practise answering the questions. Some topics in the
specification are so important that they are tested in most years with only
very minor differences in emphasis and phrasing of questions. It is because so
many questions on some topics appear to be the similar year after year, that
you must read the questions with the greatest of care to note the particular
emphasis, and the precise nature of a question.
Looking at some past mark schemes can be helpful, but you need to bear in
mind that they are mainly intended to help examiners give marks accurately,
so they are presented very concisely and do not include full answers. However,
the mark schemes for longer questions will show you what examiners are
looking for when there are 3–6 marks for part of a question.
For some questions examiners expect you to write at greater length, drawing
together and organising relevant material. No marks are allocated for the
plan you sketch out before writing your answer, but making a plan helps you
to write a well organised answer with examples and so leads to better marks.
Examiners’ reports have traditionally been written for teachers, but
increasingly they include information that students find helpful. You can
download the reports from the Pearson Edexcel website. You will find that
the report on a particular examination paper includes sample answers from
students, together with comments and tips from the examiners. Working
through some of these reports will show you how to answer questions in the
ways that gain most marks, and how to avoid common errors.
One of the most useful revision activities is to answer the questions in past
examination papers and then to check your answers against the published
mark scheme.
Question types
The Pearson Edexcel examination papers consist of a series of structured
questions. Each question is set in a particular chemical context. If the context
is familiar, it may just be introduced by a short sentence, such as: ‘This question
is about Group 2 and enthalpy changes.’ However, there are other questions
which start with much more information and data. For example, there might
be a summary of the procedure for an experiment, followed by some sample
results. Examiners do not include information that you do not need. It is
essential that you read the introduction to a question very carefully because you
need it to answer the parts of the question.
Sometimes the introductory information at the start of a question will seem
strange. This is not because the examiner has made a mistake or your teacher
has failed to cover some part of the specification. As shown by the assessment
objectives, the skills being tested in the examinations include your ability to
apply your knowledge to unfamiliar situations. You can be confident that the
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examiner has chosen a context and given data that you can be expected to
make sense of using the chemistry that you have learned during the course.
Within the structured questions there are three main question types:
multiple-choice questions, short-answer questions and questions asking for
a longer answer.
Tip
Multiple-choice questions In an examination, aim to gain
Multiple-choice questions can seem deceptively simple, because you are maximum marks for minimum
given four choices and have to choose one of them. Sometimes they test knowledge.
your knowledge of a basic idea and, if you know the answer, you will be able
● Score marks in all questions.
to answer the question quickly. However, some questions require careful
● Never leave a question blank
thinking and may involve a calculation.
because you are short of time.
If a multiple-choice question involves a calculation you should not try to do ● It is much easier to score the first
it in your head. Write down your working in rough using a blank part of the few marks in a question than the last
question paper. few marks.
● Answer the question asked.
As always, you need to read the questions carefully, especially if the words
● Make sure you don’t leave any parts
‘not’ or ‘never’ are included in the working. A questions which includes ‘not’
out.
can be confusing as in: ‘Which substance is not a product of the incomplete
● There are no marks for unnecessary
combustion of hexane?’. You have to pick the one of the four options that is
extra information.
not produced in the reaction.
You do not lose marks if you make the wrong choice, so you should always
answer every multiple-choice question. Sometimes you may be able to reject
two of the options as wrong, but then find that you are not sure which of the
remaining two options is correct. You should make an intelligent guess and
pick one of them. Tip
Watch your language. Don’t use the
Short-answer questions
wrong words.
Most of the parts of a structured question demand short answers worth 1–3
marks. Typical parts of such questions involve: ● Do you mean ions or atoms or
molecules?
● naming, stating or giving information
Don’t use the words ‘atom’ or
● writing balanced equations
‘molecule’ when discussing an ionic
● describing reactions
lattice such as sodium chloride.
● explaining the meaning of key terms
Don’t use the word chloride (for
● plotting graphs
the ion) when you mean atoms or
● interpreting data
molecules of the element chlorine
● performing calculations.
(and vice versa).
When answering these questions, you should use the space allowed for your ● Bonds or forces?
answer and the number of marks allocated to guide you when deciding how Molecules have covalent bonds
much to write. Three marks for part of a question probably means that the between atoms within/inside the
examiner is expecting three good points to be made. If asked for a chemical molecules. These do not break when
test, for example, one mark might be for the reagent to be used, one for the molecular substances melt or boil.
conditions and the third for describing the observations. Between the molecules there are
(intermolecular) forces, not bonds.
As explained in the next section, it is very important that you pay attention
These forces are overcome when
to the ‘command words’ and provide the type of answer that the examiner
molecular substances melt or boil.
has asked for.
A2.2 Exam technique 645
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Longer answer questions
Questions requiring longer answers carry 4–6 marks in the Pearson Edexcel
examination papers. Some of them ask for a description, interpretation or
explanation; others require a calculation that involves several stages. Here,
too, you must pay careful attention to the command words.
In a prose answer, you will not get credit for copying phrases from the
question. This seems obvious but students commonly lose marks by simply
repeating the question. It can help to jot down key points in rough so that
you can decide on the best order to cover them. Write short, clear sentences.
Keep your answer relevant by dealing with all the points asked for in the
question.
If you are asked to carry out an extended calculation, without any help from
the structure of the question, you must set out your working step by step
with enough words to show what you are doing at each step. The worked
examples in the main chapters of this textbook show you how to do this for
each of the types of calculation that feature in this course. Marks are awarded
for each stage of the quantitative argument, as you can see by looking at
Pearson Edexcel mark schemes.
Examiners’ terms
Every year, too many well-prepared candidates fail to score as many marks
Tip as they should because they do not answer the question set by the examiners.
Curly arrows in organic mechanisms
Examiners try very hard to set questions which are clear to candidates. Even
must start at a bond or a lone pair and
so, under examination conditions, it is all too easy to rush into writing an
end forming a new bond or a lone pair
answer before checking carefully what you have been asked to do by the
on an atom. Don’t forget to add the
examiner. You do not get marks if you fail to answer the question that the
dipoles to relevant bonds too.
examiner has set you.
A useful first step is to highlight the command words in the question. Words
such as ‘calculate’, ‘describe’ and ‘explain’ give instructions.
The command words used by examiners in Pearson Edexcel chemistry
papers are defined in Appendix 7 of the specification.
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Index
A occurrence and uses 485 names and formulae 552–3
absolute temperature 468 oxidation 488 structures 553
absorption spectra 237, 596 physical properties 487–8 ammine complexes 546
accuracy of measurements 142–3 reaction with hydrogen cyanide 491–2 ammonia
acid–base reactions 148–9 reduction 488 acid–base titrations 334
acid–base titrations 140–2, 332–5 testing for 226, 492–5 ammonia solution’s reaction with
strong acid with strong base 332 aliphatic amines 549 transition metal ions 439
strong acid with weak base 334 aliphatic hydrocarbons 177 decomposition 463
weak acid with strong base 333–4 alkali metals see Group 1 elements molecular shape 54
weak acid with weak base 335 alkaline earth metals see Group 2 elements reaction with acyl chlorides 507
acid dissociation constant (K a) 327–9 alkalis 11 reaction with halogenoalkanes 215
determination for weak acids 344 reactions 98–9 reaction with hydrogen halides 117, 286
acids 10–11, 318–21 alkanes 162, 177–89 ammonium ions 48
Arrhenius’s theory 319–20 cracking 184 shape 52
dissociation 319–20, 331 naming compounds 163–4 testing for 149
early theories of 318–20 physical properties 63, 178 amount of substance 123
as proton donors 320–1 reactions 178–81, 211 amphoteric hydroxides 439
reactions 98–9, 544–5 reforming 183, 185 anabolic steroids 23
solubilities 100 alkenes 157, 158, 190–204, 211, 574 analysis/analytical techniques 3, 593–625
strong see strong acids addition polymers 201–4 analysis of an inorganic unknown
weak see weak acids addition reactions 167, 194–5, 196–200 150, 440
see also carboxylic acids addition to unsymmetrical analysis of jasmone 608–9
activation energy 280–1, 450, 467 alkenes 199–200 chromatography 610–17
experimental determination of 469 combustion reactions 193 combined techniques 618–19
and stability 382 double bonds 190–1 infrared spectroscopy see infrared
acylation 507, 529–31 E/Z isomerism 191–3 spectroscopy
acyl chlorides 484, 507–8, 574 formation 228 mass spectrometry see mass spectrometry
Friedel–Crafts reaction 529–31 naming compounds 164–5, 190 nuclear magnetic resonance
reactions 507–8, 547 oxidation reactions 195–6 spectroscopy 597–607
acyl group 501 physical properties 190 organic compounds 241, 570, 575,
acylium ion 531 reaction mechanisms 196–8 608–9
addition polymerisation 168, 201–4 reactions 193–6 qualitative analysis 148–50
comparing addition and condensation alkylation 529–31 quantitative analysis 135–9
polymers 561 alkyl groups 163 antiseptics 519
addition reactions 167 alkynes 158 aramids 558
alkenes 167, 194–5, 196–200 allotropes 70–4 arenes 177, 519
adsorption 442, 443, 463, 610 alpha particle experiments, Rutherford’s 14 naming 524–5
air pollution 186–7 aluminium 381 physical properties 525–6
alcohols 157, 158, 219–29, 574 amides 484, 508, 574 preparation 550
combustion reactions 221 preparation 550 reactions 526
dehydration of 228 structures and names 543 see also benzene; phenol
determining the type of 225–6 amines 158, 209, 541–50, 574 arithmetic mean 630–1
names and structures 219–20 as bases 544–5 aromatic amines see arenes
oxidation reactions 224–7 as ligands 545–6 aromatic hydrocarbons 177
physical properties 221 as nucleophiles 546–7 Arrhenius equation 467–8
reaction with acyl chlorides 507 physical properties 543 Arrhenius theory of acids 319–20
reaction with carboxylic acids 505 preparation 549–50 arson, forensic investigations of 616–17
substitution reactions 211, 217, 222 reaction with acyl chlorides 508 aspirin 1, 548
aldehydes 158, 224–5, 484–6, 574 amino acids 551–6 preparation 511
formation 485–6 acid–base properties 554–5 thin-layer chromatography 613
names and structures 484 chromatography 612 asymmetric carbon atoms 477
condensation polymerisation 555–6 asymmetric molecules 477
Index 647
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atom economy 147, 636–7 hydrides 66 carbon neutral processes 188–9
addition reactions 167 ionic compounds 44 carbonates
atomic number 16–17 metals 75 of Group 1 metals 102
atomic orbitals 26–7, 30 noble gases 62 of Group 2 elements 108
atomic structure 12–16 relationship to physical properties 77 reaction with acids 99
early ideas 12–13 simple molecular structures 51 testing for 149
‘plum pudding’ model 14 Boltzmann, Ludwig 371 thermal stability 109–10
Rutherford’s model 15 bombycol 474 carbonyl compounds 484–99
atomisation, enthalpy changes of 352–3 bond angles 51–5 aldehydes see aldehydes
atoms 5 bond energies 49–50 identification of 492–5
electron structure 23–31 bond enthalpies 264–7 ketones see ketones
autocatalysis 444 bonding 41–60 oxidation 488
averages 630–1 covalent see covalent bonding physical properties 487–8
Avogadro constant 124–5 ionic 42–6, 349, 353 reaction with hydrogen cyanide 491–2
metallic 46, 74–7 reduction 490
B relationship to physical properties 77 triidomethane reaction 496
balanced equations 300, 389 bond lengths 49–50 carbonyl group 484, 487
barium 104, 105 benzene 520 carboxylate ion 494
compounds 107–9 Born–Haber cycles 354–6 carboxylic acids 158, 224, 225, 484, 485,
reactions of 106–7 and stability of ionic compounds 359 488, 500–6, 574
see also Group 2 elements boron trifluoride 52, 53 derivatives 501, 507–14
baryte 109 brass 423 esterification reactions 505–6
bases 11, 102 bromine 111, 112 names and structures 500–1
acid–base reactions 148–9 diffusion 369–70 occurrence 500
acid–base titrations see acid–base titrations reactions 113–14, 118, 181, 197–8, physical properties 502
amines as 544–5 526–7, 535 preparation 502
nitrogenous 322 see also Group 7 elements reaction with phosphorus(v)
as proton acceptors 321–2 bromoalkanes 222 chloride 504
solubilities 100 bromoethane 235 reactions as acids 503
strong see strong bases bromothymol blue 336 reduction 504
weak see weak bases Brønsted–Lowry theory 320–2 catalysts 277, 281–2, 313, 441–5
basic oxides 107 buffer solutions 340–3 development of 445
batteries 410 pH of 341–3 heterogeneous 441–2
Benedict’s solution 226, 493, 494 butane 162 homogeneous 444
benzene 519–33 see also alkanes for hydrolysis of esters 510
bonding in 521–2 catalytic converters 187, 443
delocalisation 523 C catalytic hydrogenation of alkenes 194
Friedel–Crafts reaction 529–31 calcium 104 cells see electrochemical cells
halogenation of 526–7 compounds 107–9 Chadwick, James 15–16
hydrogenation of 521, 532 reactions 106–7 chain isomerism 166, 474
nitration of 528–9 see also Group 2 elements chelates 433
physical properties 525–6 calcium fluoride 350 chemical kinetics 272
reaction with chlorine 527, 532, 533 calculations see also rates of reaction
reactivity compared with phenol 534–5 involving chemical equations 131–3, chemical shifts 599
resistance to reaction 520 634–7 chirality 473–83
stability 521 mathematical equations and amino acids 553
structure 519–22 expressions 632–3 see also optical isomerism
benzoic acid 501, 502 order of operations 626–7 chloramphenicol 569
beryllium 104, 105, 106 calomel electrode 400 chlorate ions 118
see also Group 2 elements calorimeters 249, 250–1 chlorine 111, 112
beryllium chloride 52, 53 carbocations 197 isotopes 20
best-fit line 637 primary, secondary and tertiary 199–200 reactions 102, 107, 113–14, 117–18, 291,
bias 143, 144 stability of 200 527, 532, 533
bidentate ligands 432–3 carbon 155–6 testing for 149
biofuels 188–9, 219 allotropes of 70–4 water treatment 118–19
bleach 396 carbon-13 NMR 599–601 see also Group 7 elements
blood buffers 342 carbon dioxide chloroalkanes 222, 223
boiling temperatures bonding 6 chlorofluorocarbons (CFCs) 218–19
alkanes 63 testing for 107–8, 149 chromatography 598, 610–19
checking the purity of a product 586–7 carbon monoxide 186 combined with mass spectrometry 618–19
648 Index
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gas 593, 614–19 reaction of solution with zinc 85–6 d orbitals 26–7
liquid 611–14 thermal decomposition of crystals 97 dissolving
principles of 610 cortisone 573 enthalpy change of solution 360–1, 363
R f values 612 covalent bonding 46–60 entropy changes 375–6
chromium 423 bond lengths and bond energies 49–50 distillation 581–2
oxidation states 426 dative 48, 428–9 fractional 182–4, 584
cis-platin 434 in ionic compounds 357–8 dot-and-cross diagrams 42, 349–50
cis–trans isomerism (E/Z isomerism) lone pairs of electrons 47–8 double bonds 47, 190–1
191–3, 474–5 multiple bonds 47 influence on shape of molecules and
citric acid 10, 503 polar bonds and polar molecules 56–60 ions 55
clock reactions 461–2 shapes of molecules 51–6, 639–40 dynamic equilibrium 289, 298
cobalt chloride paper 285–6, 436 covalent structures see also equilibrium
collision theory 278–82 giant lattices 6
colorimeters 452 simple molecular structures 50–1 E
cracking 183, 184 EDTA4- 433
coloured ions 426–8
crude oil 182–5 elastomers 562
combustion
use in polymer manufacture 202 electrical conductivity
alcohols 221
crystals 39–40 ionic compounds 44
alkanes 178–9, 186–7
liquid crystals 568 metals 75
alkenes 193
cumene process 536 relationship to physical properties 77
enthalpy changes 248–51
cyanide ions simple molecular structures 51
incomplete 186
reaction with halogenolkanes 214–15 electrochemical cells 397–413
standard enthalpy change of 255, 258–9
use to increase carbon chain cell diagrams 399
complex ions 428–38
length 578 cell e.m.f.s 402–8
formation 428–30
cyclohexane 521, 532 fuel cells 412–13
ligand exchange reactions 435–7
cyclohexene 521 practical investigations 404
ligand types 432–3
cytochrome oxidase 444 prediction of direction of change 402–3
naming 430–1
standard conditions 398–9
preparation 438
relative stability 437–8
D storage cells 410–11
Dalton, John 12–13 electrochemical series 405–6
shapes 431–3
data handling 629–32 electrode potentials 397–401
complementary colours 427
dative covalent bonding 48, 428–9 standard see standard electrode potentials
composites 71, 562
Davy, Humphry 319 electrolysis 44
compounds 5
d-block elements 30, 418 electrolytes 44
of metals with non-metals 7–8
see also transition metals electromotive force (e.m.f.) 398
of non-metals with non-metals 5–6
decimal places 630 calculation of 402–3
concentration 293, 327
dehydration reactions 228 cell e.m.f.s 402–8
effect on equilibrium 289–90, 310
delocalised electrons 74, 519 electron affinity 352, 353
effect on the rate of reaction 275,
in benzene 523 electron configurations 7–8
279, 454
Democritus 12 Group 1 metals 101
of solutions 134, 145
derivatives Group 2 elements 105
concentration–time graphs 451
of carboxylic acids 501, 507–14 halogens 111
for first order reactions 456
in identification of carbonyl and the periodic table 29–32
for second order reactions 457
compounds 492–3, 495 transition metals 418–20
for zero order reactions 458
desiccators 583 electronegativity 57–8
condensation polymerisation 513–14
desorption 442, 443 electron-pair repulsion theory 51–3
of amino acids 556
diamond 70–1 electrons 5, 16
comparison with addition
dibasic (diprotic) acids 321 atomic orbitals 26–7, 30
polymerisation 561
1, 2-dichlorobenzene 520 discovery of 13–14
in polyamides 557–60
diffusion 368–70 energy levels 23–9
condensation reactions 555–6
diesel engines 186 transfer in redox reactions 84–5
conductivity changes 452
dilutions, quantitative 138 electrophiles 171
conjugate acid–base pairs 323
2, 4-dinitrophenylhydrazine 493 electrophilic addition reactions 196–200
Contact process 441, 442
dipoles 59–60 addition to unsymmetrical alkenes
co-ordination compounds 429–30
dipole–dipole interactions 64 199–200
co-ordination numbers 429
displacement reactions 85–6, 115 electrophilic substitution reactions
co-polymers 562
displayed formulae 161, 572 Friedel–Crafts reaction 529–31
copper 423
disproportionation reactions 91, 117, 406–7 halogenation of benzene 526–7
copper oxide 127
dissociation of acids 319–20 nitration of benzene 528–9
copper (ii) sulfate
effect of dilution 331 electrostatic attraction 43, 51, 350
Index 649
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elements 4–5 stability constants 437, 438 relationship to equilibrium
elimination reactions 167, 168, 212, 228 equilibrium law 300–6 constant 379–80, 408
halogenoalkanes 168, 216 experimental verification of 299 and stability 380–3
empirical formulae 125–7, 159, 571 equivalence point 140, 332 free radicals 170
enantiomers 477 equivalent protons 603 free-radical substitution 179–81
endothermic changes 248, 264 errors 143–4 Friedel–Crafts reaction 529–31
equilibrium 292 in thermochemical experiments 250–1 fuel cells 412–13
feasibility of endothermic reactions esterification reactions 302–3, 505–6 fuels
368, 377 esters 484, 508–12, 574 alternative 188–9
end-point 140, 332 hydrolysis 510 from crude oil (fossil fuels) 182–8
energy levels 23–9 names and structures 509 fullerenes 72–3
energy quanta, sharing of 370–1 occurrence and uses 508–9 functional group isomerism 166, 474
enthalpy changes 246–67 physical properties 509 functional groups 156, 157–8, 573–5
of atomisation 352–3 preparation 506 acyl group 501
and bonding 264–7 ETBE 184 carbonyl group 484, 487
Born–Haber cycles 354–6 ethane 157, 162
and the direction of change 263–4 see also alkanes G
endothermic 248, 264 ethanoic acid 445 gas chromatography (GC) 593, 614–19
exothermic 246–7, 263–4 acid–base titrations 333–4 combined with mass spectrometry
and feasible reactions 368 effect of dilution on dissociation 331 618–19
of formation 255–6, 258–61, 351 ethanol 157, 161 use in forensic investigations 616–17
Hess’s law 257–63 oxidation of 227 gas constant 468
of hydration 360, 362 see also alcohols gaseous equilibria 306–8
measurement of 248–52, 253–4 ethers 158, 166, 574 gases 2, 128–30, 371–2
of neutralisation 256–7, 339–40 exam technique 644–6 collection and measurement of 452
of solution 360–1, 363 exothermic changes 246–7, 263–4, 368, 377 kinetic theory 278–82
standard see standard enthalpy changes equilibrium 292 molar volume of 130, 133
enthalpy level diagrams 247, 248 extrapolation of a graph 637–8 volume calculations 132–3, 635
entropy 367–86 E/Z isomerism 191–3, 474–5 volume measurement 132
standard molar entropies 371–2 gas laws 128
entropy changes 370–6 F geckos 61
and complex ions 437 fatty acids 512 general formulae 156
during dissolving 375–6 f-block elements 30 giant structures 40, 41, 69–74
prediction of the direction and extent of a feasible reactions 368–70 allotropes of carbon 70–4
reaction 408–9 free energy change 377–8 Gibbs, Willard 376–7
of surroundings 374 prediction of the direction and extent of a glycerol 512
of the system 373 reaction 408–9 gold 16, 381
total 374, 376 redox reactions 402–3 Gore-tex 209
equilibrium 285–317 total entropy change 374, 376 graphene 73
in buffer solutions 341 Fehling’s solution 226, 493, 494 graphite 71–2
and catalysts 313 filtration 581 graphs 637–9
dynamic 289, 298 fingerprint regions 240 Grignard reagents 578–9
effect of concentration 289–90, 310 first ionisation energies 33–4 Group 1 elements (alkali metals) 31, 100–4
effect of pressure 292, 311 first order reactions 456 compounds 102–4
effect of temperature 292, 312–13 flame colours flame colours 103–4
gaseous 306–8 Group 1 elements 103–4 melting temperatures 75
heterogenerous 294, 305 Group 2 compounds 109 reactions 102
homogeneous 294 fluorine 111 Group 2 elements (alkaline earth
influencing factors 289–92 see also Group 7 elements metals) 104–10
qualitative predictions 295 fluorite 104 compounds 107–10
reaction of iodine and chlorine 291 f orbitals 26–7 reactions 106–7
equilibrium constants 293–5, 300–4, 408 forensic investigations 616–17 thermal stability of Group 2
calculation of 303–4 formation, standard enthalpy change of carbonates 382–3
experimental determination of 301–3 255–6, 258–61, 351 Group 7 elements (halogens) 111–19
and free energy change 379–80 formulae of organic molecules 159–62, in oxidation states +1 and +5 117–18
K a 327–9, 344 571–2 reactions 113–14, 195, 197–8
Kp 308 fractional distillation 182–4, 584 solutions 112–13
Kw 325 free energy change 376–83 see also halide ions
predicting the direction and extent of prediction of the direction and extent of groups of the periodic table 29–30, 31
change 306, 408–9 a reaction 408–9 gypsum 108
650 Index
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H hydronium/hydroxonium ion (oxonium ion) Group 2 elements 105
Haber process 441 48, 428 periodic pattern 33–4
haemoglobin 298, 436 hydroxides transition metals 421
half-equations 85–6, 91–2, 388 of Group 1 metals 102 ions 7
combination of 389 of Group 2 elements 107–8 complex see complex ions
half-life of a reaction 456–7 reaction with acids 98 enthalpy changes in formation of 352
halide ions 114–15 reaction with halogenoalkanes 213–14 formation of 42
hydrogen halides 117 iron
reactions with concentrated sulfuric
I cycle of extraction and corrosion 83
ibuprofen 147 reaction with steam 286
acid 116
ice 65–6 iron ions 7
testing for 148
ideal gases 128 reactions with halogens 114
halogenation of benzene 526–7
incomplete combustion 186 isoelectric point 554
halogen carriers 527
indicators 140–1, 336–7 isoelectronic molecules and ions 52
halogenoalkanes 158, 209–19, 574
inductive effect 199 isomerism 473–4
elimination reactions 168, 216
inert substances 381 E/Z 191–3, 474–5
formation 179–81
infrared spectroscopy 237–40, 596–7, 639 optical see optical isomerism
hydrolysis 169, 212–13, 463–4
of benzene 521–2 structural 165–6, 473–4
preparation 211, 217, 222
initial-rate method 458–60 isotopes 19, 235
preparation of amines 549
integration traces 602 labelling with 172–3
reaction with amines 546–7
intermediates 172–3, 199
substitution reactions 168, 212–15
uses and impacts 218–19
and catalysts 282 J
intermolecular forces 40, 61–6 Jabir ibn-Hayyan 318
halogens see Group 7 elements
dipole–dipole interactions 64 jasmone 608–9
heat conduction 75
hydrogen bonding 64–6
Hess’s law 257–63
and the properties of alkanes 63 K
heterogeneous catalysis 277, 441–2 K a 327–9, 344
and solubility 67–8
heterogeneous equilibrium 294, 305 Kekulé, Friedrich 519–20
iodide ions
heterolytic bond breaking 170–1 kelvin temperature scale 128
reaction with hydrogen peroxide 462
high-performance liquid chromatography ketones 158, 224, 225, 486–7, 574
reaction with peroxidosulfate(vi)
(HPLC) 614 formation 487
ions 444
homogeneous catalysis 444 names and structures 486
iodine 111, 112
homogeneous equilibrium 294 occurrence and uses 486–7
molecular structure 50
homologous series 156 oxidation 488
reactions 113–14, 118, 291, 464–6
homolytic bond breaking 170 physical properties 487–8
solutions 112, 113, 287–8
hydration 69, 361 reduction 490
test for oxidising agents using iodine
of alkenes 195 testing for 226, 492–5
solution 391
enthalpy change of 360, 362 Kevlar 558–9, 560
titrations with thiosulfate 395–6
hydrocarbons 156, 177 kinetic inertness (kinetic stability) 381
see also Group 7 elements
alkanes see alkanes kinetic theory 278–82
iodoalkanes 222
alkenes see alkenes Kp 308
ionic bonding 42–6, 349
fuels from crude oil 182–8 Kw 325
lattice energies 353
hydrochloric acid 332, 334
ionic compounds 7–8, 11
hydrochlorofluorocarbons (HCFCs) 219
electron sharing 357–8
L
hydrogen, reaction with oxygen 9–10 labile protons 606
experimental and theoretical lattice
hydrogenation lactic acid 158, 327, 329
energies 356–7
of alkenes 194 lattice energies 349–66
formation 349–50
of benzene 521, 532 effects of ionic charge and ionic
properties 44–5
hydrogen bonding 64–6 radius 362
stability of 359
hydrogen chloride 116, 117 experimental and theoretical 356–7
standard enthalpy change of
hydrogen cyanide 491–2 lattices 43, 350, 639
formation 351
hydrogen halides 117 Lavoisier, Antoine 319
ionic crystals 350
hydrogen bonding 65 lead–acid cells 410
ionic equations 86, 98
reaction with alkenes 195, 197 lead(ii) nitrate, reaction with potassium
ionic precipitation reactions 99, 148
reaction with ammonia 117, 286 iodide 99
ionic product of water (Kw) 325
hydrogen–oxygen fuel cells 412–13 Le Chatelier’s principle 289–92, 312, 341
ionic radii 44–5, 46
hydrolysis 167, 168–9 levodopa 569
Group 1 elements 101
of halogenoalkanes 169, 212–13, 463–4 ligand exchange reactions 435–7
Group 2 elements 105
investigation of reaction ligands 429
ionic salts see salts
mechanism 172–3 amines as 545–6
ionisation energies 23–6, 352–3
types of 432–3
Index 651
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limewater 107–8, 149 metallic bonding 46, 74–7 compound formation with metals 7–8
limiting reagents 146 metals 4–5 compound formation with non-
linear shapes 53, 432 compound formation with metals 5–6
liquid chromatography 611–14 non-metals 7–8 reactions with halogens 114
liquid crystals 568 properties 74–7 N-phenylethanamide 588
liquids reactions of metal oxides and hydroxides nuclear magnetic resonance (NMR)
arrangement of particles 2 with acids 98 spectroscopy 597–607
miscible 68 reactions with acids 98 carbon-13 NMR 599–601
structure of 372 reactions with halogens 113 medical benefits 607
volatile 584 methane 6, 162 proton NMR 601–6
lithium 101 reaction with chlorine 179–81 nucleophiles 171, 212
compounds 102–4 shape 51, 52, 53 nucleophilic addition reactions 491–2
see also Group 1 elements methanoic acid (formic acid) 318, 500, 501 nucleophilic substitution reactions
lithium cells 402, 411 methanol fuel cells 413 212–15
lithium tetrahydridoaluminate(iii) 490, 504 methyl orange 336, 337 of amines 546–7
logarithms 24, 324 methyl red 336 and optical isomerism 479–80
and acid–base titrations 335 miscible liquids 68 SN1 reactions 464
pK a 330 mobile phase of chromatography 610 SN2 reactions 463
pKw 326 molar masses 123 nylon 557–8
London forces 61–2 gases 128–9
lone pairs of electrons 47–8, 321 molar volumes of gases 130, 133 O
influence on shape of molecules and molecular formulae 128–9, 159–60, 571 octahedral shapes 53, 431–2
ions 54–5 molecular ions 235 octane numbers 183
moles 123, 124–5 OIL RIG mnemonic 388
M monobasic (monoprotic) acids 321 optical isomerism 474, 476–81
magnesium 104, 105 monodentate ligands 432 importance to living things 481
compounds 107–9 multidentate ligands 433 and polarised light 478–9
reactions 83–4, 98, 106–7 multi-step reactions 463 and reaction mechanisms 479–80
see also Group 2 elements order of a reaction see reaction orders
magnetic resonance imaging (MRI) 607 N ordinary form 627
malleability 75 naming of compounds organic analysis 570, 575
mass calculations 131, 634–5 inorganic 89–90 infrared spectroscopy see infrared
mass number 16–17 organic 162–5, 190, 209, 219 spectroscopy
mass spectrometry 17–18, 22–3, 233–6, negative numbers 627 mass spectrometry see mass spectrometry
638–9, 593–6 neutralisation reactions 338–40 NMR spectroscopy 597–607
analysis of spectra 595 enthalpy changes 256–7, 339–40 organic chemistry 155–76
combined with gas chromatography standard enthalpy change of neutralisation analysis of compounds 241, 570, 575,
618–19 256–7, 339 608–9
high-resolution 595–6 neutrons 5, 16 empirical, molecular and structural
of molecules 22 discovery of 15–16 formulae 159–62, 571–2
in sport 23 nickel catalysts 442 functional groups 157–8
mass-to-charge ratio 18, 234 ninhydrin 612 isomerism 165–6
materials 2 nitrates naming compounds 162–5, 190,
mathematical operations, order of 626–7 of Group 1 metals 103 209, 219
Maxwell–Boltzmann distribution 279, of Group 2 elements 108 reaction mechanisms 169–73
281, 467, 638 thermal stability 109–10 reaction types 167–9
mean bond enthalpies 265 nitration see also alcohols; alkanes; alkenes; amides;
mechanisms of reactions see reaction of benzene 528–9 amines; amino acids; arenes; carbonyl
mechanisms of phenol 535 compounds; halogenoalkanes; proteins
melting temperatures nitriles 491, 574 organic synthesis 568–92
alkanes 63 hydrolysis of 502 synthetic pathways 576–80
checking the purity of a product 586–7 reduction of 549 synthetic techniques 581–8
ionic compounds 44 nitrogenous bases 322 Taxol 569–70
metals 75 nitrogen oxides outliers 630
periodicity 32–3 air pollution 187 oxidation number rules 87
relationship to physical properties 77 preparation of N2O4 94 oxidation numbers 86–90, 387, 388
simple molecular structures 51 noble gases, boiling temperatures of 62 balancing equations 93, 390
transition metals 422 non-aqueous solvents 68 in ions 86–8
Mendeléev, Dmitri 4 non-equivalent protons 604 in molecules 88
menthone 487 non-metals 4–5 and naming of compounds 89–90
652 Index
469983_Index_Chem_Y1-2_647-[Link] 652 13/04/19 10:17 PM
of the transition metals 423–6 plastics 562 see also titrations
oxidation reactions 83, 86, 387 pOH 326 quantitative dilution 138
alcohols 224–7 polar covalent bonds 56–60 quantum shells 23–9
alkenes 195–6 polarimeters 479 quantum theory 427
see also redox reactions polarisability 62, 358 quartz 6
oxidation states 88–9 polarisation 357–8 quaternary ammonium salts 547
Group 1 metals 101 polarised light 478–9
Group 2 elements 106 polarising power 110, 358 R
halogens in oxidation state -1 114–17 polar molecules 59–60, 68 racemic mixtures 478
halogens in oxidation states +1 and +5 polyamides 557–60 random errors 143
117–18 poly(chloroethene) (PVC) 204 rate constant 454
oxides of Group 2 elements 107 polyesters 513–14, 557 effect of temperature changes 467–8
oxidising agents 90–1, 391 poly(ethanol) 561 rate-determining step 463, 464
iodine–thiosulfate titrations 395–6 poly(ethene) (polythene) 156, 168, rate equations 454–66
oxoanions 85, 117, 390 201, 204 clock reactions 461–2
oxonium ions 48, 428 polymers 156, 204 for first order reactions 456
oxygen, reaction with Group 2 addition polymerisation 168, 201–4 initial-rate method 458–60
elements 106 comparison of addition and condensation investigating 465
ozone layer depletion 218–19 polymers 561 and reaction mechanisms 462–6
condensation polymerisation see for second order reactions 456–7
P condensation polymerisation for zero order reactions 457–8
palladium 313 sustainability and recycling 202–3 rates of reaction 173, 272–84, 450–72
paracetamol 508, 543, 548 poly(propene) 204 collision theory 278–82
partial pressure 307–8 p orbitals 26–7 effect of concentration 275, 279, 454
patterns in behaviour 1 position isomerism 166, 474 effect of temperature 277, 280–1,
p-block elements 29, 30 potassium 101 466–8
peptide bonds 555 see also Group 1 elements influencing factors 275–7, 450–1
peptides 555–6 potassium dichromate solution 225 measurement of 273–4, 451–4
hydrolysis of 556 potassium iodide 99 ratios 631–2
percentage composition 126 potassium manganate (vii) titrations 393– reaction mechanisms 169–71
percentages 631 4 and catalysts 277, 281–2
percentage yield 146, 585–6, 636–7 precipitation reactions 99, 148 electrophilic substitution reactions 527,
perfumes 489 precision 143, 144 529, 530–1
analysis of jasmone 608–9 pressure 306 elimination reactions 216
periodic properties 32–4 effect on equilibrium 292, 311 hydrolysis of halogenoalkanes
ionisation energies 33–4 effect on rate of reaction 279 463–4
melting temperatures 32–3 gas laws 128 investigation of 172–3
periodic table 4, 652 partial 307–8 nucleophilic addition reactions 491–2
and electron structures 29–32 primary standards 136, 393 nucleophilic substitution reactions 213–
periods and groups 29 propane 161, 162 15, 546–7
PET (polyethylene terephthalate) 513 see also alkanes and optical isomerism 479–80
petrol 183 propanoic acid 328–9 rate equations and 462–6
petrol engines, and air pollution 186–7 propanone 464–6, 486–7 reaction orders 455
pH 323–6 proportions 631–2 first order reactions 456
of blood 342 proteins 551–2 second order reactions 456–7
of buffer solutions 341–3 in the human body 551 zero order reactions 457–8
of salts 338 hydrolysis of 556 reaction profiles 280, 467, 638
phases 277 proton NMR 601–6 reaction rates see rates of reaction
phenol 533–6 coupling 604–5 recrystallisation 583
manufacture 536 labile protons 606 recycling polymers 206–7
reactions 534–5 protons 5, 16 red copper oxide 127
phenolphthalein 336 transfer of 320–2 redox reactions 83–96, 149, 387–417
phenylamine 544, 550 purification techniques 583–5 balanced symbol equations 83–4
phenyl compounds 524–5, 574 purity of a product, checking balancing equations 91–4, 389–92
pi (π) bonds 191 586–7 disproportionation reactions 406–7
pK a 330 in electrochemical cells 397–413
pKw 326 Q electron transfer 84–5
PLA (poly(lactic acid)) 514 qualitative analysis 148–50 half-equations 388
planar shapes 432 see also analysis/analytical techniques ionic half-equations 85–6
plaster of Paris 108–9 quantitative analysis 135–9 recognising 90–1
Index 653
469983_Index_Chem_Y1-2_647-[Link] 653 13/04/19 10:17 PM
redox titrations 393–6 entropy changes during formation 255–6, 258–61, 351
iodine–thiosulfate titrations 395–6 dissolving 375–6 neutralisation 256–7, 339
potassium manganate(vii) hydration of ions 361 standard form 627–8
titrations 393–4 solubility in water 68–9 standard hydrogen electrode 400
reducing agents 90–1, 391–2 sodium hydroxide 102 standard molar entropies 371–2
potassium manganate(vii) acid–base titrations 332–4 standard reference compounds 598, 602,
titrations 393–4 reaction with transition metal 603
tests for 392 ions 439–41 standard solutions 136–7, 393
reduction reactions 83, 86, 387 sodium oxide 355 stationary phase of chromatography 610
see also redox reactions solids 2 steam
reference electrodes 400–1 structure of 372 distillation with 581–2
reflux condenser 214 solubility 67–8 reaction with alkenes 195
reforming of alkanes 183, 185 acids, bases and salts 100 stereoisomerism 191–3, 473–4
relative atomic mass 19–20, 123, 595 Group 2 compounds 107–8 see also optical isomerism
relative formula mass 21 Group 7 elements 112–13 stoichiometry 389–90
relative isotopic mass 19, 595 and intermolecular forces 67–8 storage cells 410–11
relative molecular mass 21, 22 ionic compounds 44, 68–9 strong acids 141, 320, 339
retention time 615 simple molecular structures 51 acid–base titrations 332, 334
retinal 485 trends in 363 strong bases 322, 326, 339
retrosynthetic analysis 576–7 solutes 67 acid–base titrations 332–4
reversible changes 285–6 solutions 67–9 structural formulae 161, 572
equilibrium see equilibrium calculations involving 635–6 structural isomerism 165–6, 473–4
revision 641–3 concentration 134, 145 structure 39–51, 639–40
R f values 612 enthalpy change of solution 360–1, 363 relationship to physical properties 77
ROOR mnemonic 399 measurement of enthalpy changes for sub-shells of electrons 26
Rutherford, Ernest 14–15 reactions in solution 251–2, 253–4 substitution reactions 167, 168, 179–81
for quantitative analysis 135–9 alcohols 211, 217, 222
S quantitative dilution 138 halogenoalkanes 168, 212–15
salbutamol 541, 542, 569 standard 136–7, 393 sulfates
salt bridges 397–8 solvent extraction 582 Group 2 compounds 108–9
salts 11 solvents 67 testing for 148
pH of 338 non-aqueous 68 sulfuric acid 116
solubilities 68–9, 100 s orbitals 26–7 sulfur impurities in fuels 186–7
saturated compounds 177 species 124 surface area 276, 280
saturated solutions 67 specific heat capacity 249 synthesis 3
s-block elements 29, 30 spectator ions 86, 98 organic see organic synthesis
scandium 420–2 spectroscopy and spectrometry 587, 593 synthetic pathways 576–80
second order reactions 456–7 infrared see infrared spectroscopy changing the carbon chain 578–9
separation techniques 581–2 mass spectrometry see mass spectrometry changing the functional groups 577,
shapes of molecules and ions 51–6, 173, NMR 597–607 580
639–40 spin–spin coupling 604 synthetic techniques 581–8
complex ions 431–3 spontaneous changes 367–8 carrying out the reaction 581
effect of lone pairs 54–5 diffusion 368–70 identifying the product and checking
effect of multiple bonds 55 sport, detection of banned drugs in 23 purity 586–7
shielding 25, 34 stability constants 437, 438 measuring yield 585–6
sigma bonds, Σ 190–1 stability of compounds 359, 380–2 planning 581
significant figures 629–30 benzene 521 preparation of
silicon dioxide 6, 70 complex ions 436–7 N-phenylethanamide 588
silver halides 115, 148 Group 2 carbonates 382–3 purification 583–5
simple molecular structures 40, 50–1 stalactites and stalagmites 104, 285 separation techniques 581–2
skeletal formulae 161, 572 standard conditions 398, 407–8 systematic errors 143–4
SN1 reactions 464, 479–80 standard electrode potentials 399–401,
SN2 reactions 463, 479–80 405 T
soaps 512 limitations of predictive value 407–8 Taxol 569–70
sodium 101 prediction of direction and extent of a temperature
see also Group 1 elements reaction 409 effect on equilibrium 292, 312–13
sodium chloride 7, 349–50 transition metals 423 effect on rate constant 467–8
arrangement of ions 43 standard enthalpy changes 252–7, 258–63 effect on the rate of reaction 277,
crystal structure 350 atomisation 352–3 280–1, 466–8
enthalpy change of solution 360–1, 363 combustion 255, 258–9 kelvin scale 128
654 Index
469983_Index_Chem_Y1-2_647-[Link] 654 15/04/19 12:36 PM
tetrahedral shapes 53, 55, 432, 639–40 transmittance 237 reaction with amines 544
tetramethylsilane (TMS) 598, 602, 603 trends reaction with Group 1 metals 102
thalidomide 481 of Group 1 metals 101 reaction with Group 2 elements 106–7
theories 4 of Group 2 elements 105 reaction with halogenoalkanes 212–13
thermal decomposition 97 of halogens 111–12 testing for 285–6, 436
thermal stability 109–10 triglycerides 512 wavenumbers 237, 238–9
thermite reaction 374 trigonal bipyramid shape 53, 55, 640 weak acids 141, 320, 327–31
thin-layer chromatography (TLC) 611–13 trigonal planar shape 53, 640 acid–base titrations 333–4, 335
Thomson, J.J. 13–14 triiodomethane reaction 496, 579 effect of dilution on degree of
titrations 135–45 triple bonds 47 dissociation 331
acid–base titrations 140–2, 332–5 influence on shape of molecules and finding the K a value 344
calculations 140, 635–6 ions 55 neutralisation reactions 339–40
evaluating results 142–4 weak bases 327
investigation of reaction rates 453 U acid–base titrations 334–5
redox titrations 393–6 uncertainty, sources of 142–3 neutralisation reactions 339–40
standard solutions 136–7 unsaturated compounds 177, 190 weighted mean 631
TNT (trinitrotoluene) 528 unsymmetrical alkenes 199–200
Tollens’ reagent 494 X
trans-fats 194
V xenon atoms 3
vanadium 424–5 X-ray diffraction 39–40, 44
transition metals 418–49
vanadium(v) oxide catalysts 442
catalysts 441–5
characteristics 422
van der Waals, Johannes 61 Y
vapours 584 yield 146
coloured ions 426–8
variables 637 percentage yield 146, 585–6, 636–7
complex ions see complex ions
volatile liquids 584
definition 420–2 Z
VSEPR see electron-pair repulsion theory
electron configurations 418–20 zeolites 184, 272
V-shaped (bent) molecules and ions 54, 55
ion formation 420 zero order reactions 457–8
ionisation energies 421 W zinc 420–2
metallic properties 422–3 water 6, 9–10 reaction with copper(ii) sulfate
oxidation numbers 423–6 hydrogen bonding 64, 65, 66 solution 85–6
reaction of transition metal ions with ionic product of 325 zwitterions 554–5
aqueous alkalis 439–41 molecular shape 54
transition states 280, 466–7 reaction with acyl chlorides 507
Index 655
469983_Index_Chem_Y1-2_647-[Link] 655 13/04/19 10:17 PM
469983_PT_Chem_Y1-2_656-[Link] 656
The Periodic Table of Elements
656
The Periodic Table of Elements
1 2 3 4 5 6 7 0(8)
(18)
1.0 4.0
H He
hydrogen helium
(1) (2) Key 1
(13) (14) (15) (16) (17) 2
6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2
Li Be atomic symbol B C N O F Ne
lithium beryllium name boron carbon nitrogen oxygen fluorine neon
3 4 atomic (proton) number 5 6 7 8 9 10
23.0 24.3 27.0 28.1 31.0 32.1 35.5 39.9
Na Mg Al Si P S Cl Ar
sodium magnesium aluminium silicon phosphorus sulfur chlorine argon
11 12 (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) 13 14 15 16 17 18
39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
85.5 87.6 88.9 91.2 92.9 95.9 [98] 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
132.9 137.3 138.9 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 [209] [210] [222]
Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
caesium barium lanthanum hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
[223] [226] [227] [261] [262] [266] [264] [277] [268] [271] [272]
Fr Ra Ac** Rf Db Sg Bh Hs Mt Ds Rg Elements with atomic numbers 112–116 have been reported
francium radium actinium rutherfordium dubnium seaborgium bohrium hassium meitnerium damstadtium roentgenium but not fully authenticated
87 88 89 104 105 106 107 108 109 110 111
*Lanthanide series 140 141 144 [147] 150 152 157 159 163 165 167 169 173 175
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium
58 59 60 61 62 63 64 65 66 67 68 69 70 71
232 [231] 238 [237] [242] [243] [247] [245] [251] [254] [253] [256] [254] [257]
**Actinide series
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium
90 91 92 93 94 95 96 97 98 99 100 101 102 103
13/04/19 10:17 PM
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