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Chemistry FOR THE IB DIPLOMA SECOND EDITION Christopher Talbot, Richard Harwood and Christopher Coates 829055_FM_IB_Chemistry_i-x.indd 18/05/15 12:45 pm All proprietary drug names and brand names in Chapters 22–25 are protected by their respective registered trademarks 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 Hachette UK’s policy is to use papers that are natural, renewable and recyclable products and made from wood grown in sustainable forests The logging and manufacturing processes are expected to conform to the environmental regulations of the country of origin Orders: please contact Bookpoint Ltd, 130 Milton Park, Abingdon, Oxon OX14 4SB Telephone: (44) 01235 827720 Fax: (44) 01235 400454 Lines are open from 9.00 - 5.00, Monday to Saturday, with a 24 hour message answering service You can also order through our website www.hoddereducation.com © Christopher Talbot, Richard Harwood and Christopher Coates 2015 First edition published in 2010 This second edition published 2015 by Hodder Education An Hachette UK Company Carmelite House, 50 Victoria Embankment, London EC4Y 0DZ Impression number Year 2019 2018 2017 2016 2015 All rights reserved Apart from any use permitted under UK copyright law, no part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or held within any information storage and retrieval system, without permission in writing from the publisher or under licence from the Copyright Licensing Agency Limited Further details of such licences (for reprographic reproduction) may be obtained from the Copyright Licensing Agency Limited, Saffron House, 6–10 Kirby Street, London EC1N 8TS Cover photo © ESA/Herschel/PACS/MESS Key Programme Supernova Remnant Team; NASA, ESA and Allison Loll/ Jeff Hester (Arizona State University) Illustrations by Ken Vail Graphic Design and Aptara Inc Typeset in Goudy Oldstyle 10/12 pt by Aptara inc Printed in Slovenia A catalogue record for this title is available from the British Library ISBN: 978 1471 829055 829055_FM_IB_Chemistry_i-x.indd 18/05/15 12:45 pm Contents Introduction vii Acknowledgements ix Core 829055_FM_IB_Chemistry_i-x.indd Chapter Stoichiometric relationships 1.1 Introduction to the particulate nature of matter and chemical change 1.2 The mole concept 20 1.3 Reacting masses and volumes 31 Chapter Atomic structure 52 2.1 The nuclear atom 52 2.2 Electron configuration 66 Chapter Periodicity 85 3.1 Periodic table 85 3.2 Periodic trends 96 Chapter Chemical bonding and structure 114 4.1 Ionic bonding and structure 114 4.2 Covalent bonding 125 4.3 Covalent structures 129 4.4 Intermolecular forces 144 4.5 Metallic bonding 158 Chapter Energetics/thermochemistry 165 5.1 Measuring energy changes 165 5.2 Hess’s Law 178 5.3 Bond enthalpies 187 Chapter Chemical kinetics 199 6.1 Collision theory and rates of reaction 199 Chapter Equilibrium 223 7.1 Equilibrium 223 Chapter Acids and bases 250 8.1 Theories of acids and bases 250 8.2 Properties of acids and bases 256 8.3 The pH scale 261 8.4 Strong and weak acids and bases 265 8.5 Acid deposition 274 18/05/15 12:45 pm iv Contents Chapter Redox processes 283 9.1 Oxidation and reduction 9.2 Electrochemical cells 283 311 Chapter 10 Organic chemistry 322 10.1 Fundamentals of organic chemistry 10.2 Functional group chemistry 322 350 Chapter 11 Measurement and data processing 375 11.1 Uncertainties and errors in measurement and results 11.2 Graphical techniques 11.3 Spectroscopic identification of organic compounds 375 395 408 Additional higher level (AHL) Chapter 12 Atomic structure 829055_FM_IB_Chemistry_i-x.indd 435 12.1 Electrons in atoms 435 Chapter 13 The periodic table – the transition metals 451 13.1 First-row d-block elements 13.2 Coloured complexes 451 471 Chapter 14 Chemical bonding and structure 489 14.1 Further aspects of covalent bonding and structure 14.2 Hybridization 489 497 Chapter 15 Energetics/thermochemistry 522 15.1 Energy cycles 15.2 Entropy and spontaneity 522 535 Chapter 16 Chemical kinetics 552 16.1 Rate expression and reaction mechanism 16.2 Activation energy 552 575 Chapter 17 Equilibrium 585 17.1 The equilibrium law 588 Chapter 18 Acids and bases 606 18.1 Lewis acids and bases 18.2 Calculations involving acids and bases 18.3 pH curves 606 612 625 Chapter 19 Redox processes 643 19.1 Electrochemical cells 643 Chapter 20 Organic chemistry 671 20.1 Types of organic reactions 20.2 Synthetic routes 20.3 Stereoisomerism 673 692 699 Chapter 21 Measurement and analysis 719 21.1 Spectroscopic identification of organic compounds 719 18/05/15 12:45 pm Contents v Options Available on the website accompanying this book: www.hoddereducation.com/IBextras Option A Option B Option C Option D 829055_FM_IB_Chemistry_i-x.indd Chapter 22 Materials 22.1 Materials science introduction 22.2 Metals and inductively coupled plasma (ICP) spectroscopy 22.3 Catalysts 22.4 Liquid crystals 22.5 Polymers 22.6 Nanotechnology 22.7 Environmental impact – plastics 22.8 Superconducting metals and X-ray crystallography (AHL) 22.9 Condensation polymers (AHL) 22.10 Environmental impact – heavy metals (AHL) Chapter 23 Biochemistry 23.1 Introduction to biochemistry 23.2 Proteins and enzymes 23.3 Lipids 23.4 Carbohydrates 23.5 Vitamins 23.6 Biochemistry and the environment 23.7 Proteins and enzymes (AHL) 23.8 Nucleic acids (AHL) 23.9 Biological pigments (AHL) 23.10 Stereochemistry in biomolecules (AHL) Chapter 24 Energy 24.1 Energy sources 24.2 Fossil fuels 24.3 Nuclear fusion and fission 24.4 Solar energy 24.5 Environmental impact – global warming 24.6 Electrochemistry, rechargeable batteries and fuel cells (AHL) 24.7 Nuclear fusion and nuclear fission (AHL) 24.8 Photovoltaic and dye-sensitized solar cells (AHL) Chapter 25 Medicinal chemistry 25.1 Pharmaceutical products and drug action 25.2 Aspirin and penicillin 25.3 Opiates 25.4 pH regulation of the stomach 25.5 Anti-viral medications 25.6 Environmental impact of some medications 25.7 Taxol – a chiral auxiliary case study (AHL) 18/05/15 12:45 pm vi Contents 25.8 Nuclear medicine (AHL) 25.9 Drug detection and analysis (AHL) Index 737 Answers and glossary Answers and glossary appear on the website accompanying this book: www.hoddereducation.com/IBextras 829055_FM_IB_Chemistry_i-x.indd 18/05/15 12:45 pm Introduction Nature of Science 829055_FM_IB_Chemistry_i-x.indd Welcome to the second edition of Chemistry for the IB Diploma The content and structure of this second edition has been completely revised to meet the demands of the 2014 IB Diploma Programme Chemistry Guide Within the IB Diploma Programme, the chemistry content is organized into compulsory core topics plus a number of options, from which all students select one The organization of this resource exactly follows the IB Chemistry Guide sequence: ■ Core: Chapters 1–11 cover the common core topics for Standard and Higher Level students ■ Additional Higher Level (AHL): Chapters 12–21 cover the additional topics for Higher Level students ■ Options: Chapters 22–25 cover Options A, B, C and D respectively Each of these is available to both Standard and Higher Level students (Higher Level students study more topics within the same option.) These are available on the Hodder website The syllabus is presented as topics, each of which (for the core and AHL topics) is the subject of a corresponding single chapter in the Chemistry for the IB Diploma printed book The Options (Chapters 22–25) are available on the website accompanying this book, as are a comprehensive Glossary and the answers to the end-of-chapter exam and exam-style questions: www.hoddereducation.com/IBextras Special features of the chapters of Chemistry for the IB Diploma are: ■ Each chapter begins with Essential Ideas that summarize the concepts on which it is based ■ The text is written in straightforward language, without phrases or idioms that might confuse students for whom English is a second language The text is also suitable for students of all abilities ■ The depth of treatment of topics has been carefully planned to accurately reflect the objectives of the IB syllabus and the requirements of the examinations ■ Photographs and full-colour illustrations support the relevant text, with annotations which elaborate on the context, function, language, history or applications of chemistry ■ The Nature of Science is an important new aspect of the IB Chemistry course, which aims to broaden students’ interests and knowledge beyond the confines of its specific chemistry content Throughout this book we hope that students will develop an appreciation of the processes and applications of chemistry and technology Some aspects of the Nature of Science may be examined in IB Chemistry examinations and important discussion points are highlighted in the margins ■ The Utilizations and Additional Perspectives sections also reflect the Nature of Science, but they are designed to take students beyond the limits of the IB syllabus in a variety of ways They may, for example, provide a historical context, extend theory or offer an interesting application They are sometimes accompanied by more challenging, or research style, questions They not contain any knowledge which is essential for the IB examinations ■ Science and technology have developed over the centuries with contributions from scientists from all around the world In the modern world science knows few boundaries and the flow of information is usually quick and easy Some international applications of science have been indicated with the International Mindedness icon ■ Worked examples are provided in each chapter whenever new equations are introduced A large number of self-assessment questions and some research questions are also placed throughout the chapters close to the relevant theory They are phrased in order to assist comprehension and recall, and to help familiarize students with the assessment implications of the command terms ■ It is not an aim of this book to provide detailed information about experimental work or the use of computers However, our Applications and Skills icon has been placed in the margin to indicate wherever such work may usefully aid understanding ■ A selection of IB examination-style questions are provided at the end of each chapter, as well as some past IB Chemistry examination questions Answers to these are provided on the website accompanying this book 19/05/15 9:12 am viii Introduction ■ Extensive links to the interdisciplinary Theory of Knowledge (ToK) element of the IB Diploma course, including ethics, are made in all chapters ■ Comprehensive glossaries of words and terms, including IB command terms, for Core and AHL topics are included in the website which accompanies this book ■ This icon denotes links to material available on the website that accompanies this book: www.hoddereducation.com/IBextras ■■ Using this book The sequence of chapters in Chemistry for the IB Diploma deliberately follows the sequence of the syllabus content However, the IB Diploma Chemistry Guide is not designed as a teaching syllabus, so the order in which the syllabus content is presented is not necessarily the order in which it will be taught Different schools and colleges should design course delivery based on their individual circumstances In addition to the study of the chemistry principles contained in this book, IB science students carry out experiments and investigations, as well as collaborating in a Group Project These are assessed within the school (Internal Assessment) based on well-established criteria ■■ Author profiles Christopher Talbot Chris teaches IB Chemistry and ToK at a leading IB World School in Singapore He has also taught IB Biology, MYP Science and a variety of IGCSE Science courses He has moderated IB Chemistry coursework and prepared students for the Singapore Chemistry Olympiad Richard Harwood Richard was a Biochemistry researcher at Manchester Medical School and University College, Cardiff, before returning to teaching science in England and Switzerland Most recently he has been involved in projects with various Ministries of Education evaluating science courses and providing teacher training nationally, and in individual schools, in Mongolia, Kazakhstan, Zimbabwe, India and Ghana Christopher Coates Chris has previously taught in Suffolk, Yorkshire and Hong Kong at King George V School, and is currently Head of Science in the Senior School at the Tanglin Trust School, Singapore He has taught A-level and IB Chemistry as well as ToK and MYP Science ■■ Authors’ acknowledgements We are indebted to the following lecturers who reviewed early drafts of the chapters for the second edition: Dr David L Cooper, University of Liverpool (Chapters and 14), Professor Mike Williamson, University of Sheffield (Chapters 21 and 23), Professor James Hanson, University of Sussex (Chapter 20), Professor Laurence Harwood, University of Reading (Chapter 20), Professor Robin Walsh, University of Reading (Chapters and 16), Professor Howard Maskill, University of Newcastle (Chapter 20), Dr Norman Billingham, University of Sussex (Chapter 22), Dr Jon Nield, Queen Mary College, (Chapter 23), Professor Jon Cooper, University College London (Chapter 23), Dr Duncan Bruce, University of York (Chapter 22), Professor David Mankoff, University of Pennsylvania (Chapter 25), Dr Philip Walker, University of Surrey and Dr Eli Zysman-Colman (University of St Andrews (Chapter 22), and Dr Graham Patrick (Chapter 25), University of the West of Scotland I also acknowledge the contributions of Dr David Fairley (Overseas Family School, Singapore) who gave me invaluable advice and guidance on the many chemical issues I encountered when writing the book A special word of thanks must go to Mr Nick Lee, experienced chemistry and TOK teacher, workshop leader and IB examiner, for his most helpful comments on the final drafts Finally, we are indebted to the Hodder Education team that produced this book, led by Eleanor Miles and So-Shan Au at Hodder Education Chris Talbot Singapore, June 2015 829055_FM_IB_Chemistry_i-x.indd 18/05/15 12:45 pm Acknowledgements The Publishers would like to thank the following for permission to reproduce copyright material: ■ Photo credits All photos by kind permission of Cesar Reyes except: p.1 t Chris Talbot; p.6 Science photo library/Michael W Davidson; p.7 t Chris Talbot, b NASA/Johnson Space Center; p.10 t, b Andrew Lambert Photography/Science Photo Library; p.22 Chris Talbot; p.26 Reproduced with permission of the BIPM, which retains full internationally protected copyright (photograph courtesy of the BIPM); p.56 IBM Research; p.63 Tim Beddow/Science Photo Library; p.67 l Andrew Lambert Photography/Science Photo Library, c David Talbot, r Robert Balcer; p.68 Carlos Santa Maria – Fotolia; p.73 CERN; p.87 tl Andrew Lambert Photography/Science Photo Library; p.103 Prof Mark J Winter/http://www webelments.com; p.108 tr Andrew Lambert Photography/Science Photo Library; p.111 b JoLin/ istockphoto.com; p.122 Robert Balcer; p.124 Se7enimage – Fotolia; p.129 Chris Talbot; p.141 t Chris Talbot; p.143 Harry Kroto and used with the permission of The Sussex Fullerene Research Centre and photographer Nicholas Sinclair; p.144 Public Domain/Http://Commons.Wikimedia Org/Wiki/File:Graphene-3D-Balls.Png; p.162 Dirk Wiersma/Science Photo Library; p.167 t, b David Talbot; p.192 NASA/Goddard Space Flight Center; p.199 l Roger Harris/Science Photo Library, r Noaa/Science Photo Library; p.204 J C Revy /Science Photo Library; p.205 t Dr Colin Baker; p.223 t Anh Ngo – Fotolia, b Gigi200043 – Fotolia; p.224 t, b Richard Harwood; p.226 Richard Harwood; p.233 Richard Harwood; p.235 Science Photo Library; p.237 Richard Harwood; p.239 t, b Richard Harwood; p.245 Bettmann/CORBIS; p.255 Juan Gartner/Science Photo Library; p.259 l, r Richard Harwood; p.274 Leungchopan – Fotolia; p.283 b David Talbot; p.285 Phil Degginger/Alamy; p.295 b sequence Chris Talbot; p.304 t, b Chris Talbot; p.305 t Chris Talbot; p.306 Dr Colin Baker; p.307 Andrew Lambert Photography/Science Photo Library; p.309 Martyn F Chillmaid/Science Photo Library; p.317 Frank Scullion/http:// www.franklychemistry.co.uk/electrolysis_lead_bromide_video.html; p.322 Klaus Boller/Science Photo Library; p.323 t, b Chris Talbot; p.324 Richard Harwood; p.328 t Mandritoiu – Fotolia, b David Talbot; p.331 Richard Harwood; p.332 Richard Harwood; p.336 Rasmol Library/ Richard Harwood; p.339 Richard Harwood; p.342 Chris Talbot; p.346 t Geraint Lewis/Rex, b Richard Harwood; p.348 t Richard Harwood, b IBM Research; p.351 t Chris Talbot, c Full Image – Fotolia, bl Science Photo LibraryDavid Taylor/Cordelia Molly, br David Taylor/Science Photo Library; p.352 t Science Photo Library/Paul Rapson, b CSIRO/Science Photo Library; p.353 l Eye Ubiquitous/Alamy, r Robert Brook/Science Photo Library; p.354 Paul Rapson/ Science Photo Library; p.355 Chris Talbot; p.356 David Talbot; p.358 Chris Talbot; p.359 Chris Talbot; p.360 Andrew Lambert/Science Photo Library; p.361 David Talbot; p.365 l Roger Job/Science Photo Library, r Vanessa Vick/Science Photo Library; p.366 Andrew Lambert/ Science Photo Library; p.368 t Chris Talbot, b Andrew Lambert/Science Photo Library; p.370 Chris Talbot; p.375 Ted Kinsman/Science Photo Library; p.381 SciLabware; p.395 JPL/NASA; p.405 Chris Talbot; p.408 Chris Talbot; p.410 Dr Jon Hare; p.423 Dr Jon Hare; p.427 Mikhail Basov – Fotolia; p.430 James Steidl/Fotolia.Com; p.431 t Zephyr/Science Photo Library, b Dr Jon Hare; p.441 CNRI/Science Photo Library; p.458 Roger-Viollet/Topfoto; p.460 Mark A Wilson (Department Of Geology, The College Of Wooster)/Public Domain (http://Commons Wikimedia.Org/Wiki/File:Qtubironpillar.JPG); p.469 Chris Talbot; p.474 Chris Talbot; p.480 t Bruce Balick (University of Washington), Vincent Icke (Leiden University, The Netherlands), Garrelt Mellema (Stockholm University), and NASA/ESA, c Jose Ignacio Soto – Fotolia; p.481 Andrew Lambert Photography/Science Photo Library; p.516 Charles D Winters/Science Photo Library; p.526 Richard Harwood; p.537 t David Talbot; p.549 Public Domain/Http://Schneider Ncifcrf.Gov/Images/Boltzmann/Boltzmann-Tomb-3.Html; p.585 t TUDGAY, Frederick, 1841–1921, The “Dunedin” off the English Coast, 1875, oil on canvas: 487 x 790 mm, accession: 02/01, Hocken Collections, Uare Taoka o Hakena, University of Otago, b Everett Collection/Rex; p.586 Chris Talbot; p.587 Treetstreet – Fotolia; p.597 Claude Nuridsany and Marie Perennou/ Science Photo Library; p.603 Sovereign, ISM/Science Photo Library; p.628 Richard Harwood; 829055_FM_IB_Chemistry_i-x.indd 19/05/15 9:13 am 184 Energetics/thermochemistry Additional Perspectives Enthalpy changes during changes of state It is essential to specify the physical states of the substances involved when writing thermochemical equations to represent an enthalpy change This is because any change in physical state (Chapter 1) has its own enthalpy change (Figure 5.25) ■■ Figure 5.25 Changes of state and energy changes MELTING (energy absorbed) VAPORIZING (energy absorbed) FREEZING (energy released) CONDENSING (energy released) The enthalpy of fusion of ammonia would be the enthalpy change for the reaction: NH3(s) → NH3(l) and the enthalpy of vaporization of ammonia would be the enthalpy change for the reaction: NH3(l) → NH3(g) Enthalpies of vaporization and sublimation are always endothermic since intermolecular forces of attraction need to be overcome The enthalpy of sublimation for iodine would be the enthalpy change for the reaction: I2(s) → I2(g) The values of these enthalpy changes will vary with the strength of intermolecular forces (Chapter 4) Additional Perspectives Dissolving of ionic compounds When an ionic compound dissolves in water, the process may be exothermic or endothermic, depending on the substance concerned The value for the enthalpy change, ΔH, for such a reaction is the sum of two factors: energy to overcome the energy released when the ΔH = + electrostatic forces of attraction − ions attract water molecules between ions in the lattice around themselves The sign of the enthalpy change, ΔH, for the reaction depends on which of these two quantities is larger The dissolving of ionic solids is discussed in Chapters and 15 Applying Hess’s Law to calculate enthalpy changes Use the following thermochemical data to calculate the enthalpy of formation of sulfur trioxide The enthalpy of formation of sulfur dioxide is −297.4 kJ mol−1 and the enthalpy of combustion of sulfur dioxide is −97.9 kJ mol−1 Thermochemical equations can be written to describe these two enthalpy changes: S(s) + O2(g) → SO2(g) ΔH = −297.4 kJ mol−1 SO2(g) + O2(g) → SO3(g) ΔH = −97.7 kJ mol−1 Summing these two equations and cancelling sulfur dioxide gives an equation for the enthalpy of formation of sulfur trioxide: S(s) + O2(g) → SO2(g) 829055_05_IB_Chemistry_165-198.indd 184 ΔH = −297.4 + (−97.7) = −95.3 kJ mol−1 18/05/15 9:45 am 5.2 Hess’s law 185 Calculating the enthalpy change of reactions using ΔH f data The enthalpies of formation of carbon monoxide, carbon dioxide, dinitrogen monoxide and dinitrogen tetroxide are −110, −393, 81 and 9.7 kJ mol−1 Determine the value of the enthalpy of reaction for the following reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g) ΔH = ΣΔH f [products] − ΣΔH f [reactants] ΔH = [3ΔH f (CO2(g) + ΔH f (N2O(g))] − [3ΔH f (CO(g)) + ΔH f (N2O4(g))] ΔH = [(3 × −393) + 81] − [(3 × −110) + (−9.78)] = 1437.7 kJ mol−1 Determining the enthalpy change of a reaction that is the sum of multiple reactions with known enthalpy changes Calculate the enthalpy of combustion of glucose from the following thermochemical data: Equation 1: C(s) + O2(g) → O2(g) Equation 2: H2(g) + Equation 3: O (g) 2 → H2O(l) 6C(s) + 6H2(g) + 3O2(g) → C6H12O6(l) ΔH = −395.0 kJ ΔH = −269.4 kJ ΔH = −269.4 kJ The enthalpy of combustion for glucose is described by the following equation: C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) It can be obtained by multiplying equations and by and summing them: 6C(s) + 6H2(g) + 9O2(g) → 6CO2(g) + 6H2O(l) ΔH = −3986.4 kJ Subtracting equation from equation gives the enthalpy of combustion for glucose: C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔH = −2816.6 kJ ■■ Utilization: Relevance of Hess’s law in wider areas Hess’s law is relevant when discussing the energy content of foods If a bomb calorimeter is used to measure the amount of energy present in a food sample by converting it to carbon dioxide, water and nitrogen, then you obtain a certain amount of thermal energy in joules However, the body will not break down food to this extent and so this overestimates the amount of energy available to the body in food Normally, any nitrogen present in food ends up in urea, which is excreted in the urine Therefore, a more accurate measure of how much energy is present in food would be to measure how much energy would be released if all the nitrogen present ended up in urea, rather than the total nitrogen This cannot be measured in the laboratory, but if a bomb calorimeter was used to measure the amount of energy present in urea, then that could be compared with the total amount of energy present in food to get the amount of available energy This illustrates how the principles of Hess’s law can be used to determine more accurate calorific values of food Hess’s law has significance in carrying out processes in the production plant For example, a specific chemical reaction in an organic synthesis may be exothermic and release a large amount of thermal energy which would be considered too risky since the resulting temperature rise might result in decomposition of the chemicals and the risk of a runaway reaction In the research laboratory, this is less likely to happen because cooling a small-scale reaction is more efficient than a larger scale reaction The solution to this issue is to identify the different thermodynamic stages in the reaction, such as the heat of solvation of the different reagents and the heat emitted by the reaction itself Dissolving the reagents, then allowing these solutions to cool before they are mixed, means that the heat emitted during the reaction itself is less and the reaction is safer 829055_05_IB_Chemistry_165-198.indd 185 18/05/15 9:45 am 186 Energetics/thermochemistry Another aspect of Hess’s law would be in the measurement of the binding affinities of drugs to a protein target Stability is obtained due to the binding interactions formed between the drug and the binding site (hydrogen bonds, London (dispersion) forces, ionic interactions and so on), but there are also ‘energy penalties’ such as the energy required to dehydrate both the drug and the target binding site Dehydrate means to remove the surrounding water molecules (the hydration sphere) and is an example of desolvation – removal of the solvent molecules The measured binding affinity of drugs to their protein receptors is the sum of both these opposing effects To get a better idea of the strength of the binding interactions, researchers could dissolve each drug in water and measure its enthalpy of hydration, then use the negative value of those figures to get a measure of the energy penalty involved in dehydration However, this would really only be valid for rigid drugs since the effects of any entropy changes when the drug binds are not being taken into account Measuring the dehydration energy penalty of the protein is not so easy, but if the study compares different drugs, then it would be reasonable to assume that is a constant and it should be possible to get results that would show which drugs are binding most effectively without the complication of different desolvation penalties ToK Link Hess’s Law is an example of the application of the conservation of energy What are the challenges and limitations of applying general principles to specific instances? Hess’s law is always valid since the conservation of energy is a universal truth for all practical purposes It yields an enthalpy value, ΔH , but this is only related to heat if the reaction is at constant pressure and no expansion work (Figure 5.26) is done Work is the amount of energy transferred by a force: expansion work (w) = − pΔV, where p is the external pressure and ΔV is the change in volume piston velocity W ideal gas P cylinder connecting rod ■■ Figure 5.26 Expansion work An example from kinetics is simple collision theory developed from the kinetic theory of gases to account for the influence of concentration and temperature on reaction rates The theory is based on several postulates: (i) the reactant particles behave as hard spheres, (ii) there is no interaction between the reactant particles until they collide, and (iii) only collisions with a combined minimum kinetic energy greater than the activation energy, Ea, will lead to reaction The collision theory is usually able to predict, satisfactorily, the rate of reaction involving simple molecules However, difficulties arise with reactions that involve complicated molecules The observed rates tend to be lower than what the collision theory predicts, sometimes by a factor of 105 or more Therefore, another factor, called the steric factor, has to be introduced; this may be interpreted as a preference for a particular orientation (e.g direction, angle of approach) of the reacting molecules Even in its more complex form, collision theory does not accurately predict the reaction between metal atoms such as sodium or potassium and halogen molecules such as bromine, for example K(g) + Br2(g) → KBr(g) + Br(g) As the alkali metal atom approaches the bromine molecule its valence electron moves to the bromine molecule (thus providing a ‘harpoon’) There are then two ions with an electrostatic attraction between them As a result the ions move together and the reaction takes place This mechanism, which has been worked out quantitatively, explains why the reaction occurs far more readily than might be expected taking into account only mechanical collisions between the alkali metal atoms and halogen molecules ■■ Recycling The resources on the Earth are limited and it is important that the actions we take now help future generations To achieve sustainable development, there needs to be a balance between the need for economic development, where standards of living improve, and respect for the environment 829055_05_IB_Chemistry_165-198.indd 186 18/05/15 9:45 am 5.3 Bond enthalpies 187 and the resources it provides for us Recycling is an important way to help achieve sustainable development and many resources including glass, paper and especially metals can be recycled Glass is easily recycled It can be melted and re-moulded to make new objects, such as bottles The energy needed to this is less than the energy needed to make new glass from its raw materials Paper is broken up into small pieces and re-formed to make new sheets of paper This takes less energy to than making paper from trees However, paper can only be recycled a few times before its fibres become too short to be useful, and the recycled paper is often only good enough for toilet paper or cardboard But it can be used as a fuel or compost instead Metals are very important non-renewable resources, and the supplies of metals will eventually run out unless extensive recycling is practised Table 5.2 gives some approximate recent figures about selected metals ■■ Table 5.2 Recent data about selected metals Metal Iron Aluminium Copper Zinc Lead Tin Both nuclei are attracted to the same pair of electrons hydrogen molecule + H – – These are approximate figures from the past but they clearly demonstrate the need to recycle metals – that is, to melt down used metals – rather than throw them away There is also a need to reduce the environmental impact of mining, for example contamination of soil and water, as well as acid rain production and global warming A metal company will recycle metal only if it is economical The company has to work out the costs of collecting the scrap, transporting it, melting it down, getting rid of impurities and paying the workers’ wages In future, as metals get scarcer and more expensive, recycling will become a more important process but it varies in its efficiency in energy terms in different countries broken and is released when bonds are formed shared electrons energy – + H – pull atoms apart against the force of electrostatic attraction ■■ Figure 5.27 To break a covalent bond the attractive forces between the shared pair or pairs of electrons and the nuclei of the two atoms need to be overcome 829055_05_IB_Chemistry_165-198.indd 187 Approximate number of years before the metal is used up (depleted) 110 350 40 60 20 15 5.3 Bond enthalpies – energy is absorbed when bonds are + H + H Mass of metal used up each year (10 tonnes) 100 15 5 0.50 ■■ Bond enthalpies The bond enthalpy (bond energy) is the amount of energy required to break one mole of a specific covalent bond between two atoms in one mole of gaseous molecules Measurement of bond enthalpies can be performed using a mass spectrometer (Chapter 2) The concept of bond enthalpy is illustrated in Figure 5.27 using the hydrogen molecule For the hydrogen molecule, the thermochemical equation describing the bond dissociation enthalpy is: H2(g) → 2H(g) ΔH = + 436 kJ mol−1 The bond enthalpy for hydrogen is 436 kJ mol−1 Because energy is required to overcome or break the attractive forces between the shared pair of electrons and the nuclei, the bond breaking process is endothermic (that is, heat energy is absorbed from the surroundings) It should be noted that if the H−H bond had been formed, then 436 kilojoules of heat energy would have been released to the surroundings This is a simple application of Hess’s law Bond breaking is always an endothermic process; bond formation is always an exothermic process 18/05/15 9:45 am 188 Energetics/thermochemistry The strength of a covalent bond is indicated by the size of the bond enthalpy The larger the bond enthalpy, the stronger the covalent bond Also, bond enthalpy is inversely proportional to bond length Average bond enthalpies Many bond enthalpies are average bond enthalpies For example, the C–H bond enthalpy is based upon the average bond energies in methane, alkanes and other hydrocarbons Because bond enthalpies are often average values it means that enthalpy changes calculated using bond enthalpies will not be exactly equal to an accurate experimentally determined value A selection of bond enthalpies and bond lengths is given in Table 5.3 A more extensive table of bond enthalpies is given on page 11 of the IB Chemistry data booklet ■■ Table 5.3 A selection of average bond enthalpies (at 298 K) Bond H ––H C––C C==C C≡≡C N ––N N==N O ––O O==O F ––F Cl ––Cl Br––Br I ––I C––H O ––H C≡≡N H ––F N ––H O ––H C==O ■ ΔH /kJ mol−1 436 346 614 839 158 470 144 498 159 242 193 151 414 463 890 567 391 463 804 Bond length/Pm 74 154 134 120 146 125 148 121 142 199 228 267 108 97 116 92 101 97 122 Factors affecting average bond enthalpies ■ Effect of bond length The larger the atoms joined by a particular bond, the longer the bond length Large atoms have more electrons than smaller atoms and this results in an increase in repulsion between the electron shells of each atom In addition, the nucleus of each atom is more effectively shielded (Chapter 12) Both of these effects lead to a weakening of the bond For example, in the halogens the bond strength weakens in the order: chlorine, bromine and iodine Fluorine, however, has a surprisingly low bond enthalpy, which is accounted for by lone pair–lone pair repulsion ■ Effect of number of bonding electrons The more electrons present in a bond, the greater the strength of the bond This is because an increasing number of electrons leads to an increase in electrostatic forces of attraction Hence triple bonds are expected to be stronger than double bonds, which should be stronger than single bonds (for the same element) This can clearly be observed with carbon (Table 5.3) Effect of bond polarity Bonds become more polar as the difference in electronegativity (Chapter 3) between the two bonded atoms increases (Chapter 4) This increases the ionic character of the bond and increases the strength of polar covalent bonds This can be observed with N–H, O –H and F–H where bond strength increases with increasing polarity ■■ Using bond dissociation enthalpies to calculate enthalpy changes of reaction Bond enthalpies can be used to determine the enthalpy change for a particular reaction involving molecules in the gaseous state, for example the combustion of methane: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) The reaction can be regarded as occurring in two steps: first, all of the bonds in the reactants have to be broken to form atoms This is an endothermic process and heat energy has to be absorbed from the surroundings In the second step, bond formation occurs This is an exothermic process and releases heat energy to the surroundings The overall reaction is exothermic since the energy released during bond formation is greater than the energy absorbed during bond breaking (Figure 5.28) 829055_05_IB_Chemistry_165-198.indd 188 18/05/15 9:45 am 5.3 Bond enthalpies 189 C Enthalpy ■■ Figure 5.28 Breaking and forming of bonds during the combustion of methane break bonds – takes in energy H O H O H O H O make new bonds – gives out energy H C H H O O O O H O O C O H O H H H Progress of reaction Bond enthalpy data can be used to calculate the enthalpy change for this reaction Bond breaking: Breaking C––H bonds in a methane molecule = × 412 = 1648 kJ Breaking O==O bonds in two oxygen molecules = × 496 = 992 kJ Total amount of energy required to break all these bonds = (1648 + 992) = 2640 kJ Bond making: Making C–– O bonds in a carbon dioxide molecule = × 743 = 1486 kJ Making O==H bonds in two water molecules = × 463 = 1852 kJ Total amount of energy released to surroundings when these bonds are formed = (1486 + 1852) = 3338 kJ The enthalpy change for this reaction is: ΔH = ∑ energy required to break bonds −∑ energy released when bonds are formed = (2640 − 3338) = −698 kJ mol−1 Since more energy is released when the new bonds in the products are formed than is needed to break the bonds in the reactants to begin with, there is an overall release of energy in the form of heat The reaction is exothermic In a reaction which is endothermic, the energy absorbed by bond breaking is greater than the energy released by bond formation The bond breaking and making processes can be represented by a Hess’s law cycle (see Figure 5.29) This calculated value is slightly different from the real value because the bond enthalpies are averages In addition, the water that is produced (in the calculation) is not in its standard state, as a liquid The gaseous state is always used when performing bond enthalpy calculations ■■ Figure 5.29 Using a Hess’s law cycle to represent the bond breaking and bond making processes for the complete combustion of methane C(s) + 4H(g) + 4O(g) bond breaking × (C–H) × (O=O) (endothermic) CH4(g) + 2O2(g) ΔH ΔH = bond breaking + bond making bond making × (C=O) × (O–H) (exothermic) CO2(g) + 2H2O(g) Average bond enthalpies can be used to calculate the enthalpy change for any reaction involving molecules in the gaseous state This is done by assuming that an alternative route for all reactions can be achieved theoretically via the gaseous atoms involved in the compounds (Figure 5.30) 829055_05_IB_Chemistry_165-198.indd 189 18/05/15 9:45 am 190 Energetics/thermochemistry ■■ Figure 5.30 Generalized energy cycle to determine an enthalpy change from bond energies ∆H reactants products ∆H2 ∆H1 atoms in the gaseous state ΔH1 = sum of the average bond enthalpies of the reactants ΔH2 = sum of the average bond enthalpies of the products Applying Hess’s law gives: ΔH = ΔH1 − ΔH2 This leads to the expression: ΔH = Σ(average bond enthalpies of the reactants) − Σ(average bond enthalpies of the products) i.e ΔH = Σ(bonds broken) − Σ(bonds made) Bond enthalpy data can also be used to determine an unknown bond enthalpy provided that the enthalpy change and all the other bond enthalpies are known Worked example The bond enthalpies for H2(g) and HCl(g) are 435 kJ mol−l and 431 kJ mol−1, respectively, for the reaction H2(g) + Cl2(g) → 2HCl(g) and the enthalpy change of reaction is −184 kJ mol−1 Calculate the bond enthalpy of chlorine Enthalpy change = Σ (bonds broken) − Σ (bonds made) −184 = (435 + Cl−Cl) − (2 × 431) −184 = (435 + Cl−Cl) − 862 Cl−Cl = 243 kJ mol−1 ■■ Potential energy profiles Enthalpy changes can be shown using potential energy profiles (Figures 5.31 and 5.32) products Enthalpy CaO(s) + CO2(g) reactants Enthalpy CH4(g) + 2O2(g) –∆H +∆H CO2(g) + 2H2O(l) CaCO3(s) products reactant Progress of reaction ■■ Figure 5.31 Enthalpy level diagram for an exothermic reaction – the combustion of methane Progress of reaction ■■ Figure 5.32 Enthalpy level diagram for an endothermic reaction – the thermal decomposition of calcium carbonate Thermochemical equations are equations which show the associated enthalpy changes 829055_05_IB_Chemistry_165-198.indd 190 CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = −890 kJ mol−1 CaCO3(s) → CaO(s) + CO2(g) ΔH = +180 kJ mol−1 18/05/15 9:45 am 5.3 Bond enthalpies 191 For an exothermic reaction the enthalpy change, ΔH, is described as negative because potential energy has been lost from the reaction mixture to the surroundings, in the form of heat The products are at a lower energy or enthalpy level than the reactants Endothermic reactions absorb heat energy from the surroundings The products of an endothermic reaction contain more potential energy or enthalpy than the reactants For this type of reaction the enthalpy change is positive, because the chemical reactants gain heat from their surroundings Additional Perspectives Activation energy All chemical reactions, both endothermic and exothermic, have an activation energy barrier (Chapter 6) The activation energy of a reaction is usually denoted by Ea, and is given in units of kilojoules per mole The activation energy barrier (Figure 5.33) controls the rate of the reaction: the smaller the value of the activation energy, the greater the rate of the reaction ■ Figure 5.33 Matches and match box The match head consists of potassium chlorate(v), sulfur and phosphorus trisulfide; the frictional heat when the match is struck against the side of the box is sufficient to give the reactants enough combined kinetic energy to overcome the activation energy barrier Sketching potential energy profiles Sketch a potential energy profile for the following thermochemical equation: H2(g) + O2(g) → H2O(l) ΔH = −242 kJ mol H2(g) + O2(g) reactants ∆H = –242 kJ mol–1 H2O(g) products Progress of reaction ■ Figure 5.34 Potential energy profile for H2(g) + O2(g) → H2O(l); ΔH = −242 kJ mol −1 829055_05_IB_Chemistry_165-198.indd 191 Enthalpy or potential energy Enthalpy or potential energy State and explain if the reaction is exothermic or endothermic, and whether the reactants or the products are more stable The negative sign in front of the enthalpy change indicates that thermal energy (heat) flows from the system (chemicals) to the surroundings This implies that the reactants have less potential energy than the reactants and hence are more stable (Figure 5.34) 2NH3(g) + 10H2O(l) + BaCl2(s) products ∆H = 164 kJ mol–1 Ba(OH)2.8H2O(s) + 2NH4Cl(s) reactants Progress of reaction ■ Figure 5.35 Potential energy profile for Ba(OH)2.8H2O(s) + 2NH4Cl(s) → 2NH3(g) + 10H2O(l) + BaCl2(s); ΔH = +164 kJ mol−1 18/05/15 9:45 am 192 Energetics/thermochemistry Sketch a potential energy profile for the following thermochemical equation: Ba(OH)2.8H2O(s) + 2 NH4Cl(s) → 2NH3(g) + 10H2O(l) + BaCl2(s) ΔH = +164 kJ mol−1 State and explain if the reaction is exothermic or endothermic, and whether the reactants or the products are more stable The positive sign sign in front of the enthalpy change indicates that thermal energy (heat) flows from the surroundings to the system (chemicals) This implies that the products have more potential energy than the reactants and hence are less stable (Figure 5.35) For the sake of simplicity the activation energy barrier has not been drawn on the potential energy profiles ■■ Ozone depletion The chlorofluorocarbons (CFCs) have almost ideal properties for use as aerosol propellants and refrigerant heat-transfer fluids They are chemically and biologically inert (and hence safe to use and handle) and they can be easily liquefied by pressure slightly above atmospheric pressure Although fairly expensive to produce (compared to other refrigerants like ammonia), they rapidly replaced fluids that had been used before In the early 1970s, however, concern was expressed that their very inertness was a global disadvantage Once released to the environment, CFCs remain chemically unchanged for years in the atmosphere Due to their volatile nature, they diffuse upwards through the atmosphere and eventually find their way into the stratosphere (about 20 km above the Earth’s surface) Here they are exposed to the stronger ultraviolet rays of the Sun Although the carbon–fluorine bond is very strong, the carbon–chlorine bond is weak enough to be split by ultraviolet light This forms chlorine atoms (also known as radicals) which upset the delicately balanced equilibrium between ozone formation and ozone breakdown It is believed that a variety of biological consequences such as increases in skin cancer, cataracts in the eye, damage to plants, and reduction of plankton populations in the oceans may result from the increased ultraviolet exposure due to ozone depletion The social consequences involve the reduction of outdoor activities in the regions of the world most heavily affected by the two ozone holes located at the two poles and the increased need for people to use sunscreen It has been estimated that one chlorine atom can catalytically destroy over 105 ozone molecules before it eventually diffuses back into the lower atmosphere There it can react with water vapour to produce hydrogen chloride, which can be flushed out by rain, as dilute hydrochloric acid Once the ‘ozone hole’ (Figure 5.36) above the Antarctic was discovered in 1985, global agreements were signed in Montreal (1989) and London (1990) As a result, the global production of CFCs has been drastically reduced In many of their applications they can be replaced by hydrocarbons such as propane It will still take several decades for natural regeneration reactions to allow the ozone concentration to recover, but there is recent evidence that is occurring Other stratospheric pollutants such as nitrogen monoxide, NO, from high flying aircraft also destroy ozone Stratospheric ozone depletion raises a number of ethical issues Ozone depletion can harm humans, animals, plants and ecosystems Yet the harm caused by ozonedepleting chemicals was initially unknown, unintentional and indirect Producers of ozone-depleting chemicals were at first unaware that their products could harm the environment and/or people, and they had no intention to cause harm The harm was indirect because it was not the chemicals themselves, but rather the effects of the chemicals on stratospheric ozone, that caused damage Lastly, the harm was diffuse in that it was not the action of a single individual, but rather the cumulative effects of many corporations and individuals producing and using ozone-depleting chemicals, that caused the problem For example, should people be held responsible for harm that they did not know they were causing? Do corporations and nations have the same moral responsibilities as individuals? What obligations exist when the nature and size of the harm of CFCs were initially unknown? When harm is caused by the joint ■■ Figure 5.36 The ozone hole actions of many parties, how should responsibility be allocated? 829055_05_IB_Chemistry_165-198.indd 192 18/05/15 9:45 am 5.3 Bond enthalpies 193 The bond strength in ozone relative to oxygen in its importance to the atmosphere The Sun radiates a wide spectrum of radiation, ranging from the infrared to the ultraviolet (UV) region and including the visible region Part of this range of radiations corresponds to the energy required to break covalent bonds, including those in molecules such as DNA This can damage genes and lead to skin cancer, and also damage the proteins in connective tissue and lead to wrinkles The most damaging region is the ultraviolet region whose radiation has photons with the highest frequency and therefore the highest energy Some chemicals absorb this radiation, for example glass and manufactured chemicals such as sunscreens, but the best protection is the atmosphere itself The ultraviolet region is divided into three regions: UV A (wavelength 400−320 nm), UV B (wavelength 320−280 nm) and UV C (wavelength below 280 nm) UV A radiation is needed by humans for the synthesis of vitamin D but UV B and UV C are harmful Atmospheric gases in the stratosphere absorb ultraviolet radiation very well (‘strongly’) Ozone, O3, absorbs especially strongly In the stratosphere the ultraviolet radiation breaks covalent bonds in molecules to give reactive fragments called radicals The bonds in the ozone molecule absorb the harmful radiation present in UV B and UV C This leads to the destruction of the ozone: O3(g) + hν → O(g) + O2(g) O3(g) + O(g) → 2O2(g) A dynamic equilibrium is established in these reactions The ozone concentration varies due to the amount of radiation received from the Sun The bonds in ozone are slightly weaker than those in the oxygen molecule due to the presence of resonance or delocalization This means that the oxygen–oxygen bonds have partial double bond character and will have a bond energy intermediate between a double bond and a single bond The bond energy of the oxygen–oxygen bonds present in ozone is 445 kJ mol−1 while the bond energy of the double bond present in the oxygen molecule is 498 kJ mol−1 Because the bonds in the O3 molecule are weaker than those in the O2 molecule, photolysis (‘splitting by light’) is achieved with lower-energy photons Oxygen molecules also absorb ultraviolet radiation When an oxygen molecule absorbs a photon of the appropriate energy it will undergo photolysis and dissociate into oxygen atoms: O2(g) + hν → O(g) + O(g) O2(g) + O(g) → O3(g) The last reaction requires a third molecule to take away the energy associated with the free radical O and O2, and the reaction can be represented by O2(g) + O(g) + M → O3(g) + M* The principle of the conservation of energy can be used to calculate the maximum wavelength of the photon that has enough energy to break (cleave) a specific bond, for example, the O==O bond present in oxygen, O2 The energy per O==O bond is: 498 000 J mol−1 = 8.27 × 10−19 J bond−1 6.022 × 1023 bonds mol−1 The wavelength, λ, of the photons can be calculated using the expression E = hc/λ: −34 −1 λ = 6.626 × 10 J s × 3.00 × 10 m s = 2.403 × 10−7 m or 240 nm −19 8.27 × 10 J Only photons of wavelengths less than 240 nm can photolyze the O2 molecule Such high-energy photons are present in the solar spectrum at high altitude 829055_05_IB_Chemistry_165-198.indd 193 18/05/15 9:45 am 194 Energetics/thermochemistry Nature of Science H H C H C H H C H ■■ Figure 5.37 Structure of cyclopropane Models and theories Measured energy changes can be explained based on the model of bonds broken and bonds formed Since these explanations are based on a model, agreement with empirical data depends on the sophistication of the model, and data obtained can be used to modify theories where appropriate Calculations involving bond enthalpies generally give values of enthalpy changes that are close to experimentally determined values However, with a small number of molecules the calculated enthalpy change is significantly different from the real value For example, cyclopropane (see Figure 5.37) is much less energetically stable than a bond energy calculation predicts This is due to the ‘strain energy’ placed into the molecule during its formation The angles between the carbon–carbon bonds forming the ring are 60°, whereas the preferred bond angle is 109° (Chapter 4) In contrast, the benzene molecule is more energetically stable than a bond energy H H calculation would suggest This is because the molecule is a hybrid (Chapter 14) of the two C C H H H H resonance structures shown in Figure 5.38 Its C C C C carbon–carbon bonds are intermediate between C C C C single and double bonds, but because the π H H H H C C electrons (Chapter 14) move around the ring H H they stabilize the molecule ■ ■ Figure 5.38 Resonance structures of the The benzene molecule is a cyclic or ring benzene molecule system, like cyclopropane, but does not suffer from strain energy Calculating enthalpy changes from known bond enthalpy values and comparing these to experimentally measured values Calculate the expected value of the enthalpy of formation for dinitrogen monoxide and comment on the calculated value compared to its experimental value Assume N2O has one N==N bond and one O==O bond The enthalpy of formation of dinitrogen monoxide is described by the following equation: N2(g) + O2(g) → N2O(g) Enthalpy change = ∑(bonds broken) − ∑(bonds made) = (946 + × 498) − (418 + 607) = 170 kJ mol−1 The experimental value for the enthalpy of formation for dinitrogen monoxide is 82 kJ mol−1 This is a significant difference and indicates that dinitrogen monoxide is resonance stabilized The difference between the expected and observed values of the enthalpy of formation is the resonance energy (88 kJ mol−1) Additional Perspectives Enthalpies of combustion of alkanes Bond energies are by and large additive, which means that the specific bond energies are approximately constant for a range of related molecules, for example the alkanes and alcohols Consider the alkanes, a group of hydrocarbons that are derived from methane, CH4, by progressively adding methylene, – CH2– units (Figure 5.39) ■■ Figure 5.39 The formation of alkanes by the progressive addition of methylene, −CH2− units 829055_05_IB_Chemistry_165-198.indd 194 H H C H H H H H C C H H H H H H H C C C H H H H 18/05/15 9:45 am 5.3 Bond enthalpies 195 If a series of hydrocarbons is combusted, the addition of each extra methylene group will be responsible for an additional enthalpy change: –– CH2–– + O2(g) → CO2(g) + H2O(l) The enthalpy change, ΔH, for this process can be calculated using bond enthalpies: Breaking bonds: Making bonds: 1 × C–– C, × C––H, 12 × O==O × C==O, × O−H Using the values from page 11 of the IB Chemistry data booklet: Breaking bonds: × C–– C 346 × C––H × 414 1 × O==O 1 × 498 2 Total energy 1921 kJ Making bonds: × C==O × O ––H Total energy × 804 × 463 2534 kJ Enthalpy change = ∑(bonds broken) − ∑(bonds made) = (1921) + (−2534) = −613 kJ mol−1 Hence the additional enthalpy of combustion for each additional methylene unit, –– CH2––, is −613 kJ mol−1 This type of simple calculation predicts that there should be an approximate linear relationship between the enthalpy change of combustion of an alkane and the number of carbon atoms (Figure 5.40) Experimental values confirm this prediction –8000 ∆Hc /kJ mol –1 ■■ Figure 5.40 Graph of the standard enthalpies of combustion, ΔHc, of the straight-chain alkanes plotted against the number of carbon atoms in the molecule –6000 –4000 –2000 0 Number of carbon atoms 10 Utilization: Fossil fuels as energy sources A fuel is a substance that releases thermal energy (heat) that can work Most fuels release this energy during combustion with molecular oxygen An ideal chemical fuel should have a number of properties, including a high value of enthalpy of combustion Table 5.4 gives the standard enthalpies of combustion of some fuel alternatives to petrol There is no standard enthalpy change of combustion for petrol because it is a complex homogenous mixture of about 100 compounds, mainly hydrocarbons, of which the majority are alkanes However, the standard enthalpy change of combustion of octane, the major component of petrol, is −5470 kJ mol−1 Its high enthalpy of combustion is one reason why it has desirable chemical properties to be a good fuel 829055_05_IB_Chemistry_165-198.indd 195 19/05/15 9:21 am 196 Energetics/thermochemistry ■■ Table 5.4 Enthalpies of combustion of fuels Fuel Main component Formula and standard state ΔH c of main component/kJ mol –1 Hydrogen Hydrogen H2(g) −286 Compressed natural gas (CNG) 90% methane CH4(g) −890 Liquid petroleum gas (LPG) 95% propane C3H8(g) −2219 Methanol Methanol CH3OH(l) −726 Alcohol Ethanol C2H5OH(l) −1367 Another relevant property of fuels is the energy density of the fuel (Table 5.5) This is the amount of thermal energy (heat) released by kilogram of the fuel The energy density is calculated from the standard enthalpy of combustion of the fuel and the mass of one mole of the fuel Petrol has an energy density of approximately 46 000 kJ kg−1 which makes it a very concentrated energy source Liquid hydrogen has a higher density than petrol, but currently storage problems on board a vehicle are one reason why its use is limited, along with the very low temperatures required to keep it as a liquid ■■ Table 5.5 Energy densities of fuels Additional Perspectives Fuel Formula ΔHc Mass of one mole/g Energy density/kJ kg –1 Hydrogen H2(g) −286 143 000 Methane CH4(g) −890 16 27 800 Methanol CH3OH(l) −2219 32 22 700 Ethanol C2H5OH(l) −1367 46 30 000 Feasibility of reactions There are many examples of reactions which are spontaneous The vast majority of these reactions are exothermic Hence it appears that the enthalpy change, ΔH, is a reliable guide to which direction a reaction will go However, there are examples of endothermic reactions that occur without the need for heat to initiate the reaction, for example, the reaction between citric acid and a solution of sodium hydrogencarbonate Some salts dissolve endothermically in water Chapter 15 introduces a factor, known as entropy, that, in conjunction with enthalpy and temperature, determines whether or not reactions occur at a specified temperature ■■ Examination questions – a selection Paper IB questions and IB style questions Q1 When 0.3205 g of methanol is completely combusted under a water filled flame combustion calorimeter, the temperature of 1x103 cm3 of water is raised by 1.5°C (Molar mass of methanol = 32.05 g mol−1; specific heat capacity of water = 4.18 J g−1 °C1.) What is the expression for the molar enthalpy of combustion of methanol? 1×103 × 4.18 × 1.5 × 32.05 A − 0.3205 1×103 × 4.18 × (273.00 + 1.5) × 32.05 B − 0.3205 × × 103 1×103 × 4.18 × 1.5 × 32.05 C − 0.3205 × 1 × 103 0.3205 × × 103 D − 1 × 103 × 4.18 × 1.5 × 32.05 829055_05_IB_Chemistry_165-198.indd 196 Q2 Which of the following reactions would you expect to provide the largest amount of heat? A C2H6(l) + 7O2(l) → 4CO2(g) + 6H2O(g) B C2H6(l) + 7O2(g) → 4CO2(g) + 6H2O(g) C C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g) D C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) Q3 Why does the temperature of boiling water remain constant even though heat is supplied at a constant rate? A Heat is lost to the surroundings B The heat is used to break the covalent bonds in the water molecules C Heat is also taken in by the container D The heat is used to overcome the intermolecular forces of attraction between water molecules Standard Level Paper 1, Nov 2005, Q14 Q4 When 0.050 mol of nitric acid is reacted with 0.050 mol of potassium hydroxide in water, the temperature of the system increases by 13.7 °C 18/05/15 9:46 am Examination questions 197 Calculate the enthalpy of reaction in kJ mol−1 HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l) Assume that the heat capacity of the system was 209.2 J °C−1 A +57.3 kJ mol−1 C −2.87 kJ mol−1 −1 B +2.87 kJ mol D −57.3 kJ mol−1 Q5 What can be deduced about the relative stability of the reactants and products and the sign of ΔH, from the enthalpy level diagram below? Relative stability Sign of ΔH A products more stable − B products more stable + C reactants more stable − D reactants more stable + reactants ∆H products Standard Level Paper 1, May 1999, Q16 Q6 The specific heat capacities of some metals are given below Metal Specific heat capacity (J g−1 K−1) copper 0.385 magnesium 1.020 mercury 0.138 platinum 0.130 If 100 kJ of heat is added to 10.0 g samples of each of the metals above, which are all at 25 °C, which metal will have the lowest temperature? A copper C mercury B magnesium D platinum Q7 The bond energy for the H–F bond is equal to the enthalpy change for which process? A H+(g) + F− (g) → HF(g) B HF(g) → H(g) + F(g) C F (g) 2 + D HF(g) → H (g) → HF(g) 2 1 F (g) + H2(g) 2 Q8 When a sample of a pure hydrocarbon (melting point 85 °C) cools, the temperature is observed to remain constant as it solidifies Which statement accounts for this observation? A The heat released in the change of state equals the heat loss to the surroundings 829055_05_IB_Chemistry_165-198.indd 197 B The temperature of the system has fallen to room temperature C The solid which forms insulates the system, preventing heat loss D Heat is gained from the surroundings as the solid forms, maintaining a constant temperature Q9 Consider the following equation: 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) ΔH = 2824 kJ mol−1 What is the enthalpy change associated with the production of 100.0 g of C6H12O6? A 157 kJ C 508 kJ B 282 kJ D 1570 kJ Q10 N2(g) + O2(g) → 2NO(g) ΔH = 180.4 kJ mol−1 N2(g) + 2O2(g) → 2NO2(g) ΔH = 66.4 kJ mol−1 Use the enthalpy values to calculate ΔH for the reaction: NO(g) + O2(g) → NO2(g) A −57 kJ mol−1 B −114 kJ mol−1 C 57 kJ mol−1 D 114 kJ mol−1 Standard Level Paper 1, May 2000, Q18 Q11 The heating curve for 10 g of a substance is given below How much energy would be required to melt completely 40 g of the substance that is initially at 10 °C? 50 Temperature/°C 30 C mercury D platinum 10 A 4800 J B 2400 J 400 800 1200 Energy added/J 1600 C 1600 J D 800 J Q12 The bond energies for H2, I2 and HI are 432, 149 and 295 kJ mol−1, respectively From these data, what is the enthalpy change (in kJ) for the reaction below? H2(g) + I2(g) → 2HI(g) A +9 C −286 B +286 D −9 18/05/15 9:46 am 198 Energetics/thermochemistry Q13 The specific heat capacity of aluminum is 0.900 J g −1 K−1 What is the heat energy change, in J, when 10.0 g of aluminum is heated and its temperature increases from 15.0 °C to 35 °C? A +180 C +1800 B +315 D +2637 b i Using values from page 11 of the IB Chemistry data booklet, calculate the enthalpy change for the following reaction: CH4(g) + F2(g) → CH3F(g) + HF(g) [3] ii Sketch an enthalpy diagram for the reaction.[2] iii Without carrying out a calculation, suggest, with a reason, how the enthalpy change for the following reaction compares with that of the previous reaction CH3F(g) + F2(g) → CH2F2(g) + HF(g) [2] Higher Level Paper 1, May 2013, Q14 Q14 Which reaction represents the average bond enthalpy of the Si–H bond in silane, SiH4? 1 SiH4(g) SiH4(g) SiH4(g) 1 SiH2(g) + H2(g) Si(g) + H(g) Si(s) + H(g) A SiH4(g) → Si(g) + H2(g) B C D → → → Q15 Which of the following is the correct equation for the standard enthalpy change of formation of carbon monoxide? A C(s) + O2(g) → CO(g) B C(g) + O (g) 2 → CO(g) C C(g) + O(g) → CO(g) D 2C(s) + O2(g) → 2CO(g) Q16 The bond enthalpies for H2(g) and HF(g) are 435 kJ mol−1 and 565 kJ mol−1, respectively For the 2 reaction H2(g) + F2(g) → HF(g), the enthalpy of reaction is −268 kJ mol−1 of HF produced What is the bond energy of F2 in kJ mol−1? A 464 C 243 B 138 D 159 Q17 The standard enthalpy change of formation values of two oxides of phosphorus are: P4(s) + 3O2(g) → P4O6(s) ΔH f = −1600 kJ mol−1 P4(s) + 5O2(g) → P4O10(s) ΔH f = −3000 kJ mol−1 What is the enthalpy change, in kJ mol−1, for the reaction below? P4O6(s) + 2O2(g) → P4O10(s) A +4600 B +1400 C −1400 D −4600 Paper IB questions and IB style questions Q1 a i Define the term average bond enthalpy. [3] ii Explain why the fluorine molecule, F2, is not suitable as an example to illustrate the term average bond enthalpy. [1] 829055_05_IB_Chemistry_165-198.indd 198 Q2 In aqueous solution, lithium hydroxide and hydrochloric acid react as follows LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(l) The data below are from an experiment to determine the standard enthalpy change of this reaction 50.0 cm3 of a 0.500 mol dm−3 solution of LiOH was mixed rapidly in a glass beaker with 50.0 cm3 of a 0.500 mol dm−3 solution of HCl Initial temperature of each solution = 20.6 °C Final temperature of the mixture = 24.1 °C a State, with a reason, whether the reaction is exothermic or endothermic. [1] b Explain why the solutions were mixed rapidly. [1] c Calculate the enthalpy change of this reaction in kJ mol−1 Assume that the specific heat capacity of the solution is the same as that of water. [4] d Identify the major source of error in the experimental procedure described above Explain how it could be minimized. [2] e The experiment was repeated but with an HCl concentration of 0.520 mol dm−3 instead of 0.500 mol dm−3 State and explain what the temperature change would be. [2] Q3 a Define the term standard enthalpy change of formation.[2] b Define the term standard enthalpy change of combustion.[2] c State Hess’s law. [1] d Calculate the standard enthalpy change of formation of propane, C3H8, given the following standard enthalpies of combustion: ΔH c [C3H8(g)] = −2220 kJ mol−1 ΔH c [Cgraphite(s)] = −393 kJ mol−1 ΔH c [H2(g)] = −286 kJ mol−1 Draw an energy cycle and use Hess’s law to produce an equation for the enthalpy change of formation. [4] 18/05/15 9:46 am ... www.hoddereducation.com/IBextras ■■ Using this book The sequence of chapters in Chemistry for the IB Diploma deliberately follows the sequence of the syllabus content However, the IB Diploma Chemistry Guide... Welcome to the second edition of Chemistry for the IB Diploma The content and structure of this second edition has been completely revised to meet the demands of the 2014 IB Diploma Programme Chemistry. .. particles at the surface of the liquid have enough kinetic energy to overcome the forces of attraction between themselves and the other particles in the liquid and they escape from the surface to form