Oxford Cambridge and RSA This is an OCR endorsed resource.
A Level
Chemistry
for OCR
A
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resource has undergone an independent quality check. OCR has not paid for the production of this resource, nor does OCR receive any royalties from its sale. For more information about the endorsement process please visit
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Acknowledgements
Cover: EYE OF SCIENCE/SCIENCE PHOTO LIBRARY p2-3: Tischenko Irina/Shutterstock; p6-7: Science Photo Library; p9: Author; p.10: Charles D, Winters/Science Photo Library; p13: Mauro Fermaricllo} Science Photo Library; p16: Nagydodo/Shutterstock; p20: Martyn F. Chillmaid/Science Photo Library; p22: Apttone/iStockphoto; p23: Nadezda Boltaca/Shutterstock; p24(T): Fablok/Shutterstock; p24(B): Martyn F. Chillmaid/Science Photo Library; p26: Haveseen/Shutterstock; p28: David Hay Jones/Science Photo Library; p40: Author; p41(T): Martyn F, Chillmaid/Science Photo Library; p41(B): Andrew Lambert Photography/Science Photo Library; p43(T): Author; p44(B): Author; p43(B): Author; : Martyn F, Chillmaid/Science Photo Library; p44(T): Author; p50; Martyn F. Chillmaid/Science Photo Library; p51: Martyn F. Chillmaid/Science Photo Library; p60: Usas/iStockphoto; p62(L): Author; p62(R): Author; p79: Claude Nuridsany & Marie Perennou/Science Photo Library; p80: Andrew Lambert Photography/Science Photo Library; : Science Photo Library; : Ria Novosti/Science Photo Library: p87: Bizroug/Shutterstock; p88-89: Elena Moiseeva/Shutterstock; p90: Science Photo Library; p94(T): Arhip4/Shutterstock; p94(B): Stuart MonkjShutterstock; p81; Darren Begley/Shutterstock; p106: Charles
D. Winters/Science Photo Library; p107: Charles D. Winters/Science Photo Library; p108: Charles Brutlag/Shutterstock; p109(L): Author;
p109(R): Author; p110: Andrew Lambert Photography/Science Photo Library; p113(B): Petegar/iStockphoto; p111(T): Martyn F. Chillmaid/ Science Photo Library; p111({B): Andrew Lambert Photography/Science Photo Library; p115(T): Andrew Lambert Photography/Science Photo Library; p115(B): Charles D, Winters/Science Photo Library; p116: Andrew Lambert Photography/Science Photo Library; p117: Hong Xia/Shutterstock; p101: Denis Burdin/Shutterstock; p102: Mathier/ Shutterstock; p113(T): Author; p130: Martyn F. Chillmaid/Science Photo Library; p143(T): Bivdone/Shutterstock; p143(B): Trevor Clifford Photography/Science Photo Library; p126: Author; p148(B): Dorling Kindersley/Uig/Science Photo Library; p153: Charles D. Winters/Science Photo Library; p142(T): Gary718/Shutterstock; p142(B): Jean Morrison/ Shutterstock; p145(T): Andrew Lambert Photography/Science Photo Library; p145(BL): Author; p145(BR): Author; p148(T): Ssuaphotos/ Shutterstock; p155: Charles D. Winters/Science Photo Library; : Vipavienkoff/Shutterstock; p157: Hacohob/Shutterstock; p163: Jg Photography; p185: Claffra/Shutterstock; p187(T): Author; p188: Author; p177(L); Daniel Korzeniewski/Shutterstock; p177(R): Nito} Shutterstock; p166: Piccia Neri/Shutterstock; p195: Author p196(T): Zixian/Shutterstock; p187(B): Ekaterina Baranova/Shutterstock; p202(B); Author; p202(T): Stanzi/Shutterstock; p203: Martyn F. Chillmaid/Science Photo Library; p209: Hasnuddin/Shutterstock; p210: Gyvafoto/Shutterstock; p212(T): FLPA/Alamy; p211(T): Irin-K/ Shutterstock; p211(C): Maxal Tamor/Shutterstock; p211(B): Vankad/ Shutterstock; p196(BL): Thinkstock; p196(BR): Ingram; p220: Andrew Lambert Photography/Science Photo Library: p197(T): Fabio Sacchi} Shutterstock; p197(B): Pavla/Shutterstock; : Gayvoronskaya_Yana/ Shutterstock; p206: Author; p213: Satit_Srihin/Shutterstock; p212(C ): Spwidoff/Shutterstock; p212(B): Author: p216: Christian Draghici/ Shutterstock; p227: Andrew Lambert Photography/Science Photo Library; p218: Jorg Hackemann/Shutterstock; p230: Anna Omelchenko/ Shutterstock; p235(T): Author; p235(C ): Author; p235(B): Author; p236(T): Author; p236(B): Jerry Mason/Science Photo Library; p242(L): David Nunuk/Science Photo Library; p242(R): Molekuul.Be/Shutterstock: p251: Fenton One/Shutterstock; p254: Jim Varney/Science Photo Library; p246: Simon Fraser/Science Photo Library; p249: Helene Wiesenhaan/Getty Images; p164-165:; Mopic/Shutterstock; p231: Africa Studio/Shutterstock; p259: Lanych/Shutterstock; p277(T): Martyn F, Chillmaid/Science Photo Library; p299: Robert Boesch/ Corbis; p300: Hacohob/Shutterstock; p313: Costi losif/Shutterstock; p315: Photong/Shutterstock; p321: Africa Studio/Shutterstock;
p327: Charles D. Winters/Science Photo Library; p337: Sherry Yates Young/Shutterstock; p340: Science Photo Library; p444: Andrew Lambert Photography/Science Photo Library; p454(T): Rikkert Harink/ Shutterstock; p454(B); Dusan Jankovic/Shutterstock; p456: Andrew Lambert Photography/Science Photo Library; p461: Andrew Lambert Photography/Science Photo Library; p463: Artem Furman/Shutterstock; p468: Brian Lasenby/Shutterstock; p475: S Duffett/Shutterstock; p480: LittleStocker/Shutterstock; p432(R): Dionisvera/Shutterstock; p432(L): Tim UR/Shutterstock; p438: Gannet77/iStockphoto; p440: Filipw/Shutterstock; p442(C): Jon Le-Bon/Shutterstock; p450: Sovfoto/UIG/Getty Images; pS10(d): Andrew Lambert Photography/ Science Photo Library: p469: Andrew Lambert Photography/Science Photo Library; p477: Optimarc/Shutterstock; p484(T): Anukool Manoton/Shutterstock; p485: Tommaso lizzul/Shutterstock; p357: No_limit_pictures/iStockphoto; p360(L): Impactimage/iStockphoto; p360(R): Byjeng/Shutterstock; p363(T): Valentyn Volkov/Shutterstock; p363(CT): Science photo/Shutterstock; p363(CB): CAN BALCIOGLU/ Shutterstock; p366: Mangojuicy/Shutterstock; p369: Africa Studio/ Shutterstock; p372: Peticolas/Megna/Fundamental Photos/Science Photo Library; p373(L): Martyn F, Chillmaid/Science Photo Library; p373(R): Martyn F. Chillmaid/Science Photo Library; p378: Avarand/ Shutterstock; p384(C): Marco mayer/Shutterstock; p384(B): Gyvafoto/ Shutterstock; p389: Andrew Lambert Photography/Science Photo Library; p395(T): Claus Lunau/Science Photo Library; p395(B): Martin Bond/Science Photo Library; p403(C); Andrew Lambert Photography/ Science Photo Library; p415(T); Power and Syred/Science Photo Library; p491: Melinda Fawver/Shutterstock; p498(T): Monika Wisniewska/ Shutterstock; p498(B): Maks Narodenko/Shutterstock; pS08; Tanewpix/ Shutterstock; p510(a): Martyn F. Chillmaid/Science Photo Library; P510(b): Andrew Lambert Photography/Science Photo Library; p511: Andrew Lambert Photography/Science Photo Library; p512: Mauro Fermariello/Science Photo Library; pS13{B): Colin Cuthbert/Science Photo Library; p270-271: Eye of Science/Science Photo Library; p384(T): Andrew Lambert Photography/Science Photo Library; p442(B): Science Photo Library; p443: Martin Bond/Science Photo Library; p455(L): AntoinetteW/Shutterstock; p455(R): Maksimilian/Shutterstock;
p467: Africa Studio/Shutterstock; p481(L): Imageman/Shutterstock;
AS/A Level course structure
This book has been written to support students studying for
OCR AS Chemistry A and OCR A Level Chemistry A. It covers all of the modules from the OCR A Level Chemistry A specification, with modules 2, 3, and 4 also part of the OCR AS Chemistry A specification. The modules covered are shown in the contents list, which also shows you the page numbers for the main topics within each module. There is also an index at the back to help you find what you are looking for. If you are studying for OCR AS Chemistry A, you will only need to
know the content in the blue box.
Year 1 content
1 Development of practical skills in chemistry
2 Foundations in chemistry
3 Periodic table and energy
4 Core organic chemistry
AS exam
Year 2 content
5 Physical chemistry and transition elements 6 Organic chemistry and analysis
A level exam
A Level exams will cover content from Year 1 and Year 2 and will be ata higher demand. You will also carry out practical activities throughout your course.
p481(R): Valentina Proskurina/Shutterstock; p513(T): Bibiphoto/ Shutterstock; p334: Charles D. Winters/Science Photo Library; p355: MarcelClemens/Shutterstock; p429(T); Abramova Elena/Shutterstock; p429(B): Remik44992/Shutterstock; pS02: Science Photo/Shutterstock; p430-431: Eye of Science/Science Photo Library; p541: Olha Rohulya/ Shutterstock; p510(e): Andrew Lambert Photography/Science Photo Library; p419(B): Charles D, Winters/Science Photo Library;
Author Photos: p277(B), p280(R), p296, p301, p460, p464, p465, p442(T), p510{c), p510(f), p484(B), p363(B), p376(L), p376{R), p381(L), p381(C), p381(R), p382(L), p382(C), p382(R), p382(B), p394,
p400(L),p400(C), p400(R), p402, p403(T), p403(B), p405(T), p405(B), p407,
p413, p414(7), p414(C), p414(B), p415(C), p415(B), p416, p418, p419(1), p420, p494, p495(TL), p495(TC), p495(TR), p495/(BL), p495(BC), p495/BR),
p496(T), p496(BL), p496(BC), p497(T), p497(B), p506(T), p506(C), p506(B),
p507, p382(B), p486, p496(C), p496(BR), p280(L), p291, p316; lithium battery: Author;
Artwork by Q2A media
Thank you to St John Rigby college, Wigan, for the use of their laboratory facilities in the production of photographs
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in this work.
How to use this book Kerboodle
Module 1 Development of practical skills
in chemistry
Module 2 Foundations in chemistry
Chapter 2 Atoms, ions, and compounds 2.1 Atomic structure and isotopes 2.2 Relative mass 2.3. Formulae and equations Practice questions
Chapter 3 Amount of substance 3.1 Amount of substance and the mole 3.2 Determination of formulae 3.3 Moles and volumes 3.4 Reacting quantities Practice questions
Chapter 4 Acids and redox 4.1 Acids, bases, and neutralisation 4.2 Acid—base titrations 4.3 Redox Practice questions
Chapter 5S Electrons and bonding 5.1 Electron structure 5.2 lonic bonding and structure 5.3 Covalent bonding
Practice questions
Chapter 6 Shapes of molecules and intermolecular forces 6.1 Shapes of molecules and ions 6.2 Electronegativity and polarity 6.3 Intermolecular forces 6.4 Hydrogen bonding
Practice questions Module 2 summary Module 2 practice questions
Module 3 Periodic table and energy
Chapter ? Periodicity 7.1 The periodic table 7.2 lonisation energies
vii
7.3 Periodic trends in bonding and structure
Practice questions
Chapter 8 Reactivity trends
8.1 Group 2
8.2 The halogens
8.3 Qualitative analysis Practice questions
Chapter 9 Enthalpy
9.1 Enthalpy changes
9.2 Measuring enthalpy changes
9.3 Bond enthalpies
9.4 Hess’ law and enthalpy cycles Practice questions
Chapter 10 Reaction rates and equilibrium
10.1 Reaction rates
10.2 Catalysts
10.3 The Boltzmann distribution
10.4 Dynamic equilibrium and le Chatelier's principle
10.5 The equilibrium constant K_— part 1 Practice questions
Module 3 summary
Module 3 practice questions
Module 4 Core organic chemistry and analysis
Chapter 11 Basic concepts of organic
chemistry
11.1 Organic chemistry
11.2 Nomenclature of organic compounds
11.3 Representing the formulae of organic compounds
90
92 92 96 101 106
108 108 112 117 121
124 124 129 135 138 142
144 144 149 152
154 160 162 164 166
170
172 172 174
179
11.4 lIsomerism 11.5 Introduction to reaction mechanisms Practice questions
Chapter 12 Alkanes
12.1 Properties of alkanes
12.2 Chemical reactions of alkanes Practice questions
Chapter 13 Alkenes
13.1 Properties of alkenes
13.2 Stereoisomerism
13.3 Reactions of alkenes
13.4 Electrophilic addition in alkenes
13.5 Polymerisation in alkenes Practice questions
Chapter 14 Alcohols
14.1 Properties of alcohols
14.2 Reactions of alcohols Practice questions
Chapter 15 Haloalkanes 15.1 The chemistry of the haloalkanes 15.2 Organohalogen compounds in the environment Practice questions
Chapter 16 Organic synthesis
16.1 Practical techniques in organic chemistry
16.2 Synthetic routes Practice questions
Chapter 17 Spectroscopy
17.1 Mass spectrometry
17.2 Infrared spectroscopy Practice questions
Module 4 summary
Module 4 practice questions
182 184 187
190 190 193 198
200 200 203 207 211 215 220
222 222 226 229
230 230
235 238
240
240 244 250
252 252 256 262 264 266
Module 5 Physical chemistry and transition elements
Chapter 18 Rates of reactions
18.1 Orders, rate equations, and rate constants
18.2 Concentration—time graphs
18.3 Rate—concentration graphs
18.4 Rate-determining step
18.5 Rate constants and temperature Practice questions
Chapter 19 Equilibrium
19.1 The equilibrium constant Ke — part 2
19.2 The equilibrium constant Kp
19.3 Controlling the position of equilibrium Practice questions
Chapter 20 Acids, bases, and pH 20.1 Bronsted—Lowry acids and bases 20.2 The pH scale and strong acids 20.3 The acid dissociation constant Ka 20.4 The pH of weak acids 20.5 pH and strong bases
Practice questions
Chapter 21 Buffers and neutralisation 21.1 Buffer solutions 21.2 Buffer solutions in the body 21.3 Neutralisation Practice questions
Chapter 22 Enthalpy and entropy
22.1 Lattice enthalpy
22.2 Enthalpy changes in solution
22.3 Factors affecting lattice enthalpy and hydration
22.4 Entropy
22.5 Free energy Practice questions
Chapter 23 Redox and electrode potentials 23.1 Redox reactions
23.2 Manganate([VII]} redox titrations
23.3 lodine/thiosulfate redox titrations
270 272
272 277 282 287 289 292
298 302 307
23.4 Electrode potentials 386 28.3 Further synthetic routes 498
23.5 Predictions from electrode potentials 391 Practice questions 504 23.6 Storage and fuel cells 394 Chapter 29 Chromatography and Practice questions 397 spectroscopy 506 Chapter 24 Transition elements 400 29.1 Chromatography and functional 24.1 d-block elements 400 group analysis 506 24.2 The formation and shapes of 29.2 Nuclear magnetic resonance (NMR} complex ions 405 spectroscopy 512 24.3 Stereoisomerism in complex ions 409 29.3 Carbon-12 NMR spectroscopy 515 24.4 Ligand substitution and precipitation 413 29.4 Proton NMR spectroscopy 520 24.5 Redox and qualitative analysis 418 29.5 Interpreting NMR spectra 525 Practice questions 422 29.6 Combined techniques 530 Module 5 summary 424 Practice questions 534 Module 5 practice questions 426 Module 6 summary 536 Module 6 practice questions 534 Module 6 Organic chemistry and analysis 430 Unifying concepts 542 Chapter 25 Aromatic chemistry 432 Analysing and answering a synoptic question 542 25.1 Introducing benzene 432 Practice questions 54? 25.2 Electrophilic substitution reactions of benzene 437 paterence 25.3 The chemistry of phenol 442 mame - 25.4 Disubstitution and directing groups 446 index 288 Practice questions 451 Periodic table 294 Chapter 26 Carbonyls and carboxylic acids 454 26.1 Carbonyl compounds 454 26.2 Identifying aldehydes and ketones 460 26.3 Carboxylic acids 463 26.4 Carboxylic acid derivatives 466 Practice questions 472
Chapter 27 Amines, amino acids,
and proteins 474 27.1 Amines 474 27.2 Amino acids, amides, and chirality 478 27.3 Condensation polymers 483
Practice questions 488 Chapter 28 Organic synthesis 490 28.1 Carbon—carbon bond formation 490 28.2 further practical techniques 494
How to use this book
SOE EEE EEE EERE EEE EEE EEE EEE HEHE HHS
Learning outcomes > Atthe beginning of each
topic, there is a list of learning
outcomes.
> These are matched to the specification and allow you to monitor your progress.
> Aspecification reference is also included in the topic header.
Study Tips
Study tips contain prompts to help you with your understanding and revision.
Synoptic link
These highlighted the key areas where topics relate to each other. As you go through your course, knowing how to link different areas of chemistry together becomes increasingly important. Many exam questions, particularly at
A Level, will require you to bring together your knowledge from different areas.
This book contains many different features. Each feature is designed to support and develop the skills you will need for your examinations, as well as foster and stimulate your interest in chemistry.
Terms that you will need to be able to define and understand are highlighted by bold text.
Application features
These features contain important and interesting applications of chemistry in order to emphasise how scientists and engineers have used their scientific knowledge and understanding to develop new applications and technologies. There are also practical application features, with the icon @. to support further development of your practical skills.
1 Allapplication features have a question to link to material covered with the concept from the specification.
+ Extension features
These features contain material that is beyond the specification. They
are designed to stretch and provide you with a broader knowledge and understanding and lead the way into the types of thinking and areas you might study in further education. As such, neither the detail nor the depth of questioning will be required for the examinations. But this book is about more than getting through the examinations.
1 Extension features also contain questions that link the off-specification material back to your course.
Summary Questions
1 These are short questions at the end of each topic.
2 They test your understanding of the topic and allow you to apply the knowledge and skills you have acquired.
3 The questions are ramped in order of difficulty. Lower-demand questions have a paler background, with the higher-demand questions having a darker background. Try to attempt every question you can, to help you achieve your best in the exams.
LAS a, Knowledge and understanding checklist
~ | From your previous studees you shoukd be able to answer the following questions Work = through each point, using your notes and the support available on Kerboodie ~} Colcutate reection ranes tom measurement (} Cetculate an enthalpy change from
of gradients foro
ertrstion- time graphs entheipy chary
Chapters in this module 18 Rates cf reaction
Know how to investgute reecton retes LJ Apply oxidatio:
by pas collector ce mass loss over Dene Sescribe 1edon in teers OP iectrors ard
oo ‘ dott C) Write K, expeessions from an equation mneaven mune and cake late A trom egulorun tL) Recogrte tetrahedral and Introduction at the beginning
r C) Agnlyte prmupte for () Keow new to prepace a stamdbed of each module summarises Gangeata concenailon,pecanuse. and, auhdlon.to cm eukatbiadel andacie J tempersture stroightiorward titration calouidtions what you will cover. O wine nevvstsaton equations of acids (C) Monny anieos and ammonite wth Dates Qualtave anadyus
=
) Maths skills checklist ® inthis mogule, you will need to use the following maths skills. You can find support these skills on Kerboodie and through MyMeths (_} Changing the sudject of, substicuung when caloulstrg rate constants ana ys mumbert into, and setving algebraic the Amhernss equation this when (5) Finding sctthenetic means. You will
thes wnen caloutating mean Stes
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energy aed wratien calcutstens __| PRotting twe weriebies from experemental
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Gata bore mtes expeerments __) Orawing and esing the gredient of s tangent to « curve as @ measure of rate of reaction. fou wineedttte when
Achecklist helps you assess your knowledge
=.) from KS4 and earlier in your A Level course, Sass ———— before starting work on the module. There is -————— also a maths skills checklist to demonstrate the skills you will learn in that module.
Visual summaries show how some of the key concepts of that module interlink with other modules, across the entire A Level course.
Application task brings together some of the key concepts of the module in a new context.
Extension task bring together some key concepts of the module and
oe) develop them further, leading you | = | towards greater understanding and
further study.
(ae Chepter 5 Prectice qeestions
Practice questions
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Practice questions at the end of each chapter, with multiple choice questions and synoptic style questions, also covering the practical and math skills. The questions at the end
of the AS modules are also labelled according to the AS
exam structure.
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Revise with ease using
the study guides to guide you through each chapter and direct you towards the resources you need.
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dentify the lon and the amount | Fe venom om”
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centration of a coloured complex ion Investigate the concentrat on of a cole : meter in 25.4 Practical
n using a color
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eration of Practice determining the acimigyipoerr ) “= an absorption graph using 25.4 Maths skills ac pis of absorption and concentratic
Interpreting or
For teachers, Kerboodle also has plenty of further assessment resources, answers to the questions in the book, and a digital markbook along with full teacher support for practicals and the worksheets, which include suggestions on how to support and stretch students. All of the resources are pulled together into teacher guides that suggest a route through each
chapter.
MODULE 1
Development of practical skills in chemistry
Chemistry is a practical subject and experimental work provides you with important practical skills, as well as enhancing your understanding of chemical theory. You will be developing practical skills by carrying out practical and investigative work in the laboratory throughout both the AS and the
A level Chemistry course. You will be assessed on your practical skills is two different ways:
e written examinations (AS and A level]
e practical endorsement (A level only)
Practical coverage throughout this book
Practical skills are a fundamental part of a complete education in science, and you are advised to keep a record of your practical work from the start of your A level course that you Can later use as part of your practical endorsement. You can find more details of the practical endorsement from your teacher or from the specification.
In this book and its supporting materials practical skills are covered in a number of ways. By studying Application boxes and
1.1.1 Planning
Identifying variables to be controlled
Evaluating the experimental method
Skills checklist
Designing experiments O Selecting apparatus and equipment
©) Selecting appropriate techniques
Selecti ng appropriate quantities of chemicals and
~ scal
Exam-style questions in this student book, and by using the Practical activities and Skills sheets in Kerboodle you will have many Opportunities to learn about the scientific method and carry-out practical activities.
1.1 Practical skills assessed in written examinations
In the written examination papers for AS and
A level, at least 15% of the marks will be from questions that assess practical skills. The questions will cover four important skill areas, all based on the practical skills that you will develop by carrying out experimental work during your course.
e Planning — your ability to solve a chemistry problem in a practical context.
Implementing — your understanding of important practical techniques and processes.
Analysing — your interpretation of experimental results set in a practical context and related to the experiments that you would have carried out.
Evaluating — your ability to develop a plan that is fit for the intended purpose.
of working Solving chemical problems in a practical context
(©) Applying chemistry concepts to practical problems
1.1.2 Implementing
e Using a range of practical apparatus
e Carrying outa range of techniques
e Using appropriate units for measurements
SMe
e Recording data and observations in an appropriate format
1.1.3 Analysis
e Processing, analysing, and interpreting results
e Analysing data using appropriate mathematical Skills
e Using significant figures appropriately
e Plotting and interpreting graphs
————
1.1.4 Evaluation
e Evaluating results to draw
conclusions e Identify anomalies
e Explain limitations in method
e Identifying uncertainties Pafining satires and meact ey
Mast
and errors
e Suggesting improvements
1.2 Practical skills assessed in practical endorsement
You will also be assessed on how well you carry out a wide range of practical work and how to record the results of this work. These hands-on skills are divided into 12 categories and form the practical endorsement. This is assessed for A level Chemistry qualification only.
The endorsement requires a range of practical skills from both years of your course. If you are taking only AS Chemistry, you will not be assessed through the practical endorsement but the written AS examinations will include questions that relate to the skills that naturally form part of the AS common content to the A level course.
1.2.1 Practical skills
By carrying out experimental work through the course, you will develop your ability to:
design and use practical techniques to investigate and solve problems
use a wide range of experimental and practical apparatus, equipment, and materials, including chemicals and solutions
Carry out practical procedures skillfully and safely, recording and presenting results ina scientific way
e research using online and offline tools.
Along with the experimental work, these skills are covered in practical skills questions throughout the book.
1.2.2 Use of apparatus and techniques
To meet the requirements for the practical endorsement, you will be assessed in at least 12 practical experiments to enable you to experience a wide range of apparatus and techniques. These practical experiments are incorporated throughout the book in practical application boxes.
This will help to give you the necessary skills to be a competent and effective practical chemist.
Practical Activity Group (PAG) overview and Application features
The PAG labels are opportunities for activities that could count towards the practical endorsement. The table below shows where these PAG references are covered throughout this course.
PAG1—PAG3 and PAGS will be covered in Year 1, PAG6—8 and PAG10—11 in Year 2, and PAG4
and PAGS throughout the two-year course. There are a wide variety of opportunities to assess PAG12 throughout the specification.
Topic reference PAG1 Moles determination 2.1.3(d); 2.1.3[i) 3.2, 3.4, 4.2, 9.2,10.1
Specification reference
PAG2 Acid—base titration 2.1.4{d]
PAG3 Enthalpy determination 3.2.1{e)
'PAG4 Qualitative analysis of ions 3.1.4[ a) 5.3.2(a)
PAGS Synthesis of an organic liquid 4.2.3(a)
PAG6 Synthesis of anorganic solid 6.2.5{a) 6.3.1(a) 28.2, 29.1 PAG? Qualitative analysis of organic functional groups 6.3.1(c) 13.3, 15.1, 26.2, 29.1 | PAG8 Electrochemical cells 5.2.3(g) [23.4 | PAGS Rates of reaction — continuous monitoring method 3.2.2(e) 5.1.1(h ) | 10.1,18.2 PAG10 Rates of reaction — initial rates method 5.1.1[h) 18.3 /PAG14 pH measurement 5.1.3(o) 21.3 | PAG 12 Research skills |
Maths skills and How Science Works across Module 1
Maths is a useful tool for scientists and as you study your course you will learn maths techniques and equations that support the development of your science knowledge.
Each module opener in this book has an overview of the maths skills that relate to the
theory in the chapter. There are also questions using maths skills throughout the book that
will help you practice.
How Science Works skills help you to put science in a wider context, and to develop your
critical and creative thinking skills and help you solve problems in a variety of contexts. How Science Works is embedded throughout this book, particularly in application boxes and practice questions.
You can find further support for maths and how science works on Kerboodle.
MODULE 2
Foundations in Chemistry
Chapters in this module 2 Atoms, ions, and compounds
3 Amount of substance
4 Acids and redox
5 Electrons and bonding
6 Shapes of molecules and intermolecular forces
Introduction
Chemistry is about all matter and the chemical reactions that take place between atoms. This makes chemistry a vast subject, whether you are studying the 34 elements making up a human body, synthesising a new medicine, or discovering nanoparticles for a use yet to be found. This module studies the foundations of chemistry that form the building blocks for all the other modules in your A level course.
Atoms, ions, and compounds looks at some of the essential language of chemistry. You will learn about the atomic masses that you see on your periodic table. You will also learn about the special code of chemistry — the formulae and equations that allow chemists to communicate.
Amount of substance, and its unit the mole, provides chemists with an ability to convert between mass, concentration, and volume to predict how much product can be made ina chemical reaction.
Acids and redox are two important topics. Central to acids is the analysis of solutions by titration using pipettes and burettes the basis of quality testing for many materials, from washing powders to medicines. In redox you learn about oxidation numbers another essential part of the chemists toolkit for describing chemical change.
Electrons and bonding looks at the role of electrons in atoms and in chemical bonding. A good understanding of bonding and structure is essential for all further topics.
Shapes of molecules and intermolecular forces is all about the molecule. You will see how electrons determine the shape and polarity of molecule. You will learn about how intermolecular forces explain many properties of molecular compounds such as why ice floats and why water is a liquid.
Knowledge and understanding checklist From your Key Stage 4 study you should be able to answer the following questions. Work through each point, using your Key Stage 4 notes and the support available on Kerboodle.
C) Recall relative charges and approximate relative masses of protons, neutrons, and electrons.
Calculate numbers of protons, neutrons, and electrons in atoms, given atomic number and mass number.
Write formulae and balanced chemical equations.
Calculate relative formula masses of species separately and in a balanced chemical equation.
Recall that acids react with some metals and with carbonates and write equations predicting products from given reactants.
Describe neutralisation as an acid reacting with an alkali to form a salt
() Use a balanced equation to calculate masses of reactants or products. and water.
Explain reduction and oxidation in terms of gain or loss of electrons, identifying which species are oxidised and which are reduced.
C) Construct dot-and-cross diagrams for simple ionic and covalent substances.
EE Bic 4 See Maths skills checklist
In this module, you will need to use the following maths skills. You can find support for these skills on Kerboodle and through MyMaths.
Le Lh MEE
O Working with standard form and significant figures, and using appropriate units, for carrying out all calculations in this chapter.
le) Changing the subject of an equation, for carrying out structured and unstructured mole calculations.
O Using ratios, fractions, and percentages, for working with moles and equations using ratios, calculating percentage yields, and calculating atom economies.
4S
(4 Finding arithmetic means, for calculating weighted means when determining an atomic mass and when calculating mean titres.
O Using angles and shapes in regular 2-D and 3-D structures, for predicting the shapes of and bond angles in molecule and ions.
SS) MyMaths cou
Bringing |
) ATOMS, IONS, AND ' COMPOUNDS
2.1 Atomic structure and isotopes
Specification reference: 2.1.1
SPOHHHSHHHSESOHHOSSSSOSHOSOOOOSSESOOEOSE,
Learning outcomes
Demonstrate knowledge,
understanding, and application of:
> atomic structure
> isotopes. Prreerrrrerreetrrtrrirerrertirrrirrrii ty ® A third type of subatomic particle, called an electron, occupies
a region outside the nucleus, Electrons are arranged around the nucleus in shells.
Protons, neutrons, and electrons The nuclear atom
At GCSE, you learnt about the nuclear model of the atom (Figure 1).
® The atom consists of a nucleus made up of two types of subatomic particle — protons and neutrons.
proton : Properties of protons, neutrons, and electrons
Mass
Atoms and their subatomic particles have tiny masses. Instead of working in grams, chemists compare the masses of subatomic particles using relative masses (Figure 2).
neutron
4 Figure 1 The nuclear atom ® A proton has virtually the same mass as a neutron.
@ Anelectron has negligible mass, about aagih the mass of a proton. Q re) Accurate measurements show that a neutron has a slightly greater mass x z than a proton, by a factor of 1.001375. This is so close to 1 that chemists 1 neutron I proton —_ usually assume that protons and neutrons have the same mass. eo 3 Charge - ® A proton has a positive charge. 1836 electrons 1 proton
® Anelectron has a negative charge. A Figure 2 Relative masses of protons,
“ ‘ . neutwné ond elections The charge on a proton is equal but opposite to the charge on an
electron. The charges balance. ® A neutron, as its name suggests, is neutral and has no charge. The actual charge on a single proton is tiny: +1.60217733 x 10-'°C (coulombs). The charge on a single electron must balance the charge
on a proton and is —1.60217733 x 107!'°C. It is much easier to use relative charges of 1+ for a proton and 1— for an electron.
Building the atom Table 1 summarises the relative charges and masses of protons, neutrons, and electrons.
® Nearly all of an atom’s mass is in the nucleus.
® Atoms contain the same number of protons as electrons.
ATOMS, IONS, AND COMPOUNDS
¥ Table 1 Charges and masses of some subatomic particles, relative to the proton
Particle Abbreviation Relative charge Relative mass
® The total positive charge from protons is cancelled by the total negative charge from electrons.
@ The overall charge of an atom is zero — an atom is neutral.
Neutrons can be thought of as providing the glue that holds the
nucleus together despite the electrostatic repulsion between its
positively charged protons.
® Most atoms contain the same number of, or slightly more, neutrons than protons.
@ Asthe nucleus gets larger, more and more neutrons are needed.
Atomic number — the identity of an element
The number of protons in an atom identifies the element. As of 2014, the existence of 114 elements has been confirmed, and others have been tentatively reported.
@ Every atom of the same element contains the same number of protons.
@ Different elements contain atoms that have different numbers of protons.
®@ The periodic table lists elements in order of the number of protons in the nucleus. Each element is shown with the number of protons as its atomic number (or proton number).
Figure 4 shows the first 18 elements in the periodic table with their atomic numbers.
A Figure 4 Atomic numbers (proton numbers) for the first 18 elements
Isotopes
Every atom of an element has the same number of protons.
@ Every atom of nitrogen, atomic number 7, contains 7 protons. ® Every atom of oxygen, atomic number 8, contains 8 protons.
@ ... And so on.
76% Cu, 20% Zn, 4% Ni
75% Cu, 25% Ni
A Figure 3 One coin, two pounds, and three elements — next to one another in the periodic table
28 29 30 Ni Cu Zn
¢ all Niatoms have 28 protons ¢ all Cu atoms have 29 protons ¢ all Zn atoms have 30 protons
Study tip
Every periodic table shows each element labelled with its atomic number. You will always have access to a copy of the periodic table. Using the periodic table, you will always be able to work out the number of protons (and electrons) in an atom.
hydrogen deuterium 1 proton 1 proton
0 neutron 1 neutron 1 electron 1 electron
A Figure 5 Two isotopes of hydrogen with the same number of protons but different
numbers of neutrons mass number, A chemical symbol 8
atomic number, Z
A Figure 6 Isotope notation
Study tip
To work out the number of neutrons in an atom, simply subtract the atomic number, Z from the mass number A:
number of neutrons = A —Z
A Figure ? Both glasses contain water. The left-hand glass contains ice cubes made from heavy water. The right-hand glass contains normal ice. Solid D ,0 is denser than liquid water and so D0 ice cubes sink in water
10
fe Her Heavy water
2.1 Atomic structure and isotopes
Unlike protons, the number of neutrons in the atoms of an element can be different, usually within a narrow range.
@ Isotopes are atoms of the same element with different numbers of neutrons and different masses (Figure 5).
® Most elements are made up of a mixture of isotopes.
Representing isotopes Isotopes are represented using the chemical notation shown in Figure 6. ® Mass number (nucleon number) A A = number of protons + number of neutrons @ Atomic number (proton number) Z Z = number of protons
You can use this notation to work out the number of protons, neutrons, and electrons in different isotopes of an element. Table 2 shows the atomic structures of three isotopes of oxygen.
¥ Table 2 Atomic structures for isotopes of oxygen
Isotope Protons, p* Neutrons, n Electrons, e
Other ways of representing isotopes
Chemists refer to isotopes in different ways, so you may see an isotope of oxygen written as pee '6O, or simply as oxygen-16. All oxygen atoms contain eight protons, so if the ‘8’ is omitted, as in '°O and oxygen-16, you still know how many protons the isotope contains.
Isotopes and chemical reactions
Chemical reactions involve the electrons surrounding the nucleus.
® Different isotopes of the same element have the same number of electrons.
@ The number of neutrons has no effect on reactions of an element.
@ Different isotopes of an element therefore react in the same way.
There may be small differences in physical properties — with higher-mass
isotopes of an element having a higher melting point, boiling point, and density — but the chemical reactions are the same.
You may have heard of heavy water, used to control processes in nuclear reactors. The H,0 molecules in normal water nearly all contain the iH isotope of hydrogen. In heavy water, all molecules of H,0 contain the ; H isotope of
ATOMS, IONS, AND COMPOUNDS
hydrogen. The : H isotope is often referred to as deuterium and even given
its own symbol, D. The formula for heavy water is often written simply as D,0. The chemical properties of heavy water are almost identical to those of normal water. However, it has slightly different physical properties, shown in Table 3.
¥ Table 3 Properties of D,0 and H,0
Phusical Normal water Heavu water
property
melting point /°C 0.00 3.80
boiling point /°C 100.00 101.40
density/gcm™? 1.00 1.11
The greater density of D0 gives heavy water its name. If all water were heavy water, you would see ice far more often, as the water would freeze at a higher temperature.
Tritium, T, is a third isotope of water containing two neutrons in the nucleus. Tritium forms an oxide called super-heavy water. a Whatis the relative mass of a molecule of (i) H0, ii D0, and (iii) T,0? b Predict how the melting point, boiling point, and density of 1,0 would be different from H,0 and D.0.
Atomic structure of ions
An ion is a charged atom. The number of electrons is different from the number of protons.
® Positive ions, or cations, are atoms with fewer electrons than protons. Cations have an overall positive charge.
® Negative ions, or anions, are atoms with more electrons than protons. Anions have an overall negative charge.
Ions are always shown with their overall relative charge. Take the two ions *4Mg?* and *°CI-.
® Mg?* has two fewer electrons than protons.
@ Cl has one more electron than protons. Table 4 shows the number of protons, neutrons, and electrons in the ions 24Mg?* and *°CI-. 12 17 VY Table 4 Atomic structures of ions
Protons Neutrons Electrons Overall relative charge
irdeto
lons and atoms of an element have the same number of protons but a different number of electrons.
Summary questions
1 State the number of protons, neutrons, and electrons in the following isotopes:
a 2c (1 mark) b BC (1 mark) ec (1 mark)
2 lron contains a mixture of four different isotopes: 54Fe, 58Fe, 5’Fe, 58Fe. a State how these isotopes
differ. (1 mark) b State the similarity between theseisotopes. (1 mark)
3 State the number of protons, neutrons, and electrons in the following isotopes:
a '5N (1 mark) b ‘Ag (1 mark) erp (1 mark)
4 State the number of protons, neutrons, and electrons in the following ions:
het (1 mark) bas (1 mark) eo. Btr (1 mark)
5 State the difference in the number of protons, neutrons,
or electrons of the following: a ®Liand’Li (1 mark) b '80 and ‘802 (1 mark)
ce %K*and*°Ca** = (1 mark)
2.2 Relative mass
Specification reference: 2.1.1
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> relative isotopic mass and relative atomic mass
> mass spectrometry.
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VY Table 1 Masses of three subatomic particles
Particle Symbol Relative mass
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A Figure 1 Carbon-12 contains six protons (black), six neutrons (blue), and six electrons (red). An atom of carbon-12 has a mass of exactly 12u
Study tip
You need to learn the definition for relative isotopic mass.
This topic looks at how chemists use a mass system based on relative mass to compare the masses of atoms. Later you will see that this idea is extended to all chemicals.
Carbon-12
Table 1 shows the relative mass of protons, neutrons, and electrons.
To find the relative mass of an isotope, it might seem sensible to add together the relative masses of the protons, neutrons, and electrons — but things are not that simple. In fact, the strong nuclear force holding together protons and neutrons comes at the expense of the loss of a fraction of their mass. This astonishing fact was worked out by Albert Einstein over 100 years ago. The small amount of mass lost is called the mass defect. If you are studying A Level physics, you may learn more about the mass defect and its importance to the nucleus.
So how do chemists calculate the mass of atoms if some mass is lost to hold the nucleus together?
First, a standard isotope is needed on which to base all atomic masses. This role is taken by the carbon-12 isotope, which is the international standard for the measurement of atomic masses. One atom of carbon-12 has a mass of 1.992 646 538 x 10°*°kg. Working in kg would be very awkward so instead a new unit called the atomic mass unit u is used.
® The mass of a carbon-12 isotope is defined as exactly 12 atomic mass units (12 u).
@ The standard mass for atomic mass is 1 u, the mass of 75 Sth of an atom of carbon-12.
® On this scale, 1 u is approximately the mass of a proton or a neutron.
Relative isotopic mass
Relative isotopic mass is the mass of an isotope relative to > 12 | th of the
mass of an atom of carbon-12. Table 2 shows the relative isotopic mass of several isotopes relative to the mass of carbon-12. Relative isotopic mass has no units because it is a ratio of two masses.
¥ Table 2 Relative isotopic masses Relative isotopic mass
Isotope Mass number Accurate One decimal place
af
ATOMS, IONS, AND COMPOUNDS
For A Level chemistry, you will be working with masses to one
decimal place. In most cases, you can assume that the relative isotopic Study tp
mass is the same as the mass number A of the isotope (number of You may be provided with relative
protons and number of neutrons) isotopic masses, but if not you can use the mass number (the sum
Relative atomic mass of the numbers of protons and
: . : , ~ neutrons). Most elements contain a mixture of isotopes, each with a different s)
relative isotopic mass. Relative atomic mass A_ is the weighted mean
: ] mass of an atom of an element relative to —th of the mass of an atom : Z . . 12 Study tip of carbon-12.
You should learn the definition for
. > . . ss e *s ae ) ) 43 “ + The weighted mean mass takes account of Polat tami macs.
@ the percentage abundance of each isotope
® the relative isotopic mass of each isotope.
15 16
In the periodic table, in addition to the atomic number, each P Ss element is shown with its relative atomic mass, A,. Figure 2 silicon phosphorus sulfur shows six elements of the periodic table, together with their 23.1 31.0 32.1
atomic numbers and relative atomic masses.
33 34
Determination of relative atomic mass Ge As Se The percentage abundances of the isotopes in a sample BererMan aenie Selenium of an element are found experimentally using a mass | 726 74.9 79.0 spectrometer. A Figure 2 Elements in the periodic table. The
smaller number at the top is the atomic number Z, the larger number underneath is the relative atomic mass A,
Different types of mass spectrometer exist but all work to the same basic principle.
A sample is placed in the mass spectrometer.
The sample is vaporised and then ionised to form positive ions.
3 The ions are accelerated. Heavier ions move more slowly and are more difficult to deflect than lighter ions, so the ions of each isotope are separated.
4 The ions are detected on a mass spectrum as a mass-to-charge ratio m/z. Each ion reaching the detector adds to the signal, so the greater the abundance, the larger the signal.
% - mM relative mass of ion mass-to-charge ratio = = ——————$_———_—_—_ z relative charge on ion A Figure 3 A scientist using a mass For an ion with one positive charge, this ratio is equivalent to spectrometer the relative isotopic mass, which is recorded on the x-axis of
the spectrum.
75.78 Figure 4 shows part of the mass spectrum obtained from a sample of wit . . . oo rt) chlorine, with the percentage abundances for each isotope shown by £8 ° oe each peak. The mass spectrum reveals two isotopes: 4 3 ; 5 ® 75.78% of chlorine-35 % 24.22 @ 24.22% of chlorine-37. 35:37 m/z
A Figure 4 Mass spectrum of chlorine
13
2.2 Relative mass
You can work out the relative atomic mass using the method in the worked example below.
Worked example: Relative atomic mass of chlorine
As accurate relative isotopic masses have not been provided, the mass number for each isotope is used. contribution contribution from *°Cl from >*7Cl
_—— aS 75.78 x 35 + 24.22 x 37 100
Relative atomic mass = 35.4844 = 35.5 to one decimal place
Relative atomic mass =
Summary questions
1 Define the terms
casi Determination of relative isotopic mass a relative isotopic mass
The mass spectrometer can also record the accurate m/z ratio for
(1 mark) ‘ ‘ ‘h isotope so c sO ive isotopic $C & yelatiuw atomic mass aaa isotoy . so that accurate values of relative isotopic mass can re sured. (1 mark) e€ Measurec
2 Calculate the relative atomic mass of the following elements. Give your answers to two decimal places.
a Asample of potassium
+ Relative atomic masses — time for change?
Every two years, the International Union of Pure and Applied Chemistry (IUPAC) reviews values for relative atomic masses for use across the world. The review usually results in some very small adjustments, but the 2011
consisting of 93.20% of **k, review made a more fundamental change. It has been known for many years 0.07% of “°K, and 6.73% that the isotopic abundances of an element may vary slightly depending on of “1K. (1 mark) where the sample originates. In 2011, IUPAC published a new periodic table to
b Asample of antimony take into account this variation by showing the relative atomic mass of some consisting of 56.8? % elements as a range rather than a single value. Figure 4 shows an extract of Sb and 43.13% from this periodic table. Compare the relative atomic masses of silicon and of *3Sb. (1 mark) sulfur in Figure 2 and Figure 5. This change does not really affect the values A sample of neon consisting used at A Level, but atomic masses shown to greater accuracy are affected, of 91.07% of @°Ne and and future reviews may affect other elements. 8.93% of *Ne. (1 mark)
3 The accurate relative isotopic
masses for the isotopes
chlorine-35 and chlorine-3? are
34.968 852 721 69 (°Cl) and 15
36.965 902 621 1 (Cl). cl
a Calculate the relative atomic ls ald Vales il mass of chlorine, as shown 26.98 30.97 (32.05, 32.08] | (35.44, 35.46} in the worked example, A Figure 5 Extract from 2011 IUPAC Periodic Table using these accurate isotopic masses. (2 marks) Boron occurs naturally as a mixture of two isotopes, !°B and 4B. The Comment on whether the relative isotopic masses are 1B 10.00 and "*B 11.00. difference from use of Calculate the percentage abundances by mass of “B and ''B in samples mass numbers is of boron with relative atomic masses of (a) 10.80 and (b) 10.83, the limits significant. (1 mark) of the range in the new IUPAC Periodic Table.
2.3 Formulae and equations
Specification reference: 2.1.2
lonic charges ‘
; , Learning outcomes To study chemistry successfully at any level, you need to be able to write chemical formulae and construct balanced chemical equations. Demonstrate knowledge,
understanding, and application of: > writing formulae of ionic
In this topic, you will review how to write the formula of an ionic compound from ionic charges and how to balance chemical equations.
compounds Simple ions from the periodic table > prediction of ionic charge You should remember from GCSE that many atoms lose or gain from the periodic table
electrons to achieve the same electron structure as the nearest noble gas, helium (He) to radon (Rn).
names and formulae of ions
et PPP OOOOH OSE O HOHE SEO SES OOOO OEEOES,
>
> chemical equations.
® Atoms of metals on the left of the periodic table Jose electrons to Wee eeeecccccesccescecscccessscccrceseress form cations (positive ions)
Synoptic link
® Atoms of non-metals on the right of the periodic table gain
electrons to form anions (negative ions). Electron structure and ionic bonding will be developed later in Topic 5.1, Electron structure and Topic 5.2, lonic bonding and structure.
For many elements you can use the element's position in the periodic table to work out the likely charge on the ion, as shown in Figure 1.
1+ 2+ 3+ 3 2- 1-
+ ions: electrons lost — ions: electrons gained
el i ee pel I ee es ee ae
transition metals A Figure 1 The charges of some simple ions can often be deduced from their position in the periodic table
Some metals, mostly transition metals (Figure 1), can form several Study tip ions with different charges. The ionic charge is then shown with a . e : S You are expected to know the Roman numeral in the name of the ion. For example: : : charges on all the ions shown in ® Copper forms two ions — copper(I), Cu*, and copper(II), Cu?*. Figure 1. Notice that Zn@* and Ag’,
shown in green, do not fit into this pattern and you will need to learn
Binary compounds these ionic charges. A binary compound contains fwo elements only.
® fron forms two ions — iron(II), Fe2*, and iron(II), Fe**.
Synoptic link
® To name a binary compound, use the name of the first element but change the ending of the second element's name to -ide. You will learn more about using
® For ionic compounds, the metal ion always comes first. Roman numerals in names in Topic 4.3, Redox.
For example, sodium and oxygen form sodium oxide.
2.3 Formulae and equations
Polyatomic ions
Sometimes, an ion may contain atoms of more than one element bonded together. These ions are called polyatomic ions. Table | shows some common polyatomic ions and their names.
Y Table 1 Common polyatomic ions and their names
1-
ammonium hydroxide carbonate phosphate OH C0.° PO,?
3
A Figure 2 Cu** ions are responsible for the blue colour of copper(II) sulfate crystals, CuSO, nitrate sulfate
NO, s0,° Study tip nitrite sulfite
You are expected to know the NO. a names and formulae of the ions shown in blue in Table 1, but you will find it useful to learn them all. Be warned — there is no easy way manganate(VIl) to work out these formulae!
hydrogencarbonate | dichromate(V!)
HCO,” Cr,0,*
(permanganate) Mn0,-
Writing formulae from ions
An ionic compound contains a cation and an anion. The formula can be worked out from the charge on each ion.
In a correct formula: ®@ the overall charge is zero so the ionic charges must balance
@ sum ol positive charges = sum of negative charges.
Worked example: lonic formulae
Compound name Ions present Balance charges Formula zinc chloride Zn?* and Cl- 1 Zn* ions balances 2 Cl- ions ZnCl, aluminium sulfate Al** and $O,?- 2 Al** ions balance 3 SO,?- ions Al,(SO,),
The charges must balance, but it’s just a matter of multiplication tables.
1 Zn** (1 x 2+ = 2+) is balanced by 2 Cl- (2 x 1- = 2-) 2 Al’ (2 x 3+ = 6+) is balanced by 3 SO,?> (3 x 2- = 6-)
Writing the formula ® The number of each ion present is shown as a subscript after the ion.
®@ The ionic charges are usually omitted in the completed formula. ® Brackets are used if there is more than one polyatomic ion.
Aluminium sulfate contains 2 AP* ions and 3 0 i ions and so the formula is Al,(SO,),.
ATOMS, IONS, AND COMPOUNDS
Writing equations
You will have practised balancing equations at GCSE. For A Level chemistry, you will come across many more equations, but you will find that balancing them quickly becomes second nature. By the end of the course you should be able to write equations for all the reactions you have studied and also for some unfamiliar reactions.
Representing elements and compounds in equations
Elements
In equations, elements are shown simply as their symbol except
for the few elements that exist as small molecules. Most of these elements exist as diatomic molecules, containing two atoms bonded together — H,, N,, O,, F,, Cl,, Br,, and L,. The only other elements that exist as small molecules are phosphorus, P,, and sulfur, S.. (However, it is normal practice to write sulfur simply as S in equations — otherwise every formula in the equation has to be multiplied up by a factor of 8.)
Compounds
Covalent compounds do not contain ions. Most covalent compounds exist as molecules with a small number of atoms bonded together, for example, CO, and H,O. In equations, the formula of the molecule is used.
For ionic compounds, the formula worked out from the ionic charges is used in equations. This is called the formula unit.
State symbols in chemical equations
State symbols are shown in brackets after a formula to indicate the physical state. There are four state symbols:
© (g) —gas
@ (il) —-liquid
@ (s) —solid ®
(aq) — dissolved in water (aqueous)
Balancing equations
To balance an equation, you multiply each formula by a balancing number until the number of atoms of each element is the same on each side of the equation. Balancing numbers are written in front of each formula.
2 Na,O means two Na,O formula units giving 2 x 2 = 4 Na* and 2 x 1 =2 O* ions.
® When balancing an equation you must not change any chemical formula.
® Balancing numbers go in front of chemical formulae and on the line (not subscripted).
® The equation is balanced when there are the same number of atoms of each element on each side of the equation.
Study tip
Remember to use brackets. Students often lose marks by omitting brackets when writing the formula of a hudroxide.
Magnesium hydroxide contains Mg** and OH’ ions.
Mg0OH, is incorrect as it means one oxygen and two hydrogen atoms! The correct formula is Mg(OH),.
Brackets are added if there is more than one polyatomic ion in a formula.
Synoptic link
You will find out more about these different formulae in Topic 3.2, Determination of formulae.
Study tip
When balancing equations, it is a common mistake to change a formula.
The formula for water is H,0. Do NOT change the formula to H,0.,, this is hydrogen peroxide!
3 2.3 Formulae and equations
Study tip
Aluminium oxide is an ionic compound, so you will need to use ionic charges.
Study tip
Aluminium is shown simply as its symbol, Al.
Oxygen exists as diatomic molecules and so is shown as 0..
Study tip
Start with the formulae of compounds and count the number of atoms of each element.
Study tip
You are allowed to use fractions when balancing equations. If you are comfortable using fractions, this can be easier than whole numbers. For the worked example, this would give
1 2Al +150, — Al,0,
Study tip
Don't forget to add state symbols!
Worked example: Constructing a balanced equation Aluminium reacts with oxygen to form aluminium oxide. Step 1: Work out the formulae.
Aluminium oxide contains Al’* and O?- ions, so the formula is Al,O,.
Step 2: Write an equation using formulae for all reactants and products.
Formulae only: Al + 0, — Al,O,
Step 3: Balance the equation by placing balancing numbers on the line, in front of formulae.
The key here is to get the oxygen atoms equal on both sides.
4Al + 30, — 2A1,0,
As a final check,
@ Left-hand side has 4 Al and 3 x 2 (= 6) O
@ Right-hand side has 2 x Al,O, =2 x 2 (= 4) Aland 2 x 3 (= 6) O The equation is balanced.
Step 4: Finally, add state symbols to complete the equation. 4Al(s) + 30,(g) — 2Al,0,(s)
Balancing formulae with brackets
Take care when balancing formulae with brackets. Remember that the balancing number multiplies the entire formula.
3 Zn(NO,), means 3 x Zn(NO,), This gives 3 Zn, 3 x 2=6N, and3x3x2=180
Summary questions
1 Write formulae for the following:
a potassium oxide
b magnesium iodide c calcium phosphide d_iron(IIl) hydroxide e f
3 Balance the following equations: (1 mark) a NH,(g) + 0,(aq) — NO(aq) + H,0(!) (1 mark) (1 mark) b C.H,,(g) + 0,(g) — CO,(g) + H,0(I) (1 mark) (1 mark) c Al,0,(s) +H,P0,(aq) — Al,(PO,),(aq) + H,0(\)
(1 mark) (1 mark) ammonium carbonate (1 mark) d Zn(s) + HNO,(aq) — Zn(NO,),(aq) + NO(g) + H,0(!) iron(II) nitrate (1 mark) (1 mark) 2 Name each compound from the formula. 4 Write balanced equations with state symbols for the a AIN (1 mark) following reactions: b (NH,),PO, (1 mark) a Magnesium reacts with solid phosphorus c Fe,(SO,), (1 mark) to form solid magnesium phosphide. (2 marks)
b Iron reacts with aqueous copper(I) nitrate to
form copper and aqueous iron(II!) nitrate. (2 marks) c Lead(|I) nitrate decomposes to form solid
lead(II) oxide and two gases, nitrogen
dioxide and oxygen. (2 marks)
Practice questions
1
This question refers to species A-D. For each part, select the correct species.
A slp B 19R- G 1692- D 23Naqt 15 9 8 1!
a The number of protons and neutrons are the same. (1 mark)
b The number of neutrons and electrons are the same. (1 mark)
¢ The number of protons, neutrons, and electrons are all different. (1 mark)
The answer to each part of this question is a number.
a How many neutrons are in an atom of zinc-68? (1 mark)
b What is the total number of electrons in a CO,” ion? (1 mark) c What is the total number of ions is one formula unit of chromium(III) sulfate? (1 mark)
This question looks at isotopes of three elements.
a_ An isotope of element A contains the same number of neutrons as are found in an atom of *!V. The isotope of A also contains 26 protons.
(i) How many protons in an atom of >!v? (1 mark)
(ii) Write the symbol, including the mass number and the atomic number, of this isotope of A. (2 marks)
b_ An isotope of element B has half as many protons and half as many neutrons as an atom of *Ti.
Write the symbol, including the mass number and the atomic number, of this isotope of B. (2 marks)
c An isotope of element C has three more protons and four more neutrons than an atom of §'Br.
Write the symbol, including the mass number and the atomic number, of this isotope of C. (2 marks)
Neon exists as a mixture of isotopes.
a What is meant be the term isotopes? (1 mark)
b_ Define the term relative isotopic mass. (2 marks)
Species
ATOMS, IONS, AND COMPOUNDS
c A sample of gallium, A, = 69.7, was analysed and was found to consist of 65% ©°Ga and one other isotope.
Determine the mass number of the other isotope in the sample of gallium. (J mark)
d Complete the table below for two ions that have the same number of electrons as a neon atom. (2 marks)
Protons Neutrons’ Electrons
Charge
A sample of sulfur, Z = 16, was analysed in a mass spectrometer to give the following composition of isotopes.
Isotope Abundance (%)
From the results, the relative atomic mass of the sulfur sample can be calculated.
a Define the term relative atomic mass. (3 marks)
b= Calculate the relative atomic mass of the sample of sulfur. Give your answer to two decimal places. (2 marks)
c Complete the table to show the number of sub-atomic particles in a **S atom and an *4S2 jon. (1 mark)
Protons Neutrons Electrons
Write equations, with state symbols, for the following reactions.
a Magnesium reacts with nitrogen to form magnesium nitride. (2 marks)
b- Calcium reacts with water to form a solution of calcium hydroxide and hydrogen. (2 marks)
¢ Sodium hydroxide solution reacts with iron(III) sulfate solution to form an iron(II) hydroxide precipitate and sodium sulfate solution. (2 marks)
AMOUNT OF SUBSTANCE
Amount of substance and the mole
Specification reference: 2.1.3
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Learning outcomes
Demonstrate knowledge, understanding, and application of:
> the amountof substance, the mole, and the Avogadro constant
> molarmass
v
calculations involving masses and moles.
(Pee RRR RRR RRRE EERE EEE
Synoptic link
You learnt about carbon-12 in Topic 2.1, Relative mass.
Study tip The mass of 1 mole of atoms of an
element equals the relative atomic mass in grams
A Figure 1 Molar quantities for chemical elements, clockwise from top left
* 12.0gcarbon,C
* 32.1 g sulfur, S
© 55.8 iron, Fe
* 63.5 g copper, Cu
© 24.3 g magnesium, Mg.
Each sample contains the same number of atoms, but their masses
are different
Counting and weighing atoms
Chemicals are usually measured by mass or volume. Because reactions take place on an atomic scale, chemists need a method for converting a measured mass or volume into the actual number of particles involved in reactions.
Amount of substance and the mole
Chemists use a quantity called amount of substance 7 to count the number of particles in a substance, measured in a unit called the mole mo/. One mole is the amount of a substance that contains 6.02 x 107? particles. The Avogadro constant N, is 6.02 10?* mol", the number of particles in each mole of carbon-12.
The choice of 6.02 x 107? particles per mole may seem strange, but
is directly linked to the mass of carbon-12, the standard for the measurement of relative atomic masses. 12 g of carbon-12 contains 6.02 x 107* atoms. You can easily find the mass of one mole (1 mol) of atoms of any element — it is the relative atomic mass in grams.
One mole of carbon, C, atoms has a mass of 12.0g One mole of hydrogen, H, atoms has a mass of 1.0g
One mole of magnesium, Mg, atoms has a mass of 24.32
One mole of iron, Fe, atoms has a mass of 55.8¢
So if you have a sample of an element and know its mass, you now have a way of knowing the number of atoms. This is a very important idea for chemistry — it offers an easy way of counting something that cannot be seen, just by measuring the mass.
The Avogadro constant — amazingly large
It is difficult for us to comprehend the size of very large numbers. If 6.02 x 107? (1 mol) pennies were shared evenly between all humans on Earth, every person could spend £1m every hour for their whole life.
Particles matter
Amount of substance and moles can refer to anything, not just atoms. When you work in moles, it is important to use the formula or unambiguous name of a substance for clarity.
® | mol of H: 1 mol of hydrogen atoms
® 1 mol of H,: 1 mol of hydrogen molecules
AMOUNT OF SUBSTANCE
Molar mass
Molar mass, M, gives a convenient way of linking moles with mass for any chemical substance.
@ M(C) = 12.0 gmol"'!.
@ M(NO,) =14.0+ 16.0 x 2 = 46.0 gmol"'.
@ M(Na,CO,) = 23.0 x 2+ 12.0 + 16.0 x 3 = 106.0 gmol"!. Molar mass gives the mass in grams in each mole of the substance. ® Molar mass is the mass per mole of a substance.
® The units of molar mass are gmol!"!.
Amount of substance 7, mass m1, and molar mass M are linked by the equation below.
amount #7 = a ee or more simply, n= 3 Study tip m n= is a key equation for working out n, m, or M. Learn n = 47 and make sure that M you are comfortable rearranging the equation so that you can work Worked example: Amount of substance, mass, and Glas RMIT Log alld le other two. You may find this format molar mass useful to remember: 1 Calculate the amount of substance, in moles, in 96.0g of carbon, C. he SO foals das 8.0 mol
2 Calculate the mass, in g, of 0.050 mol of NO,.
rearrange n= ae tom=nx M=0.050 x 46.0 = 2.3g
If you cover n, you are left with i: If you cover m, you are left
3 Calculate the molar mass when 2.65 g contains 0.025 mol of a
saa in 265 with n x a If you cover M, you are = Ls . =—_ = = seh) n=, Tearrange: M 0.025 106.0g mol left with a Summary questions 1 Calculate the amount of substance, in mol, in the 2 Calculate the mass, in g, of the following. Use relative following. Use relative atomic masses to one decimal atomic masses to one decimal place. place. a 280 mol BeO (1 mark) a 6.00gHF (1 mark) b 0.150 mol HNO, (1 mark) b 220gN,0 (1 mark) c 0.0500 mol HPO, (1 mark) c 114gCr,0, (1 mark) = d_ 1.25 10° mol Na,CO, (1 mark) d 0.0150 gC,H,,0, (1 mark) e 4.55x 10-?mol Ca(NO,), (1 mark) -2 e 3.45x 10°g Ca(OH), (1 mark) 3 Calculate the molar mass, in g mol‘, of the following substances. Use relative atomic masses to one decimal place. a 5.00 molAhas a mass of 140g (1 mark) b 0.125 mol B has a mass of 9.25 g (1 mark)
ce 4.50x 10°? molC has amass of 3.825 (1 mark)
‘
3.2 Determination of formulae
NY > fj
Specification reference: 2.1.1, 2.1.3
CPPS OHSSEHOSSEHOSEHOSESHOOOSESOSEEOOSOS,
Learning outcomes
Demonstrate knowledge, understanding, and application of:
empirical and molecular formula formula determination hydrated salts
VuUvS
practical techniques for measuring mass.
VY Table 1 Examples of compounds with their empirical and molecular formulae
Molecular Empirical formula formula
A Figure 1 Adiamond crystal and the giant structure of carbon atoms in a diamond
Synoptic link
You learnt about relative isotopic mass, relative atomic mass, the important role of carbon-12 in
atomic mass measurements in Topic 2.2, Relative atomic mass.
Chemical formulae
In this topic, you will see how to use the results from chemical experiments to work out a chemical formula.
Molecular formulae
Some compounds are made up of small units called molecules — two or more atoms held together by covalent bonds. The molecular formula is the number of atoms of each element in a molecule.
In Topic 2.3, Formulae and equations, you looked at the elements that exist as molecules. In equations, these elements are shown as their molecular formulae — H,, N,, O,, F,, Cl,, Br,, 1, Py, and S,.
Many compounds also exist as molecules, and again the molecular formula is used in equations.
Empirical formulae
The empirical formula is the simplest whole-number ratio of atoms of each element in a compound.
The empirical formula is important for substances that do not exist as molecules. This includes metals, some non-metals (e.g., carbon, C, and silicon, Si), and ionic compounds (e.g., sodium chloride, NaCl).
These substances form giant crystalline structures of atoms or ions.
It would be impossible to base a formula on the actual number of atoms or ions — the numbers would go into billions of billions and would vary depending on the size of the crystals.
The empirical formula is the ratio of atoms or ions in the structure and will always be the same.
Figures | and 2 show giant crystalline structures of carbon and sodium chloride. You will find out more about these structures later in the course.
More relative masses
Some compounds exist as simple molecules (e.g., water, H,O, and carbon dioxide, CO,). Other compounds exist as giant crystalline structures (e.g., all ionic compounds). Two terms are needed for relative mass, one for simple molecules and another for giant structures.
Relative molecular mass
Relative molecular mass M, compares the mass of a molecule with the mass of an atom of carbon-12. You can easily calculate a relative molecular mass by adding together the relative atomic masses of the elements making up a molecule. The examples below show how to work out the relative molecular mass of molecules of water, H,O, carbon dioxide, CO,, and glucose, C,H, ,0,.
AMOUNT OF SUBSTANCE
®@ M,(H,O) = (1.0 x 2) + 16.0 = 18.0
@ M,(CH,) = 12.0 + (1.0 x 4) = 16.0
® MAC,H,,0,) = (12.0 x 6) + (1.0 x 12) + (16.0 x 6) = 180.0 Relative formula mass
Relative formula mass compares the mass of a formula unit with the mass of an atom of carbon-12. It is calculated by adding together
the relative atomic masses of the elements in the empirical formula, as shown in the examples below.
@ NaCl = 23.0 + 35.5 = 58.5 ® Ca(NO,), = 40.1 + (14.0 + 16.0 x 3) x 2 = 164.1
chloride ion, Cl~
sodium ion, Na*
Finding formulae by experiment
You can predict the formula of an ionic compound from its ions, but
if you do not know which ions are in a compound, the formula can A Figure 2 Rock salt (sodium chloride) be worked out from the results of experiments. Investigating the crystals and the giant structure with one chemical composition of a substance is called analysis. sodium ion for every chloride ion
These worked examples show two common ways to calculate empirical and molecular formulae from experimental mass readings. Notice the central role of the mole in these calculations.
Worked example: Empirical formula from mass
Study tip In an experiment, 1.203 g of calcium combines with 2.13 g of Stud : chlorine to form a compound [A,: Ca, 40.1; Cl, 35.5]. ss SUS SOPENNES ve round ratios of moles that appear Step 1: Convert mass into moles of atoms using 1 = a7 quite close to a whole-number n(Ca) = 1,203 _ 0.030 mol n(Cl) = 2.13 _ 0.060 mol ratio. If you calculate a ratio of 40.1 35.5 1: 1.67, don’t be tempted to round Step 2: To find the smallest whole-number ratio, divide by the it up to 1: 2. Instead see whether smallest whole number. you can convert the ratio into a n(Ca):n(Cl) = rh ee =3+2 whole-number ratio by multiplying -030 0.030 both sides by the same factor, in Step 3: Write the empirical formula: CaCl, this case 3 to give 3:5.
Worked example: Determination of a molecular formula
Chemical analysis of a compound gave the percentage composition by mass C: 40.00%; H: 6.67%; O: 53.33% |[A,: C, 12.0; H, 1.0; O, 16.0]. The relative molecular mass of the compound is 180.0.
Step 1: Convert % by mass into moles of atoms using 1 = + _ 40.00 _ _ 6.67 _ _ 53.33 _ n(C) = 70 3.33 mol n(H) = ao 6.67 mol n(O) = 20" 3.33 mol Step 2: Find smallest whole-number ratio and empirical formula. ; _ 3:33 6:67. 3.33 = - e n(C) : n(H):n(O) = 333 °333°3337 ee] empirical formula = CH,O
Step 3: Write the relative mass of the empirical formula CH,O: 12.0 + 1.0 x 2 + 16.0 = 30.0 180 30.0 Step 5: Write the molecular formula: CH,O x 6 = C,H,,0,
=6
Step 4: Find number of CH,O units in one molecule:
A Figure 4 Hydrated (blue) and anyhydrous (white) copper(Il) sulfate
3.2 Determination of formulae
Hydrated salts
Many coloured crystals are hydrated — water molecules are part
of their crystalline structure. This water is known as water of crystallisation. When blue crystals of hydrated copper(II) sulfate are heated, bonds holding the water within the crystal are broken and the water is driven off, leaving behind white anhydrous copper(II)
sulfate. The equation below represents the change when water is removed. The water of crystallisation is shown in the formula of hydrated copper(II) sulfate with a large dot ¢ between the compound
in Figure 4.
Formula of a hydrated salt
The method below describes how you could carry out an experiment to determine the water of crystallisation in hydrated crystals, The calculation is similar to the method described for an empirical formula.
The experiment uses hydrated copper(II) sulfate but the method would be suitable for any hydrated salt.
Step 1: Weigh an empty crucible.
Step 2: Add the hydrated salt into the weighed crucible. Weigh the crucible and the hydrated salt.
Step 3: Using a pipe-clay triangle, support the crucible containing the hydrated salt on a tripod (Figure 5). Heat the crucible and contents gently for about one minute. Then heat it strongly for a further three minutes.
Step 4: Leave the crucible to cool. Then weigh the crucible and anhydrous salt.
crucible hydrated copper(I) sulfate pipe-clay triangle
tripod
Bunsen burner
A Figure 5 Apparatus for heating crucible
formula and the five water units.
CuSO,°5H,0(s) ao hydrated
CuSO, (s) + anyhdrous
5H,O(1)
Without water, the crystalline structure is lost and a white powder remains. It is difficult to remove the last traces of water, as you can see from the very pale blue colour of the anhydrous copper(II) sulfate
The results are shown below.
mass of crucible /g 18.742 | — ReadingA
mass of crucible + hydrated salt /g 28.726 | — Reading B
mass of crucible +
anhydrous salt/g — Reading C
Step 1: Calculate the amount, in mol, of anhydrous CuSO,.
mass m of CuSO, formed = C — A= 25.126 — 18.742 = 6.384g n(Cuso,
Step 2: Calculate the mass and amount, in mol, of water.
mass m of H,0 formed = B — C = 28.726 — 25.126 = 3.6002
3.600
n(H,0) = 18.0 ~ 0.200 mol
Step 3: Find the smallest whole-number ratio. n(CuSO,):n(H,0) = 0.0400: 0.200 = 1:5
Step 4: Write down the value of x and the formula of hydrated copper sulfate. x = 5 so the formula is CuSO,,°5H,0
Use the student results below to determine the value of x and the formula of CoCl,°xH.0.
mass ofcrucible/g | _mass of crucible + hydrated salt/g
"mass of crucible + anhydrous salt/g | 17906
AMOUNT OF SUBSTANCE
How accurate is an experimental formula? S | ynoptic link
The application above gives a perfect formula for hydrated copper(II) sulfate. Some assumptions have been made, and real experiments may not always work out as well.
As part of the practical skills required for your course, you need to know how to measure mass, Assumption 1 — All of the water has been lost volumes of solutions, and volumes If the hydrated and anhydrous forms have different colours, you can of gases.
be fairly sure when all water has been removed. However you only see
age This practical application box tells the surface of the crystals and some water could be left inside. If the
: ania oi you how to measure masses. hydrated and anhydrous forms are similar colours, it is not as easy. A good solution is to heat to constant mass — the crystals are reheated Measuring mass, volumes of repeatedly until the mass of the residue no longer changes, suggesting solutions, and volumes of gases
that all water has been removed. are also covered in:
@ Topic 3.4, Reacting quantities, for how to measure the volumes of gases
Assumption 2 — No further decomposition
Many salts decompose further when heated; for example, if heated
very strongly, copper(II) sulfate decomposes to form black copper(II)
oxide. This can be very difficult to judge if there is no colour change. Topic 4.2, Acid—base titrations, for how to measure volumes of
? solutions summa ry questions Topic 9.2, Measuring enthalpy
‘ ; ; changes, for how to measure 1 a Determine the relative molecular mass of the following: (3 marks) fages i SO ii P , Topic 10.1, Reaction rates, nia, for how to measure volumes iti HCIO, : b Determine the relative formula mass of the following: (3 marks) ih i MgBr, ii NH,NO,
ili Al,(SO,),
2 Anickel compound has the formula Ni(NO,),*xH,0 and a molar mass of 290.7 g mol~!. Calculate the value of x. (3 marks)
3 Determine all possible molecular formulae and M, values less than 120 for the empirical formula CH,. (1 mark)
4 Astudent carried out an experiment to calculate the water of crystallisation of two hydrated salts. For the first salt, the student did not remove all the water of crystallisation. For the second salt, the student removed all the water of crystallisation, but unfortunately the salt decomposed further.
For each salt, explain whether the student's calculated value of x, the number of water molecules in the formula unit, would be greater or smaller than the actual value of x in data books. (4 marks)
,
\
3.3 Moles and volumes
\
Y Ly Specification reference: 2.1.3
Learning outcomes
Demonstrate knowledge,
understanding, and application of:
> concentration, solution volumes, and the mole
Using volume for measuring amount of substance
Liquids and gases are measured by volume. As with mass, the volume of a solution or a gas can be converted into amount of substance, in moles, giving us a way to count the particles present.
The volume measurements commonly used in chemistry are:
* ©» een es vows fran i eee . Das > molar gas volumes @ the cubic centimetre (cm’) or millilitre (ml): | cm?’ = | ml
® the cubic decimetre (dm?) or litre (1): 1 dm? = 1000cm? = 1000ml = 1 litre, 11.
> the ideal gas equation.
ai
A Figure 1 Laboratory glassware A 1moldm~ solution contains 1 mol of solute dissolved in each 1 dm? of solution.
You will be expected to use cm? and dm’. In practical work, you will use glassware graduated in ml and |, and you should record these readings in cm’ and dm’.
Moles and solutions
To work out the amount, in moles, of a measured volume of the solution, you need to know the concentration (moldm~*) of the solute (the dissolved compound). The concentration of a solution is the amount of solute, in moles, dissolved in each 1 dm? (1000cm>) of solution.
Converting between moles and solution volumes
For a solution, amount 7 (mol) and volume V (dm*) are linked by the concentration ¢ (moldm=?).
Study tip
You will use these equations many times, so you will need to learn them.
weex ¥
You will usually measure volumes in cm?’ so you will need to convert into dm? by dividing by 1000. The equation then becomes:
Worked example: Converting between solution volumes and moles
1 Calculate the amount of NaCl, in mol, in 30.0cm? of a 2.00 mol dm~? solution.
espe Ve fem? ie 30.0 _ n(NaCl) =c¢x 5000 2.00 x 1000 0.0600 mol
2 Calculate the volume of a 0.160moldm™? solution that contains 3.25 x 107> mol of NaCl.
peng vin 50, V= 1000 x n _ 1000 x 3.25 x 107 1000 ¢ 0.160
= 20.3cm?
AMOUNT OF SUBSTANCE
Standard solutions s ynoptic link
A standard solution is a solution of known concentration. In practical work, you will have seen bottles of standard solutions labelled with their concentration, often as | moldm~?.
Standard solutions and their
preparation are discussed in detail in Topic 4.2, Acid—base titrations.
Standard solutions are prepared by dissolving an exact mass of the solute in a solvent and making up the solution to an exact volume. Using your understanding of the mole, you can work out the mass required to prepare a standard solution.
Worked example: Standard solutions
Calculate the mass of Na,CO, required to prepare 100cm? of a 0.250 moldm~? standard solution.
Step 1: First work out the amount in moles required.
ay Vin cm?) & 100 _ n(Na,CO,) =c x 7000 0.250 x 1000 0.0250 mol
Step 2: Then work out the molar mass of Na,CO,. M(Na,CO,) = 23.0 x 2 + 12.0 + 16.0 x 3 = 106.0gmol"!
Step 3: Rearrange n = a
m=n x M=0.0250 x 106.0 = 2.65¢
to calculate the mass of Na,CO, required.
Other ways of showing concentrations
You will often see mass concentrations with units of gdm~’. For the solution of Na,CO, in the worked example above, the concentration is 0.250 moldm~>. To work out the mass concentration, you need to convert between moles and grams.
eo n= a so, m=n x M=0,250 x 106.0 = 26.502
® mass concentration of Na,CO, = 26.5gdm">.
Moles and gas volumes
In Topic 3.1, you saw how to convert between mass in grams and amount of substance in moles. It is difficult to measure the mass of a gas but easy to measure gas volumes.
At the same temperature and pressure, equal volumes of different gases contain the same number of molecules.
So when you measure a gas volume, you are indirectly counting the number of gas molecules (or the amount of gas molecules in moles).
Molar volume The molar gas volume V,, is the volume per mole of gas molecules at a stated temperature and pressure.
The volume of a gas depends on the pressure and temperature, but many experiments are carried out at room temperature and pressure (RTP).
3.3 Moles and volumes
A Figure 2 Ahelium weather balloon is released by a meteorologist. A sensor attached to the balloon will measure ozone distribution in and beyond the ozone layer, which is 20-30 km above the Earth's surface. So helium balloons are far less dense than air and they are able to rise up to the ozone layer
® RTP is about 20°C and 101 kPa (1 atm) pressure
@ AtRTP. | mole of gas molecules has a volume of approximately 24.0dm?* = 24000cm*.
® Therefore, at RTP, the molar gas volume = 24.0dm’ mol".
Converting between amount in moles and gas volumes Using the following equation, you can convert between the amount in moles of a gas, and the volume of the gas, V.
volume V molar gas volume V,, At RTP, V,, = 24.0dm’ mol', so
3 @® whenVisindn’ n= “i
wate V (cm? @ when Vis in on? gue or) saint *="34000
amount #1 (mol) =
Worked example: Converting between gas volume and amount in moles
1 Calculate the amount (mol) of hydrogen, H,(g), in 480 cm? at RTP.
V (in cm? 480 n= oe = 34000 = 0.0200 mol of H, 2 Calculate the volume, in dm’, of 0.150mol O,(g) at RTP. Rearrange to give V (dm*) =n x 24.0 so, V= 0.150 x 24.0 = 3.60dm?
The ideal gas equation
Room temperature and pressure will always be approximate, chosen to match the typical conditions that experiments are carried out in. So what do you do when carrying out experiments where the gases are at different temperatures or pressures, or if you need to be more accurate? The ideal gas equation provides a solution.
You will have come across the following assumptions for the molecules making up an ideal gas:
® random motion
e elastic collisions
® negligible size
® no intermolecular forces.
The ideal gas equation is shown below. pV = nRT
The ideal gas constant R is a pV = nRT constant and always has the same value of 8.314J mol! K~!. Temperature is in units of K (Kelvin), which starts at absolute zero (-273°C). Each 1 K rise in temperature is the same as a 1 °C rise in temperature.
ideal gas constant = 8.31) mol? kK!
AMOUNT OF SUBSTANCE
Synoptic link
If you have not met the Kelvin
scale of temperature before, you can find out more detail in Topic
V = nRT+*—\———— temperature 9.2, Measuring enthalpy changes. (K) pressure amount of gas (Pa) molecules (mol)
As long as you know three of p, V, n, and T, you can always find out the unknown variable using the ideal gas equation.
Before using the ideal gas equation, you need to convert any quantities into the correct units Pa, K, and m’, The conversions from measurements likely to be made when carrying out experiments are shown below.
® cm’ tom x 10° ® dm’ tom? x 10° ® °CtoK + 273 @ kPato Pa x 10°
Worked example: What is room temperature and pressure? You can use the ideal gas equation to find out the conditions that give a molar gas volume of 24.0dm?’ mol!. Assume that the pressure is | atm = 101 kPa and use this to calculate room temperature.
Study tip
The hardest part of calculations using the ideal gas equation,
Step 1: Convert all quantities to match the ideal gas equation. pV =nFRT, is making sure that you
p=101kPa =101x10*Pa are working in units of Pa, m?, and V=24.0dm? =24.0x107°m? K. Learn the conversion rules. n=I1mol
T = unknown
Step 2: Use the ideal gas equation to calculate the unknown.
pV = nRT rearranges to, T= Zt 3 —3 p= L101 x 10°) x (24.0 x 10™) _ 595 K = 19°C
1 x 8.314
Many people make the incorrect assumption that room temperature is 25°C (298K), the temperature often regarded as standard temperature for chemistry. Using the ideal gas equation, you can show that the molar gas volume at 25°C and an atmospheric pressure of 101 kPa is actually equal to 24.5dm’ mol"'!.
3.3 Moles and volumes
rN Finding a relative molecular mass
You can use the ideal gas equation to find the relative molecular mass of a volatile liquid. Using the method below, the unknown compound would need to be a liquid at room temperature but have a boiling point below 100°C so that it vaporises.
1 Add a sample of the volatile liquid to a small syringe via a needle. Weigh the small syringe.
Inject the sample into a gas syringe through the self-sealing rubber cap (Figure 3). Reweigh the small syringe to find the mass of the volatile liquid added to the gas syringe. Place the gas syringe ina = gas syringe boiling water bath at 100°C, ois coating as shown in Figure 3. The rubber cap boiling water liquid vaporises producing a gas. The pressure is recorded. A Figure 3 Results mass of volatile liquid =0.2245¢ volume of gas in gas syringe = 81.0cm?
atmospheric pressure = 100kPa
Follow the steps below to calculate the relative molecular mass of the volatile liquid.
Step 1: Convert all quantities to match the ideal gas equation. V=81.0cm? =81.0x 10°°m? T= 100°C = 100 + 273K=373K p=100kPa =100x10'Pa n=unknown
Use the ideal gas equation to calculate the unknown.
V
pV=nkT rearrangesto, n= “>
= (100 x 10°) x (81.0 x 10~°) - n 9314x373 0.00261 mol of X
Find the molar mass.
___massm molar mass M
_m _ 0.2245 _ “4 =n 0.00261 86.0 g mol
relative molecular mass, M. = 86.0
A0.320 g sample of a volatile liquid was heated until it vaporised. The resulting vapour then occupied 61.5 cm? at 101 kPa and 100°C.
Calculate the relative molecular mass of the volatile liquid.
AMOUNT OF SUBSTANCE
+ Real gases
The ideal gas equation relies on two key assumptions: © forces between molecules are negligible © gas molecules have negligible size compared to the size of their container.
These assumptions hold at low pressures and high temperatures when the gas molecules are far apart and moving fast.
When gas molecules are close together, the volume of the molecules compared with the volume of the container starts to become significant. Also if gas molecules move comparatively slowly, they have less energy and intermolecular forces may become significant.
Scientists have developed several improvements to the ideal gas equation for real gases. In the real gas equation, corrections have been made to take into account the volume of gas molecules and intermolecular forces.
2 real gas equation: (p + ra (V—nb) =nRT
accounts for accounts for volume intermolecular forces of gas molecules
Predict the conditions of pressure and temperature that cause the ideal gas equation to break down. Explain your answer.
Summary questions
1 Calculate the amount of substance, in moles, in: a 250cm? of a 1.00 mol dm? solution (1 mark) b 10.0cm? of a0.200 mol dm solution. (2 marks)
2 Calculate the concentration, in g dm~?, for:
a 2.00 mol of NaOH in 4.00 dm? of solution (2 marks)
b 0.500 mol of HNO, in 200 cm’ of solution. (2 marks) 3 Calculate the amount of substance, in mol, in the following gas
volumes at RTP:
a 1440dm?0,(g) (2 marks)
b 720cm?He(g) (2 marks)
c 34.0cm?H,(g) (2 marks) 4 Calculate is the volume of one mole of a gas at:
a 10°C and 100kPa (2 marks)
b 35°C and 92.0kPa (2 marks) 5 Calculate the volume in cm? at RTP of:
a 0.136 gNH,(g) (2 marks)
b 0.088 gC0.(g) (2 marks)
c 0.0175 gN.(g) (2 marks)
6 0.1565 g of X occupies 80.0 cm? at 101 kPa and 100°C. Calculate the relative molecular mass of X. (3 marks)
,
q
3.4 Reacting quantities
Y \w Ay, Specification reference: 2.1.3
, Stoichiometry Learning outcomes In a balanced equation, the balancing numbers give the ratio of Demonstrate knowledge, the amount, in moles, of each substance. This ratio is called the
understanding, and application of: > stoichiometry > quantities of reactants and
stoichiometry of the reaction.
equation 2H (g) + O,(g) —+ 2H,O(1)
products from equations amount 2 mol Imol — 2mol > percentage yield Chemists use balanced equations to find: > atom economy ® the quantities of reactants required to prepare a requried quantity > practical techniques for of a product
measuring the volume of @ the quantities of products that should be formed from certain
a gas. quantities of reactants.
These quantities can then be changed to adjust the scale of a preparation.
Quantities from amounts and equations
The two worked examples show how unknown information about a substance can be obtained using amounts and an equation together. Each example follows the same basic method:
Study tip
® Step 1: Work out the amount in moles of whatever you can. For most problems, steps 1 and 2
will follow the same method. The processing for step 3 can vary.
® Step 2: Use the equation to work out the amount in moles of the unknown chemical.
® Step 3: Work out the unknown information required.
Worked example: Reacting masses
Calculate the mass of aluminium oxide, Al,O,, formed when 8.10g of aluminium completely reacts with oxygen.
Step 1: Calculate the amount, in moles, of Al that reacts.
n{Al) = M = 37.0 = 0.300 mol
: Step 2: Use the equation to find the amount of Al,O,, in moles, Study Up that forms. You need the ratios: equation 4 Al(s) + 30,(g) — 2A1,0,(s) 4 mol Al — 2 mol Al,0, moles 4mol + 3mol — 2mol Then halve the moles of Al to get amounts 0.300mol — 0.150mol the moles of Al,0.: Step 3: Calculate the mass of Al,O, formed. 0.300 mol Al — 0.150 mol Al,0, n(Al,O,) = so, m =n x M=0.150 x (27.0 x 2 + 16.0 x 3)
= 0.150 x 102.0 = 15.3g
Worked example: Reacting mass, gas volumes, and concentration
0.552 of lithium reacts with water to form 125cm? of a solution of lithium hydroxide and hydrogen gas. Calculate
the concentration of the lithium hydroxide and the volume of hydrogen formed at room temperature and pressure (RTP).
Step 1: Calculate the amount, in mol, of lithium that reacts.
st 0552 n(Li) = i eo. 0.0800 mol Step 2: Use the equation to find the amounts of LiOH and H,
formed. equation 2Li(s) + 2H,O(l) — 2LIOH(aq) + H,(g) moles 2mol amounts 0.0800 mol
— 2mol + Imol — 0.0800 mol 0.0400 mol
Step 3: Calculate the concentration of LiOH(aq) and volume of H,(g) in cm? at RTP. . V (in cm? H = V (in cm’) n(LiOH) =c x 1000
_ 1000 xm _ 1000 x 0.0800 SS oe
S08 125
= 0.640 moldm=?
H _ V(in cm’) n(H>) = “54.000
so, V=n x 24000 = 0.0400 x 240000 = 960cm?
Identifying an unknown metal
The method below shows how you could carry out an experiment to identify an unknown Group 2 metal X. The results can then be analysed using the set method for reacting quantities.
1 Set up the apparatus shown in Figure 1.
2 Weigh a sample of the metal and add to the flask.
gas syringe
Using a measuring cylinder, add 25.0 cm? 1.0 mol dm~? HCI(aq) (an excess) to the flask and
quickly replace the bung. dilute
hydrochloric acid Measure the maximum volume of gas in the syringe. A Figure 1 Apparatus for determination of an unknown metal
unknown metal
AMOUNT OF SUBSTANCE
Synoptic link
As part of the practical skills required for your course, you need to know how to measure mass, volumes of solutions, and volumes of gases.
This practical application box tells you how to measure the volumes of gases.
Measuring mass, volumes of solutions, and volumes of gases are also covered in:
@ Topic 3.2, Determination of formulae, for how to measure mass
Topic 4.2, Acid—base titrations, for how to measure volumes of solutions
Topic 9.2, Measuring enthalpy changes, for how to measure mass
Topic 10.1, Reaction rates, for how to measure volumes of gases.
3.4 Reacting quantities
Results mass of unknown metal = 0.14 ¢ volume of H, collected = 84cm?
Step 1: From the experimental results, the amount of H.,(g) can be calculated. _ V(in cm’) Assuming RTP, n(H,) = 54000 = - ah. 24000 = 0.00350 mol
From the equation, and the result from Step 1, the amount of metal X can be determined.
X(s) + 2HCI(aq) — XCI,(aq) +H,(g) 1mol imol 0.00350 mol 0.00350 mol
Work out the unknown information. You now know the amount, in moles, and mass of X, so you are nearly there!
n(X) =a
so, M(X) = ay 0.14
~ 0.00350 =40gmol
From the periodic table, Ca has a relative atomic mass of 40.1.
The unknown metal X is calcium
The experiment was repeated with another Group 2 metal. 0.064 g produced 63 cm? of hydrogen. Analyse the results to identify the metal.
Percentage yield
So far, all our calculations have assumed that a// of the reactants are converted into products. This maximum possible amount of product is called the theoretical yield. Unfortunately this is difficult to achieve for several reasons, including:
® the reaction may not have gone to completion
® other reactions (side reactions) may have taken place alongside the main reaction
@ purification of the product may result in loss of some product.
The actual yield obtained from a reaction is usually lower than the theoretical yield. The conversion of starting materials into a desired product is expressed by the percentage yield.
actual yield
——_—_————_. x 100% theoretical yield e -
percentage yield =
AMOUNT OF SUBSTANCE
Worked example: Percentage yield
1.15g of sodium reacts with an excess of chlorine, forming 1.872 g of sodium chloride. What is the percentage yield of sodium chloride?
Step 1: Calculate the amount, in moles, of Na that reacts.
m_ 115 n(Na) = Ma 230 = 0.0500 mol Step 2: Use the equation to find the theoretical yield of NaCl, in moles. equation 2 Na(s) + Cl,(g) — 2 NaCl(s) moles 2mol — 2mol amounts 0.0500 mol — 0.0500 mol
Step 3: Calculate the actual yield of NaCl, in moles.
n(NaCl) = “A _ 1.872 58.5 = 0.0320 mol
Step 4: Calculate the percentage yield of NaCl. actual yield
% yield = + —_ x 100 % i theoretical yield ‘ _ 0.0320 rey = 00500 * 100 = 64.0% The limiting reagent
In the example above, you have used two reactants with one reactant in excess. The reactant that is of in excess will be completely used up first and stop the reaction — it is called the limiting reagent.
If you do not know which reactant is in excess, you need to find out by working out the amount in moles of each reactant and comparing with the equation. Calculations must be based on the limiting reagent.
For example, when hydrogen and oxygen gases react to form water, 2 mol of hydrogen are required for every | mol of oxygen: 2H,(g) + O,(g) + 2H,O(1)
2mol lmol
If equal amounts of hydrogen and oxygen are allowed to react, hydrogen will be used up first, and half the oxygen will be unreacted. The limiting reagent is hydrogen, so calculations must be based on hydrogen.
Atom economy
The atom economy of a chemical reaction is a measure of how well atoms have been utilised.
Study tip
Atom economy is worked out from the balanced equation. Unlike percentage yield, no experimental results are needed.
Study tip
Remember to take account of
the balancing numbers when accounting for the sum of all molar masses.
3.4 Reacting quantities
Reactions with high atom economies:
@ produce a large proportion of desired products and few unwanted waste products
® are important for sustainability as they make the best use of natural resources.
Atom economy is based solely on the balanced chemical equation for a reaction and assumes a 100% yield. sum of molar masses of desired products
x 100% sum of molar masses of a// products
atom economy =
The idea of atom economy has been developed alongside awareness of dwindling finite resources and environmental concerns about processing or disposing of harmful waste. Improving atom economy makes industrial processes more efficient, preserves raw materials, and reduces waste. In an ideal chemical process, a use would be found for all products and thus the atom economy would be 100%.
Worked example: Atom economy
Hydrogen is an important raw material and is produced from the reaction of carbon with steam.
What is the atom economy of this reaction?
Step 1: Write the equation and the molar masses of the products. C(s) + 2H,O(g) — 2H,(g) + CO,(g) —+ 2x2.0 12.0+16.0x2=44.0 Step 2: Calculate the atom economy. sum of molar masses of desired products sum of molar masses of a// products
2 x 2.0 iS a— 2x20 ___ = 8.3% TEPER nae EW eas
atom economy = x 100
How sustainable?
The worked example reveals a poor atom economy, especially as the calculation assumes a 100% yield of hydrogen — the overall yield will be even worse. Furthermore, the undesired product is carbon dioxide, one of the gases that causes global warming. So how sustainable is this process?
Atom economy only provides part of the answer.
@ The process uses reactants that are readily available, carbon from coal and steam from water. Energy will be needed to produce the steam, but costs for obtaining starting materials are low.
@ Other reactions may have a much larger atom economy but poor percentage yields. Efficiency will depend on both factors.
AMOUNT OF SUBSTANCE
Summary questions
1 a Balance the equation
Ca(s) + 0,(g) — Ca0(s). (1 mark) b State the masses of Ca and 0, that completely react
together to form 4.488 g Ca0. (2 marks) c Calculate the volume of 0,(g) at RTP that reacts
with 2.80? g Ca. (2 marks)
2 Calculate the atom economy for the following industrial processes. a NH, production:
N, + 3H, — 2NH, (2 marks) b C,H.OH production: C.H,,.0, — 2C,H.OH + 2C0, (2 marks) 3 35g of HF was prepared by reacting 112 g of CaF, with an excess of H,SO,; CaF, +H,SO, — 2HF + CaSO, Calculate the percentage yield of HF and the atom economy of this process. (3 marks) 4 a Balance the equation: SO,(g) + 0,(g) — S0,(g) (1 mark) b State the volumes of SO, and 0. at RTP, that would produce 180 cm? of SO,. (1 mark) c 150cm?SO0,(g) and 100 cm? 0,(g) react together. (2 marks)
i State which reactant is in excess. ii State the volume of SO,(g) that could form using these quantities.
5 Hydrogen can be prepared from the reaction of methane in natural gas with steam. The other product is carbon monoxide. a Write an equation for this reaction and calculate the atom economy. (2 marks) b 100g of methane react with an excess of steam, forming 324 dm? of hydrogen at RTP.
Calculate the percentage yield of hydrogen. (1 mark) 6 a Balance the equation: Cr(s) + HCI(aq) — CrCl,(aq) + H,(g) (1 mark) b Calculate the volume of H, at RTP formed by the complete reaction of 1.17 g Cr with excess HCI(aq). (2 marks)
c¢ The volume of HCI(aq) used is 150 cm?. What is the minimum concentration of HCI(aq) needed to react with all of the Cr? (2 marks)
? 0.054 g of an unknown metal X reacts with an excess of sulfuric acid to form 72 cm? of H,(g). The equation is: 2X(aq) + 3H,SO, — X,(SO,), + 3H,(g) Identify metal X. (3 marks)
Chapter 3 Practice questions
Practice questions
1 A compound has the percentage composition by mass: Ca, 30.35%; N, 21.20; 48.45% What is the empirical formula of the compound?
2 When heated, potassium chlorate(VII), KCIO,(s) decomposes to form KCl(s) and 108cm? O,(g), measured at RTP. The unbalanced equation is shown below. KCIO,(s) — KCl(s) + O,(g)
a_ Balance the equation. (1 mark) b What is the amount, in mol, of O, produced? (1 mark)
¢ What is the mass of KCIO, required to produce 108cm? of O, at RTP? Show your working (3 marks)
3 1.893g of hydrated zinc sulfate, ZnSO,°xH,O is heated to remove all water of crystallisation. The mass of anhydrous ZnSO, formed is 1.061 g.
What is the formula of the hydrated zinc sulfate?
Show your working (4 marks)
4 On heating, sodium hydrogencarbonate, NaHCO, decomposes, forming 2.48 g of sodium carbonate, Na,CO,.
2NaHCO, — Na,CO, + CO, + H,O
The percentage yield of Na,CO, is 65%.
What is the mass of NaHCO, that was heated? Show your working (5 marks)
5 Iron ore contains iron(III) oxide. In a blast furnace, the iron(II) oxide reacts with carbon monoxide to form iron and carbon dioxide.
a Write an equation for this reaction. (1 mark)
b_ Each day, a blast furnace typically reacts 10000 tonnes of Fe,O,.
Calculate the typical mass of iron produced each day from a blast furnace. (3 marks)
6 A nitrogen fertiliser, A, has the composition by mass Na, 27.1%; N, 16.5%; O, 56.4%.
On heating, 3.40 g of A decomposes into sodium nitrite, NaNO,, and oxygen gas.
a_ Calculate the empirical formula of A.
(2 marks)
b Write an equation for the decomposition
of A. (1 mark) c Calculate the volume of oxygen gas formed at RTP. (4 marks)
A chemist reacts 0.0250 mol of sodium metal with water, forming 50.0 cm? of sodium hydroxide solution and hydrogen gas.
a What mass of sodium was reacted? (1 mark)
b= Write an equation, including state symbols, for the reaction. (1 mark)
¢c Calculate the volume of hydrogen gas formed at RTP. (2 marks)
d_ Calculate the concentration of sodium hydroxide solution formed in (i) moldm~’; (ii) gdm~? (3 marks)
Chlorine gas is prepared by the electrolysis of brine, a concentrated solution of sodium chloride in water.
2NaCl + 2H,O — 2NaOH + Cl, +H,
The concentration of brine can be assumed to be 4.00 moldm~>.
a Calculate the mass of sodium chloride dissolved in 250 cm? of brine. (1 mark)
b_ Each day, the UK produces 2.5 x 10° dm? of chlorine gas, at RTP, from brine.
Calculate the volume of brine required for chlorine production each day in the UK. (3 marks)
a An organic compound X, contains carbon, hydrogen and oxygen only. Compound X has the following percentage composition by mass: 54.55%, C; 9.09% H.
(i) Calculate the empirical formula of X. (2 marks) (ii) In an experiment, 0.2103 g of a vaporised sample of X was shown to occupy 72.0cm? at 100°C and
103.0 kPa. Calculate the relative molecular mass of X. (4 marks)
(iii) Deduce the molecular formula for X. (1 mark)
10 Hydrated aluminium sulfate, Al,(SO,),°xH,O,
and chlorine, 6 are used in water treatment.
a_ A student attempts to prepare hydrated
lla
aluminium sulfate by the following method.
The student heats dilute sulfuric acid with an excess of solid aluminium oxide.
The student filters off the excess aluminium oxide to obtain a colourless solution of A/,(SO,),.
(i) State the formulae of the two main ions present in the solution of Al,(SO,),. (2 marks)
(ii) Write an equation for the reaction of aluminium oxide, A/,O,, with sulfuric acid. Include state symbols.
(2 marks)
(iii) What does ‘exH,O’ represent in the formula A/,(SO,),°xH,O? (1 mark)
(iv) The student heats 12.606 g of Al,(SO,),°xH,O crystals to constant mass. The anhydrous aluminium sulfate formed has a mass of 6.846g. Use the student's results to calculate the value of x. The molar mass of Al,(SO,), = 342.3gmol"'. (3 marks)
F321 June 2013 (2a)
Borax, Na,B,O0,°10H,O, can be used to
determine the concentration of acids such
as dilute hydrochloric acid.
A student prepares 250.cm? of a
0.0800 mol dm~? solution of borax in
water in a volumetric flask.
Calculate the mass of borax crystals,
Na,B,O,°10H,O, needed to make up
250cm? of 0.0800 moldm~> solution.
(3 marks)
The student found that 22.50cm? of
0.0800 moldm~* Na,B,O, reacted with
25.00cm? of dilute hydrochloric acid.
Na,B,O, + 2HCI + 5H,O — 2NaCl + 4H,BO,
(i) Calculate the amount, in mol, of
Na,B,O. used. (1 mark) (ii) Calculate the amount, in mol, of HCI used. (1 mark)
(iii) Calculate the concentration, in moldm~?, of HCl. (1 mark) F321 Jan 2013 5(c)(d)
12 Lithium carbonate, Li,CO, is added to an excess of dilute hydrochloric acid.
Li,CO,(s) + 2HCI(g) — 2LiCl(g) + CO,(g) + H,O(1)
13
AMOUNT OF SUBSTANCE
A student adds 1.845 g Li,CO, to 125cm?’ of 0.500 moldm~* HCl. The volume of the solution formed is 125¢cm?’.
a_ Predict two observations that you would expect to see during this reaction. (2 marks)
b- Explain what is meant by 0.500 moldm~* HCl. (1 mark)
c (i) Calculate the amount, in moles, of Li,CO, and HCI that were reacted. (3 marks)
(ii) Calculate the amount, in mol of HCl that was in excess. (2 marks)
d (i) Calculate the volume of CO,(g) that would be expected, measured at RTP. (I mark)
(ii) Suggest why the volume of CO, produced at RTP is likely to be less than your answer to (ii). (1 mark)
e Calculate the concentration, in moldm=?, of LiCl in the solution formed. (1 mark)
a_ A factory makes ethyne gas, C,H, from calcium carbide, CaC,. One of the waste products is calcium hydroxide.
CaC, + 2H,O — Ca(OH), + C,H,
Each day 1.00 x 10° grams of calcium
carbide are used and 3.60 x 10°dm? of
ethyne gas, measured at room temperature
and pressure, is manufactured.
(i) Calculate the atom economy for this process using the relative formula masses in the table below. — (2 marks)
Compound _ Relative formula mass
=)
(ii) Calculate the amount, in moles, of CaC, used each day. (1 mark)
(iii) Calculate the amount, in moles, of C,H, made each day. (1 mark) (iv) Calculate the percentage yield of C,H,. (1 mark)
(v) Comment on the percentage yield and the atom economy of this process in terms of sustainability. (2 marks)
F322 Jun 2010 I(e)
ACIDS AND REDOX 4.1
ee
Learning outcomes
Demonstrate knowledge,
understanding, and application of:
> acids and bases
=> neutralisation.
. Peete eee eee eee eeeet
VY Table 1 Common acids
Acid Formula
Synoptic link
hydrochloric acid HCI sulfuric acid H,SO, nitric acid HNO,
— a | ethanoic acid (vinegar) | CH,COOH
You will learn more and equilibrium and weak acids in Chapter 20, Acids,
bases, and pH.
A Figure 1 Some common acids —
vinegar contains ethanoic acid, orange and lemon juice and even sink cleaner contains citric acid, and apples contain
malic acid
Acids, bases, and neutralisation
Specification reference: 2.1.4
Acids
All acids contain hydrogen in their formulae (Table 1). When dissolved in water, an acid releases hydrogen ions as protons, H*, into the solution. In the equation below, hydrogen chloride gas releases H* ions as it dissolves in water.
HCl(g) + aq — H*(aq) + Cl-(aq)
In this equation + aq has been included to show that an excess of water is present. The equation is essentially hydrogen chloride gas dissolving to form an aqueous solution.
Strong and weak acids
A strong acid, such as hydrochloric acid, HCI, releases all its hydrogen atoms into solution as H* ions and completely dissociates in aqueous solution.
HCl(aq) — H*(aq) + Cl-(aq)
A weak acid, such as ethanoic acid, CH,COOH, only releases a small proportion of its available hydrogen atoms into solution as H* ions. A weak acid partially dissociates in aqueous solution.
CH,COOH(aq) = H*(aq) + CH,COO™(aq) The equilibrium sign = indicates that the forward reaction is incomplete.
It is important to realise that not all compounds that contain hydrogen atoms are acids. Each molecule of ethanoic acid contains four hydrogen atoms, but only the hydrogen atom on the COOH group is released as H*. Even then, only about one molecule in every hundred dissociates, so ethanoic acid is a weak acid. Most organic acids, like ethanoic acid, are weak acids.
Bases and alkalis
Metal oxides, metal hydroxides, metal carbonates, and ammonia, NH,, are classified as bases. A base neutralises an acid to form a salt. Table 2 and Figure 2 show some common bases.
An alkali is a base that dissolves in water releasing hydroxide ions (OH-) into the solution. The equation below shows the alkali sodium hydroxide releasing hydroxide ions as it dissolves in water.
NaOH(s) + aq — Na*(aq) + OH™~(aq)
YV Table 2 Common bases
Metal carbonates Alkalis
Metal oxides
Neutralisation
In neutralisation of an acid, H*(aq) ions react with a base to form a salt and neutral water. The H* ions from the acid are replaced by metal or ammonium ions from the base, Table 3 shows salts of common acids.
® Notice the link between the name of the acid and the salt, in blue.
® To form the salt, the hydrogen (shown in purple) in the acid is replaced by a metal or ammonium ion to form the salt.
VY Table 3 Acids and their salts Acid Salt
Name Formula Type Name Formula
oat ae fee on ia
c(h, ethanoate
Neutralisation of acids with metal oxides and hydroxides
An acid is neutralised by a metal oxide or metal hydroxide to form a salt and water only. The equations below show the neutralisation of sulfuric acid and hydrochloric acid by copper(II) oxide to form a salt and water only. Figure 3 shows solutions of the salts formed.
nitric acid
ethanoic acid (vinegar)
CuO(s) + H,SO,(aq) — CuSO,(aq) + H,O(1) CuO(s) + 2HCl(aq) + CuCl,(aq) + H,O(1)
Alkalis With alkalis, the reactants are in solution. As with metal oxides, the overall reaction forms a salt and water only:
acid + alkali — salt + water
The ionic equation, shown below, is much simpler than the overall equation: neutralisation of H*(aq) ions by OH~(aq) ions to form neutral water, H,O(1).
Full equation: — HCl(aq) + NaOH(aq) — NaCl(aq) + H,O(1)
lonic equation: H*(aq) + OH-(aq) — H,O(1)
ACIDS AND REDOX
A Figure 2 Three bases — calcium carbonate, CaC0.,, copper oxide, CuO, and sodium hydroxide, NaQH
Study tip
The reactions for the neutralisation of an acid by a metal oxide or hydroxide are all essentially the same:
acid + metal oxide/hydroxide — salt + water
Learn this to help you write the equation for any neutralisation by a metal oxide or hydroxide.
A Figure 3 Solutions of two copper salts prepared by neutralisation of two different acids with the base copper(I!) oxide: sulfuric acid has been neutralised to form blue copper(II) sulfate, CuSO, (left) © hydrochloric acid has been neutralised to form green copper(I!) chloride, CuCl,
(right)
4.1 Acids, bases, and neutralisation
Neutralisation of acids with carbonates
Like metal oxides, carbonates neutralise acids to form a salt and water. There is also a third product, carbon dioxide gas. The equations show neutralisation of two carbonates by sulfuric acid and hydrochloric acid.
Study tip
Remember the equation for the neutralisation of acids by metal carbonates.
ZnCO,(s) + H,SO,(aq) — ZnSO,(aq) + H,O(1) + CO,(g)
acid + metal carbonate — MgCO,(s) + 2HCl(aq) — MgCl,(aq) + H,O(1) + CO,(g)
salt + water + carbon dioxide(g)
It will help you write the equation
for any neutralisation of this type. + Dissociation in sulfuric acid
Sulfuric acid, H,SO,, is a strong acid, but this is true only for one of the two hydrogen atoms. When sulfuric acid is mixed with water each H,S0,, molecule dissociates, releasing just one of its two hydrogen atoms as an H™ ion: H,S0,,(aq) — H*(aq) + HSO, (aq) The resulting HSO,, (aq) ions then only partially dissociate: HSO,,-(aq) = H*(aq) + S0,°-(aq) © Sulfuric acid first behaves as a strong acid, e The HSO,” ions formed behave as a weak acid.
Other strong acids containing more than one hydrogen atom behave similarly.
Write equations to show the dissociation of the three hydrogen atoms in phosphoric acid, H,PO,.
Summary questions
1 Explain what is meant by the following terms. 4 Baking powder usually contains an organic acid and
a strong acid sodium hydrogencarbonate, NaHCO.,, also known as
b weak acid bicarbonate of soda. Sodium hydrogencarbonate is an acid salt, formed by partial neutralisation of carbonic acid, H,CO.,, with sodium hydroxide. a Write equations for the partial and complete
(1 mark) (1 mark)
2 Write equations to show the dissociation of the following acids when dissolved in water.
aeolian wire nN i Stone sen) le nantes) neutralisation reactions of carbonic acid b propanoic acid, CH,CH,COOH é : : ta weakacid) (2 marks) with sodium hydroxide. (4 marks) b Why is sodium hydrogencarbonate called 3 Write equations for the following neutralisation an acid salt? (1 mark) reactions. For each reaction, name the salt formed. c Baking powder is used to make cakes. When it is
a The reaction of MgO(s) with hydrochloric acid mixed into a batter or dough containing water and (2 marks) heated, the sodium hydrogencarbonate reacts
b The reaction of NaQH(aq) with sulfuric acid with the acid in the powder in the same way as a (2 marks) carbonate. The small quantity of baking powder
c¢ The reaction of ZnCO,(s) with nitric acid (2 marks) d The reaction of aqueous sodium hydroxide
with ethanoic acid (2 marks)
does not affect the taste, so why is it used? (2 marks)
4.2 Acid—base titrations
Specification reference: 2.1.4, 2.1.3
Titrations
A titration is a technique used to accurately measure the volume of one solution that reacts exactly with another solution. Titrations can be used for:
® finding the concentration of a solution e identification of unknown chemicals
® finding the purity of a substance.
Checking purity is an important aspect of quality control, especially for compounds manufactured for human use such as medicines, food, and cosmetics. It is essential that pharmaceuticals have a high level
of purity — just a tiny amount of an impurity in a drug could cause
a great deal of harm to a patient.
Preparing a standard solution
A standard solution is a solution of known concentration.
A volumetric flask is used to make up a standard solution very accurately. Volumetric flasks are manufactured in various sizes and can measure volumes very precisely. The volumetric flasks that you will use are manufactured to the typical tolerances below:
® a 100cm? volumetric flask: +0.20cm?
® a 250cm’ volumetric flask: £0.30cm?.
Preparing standard solutions
The solid is first weighed accurately.
The solid is dissolved in a beaker using less distilled water than will be needed to fill the volumetric flask to the mark.
This solution is transferred to a volumetric flask. The last traces of the solution are rinsed into the flask with distilled water.
The flask is carefully filled to the graduation line by adding distilled water a drop at a time until the bottom of the meniscus lines up exactly with the mark (Figure 1), Care at this stage is essential — if too much water
is added, the solution will be too dilute and must be prepared again. You should view the graduation mark and meniscus at eye level for accuracy. Finally, the volumetric flask is slowly inverted several times to mix the solution thoroughly. If this stage is omitted, titration results are unlikely to be consistent. You will be able to see the solution mixing when you invert the flask as the more dense original solution moves through the solution.
Explain the effect on the titre of the following errors. 1 The flask is filled with water above the graduation line. 2 The flask is not inverted.
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> preparation of a standard solution
> carrying outa titration
> analysing titration results by calculation
> practical techniques for
measuring the volume of solutions.
Synoptic link
Standard solutions were introduced in Topic 3.3, Moles and volumes.
A Figure 1 The volumetric flask is filled so that bottom of the meniscus just touches the graduation line
A Figure 2 The volumetric flask is slowly inverted several times to ensure that the solution is mixed evenly
43
A Figure 3 When filling a burette, run excess solution out through the tap to remove any air bubbles. If a bubble is left in the neck of the burette, the air could be released during the titration, leading to an error in the titre
A Figure 4 Take burette readings from the bottom of the meniscus, with your eye is at the level of meniscus. Read the burette to the nearest 0.05 cm?, This burette reading is 25.55 cm?
4.2 Acid—base titrations
Acid—base titrations Apparatus
In an acid-base titration, a solution of an acid is titrated against a solution of a base using a pipette and a burette, which are typically manufactured to the tolerances below:
® a 10cm’ pipette: £0.04cm?
® a 25cm’ pipette: +0.06cm*
® a 50cm? burette: +0.10cm’.
A burette reading is recorded to the nearest half division, with the bottom of the meniscus on a mark or between two marks. Each burette reading is measured to the nearest +0.05 cm? so the reading
always has two decimal places, the last place being either 0 or 5, for example, 25.40cm? or 26.25cm?.
The acid—base titration procedure
Add a measured volume of one solution to a conical flask using a pipette. Add the other solution to a burette, and record the initial burette reading to the nearest 0.05 cm?, Add a few drops of an indicator to the solution in the conical flask. Run the solution in the burette into the solution in the conical flask, swirling the conical flask throughout to mix the two solutions, Eventually the indicator changes colour at the end point of the titration. The end point is used to indicate the volume of one solution that exactly reacts with the volume of the second solution. Record the final burette reading. The volume of solution added from the burette is called the titre, which is calculated by subtracting the initial from the final burette reading. A quick, trial titration is carried out first to find the approximate titre. The titration is then repeated accurately, adding the solution dropwise as the end point is approached. Further titrations are carried out until two accurate titres are concordant — agreeing to within 0.10 cm’.
The readings from the titration are recorded in a table such as Table 1.
¥ Table 1 Table for recording titration results
final burette reading /cm?
initial burette reading / cm?
titre /cm?
mean titre /cm?
Explain the effect on the titre of the following errors.
1 The pipette has an air bubble inside. 2 The burette readings are taken from the top, rather than the bottom, of the meniscus.
ACIDS AND REDOX
The mean titre ; ; ee ee ; Synoptic link When working out the mean titre, it is important to use only your closest accurate titres. As part of the practical skills
required for your course, you need to know how to measure mass, volumes of solutions, and volumes of gases.
@ By repeating titres until two agree within 0.10cm’, you can reject inaccurate titres, ® If you were to include all the titres in the mean, you have lost the accuracy of the titration technique. The practical application boxes in
Titration calculations this topic tell you how to measure
From the results of a titration, you will know the following: the volume of solutions.
Measuring mass, volumes of solutions, and volumes of gases are also covered in:
@ both the concentration c, and the reacting volume V, of one of the solutions @ only the reacting volume V, of the other solution. @ Topic 3.2, Determination of formulae, for how to measure Step 1: Work out the amount, in mol, of the solute in the solution for mass which you know both the concentration ¢, and volume V,. Topic 3.4, Reacting quantities, for how to measure the volumes of gases
The method for analysing the results follows a set pattern.
Step 2: Use the equation to work out the amount, in mol, of the solute in the other solution.
Topic 9.2, Measuring enthalpy
changes, for how to measure
mass
Step 3: Work out the unknown information about the solute in the other solution.
Topic 10.1, Reaction rates, for
Worked example: Determination of an unknown eee : gases. concentration | Pipette: 25.00. cm? of 0.100 moldm~* KOH(aq)
Mean titre from burette: 25.70cm? of H,SO,(aq) Study tip
This method is essentially the
same as the one used to calculate Step 1: From the titration results, calculate the amount of KOH. unknown quantities using the mole
25.00 in Topic 3.4, Reacting quantities.
Unknown information: The concentration of H,SO,(aq)
serpent 2a a n( KOH) =c x 1000 = 0.100 x 1000 = 0.002 50 mol Step 2: From the equation and Step 1, determine the amount of
Bae ae Study tip
2 KOH(aq) + H,SO,(aq) + K,SO,(aq) + 2 H,0(1)
It’s all done with the ratios of the 2mol 1 mol (balancing numbers) balancing numbers and moles.
0.00250 mol 0.00125 mol
Step 3: Work out the unknown information. 3 n(H,SO,) = cx ion)
1000 _ 1000 x 2 _ 1000 x 0.00125 _ a fo a ee 0.0486 moldm
concentration of H,SO,(aq) is 0.0486 mol dm *
Study tip
You might be given a structured calculation — you will be helped through the calculation with prompts similar to the bullet points in this example.
But you might also be given an unstructured calculation with no help. You should practise both structured titration calculations and the harder unstructured problems.
Study tip
The scaling here is obvious because 250 is 10 times larger than 25. If you were scaling up a burette reading (say 29.55 cm’), the process is the same, but you would need to find the scaling
factor by dividing 250 by the titre.
4.2 Acid—base titrations
Identification of a carbonate
You can use the titration technique to work out some unknown
information about a substance, for example, you can identify an
unknown carbonate, X,CO.,.
The key steps are shown below, together with results. Prepare a solution of an unknown carbonate, X,CO., in a volumetric flask. Using a pipette, measure 25.00 cm? of your prepared solution into a conical flask. Using a burette, titrate this solution using 0.100 mol dm~? hydrochloric acid. Analyse your results to identify the carbonate.
Mass measurements
mass of weighing bottle /g
mass of weighing bottle + X,CO,/g
mass of X,CO,/g
Titration readings
final burette reading /cm? : 22.45
initial burette reading /cm? ro 00 titre /cm?
mean titre /cm?
Analysis Step 1: Calculate the amount of HCI that reacted. Use the mean titre V and — concentration of the hydrochloric acid c.
22.40 n(HCl) =¢ x a 5005 = 0.100 x SF<- = 0.002 24 mol
Step 2: Determine the amount of X,CO, that reacted. Use the equation and n(HCl). X,CO,(aq) + 2HCI(aq) — X,S0,(aq) + H,0(!) + CO,(g) 1 mol 2 mol (balancing numbers) 0.00112mol 0.00224 molHCI Step 3: Work out the unknown information. There are several stages.
1 Scale up to find the amount of X,CO, in the 250 cm? solution that you prepared, n(X,CO,) in 25.00 cm? used in the titration = 0.001 12 mol n(X,CO,) in 250.0 cm? solution = 0.001 12 x 10 = 0.0112 mol Find the molar mass of X,CO.. Use the amount, n(X,CO,), in the 250 cm? solution and the mass m of X,CO. used to prepare this solution.
= 106.25 gmol'
ACIDS AND REDOX Lk
3 Finally use M(X,CO.) to identify X in the formula X,CO.,. 106.25 = M(X) x 2 + 12.0 + (16.0 x 3) = 2M(X) + 60.0 M(x) = 208.25 — 60.0 _ 23.125 gmol-*
From the periodic table, 23.125 most closely matches Na (A_ = 23.0).
Unknown carbonate is sodium carbonate, Na,CO.,.
Use the results to determine the molar mass of an unknown acid HA. Mass readings: For preparation of a 250.0. cm? solution of HA
mass of weighing bottle/g 9.64 | mass of weighing bottle + HA/g } 12.51 |
Titration readings: Titration of 25.0 cm? volumes of the solution of HA with 0.0600 mol dm? Na,CO,
Iria final burette reading/cm? WE initial burette reading /cm? 2.00 | 0.00
Summary questions
a 25.00cm? of 0.110 mol dm~? NaQH(aq) reacts exactly with 23.30 cm? of HNO. (aq). (5 marks) i Calculate the amount, in moles, of NaOH(aq) used. ii Write the equation for the reaction and calculate the amount of HNO. (aq) in the titre. iii Calculate the concentration of the HNO,,(aq) solution. b 25.00 cm? of 0.125 mol dm~? KOH(aq) reacts exactly with 26.60 cm? of H,SO,,(aq). Write the equation and find the unknown concentration. (3 marks)
2 1.96 of an unknown hydroxide X(0H), was dissolved in water, and the solution was made up to 250.00 cm? in a volumetric flask. Ina titration, 25.00 cm? of this solution of X(OH), required 21.20 cm? of 0.250 mol dm~? HCI(aq) to reach the end point.
Equation: 2HCI(aq) + X(OH).(aq) — XCI,(aq) + 2H,0(I) Identify the unknown hydroxide X(OH),. (4 marks)
3 1.654 g of a hydrated acid H,C,0,exH,0 was dissolved in water and the solution was made up to 250. 00 ontik in a volumetric flask. 23.80 cm? of this solution required 25.00 cm? of 0.100 mol dm~? NaOH(aq) to reach the end point.
Equation: H,C,0,,(aq) + 2NaOH(aq) — Na,C,0,(aq) + 2H,0(I) Identify the value of x and hence the formula of the hydrated acid HC,0,°xH,0. (4 marks)
j NY i é
\
| 4.3 Redox
Specification reference: 2.1.5
Learning outcomes Demonstrate knowledge, understanding, and application of: > oxidation number
> oxidation and reduction
> redox reactions.
Study tip
It is a common mistake to confuse oxidation numbers of elements
and ions. The oxidation number of oxygen in 0, is zero but in H,0 is -2.
Study tip
All oxidation numbers, except zero (0), have a sign, + or -.
The sign of an oxidation number
is placed before the number: the oxidation number of Ca in a Ca?* ion is +2, not 2+.
Oxidation number
Oxidation number is based on a set of rules that apply to atoms, and can be thought of as the number of electrons involved in bonding to a different element. Use of oxidation numbers helps when writing formulae and balancing electrons as a check that all electrons have been accounted for.
Rules for elements
The oxidation number is a/ways zero for elements.
® Ina pure element, any bonding is to atoms of the same element.
® Soin H,, O,, P,, S,, Na, and Fe the oxidation number of each atom of the element is 0.
Rules for compounds and ions ® Each atom in a compound has an oxidation number.
® An oxidation number has a sign, which is placed before the number.
Table 1 shows examples of oxidation numbers of atoms in compounds and ions, including some special cases. The oxidation number of an ion of an element is numerically the same as the ionic charge but the sign comes before the number.
Y Table 1 Oxidation number rules in compounds and ions Combined element Oxidation number Examples
H,0, CaO. NH,, H.S. HF NaCl, K,0 MgCl., CaO HCI, KBr, Cal,
Special cases:
H in metal hydrides NaH, CaH,
0 in peroxides H,0,
0 bonded to F F0
Working out oxidation numbers
In addition to the rules in Table 1, there is a further rule for combined atoms:
® sum of the oxidation numbers = total charge.
ACIDS AND REDOX
Worked example: Oxidation number in compounds Study tip What is the oxidation number of sulfur in sulfuric acid, H,SO,? Remember the sum of the zi oxidation numbers must equal Step 1: Assign any oxidation numbers from the rules. the overall charge. H,SO, Total H = +1 x 2 = +2 ie 3 = Total O=-2 x 4=-8 —2
Step 2: What is the sum of oxidation numbers?
sum of oxidation numbers = total charge = 0
Step 3: Work out the unknown oxidation numbers.
sum of oxidation numbers = (+2) + (X) + (-8) =0
7
H,SO,
Oxidation number of sulfur in H,SO, = +6
Worked example: Oxidation numbers in ions
What is the oxidation number of nitrogen in NO,-?
Step 1: Assign any oxidation numbers from the rules. NO,
-2}) Total O =-2 x 3=-6
Step 2: What is the sum of oxidation numbers?
sum of oxidation numbers = total charge = -1 Step 3: Work out the unknown oxidation numbers
sum of oxidation numbers = (X) + f =-] NO,
Oxidation number of nitrogen in NO, = +5
Using Roman numerals in naming
Roman numerals are used in the names of compounds of elements that form ions with different charges. The Roman numeral shows the oxidation state (oxidation number) of the element, without a sign. The sign of the oxidation state is obvious from the overall charge:
@ iron(II) represents Fe** with oxidation number +2
@ iron(II) represents Fe** with oxidation number +3.
4.3 Redox
A Figure 1 Reduction of copper(I!) oxide with hydrogen — the green flame is excess hydrogen burning off
Study tip Remember OILRIG:
OIL Oxidation Is Loss of electrons
RIG Reduction Is Gain of electrons.
You have already seen that polyatomic ions containing oxygen, such as NO,~ and NO,~, are sometimes named using —ite and —ate. Although still in common usage, use of -ite and —ate in naming is old-fashioned and modern names use oxidation numbers shown as Roman numerals.
¥ Table 2 Naming of polyatomic ions
Oxidation Common Modern number of name : name nitrogen nitrite nitrate(Ill) NO.” nitrate +5 nitrate(V)
In common usage, the Roman numeral is often omitted for the common ion, usually with more oxygen atoms: ® nitrate is assumed to be NO,
@ sulfate is assumed to be $O,?-.
Redox reactions Reduction and oxidation
Originally the terms oxidation and reduction were used solely for reactions involving oxygen.
® Oxidation is addition of oxygen.
® Reduction is removal of oxygen.
The reaction below shows oxidation and reduction: CuO(s) + H,(g) — Cu(s) + H,O(1)
® Copper(Il) oxide has /ost oxygen and has been reduced. ® Hydrogen has gained oxygen and has been oxidised.
Redox reactions involve reduction and oxidation. If one process happens, so must the other — if something is reduced, something else must be oxidised.
The terms oxidation and reduction are now applied to many reactions that do not involve oxygen. The modern definitions are in terms of either electrons or oxidation number.
Redox in terms of electrons
@ Reduction is the gain of electrons.
® Oxidation is the loss of electrons.
The redox reaction below does not involve oxygen but does involve
gain and loss of electrons.
2Fe(s) + 3Cl,(g) — 2FeCl,(s)
FeCl, contains positive and negative ions, Fe** and Cl-.
® Iron loses electrons and is oxidised 2Fe — 2Fe** + 6e~ ® Chlorine gains electrons and is reduced 3Cl, + 6e~ — 6CI- The electrons gained and lost balance:
@® 2 Fein 2Fe each Fe loses 3e7 total of 6 electrons lost
® 6Clin 3Cl, each Cl gains le total of 6 electrons gained
Redox in terms of oxidation number @® Reduction is a decrease in oxidation number.
® Oxidation is an increase in oxidation number.
The reaction below shows oxidation and reduction in terms of
oxidation number. Each atom is assigned an oxidation number using
the oxidation number rules: Cu(s) + 2AgNO,(aq) — 2Ag(aq) + Cu(NO,),(aq) 0 — +2 oxidation +] » O reduction
The changes in oxidation number apply to each atom and the total changes in oxidation number balance:
® 1CuinCu Cu increases by +2 totalincrease = +2 @ 2Agin2AgNO, each Ag decreases by-1 total decrease =-2 Redox reactions of acids
In Topic 4.1, you saw how acids produce salts in neutralisation
reactions. Dilute acids also undergo redox reactions with some metals
to produce salts and hydrogen gas.
metal + acid — salt + hydrogen
Reaction of zinc with dilute hydrochloric acid Zn(s) + 2HCl(aq) — ZnCl,(aq) + H,(g)
0 — +2 oxidation +] > 0 reduction @ 1} ZninZn Zn increases by +2 total increase = +2 @® 2H in 2HCI each H decreases by -1 total decrease =-—2 The overall decrease in oxidation number of 2H = 2 x -1 = -2
balances the increase of Zn by +2.
Reaction of aluminium with dilute sulfuric acid 2Al(s) + 3H,SO,(aq) — Al,(SO,),(aq) + 6H,(g)
0 — +3 oxidation
+] a 0 reduction @ 2Alin2Al each Al increases by +3 total increase = +6 ® 6Hin3H,SO, each H decreases by-1 total decrease = —-6
ACIDS AND REDOX
A Figure 2 Iron reacts with chlorine gas in a gas jar, forming iron(III) chloride, FeCl,
Study tip
The oxidation number applies
to each atom of an element. In this example, each atom of Ag decreases from +1 to 0. The total change is —2 because there are 2 Ag, each decreasing by —1.
Summary questions
1 State the oxidation state of the species in the following:
a Ag® (1 mark) bE (1 mark) c NaCi0, (1 mark)
2 State the oxidation number of sulfur in the following:
a H,S (1 mark) b S0,° (1 mark) c Na,S.0, (1 mark) 3 The following reaction is a redox process: Mg + 2HCI — MgCl, +H,
a_ Identify the changes in oxidation number. (2 marks) b Explain which species is being oxidised and which is being reduced. (2 marks)
Chapter 4 Practice questions
Practice questions
1 25.0cm? sample of an aqueous H,SO,
solution of unknown concentration is titrated with 0.125 mol dm-* NaOH. 22.40cm? of NaOH are required to reach the end point. H,SO, + 2NaOH — Na,SO, + 2H,O
a Calculate the amount, in mol of
NaOH used. (1 mark) b= Calculate the amount, in mol of
H,SO, used. (1 mark) ce Calculate the concentration in
mol dm~ of the H,SO,. (1 mark)
The reaction below is a redox reaction. 3CuO + 2NH, — 3Cu + 3H,0 +N, Explain in terms of oxidation number,
what has been oxidised and what has been reduced. (2 marks)
3 This question looks at two redox reactions.
a 3Mg + 2Fe(NO,), > 3Mg(NO,), + 2Fe (i) Explain in terms of electrons, what has been oxidised and what has been reduced. (2 marks)
(ii) What is the systematic name for Fe(NO,) ,? (1 mark)
b MnO, + 4HCl — MnCl, + Cl, + 2H,O
(i) Explain, in terms of oxidation numbers, what has been reduced. (2 marks)
(ii) Use oxidation numbers to show that chlorine has only partly
been oxidised. (2 marks)
a_ A solution of calcium chloride can be prepared by neutralisation reactions of dilute hydrochloric acid with solid calcium carbonate, with solid calcium oxide and with aqueous calcium hydroxide.
(i) Write equations, with state symbols, for these methods of preparing
calcium chloride. (6 marks) (ii) Why are these reactions all neutralisation reactions? (1 mark)
(iii) Write an ionic equation for the reaction of aqueous calcium hydroxide and hydrochloric acid.
(] mark)
b= A solution of calcium chloride can be also prepared by the redox reaction of dilute hydrochloric acid and calcium metal.
(i) Explain reduction and oxidation in terms of electrons and oxidation numbers (2 marks)
(ii) Write an equation, with state symbols, for this reaction. (2 marks)
(iii) For this redox reaction, determine what had been oxidised and what has been reduced and identify the changes in oxidation numbers. (2 marks)
5 Tungsten ore contains WO,. Tungsten can
be extracted from WO, present in its ore in a redox reaction with hydrogen. The unbalanced equation is shown below.
WO, +H, ~ W+H,0
a_ Balance the equation. (J mark)
b What is meant by oxidation and reduction in terms of electrons? (J mark)
ce Using oxidation numbers show that oxidation and reduction have
taken place in this reaction. (3 marks)
d Some ore contains 2% of WO, by mass. Calculate the maximum mass of tungsten that could be obtained from the processing of 100 tonnes of ore. (4 marks)
A household cleaner containing dissolved ammonia is analysed as follows.
25.0cm? of the cleaner is diluted to 250.0cm?. 25.0cm? of the resulting solution is titrated with 0.125 mol dm~’* H,SO,(aq) and
24.40 cm? were required to reach the end point. The equation is shown below. 2NH,(aq) + H,SO,(aq) — (NH,),SO,(aq)
a Calculate the amount, in mol, of
H,SO, used in the titration. (1 mark) b= Calculate the amount, in mol, of NH, used in the titration. (1 mark)
c Calculate the concentration of ammonia in the household cleaner
(1 mark) (1 mark)
b_ The reaction between magnesium and sulfuric acid is a redox reaction.
Mg(s) + H,SO,(aq) —~ MgSO, (aq) + H,(g)
(i) in mol dm”; (ii) in g dm°>.
(i) Use oxidation numbers to identify which element has been oxidised. Explain your answer. (2 marks)
(ii) Describe what you would see when magnesium reacts with an excess of sulfuric acid. (2 marks)
c Epsom salts can be used as bath salts to help relieve aches and pains. Epsom salts are crystals of hydrated magnesium sulfate, MgSO,¢xH,O. A sample of Epsom salts was heated to remove the water. 1.57 g of water was removed leaving behind 1.51 g of anhydrous MgSO,. (i) Calculate the amount, in mol, of
anhydrous MgSO, formed. (2 marks)
(ii) Calculate the amount, in mol,
of H,O removed. (1 mark) (iii) Calculate the value of xin MgSO,,°xH,0. (1 mark)
F321 June 2009 1(b) (¢)
8 A student carries out experiments using acids,
bases and salts.
a Calcium nitrate, Ca(NO,),, is an example of a salt.
The student prepares a solution of calcium nitrate by reacting dilute nitric acid, HNO,, with the base calcium hydroxide, Ca(OH),.
(i) Why is calcium nitrate an
example of a salt? (1 mark)
(ii) Write the equation for the reaction between dilute nitric acid and aqueous calcium hydroxide. Include
state symbols. (2 marks)
(iii) Explain how the hydroxide ion in aqueous calcium hydroxide acts as a base when it neutralises
dilute nitric acid. (1 mark)
b_ A student carries out a titration to find the concentration of some sulfuric acid.
The student finds that 25.00cm? of 0.0880 moldm~ aqueous sodium hydroxide, NaOH, is neutralised by 17.60.cm? of dilute sulfuric acid, H,SO,. H,SO,(aq) + 2NaOH(aq) — Na,SO,(aq) + 2H,0(1)
ACIDS AND REDOX
(i) Calculate the amount, in moles,
of NaOH used. (1 mark) (ii) Determine the amount, in moles, of H,SO, used. (1 mark)
(iii) Calculate the concentration, in mol dm™, of the sulfuric acid. (/ mark)
ce After carrying out the titration in (b), the student left the resulting solution to crystallise. White crystals were formed, with a formula of Na,SO,exH,O and a molar mass of 322.1 g mol!.
(i) What term is given to the ‘exH,O’ part of the formula? (1 mark) (ii) Using the molar mass of the crystals, calculate the value of x. (2 marks) F321 Jan 10 Q2
9 Compound A is an organic acid containing C,
H and O only. A student analyses compound A to find its molar mass as follows.
The student dissolves 2.6432 g of compound A in water and dilutes the solution to 250cm? in a volumetric flask. The student fills a burette with this solution.
The student adds 25.0 cm? of 0.224 mol dm™? KOH to a conical flask.
31.25cm? of compound A was required for reach the end point.
The formula of compound A can be simplified as H,X. The equation in the titration is: H,X(aq) + 2KOH(aq) — K,X(aq) + 2H,O(1)
a_ Calculate the amount, in mol, of compound
A used in the titration. (2 marks) b= Calculate the molar mass of compound A. (2 marks)
c The percentage composition by mass of compound A is C, 40.68%; H, 5.08%; O, 54.24%
Calculate the empirical and molecular formulae of compound A. (3 marks)
d Write an equation for the reaction in the titration of compound A with KOH. (1 mark)
ELECTRONS AND BONDING
pe |
Electron structure
Specification reference: 2.2.1
COCO HHSSHHSEHHESHSESSESOSSESSESESEESEOSS,
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> electrons and shells
> atomic orbitals
> filling of orbitals
> electron configurations.
Feecerecceeseeesseseseceseeet”
Y Table 1 Maximum number of electrons in the first four shells
Shell Number of numbern electrons
+ Electrons in shells
Electrons and shells
You already know that the nuclear atom contains electrons in shells surrounding the nucleus. In this topic you will learn more about how electrons are arranged within shells.
Shells
For GCSE, you learnt the 2,8,8 rule for the number of electrons that can fill each shell.
For A Level, you need to know the number of electrons that fill the first four shells (#7 = 1 to 1 = 4) shown in Table 1.
What are shells?
In an atom, electron shells make up a model that helps us to visualise something that cannot be seen.
® Shells are regarded as energy levels.
® The energy increases as the shell number increases.
® The shell number or energy level number is called the principal quantum number n.
The current model for an electron is based the idea that an The maximum number of electrons in a shell n is given by electron has properties both of a wave and a particle — the formula:
wave-—particle duality. If you take A level physics, you will
learn more about wave—particle duality.
Around an atom, electrons can only fit into energy levels defined by the wave nature of an electron.
A Figure 1 An atomic orbital shown as a cluster of dots for the probability of finding an electron in a given space
© number of electrons = 2n?
1 Use the 2n? formula to check the number of electrons in Table 1. |
2 Find out the maximum number electrons for the energy levels n= 5, 6, and 2.
Atomic orbitals
Shells are made up of atomic orbitals. An atomic orbital is a region around the nucleus that can hold up to two electrons, with opposite spins.
Models visualise an atomic orbital as a region in space where there is a high probability of finding an electron (Figure 1). An electron can be thought of as a negative-charge cloud with the shape of the orbital, referred to as an electron cloud.
® An orbital can hold one or two electrons, but no more. @ There are different types of orbitals: s-, p-, d- and f-orbitals. ® Each type of orbital has a different shape.
This may seem bewildering, but you will soon see patterns emerge.
ELECTRONS AND BONDING
s-orbitals
In an s-orbital the electron cloud is within the shape of a sphere (Figure 2).
As with all orbitals, an s-orbital can hold one or two electrons.
® Each shell from # = 1 contains one s-orbital.
@ The greater the shell number 1, the greater the radius of its s-orbital.
p-orbitals
In a p-orbital, the electron cloud is within the shape of a dumb-bell. As with an s-orbital, one orbital can contain one or two electrons. There are three separate p-orbitals at right angles to one another (Figure 3). These orbitals are referred to as p,, p,, and p,.
® Each shell from » = 2 contains three p-orbitals.
@ The greater the shell number n, the further the p-orbital is from the nucleus.
d-orbitals and f-orbitals
The next orbitals, d- and [-orbitals, are more complex.
@ Each shell from » = 3 contains five d-orbitals.
® Each shell from nm = 4 contains seven f-orbitals.
Sub-shells
You may have noticed that a new type of orbital is added for each additional shell. Within a shell, orbitals of the same type are grouped together as sub-shells. The orbitals and sub-shells in the first four shells are shown in Table 2.
VW Table 2 Shells, sub-shells, and orbitals
Number of orbitals Sib-shatls Number of Number of Shell a electrons electrons resen
P in sub-shells in shell
ae ie |e ROSS oe oe | aisle
Petia Ts [7 [aecapeaicat[aesswoom] ae
Each new shell gains a new type of orbital. The number of orbitals increases with each new type of orbital s, | p, 3 d, 5 ae |
@ two electrons fit into each orbital, so the number of electrons in each sub-shell also increases s, 2 p, 6 d, 10 [, 14
S A Figure 2 Shape of an s-orbital
A Figure 3 Shapes of the three p-orbitals in a shell
Study tip
You need to know about the
existence of each type of orbital, but you only need to know the shapes of s- and p-orbitals.
5.1 Electron structure
4f n=4 > —“a4 oe i a ee 4p ‘ 3d 4 n=3,, = 3s 6a S| n=2 2p i 2s n=]
A Figure 4 Energy levels of the sub-shells
ml v
opposite spins
ni] (a) x
Same Spins
A Figure S ‘Electrons-in-box’ model showing allowed and unallowed spins
Filling of orbitals
There is a set of rules for how orbitals are occupied by electrons.
Orbitals fill in order of increasing energy
The sub-shells that make up shells have slightly different energy levels. Within each shell, the new type of sub-shell added has a higher energy. For example, in the second shell, the 2p sub-shell is the new type, and is at a higher energy than the 2s sub-shell.
® inthe = 2 shell, the order of filling is 2s, 2p.
@ inthe = 3 shell, the order of filling is 3s, 3p, 3d.
@ inthe #=4 shell, the order of filling is 4s, 4p, 4d, 4f.
Figure 4 shows the relative energy levels of the sub-shells making up the first four shells. The Aighest energy level in the third shell overlaps with the /owest energy level in the fourth shell.
@ The 3d sub-shell is at a /Aigher energy level than the 4s sub-shell. @ The 4s sub-shell therefore fills before the 3d sub-shell.
® The order of filling is therefore 3p, 4s, 3d.
Electrons pair with opposite spins
Each orbital can hold up to two electrons. Rather than drawing different orbital shapes, it is convenient to use an electrons-in-box model.
@ Electrons are negatively charged and repel one another.
@ Electrons have a property called spin — either up or down.
@ An electron is shown as an arrow indicating its spin, either | or |. ®
The two electrons in an orbital must have opposite spins, as shown in Figure 5. The opposite spins help to counteract the repulsion between the negative charges of the two electrons.
Orbitals with the same energy are occupied singly first
Within a sub-shell, the orbitals have the same energy. One electron occupies each orbital before pairing starts. This prevents any repulsion between paired electrons until there is no further orbital available at the same energy level.
Figure 6 shows how four electrons occupy the p-orbitals of a p-sub-shell.
mit) v fy] x
e With four electrons, one electron occupies e Not allowed.
each p-orbital. e One p-orbital has been left empty and e Only then can pairing take place. a second p-orbital has been paired. e The paired electrons have opposite spins,
fand¢.
A Figure 6 Allowed spins when filling p-orbitals
ELECTRONS AND BONDING
Electron configuration
Study tip Electron configuration of atoms Electron configurations eivouldhe For A Level, you are expected to be able to work out the electron shown in shell order rather than configuration of atoms up to atomic number Z = 36 (krypton, Kr,) at the order of filling. So the electron the end of Period 4 in the periodic table). The electron configuration of configuration of krypton should an atom shows how sub-shells are occupied by electrons. show the 3d sub-shell before the 4s: An atom of krypton, Z = 36, contains 36 electrons. In Figure 4, you 1s°2s*2p°3s*3p°3d4s*4p® can see the energy levels of the sub-shells in the first four shells. Table instead of
3 shows how the 36 electrons in a krypton atom fill the sub-shells 1s?2s*2p°3s*3p°4s23d 4p from the lowest energy level upwards.
¥ Table 3 Filling the sub-shells of a krypton atom (Z = 36) electrons
sub-shell ; : > R , | l 29629 3
36 electrons ‘ t f
The 4s sub-shell fills before the 3d sub-shell because 4s has a lower
energy level. A Figure 7? Electron configuration for the seven electrons in a nitrogen atom The electron configuration of krypton is therefore written as 1572s?2p°3s23p%4s73d!!4p°®.
Synoptic link Figure 7 shows the meaning of the numbers in an electron configuration, using nitrogen as an example. You will learn more about the link
; between electron configurations Table 4 shows the electron configurations of the elements across and the periodic table in Topic 2.1,
Period 2 of the periodic table, together with the much simpler electron The periodic table. structures from GCSE showing shells only.
¥ Table 4 Electron configurations across Period 2
B C N 0 F Ne
Li Be Alevel 1s*2s*2p! | 1s*2s*2p* | 1s*2s*2p? | 1s%2s*2p* | 1s*2s*2p° | 15*2s*2p®
Shorthand electron configurations
Electron configurations can be expressed more simply in terms
of the previous noble gas in the periodic table plus the outer
electron sub-shells. Table 5 shows this shorthand notation for elements in Group | of the periodic table. This notation is useful for emphasising similarities in the electron configurations of the outer shell.
¥Y Table S Shorthand notation for electron configuration
Electron configuration Shorthand notation
[ 1 Na 1s*2s*2p°3s' [Ne]3s?
5.1 Electron structure
Electron configuration of ions
® Positive ions or cations are formed when atoms /ose electrons.
Synoptic link
You will find out more about blocks peice ‘ = :
z : ean ® Negative ions or anions are formed when atoms gain electrons,
in Topic 2.1, The periodic table
Blocks and the periodic table
The periodic table can be divided into blocks corresponding to their
Synoptic link highest energy sub-shell.
You will learn more about
the electron configuration of
d-bock elements in Topic 24.1,
d-block elements.
@ s-block
highest energy electrons in the s-sub-shell (left block of two groups) ® p-block
highest energy electrons in the p-sub-shell (right block of six groups) : @ d-block Study tip highest energy electrons in the d-sub-shell (centre block of 10 groups) With the 4s and 3d sub-shells, it is a case of first in, first out.
lons of s-block and p-block elements
When forming ions, the highest energy sub-shells lose or gain
e The 4s electrons are first in. electrons. Table 6 compares the electron configurations of calcium and e The 4selectrons are first out. oxygen atoms and their common ions.
VY Table 6 Electron configurations of calcium and oxygen atoms and ions
Summary questions Number of Electron electrons configuration 1 State how many electrons the 1s?2s?2p®3s?3p%4s? | | following can hold. ; ae 3corbitl (1 mark) 2) 1s*2s*2p®3s?3p® 2 electrons /ost from 4s sub-shell b 4p sub-shell (1 mark) 1s*2s*2p®3s*3p* c n=4shell (1 mark) , 1s*2s*2p°3s*3p® 2 electrons gained by 3p sub-shell d 4d orbital (1 mark) ee ee 2 Write the electron configuration ons of d-block elements for the following atoms: For atoms of d-block elements, the 4s sub-shell is at a lower energy ac (1 mark) than the 3d sub-shell, so is filled first. The energies of the 4s and b S (1 mark) 3d sub-shells are very close together and, once filled, the 3d energy e K (1 mark) level falls below the 4s energy level. The consequence is that: d Co (1 mark) @ the 4s sub-shell fills before the 3d sub-shell
e As (1 mark) @ the 4s sub-shell also empties before the 3d sub-shell.
3 Write the electron configuration The d-block element nickel, Z = 28, forms the ion Ni?*. for the following ions:
a Mg?* (4 mark) ® Fora nickel atom, the electron configuration of the 28 electrons is b P?- (1 mark) 1572s72p°3s73p°3d54s? e Br (1 mark) ®@ To form the nickel ion, Ni**, two electrons are lost from the d Fe®* (1 mark) 4s sub-shell: 1s?2s*2p°3s?3p°3d5 4 Write the shorthand electron configurations for a Si (1 mark) b Cr (1 mark) c Mn@* (1 mark)
d Ga (1 mark)
5.2 lonic bonding and structure
Specification reference: 2.2.2
lonic bonding
Ionic bonding is the electrostatic attraction between positive and negative ions. It holds together cations (positive ions) and anions (negative ions) in ionic compounds.
Common cations include:
® metal ions (e.g., Na*, Ca**, Al’*)
® ammonium ions (NH,*). Common anions include: ® non-metal ions (e.g., Clr, 07>)
® polyatomic ions (e.g., NO, $O,*>).
lonic compounds Dot-and-cross diagrams
The simplest ionic compounds contain metal ions and non-metal ions.
At GCSE, you used a model for ionic bonding based on electron transfer.
® Outer-shell electrons from a metal atom are transferred to the outer shell of a non-metal atom.
Positive and negative ions are formed.
The ions formed often have outer shells with the same electron configuration as the nearest noble gas.
In dot-and-cross diagrams, the electrons in the original atoms are shown as either dots or crosses. It is then easy to work out the charge on each ion and to account for all electrons.
Example — potassium fluoride, KF
Figure | shows electron transfer of the one outer-shell electron in a potassium atom, K, to the outer shell of a fluorine atom, F, forming
a K* ion and an F- ion. The square brackets show that the charge is spread over each ion and that the ions are separate entities. Only the outer-shell electrons are shown because the inner shells are full and not involved in bonding. The electron structures of the K* and F~ ions formed are now the same as the nearest noble gas.
transfer of 1 electron
CO — IO]
K atom F atom K" ion F- ion 19p*,19e" 9p*,9e 19p’,i8e Y9p*,10e
2,8,8 2,8 (argon) (neon)
Electron structure
2,8,8,1 2,7
neutral 1+ 1-
Charge neutral
eth eee eeeressessceosesererere,
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> ionic bonding
> giantionic lattice structures
> properties of ionic compounds.
ORO E HEHEHE HEE EEEEEEEEES
Synoptic link
In Topic 2.3, Formulae and
equations, you looked at how ionic charges are used for writing the formulae of ionic compounds.
Study tip
Dot-and-cross diagrams often show the outer-shell electrons only. In the examples, each cation has been shown with an empty shell. This circle can be omitted for even greater clarity.
“4 Figure 1 Dot-and-cross diagram of potassium fluoride, KF
5.2 lonic bonding and structure
Example — magnesium chloride, MgCl,
Figure 2 shows electron transfer of the two outer-shell electrons in a magnesium atom to two chlorine atoms to form an Mg?* ion and two Cl- ions. The electron structures of the Mg?* and Cl- ions are the same as the nearest noble gas.
transfer of 2 electrons
O—|lO! GF le!
Cl atom Mg atom Cl atom Cl ion Mg** ion Cl ion
i? p*, ive i2p*,izge™ 17 p*,.i?e7 1?'p’,18e- i2p*,id0e” 17 p*,18e- electron 2,8,8 2,8 2,8,8
structure (argon) (argon) (argon)
charge neutral neutral neutral
A Figure 2 Dot-and-cross diagram of magnesium chloride, MgCl.
Structure of ionic compounds
Although it is convenient to look at ionic bonding acting between a small number of ions, each ion attracts oppositely charged ions in all directions.
The result is a giant ionic lattice structure containing billions of billions of ions, the actual number only determined by the size of the crystal. The giant ionic lattice of sodium chloride is shown in Figure 4. A giant lattice is a key structural feature of all ionic compounds.
@ Each Na’ ion is surrounded by 6 CI ions. ® Each Cl ion is surrounded by 6 Na* ions.
® Each ion is surrounded by oppositely charged ions, forming a giant ionic lattice.
chloride ion CI sodium ion Na™
“Figure 4 Port of the sodium chloride lattice — the regular cubic arrangement of Na* and CI ions in the giant ionic lattice structure gives the crystal its cubic shape
A Figure 3 Crystals of rock salt (sodium chloride) with a cubic shape
Properties of ionic compounds
The physical properties of ionic compounds can be explained in terms of the giant ionic lattice structure and ionic bonding.
ELECTRONS AND BONDING
Melting and boiling points
Almost all ionic compounds are solids at room temperature. At room temperature, there is insufficient energy to overcome the strong electrostatic forces of attraction between the oppositely charged ions in the giant ionic lattice. High temperatures are needed to provide the large quantity of energy needed to overcome the strong electrostatic attraction between the ions. Therefore most ionic compounds have high melting and boiling points.
Table 1 compares the melting points of several ionic compounds.
The melting points are higher for lattices containing ions with greater ionic charges, as there is stronger attraction between ions. lonic attraction also depends on the size of the ions, but in this example the Na* and Ca?* ions are similar sizes so this is not a factor here.
Solubility
Many ionic compounds dissolve in polar solvents, such as water. Polar water molecules break down the lattice and surround each ion in solution.
In a compound made of ions with large charges, the ionic attraction may be too strong for water to be able to break down the lattice structure. The compound will not then be very soluble. Table 2 compares the solubility in water of several ionic compounds. The most soluble compound.
¥ Table 2 Solubility of ionic compounds in water
Solubility at 20°C
lonic compound 3 mol dm-”
NaCl Na* and CI- 6.1
CaCl, Ca** and CI 0.67 Na,CO, Na* and CO,*" CaCO, Ca** and CO,*
Aword of caution Solubility requires two main processes:
@ the ionic lattice must be broken down ® water molecules must attract and surround the ions.
The solubility of an ionic compound in water therefore depends on
the relative strengths of the attractions within the giant ionic lattice and the attractions between ions and water molecules. For the ionic compounds in Table 2, the attractions in the giant ionic lattice have
the greater effect, and solubility decreases as ionic charge increases. But predictions of solubility should be treated with caution.
Electrical conductivity
In the solid state, an ionic compound does not conduct electricity. But once melted or dissolved in water, the ionic compound does conduct electricity (Figure 5).
¥ Table 1 Melting points of ionic compounds
lonic Melting lons compound point /°C
Na’* and F 993
Ca** and F 1423 Na* and 02° 1275
Ca**and0*-| 2614
Synoptic link
You will find out more
about polarity in Topic 6.2, Electronegativity and polarity.
5.2 lonic bonding and structure
solid state
af.
electrode electrode
ions fixed in position in the lattice @ jons cannot move @ no conductivity
In the solid state:
@ the ions are in a fixed position in the giant ionic lattice
® there are no mobile charge carriers. An ionic compound is a non-conductor of electricity in the solid state.
When /iquid or dissolved in water:
@ the solid ionic lattice breaks down
liquid or aqueous states
©- a ~.2 =e)
electrode @ a acwsde -@O-
e ions are not fixed in a lattice e ions are now free to move e electricity is conducted
A Figure 5 Electrical conductivity of e an ionic compound in solid, liquid, and aqueous states
lonic bones and teeth
You rely on ionic compounds for the skeleton framework of your body and for your teeth. So what are bones and teeth made out of? The chemistry is complex but the main ionic compound is calcium hydroxyapatite, which can be represented with the formula Ca,(P0,,),0H. This compound is also the main constituent in tooth enamel.
Unfortunately ions in tooth enamel are removed in acid conditions. Once broken down, gaps in enamel can allow tooth decay to develop beneath. Saliva helps to neutralise acidic food and also to replace ions but this may not
be enough. Fluoride ions help to replace lost ions by forming fluoropatite, Ca.(PO,),F which is stronger than hydroxyapatite and more resistant to acid conditions.
Most toothpastes contain fluoride as sodium fluoride. Your water supply may also contain fluoride depending on where you live.
® the ions are now free to move as mobile charge carriers. An ionic compound is a conductor of electricity in liquid and
aqueous states.
Summary of properties
Most ionic compounds:
e@ have high melting and boiling points
tend to dissolve in polar solvents such as water
® conduct electricity only in the liquid state or in aqueous solution.
Contains:
Sodium Fluoride (1450 ppm F~) Ingredients:
Aqua, Glycerin, Hydrated Silica, PVM/MA Copolymer,
Limonene, Ci 77891.
A Figure 6 Fluoride toothpaste
1 Show that the ionic charges of the ions in Ca.(P0,),0H and Ca,(PO,),F balance.
2 Suggest why Ca,(PO,),0H is very insoluble in water. Many people confuse fluoride with fluorine. What is the difference and why would it be very strange to find fluorine in your toothpaste?
Summary questions
1 Draw dot-and-cross diagrams, with outer shells only, for the following:
a Na,0
(1 mark)
b MgS (1mark) c AIF, (1 mark)
2 Explain why ionic compounds have high melting and boiling points.
3 Explain why ionic compounds dissolve in water.
(2 marks)
(2 marks)
5.3 Covalent bonding
Specification reference: 2.2.2
Covalent bonding Covalent compounds and molecules
Covalent bonding is the strong electrostatic attraction between a shared pair of electrons and the nuclei of the bonded atoms. Covalent bonding occurs between atoms in:
® non-metallic elements, for example, H, and O, ® compounds of non-metallic elements, for example, H,O and CO, ® polyatomic ions, for example, NH,"*.
For covalent bonding, the key feature is the sharing of a pair of electrons between the two atoms. The atoms are bonded together in a single
unit — a small molecule (e.g., H,), a giant covalent structure (e.g., SiO,), or a charged polyatomic ion (e.g., NH,*). This is very different from the model of electron transfer to form separate ions in ionic bonding.
The covalent bond Orbital overlap
A covalent bond is the overlap of atomic orbitals, each containing one electron, to give a shared pair of electrons (Figure 1).
H(g) + Hig) = ——————_ Hg)
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> single covalent bonding > multiple covalent bonding
> dative covalent (coordinate) bonding
> average bond enthalpy.
cP P EPEC PPS OCSOOSCOOOCOSS OOO C See eee eee
A Figure 1 Overlap of the two 1s orbitals of two hydrogen atoms to form a hydrogen molecule
® The shared pair of electrons is attracted to the nuclei of both the bonding atoms.
@® The bonded atoms often have outer shells with the same electron structure as the nearest noble gas.
A covalent bond is localised
In Topic 5.2, you saw that an ion attracts oppositely charged ions in all directions, resulting in a giant ionic lattice structure containing many billions of ions.
A covalent bond is very different. The attraction is /ocalised, acting solely between the shared pair of electrons and the nuclei of the two bonded atoms. The result can be a small unit called a molecule, consisting of two or more atoms, such as H, and H,O. A molecule
is the smallest part of a covalent compound that can exist whilst retaining the chemical properties of the compound.
ionic Nat ion attracts in all Ne directions in three ~~ (Na dimensions Fad %
covalent
In Ho, the attraction is solely between the shared pair of HSH
electrons in the covalent bond <> <> and the nuclei of the bonding atoms @ttraction
A Figure 2 Comparison of attraction from
anion and in a covalent bond
5.3 Covalent bonding
Single covalent bonds Dot-and-cross diagrams
As with ionic bonding, dot-and-cross diagrams are used to account for electrons in covalent bonding.
® In covalent bonding electrons are shared.
® In ionic bonding electrons are transferred.
Figure 3 shows dot-and-cross diagrams for some simple molecules. Use of dots and crosses allows the origin of each electron to be shown clearly. Each bonding atom now has the electron structure of the nearest noble gas.
Figure 3 also shows a second structure for each molecule called a displayed formula.
® A displayed formula shows the relative positioning of atoms and the bonds between them as lines.
® Paired electrons that are not shared are called lone pairs. These can also be added to displayed formulae, as in water, H,O, and ammonia, NH,, in Figure 3.
A Figure 3 Dot-and-cross diagrams and displayed formulae for hydrogen, H., water, H,0, ammonia, NH., and
methane, CH,
Study tip
You will meet covalent compounds of carbon, nitrogen, oxygen, and hydrogen frequently throughout the course, particularly in organic
chemistry.
Number of covalent bonds
Most of the covalent compounds that you will meet during the course are compounds of hydrogen, carbon, nitrogen, and oxygen. You can see in Figure 3 the number of covalent bonds usually formed by atoms of these elements:
@ carbon forms 4 bonds
® nitrogen forms 3 bonds
® oxygen forms 2 bonds oy
hydrogen forms | bond.
ELECTRONS AND BONDING
What about other elements? Cr Boron xs Boron, in Period 2 of the periodic table, has the electron configuration
1s?2s?2p', so only three outer-shell electrons can be paired. Boron Ri (J forms covalent compounds, such as boron trifluoride, BF, (Figure 4),
in which its three outer-shell electrons are paired. So a molecule of
BF, only has six electrons around the boron atom. A Figure 4 Dot-and-cross diagram of
boron trifluoride, BF, showing how the BF, shows that predictions for bonding cannot be based solely on the three boron electrons are paired noble gas electron structure,
Phosphorus, sulfur, and chlorine Table 1 shows the formulae of the fluorides of the non-metals phosphorus, sulfur, and chlorine in Period 3 of the periodic table.
¥ Table 1 Number of covalent bonds formed by phosphorus, sulfur, and chlorine
Element phosphorus Synoptic link Electron structure [Ne]3s*3p* [Ne]3s°3p* You San lise Tablet to work aikthe Outer-shell electrons i i a oxidation numbers of phosphorus,
sulfur, and chlorine in each fluoride. You studied oxidation numbers in Topic 4.3, Redox.
Formula of fluoride
Phosphorus trifluoride, PF,, sulfur difluoride, SF,, and chlorine monofluoride, CIF, follow the expected pattern of formulae, with the bonded atoms having a noble gas electron structure. But how can atoms of phosphorus, sulfur, and chlorine bond with more fluorine atoms to give the other fluorides in Table 1?
For the elements in Period 2, the 7 = 2 outer shell can hold just eight electrons. But for phosphorus, sulfur, and fluorine, the = 3 outer shell can hold 18 electrons, so more electrons are available
for bonding. Figure 5 shows how different arrangements for the
six outer-shell electrons of sulfur and the different numbers of bonds possible to bond with fluorine.
two unpaired electrons | four unpaired electrons | six unpaired electrons | Figure 5 Different numbers of
two bonds possible four bonds possible six bonds possible unpaired electrons lead to different possibilities for covalent compounds SF, SE
2 4 of sulfur
5.3 Covalent bonding
A Figure 6 Dot-and-cross diagram
of sulfur hexafluoride, SF ., showing how the six sulfur electrons all paired forming six covalent bonds
Synoptic link
You will learn more about the nature of a single and double
covalent bond when you study organic chemistry in Chapter 12, Alkanes, and Chapter 13, Alkenes.
Figure 6 shows the dot-and-cross diagram of sulfur hexafluoride, SF,. @ In SF,, six unpaired electrons from sulfur are paired.
@ The outer shell of sulfur now contains 12 electrons, far more than the nearest noble gas, argon, Ar.
This is called expansion of the octet and is possible only from the n = 3 shell, when a d-sub-shell becomes available for the expansion.
Multiple covalent bonds
A multiple covalent bond exist when two atoms share more than one pair of electrons.
Double covalent bonds
® Ina double bond, the electrostatic attraction is between two shared pairs of electrons and the nuclei of the bonding atoms.
® Figure 7 shows dot-and-cross diagrams of double bonds in molecules of oxygen (OO) and carbon dioxide (O=C=0).
@ All atoms have eight electrons in their outer shell and the electron structure of the nearest noble gas.
@ C=C and C=O double bonds are very important in organic chemistry.
oO=0 —<—{ (h-———2 A Figure ? Double covalent bonds in oxygen, 0.,, and in carbon dioxide, CO,
Triple covalent bonds
® Ina triple bond, the electrostatic attraction is between three shared pairs of electrons and the nuclei of the bonding atoms.
® Figure 8 shows dot-and-cross diagrams of triple bonds in molecules of nitrogen (N=N) and hydrogen cyanide (H—C=N).
® Again, all atoms have the electron structure of the nearest noble gas.
RE
A Figure 8 Triple covalent bonds in nitrogen, N., and in hydrogen cyanide, HCN
Dative covalent bonds
A dative covalent or coordinate bond is a covalent bond in which the shared pair of electrons has been supplied by one of the bonding atoms only. In a dative covalent bond the shared electron pair was
ELECTRONS AND BONDING
originally a Jone pair of electrons on one of the bonded atoms. For example, Figure 9 shows formation of an ammonium ion, NH,”, by reaction of ammonia, NH,, and a H* ion.
® Anammonia molecule donates its lone pair of electrons to a H* ion.
®@ The dative covalent bond in NH,* is shown by a bond with an arrowhead, —, to show that the nitrogen atom provides both electrons to the covalent bond.
® Inan NH,* ion, all four bonds are equivalent and you cannot tell which is the dative covalent bond. The arrow for the dative covalent bond (Figure 10) just helps with accounting for all electrons.
a o EE) Co — ae w wv
A Figure 9 Formation of a dative covalent bond in NH,”
+
Average bond enthalpy
Average bond enthalpy serves as a measurement of covalent bond strength. The larger the value of the average bond enthalpy, the stronger the covalent bond. Table 2 shows some average bond enthalpies.
Summary questions
|
1 State what is meant by the term covalent bond. (1 mark)
2 Draw dot-and-cross diagrams and displayed formula for the following.
a F,0 (2 marks) b PH, (2 marks) e €S; (2 marks)
3 Draw dot-and-cross diagrams and displayed formulae for the following:
a C,H, (1 mark) b H,CO (carbon is the central atom) (1 mark) c H,0* (dative bond) (1 mark) d CO (triple bond, one dative) (1 mark)
4 Draw dot-and-cross diagrams and displayed formulae for the following:
a PF, (1 mark)
b CIF, (1 mark)
c¢ SO, (two double bonds) (1 mark)
d SO, (three double bonds) (1 mark) 5 Draw displayed formulae for:
a HNO, (nitrogen is bonded to three oxygens only) (2 marks)
b H,SO, (sulfur is bonded to four oxygens only) (2 marks)
H +
A Figure 10 Displayed formula of the NH,” ion
Y Table 2 Average bond enthalpies
Average bond enthalpy kJ mol"!
cae [a
Relati Roria elative
strength
Increasing bond strength
Synoptic link
Average bond enthalpies and related calculations are covered in detail in Topic 9.3, Bond enthalpies.
In Topic 15.1 Chemistry of the haloalkanes, you will see how average bond enthalpies can explain different rates of reaction.
_3 Chapter 5 Practice questions
Practice questions f Answer the parts below with a number.
1 Using the periodic table, or the Roman (i) The number of unpaired electrons numerals in the chemical name, predict the in a sulfur atom. (1 mark) ionic charges and write formulae for the (ii) The number of electrons occupying following: p-orbitals in a germanium atom.
a_ lithium phosphide (1 mark) b lithium phosphate (iii) The number of full shells in a c chromium(III) hydroxide RED EOT Aton: mark) 3 ; 3 3 Magnesium fluoride and potassium chloride d_ iron(II) selenide BOREAS loti an are examples of compounds with ionic e titanium(III) nitride bonding. f barium sulfate a_ Explain how ionic bonding holds together g barium sulfite the particles in an ionic compound. h_ nickel(I) manganate(VI) (8 marks) (1 mark)
2 This question looks at shells, sub-shells and b Draw a dot-and-cross diagram to show orbitals. the bonding in MgF,. Show outer
electrons only. (2 marks)
a What is meant by the term orbital? (1 mark) c The diagram represents the incomplete
7 structure of potassium chloride. b How many electrons completely fill
(i) a 3p orbital (1 mark) (ii) the 3d sub-shell (1 mark) (iii) The m = 4 shell. (1 mark)
c Write the electron configuration for the following atoms: (i) What name is given to this type (i) S$ (1 mark) of structure? (1 mark) (ii) Co (1 mark) (ii) Complete the diagram by adding
d_ Electrons are arranged in energy levels. labels to each circle in the diagram for The incomplete diagram below for the the particles present in the structure. seven electrons in a nitrogen atom shows (2 marks) just the two electrons in the Is level d A student found that MgF, has different
electrical conductivities when solid and
_d SSS Si when dissolved in water.
Explain these observations. (2 marks)
energy
e The table below shows the melting points as TTL of four compounds with ionic bonding.
lonic compound | NaF | Na,O | MgF, | MgO |
Complete the diagram by Melting point/°C 12?5 2852
(i) adding arrows to the boxes — (J mark) Identify the pattern in melting points and
(ii) adding labels for the other suggest reasons for the differences. sub-shell levels. (1 mark) (3 marks) e Write the electron configuration 4 This question looks at compounds and ions for the following ions: that have covalent bonding. (i) Br (1 mark) a_ PF, has covalent bonding. (ii) Ga** (1 mark)
c
Draw a dot-and-cross diagram to show the bonding in PF,. Show outer electrons only. (1 mark)
BF, reacts with NH, to form F,BNH,, which contains a dative covalent bond.
(i) Explain how a dative covalent bond is different from a normal covalent bond. (1 mark)
(ii) Draw a dot-and-cross diagram and a displayed formula for F,BNH,, showing the dative covalent bond. (2 marks)
The nitrate(V) ion, NO,”, is a polyatomic ion, bonded by covalent bonds. The three oxygen atoms are bonded by one single covalent bond, one double covalent bond and one dative covalent bond.
(i) Draw a displayed formula for the NO, ion. (1 mark)
(ii) Draw a dot-and-cross diagram to show the bonding on NO,~. Show outer electrons only. (2 marks)
An ionic compound has the empirical formula H,N,O,. Suggest the formulae of the ions present in this compound.
(2 marks)
5 When magnesium is heated in air, it reacts
with oxygen to form magnesium oxide. 2Mg(s) + O,(s) + 2MgO(s)
a Magnesium oxide is an ionic compound.
Draw a dot-and-cross diagram for MgO. Show outer electrons only. (2 marks)
Magnesium oxide has an extremely high melting point which makes it suitable as a lining for furnaces.
Explain, in terms of its structure and bonding, why magnesium oxide has this property. (3 marks)
When magnesium oxide is added to warm dilute nitric acid, a reaction takes place forming a solution containing ions. Solid MgO does not conduct electricity but the solution formed does.
(i) Explain the different conductivities of solid MgO and the solution. (2 marks)
(ii) Write an equation for the reaction between magnesium oxide and dilute nitric acid. Include state symbols.
(2 marks)
ELECTRONS AND BONDING
(iii) State the formulae of two main ions present in this solution. (2 marks)
6 This question look at bonding involving carbon with other atoms.
a
b
Draw a dot-and-cross diagram for a carbon dioxide molecule.
One representation of the bonding in a carbon monoxide shows a triple bond, as shown in the dot and cross diagram below.
~ U (i) State the number of lone pairs and dative covalent bonds in a
CO molecule. (1 mark)
(ii) State what is meant by a covalent bond. (1 mark)
(iii) State what is meant by dative
covalent bond. (1 mark)
Ethyne, C,H, also contains a triple bond. Draw a dot-and-cross diagram of an ethyne molecule. (1 mark)
A dot-and-cross diagram for a carbon monoxide molecule can also be drawn with a double bond between the carbon and oxygen atoms.
(i) Draw this dot-and-cross diagram.
(ii) What is unusual about this dot-and-cross diagram?
The cyanide ion, CN”, has a triple
bond and is isoelectronic with carbon
monoxide,
(i) Draw a dot-and-cross diagram for a CN7 ion.
(ii) Suggest what is meant by the term isoelectronic. (1 mark)
The displayed formula for a carbonate ion, CO,*, is shown below.
0 = C
aS
6) @)
Draw a dot-and-cross diagram for a carbonate ion. (1 mark)
SHAPES OF MOLECULES AND INTERMOLECULAR FORCES
6.1 Shapes of molecules and ions
Specification reference: 2.2.2
SOPHO SHHOSOSOHHOSSSSSOEHHSOOSOOSESOOOOSE,
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> electron-pair repulsion theory > shapes of molecules and ions.
SEC EEE EOE OOCOSSOSOOOOOCOCOSS LOSES CSET Se
Electron-pair repulsion theory
An electron has a negative charge, so electron pairs repel one another. The electron-pair repulsion theory is a model used in chemistry for explaining and predicting the shapes of molecules and polyatomic ions.
® The electron pairs surrounding a central atom determine the shape of the molecule or ion.
® The electron pairs repel one another so that they are arranged as far apart as possible.
® The arrangement of electron pairs minimises repulsion and thus holds the bonded atoms in a definite shape.
® Different numbers of electron pairs result in different shapes.
Shapes of molecules
Representing molecules in three dimensions
A molecule of methane, CH,, is symmetrical with four C—H covalent bonds.
® Four bonded pairs of electrons surround the central carbon atom.
® The four electron pairs repel one another as far apart as possible in
three-dimensional (3D) space. A Figure 1 The four electron pairs and
tetrahedral shape of a methane, CH, The result is a tetrahedral shape with four equal H—C—H bond angles molecule of 109.5°, shown in Figure 1. Wedges Synoptic link eres : : It is difficult to show a three-dimensional shape on a flat sheet of paper, The shapes of molecules and ions so chemists use wedges to help visualise structures in three dimensions:
are extremely important, especially for organic chemistry and transition element chemistry, Chapter 24. ® asolid wedge —“™¥ comes out of the plane of the paper
® asolid line ~~ represents a bond in the plane of the paper
® a dotted wedge “ty, BOERS into the plane of the paper "y
Wedges are especially useful for representing bonds in organic molecules.
Bonded-pair and lone-pair repulsions
A lone pair of electrons is slightly closer to the central atom, and occupies more space, than a bonded pair. This results in a lone pair
SHAPES OF MOLECULES AND INTERMOLECULAR FORCES
repelling more strongly than a bonding pair. The relative repulsions between lone pairs and bonding pairs are shown below.
bonded-pair/bonded-pair < bonded-pair/lone-pair < lone-pair/lone-pair
increasing repulsion
Molecular shapes from four electron pairs
Methane, CH,, ammonia, NH,, and water, H,O all have four electron pairs surrounding the central atom, but in ammonia and water the electron pairs are a mixture of bonded pairs and lone pairs.
® The four electron pairs around the central atom repel one another as far apart as possible into a tetrahedral arrangement. Lone pairs repel more strongly than bonded pairs.
® Therefore, lone pairs repel bonded pairs slightly closer together, decreasing the bond angle — the angle between the bonded pairs of electrons.
® The bond angle is reduced by about 2.5° for each lone pair.
Figure 2 compares the shapes of methane, ammonia, and water molecules, and their different bond angles.
Bonded pairs 4 3 2 Lone pairs 0 1 2 Name of shape tetrahedral pyramidal non-linear
Study tip
N eve 4 tH Make sure that you learn the
1o7 4 shapes and bond angles in these Shape and : molecules. bond angle
A Figure 2 Shapes of methane, CH, ammonia, NH, and water, H,0
Molecular shapes from multiple bonds
In molecules containing multiple bonds, each multiple bond is treated as a bonding region. For example, the bonding and shape of a carbon dioxide, CO,, molecule is shown in Figure 3.
Molecule = Dot-and-cross Number of Shape and Name of diagram bonding regions bondangle shape
linear
A Figure 3 Dot-and-cross diagram and shape of a carbon dioxide, CO, molecule
6.1 Shapes of molecules and ions
® The 4 bonded pairs around the central carbon atom are arranged as two double bonds, which count as two bonded regions.
®@ The two bonded regions repel one another as far apart as possible. This gives the carbon dioxide molecule a linear shape with all three atoms aligned in a straight line.
Molecular shapes from other numbers of electron pairs
The principles of electron-pair repulsion theory can be applied to any number of electron pairs surrounding the central atom.
@ Electron pairs around the central atom repel each other as far apart as possible.
The greater the number of electron pairs, the smaller the bond angle.
Lone pairs of electrons repel more strongly than bonded pairs of electrons.
Three electron pairs
Boron trifluoride, BF,, has only three bonded pairs around the central boron atom. Electron-pair repulsion gives a trigonal planar shape with The octahedral shape resulting equal bond angles of 120° (Figure 4).
from six bonding pairs is extremely important in transition metal
chemistry, which you will study in Chapter 24, Transition elements central sulfur atom, Electron-pair repulsion gives an octahedral shape ’ *
with equal bond angles of 90° (Figure 4).
Synoptic link
Six electron pairs Sulfur hexafluoride, SF,, has six bonded pairs of electrons around the
Electron pairs/regions
Shape and bond angle
Name of shape
A Figure 4 Shapes and bond angles for molecules with different numbers of electron pairs
An octahedral shape from six bonded pairs
It may seem strange that SF,, with six bonded pairs, forms a molecule that is an octahedral shape. The reason lies with what is meant by
the shape of a molecule. The six fluorine atoms are positioned at the corners of an octahedron. From the diagrams in Figure 5, you should be able to see that an octahedral shape with eight sides is obtained by A Figure 5 In an SF, molecule, six joining together the six corners occupied by fluorine atoms.
bonded pairs give an octahedral shape
72
SHAPES OF MOLECULES AND INTERMOLECULAR FORCES
Shapes of ions
Electron-pair repulsion theory can also be used to explain and predict the shapes of ions.
The ammonium ion The ammonium ion, NH,*, has four bonded pairs surrounding the central nitrogen atom (Figure 6).
+ H + 109.5° 7
H 4
A Figure 6 The four bonded pairs and tetrahedral shape of an NH," ion
@ AnNu,- ion has the same number of bonded pairs of electrons around the central atom as a methane molecule.
@ NH,* has the same tetrahedral shape and bond angles (109.5°) as a methane molecule.
Carbonate, nitrate, and sulfate ions Figure 7 shows the shapes of carbonate, CO,*-, nitrate, NO,~, and sulfate, SO,*-, ions.
® CO,* and NO, ions have three regions of electron density surrounding the centre atom. So they have the same shape as a BF, molecule.
® SO,* ions have four centres of electron density around the central sulfur atoms and have the same shape as a methane molecule.
a) ) 0 hy i I G N ee Xu "0
0 8) trigonal planar trigonal planar tetrahedral CO,* ion NO; ion SO,? ion
A Figure ? Shapes of carbonate, nitrate, and sulfate ions
Predicting molecular shapes and bond angles
You should be able to predict the shapes of, and bond angles in, molecules and ions with different numbers of electron pairs, whether bonded pairs or lone pairs. The principles of electron-pair repulsion enable you to predict the arrangement of electron pairs around a central atom of unfamiliar molecules and ions.
Study tip
Remember that dot-and-cross diagrams can be very useful for working out the arrangement of electron pairs and therefore predicting molecular shapes.
Summary questions
1 Name the shapes and give the bond angles around the central atom in the following molecules:
a Bel, (1 mark) b BCI, (1 mark) c SiH, (1 mark) d H,C=0 (1 mark) e CS, (1 mark)
2 Name the shapes and bond angles in the following: a CH,0H (around C and 0) (1 mark) b SO, (1 mark) ce SO, (1 mark) d H,0° (1 mark)
3 BF, reacts with NH, to form the compound F,BNH.. The bond angles in BF, and NH, are different from the bond angles in F,BNH,,.
a State the bond angles in BF,andNH,. (2 marks) b Predict, with reasons, the bond angles in FBNH.. (2 marks)
6.2 Electronegativity and polarity
Specification reference: 2.2.2
Oe eececcccescccccccccccsescccoscccccoces, Electronegativity Learning outcomes
Demonstrate knowledge, understanding, and application of:
> electronegativity > Pauling electronegativity
In a covalent bond, the nuclei of the bonded atoms attract the shared pair of electrons. In molecules of elements, such as hydrogen, H,, oxygen, O,, nitrogen, N,, and chlorine, Cl,, the atoms are the same element and the bonded electron pair is shared evenly.
This changes when the bonded atoms are different elements:
values : @ the nuclear charges are different > bond polarity. . 8 See eeeeeeessesseesesseseeseseeessescese®® ® the atoms may be different sizes
@ the shared pair of electrons may be closer to one nucleus than the HS other.
The shared pair of electrons in the covalent bond may now experience more attraction from one of the bonded atoms than the other.
attraction A Figure 1 Attraction causing a covalent bond The attraction of a bonded atom for the pair of electrons in a covalent bond is called electronegativity.
How is electronegativity measured?
electronegativity increases The Pauling scale is used to compare the electronegativity of the atoms of different elements. Figure 2 shows how Pauling electronegativity values depend on an element's position in the periodic table.
Across the periodic table Mel Al Si cl ® the nuclear charge increases 121151/18] 211/125] 30 ® the atomic radius decreases. eee A large Pauling value indicates that atoms of the element 28 are very electronegative. You can see that electronegativity A Figure 2 Pauling electronegativity values in the increases across and up the periodic table. Consequently, Periodic Table fluorine is the most electronegative atom and is given a
Pauling value of 4.0. The noble gases are not included as they tend not to form compounds.
® The non-metals nitrogen, oxygen, fluorine, and chlorine have the most electronegative atoms.
@ The Group | metals, including lithium, sodium, and potassium have the least electronegative atoms.
VY Table 1 Jonic or covalent
lonic or covalent? Bond type Electronegativity If the electronegativity difference is large, one bonded atom will have difference a much greater attraction for the shared pair than the other bonded
atom, The more electronegative atom will have gained control of the electrons and the bond will now be ionic rather covalent.
Electronegativity values can be used to estimate the type of bonding as shown in Table 1.
SHAPES OF MOLECULES AND INTERMOLECULAR FORCES
Bond polarity Non-polar bonds
In a non-polar bond, the bonded electron pair is shared equally between the bonded atoms. A bond will be non-polar when:
@ the bonded atoms are the same, or @ the bonded atoms have the same or similar electronegativity.
In molecules of elements such as hydrogen, oxygen, and chlorine, the bonded atoms come from the same element and the electron pair is shared equally. The bond is a pure covalent bond (Figure 3). Carbon and hydrogen atoms have very similar electronegativities and form non-polar bonds. Hydrocarbon liquids such as hexane, C,H, ,, are non-polar solvents and do not mix with water.
Polar bonds
In a polar bond, the bonded electron pair is shared unequally between the bonded atoms. A bond will be polar when the bonded atoms are different and have different electronegativity values, resulting in a polar covalent bond.
Example: Hydrogen chloride Hydrogen chloride, HCI, has atoms of different elements.
® From Figure 2, hydrogen has an electronegativity of 2.1 and chlorine has an electronegativity of 3.0.
@ The chlorine atom is more electronegative than the hydrogen atom.
@ The chlorine atom has a greater attraction for the bonded pair of
electrons than the hydrogen atom, resulting in a polar covalent bond.
The H—C]I bond is polarised with a small partial positive charge (5+) on the hydrogen atom and a small partial negative charge (4—) on the chlorine atom (Figure 4). The delta 6 sign means small. The two charges, 5+ and 6- are partial charges, and are much smaller than a full + and — charge.
® This separation of opposite charges is called a dipole.
® The hydrogen chloride molecule is polar, with 4+ and 5- charges at different ends of the H—CI bond.
A dipole in a polar covalent bond does not change and is called a permanent dipole to distinguish it from an induced dipole, which you will meet in the next topic.
Polar molecules
Hydrogen chloride is a polar molecule, as the H—C1 bond has one permanent dipole acting in the direction of the H—ClI bond. For molecules with more than two atoms, there may be two or more polar bonds. Depending on the shape of the molecule, the dipoles may reinforce one another to produce a larger dipole over the whole molecule, or cancel out if the dipoles act in opposite directions.
HSH as a non-polar electron pair is attracted equally to each bonded atom A Figure 3 Two non-polar molecules, hydrogen, H., and chlorine, Ci,. Only the
bonding electrons have been shown for chlorine
5+ > 5- H 3c
polar bonded electron pair is attracted closer to Cl atom A Figure 4 A polar molecule of hydrogen chloride, HCI. Only the bonding electrons have been shown for chlorine
Study tip
If you are provided with Pauling electronegativity values:
@ the atom with the larger electronegativity value has the 5— charge
the atom with the smaller electronegativity value has the 5+ charge.
6.2 Electronegativity and polarity
&- 5- A water, H,O molecule is polar. a ye @ The two O—-H bonds each have a permanent dipole. Pa a , ®@ The two dipoles act in different directions but do not exactly + cecitdaenkiesiea Pesto” oppose one another. ; ® Overall the oxygen end of the molecule has a 6- charge and A Figure 5 The water, H,0, molecule y8 : ‘B is polar the hydrogen end of the molecule has a 6+ charge (Figure 5). —— 5 — A carbon dioxide, CO, molecule is non-polar. dipoles cancel : Ow d ae @ The two C=O bonds each have a permanent dipole. —_— —— —— ® The two dipoles act in opposite directions and exactly oppose ipole CO; molecule in one another. A Figure 6 The carbon dioxide, C0, ® Over the whole molecule, the dipoles cancel and the overall molecule is non-polar dipole is zero (Figure 6).
Polar solvents and solubility
Figure 7 shows a sodium chloride lattice being dissolved by water
The solubility of ionic compounds molecules to form aqueous sodium and chloride ions: by polar solvents was introduced
in Topic 5.2, lonic bonding and NaCl(s) + aq — Na*(aq) + Cl"(aq) structure. ® Water molecules attract Na* and Cl” ions.
Synoptic link
® The ionic lattice breaks down as it dissolves.
In the resulting solution, water molecules surround the Na* and Cl" ions.
You can see that:
® Na*® ions are attracted towards the oxygen of water molecules (4-)
® Cl ions are attracted towards the hydrogen of water molecules (5+).
Summary questions
1 Define: a electronegativity (1 mark) b polar covalent bond (1 mark) c dipole (1 mark)
2 This question refers to the following compounds: Br., NO. Na_0, PH,, Al,0.,, Si0., KF and LiBr Predict the type of bonding in each compound using the electronegativity values from Figure 2 in order to list the compounds from most ionic
A Figure? Solid sodium chloride, NaCl, to most covalent. (2 marks) (top) dissolving in water to give sodium
and chloride ions (bottom) surrounded 3. This question refers to the following compounds, all with polar bonds:
by water molecules HS, BeBr,, NH., BF.,, SiCl,, CHCI,, H,C=0, PF., SF,
a State which way round the dipole is in each bond. (Refer to the electronegativity values in Figure 2.)
b Classify the molecules as polar and non-polar, explaining how you have made your decisions. (You may find it useful to refer to Topic 6.1 to work out the shapes of the molecules.) (9 marks)
6.3 Intermolecular forces
Specification reference: 2.2.2
Learning outcomes
Demonstrate knowledge,
understanding, and application of:
> induced dipole—dipole interactions (London forces)
Forces between molecules Covalent bonding is strong and holds the atoms in a molecule together.
Intermolecular forces are weak interactions between dipoles of different molecules. Intermolecular forces fall into three main categories.
® induced dipole-dipole interactions (London forces)
® permanent dipole-dipole interactions > permanent dipole-dipole
© hydrogen bonding. interactions Intermolecular forces are largely responsible for physical properties such as melting and boiling points, whereas covalent bonds determine
the identity and chemical reactions of molecules.
> structure and properties of simple molecular lattices.
et PPPS OOOO OOHSESOOHEOEHSEOO OOH EESE,
meee eee eee SESS SES EEESEEESSEEESESEESESESES Table 1 compares the strengths of intermolecular forces with covalent bonds.
V Table 1 Strengths of intermolecular forces and covalent bonds
Type of bond Bond enthalpy /kJ mol~*
London forces 1-10 permanent dipole—dipole interactions 3-25 hydrogen bonds 10-40
single covalent bonds 150—S00
Induced dipole—dipole interactions (London forces)
London forces are weak intermolecular forces that exist between all molecules, whether polar or non-polar. They act between induced dipoles in different molecules. The origin of induced dipoles is described in Figure 1.
° Movement of electrons produces a changing dipole an instan| an instant in a molecule. later later
e At any instant, an instantaneous dipole will exist, instantaneous dipole constantly changing but its position is constantly shifting J
e The instantaneous dipole induces a dipole on a Q neighbouring molecule Q QQ
induced dipole on neighbouring molecule
e The induced dipole induces further dipoles on neighbouring molecules, which then attract one (=) QO Saye)
another.
induced dipole—dipole interactions
A Figure 1 Origin of induced dipole-dipole interactions (London forces)
Synoptic link
The shapes of molecules also affect the strength of London forces. This is especially important in organic chemistry (see Topic 12.1 , the properties of the alkanes).
covalent permanent bond dipole-dipole interaction
A Figure 2 Permanent dipole—dipole interactions between hydrogen chloride, HCI, molecules
78
6.3 Intermolecular forces
Induced dipoles are only temporary. In the next instant of time, the induced dipoles may disappear, only for the whole process to take place amongst other molecules.
The strength of induced dipole—dipole interactions (London forces)
Induced dipoles result from interactions of electrons between molecules. The more electrons in each molecule:
@ the larger the instantaneous and induced dipoles
®@ the greater the induced dipole-dipole interactions
@ the stronger the attractive forces between molecules.
Table 2 compares the boiling points of the first three noble gases, helium to argon.
® Larger numbers of electrons mean larger induced dipoles.
®@ More energy is then needed to overcome the intermolecular
forces, increasing the boiling point.
¥ Table 2 London forces and boiling points of the noble gases helium to argon
Noble gas Number ofelectrons Boilingpoint/°C _Relative strength
| etiumne | 2 | =268 | stronger
What ever happened to van der Waals’ forces?
The term van der Waals, forces has sometimes been used to describe induced dipole-dipole interactions. The International Union of Pure and Applied Chemistry (IUPAC), which publishes guidelines for chemical terminology, recommends that van der Waals’ forces be used for both permanent and induced dipole-dipole interactions, and London forces for induced dipole-dipole interactions. So the term van der Waals’ forces is ambiguous.
London forces is the correct term to use when describing induced dipole-dipole interactions. In other sources, you may see the term dispersion forces. Whoever said the language of science is straightforward?
Permanent dipole—dipole interactions
In Topic 6.1, you looked at polarity in bonds and molecules. Permanent dipole-dipole interactions act between the permanent dipoles in different polar molecules.
Figure 2 shows intermolecular forces arising from permanent dipole-dipole interactions in hydrogen chloride, HCL, molecules.
Table 3 compares hydrogen chloride with fluorine, F,. Molecules of hydrogen chloride and fluorine have the same number of electrons and the same shape, so the strength of the London forces in hydrogen
chloride and fluorine should be very similar.
SHAPES OF MOLECULES AND INTERMOLECULAR FORCES
Y Table 3 Induced and permanent dipoles
Study tip Permanent ms London Number of Boiling
dipole—dipole It is a common mistake to forces electrons point /°C
forget that polar molecules have induced dipole interactions as well as permanent dipole—dipole interactions.
Molecule Dipole interactions
@ Fluorine molecules are non-polar and only have London forces between molecules.
® Hydrogen chloride molecules are polar and have London forces and permanent dipole-dipole interactions between molecules.
® Extra energy is needed to break the additional permanent dipole-dipole interactions between hydrogen chloride molecules.
® The boiling point of hydrogen chloride is therefore higher than fluorine.
Simple molecular substances
e Se + ,0 <j sle ryec ae
A simple molecular substance is made up of sim] le m« lecules euccaae Srwesrvarseastibeat Sramesniinen Beteioses small units containing a definite number of atoms with a definite l, molecules break on changing state molecular formula, such as neon, Ne, hydrogen, H,, water, H,O, and carbon dioxide, CO,.
In the solid state, simple molecules form a regular structure called a simple molecular lattice. In the simple molecular lattice: @ the molecules are held in place by weak intermolecular forces ® the atoms within each molecule are bonded together strongly by covalent bonds. Figure 3 shows the different forces in the simple molecular lattice of iodine, I,.
Properties of simple molecular substances
Low melting point and boiling point strong covalent bonds between atoms in All simple molecular substances are covalently bonded. At room lz molecule do not break on changing state temperature, they may exist as solids, liquids, or gases. All simple
molecular substances can be solidified into simple molecular lattices
by reducing the temperature.
® Ina simple molecular lattice, the weak intermolecular forces can be broken even by the energy present at low temperatures.
® Simple molecular substances have /ow melting and boiling points.
When a simple molecular lattice is broken apart during melting,
@ only the weak intermolecular forces break A Figure 3 lodine, |,, is a solid at room ® the covalent bonds are strong and do not break. temperature, but can easily be turned into ~~ a purple vapour. A small Bunsen flame Solubility provides enough energy to break the Covalent substances with simple molecular structures fall into two weak intermolecular forces between |, categories — polar and non-polar. molecules in the simple molecular lattice
The solubility of non-polar substances is easier to predict. 79
i | |
— _
A Figure 4 Solubility of iodine (purple when dissolved) in polar and non-polar solvents. From left to right — cyclohexane and iodine, water and iodine, water
and cyclohexane, and layers of water (bottom) and cyclohexane (top) with iodine. lodine dissolves readily in cyclohexane, a non-polar solvent, but not in water, a polar solvent
Summary questions
1 Explain how an induced dipole forms. (2 marks)
2 Explain why simple molecular
compounds:
a_ have low melting and boiling points
b donot usually dissolve readily in water
c have poor electrical conductivity. (3 marks)
3 For each of the following, state the structure. Predict and explain the following physical properties: melting and boiling points, electrical conductivity,
solubility. ‘b CCl, (4 marks)
6.3 Intermolecular forces
Solubility of non-polar simple molecular substances
@ When a simple molecular compound is added to a non-polar solvent, such as hexane, intermolecular forces form between the molecules and the solvent.
® The interactions weaken the intermolecular forces in the simple molecular lattice. The intermolecular forces break and the compound dissolves.
Therefore, non-polar simple molecular substances tend to be soluble in non-polar solvents.
® When a simple molecular substance is added to a polar solvent, there is little interaction between the molecules in the lattice and the solvent molecules.
@ The intermolecular bonding within the polar solvent is too strong to be broken.
Therefore simple molecular substances tend to be insoluble in polar solvents.
Figure 4 compares the solubility of iodine in water (polar solvent) and in cyclohexane (non-polar solvent).
Solubility of polar simple molecular substances
Polar covalent substances may dissolve in polar solvents as the polar solute molecules and the polar solvent molecules can attract each other. The process is similar to dissolving of an ionic compound. For example, sugar dissolves in water, a polar solvent. Sugar is a polar covalent compound with many polar O—H bonds, which attract and bond with polar water molecules. This solubility can extend to liquids and gases. Hydrogen chloride is a gas with a polar H—CI bond that is extremely soluble in water, forming hydrochloric acid.
The solubility depends on the strength of the dipole and can be hard to predict. Some compounds such as ethanol, C,H,OH, contain both polar (the O—H) and non-polar (the carbon chain) parts in their structure and can dissolve in both polar and non-polar solvents.
Some biological molecules have hydrophobic and hydrophilic parts.
The hydrophilic part will be polar and contain electronegative atoms (usually oxygen) that can interact with water. The hydrophobic part will be non-polar and comprised of a carbon chain.
Electrical conductivity
® There are no mobile charged particles in simple molecular structures.
® With no charged particles that can move, there is nothing to complete an electrical circuit.
Therefore simple molecular structures are non-conductors of electricity.
6.4 Hydrogen bonding
Specification reference: 2.2.2
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> hydrogen bonding
Hydrogen bonds
A hydrogen bond is a special type of permanent dipole-dipole interaction found between molecules containing:
® an electronegative atom with a lone pair of electrons, for example, oxygen, nitrogen, or fluorine
ett teeeeersccoosseees,
. > anomalous properties of water. ® a hydrogen atom attached to an electronegative atom, for example,
H—Q, H—N, or H—F.
IPPC PPC OCC OSOOOSOOCSCOOOCOCOCOOCOSOSCOOCOO.
hydrogen bond
The hydrogen bond acts between a lone pair of electrons on an 5+ 5+ electronegative atom in one molecule and a hydrogen atom in H | H a different molecule. Hydrogen bonds are the strongest type of ee eee roe intermolecular attractions. & § ee &
F P , electronegative The hydrogen bond is shown by a dashed line. The lone pair of 0 pbs Hi
electrons on the oxygen in water, and the nitrogen in ammonia, play
a key role in a hydrogen bond. The shape around the hydrogen atom hydrogen bond
5+ involved in the hydrogen bond is linear. H R ad Hd 5+ s 5+ Is H—N:-----4-- H——N Anomalous properties of water 5 §- Hydrogen bonding has a significant influence on the properties of electronegative : N atom H N many molecules but none more so than water. Hydrogen bonding gives water some unique and anomalous (unusual) properties that 4 Figure 1 Hydrogen bonding between support the existence of life on Earth. molecules of water, H20, and ammonia, NH, The solid (ice) is less dense than the liquid (water) ® Hydrogen bonds hold water molecules apart in an open Study tip lattice structure. When you draw a diagram to show The water molecules in ice are further apart than in water. hydrogen bonding, remember
to show lone pairs and the atom
Solid ice is less dense than liquid water and floats. ae ‘ polarities. Hydrogen is always 5+!
Ice floating on water may seem obvious, but water is one of the
few substances in which the solid is less dense than water. The consequence is that ice floats instead of sinking in ponds and lakes, forming an insulating layer and preventing the water from freezing solid — good news if you are a fish. With two lone pairs on the oxygen atom and fwo hydrogen atoms, each water molecule can form four hydrogen bonds. The hydrogen bonds extend outwards, holding water molecules slightly apart and forming an open tetrahedral lattice full of holes (Figure 3). The bond angle about the hydrogen atom involved in the hydrogen bond is close to 180°.
The holes in the open lattice structure decrease the density of water
on freezing. When ice melts, the ice lattice collapses and the molecules A Figure 2 At 0°C, the density of ice is
move closer together. So liquid water is denser than solid ice. 0.917 g cm™? compared to 1.029 g cm™ for seawater. Only about 1 of the volume of the iceberg is above the surface of the woter
vy y
o- goes o>
A Figure 3 The open lattice structure of ice, in which the linear H---O—H
arrangement holds the water molecules
apart
6.4 Hydrogen bonding
Water has a relatively high melting point and boiling point
As with all molecules, water has London forces between molecules.
® Hydrogen bonds are extra forces, over and above the London forces.
@ An appreciable quantity of energy is needed to break the hydrogen bonds in water, so water has much higher melting and boiling points than would be expected from just London forces.
@ When the ice lattice breaks, the rigid arrangement of hydrogen bonds in ice is broken. When water boils, the hydrogen bonds break completely.
Without hydrogen bonds, water would have a boiling point of about -75°C and would exist as a gas at room temperature and pressure. There would be no liquid water in most places on Earth and there would be no life as we know it.
Figure 4 shows the boiling points of the hydrides of Groups 14-17 (Groups 4—7) in the periodic table.
1 Water, hydrogen fluoride, and ammonia do not follow the trend shown by the other hydrides
in each group.
a_ Estimate what the boiling points of water, hydrogen fluoride, and ammonia would be if they were to follow the group trends.
b Explain why water, hydrogen fluoride and ammonia do not follow the group trends,
boiling point/°C
2 Explain why all groups show an increase in boiling point from Period 3 to Period 6.
3 What conclusions can be drawn about the relative strengths of London forces and permanent dipole—dipole interactions for the
hydrides of Groups 14—-1? (Groups 4—7)?
Synoptic link
Hydrogen bonding is very important in organic compounds containing O—H, NH, and
C=0 groups, such as alcohols, carboxylic acids, carbonyl compounds, amines, and amino acids. You will learn about all these groups when you study organic chemistry later in the course (Chapter 11 onwards).
A Figure 4 Boiling points of hydrides of Groups 14-17 (Groups 4—7)
Other anomalous properties of water
The extra intermolecular bonding from hydrogen bonds also contributes to many more unusual properties of water. Other examples are a relatively high surface tension and viscosity, properties than result in droplets that are ‘not wet’ and allow insects to walk on pond surfaces. Detergents reduce the surface tension, making water ‘wetter’. Water has dozens of anomalous properties, so you can be relieved that you are only required to know about density, melting points, and boiling points.
SHAPES OF MOLECULES AND INTERMOLECULAR FORCES
Hydrogen bonding in DNA
The double helix structure of DNA (Figure 5) is held together by hydrogen bonds which enable a single DNA strand to create a perfect copy of itself in a process called replication.
The replication process depends on four bases adenine A, thymine T, cytosine C, and guanine G being present in the correct order.
In Figure 5, you can see that:
@ AandT pair by forming two hydrogen bonds @ CandG pair by forming three hydrogen bonds So why do the bases always pair up correctly?
The chemical structure and shape of these four bases ensure Correct pairing. A Figure 5 Hydrogen bonds in DNA hold the double helix structure together and enable DNA to replicate by ensuring
Adenine and guanine are both purine bases with two- that the correct bases are paired.
ringed structures.
Thymine and cytosine are both pyrimidine bases with 1 Suggest why pairing does not take place between Single-ringed structures. two purine bases or between two pyrimidine bases. @ Hydrogen bonding in the double helix can only take 2 Human DNA typically contains just over three billion place between a purine and a pyrimidine base. base pairs in a definite sequence. During replication, The bases must fit together so that a hydrogen atom this sequence must be copied perfectly. Ifreplication took place randomly, how many possible sequences
from one molecule and an electronegative atom ‘ of the four bases could there be in human DNA?
(either oxygen or nitrogen) from the other molecule are aligned correctly to maximize hydrogen bonding.
Figure 6 shows the pairing of A with T and C with G.
adenine e thymine cytosine A T Cc
A Figure 6 Base pairing in DNA by hydrogen bonding
Summary questions
1 State and explain two anomalous properties of water. 3 Draw diagrams showing two ways that hydrogen (4 marks) bonding can form between: a One molecule of ammonia and one molecule
2 State which of the following compounds have
¢ f water hyd bonding H,0, HS, CH,, CH.OH, NO... o BOTSGr An orn ies niger xoaB a8 (2 rans) b One molecule of water and one molecule of ethanol, CH,CH.OH. (4 marks)
3 Chapter 6 Practice questions
Practice questions 4 This question is about the molecular
1 This question looks at polarity and shapes of fhuorides, £30, CEy NEjand OF:
molecules and ions a Complete the table as follows. a_ Arrange the molecules below in order of (i) Add the number of bonded pairs and increasing polarity. lone pairs of electrons around the (i) HCl, HBr, HE, HI (1 mark) atom in bold. ii) CH.Br.. CH.L CHCLE CF ; A Ignore any inner shells. (4 marks) w) — ee af Sa (ii) Draw a 3D diagram for the shape of b_ The list of ten molecules and ions below each molecule. Include any lone pairs can be arranged into five pairs, each with (4 marks)
a different shape.
H,O, HCN, AICL,-, BF,, SO,, SCl,, H,O*,
NH,*, NH,, CO,
(i) Select the five pairs and state the shape of each pair. (5 marks)
Molecule Bondedpairs Lonepairs Shape
(ii) Which of the molecules and ions are planar? (1 mark)
= a ee ce es oe 2.4 a ee ie 2 a a Ge
b_ In their solid structures, the four molecular
z & lete the table bel follows. OMpicte Ine table DEeLow as tONOws fluorides all have London forces.
State the bond angle(s). (5 marks) ; er Explain the origin of London forces. b Name the shape of each molecule. (3 marks) (5 marks) us i 3 c Electronegativity explains why all of the Molecule Bond angle molecules have polar bonds. (i) Explain the term electronegativity. (J mark) (ii) Show all the dipoles on a molecule of CP,. (1 mark)
d= Explain which of the molecules in the table are polar and which are
3 SbCL,, exists as simple covalent molecules. non-polar. (3 marks) A ‘dot-and-cross’ diagram of SbCl, is shown 5 Much of the chemistry of water is influenced below. by its polarity and its ability for form
Pak m hydrogen bonds.
oo? a What is meant by a hydrogen bond?
* sb X ci ° (1 mark) i eC s
a hd b_ Explain why water molecules are polar. Ge (2 marks)
c Draw a diagram showing hydrogen bonding between two molecules of water. Include dipoles and lone pair of electrons. (2 marks)
a_ Predict the shape of a molecule of SbCI,. Explain your answer. (3 marks)
b SbCl, molecules are polar. Explain why. (2 marks)
d_ State the bond angle in a water F321 Jan 2014 QI(d)
molecule. (J mark)
e State and explain two properties of ice that are a direct result of hydrogen bonding. (4 marks)
SHAPES OF MOLECULES AND INTERMOLECULAR FORCES
6 Linus Pauling was a Nobel Prize winning chemist who devised a scale of electronegativity. Some Pauling electronegativity values are shown in the table.
b
Element
Electronegativity
What is meant by the term
electronegativity? (2 marks)
Show, using 6+ and 6—- symbols, the permanent dipoles on each of the following bonds. N—F; N—Br (1 mark)
Boron trifluoride, BF,, ammonia, NH,, and sulfur hexafluoride, SF,, are all covalent compounds. The shapes of their molecules are different.
(i) State the shape of a molecule of SF,. (1 mark)
(ii) Using outer electron shells only, draw ‘dot-and-cross’ diagrams for molecules of BF, and NH,. Use your diagrams to explain why a molecule of BF, has bond angles of 120° and NH, has bond angles of 107°. (5 marks)
(iii) Molecules of BF, contain polar bonds, but the molecules are non-polar. Suggest an explanation for this difference. (2 marks)
F321] Jan 11 Q3
7 Simple molecules are covalently bonded.
b
State what is meant by the term
covalent bond. (1 mark)
Chemists are able to predict the shape
of a simple covalent molecule from the number of electron pairs surrounding the central atom.
(i) Explain how this enables chemists to predict the shape. (2 marks)
(ii) The ‘dot-and-cross’ diagram of the simple covalent molecule, H,BO,, is shown below.
H . > i" ° ° 4 Os a) es te %. B ae xe = Os xe H
Predict the O—B—O and B—O—H bond angles in a molecule of H,BO,. (2 marks)
Give an example of a simple covalent molecule which has all bond angles equal to 90°. (1 mark)
F321] Jan 2013 Q2
8 Hydrogen chloride is a colourless gas which forms white fumes in moist air.
a
b
Molecules of hydrogen chloride, HCl, and molecules of fluorine, F,, contain the same number of electrons. Hydrogen chloride boils at —85 °C and fluorine boils at -188 °C.
Explain why there is a difference in the boiling points of HCl and F,. In your answer you should refer to the types of force acting between molecules and the relative strength of the forces between the molecules.
In your answer, you should use appropriate technical terms, spelled correctly. (4 marks)
Hydrogen chloride reacts with water to produce an ion with the formula H,O*. An H,O* ion has one dative covalent bond.
Draw a ‘dot-and-cross’ diagram to show
the bonding in H,O*. Show outer
electrons only. (2 marks)
F321 Jan 13 Q5(a)(b)
9 The compounds, NH,, PF, and SF, are all gases at room temperature and pressure.
a
What intermolecular forces are present in liquid samples of each compound? (3 marks)
The boiling point of NH, is much higher than PF, and SF,.
Explain why. (2 marks) Name the shapes and give the bond angles in molecules of NH, and SF,?
(4 marks)
Module 2 summary
e acid + carbonates — salt + CO. + H,0
TO) i i nase HY neutralisation
e acid metal oxides —» salt + H20
alkalis release OH” titrations e acids + alkalis —»H,0 . he acids and metals ( tect : reduction : Gane e decrease in oxidation number redox | ton electron lost ; | * increase in oxidation number —«-St@ndard solutions { @ mass:
n=m/M “ Vicm?) | 7 oR = 34000 e formula determination = vam") —« reacting quantities f 24 actual yield pV=nRT e percentage yield =
theoretical yield x 100
e solutions: n= c(dm3)xV XM,(desired products)
_ ¢xV(cm?) oe = =M,{all products) x 100 L 1000 A amount of substance a moles e relative atomic mass e mass spectrum e isotopes writing equations © ee atoms atomic structure | hydrogen bonds ions.
permanent dipole-dipole interactions London forces
"shells sub-shells orbitals
J
formulae from ions electron configuration Intermolecular forces
dot-and-cross diagrams
giant ionic lattice simple molecular lattice
— 4 electronegativity Properties ,
polarity
melting and boiling points _
Module 2 Foundations in Chemistry
Krypton
Krypton, Kr, is an element with atomic number 36. It is a colourless gas at room temperature and is used in energy-saving fluorescent lights.
Krypton is in Group 18(0) of the periodic table. It was thought that krypton was completely unreactive, However, in 1963 it was discovered that krypton could react with fluorine, under extreme conditions, to form the molecule krypton difluoride, KrF..
Kr+F, — KrF,
Krypton difluoride is unstable and very reactive — it is the most powerful oxidising agent so far discovered.
1 The main isotope of krypton is 8? Kr. State the number of protons, neutrons, and electrons in this isotope and write down the electron configuration of a krypton atom. Use your answer to suggest why krypton is usually very unreactive.
What is the oxidation state of krypton in KrF?
What mass of krypton would be needed to form 1.0 g of KrF? What would the volume in dm? of this mass of krypton (at RTP)?
Draw out a dot-and-cross diagram of the molecule KrF... Show outer electron shells only.
Extension
Research the molecule diborane, BH, and draw a dot and cross diagram of this molecule. Comment on what is unusual about its bonding and structure. Produce a summary for how to deduce whether a molecule has an overall dipole. Include details of the factors that affect bond polarity and the effect of molecular shape. In your summary, include examples of several molecules, with and without overall dipoles There are two isotopes of bromine atoms. Bromine in its standard state is a liquid consisting of Br, molecules. When a sample of bromine vapour is analysed using a mass spectrometer, three peaks are seen, with m/z values of 158, 160, and 162. a_ Use ideas about isotopes to explain why three peaks are seen and
State the mass number of the two isotopes of bromine b The percentage abundances of these peaks are as follows:
158 : 25.52%, 160 : 50.00%, 162 : 24.48%
Use these data to calculate the percentage abundance (to 4 s.f.)
of the isotopes of bromine
Module 2 Practice questions
1 What is the atomic structure of °°Se2-?
Protons Neutrons Electrons
acai =* J
d mark)
AS Paper | style question 2 Asample of ethanol, C,H,O, contains 3.00 mol of carbon atoms.
How many hydrogen atoms are in the sample of ethanol?
A 4.98x 10-74 B 1.50x 10°73 Cc 1.81x10* D 5.42x1074 = (1 mark) AS Paper | style question
3 A compound has the percentage composition by mass Ag: 71.03%, C: 7.90%, and O: 21.07%.
What is the empirical formula of the
compound? A AgCO, B AgCO, C Ag,C,O D Ag,C,O AS Paper 1 style question (1 mark)
4 14.3g of hydrated sodium carbonate, Na,CO,*10H,0 is dissolved in water and made up to 250.cm? of solution.
What is the concentration of sodium ions in the solution, in moldm-*?
A 0.0500 B 0.100 C 0.200 D 0.400
AS Paper I style question (1 mark) 5 What is the oxidation number of Mn in KMnO,?
A +2 B +4 C +6 D +7 (1 mark) AS Paper | style question
6 The atoms of an element contain seven full orbitals and two singly-occupied orbitals.
What is the element? A Si BP cs D Cl (1 mark) AS Paper | style question
7 What is the F—B—F bond angle in boron trifluoride, BF,?
A 90° B 107° C 120° D 180° (J mark)
AS Paper | style question
8 This question looks at some chemistry of
2
phosphorus and its compounds. a_ State the full electron configuration of phosphorus showing sub-shells. (J mark) b_ Phosphoric acid, H,PO,, can be made as the only product by reacting the oxide of phosphorus, P,O,,, with water. (i) Write a balanced equation for this reaction. (1 mark) (ii) What is the oxidation number of phosphorus in phosphoric acid, H,PO,? (1 mark) c A student prepares a sample of the water softener Na,PO, from 0.960 mol dm=* NaOH and 0.500 moldm” H,PO,. The equation is shown below. 3NaOH(aq) + H,PO,(aq) — Na,PO,(aq) + 3H,O(1) (i) The student measures out 150cm> of 0.960 moldm~ NaOH. Calculate the amount, in mol, of NaOH that the student uses. (1 mark) (ii) Calculate the minimum volume 0.500 moldm~ H,PO, that the student needs to add to completely react with the NaOH. (2 marks) (iii) How could the student obtain a solid sample of Na,PO, from their reaction mixture? (1 mark) (iv) What is the maximum mass of Na,PO, that the student could obtain from this experiment? (2 marks) d_ Calcium phosphate, Ca,(PO,),, contains two different ions. Suggest the formulae of the two ions. (2 marks) AS Paper | style question This question is about the simple molecular compounds water, ammonia and boron trichloride. a Complete the table below to show the numbers of bonded or lone pairs of electrons surrounding the central atom per molecule.
Molecule BCI, | NH, | H,0 |
Number of bonded pairs |
Number of lone pairs |
: G3 marks)
Module 2 Foundations in Chemistry
b Ammonia and water form hydrogen bonds. Draw a diagram, including relevant dipoles to show a hydrogen bond between one molecule of water and one molecule of ammonia. (2 marks)
ce Boron trichloride reacts vigorously with water forming a mixture of two acidic products. One of the products has a relative molecular mass of 61.8 and the following percentage composition by mass B: 17.48%, H: 4.85%, and O: 77.67%. (i) Calculate the molecular formula of compound C. (2 marks) (ii) Suggest an equation for the reaction between boron trichloride and water. AS Paper | style question (1 mark)
10 Nitrogen oxides are emitted as pollutant gases
from car exhausts. The mixture of nitrogen oxides is commonly referred to as NO, and typically consists of a mixture of nitrogen monoxide, NO, and nitrogen dioxide, NO,,. a (i) Nitrogen monoxide is formed from nitrogen and oxygen in the high temperatures in a car engine. Write an equation for this reaction. (1 mark)
(ii) Nitrogen dioxide is formed when nitrogen monoxide reacts with oxygen as the emissions cool. Write an equation for this reaction. (1 mark)
b On leaving the car’s exhaust system, NO, dissolves in water to form a mixture of acids:
2NO,(g) + H,O(l) + HNO,(aq) + HNO, (aq) This is a disproportionation reaction. Explain what is meant by disproportionation
and, using oxidation numbers, show that disproportionation has taken place. (3 marks)
c In urban traffic, a car releases 150 cm? of NO, per kilometre.
Calculate the number of molecules of NO, emitted each kilometre by this car. Give
your answer to two significant figures and in standard form. (2 marks)
d_ The mass of NO, released per kilometre in urban traffic is 0.250¢.
Using your answer to (c), calculate the relative molecular mass of the NO,.
ll
Explain whether there are more NO, molecules or more NO molecules in this sample of NO,. (3 marks) AS Paper 2 style question A student was asked to carry out an acid-base titration to find the concentration of some sulfuric acid. The student was supplied with a solution of sodium hydroxide with a concentration of 0.106moldm~?. The equation for the reaction is given below. 2NaOH(aq) + H,SO,(aq) — Na,SO,(aq) + 2H,O(1) a The aqueous sodium hydroxide was
to be added to a conical flask for the
titration. The student was supplied with a
25.00cm? pipette which may have been
left unclean from a previous experiment.
Describe how the pipette should be
prepared for use in the titration. (J mark)
b= The student carried out the titration using
a burette measuring to an accuracy of
0.05cm’.
@ The student filled the burette with the sulfuric acid and the burette reading was 0.00cm?.
@ She carried out a first titration until the burette reading was 21.55cm?.
@ The student carried out a second titration, starting at 21.55cm? and continuing until the burette reading was 42.25cm?.
@ She then added more acid to the burette so that the burette read 10.00cm?.
@ She carried out a third titration after which the burette reading was 30.75 cm’.
Calculate the mean titre to one decimal place. Use this value to calculate the concentration of the sulfuric acid. (4 marks) c_ The pipette and burette were both labelled to show the accuracy of measurements: pipette: +0.06cm* _ burette: + 0.05 cm?
Calculate the percentage error in the
volume of NaOH(aq) delivered from the
pipette and the volume of H,SO, in the titre. (2 marks) AS Paper 2 style question
MODULE 3
Periodic table and energy
Introduction
The legacy left by Mendeleev in producing the first periodic table and the work of many of other scientists have led to the development of the modern periodic table. Classifying the elements as both metal and non-metal and having a way or predicting the properties and reactivity of the elements is of upmost importance to scientific research. This module will provide the important chemical ideas for inorganic and physical chemistry.
The periodic table looks at physical trends and extends your understanding of structure and bonding.
Reactivity trends looks at group properties using Group 2 as a typical metal group and the halogens as a typical non-metal group. Redox reactions are a common theme.
Enthalpy focuses on enthalpy changes
and their determination from experimental results and data tables. You will discover how to calculate energy changes from both your own experiments and from given data.
Reaction rates and equilibrium focuses on how changing conditions affect the rate of a reaction and the position of equilibrium. The
integrated roles of enthalpy changes, rates, catalysts and equilibria are important for industrial processes, increasing yield ,and reducing energy demand whilst improving sustainability
ww
-
fe SAS
Knowledge and understanding checklist From your Key Stage 4 study you should be able to answer the following questions. Work through each point, using your Key Stage 4 notes and the support available on Kerboodle.
‘oe Describe metals and non-metals and explain the differences between them on the basis of their characteristic physical and chemical properties.
OO Explain how observed simple properties of Groups 1, 7, and 0 depend on the outer shell of electrons of the atoms and predict properties from given trends down the groups.
Distinguish between endothermic and exothermic reactions on the basis of the temperature change of the surroundings.
Calculate energy changes in a chemical reaction by considering bond making and bond breaking energies.
Interpret rate of reaction graphs.
C1) Describe the effect of changes in temperature, concentration, pressure, and surface area on rate of reaction. () Describe the characteristics of catalysts and their effect on rates of reaction.
Recall that some reactions are reversible.
. RS wd & " = ae 73 ~~ ; . _ : o A ’
Maths skills checklist
In this module, you will need to use the following maths skills. You can find support for these skills on Kerboodle and through MyMaths.
‘eo Changing the subject of an equation, for carrying out enthalpy change calculations.
Substituting numbers into algebraic equations, for carrying out enthalpy change calculations.
Solving algebraic equations, for carrying out Hess’ law calculations.
Plotting two variables from experimental or other data, for plotting graphs from collected or supplied data to follow the course of a reaction.
oOo Ob ©
Drawing and using the gradient of a tangent to a curve as a measure of rate of reaction, for when you calculate the reaction rate from a
concentration—time graph. @ MyMatt IS. co.uk
Bringing |
PERIODICITY
7.1
The periodic table
Specification reference 3.1.1
OPP Pee Pee eee eee eee eee eee eee eee eee)
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> the arrangement of elements > periodicity > groups
> trendin electron configuration.
de Focus on ekasilicon
Mendeleev was so confident in his periodic law that
The periodic table The periodic table — then
Just over 60 elements were known when Mendeleev arranged them in order of atomic mass (nothing was known about subatomic particles and atomic number at the time). He also lined up the elements in groups with similar properties. If the group properties did not fit, Mendeleev swapped elements around and left gaps, assuming that the atomic mass measurements were incorrect
and that some elements were yet to be discovered. He even predicted properties of the missing elements from group trends —a remarkable insight at the time. It was not until protons were discovered in the early 1900s that the real reason for the order in Mendeleev’s table was revealed.
V Table 1 Mendeleev's predictions for ekasilicon, Eka-Si
he predicted the properties of elements that would fill Property Eka-Si Ge
the gaps in his periodic table. In 1871, he predicted the properties of an undiscovered element that would fit in the gap below silicon. He called this element ‘ekasilicon’. density (gcm~*) 5.50
Ekasilicon was discovered in 1886 and was named formula of oxide germanium. Mendeleev's predictions are not far off, oxide density (g cm~*) considering that he had never seen the element (Table 1).
atomic mass 72.00 72.61
Look up the properties of aluminium and indium, and then predict the following properties for eka-aluminium a atomic mass b formula of chloride.
A Figure 1 Dimitri Mendeleev is one of the main chemists credited with the periodic table
92
The periodic table — now
As of 2014, the periodic table has 114 elements arranged in seven horizontal periods and 18 vertical groups (Figure 2). The periodic table is the most important organisational tool in chemistry. It is the first point of reference for chemists everywhere and most chemistry laboratories have a periodic table prominently placed on the wall.
You do not need to memorise the periodic table (although some people have done so). It is nevertheless helpful to know where the common elements are positioned, and most students know the atomic numbers and relative atomic masses of common elements such as hydrogen, carbon, nitrogen, and oxygen. It is essential that you are able to use the periodic table.
PERIODICITY
q)) (2) (3) (4) (5) (6) 7) (0)
atomic number Symbol!
relative atomic mass
i2 Mg m3 26 < Ca Fe Co Ni “ 1 55.8 58.9 2 37 42 45 re es sr Mo Rh Pd i Sn Las 95.9 102.9 ee ta 1148 ie? 74 77 Tl Ti Ww Ir rf He LL 183.8 eg oe 192.2 as 197.0 | 2006 uy he 105 108 Rg Db Hs
© 64 65 66 67 68 69 70 71 hah bee ne M442 Lie 157.2 | 1589 | 1625 | 1649 | 1673 | 1689 | 1730 | 1750 91 102 actinides igs Th Pa U No
A Figure 2 The periodic table
Arranging the elements
The arrangement, pattern, and shape of the periodic table reveal trends among the elements. The positions of the elements in the periodic table are linked to their physical and chemical properties. This makes the periodic table essential for predicting the properties of elements and their compounds.
Atomic number
Reading from left to right, the elements are arranged in order of increasing atomic number. Each successive element has atoms with one extra proton — H 1, He 2, Li 3, Be 4, and so on.
Groups
The elements are arranged in vertical columns called groups. Each element in a group has atoms with the same number of outer-shell electrons and similar properties.
Periods and periodicity
The elements are arranged in horizontal rows called periods. The number of the period gives the number of the highest energy electron shell in an element's atoms.
Synoptic link
This topic builds upon the earlier work covered in Topic 5.1, Electron structure, on electron shells, sub-shells, and energy levels.
You may find it useful to look back to Topic 5.1.
Study tip
Arepeating, periodic pattern
is called periodicity. There is a periodicity in electron configuration in the periodic table. This is where the periodic table gets its name.
Synoptic link
In Topic 8.1, Group 2, and Topic 8.2, The halogens, you will study the chemistry of a metal group and a non-metal group.
7.1 The periodic table
Across each period, there is a repeating trend in properties of the elements, called periodicity. The most obvious periodicity in properties is the trend from metals to non-metals. The topics in this chapter look at periodicity of several properties:
@ electron configuration
@ ionisation energy
@ structure @
melting points.
Periodic trend in electron configuration
The chemistry of each element is determined by its electron configuration, particularly the outer, highest energy electron shell. Figure 3 shows the electron configuration for elements in the first three periods.
1 2
1 H He
Is Is?
3 |; 47 5 | #6 7 8 | 9 | 10
2 Li Be B Cc N ) F Ne | [Hel2s! | [Hel2s* | (Hel2s*2p! | [He]2s*2p* | [Hel2s*2p° | [Hel2s2p* | {Hel2s*2p” | [Hel2s*2p® [i |i) is | 4 | 6 | @ | wti| iia
3| Na Mg Al Si P S CI Ar {Ne}3s? | [NeJ3s* | [Ne]3s*3p! | [Ne]3s*3p* | [Ne]3s73p* | [Nel3s*3p* | [Ne}3s73p° | [Ne]3s"3p°
s-sub-shell fills p-sub-shell fills
A Figure 3 Electron configuration in Periods 1-3
Trend across a period
Each period starts with an electron in a new highest energy shell.
@ Across Period 2, the 2s sub-shell fills with two electrons, followed by the 2p sub-shell with six electrons.
® Across Period 3, the same pattern of filling is repeated for the 3s and 3p sub-shells.
@ Across Period 4, although the 3d sub-shell is involved, the highest shell number is 7 = 4. From the 1 = 4 shell, only the 4s and 4p sub-shells are occupied.
For each period, the s- and p-sub-shells are filled in the same way — a periodic pattern.
Trend down a group
You will know from GCSE that elements in each group have atoms with the same number of electrons in their outer shell. Elements
in each group also have atoms with the same number of electrons in each sub-shell. This similarity in electron configuration gives elements in the same group their similar chemistry.
Blocks
The elements in the periodic table can be divided into blocks corresponding to their highest energy sub-shell. This gives four distinct
PERIODICITY
blocks, s, p, d, and f. You can follow the order of sub-shell filling by : tic link SUNOPTIC TIN
looking at the periodic table (see Figure 4). r2aesee48L 56 7 8&8 SE WN ws wD Ie 7 s-biock
p-block
Study tip
You should know the names halogens and transition elements, but you do not need to know the names of the other groups. However, you may come across them in reference books or on the Internet.
f-block
A Figure 4 Sub-shell blocks in the Periodic Table
¥ Table 3 b Names and numbers for groups a? RRP
Some groups have names. You will recognise some names such as halogens, but others such as chalcogens may be unfamiliar.
Group number Name Elements New
1 alkali metals Li, Na, K, Rb, Cs, Fr alkaline earth Be, Mg, Ca, Sr, metals Ba, Ra
Two ways of numbering groups are in use (Table 3).
@ The old numbers are the numbers that you used at GCSE. Groups 1-7 and then 0. This system is based on the s- and p-blocks. The advantage of the old numbering is that the group number matches the number of electrons in the highest energy electron shell.
@ The new numbers run from 1-18, numbering each column in the s-, d-, and p-blocks sequentially. The new numbers were approved for use by IUPAC in 1988, but it can take many years for old practices to change. Your periodic table uses both numbers with the old numbers bracketed.
Summary questions
[is | etaages [0.5.17
halogens F, Cl, Br, |, At
He, Ne, Ar, Kr,
1? 18 noble gases Xe, Rn
Old iti BE a elements ? 0
1 State the chemical symbol of: 3 Mendeleev left gaps in his periodic table that were a the second element in Period 5 (1 mark) filled later when, for example, scandium, gallium, b the third element in Group 15 (5). (1 mark) and germanium were discovered. In addition, there
2 State the outer-shell electron configuration of is an entire group in the modern periodic table that
elements in Group 15 (5). Use n for the shell number. Mendeleev omitted. State which group wes missing (1 mark) 2nd Suggest why Mendeleev was unaware of it
(2 marks)
4 ?.2 lonisation energies
Specification reference: 3.1.1
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> ionisation energy > successive ionisation energies
> making predictions from successive ionisation energies
> trends in first ionisation energies down a group
> trends in firstionisation energies across a period.
Synoptic link
You saw that electron sub-shells are filled from the lowest energy level upwards in Topic 5.1, Electron structure.
A Figure 1 An ionised gas is called
a plasma, Lightning is an example of ionisation. Neon lights work by ionising neon gas sealed in a tube at low pressure. The plasma gives out light when electricity passes through it, just as it does in lightning
What is ionisation energy?
lonisation energy measures how easily an atom loses electrons to form positive ions.
The first ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous atoms of an element to form one mole of gaseous 1+ ions. For example:
Na(g) — Na*(g) +e first ionisation energy = +496kJ mol"!
Factors affecting ionisation energy
Electrons are held in their shells by attraction from the nucleus. The first electron lost will be in the /righest energy level and will experience the /Jeast attraction from the nucleus. For example, the first electron lost from a sodium atom (1s?2s?2p%3s') is from the 3s sub-shell.
Three factors affect the attraction between the nucleus and the outer electrons of an atom, and therefore, the ionisation energy.
Atomic radius
The greater the distance between the nucleus and the outer electrons, the less the nuclear attraction. The force of attraction falls off sharply with increasing distance, so atomic radius has a large effect.
Nuclear charge The more protons there are in the nucleus of an atom, the greater the attraction between the nucleus and the outer electrons.
Electron shielding
Electrons are negatively charged and so inner-shell electrons repel outer-shell electrons. This repulsion, called the shielding effect, reduces the attraction between the nucleus and the outer electrons.
Successive ionisation energies
An element has as many ionisation energies as there are electrons. For example, helium has two electrons and two ionisation energies:
He(g) + He*(g)+e- first ionisation energy
He*(g) » He**(g)+e> second ionisation energy
The second ionisation energy of helium is greater than the first ionisation energy. In a helium atom, there are two protons attracting two electrons in the Is sub-shell. After the first electron is lost, the single electron is pulled closer to the helium nucleus. The nuclear attraction on the remaining electron increases and more ionisation energy will be needed to remove this second electron.
Successive ionisation energies are defined in the same way as the first ionisation energy. Just be careful with the species losing the electron.
The second ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous 1+ ions of an element to form one mole of gaseous 2+ ions.
Successive ionisation energies and shells
Successive ionisation energies provide important evidence for the different electron energy levels in an atom. Figure 2 shows a graph for the successive ionisation energies of fluorine.
Two electrons from the 1st shell, closest to the nucleus.
Seven electrons from 2nd shell, furthest from the nucleus.
Large difference in ionisation | energy.
Change from n= 2 shell to n=1 shell, which is much closer to the nucleus.
ionisation energy
1 2 3 4 5 6 7 8 9 ionisation number
A Figure 2 Successive ionisation energies of fluorine
The large increase between the seventh and eighth ionisation energies suggests that the eighth electron must be removed from a different shell, closer to the nucleus and with less shielding.
@ The first shell (7 = 1, closer to the nucleus) contains two electrons.
@ The second shell (7 = 2, the outer shell) contains seven electrons.
Making predictions from successive ionisation energies
Successive ionisation energies allow predictions to be made about:
@ the number of electrons in the outer shell @ the group of the element in the periodic table @ the identity of an element.
The ionisation energies shown in Table 1 steadily increase but then there is a large increase between the third and fourth ionisation energies. This shows that the fourth electron is being removed from an inner shell. Therefore there are three electrons in the outer shell and the element must be in Group 13 (3). Since it is in Period 3, the element must be aluminium.
Trends in first ionisation energies
Periodic trends in first ionisation energies provide important evidence for the existence of shells and sub-shells.
PERIODICITY
Study tip
Don’t get your ionisation energies muddled. The number of the ionisation energy is the same as the charge on the ion produced. For helium, the second ionisation energy produces a 2+ ion from
a i+ion.
Study tip
The key to analysing graphs of successive ionisation energies is to look for any large jumps in ionisation energy. This marks a change from one shell to another.
Study tip
Make sure that you get the shells the right way round.
The electrons with the largest ionisation energy are from the shell closest to the nucleus.
VY Table 1 Successive ionisation energies for an element from Period 3
lonisation lonisation
number energy /kJ mol”!
7
7.2 lonisation energies
Figure 3 shows the first ionisation energies for the first 20 elements in the periodic table. There are two key patterns: ® a general increase in first ionisation energy across each period (H — He, Li — Ne, Na -— Ar) @ a sharp decrease in first ionisation energy between the end of one period and the start of the next period (He — Li, Ne — Na, Ar — K).
These trends can be explained in terms of atomic radius, electron shielding, and nuclear charge.
. : Trend in first ionisation energy down a group First ionisation energies decrease down a group. You can see this trend in Figure 3 and Table 2 by comparing the three noble gases helium, neon, and argon.
each ‘peak’ is a noble gas
i Ne, Ar) at the end of each period He 4
first ionisation energy
Synoptic link
You will see how the trend in ionisation energy down a group affects the reactivity in Topic 8.1, Group 2.
0 2 4 6 8 10 12 14 16 18 20 atomic number Z
Study tip A Figure 3 Trend in first ionisation energy Table 2 explains the decrease in first ionisation energies down a group. First ionisation energy decreases down every group in the periodic table for the same reasons. Although the nuclear charge increases, its effect is outweighed by the increased radius and, to a lesser extent, the
Down a group, the increased atomic radius and shielding are the important factors for the decrease in first ionisation energy.
increased shielding.
VY Table 2 Trend in first ionisation energy down a group
Number of
Noble gas Atomicradius . inner shells
helium, He atomic radius increases
more inner shells so shielding increases
= nuclear attraction on outer electrons decreases
first ionisation energy decreases argon, Ar 2
Trend in first ionisation energy across a period
Figure 3 shows the general increase in first ionisation energy across the first three periods. Table 2 explains the general increase in first ionisation energies of the elements across Period 2.
VY Table 3 General trend in first ionisation energy across Period 2
Atomic radius
Element
ote tefel*|ol sw! Se 2 | +e | se | oe | ee | om | 9m | som
Nuclear charge increases Same shell: similar shielding Nuclear attraction increases Atomic radius decreases
First ionisation energy increases
Sub-shell trends in first ionisation energy
Although first ionisation energy shows a general increase across both Period 2 and Period 3, it does fall in two places in each period. The drops occur at the same positions in each period, suggesting that there might be a periodic cause. The reason is linked to the existence of sub-shells, their energies, and how orbitals fill with electrons.
Across Period 2, the first ionisation energy graph (Figure 4) shows three rises and two falls:
@ arise from lithium to beryllium
@ a fall to boron followed by a rise to carbon and nitrogen
® a fall to oxygen followed by a rise to fluorine and neon.
Figure 4 links the pattern to the filling of the s- and p-sub-shells.
Ne
pairing of 2p electrons
first ionisation energy
Filling of 2s ep apette adding one electron to each , 2p orbital
ti
1 2 3 4 5 6 7 8 9 10
atomic number Z
PERIODICITY
Study tip
Across a period, the increased nuclear charge is the most important factor for the general increase in first ionisation energy.
Synoptic link
Review Topic 5.1, Electron structure,
for details about the electron structure, shells, sub-shells, and orbitals.
Figure 4 Filling the sub-shells across Period 2
Summary questions
Write the equations to represent the first two ionisation energies of sulfur. (2 marks)
Write an equation, with
state symbols, to represent the first ionisation energy of aluminium. (1 mark)
Explain why successive ionisation energies
always increase. (2 marks)
a Explain the general trend in first ionisation energy fromNatoAr. (3 marks) b Explain the sharp drop in first ionisation energy between Ne and Na. (3 marks) c Explain the trend in first ionisation energy shown by He, Ne, and Ar. (3 marks)
The first six ionisation energies of an element in Period 3 are 78?, 157?, 3232, 4356, 16091, 19 805 kJ mol”!. Identify
the element and explain yourreasoning. (3 marks)
6 a Explain why Al has
a lower first ionisation
energy than Mg. (2 marks) b Explain why S has
a lower first ionisation
energythanP. (2 marks)
7.2 lonisation energies
Comparing beryllium and boron The fall in first ionisation energy from beryllium to boron marks the start of filling the 2p sub-shell (Figure 5).
The 2p sub-shell in boron has a higher energy than the 2s sub-shell
in beryllium. Therefore, in boron the 2p electron is easier to remove than one of the 2s electrons in beryllium. The first ionisation energy of boron is less than the first ionisation energy of beryllium.
energy beryllium energy boron wo— n=2—C
\ 2s
B: outermost electron in 2p
Vy 2s
Be: outermost electron in 2s A Figure § Quter-shell electrons in beryllium and boron atoms
Comparing nitrogen and oxygen The fall in first ionisation energy from nitrogen to oxygen marks the start of electron pairing in the p-orbitals of the 2p sub-shell (Figure 6).
@ In nitrogen and oxygen the highest energy electrons are in a 2p sub-shell.
@ In oxygen, the paired electrons in one of the 2p orbtals repel one another, making it easier to remove an electron from an oxygen atom than a nitrogen atom.
@ Therefore the first ionisation energy of oxygen is less than the first ionisation energy of nitrogen,
energy nitrogen energy oxygen n=2—c n=2—f¢ \ t | |2s \ t | {2s y
Three 2p electrons: 2p,'2p,'2p.' Four 2p electrons: 2p,?2p,'2p,' ® one electron in each 2p orbital © two electrons in one 2p orbital @ spins are at right angles — @® 2p electrons start to pair
equal repulsion as far apart @ the paired electrons repel as possible
A Figure 6 Quter-shell electrons in nitrogen and oxygen atoms
?.3 Periodic trends in bonding
and structure
Specification reference: 3.1.1
Metals and non-metals
One of the key trends in the periodic table is the change from metals to non-metals from left to right across each period. The changeover from metal to non-metal takes place on a diagonal line from the top of Group 13 (3) to the bottom of Group 17 (7) (Figure 1). Elements near to the metal/non-metal divide (e.g., boron, B, silicon, Si, germanium, Ge, arsenic, As, and antimony, Sb) can show in-between properties and are called semi-metals or metalloids.
Going down these groups, there is a trend from non-metal to metal. The divide is clearest in Group 14 (4), carbon (non-metal) to lead (metal). The two elements on either side of the divide, silicon and germanium, are semi-metals.
There are far more metallic than non-metallic elements — 92 metals to 22 non-metals. However, despite being in a minority in the periodic table, non-metals are extremely important, especially the elements carbon, hydrogen, nitrogen, and oxygen in organic chemistry and biochemistry.
Metallic bonding, structures, and properties
At room temperature, all metals except mercury are solids. The 92 known metals have a wide range of properties — some are strong
and hard like tungsten, W, some soft like lead, Pb, some light like aluminium, Al, and some very heavy like osmium, Os, which is twice as dense as lead.
The one constant property of all metals is their ability to conduct
electricity. This is a remarkable property for a solid, as charge must be able to move within a rigid structure for conduction to take place.
Metallic bonding and structure
You have already met two main types of chemical bonding — ionic and covalent. Metallic bonding is a special type of bonding for metals.
@ Ina solid metal structure, each atom has donated its negative outer-shell electrons to a shared pool of electrons, which are delocalised (spread out) throughout the whole structure.
®@ The positive ions (cations) left behind consist of the nucleus and the inner electron shells of the metal atoms.
Metallic bonding is the strong electrostatic attraction between cations (positive ions) and delocalised electrons (see Figure 2).
@ The cations are fixed in position, maintaining the structure and shape of the metal.
® The delocalised electrons are mobile and are able to move throughout the structure. Only the electrons move.
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> metallic bonding and structure
> giant covalent structures
> periodic trend in melting points.
ett PPP eeeseresesereseseseseseres,
IPE CEES EOC OE SECO COSCO COSCO S Eee eee eee
(3) (4) (5) (6) (7) (©) 13 14 15 16 17 18
A Figure 1 The division between metals and non-metals in the p-block of the periodic table. The blue elements are metals and the yellow elements are non-metals
101
O08 OcO 2.0,0, 0 20 0.0
A Figure 2 Metallic 2
A Figure 3 Metal eee. —all the delocalised electrons will move towards the positive terminal whilst the negative terminal donates electrons that move into the structure
Synoptic link
Review Topic 5.2, lonic bonding for details about the conductivity of ionic compounds.
102
7.3 Periodic trends in bonding and structure
In Figure 2, there are 12 cations, each with a 1+ charge, and
12 electrons, each with a 1— charge. This balances the charge. For metals containing 2+ cations, twice as many negatively charged electrons are present to balance the charge.
In a metal structure, billions of metal atoms are held together by metallic bonding in a giant metallic lattice.
Properties of metals
Most metals have:
® strong metallic bonds — attraction between positive ions and delocalised electrons
@ high electrical conductivity @ high melting and boiling points.
The physical properties of metals can be explained in terms of the giant structure of the lattice and metallic bonding.
Electrical conductivity
Metals conduct electricity in solid and liquid states. When a voltage is applied across a metal, the delocalised electrons can move through the structure, carrying charge, as shown in Figure 3. Contrast this ability with the conductivity of ionic compounds, which have no mobile charge carriers in the solid state.
Melting and boiling points
Most metals have high melting and boiling points. In Figure 4 coloured shading is used to compare the melting points of the transition metals. Tungsten, W, has the highest melting point at 3422 °C, which is why it is used in the filaments of halogen lamps; other metals would melt. In contrast, mercury, Hg, melts at —39°C. Other metals that melt at low temperatures include those in Group | of the periodic table, which all have melting points below 200°C.
A Figure 4 Comparison of the melting points of transition metals
The melting point depends on the strength of the metallic bonds holding together the atoms in the giant metallic lattice.
@ For most metals, high temperatures are necessary to provide the large amount of energy needed to overcome the strong electrostatic attraction between the cations and electrons.
This strong attraction results in most metals having high melting and boiling points.
Solubility Metals do not dissolve. It might be expected that there would be some interaction between polar solvents and the charges in a metallic lattice,
as with ionic compounds, but any interactions would lead to a reaction, rather than dissolving, as with sodium and water.
Giant covalent structures
You already know that many non-metallic elements exist as simple covalently bonded molecules. In the solid state, these molecules form a simple molecular lattice structure held together by weak intermolecular forces. These structures therefore have low melting and boiling points.
The non-metals boron, carbon, and silicon have very different lattice structures. Instead of small molecules and intermolecular forces, many billions of atoms are held together by a network of strong covalent bonds to form a giant covalent lattice.
Carbon and silicon are in Group 14 (4) of the periodic table and their atoms have four electrons in the outer shells. Carbon (in its diamond form) and silicon use these four electrons to form covalent bonds to other carbon or silicon atoms. The result is a tetrahedral structure, as shown in Figure 5 for carbon (diamond).
@ Figure 5 shows the tetrahedral arrangement of atoms in the diamond form of carbon.
@ The bond angles are all 109.5° by electron-pair repulsion.
@ The dot-and-cross diagram shows part of the covalently bonded network of carbon atoms.
Properties
Typical properties of substances with a giant covalent lattice structure are shown below. The properties are dominated by the strong covalent bonds, which make for very stable structures that are very difficult to break down.
Melting and boiling points
Giant covalent lattices have high melting and boiling points. This is because covalent bonds are strong. High temperatures are necessary to provide the large quantity of energy needed to break the strong covalent bonds,
Solubility
Giant covalent lattices are insoluble in almost all solvents. The covalent bonds holding together the atoms in the lattice are far too strong to be broken by interaction with solvents.
Electrical conductivity Giant covalent lattices are non-conductors of electricity. The only exceptions are graphene and graphite, which are forms of carbon.
® Incarbon (diamond) and silicon, all four outer-shell electrons are involved in covalent bonding, so none are available for conducting electricity.
@ Carbon is special in forming several structures in which one of the electrons is available for conductivity. Graphene and graphite are able to conduct electricity.
PERIODICITY 7
Synoptic link
See Topic 6.3, Intermolecular forces, if you need to review the simple molecular lattice structures of covalently bonded molecules.
A Figure 5 Structure and bonding in carbon (diamond)
Synoptic link
Review electron-pair repulsion in Topic 6.1, Shapes of molecules and ions.
A Figure 6 Sand dunes in the Sahara desert. Sand is mainly silicon dioxide, Si, which has a giant covalent lattice structure similar to diamond. The strong covalent bonds make giant covalent substances stable and generally unreactive
103
Graphene and graphite
Apart from diamond, carbon forms giant covalent structures based on planar hexagonal layers.
You can see from the dot-and-cross diagram in Figure 7 that only three electrons of the four outer-shell electrons are used in covalent bonding. The remaining electron
is released into a pool of delocalised electrons shared by all atoms in the structure. Structures of carbon containing planar hexagonal layers are therefore good electrical conductors.
Graphene and graphite are both giant covalent structures of carbon based on planar hexagonal layers with bond angles of 120° by electron-pair repulsion.
Graphene
Graphene is a single layer of graphite, composed of hexagonally arranged carbon atoms linked by strong covalent bonds (Figure 8). Graphene has the same electrical conductivity as copper, and is the thinnest and strongest material ever made.
Graphene was discovered in 2004 by Andre Geim and Konstantin Novoselov from the University of Manchester. They were awarded a Nobel prize in 2010. Geim famously made graphene by using sticky tape to pull single layers of carbon atoms from the surface of graphite. Such a simple idea to start work that won a Nobel prize!
a > re a a a a a
—
A Figure 8 Graphene
7.3 Periodic trends in bonding and structure
A Figure ? Planar hexagonal layer (top) and dot-and-cross diagram (bottom) in graphene and graphite layers
Graphite
Graphite is composed of parallel layers of hexagonally arranged carbon atoms, like a stack of graphene layers (Figure 9). The layers are bonded by weak London forces.
weak force between layers
strong covalent bond A Figure 9 Structure of graphite
The bonding in the hexagonal layers only uses three of carbon’s four outer-shell electrons. The spare electron is delocalised between the layers, so electricity can be conducted as in metals.
Carry out some research to find these answers.
Many materials are made out of carbon fibre.
a Howis carbon fibre linked to the structures of graphite and graphene.
b Howare carbon fibre materials strengthened?
Periodic trend in melting points
The melting points of the elements in Period 2 and Period 3 are shown in Figure 10.
Period 2. ant é Period 3
structures giant
structures Si
simple molecules <>
melting point melting point
simple molecules <>
NO F Ne
p S
2 4 6 8 10 10 12 14 16 18 atomic number/ Z atomic number/Z
A Figure 10 Trend in melting points across Periods 2 and 3
Across Period 2 and Period 3, @ the melting point increases from Group | to Group 14 (4)
@ there is a sharp decrease in melting point between Group 14 (4) and Group 15 (5)
@ the melting points are comparatively low from Group 15 (5) to Group 18 (0).
The sharp decrease in melting point marks a change from giant to simple molecular structures, shown in Figure 11. You can also see the start of the diagonal divide between metals and non-metals.
On melting, giant structures have strong forces to overcome so have high melting points. Simple molecular structures have weak forces to overcome, so have much lower melting points.
The trend in melting points across Period 2 is repeated across Period 3, and continues across the s- and p-blocks from Period 4 downwards.
Giant metallic structure Strong metallic bonds
between cations and delocalised electrons
Simple molecular structure
A Figure 11 Trend in structure across Periods 2 and 3
PERIODICITY 7
Study tip
Substances with giant structures have high boiling points as strong forces are broken on boiling.
Study tip
Substances with simple molecular structures have low boiling points as weak forces are broken on boiling.
Summary questions
1 Explain what is meant by metallic bonding and why this type of bonding enables metals to conduct electricity. (3 marks)
2 Explain how the bonding ina simple molecular lattice differs from that in a giant covalent lattice. (2 marks)
3 Across Period 4, the trend in properties is not exactly the same as across Periods 2 and 3. Suggest explanations for the following.
a Germanium is a good conductor of electricity. (2 marks) b Arsenic has a much higher melting point than nitrogen and phosphorus. (2 marks)
Chapter ? Practice questions
Practice questions 1 This question looks at ionisation energies a_ For the third ionisation energy of Mg,
(i) write an equation, with state symbols for the third ionisation energy of magnesium. (1 mark)
(ii) which sub-shell loses this electron? (1 mark)
b_ The Ist to 8th successive ionisation energies, in kJ mol-', of an element in Period 3 are listed below.
1012, 1903, 2912, 4957, 6274, 21269, 25398, 29855
What is the element? Explain your reasoning. (3 marks)
c Suggest, with a reason, which successive ionisation energy is being described below.
The energy required to remove one electron from each ion in one mole of gaseous 4+ ions. (1 mark)
2 This question looks at trends across a period the periodic table. Four sequences of elements across Period 3 are shown below.
Na,Mg, Al ALSiiP SLPS PS,Cl
a Which sequence shows the melting point increasing across Period 3? (1 mark)
b Which sequence shows the first ionisation energy increasing across Period 3? (1 mark)
c¢ Which sequence shows the second ionisation energy increasing across Period 3? (J mark)
d Two sequences contain elements that have the same structure in the solid state.
Identify these two sequences and state the structure for each. (2 marks)
3 Solid graphite and iodine contain covalent bonds but their melting points and electrical conductivities are very different.
a_ State and explain the difference in melting points. (4 marks)
b= State and explain the difference in electrical conductivity. (3 marks)
Ionisation energies of the elements H to K are shown below.
| $——-0—$-— 9 — 9 — § — 9 ge dng og
8
4 —-$ —4— 9 —-$— 4 — $F — fp —
Bs
:
first ionisation
energy /kJ mol ~}
0 2 4 6 8 10 12 14 16 18 20 atomic number
a_ Define the term first ionisation energy. (3 marks)
b- Explain why the first ionisation energies show a general increase across Period 3
(Li to Ne). (3 marks) ¢ Explain why the first ionisation energy of B is less than that of Be. (2 marks) d= Explain why the first ionisation energy of O is less than that of N. (2 marks)
e State and explain the trend in first ionisation energies shown by the elements with atomic numbers 2, 10, and 18.
(3 marks)
c The Ist to 3rd ionisation energies, in kJ mol! of nitrogen are 1402, 2856, and 4578.
(i) Write an equation, with state symbols, to represent the fourth ionisation energy of nitrogen.
(1 mark)
(ii) Suggest why the successive ionisation energies of nitrogen increase in value.
(1 mark)
5 This question is about aluminium oxide, Al,O,.
a_ Successive ionisation energies provide evidence for the arrangement of electrons in atoms. The graph below shows 8 successive ionisation energies of oxygen.
ionisation energy
1 2 3 4 5 6 7 8 ionisation number
(i) Write an equation, including state symbols, to represent the second ionisation energy of oxygen. (2 marks)
(ii) How does this graph provide
evidence for the existence of two
electron shells in oxygen? (2 marks)
Write the electron configuration for
an aluminium atom. (1 mark)
(ii) Sketch a graph to show the thirteen successive ionisation energies of aluminium. (3 marks)
OCR 2811 Jun 2002 Q3(a)(b)
6 Solids exist as lattice structures.
a_ Giant metallic lattices conduct electricity. Giant ionic lattices do not. If a giant ionic lattice is melted, the molten ionic compound will conduct electricity.
Explain these observations in terms of bonding, structure, and particles present. (3 marks)
b_ The solid lattice structure of ammonia, NH,, contains hydrogen bonds.
(i) Draw a diagram to show hydrogen bonding between two molecules of NH, in a solid lattice. Include relevant dipoles and lone pairs. (2 marks) (ii) Suggest why ice has a higher melting point than solid ammonia. (2 marks) c Solid SiO, melts at 2230 °C. Solid SiCl, melts at —-70 °C. Neither of the liquids formed conducts electricity. Suggest the type of lattice structure in solid SiO, and in solid SiCl, and explain the difference in melting points in terms of bonding and structure. (5 marks)
OCR F321 Jun 11 Q5
PERIODICITY
7 The table below shows the melting points of
the elements Na to Cl in Period 3.
Na} Mg] w | si [P| s | |
Melting point °C
coco
Bonds/forces broken on boiling
Complete the structure row of the table using M for giant metallic, C for giant covalent, and S for simple molecular.
(3 marks)
(ii) Complete the final row by using MB for metallic bonds, CB for covalent bonds,
LF for London forces (1 mark)
b= State what is meant by metallic bonding. Use a diagram as part of your answer. (3 marks)
c Suggest why the melting point increases from Na to Al. (2 marks)
d_ Explain why the melting point of phosphorus is much lower than that of silicon. (3 marks) e Explain why the melting point of sulfur is higher than that of chlorine (2 marks) This question compares the electrical conductivity of sodium, sodium chloride, and chlorine in their solid and molten states.
a State the structure and particles making up the structure in the solid state for
(i) sodium (2 marks) (ii) sodium chloride (2 marks) (iii) chlorine (2 marks)
b Compare and explain the electrical conductivities of sodium, sodium chloride and chlorine in the solid state. (3 marks)
¢ Compare and explain any differences in electrical conductivities in the liquid state (1 mark)
compared with the solid state.
REACTIVITY TRENDS
8.1
Group 2
Specification reference: 3.1.2
ee
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> redox reactions of Group 2 elements
> trends in reactivity and ionisation energies
> action of water on Group 2 oxides
> uses of some Group 2 compounds.
VY Table 1 OQuter-shell electron configurations of Group 2 atoms and Group 2 ions
Be | [He] 2s* Be** | [He]
A Figure 1 Group 2 elements, from left to right — calcium, beryllium, and magnesium
If you are unsure about redox reactions and oxidation numbers, look back at Topic 4.3, Redox.
108
Characteristic physical properties
The elements in Group 2 of the periodic table are metals, sometimes named the alkaline earth metals. The name comes from the alkaline properties of the metal hydroxides. The elements are reactive metals and do not occur in their elemental form naturally. On Earth, they are found in stable compounds such as calcium carbonate, CaCO,,.
Redox reactions and reactivity Reducing agents
Each Group 2 element has two outer shell electrons, two more than the electron configuration of a noble gas. The two electrons are in the outer s sub-shell.
Redox reactions are the most common type of reaction of Group 2 elements. Each metal atom is oxidised, losing two electrons to form a 2+ ion with the electron configuration of a noble gas.
Ca > Ca**+2e Ca is oxidised
[Ar]4s? [Ar] @ Another species will gain these two electrons and be reduced. @ The Group 2 element is called a reducing agent because it has
reduced another species.
Table 1 compares the outer shell electron configurations of Group 2 atoms with Group 2 ions.
Redox reactions with oxygen The Group 2 elements all react with oxygen to form a metal oxide
with the general formula MO, made up of M** and O?- ions.
You will have seen the reaction of magnesium with oxygen in the air in the laboratory. The magnesium burns with a brilliant white light and forms white magnesium oxide.
2Mg(s) + O,(g) — 2MgO(s) 0 = <2 oxidation 0 2 reduction The total changes in oxidation number balance: @ two Mgin2Mg — each Mg increases by +2 _ total increase = +4
® two OinO, each O decreases by —2 total decrease = —4
Redox reactions with water
The Group 2 elements react with water to form an alkaline hydroxide, with the general formula M(OH),, and hydrogen gas. Water and magnesium react very slowly, but the reaction becomes more and more vigorous with metals further down the group — reactivity increases down the group.
Sr(s) +2H,O(l) — Sr(OH),(aq) + H,(g) 0 — +2 +] : 0
oxidation
reduction Not all the hydrogen atoms are reduced:
@ | SrinSr total increase = +2
@ 4Hin 2H,0
Sr increases by 2
two H decrease by 1 forming H, two H do not change forming Sr(OH),
Redox reactions with dilute acids
Many metals take part in redox reactions with dilute acids to form a salt and hydrogen gas.
metal + acid — salt + hydrogen
All Group 2 elements react in this way. Again, the reactivity increases down the group.
Figure 2 shows the reaction of magnesium with dilute hydrochloric acid. Mg(s) + 2HCl(aq) — MgCl,(aq) + H,(g)
0 — +2 +] > 0
oxidation reduction The total changes in oxidation number balance:
@ | Mgin Mg total increase = +2
@® 2Hin 2HCl
Mg increases by 2
i | No
each H decreases by 1 total decrease
Trend in reactivity and ionisation energy
When the redox reactions above are carried out with each Group 2 element, the reactivity increases down Group 2 (Figure 2).
So why does the reactivity increase?
The atoms of Group 2 elements react by losing electrons to form +2 ions. The formation of +2 ions from gaseous atoms requires the input of two ionisation energies: @ M(g) — M*(g) +e
@ M*(g) > M?*(g) +e
first ionisation energy second ionisation energy
Figure 4 shows the first and second ionisation energies for the Group 2 elements. The ionisation energies decrease down the group because the attraction between the nucleus and the outer electrons decreases as a result of increasing atomic radius and increasing shielding.
REACTIVITY TRENDS
total decrease = —2
Be least reactive Mg Ca Sr
Ba Rn A Figure 2 The reactivity of Group 2 elements
most reactive
A Figure 3 Magnesium ribbon reacts with dilute hydrochloric acid to give off tiny bubbles of hydrogen
Synoptic link
You can review ionisation energies in Topic 2.2, lonisation energy.
Wilst 2nd Be Mg
Ca
Element
Sr Ba | “BBE BS & ~ -~ N N ap) lonisation energy/kJ mol”!
A Figure 4 First and second ionisation energies of Group 2 elements
109
8.1 Group 2
VY Table 2 Trend in alkalinity Hydroxide
solubility increases
pH increases
alkalinity increases
A Figure 5 Spreading lime (Ca(OH).) on a field to reduce soil acidity
110
Although other energy changes take place when Group 2 elements react, the first and second ionisation energies make up most of the energy input. From Figure 4, it is clear that the total energy input from ionisation energies to form 2+ ions decreases down the group.
The Group 2 elements become more reactive and stronger reducing agents down the group.
Reactions of Group 2 compounds
Group 2 oxides
Reactions with water The oxides of Group 2 elements react with water, releasing hydroxide ions, OH~, and forming alkaline solutions of the metal hydroxide.
CaO(s) + H,O(1) —— Ca**(aq) + 2OH™(aq)
The Group 2 hydroxides are only slightly soluble in water. When the solution becomes saturated, any further metal and hydroxide ions will form a solid precipitate:
Ca**(aq) + 2OH~ (aq) ——> Ca(OH), (s)
Solubility of hydroxides The solubility of the hydroxides in water increases down the group, so the resulting solutions contain more OH~(aq) ions and are more alkaline.
@ Mg(OH),(s) is only very slightly soluble in water. The solution has a low OH~(aq) concentration and a pH ~ 10.
@ Ba(OH),(s) is much more soluble in water. The solution has a greater OH~(aq) concentration and a pH ~ 13.
The trend is shown in Table 2. You can easily show this trend by carrying out the following experiment.
Add a spatula of each Group 2 oxide to water in a test tube.
2 Shake the mixture. On this scale, there is insufficient water to dissolve all of the metal hydroxide that forms. You will have a saturated solution of each metal hydroxide with some white solid undissolved at the bottom of the test-tube.
3 Measure the pH of each solution. The alkalinity will be seen to increase down the group.
Uses of Group 2 compounds as bases
The Group 2 oxides, hydroxides, and carbonates have many uses related to their basic properties and ability to neutralise acids.
Group 2 compounds in agriculture
Calcium hydroxide, Ca(OH),, is added to fields as lime by farmers
to increase the pH of acidic soils. You may have seen the white lime powder on fields (Figure 5). The calcium hydroxide neutralises acid in the soil, forming neutral water:
Ca(OH),(s) + 2H*(aq) — Ca?*(aq) + 2H,O(1)
REACTIVITY TRENDS
Group 2 compounds in medicine Synoptic link
Group 2 bases are often used as antacids for treating acid indigestion.
Many indigestion tablets use magnesium and calcium carbonates as If you are unsure about
the main ingredients, whilst ‘milk of magnesia’ is a suspension of neutralisation reactions look back white magnesium hydroxide, Mg(OH),, in water. Remember that at Topic 4.1, Acids, bases, and magnesium hydroxide is only very slightly soluble in water. neutralisation.
The acid in your stomach is mainly hydrochloric acid and the equations below show the neutralisation reactions that take place with Mg(OH), and CaCo,,.
Mg(OH),(s) + 2HCl(aq) ——> MgCl, (aq) + 2H,O(I)
CaCO, (s) + 2HCl(aq) —— CaCl, (aq) + H,O(1) + CO,(g)
GAVISCON , Rennie
| PEPPERMINT
Calcium Carbonate, Magnesium Carbonate
pe Be Heartbum & Indie, 72 Fast effective relief from Lasts up tablets indigestion and heartburn
Sodium alginate Sodium bicarbonat® Calcium carbonate
A Figure 6 Common indigestion remedies made with Group 2 compounds, such as calcium carbonate, CaCO,
Summary questions 1 Explain why Group 2 elements are reducing agents. (2 marks)
2 The following reaction is a redox process: Mg + 2HCI — MgCl, + H,
a Identify the changes in oxidation number. (2 marks) b State which species is being oxidised and which is being reduced. (1 mark)
3 Explain why the Group 2 elements become more reactive down the group. (4 marks)
4 State and explain the trend in alkalinity of the solution formed when Group 2 oxides are added to water. (3 marks)
8.2 The halogens
Specification reference: 3.1.3
Learning outcomes
Demonstrate knowledge,
understanding, and application of:
> characteristic physical properties of halogens
Characteristic physical properties
The halogens, Group 17 (7) of the periodic table, are the most reactive non-metallic group. The elements do not occur in their elemental form in nature. On Earth, the halogens occur as stable halide ions (Cl, Br-, and I~) dissolved in sea water or combined with sodium or potassium as solid deposits, such as in salt mines containing common
> redox reactions and reactivity salt, NaCl.
of halogens
> the trend in reactivity of the halogens Cl,, Br,, and I,
Trends in boiling points
At room temperature and pressure (RTP), all the halogens exist as diatomic molecules, X,. The group contains elements in all three physical states at RTP, changing from gas to liquid to solid down the group (Figure 1). In their solid states the halogens form lattices with simple molecular structures. Table | explains the trend in boiling points of the five halogens — fluorine to astatine.
> disproportionation > the use of chlorine in water purification and bleach.
TEPER EECOEOSOOSOSOOCSOOOOOCOCOSOOOCOOO Se
Sunoptic link
Figure 1 Chlorine is a pale green gas at RIP Bromine liquid is extremely toxic, and vaporises readily at room temperature, as can be seen from the
Topic 6. ienaesiar — orange gas above the red-brown liquid. h — lodine is a solid with grey-black crystals
Y Table 1 Boiling points of the halogen. Astatine only exists as short-lived isotopes and its boiling point has been estimated from the group trend
Halogen Number of Boili & we Ing Appearance and state at RTP molecule electrons point /°C
| 8 | 188 | paleyetow ges sale more electrons pale green gas stronger London forces red-brown liquid more energy required to break shiny grey-black solid the intermolecular forces boiling point increases never been seen
Redox reactions and reactivity of halogens Redox reactions
Each halogen has seven outer-shell electrons, just one electron short of the electronic configuration of a noble gas. Two electrons are in the outer s sub-shell and five in the outer p sub-shell — s*p”
REACTIVITY TRENDS ae
Redox reactions are the most common type of reaction of the VY Table 2 Comparison of outer-shell halogens. Each halogen atom is reduced, gaining one electron to form electron configurations of halogen a 1— halide ion with the electron configuration of the nearest noble atoms and halide ions
gas (see Table 2).
Halogen atom Halide ion
Cl, + 2e~ — 2CI chlorine is reduced
Another species loses electrons to halogen atoms — it is oxidised. The halogen is called an oxidising agent because it has oxidised another species.
Halogen—halide displacement reactions
Displacement reactions of halogens with halide ions can be carried out on a test-tube scale. The results of the displacement reactions show Y Table 3 Halogen solutions in water that the reactivity of the halogens decreases down the group. and cyclohexane
A solution of each halogen is added to aqueous solutions of the other Br halides. For example, a solution of chlorine (Cl,) is added to two aqueous solutions containing bromine (Br~) and iodine (I>) ions. If the halogen added is more reactive than the halide present pale green Solutions of iodine and bromine in water can appear a similar orange- - J
brown colour, depending on the concentration. To tell them apart,
an organic non-polar solvent such as cyclohexane can be added and solution in cyclohexane (top layer) the mixture shaken. The non-polar halogens dissolve more readily in cyclohexane than in water. In cyclohexane their colours are much
easier to tell apart, with iodine being a deep violet. The colours are shown in Table 3.
solution in water
@ a reaction takes place, the halogen displacing the halide from solution
@ the solution changes colour.
pale green orange violet
The results and conclusions for these displacement reactions of aqueous solution of halogens and halides are shown in Table 4.
VY Table 4 Halogen displacement reactions
Halogen uk Br,(aq) 1,(aq) Ci-(aq) no reaction
Br-(aq) no reaction
From the results:
@ chlorine has clearly reacted with both Br- and I- @ bromine has reacted with I> only
@ jodine has not reacted at all.
113
8.2 The halogens
Synoptic link
If you are unsure about redox
reactions and oxidation numbers, look back at Topic 4.3, Redox.
chlorine most reactive bromine iodine least reactive
A Figure 2 The order of reactivity of chlorine, bromine, and iodine
114
Reaction of chlorine with bromide ions
The equation and oxidation number changes are shown in full for the redox reaction between aqueous solutions of chlorine and sodium bromide.
full equation Cl,(aq) + 2NaBr(aq) — 2NaCl(aq) + Br,(aq) ionic equation = Cl,(aq)+ 2Br(aq) ~ 2Cl-(aq) + Br,(aq)
0 — —] reduction
-1 — 0 oxidation
@ 2Brin 2Br each Br increases by +1 total increase = +2 @ 2ClinCl, each Cl decreases by -1 total decrease = —2
Figure 2 shows the order of reactivity of the three halogens chlorine, bromine, and iodine.
What about fluorine and astatine? Fluorine is a pale yellow gas, reacting with almost any substance that it comes in contact with.
Astatine is extremely rare because it is radioactive and decays rapidly, and the element has never actually been seen. It is predicted to be the least reactive halogen.
Trend in reactivity
In redox reactions, halogens react by gaining electrons. Down the group, the tendency to gain an electron decreases and the halogens become less reactive (Table 5).
VY Table 5 Trend in reactivity of the halogens
Halogen Atomic Number of molecule radius innershells
Atomic radius increases
Oo More inner shells so shielding increases
an electron from another species
oO Less nuclear attraction to capture
At, Reactivity decreases
In the halogens, fluorine is the strongest oxidising agent, gaining electrons from other species more readily than the other halogens. The halogens become weaker oxidising agents down the group.
Disproportionation
Disproportionation is a redox reaction in which the same element is both oxidised and reduced. The reaction of chlorine with water and with cold, dilute sodium hydroxide are two examples of disproportionation reactions.
REACTIVITY TRENDS
The reaction of chlorine with water
You will know that chlorine is used in water purification. Chlorine began to be widely used as a disinfectant for drinking water treatment over 100 years ago, revolutionising public health by reducing the incidence of waterborne diseases by killing harmful bacteria.
When small amounts of chlorine are added to water, a disproportionation reaction takes place. For each chlorine molecule, one chlorine atom is oxidised and the other chlorine atom is reduced.
Cl,(aq) + H,O(1) + HClO(aq) + HCl(aq) 0 > —] reduction
0 = +] oxidation
The two products are both acids, chloric(1) acid, HCIO, and hydrochloric acid, HCI. The bacteria are killed by chloric(1) acid and chlorate(I) ions, ClO, rather than by chlorine. Chloric(I) acid also acts as a weak bleach. You can demonstrate this by adding some indicator solution to a solution of chlorine in water. The indicator first turns red, from the presence of the two acids. The colour then disappears as the bleaching action of chloric(I) acid takes effect.
The reaction of chlorine with cold, dilute aqueous sodium hydroxide
The reaction of chlorine with water is limited by the low solubility of eee ih etane. eiea’ chlorine in water. If the water contains dissolved sodium hydroxide, A Figure 3 Household bleach contains much more chlorine dissolves and another disproportionation reaction _ sodium chlorate(I), NaCl0, made by
takes place. reacting chlorine with sodium hydroxide
Cl,(aq) + 2NaOH(aq) — NaClO(aq) + NaCl(aq) + H,O(1) 0 > -| reduction
0 — +i oxidation Sunoptic link
The resulting solution contains a large concentration of chlorate(I), ClO~, ions from the sodium chlorate(I), NaClO, that is formed. This solution finds a use as household bleach, which is made by reacting chlorine with cold dilute aqueous sodium hydroxide (See Figure 3).
Benefits and risks of chlorine use
Although chlorine is beneficial in ensuring that our water is fit to drink and that bacteria are killed, chlorine is also an extremely toxic gas. Chlorine is a respiratory irritant in small concentrations, and large concentrations can be fatal.
Chlorine in drinking water can react with organic hydrocarbons such as methane, formed from decaying vegetation. Chlorinated hydrocarbons
are formed, which are suspected of causing cancer. However, the overall risk to health of not adding chlorine to the water supply is far greater than —_@ Figure 4 In water, chlorine tablets
the risk posed by the chlorinated hydrocarbons. The quality of drinking release chlorine in low concentrations water would be compromised and diseases such as typhoid and cholera at a steady rate. The tablets are used might break out. Before safeguarding against what might be a minimal to purify water for drinking and in risk, you should also consider why chlorine is added to drinking water swimming pools and can be used to in the first place. After a natural disaster one of the very first, life-saving purify water supplies after a natural
tasks is to ensure that the survivors have a safe water supply (Figure 4). disaster
8.2 The halogens
Tests for halide ions Precipitation reactions with aqueous silver ions
Aqueous halide ions react with aqueous silver ions to form precipitates
of silver halides, as shown by the general equation below. X~(aq) represents an aqueous solution of any halide.
Ag*(aq) + X"(aq) — AgX(s)
This reaction forms the basis for a test for the presence of halides. Halide tests are discussed in detail in Topic 8.3, Qualitative analysis.
+ Halide ions as reducing agents
In the displacement reactions between halogen and halide ions, the halogen gained electrons and the halide lost electrons. So halogens are oxidising agents and halide ions are reducing agents.
The reducing ability of halide ions can be shown by their reactions with sulfuric acid, H,SO.,, which is a strong oxidising agent. If a reaction takes place, the halide ions will be oxidised to form the halogen.
Chloride ions are not powerful enough to reduce H.SO,,.
Bromide ions are more powerful and can reduce H,SO, to sulfur dioxide, SO.: 2H* +H,SO0, + 2Br- — SO, + Br, + 2H,0 Notice that the both the atoms and the charges balance.
lodide ions are even more powerful and reduce the sulfur dioxide formed to sulfur, S, which is then reduced further to hydrogen sulfide, H.S.
The unbalanced equations below are for the stepwise reduction, by I" ions, of H,S0, to HS.
Stepi H*+H,S0, +!" + SO, +1, +H,0
Step2 H*+S0,+1° +S+1,+H,0
Step3 H*+S+I—H,S+l,
a_ Balance the equations
b What are the oxidation number changes in each stage?
¢ Write an overall equation for the conversion of H,SO, to H,S
Summary questions
1 State and explain the trend in boiling points of the halogens fluorine to iodine. (3 marks)
2 Write full and ionic equations for displacement reactions of Cl,(g) with a KBr(aq) b Mgl.(aq). (4 marks)
3 Chlorine reacts with hot concentrated NaQH(aq) as below. 3CI, (aq) + 6NaOH(aq) — NaCi0,(aq) + SNaCI(aq) + 3H,0(\) Show that this is a disproportionation reaction. (3 marks)
8.3 Qualitative analysis
Specification reference: 3.1.4
Qualitative analysis of ions
You have already learned about titration as a quantitative analysis technique (a technique with numerical results). Qualitative analysis relies on simple observations rather than measurements, and can often be carried out quickly on a test-tube scale. The observations may be gas bubbles, precipitates, colour changes, or identification of gases.
Tests for anions
Tests based on gases
Carbonate test
Carbonates react with acids to form carbon dioxide gas. The equation below shows the reaction of dilute nitric acid with aqueous sodium carbonate.
Na,CO,(aq) + 2HNO,(aq) — 2NaNO,(aq) + CO,(g) + H,O(1) This reaction forms the basis for a test for the carbonate ion, CO,*-. 1 Ina test tube, add dilute nitric acid to the solid or solution to be tested. If you see bubbles, the unknown compound could be a carbonate.
To prove that the gas is carbon dioxide, use the test that you will remember from GCSE.
@ Bubble the gas through lime water — a saturated aqueous solution of calcium hydroxide, Ca(OH),.
® Carbon dioxide reacts to form a fine white precipitate of calcium
carbonate, which turns the lime water cloudy (milky) (Figure 1).
CO,(g) + Ca(OH),(aq) + CaCO, (s) + H,O(1)
Tests based on precipitates
Sulfate test
Most sulfates are soluble in water, but barium sulfate, BaSO,, is very insoluble. The formation of a white precipitate of barium sulfate is the basis for the sulfate test, in which aqueous barium ions are added to a solution of an unknown compound. The ionic equation
is shown below.
Ba?*(aq) + SO,*-(aq) — BaSO,(s)
Usually the Ba**(aq) ions are added as aqueous barium chloride or barium nitrate. If you intend to carry out a halide test afterwards, use barium nitrate — with barium chloride, you are introducing chloride ions to your solution.
ett PP SPSO SOHO OE SEOSOSEHOOEESEOESESESES OOS
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> qualitative analysis of ions > test-tube tests
> analysing unknown compounds.
Synoptic link
All carbonates react with acids in a similar way, as described in Topic 4.1, Acids, bases, and neutralisation.
A Figure 1 Lime water test for carbon dioxide
A Figure 2 Adense white precipitate forms when Ba** (aq) ions are added to a solution containing sulfate ions, S0,°- - the heavy precipitate of BaSO, sinks to the bottom of the test tube
117
8.3 Qualitative analysis
Halide tests
Most halides are soluble in water, but silver halides are insoluble. Aqueous silver ions react with aqueous halide ions to form precipitates of silver halides, as shown by the general equation below. X~(aq) represents an aqueous solution of any halide.
Ag*(aq) + X"(aq) — AgX(s)
This reaction forms the basis for a halide test, which was briefly introduced in Topic 8.2, The halogens.
1 Add aqueous silver nitrate, AgNO,, to an aqueous solution of a halide.
2. The silver halide precipitates are different colours — silver chloride is white, silver bromide is cream-coloured, and silver iodide is yellow (Figure 3).
3 Add aqueous ammonia to test the solubility of the precipitate. This stage is very useful because the three precipitate colours can be difficult to tell apart.
A Figure 3 From left to right, the test tubes contain — a precipitate of silver chloride, silver chloride after addition of dilute aqueous ammonia, a precipitate of silver bromide, silver bromide after addition of concentrated aqueous ammonia, a precipitate of silver iodide, and silver iodide after addition of concentrated aqueous ammonia, which fails to dissolve the silver iodide precipitate
The reactions with Ag*(aq) ions and the solubilities of the silver halide precipitates in aqueous ammonia are summarised in Table 1. V Table 1 Halide tests using aqueous silver ions, Ag* (aq)
Halide lonic equation Colour of precipitate Solubility in NH,(aq)
chloride, CI- Ag*(aq) + Ci-(aq) — AgCi(s) white | soluble in dilute NH, (aq) ee) ab) tone Wd
nea) -Foa) =A) | vow [ince non)
REACTIVITY TRENDS
A barium meal — making use of precipitation reactions
You may have heard of barium meals — not some kind of food, but an application of precipitation in medicine.
Barium meals are used to enable doctors to see the outline of the gullet, stomach, and upper small intestine in order to identify abnormalities such as ulcers or tumours. The patient swallows water that has been shaken with barium sulfate, the insoluble compound that forms during the sulfate test. The white precipitate coats the inner lining of the gut. An X-ray image is then
taken that disp! the bari lfat ting th t. aken that displays the barium sulfate coating the gu ia Wigari'd. iicupincmn ofa pertane wha
has taken a barium meal to coat the upper gastrointestinal tract with barium sulfate, which shows up white in the picture
Barium ions in solution are extremely toxic. Why are patients not poisoned by this treatment?
Sequence of tests
If you are asked to analyse an unknown inorganic compound,
you will need to carry out the tests for anions in the correct order. Otherwise you could obtain confusing results and make an incorrect identification. For anions, the correct order for tests is:
1 carbonate, CO,*- 2 sulfate, SO,?-
3. ~=halides, Cl-, Br-, and I-
Why is there a correct order?
To understand the reasons behind the correct order, you need to think about the chemistry involved in each test.
Carbonate test In the carbonate test, you add a dilute acid and are looking for effervescence from carbon dioxide gas.
Neither sulfate nor halide ions produce bubbles with dilute acid. The carbonate test can be carried out without the possibility of an incorrect conclusion. If the test produces no bubbles, then no carbonate is present and you can proceed to the next test.
Sulfate test In the sulfate test, you add a solution containing Ba?*(aq) ions and are looking for a white precipitate of BaSO,(s).
Barium carbonate, BaCO,, is white and insoluble in water. So if you carry out a sulfate test on a carbonate, you will get a white precipitate too. Therefore it is important to carry out the carbonate test first and only proceed to the sulfate test when you know that no carbonate is present.
Halide test In the halide test, you add a solution containing Ag*(aq) ions, as AgNO, (aq), and are looking for a precipitate.
119
Summary questions
1 How could you distinguish between NaCl, NaBr, and Nal by a simple test?
2 Explain why it is important to carry out the carbonate test before carrying out a sulfate test on an unknown chemical.
(2 marks)
a mixture, it is important to use dilute nitric acid, rather than sulfuric or hydrochloric acid, for the carbonate test?
(2 marks)
(3 marks)
Explain why, if you are testing
8.3 Qualitative analysis
Silver carbonate, Ag,CO,, and silver sulfate, Ag,SO,, are both insoluble in water and will form as precipitates in this test. It is therefore important to carry out the halide test last, after carrying out carbonate and sulfate tests to rule out those possibilities.
What about a mixture of ions?
If you are asked to analyse a mixture of chemicals, you carry out the tests in the same sequence and on fhe same solution.
1 Carbonate test @ If you see bubbles, continue adding dilute nitric acid until the bubbling stops.
@ All carbonate ions will then have been removed and there will be none left to react in the next tests.
If you intend to test for sulfate or halide ions, it is important to use dilute nitric acid, HNO,, for this test. Sulfuric acid contains sulfate ions and hydrochloric acid contains chloride ions, which will show up in the sulfate and halide tests.
2 Sulfate test
@ To the solution left from the carbonate test, add an excess of Ba(NO,),(aq). Any sulfate ions present will precipitate out as barium sulfate.
@ Filter the solution to remove the barium sulfate.
If you intend to test for halide ions, it is important not to use BaCl,(aq), because the chloride ions will show up in the halide test.
3 Halide test @ To the solution left from the sulfate test, add AgNO,(aq).
@ Any carbonate or sulfate ions initially present have already been removed. Therefore any precipitate formed must involve halide ions.
@ Add NH,(aq) to confirm which halide you have.
Tests for cations Test for ammonium ion, NH,*
When heated together, aqueous ammonium ions and aqueous hydroxide ions react to form ammonia gas, NH,. NH,*(aq) + OH-(aq) > NH,(g) + H,O(I) This reaction forms the basis for a test of the ammonium ion. 1 Aqueous sodium hydroxide, NaOH, is added to a solution of an ammonium ion.
2 Ammonia gas is produced. You are unlikely to see gas bubbles as ammonia is very soluble in water. The mixture is warmed and ammonia gas is released. You may be able to smell the ammonia, but it is easy to test the gas with moist pH indicator paper. Ammonia is alkaline and its presence will turn the paper blue.
Practice questions 1 The equation for the redox reaction between
3
chlorine and hot concentrated NaOH(aq) is shown below.
Cl,(g) + 6NaOH(aq) — 5NaCl(aq)
+ NaClO,(aq) + 3H,O(1) This redox reaction is an example of disproportionation.
a_ Explain what is meant by
disproportionation. (1 mark)
b_ Using oxidation numbers, show that this reaction is an example of disproportionation. (3 marks)
ce Chlorine reacts with hot iron to form iron(II) chloride.
(i) Write the full equation, with state symbols, for the reaction. (2 marks)
(ii) Write a half equation to show the reduction of chlorine in this reaction. (1 mark)
(iii) The reaction is repeated using bromine instead of chlorine.
What difference would you expect in the rate of the reaction. Explain your answer. (1 mark)
In Group 2, reactivity increases down the
group.
a_ Explain, in terms of first ionisation energy, this trend in reactivity. (4 marks)
b_ The second ionisation energy is important when explaining trends in reactivity in Group 2.
(i) Write an equation, with state symbols, to represent the second ionisation energy of calcium.
(1 mark)
(ii) Why is the 2nd ionisation enthalpy important when explaining
Group 2 reactivity? (1 mark)
The Group 2 element barium was first isolated by Sir Humphrey Davy in 1808.
Barium has a giant metallic structure and a melting point of 725 °C.
a_ Describe, with the aid of a labelled diagram, the structure and bonding in barium and explain why barium has a high melting point.
REACTIVITY TRENDS
Include the correct charges on the metal particles in your diagram.
In your answer, you should use appropriate technical terms, spelled correctly. (3 marks)
A chemist reacts barium with water. A solution is formed which conducts electricity.
(i) Write the equation for the reaction of barium with water. Include state symbols. (2 marks)
(ii) Predict a value for the pH of the resulting solution. (1 mark)
(iii) Give the formula of the negative ion responsible for the conductivity of the solution formed. (1 mark)
Heartburn is a form of indigestion caused by an excess of stomach acid.
State a compound of magnesium that could be used to treat heartburn. (1 mark)
In an experiment, a student makes a solution of strontium chloride, SrCl,, by adding excess dilute hydrochloric acid to strontium carbonate.
(i) Describe what the student would observe and write the equation for the reaction.
(2 marks)
(ii) Draw a dot-and-cross diagram to show the bonding of strontium chloride. Show outer electrons only. (2 marks)
In another experiment, a student attempts to make a solution of strontium chloride by adding chlorine water to aqueous strontium bromide. (i) Describe what the student would observe. (1 mark) (ii) Write the ionic equation for the reaction which takes place. (1 mark) (iii) Chlorine is more reactive than bromine. Explain why. (4 marks)
OCR F321 Jan 2013 Q4
Chapter 8 Practice questions
b_ The student also discovered that chlorine, Cl,, is used in the large-scale treatment of
4 Group 2 elements react with the halogens a_ Describe and explain the trend in
reactivity of Group 2 elements with chlorine as the group is descended.
In your answer you should use appropriate technical terms, spelled correctly. (5 marks)
A student was provided with an aqueous solution of calcium iodide. The student carried out a chemical test to show that the solution contained iodide ions. In this
water.
(i) State one benefit of adding chlorine to water. (1 mark)
(ii) Not everyone agrees that chlorine should be added to drinking water.
Suggest one possible hazard of adding chlorine to drinking water. (1 mark)
c The equation for the reaction of chlorine test, a precipitation reaction took place. with water is shown below. (i) State the reagent that the student Cl,(g) + H,O(l) > HCl(aq) + HC1O(aq) would need to add to the solution ; Pye of caldum sadide (mark) (i) State the oxidation number of ara , Aa chlorine in (ii) What observation would show that ; R ay Cl; HCl Hclo 1 mark the solution contained iodide ions? : : ( (1 mark) (ii) The reaction of chlorine with water is ee : nes: 7 : ; a disproportionation reaction. (iii) Write an ionic equation, including ies ee er took place. (1 mark) to explain why. (2 marks) (iv) The student is provided with an (iii) Chlorine reacts with sodium aqueous solution of calcium bromide hydroxide to form bleach in another that is contaminated with calcium disproportionation reaction. iodide. The student carries out the Write an equation for this reaction. same chemical test but this time (1 mark) needs to add a second reagent to d= Two other compounds of chlorine are
show that iodide ions are present.
State the second reagent that the student would need to add. (1 mark)
F321 June 2013 Q4
chlorine dioxide and chloric(V) acid.
(i) Chlorine dioxide, ClO,, is used as a bleaching agent in both the paper and the flour industry. When dry,
ClO, decomposes explosively to form oxygen and chlorine.
5 Astudent used the internet to research chlorine and some of its compounds.
a_ They discovered that sea water contains Construct an equation for the
chloride ions. The student added aqueous silver nitrate to a sample of sea water.
(i) What would the student see? (1 mark)
(ii) Write an ionic equation, including state symbols, for the reaction that would occur. (2 marks)
(iii) After carrying out the test in (i), the student added dilute aqueous ammonia to the mixture.
What would the student see? (1 mark)
decomposition of ClO,. (1 mark)
(ii) Chlorie(V) acid has the following percentage composition by mass: H, 1.20%; Cl, 42.0%; O, 56.8%.
Using this information, calculate the empirical formula of chloric(V) acid. Show all of your working. (2 marks) (iii) What does (V) represent in chloric(V) acid? (1 mark)
F321 Jan 2009 Q3
6. This question is about properties of the
ike
halogens
a_ Describe and explain the trend in boiling points in the halogens. (4 marks)
b Bromine reacts with calcium to form an ionic compound.
State the electron configuration, in terms of sub-shells, for:
(i) Ca and Br atoms. (2 marks)
(ii) the ions formed in the reaction. (2 marks)
ce lodine and strontium are reacted together.
Explain why it is difficult to predict whether this reaction is more or less reactive than the reaction of bromine with calcium. (3 marks)
The Group 2 element barium, Ba, is silvery white when pure but blackens when exposed to air. The blackening is due to the formation of both barium oxide and barium nitride. The nitride ion is N*. a_ Predict the formula of:
(i) barium oxide
(ii) barium nitride (2 marks)
b AO.11g sample of pure barium was added to 100cm? of water.
Ba(s) + 2H,O(l) — Ba(OH),(aq) + H,(g)
(i) Show that 8.0 x 10~4 mol of Ba were added to the water. (1 mark)
(ii) Calculate the volume of hydrogen, in cm’, produced at room temperature and pressure. (1 mark)
(iii) Calculate the concentration, in mol dm~’, of the Ba(OH),(aq)
solution formed. (1 mark)
(iv) State the approximate pH of the Ba(OH),(aq) solution. (1 mark) c A student repeated the experiment in (b) using a 0.11 g sample of barium that had blackened following exposure to the air.
Suggest why the volume of hydrogen produced would be slightly less than the volume collected using pure barium.
(1 mark)
REACTIVITY TRENDS
d_ Describe and explain the trend, down the group, in the reactivity of the Group 2 elements with water. (5 marks)
F321 June 2009 Q5
A student is provided with a solution
containing two different sodium compounds.
The student carries out the series of tests
below on the mixture.
Test I
Observation: Effervescence.
Add dilute nitric acid.
Test 2 Add aqueous barium nitrate to the resulting mixture. If a precipitate forms, filter off the precipitate before carrying out Test 3.
Observation: No observable change Test 3 Add aqueous silver nitrate
Observation: Cream precipitate which dissolves in concentrated aqueous ammonia.
a What conclusions can be drawn from the tests? Include the formulae of any ions identified and equations for any
reactions that take place. (4 marks)
b A second student carries out the same sequence but added dilute sulfuric acid in the first step rather than dilute nitric acid. The student continues with the series of tests and obtains a different result and comes to an incorrect conclusion.
State and explain the second student's different result and conclusion. Include an equation for any different reaction.
(3 marks)
You are provided with unlabelled solutions of sodium bromide, ammonium bromide, ammonium iodide, and ammonium iodide.
Describe test-tube tests that would allow
you to identify which solution is which. For each test, state the observation for a positive result and write an equation to illustrate a positive test. (7 marks)
ENTHALPY
9.1
Enthalpy changes
Specification reference: 3.2.1
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> exothermic and endothermic changes
enthalpy profile diagrams activation energy standard enthalpy changes.
Peccceccceeeeeeaseseseeseseeesse®
vous
universe = system + surroundings
surroundings system
(contents of flask)
A Figure 1 The system and the surroundings
Enthalpy and enthalpy change Enthalpy
Enthalpy #H is a measure of the heat energy in a chemical system. The chemical system refers to the atoms, molecules, or ions making up the chemicals.
Enthalpy is sometimes thought of as the energy stored within bonds. Enthalpy cannot be measured, but enthalpy changes can.
Enthalpy change
In a chemical reaction, the reactants and products are likely to have different enthalpies. The difference in the enthalpies is the enthalpy change AI:
AH = H(products) — H(reactants)
AH can be positive or negative, depending on whether the products contain more or less energy than the reactants.
Conservation of energy The law of conservation of energy is one of the fundamental rules of science and states that energy cannot be created or destroyed.
When a chemical reaction involving an enthalpy change takes place, heat energy is transferred between the system and the surroundings. @ The system is the chemicals — the reactants and products.
@ The surroundings are the apparatus (e.g., the thermometer and apparatus), the laboratory, and everything that is not the chemical system.
@ The universe is everything, and includes both system and surroundings, as shown in Figure 1.
Heat in or heat out?
Unlike enthalpy, an enthalpy change AH can be determined experimentally by measuring the energy transfer between the system and the surroundings.
Energy transfer can be in either of two directions:
@ from the system fo the surroundings - an exothermic change @ from the surroundings to the system — an endothermic change.
In Figures 2 and 3, the diagrams on the left are called enthalpy profile diagrams. Each diagram shows the relative enthalpies of the reactants and products and the enthalpy change AH.
ENTHALPY
Exothermic — heat out of the system : Study tip
The conservation of energy means that: Remember the directions of
energy transferred energy transfer:
from the system
Exothermic @ The chemical system releases heat energy to the surroundings. @ AHis negative
@ Any energy loss by the chemical system is balanced by the same energy gain by the surroundings.
chemical system loses energy
e @ surroundings gain energy e
@ <Al/is negative. temperature of surroundings
@ The temperature of the surroundings increases as they gain energy. increases.
SURROUNDINGS
reactants
= Chemical system
r=. loses energy
= AH is negative
| ond
S temperature products rise
SURROUNDINGS ; Surroundings gain energy progress of reaction temperature increases
A Figure 2 Exothermic energy change
Endothermic — heat into the system
Study tip The conservation of energy means that: Remember the directions of energy transferred < energy transfer: to the system Endothermic The chemical system takes in heat energy from the surroundings. @ AHis positive Any energy gain by the chemical system is balanced by the same @ chemical system gains energy
energy loss by the surroundings. @ surroundings lose energy ®@ AH is positive. @ temperature of surroundings @ The temperature of the surroundings decreases as they lose energy. decreases.
SURROUNDINGS
enthalpy H
temperature
reactants fal
SURROUNDINGS
progress of reaction Surroundings lose energy temperature decreases
A Figure 3 Endothermic energy change
9.1 Enthalpy changes
Synoptic link
You will learn more about activation energy and its relevance
to catalysis in Chapter 10, Reaction rates and equilibrium.
2Mg(s) + 0.(g)
AH is negative
enthalpy H
progress of reaction
A Figure 4 Exothermic enthalpy profile diagram
AH is positive
enthalpy H
progress of reaction
A Figure S Endothermic enthalpy profile diagram
synoptic link You will learn about the Kelvin scale of temperature in Topic 9.2, Measuring enthalpy changes.
4)
Activation energy
Atoms and ions are held together by chemical bonds. During chemical reactions, the bonds in the reactants need to be broken by an input
of energy. New bonds in the products can then form to complete the reaction. The energy input required to break bonds acts as an energy barrier to the reaction, known as the activation energy E,. Activation energy is the minimum energy required for a reaction to take place.
Figures 4 and 5 show full enthalpy profile diagrams for exothermic and endothermic reactions. Note that these diagrams show the reactants and products, together with labels for AH and E,.
In general, reactions with small activation energies take place very rapidly, because the energy needed to break bonds is readily available from the surroundings. Very large activation energies may be present such a large energy barrier that a reaction may take place extremely slowly or even not at all.
Standard enthalpy changes
The enthalpy change for a reaction can vary slightly depending on the conditions used. Chemists use standard conditions for physical measurements such as enthalpy changes, close to typical working conditions of temperature and pressure. Tables of data always include values taken under standard conditions.
Standard conditions
A standard physical value, such as an enthalpy, is shown in data tables using a special standard sign”. A standard enthalpy change AH” refers to an enthalpy H change A under standard conditions”.
Units are usually kJ mol“', with mol"! referring to the amount in mol given by the balancing numbers of the chemicals in a stated equation for the reaction (see enthalpy change of reaction below).
@ Standard pressure is 100kPa. This is very close to a pressure of one atmosphere, 101 kPa.
@ Standard temperature is a stated temperature, usually 298K (25°C). In this book, standard temperature refers to 298 K.
@ Standard concentration is | moldm~? (this is relevant for solutions only).
@ Standard state is the physical state of a substance under standard conditions. Most data tables show the standard state at 100 kPa and 298K.
Enthalpy change of reaction
The standard enthalpy change of reaction AH” is the enthalpy change that accompanies a reaction in the molar quantities shown in a chemical equation under standard conditions, with all reactants and products in their standard states.
A.H® always refers to a stated equation, and its value depends on the balancing numbers. For example, the equation for the reaction of magnesium with oxygen to form magnesium oxide can be written using a fraction to balance O,,.
Mg(s) + 50,18) —+ MgO(s)
1 mol Smol 1 mol
A.H® = -602kJ mol"!
If the equation is balanced with whole numbers, the amounts are doubled and the enthalpy change is doubled.
2Mg(s) + O,(g) — 2MgO(s) A.H® =-1204kJ mol"!
2 mol lmol 2mol
Both A, H™ values are correct, but only for the quantities given in each equation written for each value.
Enthalpy change of formation
The standard enthalpy change of formation A,H”™ is the enthalpy change that takes place when one mole of a compound is formed from its elements under standard conditions, with all reactants and products in their standard states.
Compounds
The fractional equation for the reaction of magnesium and oxygen above gives the enthalpy change for the formation of | mol of a compound. By definition, this is 4,H” for MgO(s).
Mg(s) + 50,(8) — MgO(s) A,H™ = -602kJ mol"!
elements — | mol
The equation could have been written using whole balancing numbers, but it would then not match the definition for A,H”, which requires formation of one mole of MgO.
Elements
From its definition, 4,H~ for an element refers to the formation of one mole of an element from its element. This is clearly no change, so all elements have an enthalpy change of formation of 0kJ mol!"'.
Enthalpy change of combustion
The standard enthalpy change of combustion A_//” is the enthalpy change that takes place when one mole of a substance reacts completely with oxygen under standard conditions, with all reactants and products in their standard states.
When a substance reacts completely with oxygen the products are the oxides of the elements in the substance.
The equation for the combustion of one mole of butane, C,H)9, is shown below.
C,H,(g) + 6503(8) + 4CO,(g) + 5H,0(1)
1 mol —» combustion products
A.H® = -2877kJ mol!
ENTHALPY
Study tip
Learn the definition for A,H~.
Study tip
When balancing equations for enthalpy changes of formation, you must not add a balancing number in front of the product that has formed. Balance the equation to give one mole of the product.
Remember that A,H “is for the formation of one mole of a substance.
Study tip
Learn the definition for AH”.
Study tip
Learn the definition for A. .H™.
neut
Summary questions
1 Sketch enthalpy profile
diagrams, including E “3 for the following reactions: a N.(g) + 3H,(g) — 2NH,(g) AH = -92kJ mol“! (2 marks) b N,0,,(g) =a 2N0.(g) AH =+58kJ mol"! (2 marks)
Write equations, including state symbols, that give A, H™ for:
a CH,
b NO,
(1 mark) (1 mark)
Write equations, including state symbols, that give
A.H* for: a H,S(g) (1 mark) b Al (1 mark)
9.1 Enthalpy changes
The equation could have been balanced without fractions, but it would then not match the definition, which requires combustion of one mole of C,H, 9.
Enthalpy change of neutralisation
The standard enthalpy change of neutralisation A... H™ is the energy change that accompanies the reaction of an acid by a base to form one mole of H,O(1), under standard conditions, with all reactants
and products in their standard states.
H®* =-57kJmol!
neut
H*(aq)
acid
+ OH (aq) —
base —
H,O(l) A
1 mol
The neutralisation of hydrochloric acid by sodium hydroxide to form one mole of H,O(1) is shown below:
HCl(aq) + NaOH(aq) — H,O(l) + NaCl(aq) acid base — I1mol For A... 4”, neutralisation involves the reaction of H*(aq) with
neur/? is the same
OH (aq) to form one mole of H,O(l). The value of A for all neutralisation reactions.
Hex What about calories?
Calories are in the news and the importance of a calorie-controlled diet is constantly talked about. You may know that calories have something to do with energy but how are they related to joules? The calorie content of many foods is stated on the packaging in the nutrition label, usually on the back or side of packaging. This information will appear under the Energy heading. Strangely the food calorie is really a kcal.
Figure 6 shows the label from a chocolate bar. You can see the energy information but how many people read this? The label also shows how much this single bar contributes to daily energy requirement! A Figure 6 Chocolate bar with energy information
1 Acalorie is equal to 4.18 kJ. a Where have you seen this number before? b How many kJ of energy are available in a Mars bar?
2 Use kitchen scales to find the mass of a cup of tea. Then calculate how many cups of tea could be made from the energy available in a chocolate bar. Assume that the water from the tap is at 15°C
9.2 Measuring enthalpy changes e"
Specification reference: 3.2.1, 2.1.3
Measuring energy changes
In Topic 9.1, you looked at the distinction between the chemical system and the surroundings. When you carry oul experiments
to determine enthalpy changes, the thermometer is part of the surroundings and you will be measuring the temperature change of the surroundings.
The Kelvin scale of temperature
The Kelvin scale of temperature is commonly used in science. It starts at absolute zero, 0K and is equivalent to —-273°C. The Kelvin scale
is part of the International System of Units (SI units). On the Kelvin scale, ice melts at 273 K (0°C) and water boils at 373K (100°C). Soa 1K rise in temperature is the same as a 1 °C rise in temperature. If you record temperatures using a thermometer graduated in °C, the value of the temperature change is exactly the same in °C and K.
Calculating an energy change
The energy change of the surroundings is calculated from three quantities — mass, specific heat capacity, and temperature change.
The mass of the surroundings m
The mass is measured simply by weighing. You have to identify the materials that are changing temperature. Mass is usually measured in grams (g) to match the scale often used in experiments.
The specific heat capacity of the surroundings c
Different materials require different quantities of energy to produce the same temperature change. The specific heat capacity c is the energy required to raise the temperature of | g of a substance by 1 K.
Every substance has a specific heat capacity. Good conductors of
heat, such as metals, have small values of ¢. Insulators of heat such as foam plastic, have large values of c. In most experiments, you will be measuring the temperature change of water or aqueous solutions. For water, c= 4.18Jg-' K"!.
The temperature change of the surroundings AT The temperature change AT is determined from the thermometer readings:
AT = T(final) — T(initial).
Calculating an energy change
Heat energy is given the symbol g. Once you have values for m1, ¢, and AT from an experiment, it is very easy to calculate the energy in joules (J) using the simple equation below.
q = mcAT
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Learning outcomes
Demonstrate knowledge, understanding, and application of:
> calculating energy changes
from experiments > determination of enthalpy changes directly > practical techniques for measuring mass. a water 400K - 350K 50 °C = 300K oe water | 0 Be ia freezes afon 50°C i 250K S dry ice 200K ~78 °C solid CO, ag 195K ~100 °C 150K -150 °C -~200 °C air 50K -250 °C - absolute “273"C zero OK Celsius Kelvin
A Figure 1 Comparison of Celsius and Kelvin scales of temperature
Study tip
It is easy to convert between Kelvin and Celsius. Add 273 to a Celsius reading to get the Kelvin reading. Subtract 273 from a Kelvin reading to get the Celsius reading.
Study tip
Learn g =mcAT,
Remember that m is the mass that changes temperature not the mass of the reactants.
thermometer
beaker clamp water spirit burner wick methanol
A Figure 2 Apparatus for determination of A_H — of a liquid fuels such as methanol
Study tip
For Step 1, you are calculating the energy change in the surroundings — where the thermometer is.
You always use q = mcAT Joules
Finally divide by 1000 to convert to kJ
synoptic link
You learnt about this equation, n= 7, in Topic 3.1, Amount of substance and the mole.
9.2 Measuring enthalpy changes
: : Determination of an enthalpy change of combustion AH Combustion is simply a reaction of a substance with oxygen (see Topic 9.1) and is one of the easiest enthalpy changes to determine
experimentally. The equation for the combustion of methanol is shown below.
CH,OH(l) + 1%0,(g) + CO,(g) + 2H,0(1)
Spirit burners
Liquid fuels, such as methanol, can easily be burnt using small spirit burners. The experimental method and the calculation are outlined below.
1 Using a measuring cylinder, measure out 150 cm? of water. Pour the water into the beaker. Record the initial temperature of the water to the nearest 0.5°C.
2 Add methanol to the spirit burner. Weigh the spirit burner containing methanol.
3 Place the spirit burner under the beaker as shown in Figure 2. Light the burner and burn the methanol whilst stirring the water with the thermometer.
4 After about three minutes extinguish the flame. Immediately record the maximum temperature reached by the water.
5 Re-weigh the spirit burner containing the methanol. Assume that the
wick has not been burnt.
Worked example: Determination of A_H of methanol
Results Mass of spirit burner and methanol before burning
196.978
Mass of spirit burner and methanol after burning 195.37¢
Mass of fuel burnt 1.60g
Initial temperature of water = 215°C
Final temperature of water = 62.5°C
Temperature change AT of water = 41.0°C
For water, density = 1.00gem™?>, c= 4.18Jg"' K"!
Calculation Step 1: Calculate the energy change q of the water in kJ.
Density = 1.00gcm~* so 150.cm? of water has a mass of 150g.
q = mcAT = 150 x 4.18 x 41.0 = 25707 J = 25.707 kJ Step 2: Calculate the amount, in mol, of CH,OH burnt.
n(CH,OH) = = ih = 0.0500 mol
Step 3: Calculate AH in kJmol"'.
A.H is the enthalpy change for the complete combustion of 1 mol CH,OH.
In the experiment, 0.0500 mol CH,OH transfers 25.707 kJ of energy to the water.
The water gained 25.707 kJ of energy from the combustion of 0.050 mol CH,OH
25.707
1 mol CH,OH has lost mol CH, las los 0.0500
= 514.14kJ of energy on
combustion A-H(CH,OH) = -514kJ mol".
In the calculation, rounding was left until the final answer. This will always give the most accurate result. Work out the error that is introduced in the final answer by rounding the answer in Step 1 to:
a three significant figures
b two significant figures.
How accurate is the experimental A_H value?
The data book value for A.H of methanol is -726 kJ mol"!. This seems very different from the experimental value of -514kJmol-! shown above. Data book values are obtained using much more sophisticated apparatus than a spirit burner and a beaker of water. Clearly much less heat was transferred to the water in our experiment than expected. Possible reasons are listed below.
@ Heat loss to the surroundings other than the water. This includes the beaker but mainly the air surrounding the flame.
@ Incomplete combustion of methanol. There may be some incomplete combustion, with carbon monoxide and carbon being produced instead of carbon dioxide. You would see carbon as a black layer of soot on the beaker.
@ Evaporation of methanol from the wick. The burner must be weighed as soon as possible after extinguishing the flame. Otherwise some methanol may have evaporated from the wick. Spirit burners usually have a cover to reduce this error.
® Non-standard conditions. The data book value is a standard value. The conditions for this experiment are unlikely to be identical to standard conditions.
All but the last of these reasons would lead to a value for A_H that is less exothermic than expected.
ENTHALPY
Study tip
AH only makes an appearance right at the end. This is the
time to decide on the sign. As water has gained energy, the combustion of methanol must have lost the same quantity of energy. So AH is negative and the reaction is exothermic.
Synoptic link
As part of the practical skills required for your course, you need to know how to measure mass, volumes of solutions, and volumes of gases.
This practical application box tells you how to measure masses.
Measuring masses, volumes of solutions, and volumes or gases are also covered in:
@ Topic 3.2, Determination of formulae, for how to measure mass
@ Topic 3.4, Reacting quantities, for how to measure the volumes of gases
@ Topic 4.2, Acid—base titrations, for how to measure volumes of solutions
@ Topic 10.1, Reaction rates, for how to measure volumes of gases.
Study tip
Be very careful when comparing negative numbers.
The experimental and data book values for AH are —514 kJ mol and ~726 kJ mol”!. Correct comparisons for the experimental value would be less exothermic or less negative. Do not be tempted to state simply that the experimental value is less.
—514 is more than —726 as -514 is less negative!
A Figure 3 Asimple experiment to determine the enthalpy change of a reaction
132
9.2 Measuring enthalpy changes
Use of draught screens and an input of oxygen gas could minimise errors from heat loss and incomplete combustion.
Determination of an enthalpy change of reaction AH
Many reactions take place between two solutions, or between a solid and a solution. The enthalpy change of these reactions can be determined using plastic cups made of polystyrene foam. These are
cheap, waterproof and light weight, and offer some insulation against
heat loss to the surroundings.
When carrying out reactions between aqueous solutions, the solution
itself is the immediate surroundings. The chemical particles within
the solutions may react when they collide, and any energy transfer is
between the chemical particles and water molecules in the solution. A thermometer in the solution will record any temperature change, allowing the heat energy change to be calculated using mcAT.
Worked example: Determination of A H fora solid and a solution
An excess of zinc powder is added to 50.0cm? of 1.00 moldm~* copper(II) sulfate. The mixture is stirred until a maximum temperature is obtained.
Find AH for Zn(s) + CuSO,(aq) — Cu(s) + ZnSO,(aq)
Results
Initial temperature of solution = 22.5°C Final temperature of solution = 60.5°C Temperature change AT of solution = 38.0°C
For the solution, density and specific heat capacity are close to those of water (density = 1.00gcm~?*; c= 4.18Jg-'! K"')
Calculation Step 1: Calculate the energy change gq in the solution in kJ.
Density = 1.00gcm~? so 50.0cm? of the solution has a mass of 50.0¢.
q = mcAT = 50.0 x 4.18 x 38.0 = 7942 J = 7.942kJ
Step 2: Calculate the amount, in mol, of CuSO, that reacted (Zn is in excess). Vii “ry > (in cm’) = T100'% 50.0 1000 1000 Step 3: Calculate A.H in kJmol"!.
= 0.0500 mol
n(CuSO,) =c x
AH is for the reaction Zn(s) + CuSO,(aq) — Cu(s) + ZnSO,(aq)
Balancing numbers Imol Imol —Imol Imol give amounts:
ENTHALPY
In the experiment, 0.0500 mol CuSO, transfers 7.942 kJ of energy
to the solution.
7.942 1 mol CuSO, has lost 0.0500
A,H =-159kJmol".
Cooling curves
This experiment measures the enthalpy change for the reaction as on the previous page but the method has been adapted to correct for heat loss by use of a cooling curve correction.
This method can be used to correct for heat loss in other similar enthalpy experiments.
Method
1 Pipette 25.0. cm? of 1.00 mol dm-? CuSO, into a polystyrene cup. Weigh out an excess of zinc powder. Start a stop-clock and take the temperature of the solution every 30 s until the temperature stays constant.
Add the zinc to the solution and stir the mixture. Record the temperature every 30 seconds until the temperature has fallen for several minutes.
4 Plot a graph of temperature against time (Figure 4).
To correct for cooling, extrapolate the cooling curve section of the graph back to the time when the zinc was added. Draw a vertical line from the time that the solutions were mixed to the extrapolated cooling curve (Figure 4).
= 159kJ of energy to the solution.
temperature/°C
SDSSESVSSERKSSESSECLRKSR
time /min A Figure 4 Cooling curve method for heat loss Using the results shown on Figure 4, A H = -186 kJ mol~?.
This compares with —159 kJ mol! for the method on the previous page. You should be able to check this.
The table below gives results for the reaction of excess Mg with 100.0 cm?
0.500 mol dm~? CuSO,(aq)
Plot a graph of temperature against time and make a correction for heat loss based on the cooling curve. Hence obtain a corrected value for AT.
From your results, calculate the enthalpy change for the reaction of
1 mol CuSO, (aq) with Mg.
9.2 Measuring enthalpy changes
Determination of an enthalpy change of neutralisation A. ...H
This procedure is very similar to the previous example. The only difference is that two solutions react, rather than a solution and a solid.
Worked example: Determination of A. .H
neut A student measures out and mixes 35.0cm? of 2.40 moldm=*> NaOH and 35.0cm? of 2.40 moldm~?* HCl. The temperature rises by 16.5°C.
Specific heat capacity of the mixture is 4.18Jg~'K~'. The density of the mixture is 1.00gcm>>. Calculate the enthalpy change of neutralisation, in kJ mol". Step 1: Calculate the energy change q in the solution inkJ. Total volume of solution changing temperature = 35.0 + 35.0 = 70.0cm’. Density of mixture is 1.00gcm~*, so 70.0cm? has a mass of 70.0. gq = mcAT = 70.0 x 4.18 x 16.5 = 4827.9 J = 4.8279 kJ Step 2: Calculate the amount, in mol, of NaOH and HCI that reacted.
V (in cm?) _ 35.0 _ TO00 = 2:40 x Fohg = 0.0840mol
Step 3: Calculate A... Hin kJmol".
neul
n(NaOH) = n(HCl) =c x
A... fl is defined as the enthalpy change required for the neutralisation of an acid by an alkali
neut to form 1 mol H,O(1)
Reacting quantities NaOH(aq) + HCl(aq) — NaCl(aq) + H,O(1)
1 mol I mol — Imol I mol Experiment 0.0840mol 0.0840mol — 0.0840 mol Formation of 0.0840 mol H,O transfers 4.8279 kJ of energy to the solution.
4.8279 0.0840
Formation of 1 mol H,O loses = 57.5kJ of energy to the solution.
A... H=-57.5kJmol"!.
neut
Summary questions
1 Calculate the energy change, in kJ, of the following 3 Combustion of 1.656 g of ethanol, C,H.OH, raised the
(c=4.18 Jg-*K-*): temperature of 150 g of water from 22.5 °C to 74.5°C. a 50cm? of water decreases in temperature from a Write the equation to represent the enthalpy
62°C to 19°C; (2 marks) change of combustion of ethanol. (2 marks) b 50cm? of one aqueous solution is mixed with b Calculate the enthalpy change of combustion
75 cm? of a second solution and the temperature of ethanol to three significant figures. (2 marks)
increases from 23°C to 39°C. (1 mark)
2 Combustion of 1.29g C.H,, releases 34.2 kJ of energy. Calculate A_H for C,H,,. (2 marks)
9.3 Bond enthalpies
Specification reference: 3.2.1
Average bond enthalpy
In this topic you will look at how average bond enthalpies can be used to calculate enthalpy changes of reaction without carrying oul any experiments, as well as the limitations of this approach.
Average bond enthalpy is the energy required to break one mole of a specified type of bond in a gaseous molecule.
@ Energy is always required to break bonds.
@ Bond enthalpies are always endothermic.
@ Bond enthalpies always have a positive enthalpy value.
Table 1 lists average bond enthalpies for some of the more common covalent bonds. You will need to refer to this table during this topic.
Limitations of average bond enthalpies
The actual bond enthalpy can vary depending on the chemical environment of the bond. Figure 1 shows examples of the actual bond enthalpy of a C—H bond in different environments.
H H H H HH | Zz Ba H—C—H H—C—C—H H—C—C—C—H | | | rt} | H H H H ~ -
C—H 422kJmol™ C—H 439kJmol C—H 420kJmol™ C—H 411kJmol!
A Figure 1 Bond enthalpies of C—H bonds in different environments
An average bond enthalpy is calculated from the actual bond enthalpies in different chemical environments. In calculations, you will usually be provided with average bond enthalpies, but sometimes you may be provided with the actual bond enthalpy of an individual bond.
Bond breaking and bond making In chemical reactions, bonds break and new bonds are formed. @ Energy is required to break bonds bond breaking is endothermic AH is positive @ Energy is released when bonds form bond making is exothermic AH is negative
The difference between the energy required for bond breaking and the energy released by bond making determines whether an overall reaction is exothermic or endothermic (Figure 2).
oO OOOO OO 8OO OOO OOOOH OOOOGE,
Learning outcomes
Demonstrate knowledge,
understanding, and application of:
> average bond enthalpy
> calculating enthalpy changes from bond enthalpies.
Average bond enthalpy (kJ mol-*)
9.3 Bond enthalpies
exothermic endothermic
making bonds
breaking H bonds making H breaking
bonds bonds
The energy released when making bonds The energy required when breaking bonds is greater than the energy required when is greater than the energy released when breaking bonds. making bonds.
A Figure 2 Enthalpy profile diagrams for bond breaking and bond making in exothermic and endothermic reactions
Calculating enthalpy changes from average
bond enthalpies The enthalpy change of reaction AH can be found by calculating the Study tip bond enthalpies of the bonds in the reactants and the products. X (a capital Greek letter sigma) For a reaction involving gaseous molecules of covalent substances:
is shorthand for sum of.
A.H = &{(bond enthalpies in reactants) — &(bond enthalpies in products)
Worked example: Combustion of propane
Using average bond enthalpies, calculate AH for the reaction of propane with oxygen.
C,H,(g) + 50,(g) > 3CO,(g) + 4H,0(g)
VW Table 2 A H calculation for the reaction of propane with oxygen Equation 3CO,(g) + 4H,O(g)
stelecmecae &(C—H) 2(C—G) 5(O=0)
8 x 413 2 x 347 5 x 498 Energy / ; (3304 + 694 2490) kJ mol™* 6488
AH -—2054kJ mol"!
Limitations
As you are using average bond enthalpies, the actual energy involved in breaking and making individual bonds would be slightly different (the bonds may be in different environments).
ENTHALPY
Despite this limitation, the calculated enthalpy change of reaction should be in general agreement with the actual enthalpy change of reaction.
Calculations using average bond enthalpies need all species to be gaseous molecules. In the worked example above, you produced H,O(g) rather than H,O(1). This means that your calculated A.H is not a standard enthalpy change. You could still work out the standard enthalpy change but you would need to also take into account the enthalpy change for H,O(g) condensing into H,O(1).
Study tip
+ Bond enthalpies and combustion
In a question, if you are only provided with an equation, always draw out all the bonds. It is then much easier to get the
Values for the enthalpy change of combustion of alcohols can be calculated from bond enthalpies. AH values for three alcohols are shown below together with their equations.
CH,0H + 1.50, > CO, + 2H,0 A.H = -658kJ mol” C,H.OH + 30, — 2C0, + 3H,0 A.H =-12?6 kJ mol”! C,H,0H + 4,50, — 3C0, +4H,0 A.H = —1894 kJ mol”!
The enthalpy changes increase by a constant quantity for each increase in the carbon chain length. The reason is all linked to bond breaking and bond making.
number of each type of bond correct. The calculations are not difficult, but many students slip up by short-cutting and getting the wrong number of bonds.
1 Show that the increase in enthalpy change matches the extra bonds broken and made during combustion in progressing from one alcohol?
2 What would be the calculated enthalpy change of combustion of the alcohol with 20 carbon atoms?
3 Give two reasons why enthalpy changes calculated from average bond enthalpies are not standard enthalpy changes.
Summary questions
1 State what is meant by the term average bond 3 Use Table 1 and the following data: enthalpy. Explain why you would expect the actual 2S0,(g) + 0,(g) — 280,(g) AH=-192kJ mol? C=0 bond enthalpy to be different in CO, and Assume that SO, and SO, contain only S=0 bonds. in H,C=0. (2 marks) a Calculate the average bond enthalpy for 2 You are provided with data for the following reaction. tes bone: (2 marks) aT Beene in SO, is 531 kJ mol-*, Calculate the actual E, = +183kJ mol”! in SO, is mol-*. Calculate the actua
a Assuming that the activation energy breaks bond enthalpy forthe s==0 bondin 30: (2 marks)
all the bonds in the reactants, calculate how much energy is released during bond formation. (2 marks) b Explain what can be deduced about the relative strengths of the bonds that are broken and the bonds that are formed. (2 marks)
py , 9.4 Hess’ law and enthalpy cycles
Specification reference: 3.2.1
eae eset Hess’ law a eee The techniques described in Topic 9.2 allow enthalpy changes to Demonstrate knowledge, be determined directly in a single experiment. Unfortunately, the
understanding, and application of: > enthalpy cycles > indirect determination
enthalpy changes of many reactions are very difficult to determine directly. Hess’ law comes to the rescue, allowing enthalpy changes to be determined indirectly.
of enthalpy change from Hess’ law states that, if a reaction can take place by two routes, and enthalpy changes of the starting and finishing conditions are the same, the total enthalpy formation change is the same for each route. Ss tai sate paiect derernyrayen Hess’ law comes from the idea of conservation of energy (see Topic 9.1) of enthalpy change from . near : ‘ j and is easy to visualise with a diagram. Figure | shows an enthalpy enthalpy changes of A ; ; : cycle with two routes for converting reactants into products. combustion > unfamiliar enthalpy cycles. Following the arrows from reactants to COCO e eee ee eee eee ees ee eeseseeeeseeseset” products for the two routes: Route 1:A+B Route 2: C A By Hess’ law, the total enthalpy change is the same for each route. reactants
A+B=C If two of A, B, and C are known, the third can be calculated.
A Figure 1 Enthalpy cycle illustrating Hess’ law
Worked example: An enthalpy cycle
For the enthalpy cycle in Figure 1, A = +110kJmol"!; = -150kJmol'. Find C.
Step 1: Substitute values for A and B
into the cycle. Step 2: Calculate C. reactants -150 +110 + (-150) =C c C€ =-40kJ mor!
A Figure 2
This principle can be extended for any number of enthalpy changes. Provided that all enthalpy changes are known except for one, the unknown enthalpy change can always be determined.
ENTHALPY
Indirect determination of enthalpy changes Study tip
In Topic 9.1, you learnt about several standard enthalpy changes. Chemists have determined the values of many of these enthalpy changes and have listed them in data books. The most useful values are for standard enthalpy changes of formation A,H~ and combustion Usually you will not be provided AH”. The worked examples show how known enthalpy changes can with a value of zero for elements. be used with Hess’ law to determine enthalpy changes indirectly.
Remember, for elements A,H © is
always zero (Topic 9.1).
Worked example: Enthalpy changes from A,H®
You can work out the standard enthalpy change of any reaction from the standard enthalpy changes of formation A,H™ of the reactants and products.
Calculate the standard enthalpy change of reaction for the reaction shown below. Fe,O,(s) + 3Ca(s) — 2Fe(s) + 3CaO(s) The standard enthalpy changes of formation of the reactants and products are listed in Table 1.
VY Table 1 Standard enthalpy changes of formation
substance JOC) ca0(9)
Step 1: Construct the enthalpy cycle between the reactants, the products, and their elements. In the enthalpy cycle, the elements, Fe(s), Ca(s), and O,(g), form the common link between reactants and products. Using A,H”, the reactants and products of the original reaction are formed from their elements and the arrow points upwards.
A Fe,O,(s) + 3Ca(s) ==> 2Fe(s) + 3CaO(s) Following the arrows from reactants
to products for the two routes:
Route 1: B+A YAH” reactants B C YA,H*products Route 2: C
elements
Fe(s), Ca(s), 0,(g) By Hess’ law, B+ A= C
: so, A=C-B A Figure 3 Construction of the enthalpy cycle Step 2: Add A,H values and calculate the unknown enthalpy change.
A
Fe,0,(s) + 3Ca(s) 2Fe(s) + 3CaO(s) A=(3 X —635) — (-824)
=-1081 kJ mol"!
(-824)+0 B C 0+3 X -635
elements Fe(s), Ca(s), O,(g) A Figure 4 Calculation using the enthalpy change
9.4 Hess’ law and enthalpy cycles
Worked example: Enthalpy changes from A.H* The equation for the formation of butane, C,H,,(g), is shown below. 4C(s) + 5H,(g) + C,H, ,(g)
It would be impossible to measure the enthalpy change of this reaction directly. Carbon and hydrogen form so many compounds that C,H,, would be formed alongside dozens of other compounds.
However, the enthalpy changes of combustion of the reactants C(s) and H,(g) and of the product C,H, ,(g) can all be measured directly. The enthalpy change of reaction can then be found indirectly using the A-H™ values shown in Table 2.
VY Table 2 Standard enthalpy changes of combustion
Substance CoH
Step 1 Construct the enthalpy cycle between the reactants, the products, and their common combustion products, CO,(g) and H,O(l). Using AH”, both the reactants and the products of the original reaction react to form combustion products and the arrows point downwards.
A 4C(s) + 5H,(g) => C,H, ,(g) Following the arrows from reactants
to products for the two routes:
Route 1: A+C
AH” B C YAH” products reactants Route 2: B combustion products By Hess’ law, A+ C=B CO,(g), H,O(1) so, A=B-C
A Figure 5 Construction of an enthalpy cycle
Step 2 Add A_H values and calculate the unknown enthalpy change.
A AC(s) + 5H,(g) ——_—==»> C,H, o(a) A = (4 x -394) + (5 x -286) — ( -2877)
= -129kI mol"!
(4 x -394) +
aa C (-2877)
combustion products CO,(g), H,O(1)
A Figure 6 Calculation of the enthalpy change
Summary There are two rules that you can use to help you with your calculations using enthalpy changes of formation and combustion
Using enthalpy changes of formation, A,H, @ AH=* AH products — ¥ AH reactants Using enthalpy changes of combustion, AH,
@ AH=2 AH reactants - © AH products
ENTHALPY
Unfamiliar enthalpy cycle
The energy cycle below and enthalpy change information can be used to calculate an unfamiliar enthalpy change X.
Y Table 3 Enthalpy information
Enthal Reaction 4 PY ‘ change / kJ mol
C(s) + 0,(e) ~ C0,(s) CaCO,(s) + 2HCI(aq) > CaCl(aq) +C0,(g)+H,0(1) | 54 Ca(s) + 2HCI(aq) — CaCl, (aq) + H,(g)
AH Ca(s) + 2HCI(aq) + C(s) + 150,(g) ——> CaC0,(s) + 2HCI(aq) A 1
CaCl,(s) + H,(aq) + C(s) + 15 0,(g)
|e CaCI,(s) + H,0(1) + C(s) + 0,(g) —» CaCl, (s) + H,0(!) + CO, (g) To calculate and identify the unknown enthalpy change, match what has changed at each stage in the cycle with the enthalpy changes provided. Then work out the two routes to solve the unknown enthalpy change.
1 Use Table 3, to identify the values for enthalpy changes A—D.
2 Using the values for A—D and the enthalpy cycle, calculate the enthalpy change of formation A,H of calcium carbonate, CaC0,,.
Summary questions
1 Explain why A_H°(C(s)) and A,H® (CO.(g)) have the same value: -394 kJ mol”?. (1 mark)
2 You are provided with the following data.
Calculate A H “ for the following reactions.
a NH,(g) +HCI(g) > NH,Ci(s) (1 mark) b 2NH,Ci(s) + Ca(OH),(s) — CaCl,(s) + 2NH,(s) + 2H,0(() (1 mark) c 2NH,(g) + 3CuO(s) — N,(g) + 3Cu(s) + 3H,0(|) (1 mark)
3 You are provided with the following data.
Substance C.H,,(I) C,H.OH(I)
Calculate the enthalpy change of formationof a C.H., (1 mark) b C,H.OH((). (1 mark)
— Chapter 9 Practice questions
Practice questions
1 Write equations to represent the following
enthalpy changes.
a_ The standard enthalpy change of formation for ethanol, C,H,OH(I).
b_ The standard enthalpy change of combustion for hexane, C,H,,(1)
c The standard enthalpy change of neutralisation.
d_ The bond enthalpy of H—Br. The table below shows enthalpy changes of
formation. | .05(s) | cole) | c0.(e)
Compound
4/100 TC
a_ Define the term standard enthalpy change of formation.
Include the standard conditions in
your answer. (3 marks) b Calculate the value of A.H, in kJmol"', for
the reaction in the following equation?
1,0,(s) + 5CO(g) = L(s) + 5CO,(g)
(3 marks)
The equation for the reaction of nitrogen and hydrogen to form ammonia is shown below. N,(g) + 3H,(g) — 2NH,(g) AH® =-93 kJmol!. The H—H and N=N bond enthalpies are +436 and +941 respectively. Calculate the N—H bond enthalpy in kJ mol"!. (3 marks)
0.766 g of Mg are added 100 cm? (an excess)
of 1.00 moldm~* HCl. The temperature
changed from 22.0 °C to 44.5 °C.
a_ Write the equation for the reaction that takes place. (1 mark)
b Show that the HCI was in excess. (2 marks)
Show your working
c Calculate the energy change in the reaction. Assume that the specific heat capacity of the solution is the same as water. (] mark) d_ Calculate the enthalpy change, in kJ mol"! for the reaction of 1 mol of Mg. (2 marks)
bond enunalpy kJ mol”*
5 Enthalpy changes can be calculated from
average bond enthalpies. You are supplied with some average bond enthalpies.
SM [cc [0-n | mo | mo
a_ Explain why bond enthalpies are always endothermic. (1 mark)
b= The equation for the combustion of pentane is: C5H,,(g) + 80,(g) — 5CO,(g) + 6H,O(g) (i) Using the bond enthalpies, calculate the enthalpy change of combustion of pentane (3 marks)
(ii) Explain why this calculated enthalpy change is not a standard enthalpy change. (1 mark)
(iii) What are the limitations of using average bond enthalpies for calculating enthalpy changes?
(1 mark)
6 The enthalpy change of neutralisation can be
determined from experimental results.
a_ Define the term enthalpy change of neutralisation. (1 mark)
b_ Write the ionic equation for the change that represents enthalpy change of neutralisation. (1 mark)
ce 25.0 cm? of 2.00moldm~ HCl is placed in a plastic cup and its temperature is recorded. 25.0cm? of 2.00 moldm~* KOH is placed in a different plastic cup and its temperature is recorded.
The initial temperature of both solutions is 19.0 °C. After mixing the maximum temperature of the reaction mixture was B12 6:
Density of solution = 1.00gcm~*.
The specific heat capacity of the solution is the same as water.
Calculate the enthalpy change of
neutralisation. (4 marks)
d The same experiment was repeated using 100cm? of HCI and NaOH.
What would be the difference, if any in the temperature change and the enthalpy change of neutralisation? Explain your answer. (2 marks)
7 Hydrocarbons such as heptane, C,H, ,(1), are
used as fuels, making use of their combustion reaction with oxygen to form carbon dioxide and water.
a_ Define the term standard enthalpy change of combustion. Include the standard conditions in your answer. (3 marks)
b_ Write the equation, with state symbols for the equation that represents the enthalpy change of combustion of heptane.
(2 marks)
e Calculate the enthalpy change of combustion of heptane from the A,H values in the table below. (3 marks)
Substance AH/kJ mol
co. oe
Enthalpy changes of combustion, AH, are amongst the easiest enthalpy changes to determine directly.
a_ Define the term enthalpy change of combustion. (2 marks) b= A student carried out an experiment to determine the enthalpy change of combustion of pentan-1-ol, CH,(CH,),OH.
In the experiment, 1.76 g of pentan-1-ol was burnt. The energy was used to heat 250 cm? of water from 24.0°C to 78.0°C.
(i) Calculate the energy released, in kJ, during combustion of 1.76g pentan-1-ol. The specific heat capacity of water = 4.18Jg"! K"!. Density of water = 1.00gcm~’. (1 mark) (ii) Calculate the amount, in moles, of pentan-1l-ol that was burnt. (2 marks)
(iii) Calculate the enthalpy change of combustion of pentan-1-ol. Give your answer to three significant figures.
(3 marks)
ENTHALPY
The standard enthalpy change of formation of hexane can be defined as:
The enthalpy change when | mol of hexane is formed from its constituent elements in their standard states under standard conditions.
Hexane melts at —95 °C and boils at 69 °C. (i) What are standard conditions? (1 mark)
(ii) An incomplete equation is shown below for the chemical change that takes place to produce the standard enthalpy change of formation of hexane.
Add state symbols to the equation to show each species in its standard state.
BE csnee) FH FH ecennt) SP OLB lisssee) (1 mark)
(iii) It is very difficult to determine the standard enthalpy change of formation of hexane directly. Suggest a reason why. (1 mark)
(iv) The standard enthalpy change of formation of hexane can be determined indirectly.
Calculate the standard enthalpy change of formation of hexane using the standard enthalpy changes of combustion below. (3 marks)
Substance AH/kJ mol*
OCR F322 Jun 09 Q2
| REACTION RATES AND ’ EQUILIBRIUM
10.1
Reaction rates
Specification reference: 3.2.2, 2.1.3
SHEER EERE EEE EHH EERE EEE EEEEEEEEE EES.
Learning outcomes
Demonstrate knowledge, understanding, and application of:
.
—> the effect of concentration on reaction rate
> the effect of pressure on reaction rate
> reaction rates from the gradients of graphs.
POR
. DOP POPP PPR RRR REE REE EERE EEE
A Figure 1 Firework display over Tower Bridge in London. Fireworks are an example of a fast reaction
A Figure 2 The Forth Rail Bridge takes many years to paint but fortunately, longer to rust
How fast is a reaction?
Some chemical reactions are complete within a fraction of a second, whereas others may take centuries. High explosives detonate immediately and fireworks shoot upwards producing amazing displays of colour the instant the fuse burns down (Figure |), whereas iron rusts relatively
slowly — good news if you want to cross the Forth Rail Bridge (Figure 2).
What is meant by rate of reaction?
The rate of a chemical reaction measures how fast a reactant is being used up or how fast a product is being formed. The rate of a reaction can be defined as the change in concentration of a reactant or a product in a given time.
rate =
change in concentration ... moldm=> 3 units - = moldm~’s
time S
® The rate of a reaction is fastest at the start of the reaction, as each reactant is at its highest concentration.
@ The rate of reaction slows down as the reaction proceeds, because the reactants are being used up and their concentrations decrease.
@ Once one of the reactants has been completely used up, the concentrations stops changing and the rate of reaction is zero.
A Figure 3 Concentration—time graphs can be used to monitor the rate of a chemical reaction
REACTION RATES AND EQUILIBRIUM
Figure 3 shows the formation of a product over the course of a chemical reaction. The slope of the curve is steepest at the start of the reaction, when the rate is greatest. The curve becomes less steep as the reaction proceeds. Eventually the curve becomes a straight line parallel to the x-axis when the reaction is complete and the rate is zero.
Altering the rate of a chemical reaction
You will remember from GCSE that a number of factors can change the rate of a chemical reaction:
@ concentration (or pressure when reactants are gases)
@® temperature
@ use of a catalyst c)
surface area of solid reactants.
To understand why the rate changes, you need to think about A Figure 4 You can think of the effect reactions in terms of the particles involved. The collision theory of concentration as like a crowded states that two reacting particles must collide for a reaction to occur. street. The greater the number of people
in a crowd, the more chance there is of
Usually only a small proportion of collisions result in a chemical be a collision taking place
reaction. In most collisions, the molecules collide but then bounce off each other and remain chemically unchanged.
Why are some collisions effective and others ineffective? Synoptic link An effective collision is one that leads to a chemical reaction (Figure 5).
A collision will be effective if two conditions have been met: You learnt about activation energy
is in Topic 9.1, Enthalpy changes. @ the particles collide with the correct orientation We will look in more detail at @ the particles have sufficient energy to overcome the activation activation energy and reaction rate energy barrier of the reaction. in Topic 10.2, Catalysts.
Ineffective collision —> Effective collision
ad
A Figure 5 Fora reaction to occur, collisions must take place with the correct orientation and with sufficient energy for a reaction to occur
No reaction — the two biue atoms must collide for a reaction to occur
eo # %
Reaction takes place, as the two blue atoms collide to form a new product
oe” i
How does increasing the concentration affect the rate of reaction? When the concentration of a reactant is increased, the rate of reaction generally increases. An increase in concentration increases the number
A Figure 6 Marble chips in different concentrations of acids — the highest concentration on the left and the
of particles in the same volume. The particles are closer together and lowest concentration on the right. The collide more frequently. In a given period of time there will therefore more concentrated the acid, the more be more effective collisions (correct orientation and sufficient energy) frequent the collisions of H* (aq) ions and an increased rate of reaction (Figure 6). with the marble chips and the faster
hydrogen gas is produced
145
10.1 Reaction rates
Study tip
When asked to identify a method for measuring the rate of a reaction, the equation may give you a clue.
If a product has a (g) as its state symbol then the method could involve gas collection.
Study tip
When setting up this experiment you should ensure the water level starts close to zero on the scale.
How does increasing the pressure of a gas affect the rate of reaction? When a gas is compressed into a smaller volume the pressure of a gas is increased and the rate of reaction increases. The concentration of the gas molecules increases as the same number of gas molecules occupy a smaller volume. The gas molecules are closer together and collide more frequently, leading to more effective collisions in the same time.
Methods for following the progress of a reaction
The progress of a chemical reaction can be followed by:
@ monitoring the removal (decrease in concentration) of a reactant ®@ following the formation (increase in concentration) of a product. The method chosen will depend on the properties and physical states of the reactants and products in the reaction. In addition to concentration,
measurable properties that might change as the reaction proceed include gas volume, mass of reactants or products, and colour.
Reactions that produce gases
If a reaction produces a gas, two methods that can be used to
determine the rate of the reaction are:
@ monitoring the volume of gas produced at regular time intervals using gas collection
@ monitoring the loss of mass of reactants using a balance.
Volume of gas produced and mass loss are both proportional to the
change in concentration of a reactant or product. So the change in
volume with time or the mass loss with time both give a measure of the rate of reaction.
=-Y Monitoring the production of a gas using gas collection The rate of reaction for the decomposition of hydrogen peroxide, H,0., can be measured using the apparatus shown in Figure ?. The equation for the reaction is shown below. 2H,0,(aq) 2 AS 2 (1) + 0,(9)
1 Hydrogen peroxide is added to the conical flask and the bung is replaced. 2 The initial volume of gas in the measuring cylinder is recorded.
clamp
delivery tube measuring cylinder
conical flask
i r \
A Figure ? Apparatus to measure the rate of a reaction in which a gas is produced
reactants
REACTION RATES AND EQUILIBRIUM
3 Manganese dioxide, MnO.,, catalyst is then quickly added to the conical flask and the bung is replaced. A stop clock is started, The volume of gas produced in the measuring cylinder is recorded at regular intervals until the reaction is complete.
5 Thereaction is complete when no more gas is produced.
Alternatively, a gas syringe can be used instead of a measuring cylinder.
A graph is plotted of total volume of gas produced
against time. To calculate the initial rate of the reaction,
a tangent is drawn to the curve at t = 0 (Figure 9). The
gradient of the tangent gives the reaction rate. A Figure 8 Apparatus for collecting a gas with a syringe
Worked example: Calculating reaction rates from gas produced
8
Step 1: Plot a graph of volume of gas produced against time (Figure 8).
Step 2: Draw a tangent at f= 0 (red line in Figure 8). This is the initial rate.
> o
Step 3: Calculate the rate from the gradient of the tangent.
volume of gas produced /cm? 3
oO
rate from the gradient == =
= 80 ~ 5 O¢m3s"! Time/s l
A Figure 9 Graph of volume of gas produced against time
Calculate the rate after: a 24s b 40s
Monitoring the loss of mass of reactants using a balance
The rate of reaction between calcium carbonate and hydrochloric acid can also be determined by monitoring the loss in mass of the reactants over a period of time. The equation for the reaction is shown below.
CaCO, (s) + 2HCl(aq) — CaCl,(aq) + CO,(g) + H,O(1)
The carbonate and the acid are added to a conical flask on a balance. The mass of the flask and contents is recorded initially and at regular time intervals. The reaction is complete when no more gas is produced so no more mass is then lost. A graph of mass lost against time is plotted.
A Figure 10 Monitoring mass loss using a balance. The two readings show that mass has been lost as gas is released
— 10.1 Reaction rates
Worked example: Calculating reaction rates from mass loss Step 1: Plot a graph of mass lost against time (Figure 10).
Step 2: Draw a tangent to the curve at f= 0 (red line in Figure 11). This is the initial rate.
Step 3: Calculate the rate from the gradient of the tangent.
0.6
= =3g¢-1 0 = 8.6 x 10-°gs
rate from the gradient = 7 = Step 4: To calculate the rate at a specific time, the same tangent method is used. In Figure 11, the orange tangent is used
to find the rate of reaction after 100s.
j = — 9.36 _ 37! gradient = © = S10 = 1-7 x 10> gs
mass lost/g So Oo 6 & > Ww NM _ Oo
-0.6
o
50 100 150 200 250 300 350 time/s
A Figure 11 Graph of mass loss against time
Summary questions
1 Changing the concentration of a reactant alters the rate of a reaction. State three other factors that can affect the rate of a chemical reaction. Explain how an increase in concentration increases the rate. (5 marks)
2 State two possible methods for monitoring the rate of the following reaction: MgC0.(s) + 2HCI(aq) — MgCl,(aq) + CO,(g) + H,0(!) (2 marks)
3 The reaction 2N,0,(g) — 4NO,(g) + 0,(g) was carried out. The concentration of the N,0.. was recorded every 200s and the following data obtained.
Nolen | 100 | O88 | O78 | 069 | O61 | Om | oA | 049 | O38 | om [time/s | 0 | 200 | 400 | 600 | @00 | i000 | s200 | 1400 | 1600 | 1000 |
a Plot a graph of (N,0.(g)) on the y-axis against time on the x-axis. (2 marks) b Calculate the initial rate of reaction and the rate of reaction after 1000s. (2 marks)
148
10.2 Catalysts
Specification reference: 3.2.2
What does a catalyst do? et P PP SOSOSOHOOEOOO SOOO OHO SE OEOOESEOOSOE® Learning outcomes
Demonstrate knowledge,
understanding, and application of:
> the role of a catalyst > enthalpy profile diagrams
> homogeneous and heterogeneous catalysts.
A catalyst increases the rate of a chemical reaction by providing Viudeveveteusetesveuecseoseuveeveveueseves
an alternative reaction pathway of lower activation energy. See the
enthalpy profile diagrams in Figure 1 and Figure 2 for catalysis in
exothermic and endothermic reactions (Figure 1).
A catalyst is a substance that changes the rate of a chemical reaction without undergoing any permanent change itself. @ The catalyst is not used up in the chemical reaction,
@ The catalyst may react with a reactant to form an intermediate or may provide a surface on which the reaction can take place.
@ Atthe end of the reaction the catalyst is regenerated.
_ Study tip E,: without catalyst When drawing enthalpy profile E.: with catalyst diagrams: > > * Reactants and products should s Fs] be shown at the correct levels = 5 with respect to each other AH should be shown with an = arrow pointing in the correct direction progress of reaction progress of reaction Kee ; sibake A Figure 1 Exothermic reaction, with and A Figure 2 Endothermic reaction, with ee ih si without a catalyst and without a catalyst iad ; Types of catalyst Homogeneous catalysts A homogeneous catalyst has the same physical state as the reactants. unoptic link
The catalyst reacts with the reactants to form an intermediate. The intermediate then breaks down to give the product and regenerates the catalyst.
Two examples of the many reactions of gases and liquids that use homogeneous catalysis are shown below:
1 Making esters with sulfuric acid as a catalyst The equation below shows the preparation of the ester, CH,COOC,H.,, from ethanoic acid, CH,COOH, and ethanol, C,H,OH. Sulfuric acid, H,SO,, is the catalyst. H,SO,(1)
Sunoptic it
C,H,OH(l) + CH,COOH(I) CH,COOC,H, (I) + H,O(1)
The reactants (ethanol and ethanoic acid) and the catalyst (sulfuric acid) are all liquids.
10.2 Catalysts
2 Ozone depletion (Cle radicals as catalyst)
Synoptic link The equation below shows the depletion of ozone, O,, in the You will learn more about the presence of chlorine radicals, Cle, which act as a catalyst. catalytic breakdown of ozone Cle(g)
saint 20,(g) 30,(g)
in Topic 15.2, Organohalogen
compounds in the environment. The reactant (O,) and the catalyst (Cle) are both gases.
Heterogeneous catalysts
A heterogeneous catalyst has a different physical state from the reactants. Heterogeneous catalysts are usually solids in contact with gaseous reactants or reactants in solution. Reactant molecules are adsorbed (weakly bonded) onto the surface of the catalyst, where the reaction takes place. After reaction, the product molecules leave the surface of the catalyst by desorption.
Some of the many common industrial processes that use heterogeneous catalysis are listed in Table 1. Y Table 1 /ndustrial processes involving heterogeneous catalysts
Process Catalyst Equation
N,(@) + 3H,(@) = 2NH,(e)
oO C.H,4(g) > C,H,.(g) + H,(g)
C,H,(g) + H.(g) = C3H,(g)
v,0,(s) 2S0.(g) + 0,(g) rs 2S0,(g)
Since 1992 all petrol vehicles manufactured for road use in the UK must be fitted with a catalytic converter by law to pass the MOT test.
. Catalytic converters contain a catalyst made of platinum, rhodium, and A Figure 3 Fumes from car exhaust pipes palladium supported on a honeycomb mesh that provides a large surface area on which the reactions can take place (Figure 4). The hot exhaust gases are passed over this heterogeneous catalyst, and harmful gases are converted into less harmful products.
Combustion in a petrol engine forms the toxic gases carbon monoxide and nitrogen monoxide. In the catalytic converter, carbon monoxide is oxidised to carbon dioxide, and nitrogen monoxide is reduced to nitrogen gas.
The carbon dioxide and nitrogen products are both non-toxic and can be released into the atmosphere. In addition, any unburnt hydrocarbons are oxidised to water and carbon dioxide.
A Figure 4 Catalytic converters have
a large surface area for heterogeneous Write a balanced equation for the reaction of carbon monoxide and
catalysis to convert harmful exhaust nitrogen monoxide in a catalytic converter. gases into less harmful gases that can
be released into the atmosphere
150
REACTION RATES AND EQUILIBRIUM
Catalysis — sustainability and economic importance
It is estimated that 90% of all chemical materials are produced using a catalyst. Catalysts increase the rate of many industrial chemical reactions by lowering the activation energy. This then reduces the temperature needed for the process and the energy requirements
Synoptic link
You learnt about atom economy in Topic 3.4, Reaction quantities.
If a chemical process requires less energy, then less electricity or fossil
fuel is used. Making the product faster and using less energy can cul costs and increase profitability. The economic advantages of using a catalyst outweigh any costs associated with developing a catalytic process.
The modern focus on sustainability requires industry to operate processes with high atom economies and fewer pollutants. Using less fossil fuel will cut emissions of carbon dioxide, a gas linked to global warming.
+ Autocatalysis
A chemical reaction is said to have undergone autocatalysis if a reaction product acts as a catalyst for that reaction. Typically the reaction starts slowly and then speeds up as the products are formed.
An example of autocatalysis is shown in the equation below 2Mn0,” + 16H’ + SC,0,2" —- 2Mn*" + 8H,0 + 10€0,
This reaction is very slow in the absence of a catalyst. However Mn°* ions can act as a Catalyst because manganese easily changes between the oxidation states, Mn®* and Mn**.
In the first step of the autocatalysis, the Mn** formed reduces MnO, to Mn*", as shown in the equation below.
4Mn?* + MnO,” + 8H* — SMn?* +4H,0
The Mn3* then oxidises the C,0,°" to CO, reforming Mn°”.
1 Whatis the catalyst? 2 Write an equation for the reforming of Mn2*.
Summary questions
1 State the difference between a homogeneous catalyst and a heterogeneous catalyst. (1 mark)
2 Describe the effect of a catalyst on the activation energy of a chemical reaction and on the enthalpy change of reaction. (2 marks)
3 Methanol can be manufactured by the reaction of carbon dioxide with hydrogen as shown in the equation: 3H,(g) + CO,(g) = CH,0H(g) +H,0(g) AH=-49kJ mol The activation energy of the forward reaction is +225 kJ mol”.
a Draw an enthalpy profile diagram for this reaction. (2 marks) b Calculate the activation energy of the reverse reaction. (2 marks)
151
4 10.3 The Boltzmann distribution
Specification reference: 3.2.2
SCHHSHHHHHHAHHHHHHHHHHHHHSHHHHOHOHOOOOOOS,
Learning outcomes Demonstrate knowledge, understanding, and application of: > the Boltzmann distribution
> the Boltzmann distribution and activation energy
> the Boltzmann distribution, temperature changes, and catalysts.
Study tip
When sketching the Boltzmann curve, make sure it starts at the origin and the curve never crosses the x-axis, even at high energy.
The energy of moving particles
You learnt at GCSE that molecules in a gas move at high speed, colliding with each other and with the walls of the container they are held in. These collisions are said to be elastic; the molecules do not slow down as a result of a collision and no energy is lost.
In a gas, a liquid, or a solution, some molecules move slowly with
low energy and some molecules move fast with high energy. Most molecules move close to the average speed and have close to the average energy. This spread of molecular energies in gases is known
as the Boltzmann distribution (Figure 1). The graph is marked with a line, £,, that represents the activation energy of a reaction. You can see from the shaded area that only a small proportion of the molecules have more energy than E,, that is, enough energy to react.
number of molecules with a given energy
energy A Figure 1 The Boltzmann distribution of molecular energies
There are a number of features of the Boltzmann distribution:
@ No molecules have zero energy — the curve starts at the origin. @ The area under the curve is equal to the total number of molecules.
@ There is no maximum energy for a molecule — the curve does not meet the x-axis at high energy. The curve would need to reach infinite energy to meet the x-axis.
The Boltzmann distribution and temperature
The effect of temperature on a Boltzmann distribution curve is shown in Figure 2. As the temperature increases, the average energy of the molecules also increases. A small proportion molecules will still have low energy, but more molecules have higher energy. The graph is now stretched over a greater range of energy values. The peak of the graph is lower on the y-axis and further along the x-axis — the peak is at a higher energy. The number of molecules is the same, so the area under the curve remains the same.
REACTION RATES AND EQUILIBRIUM
At higher temperature, Tp, the peak is lower
Cc
g and shifted to the right
oD
oo
A
= At higher temperature, — Tp, a greater proportion > S E of molecules can overcome 8 s . the activation energy.
E
S
gs
E
=
c
energy
A Figure 2 The effect of temperature on the Boltzmann distribution
At higher temperature:
@® More molecules have an energy greater than or equal to the activation energy.
@ Therefore a greater proportion of collisions will lead to a reaction, increasing the rate of reaction.
@ Collisions will also be more frequent as the molecules are moving faster, but the increased energy of the molecules is much more important than the increased frequency of collisions.
The Boltzmann distribution and catalysts
In Topic 10.2 you learnt that a catalyst lowers the activation energy of a reaction. The effect of a catalyst on activation energy is shown on a
Boltzmann distribution curve in Figure 3.
In the presence of a catalyst a greater proportion of molecules exceeds the new lower activation energy.
activation energy with catalyst
activation energy without catalyst
number of molecules with a given energy
E. E
energy :'
activation energy reduced A Figure 3 The effect of a catalyst on the number of molecules with enough energy to react.
A catalyst provides an alternative reaction route with a lower activation energy (E. on the graph). Compared to E,, a greater proportion of molecules now have an energy equal to, or greater than the lower activation energy, E.. On collision, more molecules will react to form products. The result is an increase in the rate of reaction.
Study tip
You may be asked to explain how increasing the temperature or
using a catalyst increases the rate of a chemical reaction. You will need to include ideas about the Boltzmann distribution.
Summary questions
1 Explain what is meant by the term activation
energy.
(1 mark)
2 Describe and explain how the rate of reaction is affected by a decrease
in temperature
(2 marks)
Sketch a Boltzmann distribution curve for
a volume of gas at temperature T,. Add a second curve for the distribution at a higher temperature, T,. (2 marks) Explain how raising the temperature increases
the rate of reaction,
using your answer
to(a) above. (2 marks) Using the Boltzmann distribution, explain how the presence of a catalyst increases the rate
of reaction. (2 marks)
153
10.4 Dynamic equilibrium and
le Chatelier’s principle
Specification reference: 3.2.3
Learning outcomes
Demonstrate knowledge, understanding, and application of:
> dynamic equilibrium > le Chatelier's principle
> the effect of temperature, concentration, and pressure on the position of equilibrium > catalysts and equilibrium.
Introducing reversible reactions
When ignited, hydrogen reacts with oxygen in the air to form water. At the end of the reaction all the hydrogen has been used up. The reaction has gone to completion. The equation contains an arrow pointing from reactants to products.
2H,(g) + O,(g) — 2H,O(1) reactants products
In this topic, you will look at some reversible reactions, reactions that take place in both ‘forward’ and ‘reverse’ directions. Many of these reactions are important industrial processes, for example, the Haber process for manufacturing ammonia:
N,(g) + 3H,(g) = 2NH,(g)
The = symbol indicates that the reversible reaction is in equilibrium.
Dynamic equilibrium In an equilibrium system:
@ the rate of the forward reaction is equal to the rate of the reverse reaction
@ the concentrations of reactants and products do not change.
Equilibrium systems are dynamic. At equilibrium both the forward and reverse reactions are taking place. As fast as the reactants are becoming products, the products are reacting to become reactants. Therefore in an equilibrium system the concentrations of the reactants and products remain unchanged even though the forward and reverse reactions are still taking place.
For a reaction to remain in equilibrium, the system must be closed.
A closed system is isolated from its surroundings, so the temperature, pressure, and concentrations of reactants and products are unaffected by outside influences.
le Chatelier’s principle
The position of equilibrium indicates the extent of the reaction. In a reversible reaction, if the temperature, pressure (for reactions involving gases), or concentration of the reactants or products is changed, then the position of equilibrium may change.
le Chatelier’s principle states that when a system in equilibrium is subjected to an external change the system readjusts itself to minimise the effect of that change.
REACTION RATES AND EQUILIBRIUM
Figure | illustrates a system in which the equilibrium has been disrupted by adding more reactant molecules. The position of equilibrium shifts to the right of the equation. More products are made than reactants until a new equilibrium is established.
system in equilibrium equilibrium disrupted system readjusted by the addition of more reactants
A Figure 1 When an equilibrium system is subjected to a change the position of equilibrium moves in such a way as to minimise the change
The effect of concentration changes on equilibrium
Changing the concentration of a reactant or a product in an equilibrium
system will change the rate of the forward or reverse reactions. The
position of equilibrium will then change. Figure 2 shows the effect increase in concentration of A or B of changing the concentration of reactants or products. decrease in concentration of C or D
|
When an equilibrium system adjusts as a result of a change:
@ if there are more products formed, the position of the Qa
equilibrium has shifted to the right ‘ ba increase in concentration of C or D @ if there are more reactants formed, the position of the
equilibrium has shifted to the /eft.
decrease in concentration of A or B
A Figure 2 The effect of concentration If you choose equilibria where the reactants and products have on the position of the equilibrium
different colours, simple experiments can illustrate how the position of equilibrium changes with an external change.
Investigating changes to the position of equilibrium with concentration The equilibrium between aqueous chromate ions, CrO,?~, and dichromate ions, Cr,0,*-, is sensitive to changes in acid concentration. Solutions of chromate and dichromate ions have different colours
so it is easy to see any shift in the position of equilibrium (Figure 3).
2CrO,?(aq) + 2H*(aq) = Cr,0,7-(aq) + H,O(1) yellow orange
A Figure 3 The chromate/dichromate equilibrium
add acid concentration of H*(aq) increases
——==_
2Cr0,?-(aq) + 2H*(aq) = Cr,0,2-(aq) + H,0(!)
yellow orange add alkali
concentration of H*(aq) decreases
A Figure 4 The effect of changing the concentration of H*(aq) on the CrO,,2-(aq)/Cr,0,7- (aq) equilibrium
>ynoptic link
You met in Topi, Ents
changes.
10.4 Dynamic equilibrium and le Chatelier’s principle
Looking to Figure 3, you would expect that the position of equilibrium could be changed by altering the concentrations of the reactants or products. You can carry out a simple experiment to show this.
1 Adda solution of yellow potassium chromate, K,CrO,, to a beaker.
2 Add dilute sulfuric acid, H,SO,, dropwise until there is no further change. The solution turns an orange colour.
3 Add aqueous sodium hydroxide, NaOH(aq), until there is no further change. The solution changes back to a yellow colour.
You can repeat steps 2 and 3 many times and the colour will change each time from yellow to orange and back to yellow again. So how does it work?
When you add dilute sulfuric acid, H,SO,, you are increasing the concentration of H*(aq) ions. This increases the rate of the forward reaction and so causes the position of equilibrium to shift to minimise the change in H*(aq) concentration.
This shift decreases the concentration of the added reactant, H*(aq).
2 The position of equilibrium shifts to the right of the equation, making more products
3 Anew position of equilibrium is established towards the products. @ The solution turns orange as Cr,0,*> forms.
When you add aqueous sodium hydroxide, NaOH(aq), the added OH (aq) ions react with H*(aq) ions, decreasing the concentration of H*(aq) ions.
H*(aq) + OH™(aq) — H,O(1)
The decreased concentration of the reactant, H*(aq) decreases the rate of the forward reaction and so causes the position of equilibrium to shift to minimise the change in concentration.
1 The shift increases the concentration of the reactant that has been removed, H*(aq).
2 The position of equilibrium shifts to the left, making more of the H*(aq) reactant.
3 Anew position of equilibrium is established.
@ The solution turns yellow as CrO,?-(aq) forms. Figure 4 summarises these changes.
Investigating changes to the position of equilibrium with temperature Changing the temperature of a system in equilibrium will result in the position of equilibrium changing.
The direction in which the equilibrium shifts depends on the sign of AH. @ Forward and reverse directions have the same value for the enthalpy change — but the signs are opposite.
@ An increase in temperature shifts the equilibrium position in the endothermic direction (AH is positive)
@ An decrease in temperature shifts the equilibrium position in the exothermic direction (AH is negative)
REACTION RATES AND EQUILIBRIUM
Cobalt chloride, CoCl,, dissolves in water to form a pink solution. The dissolving process actually produces an equilibrium between two complexes of cobalt that are different colours: When discussing changes in AH [Co(H,0),}?*(aq) + 4Cl-(aq) = CoCl2-(aq) + 6H,O(l) AH lb a Rslia alti Aken hiddod cl Z baaag— wae the forward reaction is exothermic is negative a is positive :
pink blue or endothermic.
Study tip
A Figure 5 The [Co(H,0),]**(aq)/CoCl,?- (aq) equilibrium
This equilibrium is sensitive to changes in temperature and the different colours makes it easy to follow any change in equilibrium position. You can carry out a simple experiment to show this. A small amount of hydrochoric acid can be added to provide more CI-(aq) ions. This shifts the
Study tip
1 Dissolve cobalt chloride in water in a boiling tube. Add a small quantity of hydrochloric acid. Place the boiling tube in some iced water. The solution is a pink colour.
equilibrium slightly towards the right and helps to achieve the colour changes. (See effect of concentration above)
2 Set up a boiling water bath and transfer the boiling tube into the boiling water. The solution turns a blue colour.
3 Transfer the boiling tube back to the iced water. The solution changes back to a pink colour.
You can repeat steps 2 and 3 many times and the colour will change each time from pink to blue and back to pink again. So how does it work?
In the boiling water, you are increasing the heat energy of the system. This causes the position of equilibrium to shift to minimise the change.
@ As the forward reaction is endothermic (AH is positive), the position of equilibrium shifts to the right in the endothermic direction, to take heat energy in and minimise the increase in temperature.
@ The solution turns a blue colour.
Decreasing the temperature shifts the position of equilibrium in the opposite direction, in the direction that gives out energy, to the reverse exothermic side (AH is negative) on the left.
Figure 6 summarises these changes.
Increase temperature Shift in endothermic direction
AH [Co(H,O),]?*(aq) + 4Cl-(aq) = CoCl?-(aq) + 6H,O(1) AH is negative pink blue is positive
decrease temperature Shift in exothermic direction
A Figure 6 The effect of changing temperature on the [Co(H,0),]°* (aq) /CoCl,?-(aq) equilibrium
157
Study tip
When discussing changes in temperature, always state whether the forward reaction is exothermic or endothermic. An exothermic change has a negative AH value, and an endothermic change has a positive AH value.
Study tip
When discussing changes in
pressure, always state whether there are more moles of gas on the left or right of the equation.
increase pressure shift towards fewer gaseous molecules
———
2NO,(2) —* N,0,(g) brown colourless
=
decrease pressure shift towards fewer gaseous molecules
A Figure ? Effect and increasing and decreasing the pressure on the equilibrium 2NO,(g) = N,0,(g)
10.4 Dynamic equilibrium and le Chatelier’s principle
The effect of temperature on the position of equilibrium for exothermic and endothermic reactions is summarised in Table 1.
V Table 1 The effect of temperature on the position of equilibrium
Forward reaction Increase temperature Decrease temperature
exothermic Position of equilibrium Position of equilibrium
(AHis negative) shifts to the left. shifts to the right.
More reactants are made. | More products are made.
endothermic Position of equilibrium Position of equilibrium shifts to the right. shifts to the left.
(AH is positive)
More products are made. More reactants are made.
The effect of pressure changes on equilibrium
Changing the pressure of a system containing gases in equilibrium may result in the position of equilibrium changing, but only if there are more gaseous molecules on one side of the equation than the other.
The gases nitrogen dioxide, NO,(g), and dinitrogen tetroxide, N,O,(g), have different colours. The gases form the equilibrium below.
2NO,(g) N,0,(g) brown colourless 2mol 1 mol
The pressure of a gas is proportional to its concentration. In the same container, two moles of NO,(g) would have twice the concentration and twice the pressure as the same container holding one mole of N,O,(g)
Increasing the pressure of the system will shift the position of equilibrium to the side with the fewer molecules, reducing the pressure of the system,
@ As there are fewer gaseous moles on the right-hand side of the equilibrium, the position of equilibrium shifts to the right reducing the number of gaseous moles to minimise the increase in pressure.
®@ More colourless N,O,(g) is formed and the brown colour fades.
Decreasing the pressure shifts the position of equilibrium in the opposite direction, to the side with more gaseous moles on the left and making the brown colour deeper (Figure 7).
The effect of a catalyst on equilibrium
A catalyst does not change the position of equilibrium; it merely speeds up the rates of the forward and reverse reactions equally. A catalyst will, however, increase the rate at which an equilibrium