MTG Chemistry Eng

i e n c e Pa r t 2 Grade Physical Sc Chemistry 12 Study Guide

Colour Diagrams: Acid-base Indicators* * Refer to page 136

© Department of Basic Education 2015 This content may not be sold or used for commercial purposes. Curriculum and Assessment Policy Statement (CAPS) Grade 12 Mind the Gap study guide for Physical Science Part 2: Chemistry ISBN 978-1-4315-1937-8 This publication has a Creative Commons Attribution NonCommercial Sharealike license. You can use, modify, upload, download, and share content, but you must acknowledge the Department of Basic Education, the authors and contributors. If you make any changes to the content you must send the changes to the Department of Basic Education. This content may not be sold or used for commercial purposes. For more information about the terms of the license please see: http://creativecommons.org/licenses/by-nc-sa/3.0/. Copyright © Department of Basic Education 2015 222 Struben Street, Pretoria, South Africa Contact person: Dr Patricia Watson Email: watson.p@dbe.gov.za Tel: (012) 357 4502 http://www.education.gov.za Call Centre: 0800202933 The first edition, published in 2012, for the Revised National Curriculum Statement (RNCS) Grade 12 Mind the Gap study guides for Accounting, Economics, Geography and Life Sciences; the second edition, published in 2014 , aligned these titles to the Curriculum and Assessment Policy Statement (CAPS) and added more titles to the series in 2015, including the CAPS Grade 12 Mind the Gap study guide for Physical Science Part 2: Chemistry ISBN 978-1-4315-1937-8 Mind the Gap team Series managing editor: Dr Patricia Watson Production co-ordinators: Lisa Treffry-Goatley and Radha Pillay Production assistants: Nomathamsanqa Hlatshwayo and Motshabi Mondlane Authors: Physical Science: Mariaan Bester, Nkosinathi Michael Khathi, Karen Yvonne Kornet, Jenny Koster, Mosala Daniel Leburu, Karen Reynecke, Lerato Mabogatsu Mphachoe, Thabiso Precious Ndaba, Gilberto Izquierdo Rodriguez, Meshack Jabu Sithole, Kumar Kuttan Pillai Suresh, Sehubedu Jerry Tladi Expert readers: Morongwa Masemolo, Prof Mundslamo Fathuwani, Dr T.D.T Sedumedi, Veena Maharaj, John Ostrowick, Karen van Niekerk Proofreaders: John Ostrowick, Sehubedu and Angela Thomas Designers: Sonja McGonigle and Erika van Rooyen Illustrators: Michele Dean, Vusi Malindi, Khosi Pholosa, John Ostrowick, Kenneth Tekane Cover illustration: Alastair Findlay Afrikaans Translation: Johan Steenkamp Onsite writers’ workshop IT support: Wayne Cussons Special thanks to Winning Teams, CEO Denzil Hollis, for the organisation’s subject expertise and workshop support. Winning Teams board game facilitators: Mantse Khoza and Sue Jobson

© Department of Basic Education 2015 Ministerial foreword The Department of Basic Education (DBE) has pleasure in releasing the second edition of the Mind the Gap study guides for Grade 12 learners. These study guides continue the innovative and committed attempt by the DBE to improve the academic performance of Grade 12 candidates in the National Senior Certificate (NSC) examination. The study guides have been written by teams of exerts comprising teachers, examiners, moderators, subject advisors and coordinators. Research, which began in 2012, has shown that the Mind the Gap series has, without doubt, had a positive impact on grades. It is my fervent wish that the Mind the Gap study guides take us all closer to ensuring that no learner is left behind, especially as we celebrate 20 years of democracy. Matsie Angelina Motshekga, MP The second edition of Mind the Gap is aligned to the 2014 Curriculum and Minister of Basic Education Assessment Policy Statement (CAPS). This means that the writers have considered the National Policy pertaining to the programme, promotion requirements and protocols for assessment of the National Curriculum Statement for Grade 12 in 2014. The CAPS aligned Mind the Gap study guides take their brief in part from the 2013 National Diagnostic report on learner performance and draw on the Grade 12 Examination Guidelines. Each of the Mind the Gap study guides defines key terminology and offers simple explanations and examples of the types of questions learners can expect to be asked in an exam. Marking memoranda are included to assist learners to build their understanding. Learners are also referred to specific questions from past national exam papers and examination memos that are available on the Department’s website – www.education.gov.za. The CAPS editions include Accounting, Economics, Geography, Life Sciences, Mathematics, Mathematical Literacy and Physical Sciences Part 1: Physics and Part 2: Chemistry. The series is produced in both English and Afrikaans. There are also nine English First Additional Language (EFAL) study guides. These include EFAL Paper 1 (Language in Context); EFAL Paper 3 (Writing) and a guide for each of the Grade 12 prescribed literature set works included in Paper 2. These are Short Stories, Poetry, To Kill a Mockingbird, A Grain of Wheat, Lord of the Flies, Nothing but the Truth and Romeo and Juliet. (Please remember when preparing for EFAL Paper 2 that you need only study the set works you did in your EFAL class at school.) The study guides have been designed to assist those learners who have been underperforming due to a lack of exposure to the content requirements of the curriculum and aim to mind-the-gap between failing and passing, by bridging the gap in learners’ understanding of commonly tested concepts, thus helping candidates to pass. All that is now required is for our Grade 12 learners to put in the hours required to prepare for the examinations. Learners, make us proud – study hard. We wish each and every one of you good luck for your Grade 12 examinations. _______________________________ Matsie Angelina Motshekga MP Minister of Basic Education 2015

© Department of Basic Education 2015

© Department of Basic Education 2015 Table of contents Dear Grade 12 learner. ..............................................................viii How to use this study guide.........................................................ix Top 10 study tips............................................................................x Mnemonics. ...................................................................................xi Mind maps. ...................................................................................xii On the day of the exam..............................................................xiii Question words to help you answer questions. ...................... xiv Vocabulary..................................................................................... xv General terms............................................................................... xv Technical terms......................................................................... xviii The maths you need. .............................................................. xxxiv Resource sheets..........................................................................xlii Formulas. .................................................................................... xliv Standard reduction potentials. ................................................. xlv Unit 1: Organic compounds and macromolecules.................. 1 1.1 Organic compounds..................................................................... 1 1.2 Physical properties and structure............................................. 15 1.3 Physical properties of organic compounds. ............................. 18 1.4 Factors that have an influence on the physical properties of organic compounds. .............................................................. 19 1.5 Solids, liquids and gases........................................................... 24 1.6 Chemical properties of organic compounds. ........................... 25 1.7 Reactions of organic compounds. ............................................ 29 1.8 Reactions of different homologous series................................ 31 1.9 Creating one hydrocarbon from another. ................................. 37 1.10 Plastics and polymers................................................................ 40 1.11 Plastics and pollution................................................................. 46 Unit 2: Rate and extent of reactions........................................49 2.1 Summary. ................................................................................... 49 2.2 Energy changes during chemical reactions. ............................ 49 2.3 Activation energy and the activated complex........................... 50 2.4 Catalysts. .................................................................................... 50 2.5 Energy changes during chemical reactions: ∆H. ..................... 50 2.6 Endothermic and exothermic reactions.................................... 51 2.7 Rates of reactions...................................................................... 54 2.8 Factors which affect reaction rate. ........................................... 55 © Department of Basic Education 2015 CONTENTS v Mind the Gap Chemistry

© Department of Basic Education 2015 2.9 The Collision Theory................................................................... 55 2.10 The mechanism of reactions..................................................... 56 2.11 Measuring rates of reactions. ................................................... 59 Unit 3: Chemical equilibrium....................................................78 3.1 Summary. ................................................................................... 78 3.2 Key concepts. ............................................................................. 78 3.3 Factors that influence chemical equilibrium position.............. 82 3.4 Le Châtelier’s Principle.............................................................. 82 3.5 The Equilibrium Constant (Kc) (The Law of Mass Action). ....... 90 3.6 Interpretation of graphs for chemical systems in dynamic equilibrium................................................................................103 3.7 Applications of equilibrium principles in the chemical industry..................................................................................... 117 Unit 4: Acids and bases...........................................................122 4.1 Summary. .................................................................................122 4.2 Properties of acids and bases................................................. 124 4.3 Common acids..........................................................................125 4.4 Common bases. .......................................................................126 4.5 Mono- and polyprotic acids. .................................................... 127 4.6 Conjugate acid-base pairs....................................................... 127 4.7 Ampholyte (amphiprotic) substance. ...................................... 131 4.8 Salt hydrolysis...........................................................................132 4.9 Acid-base indicators.................................................................134 4.10 Acid-base titrations (volumetric analysis)............................... 137 4.11 Preparing a standard solution................................................. 137 4.12 Dilution of solutions.................................................................139 4.13 Acid-base titration calculations...............................................140 4.14 The Equilibrium Constant (Kc) (The Law of Mass Action). .....144 4.15 Ka and Kb values.......................................................................145 4.16 The relationship between Ka and Kb for a substance............146 4.17 Auto-ionisation of water........................................................... 147 4.18 Equilibrium constant for water (Kw)......................................... 147 4.19 The pH scale............................................................................. 147 4.20 pH Calculations for strong acids and bases...........................149 4.21 Summary: Strong and weak acids and bases........................150 4.22 Summary: Concentrated and dilute acids and bases. .......... 151 Unit 5: Electrochemistry..........................................................155 5.1 Summary. .................................................................................155 5.2 Key concepts: Definitions and terminology. ...........................155 5.3 Electrochemical cells. .............................................................. 157 5.4 Electrolytic cells........................................................................159 5.5 Application of electrolysis........................................................ 161 5.6 Voltaic (Galvanic) cells.............................................................169 5.7 Cell notation. ............................................................................ 171 5.8 Standard electrode potentials................................................. 173 5.9 The standard hydrogen electrode. .......................................... 178 5.10 The emf of an electrochemical cell.........................................180 vi CONTENTS  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 Unit 6: The chlor-alkali industry. ............................................193 6.1 Summary. .................................................................................193 6.2 Definitions and terminology.....................................................194 6.3 The chlor-alkali industry – reactants and products. ..............195 6.4 The chlor-alkali industry – industrial process. .......................196 Unit 7: The fertiliser industry. .................................................202 7.1 Nutrients required by plants....................................................202 7.2 Functions and sources of the primary mineral nutrients. .....203 7.3 The industrial production of fertilisers....................................206 7.4 Flowchart: The industrial production of fertilisers. ................ 207 7.5 N:P:K fertilisers. ....................................................................... 210 7.6 Excessive use of fertiliser and the environment.................... 213 7.7 Alternatives to inorganic fertilisers. ........................................ 215 © Department of Basic Education 2015 CONTENTS vii Mind the Gap Chemistry

© Department of Basic Education 2015 We are confident that this Dear Grade 12 learner Mind the Gap study This Mind the Gap study guide helps you to prepare for the end-of-year guide can help you to prepare well so that you CAPS Grade 12 exam. pass the end-of-year The study guide does NOT cover the entire curriculum, but it does focus on exams. core content of each knowledge area and points out where you can earn easy marks. You must work your way through this study guide to improve your understanding, identify your areas of weakness and correct your own mistakes. To ensure a good pass, you should also cover the remaining sections of the curriculum using other textbooks and your class notes. Overview of the Grade 12 exam The following topics make up each of the TWO exam papers that you write at the end of the year: Cognitive level Description Paper 2 (Chemistry) 1 Remembering/Recall 15% 2 Understanding/Comprehension 40% 3 Applying and analysing 35% 4 Evaluating and creating (synthesis) 10% Paper Type of questions Duration Total Date Marking Chemistry 10 multiple-choice 2 questions – 20 marks 3 hours 150 October/November External Structured questions – 130 marks Paper 2: Chemistry Focus Content Marks Total Duration Weighting of cognitive levels Chemical change 84 150 3 hours 15 40 35 10 marks viii introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 How to use this study guide Look out for these icons in the This study guide covers selected parts of the different topics of the CAPS study guide. Grade 12 curriculum in the order they are usually taught during the year. The selected parts of each topic are presented in the following way: • An explanation of terms and concepts; NB PAY SPECIAL ATTENTION NB • Worked examples to explain and demonstrate; PAY SPECIAL ATTENTION NB NB • Activities with questions for you to answer; and • Answers for you to use to check your own work.PAY SPECIAL ATTENTION NB NB HINT! HINT! hint hint Hints to help you remember a concept NB PAY SPECIAL ATTENTION Pay special attention NBhint HINT! orEG - guide you in e.g. EG - worked examples Worked examples e.g. worked examples solving problems exams HINT! Step-by-step e.g. exams EG - worked examples Refers you to the Activities with hint questions for you instructions exemplar paper to answer exams e.g. EG - worked examples activity • The activities are based on exam-type questions. activity Cover the answers Use this study ACTIVITIES boy provided and do each activity on your own. Then check your answers. and girl-left and ACTIVITIES boy guide as a workbook.right of page exams Make notes, draw pictures, and girl-left and Reward yourself for things you get right. If you get any right incorrect of page activity answers, make sure you understand where activity you went wrong before and highlight important activity concepts. moving on to the next section. ACTIVITIES boy and girl-left and right of page • In these introduction pages, we will go through the mathematics that youactivity need to know, in particular, algebra activity and graphs. These are Step by step comment comment comment crucial skills that you willACTIVITIES need boy for any subject that makes use of Step by step comment mathematics. Make sureright of you page understand these pages before you go and girl-left and any further. activity Step by step comment comment • Go to www.education.gov.za to download past exam papers for you to practice. Step by step comment comment © Department of Basic Education 2015 INTRODUCTION ix Mind the Gap Chemistry

© Department of Basic Education 2015 Top 10 study tips 1. Have all your materials ready before you begin studying – pencils, pens, highlighters, paper, etc. Try these study tips to make learning easier. 2. Be positive. Make sure your brain holds on to the information you are learning by reminding yourself how important it is to remember the work and get the marks. 3. Take a walk outside. A change of scenery will stimulate your learning. You’ll be surprised at how much more you take in after being outside in the fresh air. 4. Break up your learning sections into manageable parts. Trying to learn too much at one time will only result in a tired, unfocused and anxious brain. 5. Keep your study sessions short but effective and reward yourself with short, constructive breaks. 6. Teach your concepts to anyone who will listen. It might feel strange at first, but it is definitely worth reading your revision notes aloud. 7. Your brain learns well with colours and pictures. Try to use them whenever you can. 8. Be confident with the learning areas you know well and focus your brain energy on the sections that you find more difficult to take in. 9. Repetition is the key to retaining information you have to learn. Keep going – don’t give up! 10. Sleeping at least 8 hours every night, eating properly and drinking plenty of water are all important things you need to do for your brain. Studying for exams is like strenuous exercise, so you must be physically prepared. If you can’t explain it simply, you don’t understand it well enough. Albert Einstein x introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 Mnemonics A mnemonic code is a useful technique for learning information that is difficult to remember. Here’s the most useful mnemonic for Mathematics, Mathematical Literacy, and Physical Science: BODMAS: B – Brackets O – Of or Orders: powers, roots, etc. D – Division Mnemonics encode information and make it easier to remember M– Multiplication A – Addition S – Subtraction Throughout the book you will be given other mnemonics to help you remember information. The more creative you are and the more you link your ‘codes’ to familiar things, the more helpful your mnemonics will be. Education helps one cease being intimidated by strange situations. Maya Angelou © Department of Basic Education 2015 INTRODUCTION xi Mind the Gap Chemistry

© Department of Basic Education 2015 Mind maps There are several mind maps included in the Mind the Gaps guides, summarising some of the sections. Mind maps work because they show information that we have to learn in the same way that our brains ‘see’ information. As you study the mind maps in the guide, add pictures to each of the branches to help you remember the content. You can make your own mind maps as you finish each section. How to make your own mind maps: 1. Turn your paper sideways so your brain has space to spread out in all directions. 2. Decide on a name for your mind map that summarises the information you are going to put on it. Mind mapping 3. Write the name in the middle and draw a circle, bubble or picture your notes makes around it. them more interesting 4. Write only key words on your branches, not whole sentences. and easier to remember. Keep it short and simple. 5. Each branch should show a different idea. Use a different colour for each idea. Connect the information that belongs together. This will help build your understanding of the learning areas. 6. Have fun adding pictures wherever you can. It does not matter if you can’t draw well. xii introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 On the day of the exam 1. Make sure you have all the necessary stationery for your exam, i.e. pens, pencils, eraser, protractor, compass, calculator (with new batteries). Make sure you bring your ID document and examination admission letter. 2. Arrive on time, at least one hour before the start of the exam. 3. Go to the toilet before entering the exam room. You don’t want to waste valuable time going to the toilet during the exam. 4. Use the 10 minutes reading time to read the instructions carefully. This helps to ‘open’ the information in your brain. Start with the question you think is the easiest to get the flow going. 5. Break the questions down to make sure you understand what is being asked. If you don’t answer the question properly you won’t get any marks for it. Look for the key words in the question to know how to answer it. Lists of difficult words (vocabulary) is given a bit later on in this introduction. 6. Try all the questions. Each question has some easy marks in it so make sure that you do all the questions in the exam. 7. Never panic, even if the question seems difficult at first. It will be linked with something you have covered. Find the connection. GOOD LUCK! 8. Manage your time properly. Don’t waste time on questions you are unsure of. Move on and come back if time allows. Do the questions that you know the answers for, first. 9. Write big and bold and clearly. You will get more marks if the marker can read your answer clearly. 10. Check weighting – how many marks have been allocated for your answer? Take note of the ticks in this study guide as examples of marks allocated. Do not give more or less information than is required. © Department of Basic Education 2015 INTRODUCTION xiii Mind the Gap Chemistry

© Department of Basic Education 2015 Question words to help you answer questions It is important to look for the question words (the words that tell you what to do) to correctly understand what the examiner is asking. Use the words in the table below as a guide when answering questions. Question word What is required of you Analyse Separate, examine and interpret Calculate This means a numerical answer is required – in general, you should show your working, especially where two or more steps are involved Classify Group things based on common characteristics Compare Point out or show both similarities and differences between things, concepts or phenomena Define Give a clear meaning Describe State in words (using diagrams where appropriate) the main points of a structure/process/phenomenon/ investigation Determine To calculate something, or to discover the answer by examining evidence Differentiate Use differences to qualify categories Discuss Consider all information and reach a conclusion Explain Make clear; interpret and spell out Identify Name the essential characteristics PAY SPECIAL ATTENTION In every exam Label Identify on a diagram or drawing question, put a CIRCLE around the question word and List Write a list of items, with no additional detail underline any other important Mention Refer to relevant points key words. These words tell you exactly what is Name Give the name (proper noun) of something being asked. State Write down information without discussion Suggest Offer an explanation or a solution Tabulate Draw a table and indicate the answers as direct pairs xiv introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 Vocabulary approximate (v. & adj.). Come close to (v); roughly, almost, not perfectly accurate, close but not exact. The The following vocabulary consists of all the verb is pronounced “approxi-mayt” difficult words used in Mind the Gap Mathematics, and the adjective is pronounced Mathematical Literacy, and Physical Science. We “approxi-mitt”. suggest that you read over the list below a few aquatic (adj). Growing or living near or in times and make sure that you understand each water. term. Tick next to each term once you understand arbitrary (adj). Based on random choice; it so you can see easily where the gaps are in your unrestrained and autocratic. knowledge. KEY C Abbreviation Meaning (v) verb: doing-word or action word, category (n). Class or group of things. such as “walk” cause (v). Make something happen. (n) noun: naming word, such as cause (n). The person or thing that “person” makes something happen; an aim (adj) adjective: describing word such as or movement to which a person is “big” committed. (adv) adverb: describing word for verbs, causality (n). Someone or something such as “fast” responsible for a result. (prep) preposition: a word describing a collide (v). To crash into; to hit. position, such as “on”, “at” complex (adj). Consisting of many different (sing) singular: one of parts; not easy to understand (n) a group or system of things (pl) plural: more than one of connected in a complicated way. (abbr) abbreviation component (n). A part. General terms compose (v). To make up from parts. composite (n). Something made up of parts; (adj) made up of several parts. Term Meaning condition (n). The state something is in; the A situation that must exist before something else is possible. abbreviate (v). Make shorter. conjunction (n). When two or more things abundant (adj). Having plenty of something. come together at the same point; accept (v). Agree to receive something; in grammar, a part of speech say yes to; believe; tolerate that connects words, sentences, something. phrases or clauses, e.g.: “and” account for (v). Explain why. consider (v). Think about. adjacent (adj). Next to something. contrast (v). Show the difference between; (n). something that is very affect (v). Make a difference to; touch different from what it is being the feelings of. Do not confuse compared with. with effect. See effect. conversely (adv). The opposite of. analyse (v). Examine something in detail. counteract (v). Act against something in order ante- (prep). Before (e.g., ante-natal – to stop it. before birth) D anti- (prep). Against (e.g., anti- apartheid – against apartheid). apparent (adj). Clearly visible; the way data (pl), (n). Information given or found. something seems to be or the way datum (sing) it appears. deduce (v). To work something out by appear (v). Come into sight; seem to be. reasoning. © Department of Basic Education 2015 INTRODUCTION xv Mind the Gap Chemistry

© Department of Basic Education 2015 deduction (n). Conclusion or idea that find (v). Discover or locate. someone has worked out. find (n). Results of a search or define (v). Give the meaning of a word or discovery. words. finding (n). Information discovered as the definition (n). The meaning of a word or result of an inquiry. words. fixed (adj). Not able to move, attached; deliver (v). To bring and hand over. or repaired, not broken. denote (v). To refer to or mean something. format (n). Layout or pattern; the way determine (v). Work out, usually by something is laid out. experiment or calculation. discreet (adj). Careful, polite. G discrete (adj). Single, separate, distinct, a part. global (adj). Found all over the world (globe). E H effect (n). Result. effect (v). Carry out, do, enact. hazard (n). Something dangerous. eject (v). Force or throw something heterogeneous (adj). Made up of many different or someone out violently or parts. suddenly. homogeneous (adj). Uniform, made up of the elapse (v). Pass by or finish, e.g., time. same types of parts. establish (v). Show or prove, set up or hypothesis (n). A theory or proposed create. explanation. exceed (v). Go beyond. hypothetical (adj). Theoretical or tentative; excess (n). More than necessary. waiting for further evidence. excluding (prep). Not including. exclusive (adj). Excluding or not admitting I other things; reserved for one particular group or person. identify (v). Recognise or point out. exemplar (n). A good or typical example. illustrate (v). Give an example to show what exempt (v). To free from a duty. is meant; draw. exempt (adj). Be freed from a duty. impair (v). Weaken or damage. exemption (n). Being freed from an imply (v). Suggest without directly saying obligation. what is meant. exhibit (v). To show or display. indicate (v). Point out or show. exhibit (n). A part of an exhibition. initial (n). First. expel (v). Force someone or something initiation (n). The action of beginning to leave a place. Eject. something; the action of admitting extent (n). The area covered by somebody into a group or something. organisation. insufficient (adj). Not enough. F interchange- (adj). Can be swapped or able exchanged for each other. investigate (v). Carry out research or a study. factor (n). A circumstance, fact or influence that contributes to a issues (v). Comes out of. result; a component or part. issues (n). An important problem or a factory (n). A place where goods are topic for debate. made or put together from parts. xvi introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 M Q macroscopic (adj). Visible without being made quality (n). The standard of something bigger. compared to other similar things; manipulate (v). Handle or control (a thing or a a characteristic of someone or person). something. microscopic (adj). Very small, not visible R without being made bigger. motivate (v). Give someone a reason for reciprocal (adj). Given or done in return. doing something. record (v). Make a note of something multiple (adj). Many. in order to refer to it later (pronounced ree-cord). N record (n). A note made in order to refer to it later; evidence of something; a copy of something (pronounced negligible (adj). Small and insignificant; rec-cord. can be ignored. From “neglect” relative (adj). Considered in relation to (ignore). something else; compared to. numerical (adj). Relating to or expressed as relative (n). A family member. a number or numbers. represent (v). Be appointed to act or speak numerous (adj). Many. for someone; amount to. resolve (v). Finalise something or make O it clear; bring something to a conclusion. observe (v). Look at; watch carefully. respect (v). Admire something or someone; consider the obtain (v). Get. needs or feelings of another occur (v). Happen. person. operate (v). Work; drive; control. respectively (adj). In regards to each other, optimal (adj). Best; most favourable. in relation to items listed in the optimum (adj). Best; (n) the most same order. favourable situation for growth or success. S overabundance (n). More than enough; too much. simultaneously (adv). At the same time. P site (n). Place. suffice (v). Be enough. phenomenon (n). A fact or situation that is seen surplus (adj). More than is needed. to exist or happen. survey (n). A general view, examination, phenomena (n). Plural of phenomenon. or description of someone or prefix (n). Part of a word that is attached something. to the beginning of many different survey (v). Look closely at or examine; words, changing their meaning, consider a wide range of opinions e.g., prehistoric – before written or options. records were kept. prepare (v). Make ready before an event; T set things up. principal (n). Head of a school. tendency (n). An inclination to do something principal (adj). Main or most important. in a particular way; a habit. principle (n). A basic truth that guides the tertiary (adj). Third level. way a person behaves. provide (v). Make available for use; supply. © Department of Basic Education 2015 INTRODUCTION xvii Mind the Gap Chemistry

© Department of Basic Education 2015 U affinity (n). A liking for something; an attraction to something; a tendency to react with something uniform (n). Standardised clothing. (chemistry). See also paraffin. uniform (adj). Remaining all the same at alcohol (n). In common usage, ethanol all times; unchanging. C2H5OH. Technically, any organic substance or molecule containing V an -OH group. See organic. aldehyde (n). Any organic molecule containing -CHO, formed by oxidising alcohols. verify (v). Show to be true; check for See alcohol and organic. truth; confirm. algae (n). Adjective: algal (pertaining to vice versa (adv). The other way round. algae); an aquatic plant (lives in versus (prep). Against. Abbreviated “vs” water), which lacks leaves, stems, and sometimes “v”. roots. algebra (n). A mathematical system Technical terms where unknown quantities are represented by letters, which A can be used to perform complex calculations through certain rules. alkali (n). See base. absorption (n). To take into; the process of taking something in. alkane (n). An organic molecule or compound or substance which account (n. & v.). Finance: A record of contains only single bonds income and expenditure. To between carbons. See organic. explain (v), e.g. “Account for why the sky is blue”. alkene (n). An unsaturated organic molecule, compound or substance, acetic (adj). Pertaining to vinegar; an which contains at least one double organic molecule containing two bond between carbons. See carbons. See organic, eth- for organic and unsaturated. more. alkyl (n). A prefix (word part) which shows acetone (n). Propanone. CH3COCH3; the that the word after it has an alkane ketone of acetic acid. See ketone group attached to it, by removing for more. one hydrogen from the alkane. acetylene (n). Ethyne, C2H2. Used in welding alkyne (n). An unsaturated organic torches (blowtorches). See also molecule, compound or substance, alkene. which contains at least one triple acid (n). A proton donor or substance bond between carbons. See that ionises into H+ or H3O+ when organic and unsaturated. dissolved in water; sour-tasting amalgam (n). General use: a mixture. In substance; corrosive; pH below 7. chemistry, specifically a mercury See also base and alkali. alloy. acidified (adj). To have been made acidic. amide (n). An organic compound acidity (n). How acid something is. containing the group -C(O)NH2; an activated (adj). Made to function. Chemistry: inorganic compound containing something moved into an unstable the group NH2–. higher-energy level or state. amine (n). The same as an amide except Usually “activated complex”, the the -NH2 can be attached to combination of reactants just anything, and does not have the before they turn into products. See CO group shown above. also reactant, reagent, product. ammonia (n). NH3. activation (n). The process of activating ammonium (adj). NH4+, found as a cation or as something. “Activation energy”, part of a salt. See salt and cation. the required energy to create an activated complex. amphiprotic (n). See ampholyte. aerosol (n). A solution of substances in air or ampholyte (n). A substance that can act as an other gas, e.g. as in an aerosol can. acid or base. See acid, base. xviii introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 anions (n). A negative ion. See cation, ion. axis (sing), (n). A line along which points can be anode (n). The negative electrode of axes (pl, plotted (placed), showing how far a cell or current supplier; the pronounced they are from a central point, called positive electrode of an electrolytic “akseez”) the origin. See origin. “Vertical axis” system; attracts negative ions. See or “y-axis” refers to how high up a electrode, cathode. point is above the origin (or how far below). “Horizontal axis” or “x-axis” antacid (n). A substance used to neutralise refers to how far left or right a point (react with) acid. E.g. chalk. See is away from the origin. acid, neutralise. apparatus (n). Equipment; parts of a scientific experiment. B aqueous (n). Dissolved in water. Arrhenius (n). Arrhenius’ theory of acids and bakelite (n). A type of hard, brittle plastic bases: That acids produce H+ or that can’t melt once it has set H3O+ in water, and bases produce or taken shape (thermosetting), OH–. See Brønsted-Lowry. made from phenol C6H5OH, and formaldehyde (methanal), CH2O. asbestos (n). A fireproof fibrous substance containing silicon used for balance (v. & n.). To make two things equal fireproofing. (v); a scale to weigh objects (n). Chemistry: to compare two sides asbestosis (n). A lung disease caused by of a chemical equation and make inhaling asbestos fibres, can lead sure that there are the same to cancer. numbers of atoms on both sides. atm (n). Abbreviation: atmospheres base (n). A proton acceptor, or substance of pressure (1 atm = 101,3 kPa). that ionises into OH– when The pressure of the air at sea dissolved in water; a bitter-tasting level. Same as “bar” (barometric substance, corrosive, pH above 7. pressure). See also acid. Do not confuse with ATM (n). Abbreviation: automatic teller common everyday use, meaning machine. “the bottom” or “low”. atmosphere (n). The air or the gases basic (n). Bitter or made of a base. Do not surrounding a planet; the sky; as a confuse with popular use, meaning unit of measurement, see atm. “low” or “simple” or “crude”. atmospheric (adj). To do with the atmosphere. battery (n). A collection of cells connected atom (n). The smallest unit of a chemical in series (end-to-end). See cell. In element, which, if broken down common use, “battery” is used to further, no longer behaves in the mean the same as “cell” (e.g. a same way chemically. Consists of penlight or AA cell), but this use is a nucleus or centre part which is incorrect except for a car battery, positively charged, and an electron which consists of a series of cells. cloud (negatively charged) which benzoic (adj). Contains benzene or a surrounds the nucleus. See nuclear. benzene ring. attract (v). To bring something closer. bi- (prefix). Two. average (n). Mathematics: The sum of parts bicarbonate (n). Any salt containing the ion divided by the quantity of parts. In -HCO3–. So called because the common use: neither very good, carbonate (CO32–) attaches to strong, etc., but also neither very another ion and the hydrogen weak, bad, etc; the middle. In (two bonds). The name “hydrogen Physical Science and Mathematics: carbonate” is now preferred. if you are asked to find the average, biodegradable (adj). Can be broken down by you always have to calculate it natural processes e.g. involving using the information you have. For bacteria, moulds, fungus, etc. example, the average of (1;2;3) is 2, because (1+2+3)/3 = 2. See also biodiesel (n). Diesel (a type of petrol) made mean, median and mode. from plants rather than fossil fuels (coal, oil). avogadro (n). 6,023 × 1023 particles; one (constant or mole. See mole and mol. number) © Department of Basic Education 2015 INTRODUCTION xix Mind the Gap Chemistry

© Department of Basic Education 2015 BODMAS (abbr). Brackets, of/orders carbohydrate (n). Organic compounds containing (powers, squares, etc), division, carbon and hydrogen, occurring multiplication, addition, subtraction. in foods and living tissues and A mnemonic (reminder) of the including sugars, starch, and correct order in which to do cellulose. They contain hydrogen mathematical operations. and oxygen in the same ratio boil (v). Physics: to cause a liquid’s as water (2:1). Not the same vapour pressure to exceed as hydrocarbons, which are any the pressure of the gas in the substances containing mostly container, usually by heating it, hydrogen and carbon. but it can be done by lowering carbonate (n). -CO32– the pressure of the gas in the carbonic (adj). Anything containing carbon, container, too. See vapour or more specifically, CO2 pressure. In common usage, to make a liquid hot until it bubbles. carbonyl (adj). Containing double bonded bond (n). A connection. In physics and carbon and oxygen: =C=O. chemistry, between atoms and carboxyl (adj). Containing -COOH. molecules. carboxylic (adj). Carboxyl-containing. breadth (n). How wide something is. From Cartesian (adj). Anything believed or the word “broad”. proposed by Rene Descartes. brine (n). A saturated salt solution (a In particular, the x-and-y axis mixture of water and salt which coordinate system. can’t dissolve any more salt). catalyst (n). A substance that alters the bromide (n). Something containing bromine, rate of a chemical reaction without usually one ion. See ion. itself being consumed in the bromo- (prefix). Something containing reaction. Without qualification, or bromine. as “positive catalyst”, something bromothymol (n). A type of acid-base indicator that starts or speeds up a reaction. (blue) used to tell whether something A “negative” catalyst slows down a is an acid or base. Turns blue (in reaction. base) or yellow (in acid). catalytic (n). Containing or using a catalyst. Brønsted-Lowry (n). A theory of acids and bases cathode (n). The positive electrode of a cell which says that acids are proton or current supplier; the negative donors (they give away protons), electrode of an electrolytic and bases are proton acceptors system; attracts positive ions. See (they take protons). Since H+ is electrode, anode, ion. just a proton, this does not mean something different from the cathodic (adj). Involving a cathode. Cathodic Arrhenius theory that an acid is a protection: To use a more reactive substance that dissolves into H+ in metal to protect a less reactive water. See proton, Arrhenius. metal from oxidation. See anode, but- (prefix). Four carbons. Pronounced cathode, oxidise. “beaut”. E.g. butane is a four- cation (n). A positively charged ion. See carbon alkane. anion, ion. caustic (adj). Basic; a base. C cell (n). An apparatus that generates electricity using electrochemistry. calibrate (v). To adjust a measuring tool An AA or Penlight battery, as it is or measurement against a commonly called, is a cell. A car known accurate measurement to battery consists of a number of ensure that the measuring tool or cells inside a single container. measurement is accurate; to check CFC (n). A chlorofluorocarbon. A a measurement or measuring substance containing carbon, tool’s accuracy; to mark with chlorine and fluorine. Responsible accurate measurements using a for breaking down ozone (O3) which standard scale like cm, mm, mℓ, protects us from too much UV etc. Common use: to assess or radiation from the sun. evaluate carefully. xx introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 chain (n). Chemistry: a long series of compound (n). A substance made up of atoms bonded together, usually molecules consisting of more carbon. than one different type of atom, charge (n). Chemistry: having too chemically bonded in a constant many or too few electrons ratio. E.g. Water (H2O) is a (most commonly), resulting in compound, but Sulphur powder a substance ionising. A positive (S) is not. In a compound, the charge results from too few original chemicals (reactants) have electrons, and a negative charge reacted or merged to form a new from too many electrons. Physics: substance. Compare to mixture. a basic feature of all physical compressed (adj). Subjected to pressure, electromagnetic particles, except, squashed. e.g. neutrons and photons, which concentration (n). The number of moles of have zero charge. All protons have substance per unit volume. a positive charge, all electrons See mol, moles. How “strong” a have a negative charge. solution is. See solution. chart (v). To draw a diagram comparing condensation (n). When a vapour or gas cools values on Cartesian axes. down and starts to collect into Le Châtelier’s (n). That in reversible reactions, larger droplets; changing phase Principle chemical systems will favour the from vapour or gas to liquid. forward or reverse reaction to Condensation reaction: to produce minimise the change imposed a larger molecule from two smaller on the system. If a chemical ones. equilibrium is disturbed by conditions (n). Physics and Chemistry: how changing the conditions, the (STP) the environment is: temperature position of equilibrium moves to and pressure. STP (Standard counteract the change. Temperature and Pressure is 25°C chloro- (prefix). Containing chlorine. and 1 atm). chlorofluoro- See CFC. conjugate (n). To join together. Chemistry: two carbons things that belong together, e.g. chloroform (n). CHCℓ2. A liquid formerly used conjugate acid-base pairs. as an anaesthetic. conservation (n). A law which describes chlorophyll (n). A green substance found something that does not change. in plants which enables E.g. the conservation of matter- photosynthesis (broadly, energy says that matter-energy generating food from CO2). See cannot be created or destroyed, photosynthesis. only transformed from one form into another. There are a number coefficient (n). A constant value placed of other conservations, e.g. next to an algebraic symbol as a momentum and torque. multiplier. Same as constant (see below). Or: a multiplier or factor constant (n). See coefficient. Means that measures a property, e.g. “unchanging”. coefficient of friction. contaminate (v). Chemistry: to introduce combustion (n). The process of burning, usually impurities or other substances in oxygen. Rapid oxidation. which are not meant to be part of a reaction. completion (n). Chemistry: when a reaction no longer proceeds (continues) control (n. and v.). To ensure something because it has run out of one does not change without or more of the reactants. See being allowed to do so (v); reactant, reaction. an experimental situation to which nothing is done, in order complex (n). See activated. to compare to a separate (activated) experimental situation, called the ‘experiment’, in which a change is attempted. The control is then compared to the experiment to see if a change happened. © Department of Basic Education 2015 INTRODUCTION xxi Mind the Gap Chemistry

© Department of Basic Education 2015 control variable (n). A variable that is held constant degradable (adj). Capable of breaking down or in order to discover the relationship being broken down. between two other variables. dehydrating (adj). To remove water from. A “Control variable” must not be (agent) “dehydrating agent” is a substance confused with “controlled variable” which can remove water from (see independent variable). another substance. E.g. H2SO4, coordinate (n). The x or y location of a point on ethanol. a Cartesian graph, given as an x or dehydration (n). The process of removing water y value. Coordinates (pl) are given from a substance. as an ordered pair (x, y). dehydrogena- (n). The process of removing correlate (v). To see or observe a tion hydrogen from a substance. relationship between two things, without showing that one causes dehydrohalo- (n). To remove hydrogen and a the other. genation halogen from a substance. See halogen. correlation (n). That there is a relationship between two things, without denominator (n). See divisor. In popular speech: showing that one causes the other. a common factor. correspond (v). To pair things off in a depend (v). To be controlled or determined correlational relationship. For two by something; to require something things to agree or match. E.g. A to happen or exist first. corresponds to 1, B corresponds to dependent (adj/n). A variable whose value 2, C corresponds to 3, etc. (variable) depends on another; the thing corrode (v). Chemistry: to destroy by that comes out of an experiment, gradual chemical action. Usually the effect; the results. See also refers to acidic action. Compare independent variable and control to erode. General use: to destroy variable. The dependent variable gradually. has values that depend on the independent variable, and we plot corroding (adj). A process or substance it on the vertical axis. that corrodes; to be subject to corrosion. depleted (adj). Having been used up; run out of. corrosion (n). To corrode. deposit (n). Finance: to place money into corrosive (adj). To be capable of corroding an account. Physical Science: to something. cover a surface of one substance counteract (v). Oppose or resist. with another substance, e.g. metal covalent (adj). Chemistry: a bond which plating on an electrode. results from sharing electrons determine(s) (v). To cause; to ensure that; to set between atoms. Compare ionic (causation) up so that; to find out the cause of. bond. di- (prefix). Two. cracking (n). A process of breaking a diamine (n). A substance containing two complex organic molecule into amine groups. See amine. simpler parts using heat and pressure. diammonium (n). Having two ammonium (NH4) groups. See ammonium. cubed (adj). The power of three; diaphragm (n). A thin sheet of any substance multiplied by itself three times. covering a gap. Biology: the muscle cubic (adj). Shaped like a cube; having below the lungs which moves to been multiplied by itself three cause breathing. Electrochemistry: times. a thin sheet inside a cell which current (n). Flowing electrons. separates the electrodes. It is porous and allows solutions containing ions through, but serves D to separate gas products. difference (n). Mathematics: subtraction. decompose (v). To break down into Informally: a dissimilarity. How components. things are not the same. xxii introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 dilute (adj. & v). To lower the domain (n). The possible range of x-values concentration of a solution. See for a graph of a function. See solution and concentration (v): range. a solution which has had its durable (adj). Tough; something that can concentration lowered (adj). endure. dilution (n). The opposite of concentration; dynamic (adj). Changing often. Relating how low a concentration is, to forces that produce motion. measured in mol/dm3; the process Opposite of static. See static and of diluting. electrostatic. dimer (n). A molecule made of two identical parts. See also polymer. E diode (n). A semiconductor device with two terminals (electrodes), usually allowing current to flow in one ecosystem (n). An integrated, complex, direction only. interacting, mutually dependent diol (n). A molecule with two alcohol/ living system or environment. hydroxyl (OH) groups. electric (adj). Containing electricity dipole (n). A polarised molecule with a (electrons). distinctly positive and distinctly electro- (adj). Where chemical reactions negatively-charged end. chemical cause the release of electrons, diprotic (n). Having two protons. usually into a circuit. General use: anything relating to electrical and displace (v). To move or relocate something. chemical phenomena. dissipate (v). To disperse or scatter (e.g. electrode (n). General use: the point where gas). Thermodynamics (Energy): electrons enter or exit a power to cause energy to be lost as heat. source or a circuit. Specifically Popular use: to disappear. (Electrochemistry): Part of a circuit dissociate (v). To break apart; to no longer be dipped into a solution to receive or associated with. release electrons. See anode and dissolve (v). To break up into ions within a cathode. solution (usually water); to mix a electrolysis (n). The splitting of a chemical solid (usually powder) into a liquid, into ions. (The chemical is usually to form a solution. See solution, dissolved in water or another ion. Alternative popular use: to solution.) It is done by means of bring to an end. electricity. See electrochemical. distil(l) (v). To purify through repeated electrolyte (n). A substance (usually liquid heating of a liquid and collection or gel solution) which contains of condensation. The heating a compound that will be split by process causes the liquid to form electricity. Ionisable solutions or gas or vapour, which condenses components. on the side of the heating vessel electromotive (adj). Usually electromotive force (container) or an exit tube, leaving or emf. The potential difference impurities behind in the heating caused by electromagnetism, vessel. which causes current to flow. distribution (n). How something is spread Producing a current with out. Mathematics: the range and electromagnetism. See emf. variety of numbers as shown on a electron (n). A fundamental physical graph. particle bearing a negative charge, disturb (v). Chemistry: to mix or stir a liquid weighing approximately 9 × 10−28g, or solution; to shake it up. which is found around atomic divisor (n). The number below the line nuclei in areas called ‘orbitals’. in a fraction; the number that is Responsible for electricity and dividing the other number above chemical reactions. Symbol e–. See the fraction line. See numerator, proton, nucleus. denominator. © Department of Basic Education 2015 INTRODUCTION xxiii Mind the Gap Chemistry

© Department of Basic Education 2015 electroplate (v). To cover a surface of a less energy (n). Work or the ability to do valuable substance with a more work. There are various forms of valuable metal, using electrolysis. energy: motion (Ek), light energy element (n). Mathematics: part of a set (photons), electrical energy, heat, of numbers. Physics: a pure etc. Energy can neither be created substance made only of atoms of nor destroyed, but only converted one type, with the same number from one form to another. See of protons in each nucleus. An conservation. element cannot be broken down enthalpy (n). The total heat content of a further without losing its chemical system, H, including the chemical properties. Each element has a energy. unique atomic number which is the equilibria (pl), (n). The state of being in balance. number of protons in the nucleus. equilibrium Chemistry: when the forward See nucleus, atom, isotope. (sing) reaction rate is equal to the Popular use: part of. reverse reaction rate. See Le eliminate (v). To remove or get rid of. Châtelier’s Principle. Mathematics: to cancel a factor erode (v). To wear away by means of out of one side of an equation by friction (rubbing). dividing by that factor throughout, or by substituting in another ester (n). An organic compound formula or value that is equal. produced by bonding an alcohol Chemistry: to produce a smaller to a carboxylic (organic) acid, substance as a by-product by means of dehydration. See from reacting more complex carboxylic, organic, alcohol, substances, usually water or CO2; dehydration. Responsible for fruit in the sense of: to remove those flavours and many pleasant odours molecules from the reaction. (smells). emf (abbr). Same as electromotive esterfication (n). The production of esters. force. Always written in lowercase estimate (n., v.). To give an approximate (small letters). value close to an actual value; an emission (n). Something released, e.g. gas, imprecise calculation. light, heat. eth- (prefix). Containing two carbons. emit (v). To release. eutrophication (n). Excess nutrients in water empirical (adj). Relating to the senses or to causing excessive plant growth and things that you can see, touch, strangulation of a waterway. taste, etc. Chemistry: empirical evaporate (v). To change phase of matter formula: a formula giving the from liquid to gas. Compare proportions of the elements sublimate and boil. present in a compound but not the excited (n). The state of being in a higher actual numbers or arrangement energy level (higher than ground of atoms; the lowest ratio of state). elements without giving structure exo- (prefix). Outside of. or quantities. exothermic (adj). Giving off heat, ΔH < 0. See emulsion (n). Small particles or droplets of enthalpy, endothermic. a substance or liquid which do not dissolve in a different liquid; exponent (n). When a number is raised to a suspended or floating within that power, i.e. multiplied by itself as liquid, e.g. to mix oil and water by many times as shown in the power shaking them up. (the small number up above the base number). So, 23 means endothermic (adj). Taking in heat, ΔH > 0. See 2 × 2 × 2. See also cubed. enthalpy, exothermic. exponential (adj). To multiply something many energetic (adj). Having a lot of energy; times; a curve representing an performing a lot of work. exponent. xxiv introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 extrapolation (n). To extend the line of a graph formic (prefix). Same as meth-. Contains further, into values not empirically one carbon. From Latin “formica”: documented, to project a future ant, referring to formic acid, which event or result. In plain language: is found in an ant’s sting. to say what is going to happen fraction (n). Mathematics: Not a whole based on past results which were number; a representation of a obtained (gotten) by experiment division. A part. E.g. the third and measurement. If you have fraction of two is 0,666 or ⅔. a graph and have documented meaning two divided into three certain results (e.g. change vs parts. Chemistry: a part of a time), and you draw the line further solution or mixture separated out in the same curve, to say what by distillation. See distil. future results you will get, that is called ‘extrapolation’. See predict. function (n). Mathematics: when two attributes or quantities correlate. extrude, (v., n.). To push something through If y changes as x changes, then extrusion a mould or shape, usually a liquid y = f(x). See correlate, graph, or gel through a hole, to create a Cartesian, axis, coordinate. Also: new shape. a relation with more than one variable (mathematics). Chemistry: F functional group: part of a molecule that gives the substance its chemical properties in common fahrenheit (n). A temperature scale based with other similar chemicals. on human body temperature. Water freezes at 32°F and boils at 212°F under standard G conditions. The conversion formula to centigrade/celcius is: galvanic (adj). Relating to currents caused (˚F – 32) × 5/9 = ˚C by a chemical reaction. See favoured (adj). Preferred. Chemistry: in a electrochemistry. chemical reaction, the direction of galvanising (n). To electroplate so as to protect, the reaction after the equilibrium e.g. cover iron with zinc to prevent is broken. rust. fermentation (n). The conversion of a sugar or gas (n). The third phase of matter. carbohydrate to an alcohol, usually When a solid is heated it turns into by yeast or bacteria. liquid, and when a liquid is heated fertilisation (n). The process of adding it turns into gas. nutrients to soil. Biology: the fusion gaseous (adj). In a gas form. of male and female gametes (sex gradient (n). A slope. An increase or cells) into a zygote (group of cells decrease in a property or that will become a foetus or other measurement. Also the rate of such living organism). a change. In the formula for a line fertiliser (n). A substance added to soil for graph, y = mx + c, m is the gradient. nutritional reasons. gradually (adv). To change or move slowly. fixation (n). The process of fixing on graph (n). A diagram representing something or someone. experimental or mathematical fixed (n). The process of attaching one values or results. See Cartesian. (chemistry) chemical or compound to another. graphic (n., adj.). A diagram or graph E.g. nitrogen fixing is the bonding (n). Popular use: vivid or clear or of free N2 to organic chemicals. remarkable (adj.). fluid (n). Any substance that can flow graphically (adv). Using a diagram or graph. and take the shape of a container; Popular use: to explain very clearly. liquid, some gels, and gas. groundwater (n). Water held in the earth fluorescent (n). Bright, colourful, due to (underground). changes in energy levels of electrons, as seen in fluorescent gypsum (n). Calcium sulphate. tubes, ink, etc. See reflective. © Department of Basic Education 2015 INTRODUCTION xxv Mind the Gap Chemistry

© Department of Basic Education 2015 H hydrate (n). To add water to. hydration (n). Having had water added. Haber (process) (n). An industrial process to produce Some salts are hydrated, meaning ammonia from nitrogen and that they have a number of water hydrogen, using an iron catalyst at molecules bonded to them, which high temperature and pressure. can be removed by heat. Heating a hydrated salt changes its colour haemoglobin (n). A compound containing but not the chemical reactions it iron, found in red blood cells, will undergo. See salt. responsible for carrying oxygen. hydrocarbon (n). Any compound consisting half-cell (n). One of the sides of an mainly of hydrogen and carbon. electrochemical cell; one of the Compare carbohydrate. electrodes and the chemical solution around it. hydrochloric (adj). Any chemical containing HCℓ. half-reaction (n). The equation for the chemical hydrohalogen- (n). Adding a hydrogen and halogen reaction occurring in a half-cell. ation atom to a molecule. halide (n). A compound containing a single hydrolysis (n). Splitting by reacting with halogen, e.g. NaCℓ. See halogen. water. Applies to salts and organic chemicals. E.g. A haloalkane plus halo- (prefix). Containing a halogen. See water or a dilute NaOH gives an halogen. alcohol and either a hydrohalide haloalkane (n). An alkane bonded to a or sodium salt. Dissolving a halogen. See halogen. salt in water can be considered halogen (n). Any of the elements fluorine hydrolysis. (F), chlorine (Cℓ), bromine (Br), hydronium (n). H3O+ ion. iodine (I), and astatine (At), in hydroxide (n). OH– ion. This usage applies group VIIA (17) of the periodic specifically to bases (see acid, table. They combine with metals to base). In organic molecules, OH– is produce salts. See salt. an alcohol functional group called halogenation (n). Adding a halogen. the hydroxyl group. See alcohol, hardness (n). (Water). Containing salts, usually hydroxyl, diol. calcium carbonate. If water contains hydroxyl (n). See hydroxide. too many such salts, soap does hydroxylamine (n). An amine with a hydroxyl not function properly and doesn’t group. produce bubbles or foam. “Water’s hardness is determined by the concentration of multivalent cations I in the water. Multivalent cations are cations (positively charged metal ideal (adj). Not as seen in real complexes) with a charge greater life; theoretical. Ideal gas: a than 1+. Usually, the cations have hypothetical gas whose molecules the charge of 2+. Common cations occupy negligible space and found in hard water include Ca2+ and Mg2+.” (Wikipedia). have no interactions, and which consequently obeys the gas laws heat (n). Physics: a measure of the (PV = nRT) exactly. average kinetic energy of the molecules or atoms in a substance; ignition (n). The start of a combustion enthalpy; the energy of an object reaction. Common use: to start as molecular motion. Alternatively, a car (which has an internal infra-red radiation (heat radiation) combustion engine). See engine, coming off a body. See body. combustion. homologous (n). Belonging to the same group illuminate (v). To explain or light up. of things; analogous. Biology: immerse (v). To cover in liquid. a flipper is homologous with a impair (v). Prevent; hinder; slow down. leg or arm. Chemistry (organic): impure (adj). Containing a variety of belonging to the same series of additional chemicals in smaller molecules, e.g. alkanes: methane, amounts in addition to the main ethane, propane; having the same chemical. functional group. xxvi introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 incandescent (adj). Giving off light as a result of irreversible (adj). Cannot be reversed. Said being heated. of certain chemical reactions, independent (n). The things that act as input in which case it specifically (variable) to the experiment, the potential means that the reaction does not causes. Also called the controlled spontaneously reverse (not an variable. The independent variable equilibrium reaction). The reaction is not changed by other factors, and only proceeds in one direction. we plot it on the horizontal axis. See Example: combustion. control, dependent variable. isolate (v). To separate from. Usual use indicator (n). Chemistry: a substance used in Physical Science means to to check for pH levels, which separate one chemical from changes colour according to pH. another. Compare insulate, distil. See acid, base, pH. isomer (n). A substance with the same indigo (n). The colour between violet and empirical formula but a different blue; purplish-blue. structural formula. See empirical, inert (adj). Chemistry: a chemical or structural. element which does not react or is isotope (n). An element which has a difficult to cause to react, e.g. Ne, different number of neutrons from Xe, He. N2 is sometimes described the usual number of neutrons in the as inert but it’s not in the group of element. E.g. 12C has 6 protons and Noble Gases. Common use: lazy, 6 neutrons, but 14C has 8 neutrons unwilling to move. and 6 protons, and is radioactive. inflammable (adj). Same as flammable; easily IUPAC (abbr). International Union of Pure set on fire (combustion). and Applied Chemistry. Standardised inhibitor (n). Something that slows down or naming conventions for chemicals. prevents. inorganic (adj). Not containing carbon; J mineral. Exceptions are C, CO, CO2, which, whilst they contain carbon, are not considered organic as they joule (n). Unit of energy. can be produced during inorganic chemical reactions. K insoluble (adj). Not able to dissolve. insufficient (adj). Not enough. kelvin (n). Unit of temperature, with interact (v). To affect each other, to be absolute zero being the point where directly involved with or act on no molecular motion occurs, at each other. -273,15˚C. Hence, the freezing point intermediate (adj). A state in between. of water is 273,15 K. Note that there intermolecular (adj). Between molecules. See is no degree sign before K. molecule, intramolecular. ketone (n). An organic compound with the intramolecular (adj). Within or inside a molecule. carbonyl group = C = O. Made by See molecule, intermolecular. oxidising secondary alcohols. E.g. acetone. inversion (n). Chemistry: turning something upside down. ion (n). An atom or molecule or part of L a molecule which has an electrical charge due to gaining or losing one law (n). In Physical Science, a or more electrons. formula or statement, deduced ionic (bond) (adj.). A bond in which electrons (discovered) from observation have been transferred from one (watching). The formula or side of the molecule to another statement will then predict that resulting in a cation and anion, under the same conditions the which then attract. E.g. NaCℓ. same thing will always happen. E.g. ionisation (n). The process of ionising. See the first law of thermodynamics ionise. says that matter and energy ionise (v). To turn into an ion. See ion. cannot be destroyed, but only changed from one form to another. © Department of Basic Education 2015 INTRODUCTION xxvii Mind the Gap Chemistry

© Department of Basic Education 2015 leach (v). When a substance drains out minimum (n). The smallest amount possible. of soil, e.g. into rivers. Similar to mixture (n). When you mix or combine “leak”. substances without them litmus (n). A type of acid/base indicator. undergoing a chemical reaction. In See indicator. It is red when other words, the substances mixed exposed to acid and blue when stay separate (chemically) and do exposed to a base. not bond. Different to compound. See compound, reaction. M modal (adj). Pertaining to the mode, or method. Can mean: about the mathematical mode or about the macromolecule (n). A large molecule, usually a method used. See mode. polymer or protein. mode (n). Mathematics: the most macroscopic (adj). Large enough to be visible common number in a series of to the unaided human eye; big numbers. See also mean, median. enough to be seen. mol (abbr). Mole. magenta (n). A bright purple/pink colour. molar (adj). About a mole. See mole. manipulate (v). To change, or rearrange mole (n). A unit describing an amount something. Usually in Mathematics of substance. 6,023 × 1023 it means to rearrange a formula to molecules or atoms of the solve for (to get) an answer. substance. material (n). Any substance, not just cloth. E.g. 18 g of water is 1 mol of matter (n). Substance; stuff. Opposite of water (H = 1, O = 16, H2O = 18). vacuum (nothing). molecular (adj). About molecules. See mean (n). See average. molecule. median (n). Mathematics: the number in molecule (n). The smallest amount of a the middle of a range of numbers compound; a single particle written out in a line or sequence. composed of the elements that make up the compound. E.g. in metal (n). A substance which is malleable water, a single particle consisting (can be hammered flat), is ductile of two hydrogen atoms and one (can be drawn into a wire), which oxygen atom. conducts electricity and heat well and which is reflective (most light mono- (prefix). One. striking it is emitted again). Most monomer (n). Part of a macromolecule; the elements are metals except the simplest repeating unit. Monomers few on the right hand side of the bond to form polymers. See periodic table starting at Boron (B) polymer, isomer. and running diagonally down to monoprotic (adj). Having one proton. Astatine (At). meth- (prefix). Having one carbon. See formic. N methylated (adj). Having had a single carbon or methyl group added. neutral (adj). Chemistry: pH 7,0. Neither metric (adj). A measurement system, acid nor base. E.g. water. In using a base of 10 (i.e. all the common use: not biased. See bias. units are divisible by 10). The USA neutralise (v). To make something neutral; to uses something known as the complete an acid/base titration. Imperial system, which is not used neutron (n). A subatomic particle with in science. The Imperial system is no charge, mass approximately based on 12. Examples: 2,54 cm the same as a proton, found in the (metric) = 1 inch (imperial). nucleus of an atom. Symbol n0. 1 foot = 12 inches = approx. If there are too many protons in 30 cm; 1 metre = 100 cm. 1 Fl.Oz a nucleus, the substance will be (fluid ounce) = approx 30 mℓ. radioactive as it releases microscopic (adj). Too small to be seen by the alpha particles (helium nuclei, unaided human eye. 2p+ + 2n0). xxviii introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 nitrate (n). Containing NO3−. parallel (adj). Keeping an equal distance nitrite (n). Containing NO2−. along a length to another item (line, object, figure). Mathematics: nomenclature (n). A system of describing things; two lines running alongside each a naming system, designed to other which always keep an equal make a name unambiguous or distance between them. unique. particle (n). Any small part, e.g. a proton, nonmetal (n). Any of the elements that are an atom, a molecule. not metals, e.g. Boron (B), Silicon (Si), Sulphur (S), Oxygen (O), etc. pascal (n). The unit of pressure, abbreviated Pa, units: N/m2 nucleus (sing), (n).The centre of something nuclei (pl), (generally), specifically the centre pent- (prefix). Five. nuclear (adj) of an atom, consisting of at least per (prep). For every, in accordance one proton (hydrogen), or two with. Chemistry: the maximum protons and two neutrons (helium). amount of an element possible Plural nuclei is pronounced “noo- for the number of bonds available. klee-eye”. See e.g. peroxide. numerator (n). The opposite of a denominator; peroxide (n). H2O2. the number on top in a fraction. perpendicular (adj). Normal; at right angles to (90˚). O phase (n). Time, period; a state of matter (solid, liquid, gas); the relationship in time between the cycles of a -oate (suffix). An ester. system (such as an alternating odour (n). A smell. electric current or a light or sound optimal (adj). Best, most. wave) and either a fixed reference point or the states or cycles of organic (adj). Containing carbon, except C, another system with which it CO, CO2. may or may not be synchronised origin (n). Mathematics: the centre of (simultaneous). I.e. if two systems a Cartesian coordinate system. vibrate at the same time at the General use: the source of same rate, they’re “in phase”. anything, where it comes from. photosynthesis (n). The process of converting outlier (n). Statistics: a data point which CO2 into carbohydrates using lies well outside the range of atmospheric CO2, chlorophyll, and related or nearby data points. light. oxalic (acid) (n). Ethanedioic acid; chem. pi (n). π, the Greek letter p, the formula: (COOH)2. ratio of the circumference of a oxidation (n). Specifically, adding an oxygen circle to its diameter. A constant atom to a molecule, but more without units, value approximately general use: losing electrons from 3,14159. any substance in a redox reaction. plastic (n., adj.). An artificial substance See redox. made from hydrocarbon polymers oxide (n). A compound containing which is often flexible and able to an oxygen atom, especially if it be moulded and is often a poor previously did not contain one, e.g. electrical conductor (n); flexible iron (metal), vs iron oxide (rust). (adj). oxidise (v). To add an oxygen or remove plot (v). To place points on a Cartesian electrons from a substance. coordinate system; to draw a graph. P poly- (prefix). Many. polyester (n). A polymer made from esters. paraffin (n). Any waxy organic substance, polymer (n). A synthetic substance made previously the general name for from many monomers (repeating alkanes. Formula ranges from units). See monomer. C20H42 to C40H82. polymerisation (n). Making a polymer. polyprotic (adj). Having many protons. © Department of Basic Education 2015 INTRODUCTION xxix Mind the Gap Chemistry

© Department of Basic Education 2015 porous (adj). Having many holes that allow proton (n). The positively-charged particle fluids through. that forms the centre of an atomic positive (adj). Having many protons not nucleus, weighing 1 836 times as paired with electrons; a lack of much as an electron, but having electrons. the same and opposite charge. Symbol p+. See also nucleus, potential (n). Having the ability to do work, in neutron, electron. particular, Ep (potential energy, the tendency to fall or start moving, pump (n). A machine that uses energy as in a spring), or emf (voltage). to transfer a fluid from one General use: potential exists when place to another. In Biology one there is an energy difference finds cellular pumps, which are between two points, e.g. due to biological machines for transferring gravity or electrical charge. In the ions and nutrients. context of electricity, read it as pure (adj). Containing only the “voltage”. compound or element in question, precipitate (n). Chemistry: A product of a without any other compounds or reaction that cannot dissolve in the elements mixed in. See impure. solution and settles at the bottom purification (n). The process of removing of the reaction vessel (container). impurities. See distil. predict (v). General use: to foresee. Physical Science: to state what will Q happen, based on a law. See law. pressure (n). A continuous force exerted on an object over a certain area, in qualitative (adj). Relating to the quality pascals, Pa. N/m2. See pascal. or properties of something. A qualitative analysis looks at product (n). Chemistry: the substance or changes in properties like colour, compound made as a result of a that can’t be put into numbers. chemical reaction. See reaction. Often contrasted with quantitative. Mathematics: the result of multiplying two numbers. quantitative (adj). Relating to, or by comparison to, quantities. Often contrasted project (n. & v.). A project (n., pronounced with qualitative. A quantitative PRODJ-ekt) is a plan of action analysis is one in which you or long-term activity intended to compare numbers, values and produce something or reach a goal. measurements. To project (v., pronounced prodj- EKT), is to throw something, or to quantity (n). Amount; how much. guess or predict (a projection). To project a result means to predict a R result. See extrapolate. prop- (prefix). Three carbons. rancid (n). Having an unpleasant smell due Pronounced “prope” (rhymes with to having started to ferment or rot, “rope”). usually said of meat, oil or butter. proportion (n). To relate to something else random (n). Unpredictable, having no cause in a regular way, to be a part of or no known cause. Done without something in relation to its volume, planning. size, etc; to change as something else changes. See correlate and range (n). Mathematics: the set of values respectively. that can be supplied to a function. The set of possible y-values in a protein (n). A large, complex organic graph. See domain. molecule containing nitrogen, usually making up structural rate (n). How often per second (or per elements of living things (building any other time period). Physics: blocks of cells, antibodies, etc). number of events per second; see frequency. protolytic (adj). Capable of removing a proton; proton transfer. Compare acid. xxx introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 ratio (n). A fraction; how one number S relates to another number; exact proportion. If there are five women for every four men, the ratio of salt (n). In common usage, NaCℓ. women to men is 5:4, written Chemistry: any compound formed with a colon (:). This ratio can be from the reaction of an acid with represented as the fraction 5/4 or a base, with the hydrogen of the 1¼ or 1,25; or we can say that acid replaced by a metal or other there are 25% more women than cation. A non-metal ion bonded to men. a metal ion. react (v). Chemistry: when two or more saturated (adj). Organic chemistry: Having elements or compounds are no available bonds or only single brought into a mixture and form bonds. Common use: cannot take chemical bonds, creating new any more, usually said of a cloth compounds. and liquid. reactant (n). A chemical before it bonds with SI (abbr). Système International. The another chemical. See reagent. international system of metric units used by scientists. See reaction (n). Chemistry: The process metric, IUPAC. of reacting; a state in which chemicals react. See endothermic simplify (v). To make simpler. Mathematics: and exothermic. Physics (nuclear): to divide throughout by a common When a nucleus of an atom breaks factor (number or algebraic letter) down and subsequently releases that will make the equation easier energy and/or bonds with another to read and calculate. nucleus. In the first case, it is a slaked (lime) (n). Calcium Hydroxide, Ca(OH)2. fission (splitting) reaction, in the solubility (n). How easily something dissolves second case a fusion (joining) (mixes into a liquid). reaction. soluble (adj). See solubility. reactive (adj). Tending to react easily. solute (n). The substance that you place reactivity (n). How reactive a substance is in a liquid (the solvent) so as to (unreactive or reactive). dissolve it. E.g. salt. See solvent. reagent (n). A reactant when it is still in its solution (n). A mixture of a solute and a bottle or container, before being solvent. A liquid which has had mixed. something dissolved in it (mixed). redox (abbr). Chemistry: reduction- Mathematics: the step-by-step oxidation reaction; a chemical displaying of calculations to reaction in which one substance arrive at answers. Common use: is reduced (gains electrons), and the answer to a problem, in the another is oxidised (loses electrons). sense of dissolving (removing) a See reduce, oxidise, anode, cathode, problem. electrode, electrochemistry. solvent (n). The liquid that dissolves the reduce (v). To make smaller. Chemistry: to substance placed into it. E.g. water. gain electrons (negative charges). spectator (n). Chemistry: a compound or reflux (n., v.). A substance that flows back chemical which does not get into its container after coming out. involved in a chemical reaction. Chemistry: the process of boiling a spontaneous (adj). Randomly, without liquid so that any vapour is liquefied provocation or cause or prior and returned to the stock (source). planning. refraction (n). Bending light when it travels stable (adj). Chemistry and Nuclear from one medium (e.g. air) into Physics: not likely to break down or another medium (e.g. water or react further. glass). Changing the direction of standardised (adj). Chemistry: a solution of propagation of any wave as a result known concentration, e.g. 1 molar of its travelling at different speeds at (1 mol/dm3). different points along the wave front. See Huygens’ principle, diffraction. © Department of Basic Education 2015 INTRODUCTION xxxi Mind the Gap Chemistry

© Department of Basic Education 2015 steam (n). Water vapour, microscopic T droplets of water. Not a gas, a suspension of water droplets in air. See suspension, gas, liquid, phase, terminal (n). Final; end point. aerosol. termination (n). Coming to an end. STP (abbr). Standard temperature and tetra- (prefix). Four. pressure; 101,3 kPa and 25˚C. theory (n). A usually mathematical structural (adj.). Pertaining to structure; representation of an explanation (isomer) a series of molecules whose for something in the sciences, structures are different but their which does not depend on the chemical or empirical formulae are thing being explained. A theory the same. differs from a law in that theories sublimate (v). To change phase of matter are prone to empirical (visible from solid straight to gas without or measurable) refutation the intermediate phase of liquid. (rejection); meaning that they See the case of dry ice (CO2). can be discarded if evidence comes in that they are wrong. subscript (n). A number placed below the See law. rest of the line, e.g. CO2. thermoplastic (n). Chemistry: will melt if heated. substance (n). Matter. Physical things. thermoset (n). Once set into a shape it cannot substituent (n). Chemistry: an ion or functional melt again. group or group of atoms that replaces a hydrogen on an organic thiosulphate (n). A salt containing the anion molecule. S2O32−. substitute (v). To replace. threshold (n). Physical Science: the magnitude or limit of something, substitutents (n). Something that gets replaced. which, if exceeded, will cause substitution (n). The process of substituting. something else, e.g. release of Mathematics: to replace an radiation, a chemical reaction, etc; algebraic symbol in a formula with the minimum amount of energy a known value or another formula, required to cause something. so as to simplify the calculation. Medicine: the maximum safety See simplify. Chemistry: to level of a dose. cause a substituent to bond to a titrate (v). To measure off a reagent substance. precisely drop by drop into a vessel superscript (n). A number placed above the (container) containing another rest of the line, e.g. πr2. reagent, so as to work out the synthesis (n). The process of concentration of the reagent in the manufacturing (making) vessel. something. Chemistry: to bond toxic, toxin (adj., n.). Poisonous, poison. smaller molecules together transfer (n). To move from one place to to create a larger molecule, another. Chemistry: usually refers e.g. methanol from CO and H2. to moving an electron from one Synthesis gas: a gas mixture compound to another. Finance: (e.g. CO, H2), which when heated usually refers to a payment produces a new compound, e.g. or credit. See credit, debit, methanol. transaction. synthetic (adj). Artificial, man-made. trends (n). Mathematics: regular patterns system (n). Any closely associated and within data. inter-related or inter-dependent tri- (prefix). Three. group of things; a set of things working together. Chemistry: a trial (n). Chemistry: to repeat an vessel (container) which contains a experiment, an iteration, chemical reaction. or particular attempt at an experiment. (From “try”, to try once). xxxii introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 triprotic (n). Having three protons. vessel (n). Any container. Common use: a turbidity (n). How muddy, muddled or container or ship. opaque or disturbed a liquid is. visible (adj). Able to be seen by the human eye, opposite of invisible. Compare microscopic, macroscopic. U viscosity (n). The thickness of a fluid. A viscous fluid flows slowly, e.g. unit (n). A subdivision of a scale. See syrup. Pronounced “viss-KOSS- scale. itee” and “viss-k’s”. universal (n). A chemical which can indicate volatility (n). How easily something (indicator) how acid or basic a solution is, evaporates. E.g. Ether (C2H5OC) is ranging from reds (acidic) to violets more volatile than water. (basic), including most of the volt (n). Unit of potential difference colour spectrum. in electricity. The difference of unsaturated (n). Organic chemistry: having potential (Ep) that would carry one double or triple bonds present. ampere of current against one unstable (n). Chemistry or Nuclear Physics: ohm resistance. Same as emf. See prone to disintegrating or reacting. emf, resistance, ohm, ampere. urea (n). CO(NH2)2. The substance used voltage (n). The measurement of volts. to remove excess nitrogen from voltaic (n). The production of electricity animals via urination. Useful as a in a cell. See battery, cell, fertiliser. electrochemistry, electrode, cathode, anode, galvanic. V volume (n). A measure of the space occupied by an object, equal to length x breadth x height. vapour (n). The pressure above a liquid pressure caused by molecules evaporating from the surface of its liquid form, Y when in phase equilibrium (i.e. as many molecules leaving the liquid yield (n., v.). The amount of substance surface are condensing back into produced in a chemical reaction the liquid). (n); products of a process (n); to variable (n., adj.). A letter used to represent hand over or give up (v). an unknown quantity in algebra (n); a quantity that changes (n); subject to change (adj). © Department of Basic Education 2015 INTRODUCTION xxxiii Mind the Gap Chemistry

© Department of Basic Education 2015 The maths you need This section gives you the basic mathematical skills that you need to pass any subject that makes use of mathematics. Whether you’re studying for Physical Science, Mathematics, or Mathematical Literacy, these basic skills are crucial. Do not go any further in this book until you have mastered this section. 1. Basic pointers • If a formula does not have a multiplication (×) sign or a dot-product (·), and yet two symbols are next to each other, it means “times”. So, m1m2 means mass 1 times mass 2. You can also write it as m1 × m2, or m1·m2 • Comma means the same as decimal point on your calculator (i.e. 4,5 = 4.5). Do not confuse the decimal point with dot product 1 (multiply): 4.5 = 4​ __ 2 ​but 4·5 = 20. Rather avoid using the dot product for this reason. • In science it is common to write divisors with an exponent. This means, for example, that 0,5 metres per second is usually written 0,5 m·s–1 rather than 0,5 m/s. Both notations are perfectly correct, however, and you may use either. It is important, however, that you either use –1 or / . If you just put 0,5 ms, that means 0,5 milliseconds, which is not a velocity (speed in a direction); it is a time. • A variable is something that varies (means: changes). So, for example, the weather is a variable in deciding whether to go to the shops. Variables in science and mathematics are represented with letters, sometimes called algebraic variables. The most common you see in maths is x, probably followed by y, z. In science, variables are given their letter symbols specifically depending on what they stand for; so, for example, M or m are used for mass (amount of substance in kilograms); v is used for velocity (speed in a certain direction); a is used for acceleration (change in velocity), etc. You can guess for the most part what a variable is for by what its letter is; so V is voltage, R is resistance, P is pressure, and so on. 2. Subject of formula or solving for Very often you have to “make something the subject of a formula” or “solve for something”. This refers to finding the value of an unknown quantity if you have been given other quantities and a formula that shows the relationship between them. xxxiv introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint © Department of Basic Education 2015 e.g. Worked example 1 EG - worked examples If John has 5 apples, and he gives some to Joanna, and he has two apples left, how many did he give to Joanna? Well, the formula would be something like this: exams 5–x=2 To solve for x, we simply have to swap the x and the 2. What we’re actually doing is adding “x” to both sides: 5–x+x=2+x this becomes: 5activity =2+x then we subtract 2 from both sides PAY SPECIAL to move ATTENTION the boy ACTIVITIES 2 over: NB 5 – 2 = 2 – 2 + x and girl-left and right of page NB 5–2=x 3activity =x ... so John gave Joanna three apples. The same proceduresHINT! apply no matter how complex the formula looks. Just either add, hint subtract, square, square root, multiply, or divide throughout to move the items around. Step by step comment comment e.g. Worked example 2 EG - worked examples Let’s take an actual example from Electricity: V = IR. This means, the voltage in a circuit examsis equal to the current in the circuit times the resistance. Suppose we know the voltage is 12 V, and the resistance is 3 Ω. What is the current? V = IR 12 = 3 × I activity divide throughout by 3 so as to isolate the I (3) 3 12 ​= ​ ​ _ ​ __ 3 ​  ​ I ACTIVITIES boy and girl-left and remember that anything divided by itself right is 1, so: of page ​  12 ​= (1) × I ... and __ __ ​  12 ​= 4 ... so 3 4 = I or activity 3 I = 4 A ... The circuit has a current of 4 amperes. It is possible to remember how to solve for these equations using a triangle mnemonic Step by step comment comment as follows: V If you’re solving for V, cover V with your hand. Then, I next to R means I times R, or IR. So, V = IR. If you’re solving for R, cover R with your hand. V is over I. So R = _​  V ​. While I this is an easier way to do it, remember that many formulas do not consist of only three I R parts, so it is better to know how to make something the subject of a formula, or solve for something. © Department of Basic Education 2015 INTRODUCTION xxxv Mind the Gap Chemistry

HINT! © Department of Basic Education 2015 hint e.g. Worked example 3 EG - worked examples Here’s a more tricky example. Given exams Kc = 4,5 [SO3] = 1,5 mol/dm3 [SO2] = 0,5 mol/dm3 (x – 48) [O2] = ​  ______​ mol/dm3 64 solve for x. activity _________ (1,5)2 [SO3]2 ​  2 ​ (0,5) (x - 48) Kc = ​ _______ ​ ________ ∴ 4,5 = ​  ACTIVITIES and 64 boy​ [SO2]2[O2] girl-left and right of page ∴ x = 176 g How did we activity get that answer? Step by step Step by step comment comment Let’s see how it works. First, solve for the exponents (powers): 2,25 4,5 = _________ ​  (x – 48) ​ (0,25) ​ ______​ 64 Now, we can see that 2,25 and 0,25 are similar numbers (multiples of five), so let’s divide them as shown. 4,5 = (2,25 / 0,25) × ( (x – 48) / 64 ) That leaves us with 4,5 = 9 × ( (x – 48) / 64 ) But if we’re dividing a divisor, that second divisor can come up to the top row. Here’s a simple example: 1 ÷ (2 ÷ 3) = ​ ___12 ​ ​  ​ 3 ×3 = ​  1____ ​ 2 = ​ _3 ​ = 1,5 2 If you doubt this, try it quickly on your calculator: 1 ÷ (2 ÷ 3) ... this means, one, divided by two-thirds. Well, two-thirds is 0,6667, which is almost one. So how many “two- thirds” do you need to really make up one? The answer is one and a half “two-thirds”... i.e. 0,6667 + (0,6667 ÷ 2) = 1. Hence the answer is 1,5. So, back to the original problem, we can bring the 64 up to the top line and multiply it by nine: 4,5 = 9 × ( (x – 48) / 64 ) 4,5 = ​  9 × _____64 ​ x – 48 4,5 = ​  576 ​ _____ x – 48 Now we can inverse the entire equation to get the x onto the top: 48 1 ​ = x – ​ ___ ​ ___ ​ 4,5 576 xxxvi introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 Now we multiply both sides by 576 to remove the 576 from the bottom row 576 ________ (x – 48) 576 ​ ___ ​ = ​  ​ 4,5 576 and we cancel the 576’s on the right hand side as shown above. Now, if 576 ÷ 4,5 = 128, then 128 = x – 48 Now we add 48 to both sides to move the 48 across 128 + 48 = x – 48 + 48 ... hence, 128 + 48 = x = 176. 3. Statistics Many experiments in science use statistics. You should therefore at least know the following: Dependent variable: The thing that comes out of an experiment, the effect; the results. Independent variable(s): The things that act as input to the experiment, the potential causes. Also called the controlled variable. Control variable: A variable that is held constant in order to discover the relationship between two other variables. “Control variable” must not be confused with “controlled variable”. It is important to understand that in science, correlation does not mean causation. That is, if two variables seem to relate to each other (they seem to co-relate), it doesn’t mean that one causes the other. A variable only causes another variable if one of the variables is a function f(x) of the other. We will see more about this when we look at graphs, below. Mean: The average. In the series 1, 3, 5, 7, 9, the mean is 1 + 3 + 5 + 7 + 9 divided by 5, since there are 5 bits of data. The mean in this case is 5. Median: The datum (single bit of data) in the precise middle of a range of data. In the series 1, 3, 5, 7, 9, the median value is 5. Mode: The most common piece of data. In the series 1, 1, 2, 2, 3, 3, 3, 4, 5, the mode is 3. Often in scientific formulas it is said that things are proportional to each other. However, we cannot calculate the value of a force or energy output or mass etc., if we only know what things are in proportion (i.e. which things correlate). Let’s take momentum for example. Momentum (how forcefully something moves, more or less), is proportional to velocity (speed in a direction). So the faster something’s moving, the more momentum it has. But p (momentum) can’t be calculated if we only know velocity; we need to know mass as well. Why? Because momentum is also proportional to mass; the more massive something is, the more momentum it has. Thus, to get rid of the proportionality sign (∝), we have to come up with a formula. Many experiments in science serve to find out what the relationship is between two variables, i.e. if they’re merely correlated — proportional — or if they’re causally related. In the case of momentum, it’s easy, because © Department of Basic Education 2015 INTRODUCTION xxxvii Mind the Gap Chemistry

© Department of Basic Education 2015 there are no further variables: p = mv. However, in the case of gravity or electric or magnetic field strengths, it’s not that easy. In those cases, we have to introduce something called a “constant”. A constant is a fixed value that is always multiplied into an equation. Constants are often written k. However, some specific constants, such as in the Law of Gravity, have their own symbol, in this case, G. These constants are given in the tables later in this book. 4. Graphs A lot of work in science involves interpreting graphs. You get graphs of motion, graphs of rates of chemical reactions, graphs of distance-relative strengths of force fields, and so on. Before you can understand these graphs, it’s probably PAY best SPECIALto start from scratch with Cartesian Coordinates. ATTENTION NB NB “Coordinates” are numbers that refer to the distance of a point along a line, or on a surface, or in space, from a central point called the “origin”. Graphs that you will use have only two dimensions (directions). The positions of pointsHINT! on these graphs are described using two coordinates: how far across (left-to-right) the point is, called the x-coordinate, and how far hint up-or-down on the page the point is, called the y-coordinate. e.g. Worked example 4 EG - worked examples Consider the following graph. It shows six points in a straight line. exams y 3 2 1 activity -x x –2 –1 ACTIVITIES boy1 2 3 and girl-left and right of page –1 activity –2 –3 -y The coordinates shown canstep Step by be described using comment what are called comment “ordered pairs”. For example, the furthest point in this graph is 3 units across on the “x-axis” or horizontal line. Likewise, it is also 3 units up on the y-axis, or vertical (up and down) line. So, its coordinates are (3;3). The point just below the midpoint or “origin”, is one unit down of the x-axis, and one unit left of the y-axis. So its coordinates are (-1;-1). Note that anything to the left or below of the origin (the circle in the middle), takes a minus sign. In most cases in science, you’ll only have graphs showing positive axes (plural of axis, pronounced aks-eez), since most graphs are of time. xxxviii introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 This series of dots look like they’re related to each other, because they’re falling on a straight line. If you see a result like this in an experimental situation, it usually means that you can predict what the next dot will be, namely, (4;4). This kind of prediction is called “extrapolation”. If you carry out the experiment, and find that the result is (4;4), and then (5;5), you’ve established that there is a strong relation or correlation. You can therefore start thinking about a formula to describe your findings. For example, this might be a graph showing a measurement of voltage (x) against a measurement of resistance (y). Now, another way of saying that x relates to y, or x is proportional to y, is to say that PAY SPECIAL ATTENTION y is aNB NB given above, voltage function of x. This is written y = f(x). So, in the example is a function of resistance. But how is y related to x in this graph? Well, it seems to be in a 1 to 1 ratio: y = x. So the formula for this graph is y = x. In this case, we’re only dealing with two factors; x and y. In other graphs you’ll find that sometimes more factors are involved, such as acceleration graphs, which have units of m/s2. Don’t worry HINT! about that; you treat them the same way (for example, m/s2 vs. time). hint e.g. Worked example 5 EG - worked examples Now, let’s take a slightly more complex case, illustrated next to this paragraph. In exams this graph, we see that wherever x is equal to something, y is one more. So, trace your finger from the bottom left dot upwards. It meets the x-axis at the point -3. Do the same for the same point towards the y-axis. You’ll see it meets the y-axis at -2. You’ll see the next coordinates are (–2;–1), then (–1;0), then (0;1), (1;2), and finally (2;3). From this we can see that whatever x is, y is one more. So, y = x + 1 is the formula for this line. activity y 3 ACTIVITIES boy and girl-left and right of page 2 activity 1 –xx –3 –2 –1 1 2 3 Step by step comment comment –1 –2 –3 –y © Department of Basic Education 2015 INTRODUCTION xxxix Mind the Gap Chemistry

HINT! © Department of Basic Education 2015 hint e.g. Worked example 6 EG - worked examples Let’s take another case. In this next case, we see the following values: where x has a certain exams value, y has double that value. Let’s tabulate it. y 3 2 1 activity –xx ACTIVITIES boy –3 –2 and girl-left and 1 2 3–1 right of page –1 activity –2 –3 –y Step by step comment comment x y 1,5 3 1 2 0,5 1 0 0 –0,5 –1 1 –2 –1,5 –3 So, when x is 1,5, y is 3, when x is 1, y is 2. Thus, the formula for this line is: y = 2x. This value next to x is called the “gradient” or “slope” of the line. The larger the value next to x is, i.e. the larger the gradient, the steeper the slope. The gradient is usually abbreviated as “m” when it is unknown. Now, how this applies to science is simple: if we are looking, for example, at a case of a graph of a chemical reaction, we will usually have the x-axis as time. And the y-axis will usually be the quantity (amount) of substances produced. So, if we have a graph of a chemical reaction with a large gradient, it means that the reaction is fast; a lot of substance (y) i s produced in a short time (x). If, for example, we heated the reaction, and saw that the gradient increased even more, that would show that the chemical reaction was sped up by heat, or, that reaction rate is proportional to heat. Likewise, if the gradient sloped downwards, it would show that the reaction slowed down over time, because y, the amount of substance produced, was decreasing, as x (time) increased, e.g. because the reactants were being used up. xl introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint © Department of Basic Education 2015 e.g. Worked example 7EG - worked examples Let’s do one more case. In this case, we can see that y is a function of x, since it’s a straight exams line graph. However, it’s not that easy to see the relationship between x and y. We can see that the slope is the same as the previous graph, so it must be something like y = 2x. However, it doesn’t quite make sense, since 2(–1,5) is not –2. We see that where x is zero (at the origin), y is at 1. But the slope is the same, so it must be y = 2(0) + 1. So the formula is: y = 2x + 1. y 3 activity ACTIVITIES2 boy and girl-left and right of page activity 1 –xx –3 –2 –1 1 2 3 Step by step comment –1 comment –2 –3 –y x y 2x + 1 –1,5 –2 2(–1,5)+1 = –3+1 = –2 –1 –1 2(–1)+1 = –2+1 = –1 As you can see, this is –0,5 0 2(–0,5) +1 = –1+1 = 0 where mathematicians get the general equation for a 0 1 2(0)+1 = 0+1 = 1 straight line of y = mx + c. (“c” stands for “constant”). 0,5 2 2(0,5)+1 = 1+1 = 2 1 3 2(1)+1 = 2+1 = 3 © Department of Basic Education 2015 INTRODUCTION xli Mind the Gap Chemistry

© Department of Basic Education 2015 Resource sheets The following information sheets will be supplied to you in the exam. You do not need to memorise them. SI Units: Multipliers Prefix Symbol Value Value written in full tera T 10 12 1 000 000 000 000 giga G 109 1 000 000 000 mega M 106 1 000 000 kilo k 10 3 1 000 hecto h 10 2 1 00 deka da 10 1 1 deci d 10 –1 0,1 centi c 10–2 0,01 milli m 10–3 0,00 1 micro µ 10 –6 0,00 000 1 nano n 10 –9 0,00 000 000 1 pico p 10 –12 0,00 000 000 000 1 femto f 10 –15 0,00 000 000 000 000 1 Constants Name SI Unit Symbol Approximate Value Easier to Understand STP (Standard not applicable, two 1 ATM (101,3 kPa), 25°C You generally put two Temperature and conditions (298 Kelvin (K)) ATM (bar) pressure in car Pressure), (in Physics). tyres, i.e. the pressure in a car tyre is twice atmospheric pressure Standard Conditions not applicable, three STP plus 1 mol/dm3 As above (Chemistry) conditions Gas constant R 8,3 J / mol⋅K The R in PV = nRT Molar gas volume at STP V0 22,4 dm3 / mol 22,4 Litres of gas is made by a mole of a substance Avogadro’s constant NA 6,022045 × 1023 6 022 000 000 000 000 units/mol 000 000 00 particles is one mole xlii introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 © Department of Basic Education 2015 INTRODUCTION xliii Mind the Gap Chemistry

© Department of Basic Education 2015 Formulas Moles, Gas Laws, Chemical Equilibria: R = gas constant n = number of moles m = mass T = temperature V = volume c = concentration, also [ ] m n m n = ​ __ M ​ c = ​ __ V ​ c = ​ ___ MV ​ p1V1 p2V2 PV = nRT ​ ____ ____ T ​= ​  T ​ 1 2 2P + 3R ⇋ 2S + 4T [S]2 × [T]4 Kc = ________ ​  2 [P] × [R]3 ​ Where [S] is the concentration of S in mol/dm3 Notes: • In most cases in chemistry, subscripts refer to how many atoms there are; e.g. H2O = two atoms of H & 1 atom of O. xliv introduction  © Department of Basic Education 2015  Mind the Gap Chemistry

© Department of Basic Education 2015 Standard reduction potentials Notes: • Eθ means the same as E0. • There are two versions of this table; they are identical except one is upside-down. Just memorise that Fluorine (F) has the greatest oxidising ability. • A strong reducing agent will displace a weaker reducing agent from its salt. • Always start with the oxidation half reaction. • A redox reaction will take place when a reducing agent reacts with an oxidising agent. • Balance the electron charge of each half- reaction by multiplying each with a suitable coefficient. • Add the two half-reactions together, eliminating electrons from both sides. • Eliminate common ions or molecules from both sides of the equation e.g. H+ and H2O. • You can then combine the E0 voltages to get the total voltage of a cell. Use the values exactly as they are; do not round off. E0cell = E0cathode – E0anode OR E0cell = E0oxidising agent – E0reducing agent • A positive answer means that the reaction will proceed spontaneously from left to right. A negative value means that it is not spontaneous. © Department of Basic Education 2015 INTRODUCTION xlv Mind the Gap Chemistry

© Department of Basic Education 2015

Unit 1 Organic compounds and macromolecules Organic chemistry is the chemistry of carbon compounds in living and non-living systems. Organic molecules: molecules containing carbon atoms Note: Carbon dioxide (CO2), carbon monoxide (CO), carbonates (CO32–) and cyanides (CN–) are exceptions because they are considered inorganic compounds. Organic compounds typically have a backbone of linked carbon atoms that other atoms attach to. 1.1 Organic compounds 1.1.1 Chemical bonding The organic compounds that we study consist of Carbon, Hydrogen, Oxygen atoms and the Halogens ( -Br, -Cl, -I ), which are often represented collectively as –X). PAY SPECIAL ATTENTION NB NB Carbon is always surrounded by EXACTLY four bonds HINT! hint Oxygen is always surrounded by EXACTLY two bonds Hydrogen is always attached to EXACTLY one bond e.g. EG - worked examples Halogen is always attached to EXACTLY one bond exams activity ACTIVITIES boy and girl-left and right of page activity © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 1  Mind the Gap ChemistryStep comment by step comment

1 Unit 1.1.2 The homologous series PAY SPECIAL ATTENTION NB NB DEFINITIONS A homologous series is a series of compounds (molecules) that has the same general HINT! formula and the same functional group. Each member of a homologous series differs hint from the previous member by a – CH2 group. Saturated: compounds in which there are no multiple bonds between carbon atoms in their hydrocarbon chains Unsaturated: compounds with one or more multiple bonds between carbon atoms in e.g. EG - worked examples their hydrocarbon chains. A functional group is a bond, an atom or a group of atoms that determine(s) the physical and chemical properties of a group of organic compounds. exams Classification of organic molecules according to homologous series Homologous General Functional Suffix Example name Structural Condensed Molecular series formula Group formula structural formula formula activity | | Alkanes CnH2n+2 ─c─c ACTIVITIES boy─ -ane propane CH3CH2CH3 C3H8 and girl-left and right of page| | Hydrocarbons activity Alkenes CnH2n -ene propene CH2=CHCH3 C3H6 Alkynes CnH2n–2 -yne propyne CH≡CCH3 C3H4 Step by step comment comment CnH2n+1X Haloalkanes/ -ane 2-bromopropane CH3CHBrCH3 C3H7Br alkyl halides (X = F, Cℓ, Br, I) Alcohols CnH2n+1OH -ol propan-2-ol CH3CHOHCH3 C3H7OH CnH2nO propanal Aldehydes -al CH3CH2CHO C3H6O n =1, 2, ... CnH2nO Ketones -one propanone CH3COCH3 C3H6O n = 3, 4, ... CnH2nO2 Carboxylic -oic acid propanoic acid CH3CH2COOH C3H6O2 acids n = 1, 2, ... CnH2nO2 ethyl Esters -oate methanoate CH3CH2OOCH C3H6O2 n = 2, 3, ... 2 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 Note: Alkanes have the general formula CnH2n+2 In propane n = 3 and therefore the general formula is C3H2(3)+2 = C3H8 In a structural formula all the bonds in the compound are shown. If any of these bonds are omitted, it is called a condensed structural formula. E.g. structural formula for propane possible condensed structural formula for propane 1.1.3 Isomers PAY SPECIAL ATTENTION In isomers, the same atoms are used in different compounds, but they NB are assembled or connected in different ways. There are different kinds of isomers in organic chemistry. Structural isomers refer to the order in which DEFINITIONS different atoms in a compound are combined. 1. Structural isomers are We will focus on three types of structural isomers, namely chain isomers, compounds with theHINT! positional isomers and functional isomers. hint same molecular formulae, but different structural formulae. 2. Chain isomers: EG - worked examples Structural formula: structure 1 Structural formula: structure 2 e.g. Molecules with the same molecular formula and in the same homologous series but different types exams Chain isomers of chains e.g. butane and methyl propane. IUPAC name: butane 3. Positional isomers: Molecules with the same Molecular formula: C4H10 IUPAC name: 1-methylpropane molecular formula and Molecular formula: C4H10 in the same homologous activity series but the position of the functional group ACTIVITI and girl- Structural formula Structural formula differs (e.g. butan-1-ol right of p and butan-2-ol) or the activity substituent or side chains differ. 4. Functional isomers: Positional isomers Molecules with the same molecular formula but Step by step com in different homologous name: butan-1-ol IUPAC name: butan-2-ol series or have different Molecular formula: C4H10O Molecular formula: C4H10O functional groups e.g. ketones and aldehydes. © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 3 Mind the Gap Chemistry

1 Unit AL ATTENTION Structural formula Structural formula NB The numbering of the carbon chain: Functional isomers In the naming of the organic compound, the longest carbon Homologous series: ketone Homologous series: ketone chain is numbered in such a way that the –OH IUPAC name: butanone IUPAC name: butanal functional group is attached ed examples to the carbon that is assigned Molecular formula: C 4 H8 O Molecular formula: C 4 H8 O the smallest number. Therefore, in this example, In this example both organic compounds consist of the same number the first organic compound is of carbon, hydrogen and oxygen atoms, but in the structure of the first called butan-1-ol. compound, the (=) is attached to the middle of the carbon chain, which makes it a ketone. In the second example, the (=O) is attached to the beginning of the carbon chain, which makes it an aldehyde. Activity 1 ACTIVITIES boy and girl-left and right of page For each of the following isomers: a) Write down the structural formula b) Identify the type of isomer 1. 2-methylpentane and 3-methylpentane. 2. 1-chlorobutane and 1-chloro-2-methylpropane p comment comment 3. Propanoic acid and ethyl methanoate  (13) [13] Solutions 1. a) 3 3 3 3 2-methylpentane 3-methylpentane 1. b) Positional isomers 3 2. a) 3 3 3 1-chlorobutane 1-chloro-2-methylpropane 4 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 2. b) Chain isomers 3 PAY SPECIAL ATTENTION NB 3. a) 3 3 HINT! 3 hint propanoic acid ethylmethanoate Structural isomers have different IUPAC names and 3 b) Functional isomers 3 [13] e.g. different physicalEGproperties. - worked examples 1.1.4 Nomenclature (naming conventions exams or rules) Each IUPAC name consists of three parts: PAY SPECIAL ATTENTION Prefix: Root: Suffix: NB Position and names Number of C-atoms in Determined by the of substituents the main C-chain. homologous series. activity Determine the root of the (side chains), listed name from the number of ACTIVITIES boy and girl-left and alphabetically. right of page C-atoms in the longest (main) HINT! Organic compounds are named according to the IUPAC (International hint C-chain. activity 1 = meth- Union of Pure and Applied Chemistry) system. 2 = eth- 3 = prop- General steps for the IUPAC 4 = but- 5 = pent- e.g. EG - worked examples Step by step comment 6 = hex- naming of all organic compounds 7 = hept- Step 1: exams 8 = oct- 9 = non- Identify the functional group in the compound and the homologous series Note: old names: 1 = form- ; it belongs to. This determines the suffix (ending). Example: Naming of 2 = acet- 2-methylprop-1-ene activity ACTIVIT and girl In this case there are only carbons and hydrogens, with a double bond right of between the carbons. This indicates that this compound is an alkene activity and will end with the suffix -ene Step 2: • Find the longest chain of carbon atoms. It must include the functional group, and need not be in a straight line. Step by step com • Number the carbon atoms in this chain from the side nearest to the functional group e.g. a double or triple bond, a hydroxyl group, a carbonyl group or a carboxyl group. • In the case of alkanes or haloalkanes, start numbering from the carbon nearest to a substituent (side chain) e.g. an alkyl group or halogen atom. © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 5 Mind the Gap Chemistry

1 Unit • Indicate the position of the functional group (except in the case of the alkanes). For alkenes and alkynes, give the smaller of the numbers of the C-atoms between which the double or triple bond exists. Example: Naming of 2-methylprop-1-ene Example: The longest chain that includes the functional group, is numbered in the example above. Since the double bond is to the right of this example, we number the longest chain from right to left. The longest chain with the functional group contains three carbons and we know it must be a propene. Since the double bond is attached to the first carbon, it must be a prop-1-ene. Step 3: • Determine whether there are any substituents. • Count the number of C-atoms in the substituent to determine the prefix and end it on –yl. • Write the number of the C-atom on the main chain where the alkyl group is attached, in front of the alkyl name. • Separate the number and the name with a hyphen, e.g. 2-methyl. • If a substituent occurs more than once, use the appropriate prefix: twice – di; three times – tri; four times – tetra e.g. 3,4-dimethyl • If there is more than one substituent, write their names and positions in alphabetical order. Example: The methyl group is attached to the second carbon, hence the name 2-methyl prop-1-ene Step 4: Naming the organic compound with substituent(s) in front of the parent chain. Example: the IUPAC name of the compound discussed above: 2-methylprop-1-ene 6 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 1 e.g. Worked example 1EG - worked examples Give the IUPAC name for the following organic compound: exams Step 1: Identify the functional group, and the homologous series to which this compound belongs. There are only C-atoms and H-atoms present, so this is a hydrocarbon. There are only single bonds between the C-atoms, therefore this is an alkane and the suffix will be –ane. activity ACTIVITIES boy and girl-left and right of page activity Step 2: Find the longest C-chain and countcomment Step by step the C-atoms in it, starting at the side closest to the functional group. This is comment an alkane, so we start numbering from the side closest to a substituent Step 3: Look at the substituents. Count the C-atoms in each substituent, determine the prefix and end it on –yl. The alkyl groups must be listed alphabetically, so we write the ethyl before the methyl PAY SPECIAL ATTENTION NB NB Final name•of Numbers compound: HINT! and letters of the alphabet are separated by a hyphen. hint 3-ethyl 2-methyl(e.g pentane 2-methylpropane) • Numbers are separated by the comma, and no space between the substituent(s) and the parent chain. ( e.g. 2,2-dichloropentane) e.g. EG - worked examples © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 7 exams Mind the Gap Chemistry

HINT! 1 Unit hint e.g. Worked example 2 EG - worked examples Give the IUPAC-name of the following haloalkane: exams activity ACTIVITIES boy and girl-left and right of page activity PAY SPECIAL ATTENTION NB NB Step by step comment comment HINT! hint e.g. Worked example 3 EG - worked examples Give the IUPAC-name of the following alkene: exams activity ACTIVITIES boy and girl-left and right of page activity Solution: The compound is therefore 2-methylbut-1-ene. Step by step comment comment 8 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 1 e.g. Worked example 4EG - worked examples exams activity ACTIVITIES boy and girl-left and right of page activity Solution: butan–1–ol Step by step comment comment Solution:: butan–2–ol © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 9 Mind the Gap Chemistry

HINT! 1 Unit hint e.g. Worked example 5 EG - worked examples Give the IUPAC name for the following carboxylic acids: exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment Solution: Ethanoic acid Solution: 2–methylpropanoic acid 10 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 1 e.g. Worked example 6EG - worked examples Give the IUPAC name for the following aldehyde: exams PAY SPECIAL ATTENTION NB NB activity ACTIVITIES boy and girl-left and HINT! right of page hint propanal Solution: activity e.g. Worked example 7EG - worked examples Step by step comment comment Give the IUPAC name for the following ketone exams activity ACTIVITIES boy Solution: pentan–2–one and girl-left and right of page activity Step by step comment comment © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 11 Mind the Gap Chemistry

HINT! 1 Unit hint e.g. Worked example 8 EG - worked examples Give the IUPAC name for the following ester: exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment Activity 2 Write down the IUPAC names for the following organic molecules. 1. CH3–CH2–CH2–CH2–C≡C–CH–CH3 | CH3 (1) 2. (1) 3. C3H7COC2H5 (1) 4. (1) 12 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 5. (1) 6. (1) 7. (1) [7] Solutions 1. 2-methyloct-3-yne 3 (1) 2. butan-2-ol 3 (1) 3. butanal 3 (1) 4. hexan-3-one 3 (1) 5. 3-methyl-butanoic acid 3 (1) 6. methyl butanoate 3 (1) 7. 1-bromo-2,2-dimethyl-propane 3 (1) [7] Important to remember: • When naming haloalkanes, the halogen atoms do not get preference over alkyl groups – numbering should start from the side nearest to the first substituent, either the alkyl group or the halogen. In haloalkanes, where e.g. a Br and a Cℓ have the same number when numbered from different sides of the chain, Br gets alphabetical preference. • When writing IUPAC names, substituents appear as prefixes written alphabetically (bromo, chloro, ethyl, methyl), but the prefixes di- and tri- should not be used to determine the alphabetical order. • In molecules where the functional group is ALWAYS on the first carbon (such as in the case of carboxylic acids, aldehydes there is NO number added to indicate the position of the fuctional group. (e.g. Ethanoic acid or pentanal.) © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 13 Mind the Gap Chemistry

1 Unit Activity 3 1. Which ONE of the following formulae represents an alkane? A C2H2 B C3H4 C C3H6 D C3H8(2) 2. An organic compound has the structural formula shown below: The correct systematic (IUPAC) name for the compound is … A but-1-ene. B but-2-ene. C methylpropene. D methylpropane. (2) 3. Which ONE of the following compounds can exist as an isomer? A A C D  (2) 4. An example of an unsaturated hydrocarbon is: A C2HCℓ3 B C3H6 C C2H6 D C2H5OH  (2) [8] Solutions 1. D33 (2) 2. C33 (2) 3. B33 (2) 4. B33 (2) [8] 14 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 Activity 4 Consider the following list of organic compounds: 1. Using structural formulae, write an equation for the preparation of an ester. Choose the reactants from the above list.  (5) 2. Write the letters representing TWO compounds in this list that are isomers.  (2) 2.1 C 2.2 D  (2) 3. Write down the letter that represents the compound that is formed when ethanol is oxided.  (1) 4. Give one use of esters.  (1) [11] Solutions 1. 3 3 33 3 2. A 3 & F 3 (5) 2.1 ethylmethanoate 3 (2) 2.2 but-2-ene 3 (2) 3. B 3 (1) 4. Manufacturing of perfumes/flavourants 3 (1) [11] 1.2 Physical properties and structure The structure of an organic compound — that includes the order in which atoms are connected in the compound, as well as the forces that is present between different molecules — has an influence on the physical properties of the compound. These physical properties include boiling points, melting points and the vapour pressure above a liquid organic compound. In the following section the influence of the structure of molecules on its physical properties will be discussed. © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 15 Mind the Gap Chemistry

1 Unit 1.2.1 Intermolecular forces in organic compounds Key for this section: Carbon Oxygen Hydrogen Intermolecular forces that act between DIFFERENT molecules of a compound are called Van der Waals forces. In the organic chemistry that we study, there are primarily two kinds of forces, namely Hydrogen bonds and Dipole-dipole forces. Induced dipole force: When two molecules come very close to each other, the electrons move and the distribution of electrons in the molecules temporarily shifts so that there is a slightly positive side and a slightly negative side on each molecule. The two molecules attract each other with a weak force. Note: The sketches that represent the forces between atoms relative to the size of the atoms are NOT drawn to scale. In reality, molecules are very close to each other and the intermolecular forces Dipole-dipole forces: When a molecule has a slightly positive side and act over small distances. another molecule has a slightly negative side, the two molecules will attract each other with a weak force. Intermolecular bonds form between molecules so that they form liquids or solids. Hydrogen bonds: A stronger type of dipole-dipole force. The definition of hydrogen bonding was discussed thoroughly in Grade 11, but for the sake of the organic structures studied in Grade 12 it is helpful to observe that hydrogen bonds are mostly found between different molecules where at least one molecule has an –OH group (hydroxide group). Hydrogen bonds are present between the -O atom connected to one molecule (which is slightly negative) and the -H atom of a different molecule (that is slightly positive). 16 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 PAY SPECIAL ATTENTION NB NB HINT! . hint e.g. Worked example 9EG - worked examples Identify the intermolecular forces found in propane. exams activity ACTIVITIES boy and girl-left and right of page Solution PAY SPECIAL ATTENTION NB NB activityforces found between the molecules of propane are The intermolecular induced dipole forces. The Van der Waals forces (induced dipole forces in this case) between the molecules are very weak, and are easily broken. This leads to the HINT! hint relativelySteplow alkanes’ boiling points. by step comment comment e.g. Worked example 10 EG - worked examples Identify the intermolecular forces found in water. exams H H O H O 0 H H hydrogen bonds H strongest intermolecular force Solution activity ACTIVITIES boy The intermolecular forces foundandbetween girl-left and the molecules of water are right of page hydrogen bonds. activityare strongly attracted to each other by the hydrogen The water molecules bonds. Water has a relatively high boiling point due to the presence of the hydrogen bonds. Step by step comment comment © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 17 Mind the Gap Chemistry

HINT! 1 Unit hint e.g. Worked example 11 EG - worked examples Identify the intermolecular forces found between the molecules of ethanol. exams strong hydrogen bonds H O H O H–C–H H–C–H H–C–H H–C–H H weak Van der Waals forces H activity ACTIVITIES boy Solution and girl-left and right of page There are hydrogen bonds between the hydroxyl groups (–OH) of PAY SPECIAL ATTENTION the NB activity different NB forces between the molecules. There are dipole-dipole hydrogen molecules connected to the different molecules. Both hydrogen bonds and dipole-dipole forces exist between alcohol molecules. The hydrogen bonds exist between the polar OH– groups and the dipole-dipole HINT! forces exist between the non-polar hydrocarbon Step by step comment comment IAL ATTENTION hint parts. NB DEFINITIONS Vapour pressure: The vapour e.g. Worked example 12 EG - worked examples pressure of a liquid is the equilibrium pressure of a Identify the intermolecular forces present in methanoic acid. vapour above its liquid; that exams is, the pressure of the vapour O H–O– resulting from evaporation of = ed examples a liquid above a sample of the H–C– C–H liquid in a closed container. O–H O= Boiling point: The temperature at which Solution the vapour pressure of a There areactivity hydrogen bonds between the molecules of methanoic acid. substance equals atmospheric ACTIVITIES boy pressure. The stronger the Intermolecular forces between and carboxylic girl-left andacids are stronger than those intermolecular forces, the in alcohol because the molecules rightare bonded by two hydrogen bonds. of page higher the boiling point. The boiling point of a substance activity is the temperature at which it changes from a liquid to a ACTIVITIES boy 1.3 Physical properties of organic gas and girl-left right of page andand is reached when the vapour pressure above the compounds comment Step by step comment liquid equals the atmospheric • Each chemical compound has physical properties such as its phase, pressure. melting point, boiling point, vapour pressure, viscosity, density and Melting point: The melting solubility. Note that the type of bonds affect all these properties. point of a substance is the • These properties are affected by the strength of the intermolecular temperature at which it forces between the molecules in the compounds. ep changescomment comment from a solid to a liquid. 18 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 When the intermolecular forces between the molecules of a substance are strong: PAY SPECIAL ATTENTION NB • a lot of energy is required to overcome these intermolecular forces of attraction; REMEMBER • the molecules aren’t easily separated from one another. In general: As the strength of the HINT! hint forces intermolecular 1.4 Factors that have an influence INCREASES: • Less vapour is produced on the physical properties of hence vapour pressure e.g. above the liquid EG - worked examples organic compounds DECREASES; • Melting and boiling points In the Grade 12 exam you will be asked to explain why the physical INCREASE. properties (melting point, boiling point, vapour pressure) of different exams organic compounds, differ. In order to determine why the physical properties differ, you will need to find answers to one or more of the following questions: 1. What is the influence of the type of intermolecular forces present between the molecules of the organic compound on its physical properties? activity ACTIVITIE and girl-l Little energy needed to break right of p intermolecular forces and let Low boiling point/ Weak intermolecular forces molecules move away from each other melting point activity (for example induced dipole, dipole-dipole forces) Little energy needed for molecules to break out of liquid and form High vapour vapour pressure Step by step com Lots of energy needed to break intermolecular forces and let High boiling point/ Strong intermolecular molecules move away from each melting point forces other (for example hydrogen bonds) Lots of energy needed for molecules to break out of liquid Low vapour and form vapour pressure 2. What is the influence of the type of functional group that is found in the molecule on its physical properties? In organic compounds where hydroxyl groups (-OH) are present (for example alcohols, carboxylic acids) you will find hydrogen bonds and induced dipole forces. In organic compound where hydroxyl groups are not present (for example alkanes, alkenes, alkynes, halo-alkanes, aldehydes, ketones and esters) you will find induced dipole forces or dipole-dipole forces. © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 19 Mind the Gap Chemistry

1 Unit Lower boiling/ Alkanes Less energy to break bonds melting point Alkenes Weak Alkynes intermolecular Haloalkanes forces Less energy for molecules to Higher vapour Esters evaporate pressure Ketones Aldehides Higher boiling/ Strong More energy to break bonds melting point intermolecular Alkohols forces Carboxylic acids More energy for molecules to Lower vapour evaporate pressure 3. What is the influence of the chain length of the molecule on its physical properties? As the length of the carbon chain in the organic compound increases, the number of sites where you will find the intermolecular forces will increase. Therefore, the longer the carbon chain, the stronger the intermolecular forces between the molecules. Less energy to Lower boiling/ break bonds melting point Weak Shorten carbon intermolecular chains forces Less energy for molecules to Higher vapour evaporate pressure More energy to Higher boiling/ break bonds melting point Strong Longer carbon intermolecular chains forces More energy for molecules to Lower vapour evaporate pressure 20 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 4. What is the influence of the molecular mass of an organic compound on its physical properties? Low boiling/ Small More energy needed to move melting point molecular molecules away from each other mass and break intermolecular forces Higher vapour presser Higher boiling/ melting point Greater More energy needed to move molecular molecules away from each other mass and break intermolecular forces Lower vapour presser 5. What will the influence on the physical properties of an organic compound be if the compound is made up of branched carbon- chains? As the number of branches (alkyl groups) increases, the shape of the molecule changes to a more compact, spherical shape with a smaller surface area, resulting in a smaller contact area between the molecules and weaker net Van der Waals forces. Lower boiling/ Less energy to melting point break bonds Molecules where Weak carbon chains are intermolecular branched forces Less energy for Higher vapour molecules to pressure evaporate More energy to Molecules with Higher boiling/ break bonds straight carbon Strong melting point chains intermolecular forces More energy for molecules to Lower vapour evaporate pressure © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 21 Mind the Gap Chemistry

1 Unit Activity 5 1.1 The boiling points of branched alkanes are lower than those of straight chain alkanes containing the same number of carbon atoms because branched alkane chains have A Large molecular masses B Longer chain lengths C More electrons D Smaller effective molecular surface areas (2) 1.2 Which ONE of these compounds has the highest vapour pressure at room temperature? A Propane B Ethane C Ethanol D Fluoroethane(2) 2. Will the boiling point of hexane be HIGHER THAN or LOWER THAN that of pentane? Refer to MOLECULAR STRUCTURE, INTERMOLECULAR FORCES and ENERGY needed to explain the answer.(4) 3. How will the boiling point of an ISOMER of Butane compare to that of Butane? Write down HIGHER THAN, LOWER THAN or EQUAL TO. Refer to MOLECULAR STRUCTURE, INTERMOLECULAR FORCES and the ENERGY needed to explain the answer. (4) 4. Consider the boiling points of compounds propan-1-ol (97˚C) and ethanoic acid (118˚C). 4.1 Give a reason for this difference in boiling points by referring to the intermolecular forces present in EACH of these compounds.(3) 4.2 Which ONE of the compounds propan-1-ol or ethanoic acid has a higher vapour pressure? Refer to their boiling points to give a reason for the answer. (3) [18] 22 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 Solutions 1.1 D 3 3 (2) 1.2 A 3 3 (2) 2 Higher 3 Structure: Hexane has a longer chain length than pentane. OR Hexane contain more C- atoms than pentane. OR Hexane has a greater molecular size than pentane. OR Hexane has a larger surface area than pentane. 3 Intermolecular forces: Between the different molecules of hexane you will find stronger or more intermolecular forces than between the different molecules of pentane. 3 Energy: More energy needed to overcome or break intermolecular forces between hexane molecules than you need between pentane molecules. 3(4) 3. Lower than 3 Isomers of Butane: More branching / Smaller surface area (over which the intermolecular forces act.) 3 Weaker/less intermolecular forces. 3 Less energy needed to overcome intermolecular forces. 3(4) 4.1 Ethanoic acid: Two sites for hydrogen bonding/forms dimers 3 Propan-1-ol: One site for hydrogen bonding. 3 Therefore it will require more energy to break the bonds between the ethanoic acid molecules than to break the bonds between the propan-1-ol molecules, 3 which explains why ethanoic acid has the lower boiling point. (3) 4.2 Propan-1-ol has the highest vapour pressure. 3Because propan-1-ol has a lower boiling point it means less energy is required to break the bond between the particles of propan-1-ol 3and therefore more molecules will be present in the vapour state and therefore vapour pressure will increase. 3(3) [18] © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 23 Mind the Gap Chemistry

1 Unit 1.5 Solids, liquids and gases Solids, liquids and gases are all phases of matter. SOLIDS LIQUIDS GASES Strong Weaker Very weak intermolecular intermolecular intermolecular forces. forces forces. Have a specific Particles are freer Particles move freely, shape. to move, liquid body of gas takes the takes the shape its shape its container. container. Alkanes of C18 and Alkane members The first four members bigger are waxy from C5 – C17 are of the alkanes C1 – C4 solids at STP liquids at STP (methane to butane) are gases at STP Alkenes of C15 and Alkene members The first three bigger are solids at from C5 – C15 are members of the STP liquids at STP alkenes C2 – C4 (ethene to butene) are gases at STP Haloalkanes are mostly liquids and solids at A few haloalkane STP. gases exist at room temperature. SOLIDS LIQUIDS The bigger molecules of alcohols, The smaller molecules of alcohols, aldehydes, ketones, ethers, aldehydes, ketones, ethers, carboxylic acids and esters are carboxylic acids and esters are solids, as a result of the increasing liquids at STP. strength of the Van der Waals forces. 24 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 1.6 Chemical properties of organic compounds Alkanes our most important fossil fuel and are good sources of heat and energy. 1.6.1 HYDROCARBONS ALKANES ALKENES ALKYNES CnH2n+2 CnH2n CnH2n−2 • Are saturated • Are unsaturated • Are highly hydrocarbons and hydrocarbons and unsaturated therefore relatively therefore reactive. hydrocarbons and unreactive. therefore very reactive. • Have low boiling • Ethene, C2H4, is the • Ethyne points and evaporate simplest alkene. easily. • Most are produced • Cause suffocation if by cracking of inhaled. alkanes. Cracking is • Have characteristic defined further on. odours. • Previously known as the ‘paraffins’ (meaning without affinity or reactivity). • Weak Van der Waals intermolecular forces • Non-polar • All are volatile, explosive and flammable and are used as fuels because their oxidation (with oxygen) is exothermic, and releases large amounts of energy. • Butane is used in • Ethene is used in • Ethyne used to be cigarette lighters. the manufacturing known as acetylene • ‘Handigas’ (in of plastic (polymer of and is used on a camping gas bottles) ethane). large scale as fuel is a mixture of for the oxyacetylene propane and butane. blowtorch used to cut and weld metals. • Petrol is a mixture When ethyne burns of different alkanes, in the presence mainly iso-octane of oxygen, a flame and heptane. temperature of 2 800 °C can be reached. © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 25 Mind the Gap Chemistry

1 Unit 1.6.2 HALOALKANES Are • alkanes with halogen (Br, Cℓor I) substituents; • saturated and therefore • relatively unreactive. Are • bonded by weak Van der Waals intermolecular forces • poisonous • mostly liquids at room temperature. The more halogen atoms on a haloalkane, • the more poisonous the substance and • the less flammable e.g. CH3Cℓ → CH2Cℓ2 → CHCℓ3 → CCℓ4 Tetrachloromethane CCℓ4 is Other haloalkanes are • a non-polar molecule; • very slightly polar and have • used as a non-polar solvent; • slightly stronger dipole-dipole Van • insoluble in water; der Waals intermolecular forces. • inflammable and has • very weak London Van der Waals intermolecular forces. • Tetrachloromethane CCℓ4 used to be used in dry cleaning but is toxic, also called “carbon tetrachloride”. • Trichloromethane (chloroform) CHCℓ3 used to be used as an anaesthetic • Tetrachloroethene (persolve), C2H4, is a haloalkane that is used in the dry cleaning industry. DID YOU KNOW? Chlorofluorocarbons (CFCs) like dichlorodifluoromethane (CCℓ2F2) were previously used for cooling and as propellants in aerosol cans, because of their low boiling point. CFCs contribute to the depletion of the ozone layer that protects us from the sun’s UV radiation; which has contributed to global warming; their use is therefore being phased out. Chlorotrifluoroethene Dichlorodifluoroethene By Jynto and Ben Mills [Public domain], via Wikimedia Commons 26 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 1.6.3 ALCOHOLS CnH2n+1OH Are • saturated and • have a hydroxyl functional group. Are • bonded by strong hydrogen bonds and by weak Van der Waals forces; • less polar as the number of C-atoms in the molecule increases; • generally soluble in polar and in non-polar solvents; • flammable; • mostly liquids at room temperature. If the carbon that is bonded to the hydroxyl group (-OH) is bonded to: ONE other C-atom TWO other C-atoms THREE other C-atoms it is a it is a it is a PRIMARY ALCOHOL SECONDARY ALCOHOL TERTIARY ALCOHOL CH3 propan-1-ol propan-2-ol methyl propan-2-ol • Methanol (CH3OH): • a poisonous liquid that can cause blindness; • used as a solvent and sold as methylated spirits. • Ethanol (CH3CH2OH) • a colourless liquid found in alcoholic beverages like beer and wine; • prepared through the fermentation of starches (beer) or sugars (wine): C6H12O6 2C2H5OH + 2CO2 1.6.4 KETONES 1.6.5 ALDEHYDES CnH2nO Are saturated molecules with Are saturated molecules with • a carbonyl functional group • a carbonyl functional group (–C–) • on a central C-atom. In these structures, you will often see an R attached to part of the • on the first C-atom. carbon chain. R refers to the rest of the molecule. It can be (but is not limited to) a chain of carbon atoms or substitutents. © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 27 Mind the Gap Chemistry

1 Unit Are Are • functional isomers of the • functional isomers of the ketones; aldehydes; • liquids at room temperature except • liquids at room temperature; for methanal which is gaseous; • abundant in nature and often have • responsible for pleasant smells pleasant scents (smell); (long C-chain aldehydes) and • used to produce perfumes. odours (unpleasant smells) in short chain aldehydes. Intermolecular forces: Intermolecular forces: • weak induced dipole forces and • weak induced dipole forces and • stronger dipole-dipole forces. • stronger dipole-dipole forces. Solubility in H2O: Solubility in H2O: • Short C-chain ketones form • Short C-chain aldehydes form hydrogen bonds with H2O hydrogen bonds with H2O molecules ∴ are soluble in H2O; molecules ∴ are soluble in H2O; • Long C-chain ketones are insoluble • Long C-chain aldehydes are in H2O. insoluble in H2O. Propanone is also known as acetone and is used as nail polish remover. 1.6.6 CARBOXYLIC ACIDS 1.6.7 ESTERS CnH2nO2 Are saturated molecules Are saturated molecules with with • an ester functional group. • a carboxyl functional group. Are Are • functional isomers of the esters; • functional isomers of the carboxylic • liquids at room temperature; acids; • weak acids; • liquids at room temperature; • very polar molecules. • produced when an alcohol reacts with a carboxylic acid; • responsible for pleasant smells (long C-chain esters); • volatile (evaporate easily). Intermolecular forces: Intermolecular forces: • weak induced dipole forces and • weak induced dipole forces and • hydrogen bonds. • stronger dipole-dipole forces but • Soluble in H2O due to hydrogen • NO hydrogen bonds. bonds that form between the • Oily and become more so as the carboxylic acid and H2O molecules. C-chain increases in length, due to stronger net intermolecular forces. • Only short C-chain esters are soluble in H2O. 28 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 Boiling points: Boiling points: • Similar to those of aldehydes and • Similar to those of aldehydes and ketones with similar molecular ketones with similar molecular mass. mass; • Lower than those of alcohols and carboxylic acids with similar molecular mass due to absence of hydrogen bonds. Usage and Occurrence: Used in the manufacture of: • Methanoic acid (CHOOH) is • perfumes (pleasant smell) • known as formic acid (Latin: • solvents Formica = ant); • artificial fruit flavours. E.g. butanoic • found in sting nettles and in the acid + ethanol catalysed by H2SO4 PAY SPECIAL ATTENTION poison of ants when they bite. yields ethyl butanoate, a.k.a apple NB • Ethanoic acid (CH3COOH), also flavour. known as acetic acid, is vinegar. VOCABULARY (Latin: Acetum = vinegar). • rancid: old, stale and with • Butanoic acid (CH3CH2CH2COOH) is an unpleasant smell HINT! • also known as butyric acid and • hint preservative: prevents • gives rancid butter its food from going off or unpleasant smell. decaying • Benzoic acid is used as a food • readily: easily preservative. e.g. EG - worked examples exams 1.7 Reactions of organic compounds 1.7.1 Types of chemical reactions Organic compounds take part in different types of reactions which can be grouped into: activity ACTIVITI • Oxidation and girl- • Substitution right of p • Addition and • Elimination reactions. activity 1.7.1.1 Oxidation reactions: Reaction with oxygen (O2) Step by step com Exothermic (release energy) Excess O2 alkane + O2(g) → CO2(g) + H2O(g) + energy 1.7.1.2 Substitution reactions: • Take place when saturated compounds react and an atom bonded to the carbon chain is substituted by another atom or another group of atoms. • Produce products which are also saturated. • Are slow. • Are not spontaneous – additional energy (e.g. sunlight hf or heat Δ) is needed for the reactions to take place. © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 29 Mind the Gap Chemistry

1 Unit SUBSTITUTION Reaction Reaction equations REACTIONS conditions Endothermic: Halogenation Sunlight or heat hf or ∆ alkane + halogen ____→ haloalkane + HX hf or ∆ Endothermic haloalkane + H2O → alcohol + HX VERY slow Hydrolysis Slow haloalkane + NaOH (diluted) → alcohol + NaX AL ATTENTION NB 1.7.1.3 Addition reactions: • Take place when atoms attach to the double or triple bond of an REMEMBER unsaturated compound (alkenes or alkynes), breaking the double or Markovnikov’s rule: triple bond during the reaction. Two products are formed • Form products which are more saturated than the reactants. during addition of water or of • Are faster than substitution reactions. HX to an alkene. • Are usually spontaneous. The major product is formed ed examples when the H-atom from the ADDITION Reaction added molecule bonds to Reaction equations REACTIONS conditions the C-atom which is already bonded to the most other Catalyst: Hydrogenation alkene + H2(g) Pt → alkane H-atoms. Pt, Pd or Ni The secondary product is formed when the H-atom Halogenation No catalyst alkene + X2 → haloalkane from the added molecule bonds to the C-atom which Hydrohalogenation No catalyst. alkene + HX → haloalkane is bonded to the least other H-atoms. Concentrated H2SO4 (conc) Hydration alkene + H2O alcohol H2SO4 ACTIVITIES boy and girl-left and right of page 1.7.1.4 Elimination reactions: • Occur when two atoms or groups of atoms are removed from adjacent carbon atoms in a saturated compound (like an alkane, a haloalkane or an alcohol) to form two compounds; • are always endothermic i.e. the reactants must be heated. p comment comment ELIMINATION Reaction Reaction equations REACTIONS conditions Catalyst (Pt), Pt∆ Cracking alkane alkene + alkane Heat (long chains form shorter structures) Catalyst (Pt), Pt∆ Dehydrogenation alkane alkene + H2 Heat Concentrated Dehydrohalogenation strong base haloalkane NaOH(c)∆ alkene + HX (NaOH), Heat Concentrated H2SO4(c)∆ Dehydration alcohol alkene + H2O H2SO4, Heat 30 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 1.8 Reactions of different homologous series 1.8.1 Reactions of the ALKANES As alkanes are saturated hydrocarbons, they are relatively unreactive. PAY SPECIAL ATTENTION NB Alkanes burn in excess oxygen to produce carbon dioxide and in limited oxygen, to produce carbon monoxide. In both cases water vapour is Zaitsev’s* rule: produced and both reactions are exothermic (∆H < 0). Alkanes are therefore sources of energy and can be used as fuel. If more than one elimination product is possible HINT! STEPS to balance reaction equations for the oxidation of alkanes, alkenes and • thehint major product is alkynes. formed when the H-atom 1. Write the standard reaction equation, with the correct formula for the alkane. is removed from the C-atom with the least NB alkane PAY + (excess) SPECIAL O2(g) → CO2(g) + H2O(g) + energy ATTENTION NB e.g. EG - worked examples H-atoms bonded to it and 2. Put a 2 in front of the alkane, alkene or alkyne. • the secondary product 3. Balance the C-atoms on the right hand side. is formed when the 4. Balance the H-atoms on the right hand side. H-atom examsis removed from 5. Balance the O-atoms HINT! on the left hand side. the C-atom with the most H-atoms bonded to it. hint 6. Check whether the balancing ratio is in the simplest form. * If you look this rule up online, you might find it is spelled ‘Saytseff’ as well. e.g. Worked example 13 EG - worked examples activity ACTIVIT and girl Write the balanced reaction for the reaction of ethane burning in excess oxygen, using right of molecular examsformulae. activity Solution C2H6(g) + O2(g) → CO2(g) + H2O(g) + energy (unbalanced) 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g) + energy (balanced) Step by step com activity • Alkanes react with halogens ACTIVITIES boy X2 (X = Br, Cℓ) when heated (indicated with and girl-left and ∆) or in the presence of light right of page (indicated with hf). activity • The reaction adds a halogen atom to the molecule, so it is known as halogenation. halogenation SUBSTITUTION Example: 1. Write a balanced reaction for the reaction of methane ( + X2, hf, ∆ ) with chlorine gas, using molecular formulae. Name the products. comment Step by step comment Solution: hf or ∆ CH4(g) + Cℓ2(g) CH3Cℓ(g) + HCℓ(g) methane chloromethane hydrochloric acid © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 31 Mind the Gap Chemistry

1 Unit • This involves breaking up large alkane molecules into smaller and more useful molecules. • Reaction conditions: • high pressure and • high temperature without a catalyst (thermal SUBSTITUTION cracking), or CRACKING • lower temperatures and pressures in the presence of a catalyst (Pt) (catalytic cracking). ( ∆, Pt ) Examples: Pt∆ • CH3CH2CH2CH3 H2 + CH2=CHCH2CH3 butane hydrogen + but-1-ene Pt∆ • CH3CH2CH2CH3 CH2=CH2 + CH3CH3 butane ethene + ethane 1.8.2 Reactions of the ALKENES Pt∆ alkene + H2(g) alkane • The alkene must first be dissolved in a non-polar solvent (e.g. CCℓ4) • Catalyst: Pt, Pd or Ni • Margarine is manufactured by the hydrogenation of unsaturated plant oils such as sunflower oil in the presence of platinum as the catalyst. These unsaturated oils are then changed to saturated soft solids. Examples: HYDROGENATION AL ATTENTION 1. Write an equation for the hydrogenation of ethene NB ADDITION using condensed structural formulae. Name the product. ( + H2, Pt ) Remember Solution: • The reaction of a Pt∆ CH2=CH2 + H2 CH3–CH3 haloalkane with water to form an alcohol is called ethene ethane hydrolysis. Any splitting 2. Write an equation for the hydrogenation of propene, of a molecule by water is using structural formulae. Name the product. ed examples also called hydrolysis. • However other reactions Solution: between organic compounds and water are called hydration reactions as the hydrogen from the water becomes part of the propene propane product. PAY SPECIAL ATTENTION NB Remember NB • Saturated compound elimination unsaturated compound alkane / haloalkane / alcohol alkene ACTIVITIES boy Unsaturated compound addition saturated compound and girl-left and right of page • HINT! alkene alkane / haloalkane / alcohol hint 32 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  comment e.g.  examples EG - worked © Department of Basic Education 2015 Mind the Gap Chemistry ep comment

Unit 1 alkene + X2 → haloalkane (X = Cℓ, Br, F, I) • NO catalyst is needed – the reaction is spontaneous. • The reaction with fluorine is highly exothermic and explosive. Examples: 1. Write an equation for the addition of bromine to ethene, using condensed structural formulae. Name the product. Solution: HALOGENATION CH2=CH2 + Br2 → CH2Br–CH2Br ADDITION ethene 1,2-dibromoethane ( + X2 ) 2. Write an equation for the addition of chlorine to propene, using structural formulae. Name the product. Solution: propene 1,2-dichloropropane alkene PAY + HX SPECIAL → haloalkane ATTENTION (X = F, Cℓ, Br, I) NB NB • NO catalyst is needed – the reaction is spontaneous. • The reaction must take place in the absence of water. HINT! Apply Markovnikov’s rule as discussed earlier, in order to determine the hint major product! Examples: e.g. EG - worked examples 1. Write an equation for the reaction between ethene and hydrogen fluoride, using condensed structural formulae. Name the product. HYDROALOGENATION Solution: ADDITION exams CH2=CH 2 + HCℓ → CH3CH2Cℓ ethene chloroethane ( + HX ) 2. Write an equation for the reaction of propene with hydrogen chloride (hydrochloric acid), using structural formulae. Name the two products and identify the major product. Solution: activity ACTIVITIES boy and girl-left and right of page activity propene 2-chloropropane 1-chloropropane Major product Step by step (apply comment comment Markovnikov) © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 33 Mind the Gap Chemistry

1 Unit H2SO4 (conc) alkene PAY SPECIAL + H2O ATTENTION alcohol NB NB • Catalyst: concentrated strong acid: H2SO4 or H3PO4 • The reaction must take place in excess H2O HINT! Apply Markovnikov’s rule as discussed earlier, in order to hint determine the major product! HYDRATION Examples: e.g. 1. Write an equation for the EG - worked reaction between ethene and water, using condensed examples structural formulae. Name the product. ADDITION Solution: H SO (conc) ( + H2O, H2SO4 concentrated ) CH2=CH2 + H2O 2 4 CH3 CH2OH exams 2. Write an equation for the hydration of propene, using structural formulae. Name the two products and identify the major product. Solution: activity ACTIVITIES boy and girl-left and right of page propene propan-2-ol propan-1-ol activity Major product 1.8.3 Reactions of the HALOALKANES haloalkane + NaOH (dilute) → alcohol Step by step comment + NaX comment • Reactions with dilute strong bases (KOH or NaOH) produce alcohols. • The haloalkane must be dissolved in ethanol before treatment with dilute sodium hydroxide (aq) or potassium hydroxide (aq). • The halogen atom is removed from the haloalkane and is replaced by the OH– -ion from the KOH or NaOH. WITH DILUTE STRONG BASE (NOH oe KOH) • The K+ or Na+ -ion bonds to the substituted halogen atom, forming NaX as a by- product. Examples: 1. Write an equation for the reaction of bromo methane with a dilute sodium hydroxide solution, using condensed structural formulae. Name the products. SUBSTITUTION Solution: CH3Br + NaOH → CH3OH + NaBr TO PRODUCE AN ALCOHOL bromo methane methanol sodium bromide 2. Write an equation for the reaction of 2-bromopropane with a dilute potassium hydroxide solution, using structural formulae. Name the products. Solution: 34 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 AL ATTENTION NB Remember The reaction of a haloalkane with water to form an PAY SPECIAL ATTENTION alcohol is called hydrolysis. Generally, splitting any NB NB molecule with water is also called hydrolysis. VOCABULARY However other reactions between organic compounds and water are called hydration Heated under reflux: A solution is heated and the reactions, particularly when OH is added to the vapour is cooled down soHINT! that it condenses and ed examples compound. hint return to the reaction vessel. Haloalkane + H2O → alcohol + HX • Reactions with water (+ heat) produce alcohols. e.g. EG - worked examples • The haloalkane must first be dissolved in ethanol • The reaction is slower than the reaction between a haloalkane and a strong HYDROLYSIS base. exams • The halogen atom is removed from the haloalkane and is replaced by the OH– from the water molecule. • The H+ -ion from the water then bonds to the substituted halogen atom, forming HX as a byproduct. SUBSTITUTION ACTIVITIES boy and girl-left and Example: right of page activity 1. Write an equation for the reaction of chloroethane with a dilute sodium WITH WATER (H2O) TO PRODUCE ALCOHOLS hydroxide solution, using condensed structural formulae. Name the products. ACTIVITIES boy Solution: and girl-left and right of page CH3CH2Cℓ + NaOH(dil) → CH3CH2OH + NaCℓ ethanol activity sodium chloride 2. Write an equation for the reaction of 1-bromopropane with water, using p comment comment condensed structural formulae. Name the products. Solution: CH3CH2CH2Br + H2O → CH3CH2 CH2OH + HBr Step by step comment comment propan-1-ol hydrogen bromide NaOH(c)∆ Haloalkane alkene + HX (X = Cℓ, Br, I) • Heating a haloalkane under PAY SPECIAL reflux in a concentrated solution of NaOH or KOH in ATTENTION NBethanol as the solvent (hot ethanolic NaOH pure NB / KOH) produces an alkene. • The halogen atom is eliminated from the haloalkane. • A hydrogen atom is eliminated from a neighbouring C-atom. DEHYDROHALOGENATION HINT! Apply Zaitsev’s rule as discussed earlier, in order to determine the major hint ELIMINATION product! Example: (NaOH (c), ∆) e.g. Write an equationEGfor the reaction - worked examples of 2-chlorobutane with concentrated sodium hydroxide when heated under reflux, using structural formulae. Show the major organic product only and name it and the by-product. Solution: exams H H H H H H H H NaOH (c), ∆ H C C C C H → C=C C C H + H Cℓ hydrogen chloride H Cℓ H H H H H 2-chlorobutane but-1-ene activity © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 35 ACTIVITIES boy Mind the Gap Chemistry and girl-left and right of page

1 Unit 1.8.4 Reactions of ALCOHOLS alcohol + NaX → haloalkane + H2O 1. Tertiary alcohols react with HX (X = Cℓ, Br) to produce haloalkanes and water. 2. Secondary and primary alcohols react slowly and at high temperatures. Example: Write an equation for the reaction of propan-2-ol with hydrogen bromide, using structural HALOGENATION SUBSTITUTION formulae. Name the products. Solution: (+ NaX) propan-2-ol 2-bromobutane water H SO (c)∆ alcohol 2 4 alkene + H2O • Acid catalysed dehydration takes place when an alcohol is heated in the presence of a dehydrating catalyst. PAY SPECIAL ATTENTION • NB NB H2SO4. Dehydrating catalyst: Concentrated sulphuric acid, • The hydroxyl group is eliminated from the alcohol. DEHYDRATION • A hydrogen atom is eliminated from a neighbouring C-atom. ELIMINATION HINT! Apply Zaitsev’s rule as discussed earlier, in order to determine the major product! hint (H2SO4 concentrated) Example: Write an equation for the elimination reaction of butan-2-ol, using condensed structural e.g. EG - worked examples formulae. Show and name only the major organic product and the by-product. Solution: H SO (c)∆ CH3 CH2CHOHCH3 2 4 CH3 CH=CHCH3 + H2O exams butan-2-ol but-2-ene water 1.8.5 Reactions of ESTERS • Esterfication takes place when an alcohol and a carboxylic acid are heated in the presence of a concentrated H2SO4 (c). activity • The reaction is a acid catalysed condensation reaction. ESTERFICATION • Water is produced during the reaction. ACTIVITIES boy • The product is an ester which is named and girl-leftas andan alkyl alkanoate. right of page • The alkyl is formed from the C-chain in the alcohol. (alcohol + carboxylic acid, H2SO4 activity • The alkanoate is formed from the C-chain in the carboxylic acid. concentrated) Step by step comment comment • 36 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 Examples: 1. What ester will form during the acid catalysed reaction of ethanol and propanoic acid? Solution: PAY SPECIAL ATTENTION Ethyl propanoate NB NB Which two compounds need to be heated in the presence of concentrated sulphuric acid in order to form butyl methanoate? Solution: HINT! Remember butanol + methanoic acid hint • Reaction conditions in Addition, Elimination and Substitution • The role of Sulphuric acid in Esterification, e.g. Addition EG & examples - workedElimination 1.9 Creating one hydrocarbon from another exams The diagram below illustrates how to use Elimination, Substitution and Addition reactions to create one hydrocarbon from another. activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 37 Mind the Gap Chemistry

1 Unit Activity 6 1. Unsaturated vegetable oils are hardened to make margarine. The reaction that takes place during the hardening process can be described as: I Addition II Hydrogenation III Substitution A I and II B II and III C Only I D Only II (2) 2. Which ONE of the following compounds has the highest melting point? A Butane B Butene C 1-bromobutane D 1-methylpropane (2) 3. Which ONE of the following pairs of compounds correctly represents the products formed during the COMPLETE combustion of Octane? A CO and H2O B CO and H2 C CO2 and H2 D CO2 and H2O (2) 4. Most organic compounds can undergo substitution or addition or elimination reactions to produce a variety of organic compounds. Some incomplete reactions are shown below: 38 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

4.1 Name the type of reaction represented by Reaction Ill. (1) 4.2 Both Reactions | and Il are examples of addition reactions. Name the type of addition reaction that is represented by each reaction. (2) 4.3 Write down the IUPAC name of the major product formed in Reaction I. (1) 4.4 Reaction | only takes place in the presence of a catalyst. Write down the formula of the catalyst used in Reaction |. (1) 4.5 Write down the structural formula and the IUPAC name of the major product in Reaction Il. (3) 4.6 To which homologous series does the organic product formed in Reaction III belong? (1) [15] Solutions 1 AVY (2) 2. CVV (2) 3. DV (2) 4.1 Elimination ¥ (1) 4.2 |- hydration / ll- hydrohalogention / (2) 4.3 Butan-2-ol / (1) 4.4 H,SO, / (1) 4.5 2-bromo-2-methylpentane / (3) H H H Br H Y | I | | | H-C-C-C-C-C-H 14 | | H HH H H-C-H | v H 4.6 Alkene / (1) [15] © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 39 Mind the Gap Chemistry

1 Unit 1.10 Plastics and polymers 1.10.1 Polymers PAY SPECIAL ATTENTION NB NB DEFINITION Polymers: HINT! • are a type of macromolecule which hint • form when monomers bond to each other chemically • to produce large molecules that are built up of repeating units in chains • that vary in length depending on the number of monomers that bond. Many polymers occur in nature (wood, carbohydrates, proteins etc.) while a large e.g. EG - worked examples number are produced synthetically (most are plastics) as they are very useful. exams 1.10.2 Synthetic polymers: Plastics • All contain carbon in their basic structure. • Are light because they have a less dense structure than other solids. • Longer polymers have stronger intermolecular forces between them. • Cross linking between molecules increases their strength. • Do not conduct heat or electricity. activity • Are versatile and can be used as foam, can be moulded or extruded into fibres. ACTIVITIES boy and girl-left and right of page Synthetic polymers (plastics) can be divided into two main types, namely activity thermoplastic and thermoset polymers. THERMOPLASTIC POLYMERS THERMOSET POLYMERS Can be softened by heating and Are melted when they are made Step by step comment hardened by comment cooling repeatedly and therefore can be remoulded into set into a hard insoluble mass different shapes and so can be but can’t be softened again and recycled. therefore can’t be remoulded into different shapes, so can’t be recycled. • Consist of long chains • Cross-links in the molecule • that slide past each other when • prevent molecules from sliding melted. past each other. Source: Wikimedia Source: Wikimedia 40 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 Examples: Examples: Melamine http://en.wikipedia.org/wiki/Melamine http://en/wikipedia.org/ wiki?File:Recyclables.JPG Bakelite http://en.wikipedia.org/wiki/Bakelite 1.10.3 Properties of plastics, nylons and polyesters Plastics Nylons Polyesters Are formed by addition Are formed by Are formed by polymerisation condensation condensation polymerisation polymerisation between a dicarboxylic between a dicarboxylic acid and a diamine. acid and a diol. • If the chains are • Molecular chains • Depending on the lined up in a regular are drawn into thin chain they may be way, the plastic is fibres. stronger or weaker hard and strong. • Are very strong. than nylon. • If the chains are irregular or there are many side chains the plastic is softer and can bend. Examples: Examples: Examples: • Polyethene • Nylon • Used for clothing (plastic bags, (ropes, carpets, fibres shopping bags, clothes) • cold drink bottles plastic containers) • Polypropene Hard (Piping, bottle crates) • PVC (water pipes, artificial leather) © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 41 Mind the Gap Chemistry

1 Unit Properties of polymers are determined by the intermolecular forces between the chains. Polymers • with carbonyl groups can form strong hydrogen bonds and are strong with high melting points; • like polyesters have dipole-dipole bonds (that are weaker than hydrogen bonds) between the chains, so have lower melting points and are more flexible (bendable); • with only Van der Waals forces between the chains have low melting points and are flexible. 1.10.4 Polymerisation Polymerisation • is a chemical process in which monomers combine to form polymers; • takes place in three steps: –– initiation (the step that starts the reaction) –– propagation (the steps that allow the reaction to continue) and –– termination (the step that ends the reaction). • is either addition polymerisation or condensation polymerisation. ADDITION POLYMERISATION CONDENSATION POLYMERISATION • The monomer must be • Instead of double bonds, these unsaturated – it must contain monomers have functional a double bond or a triple bond, groups like an alcohol (-diol), or e.g. an alkene like ethene. dicarboxylic acid. • Is the direct addition of the • Each monomer must have two or same monomer molecules to more active groups – to link the form a single product. molecules. • No atoms are lost or gained • Is the reaction between during the reaction. monomers which are not always • The empirical (simplest) the same to form two products – formulae of the monomer and of the polymer and a molecule like the polymer are the same. H2O or HCℓ. • The chain is bonded by strong • The polymers that form are covalent bonds. • polyesters or • nylon. • Reaction conditions: • Reaction conditions: • High temperature (± 200 °C) • High temperature • High pressure (± 1000 atm) • Concentrated sulphuric acid, • A small amount of a very H2SO4 reactive initiator. 42 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 Example: Example: Polimerisation of ethene to form Polymerisation to produce polythene polyesters is called esterfication • N(CH2═CH2) → (−CH2−CH2−) • during which a water molecule H2O is eliminated and • the organic product contains an ester functional group. • The name of the polymer is • The name of the polymer is derived from the monomer used. deduced from the names of • E.g. If the monomer is ethene the 2 bonding molecules – an the polymer will be polyethene. organic alcohol and a carboxylic acid. 1.10.5 Addition polymerisation of ethene to produce polyethene Example 1: (Note: This is the only addition polymerisation example you need to learn.) Polyethene is also known as polyethylne • The length of the polyethene macromolecule depends on the number of ethene monomers that bond. • The structural formula of the polymer polyethene can therefore be abbreviated as follows: where n is the number of monomers that joined. © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 43 Mind the Gap Chemistry

1 Unit There are two types of polyethene Low density polyethene High density polyethene LDPE HDPE Relative strength Weak Strong Number of branches on the Many Few polymer Maximum usable temperature (°C) 85 120 Freezer bags Shopping bags Soft bottles Plastic crates Bendable cables Plastic pipes Toys Uses: http://en.wikipedia.org/wiki/ http://commons.wikimedia.org/wiki/ Hose_%28tubing%29 File:Plastic_beads2.jpg • Easily moulded • Can be made in any colour • Light • Hardness can be manipulated during manufacture • Not affected by water, air etc. • Can be used to pack food – hygenic 44 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 1.10.6 Condensation polymerisation to produce a polyester • The alcohol has a hydroxyl group on each end of its carbon chain, so it is a -diol. • A H-atom is removed from the hydroxyl group on the alcohol and a hydroxyl group is lost from the dicarboxylic acid. • The −H and the −OH bond to form a water molecule, H2O. • A new bond forms between the O-atom on the alcohol and the C-atom on the carboxylic acid. This is an ester linkage. • The carboxyl group on the other end of the dicarboxylic acid can now react further with another -diol, while the hydroxyl on the other end of the -diol can now react with another dicarboxylic acid. • This results in a long chain-like compound – a polymer. • This polymer is known as polyester. Example: • There are many different polyester fibres with a wide range of uses. Kevlar and Mylar are two examples. KEVLAR MYLAR Properties • Much stronger than nylon or steel • Stretched polyester film • Fire resistant • High tensile strength • Transparent Uses • Bulletproof vests • Helium balloons • Boat hulls • Space / rescue blankets • Aeronautical industry • Isolators http://en.wikipedia.org/wiki http://en.wikipedia.org/wiki/ /File:Balloonsanimals.jpg File:Kevlar-vest-ct-wiki.jpg http://en.wikipedia.org/wiki/ http://en.wikipedia.org/wiki/File: File:Kevlar_canoe_algonquin.JPG RescueFoil.jpg © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 45 Mind the Gap Chemistry

1 Unit 1.11 Plastics and pollution 1.11.1 Disadvantages of polymer usage Although synthetic plastics are light, easy to shape and relatively cheap, there is widespread opposition to the use of these materials. • Most are not bio-degradable. • When burned to dispose of them, poisonous gases may be released. • Landfill sites use up land and fill quickly. • Litter causes ecological problems. 1.11.2 Recycling • Biodegradable plastics would solve most problems, but are expensive. • Responsible recycling is essential. http://en.wikipedia.org/wiki/ File:Canal-pollution.jpg http://www.flickr.com/photos/dave- goodman/3609172860/ Problems with Recyclability: • Not possible in very rural areas. • High costs. • Not all plastics can be recycled. • Communities are not all aware of the problems. High density polythene (HDPE) Low density polythene (LDPE) Recycled to produce Recycled to produce • plastic wood for garden furniture • fibres for clothing • rubbish bins • plastic liners for rubbish bins • rulers 46 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 1 Activity 7 1. Give the definition of a macromolecule (1) 2. Explain the difference between a monomer and a polymer. (2) 3. What is the chemical reaction in which a monomer molecules join to form a polymer called? (1) 4. The polymer below is the product of a polymerisation reaction 4.1 What is the IUPAC name of the monomer used to form this polymer?(1) 4.2 Give the structural formula of the monomer used to form this polymer.(2) 4.3 Is this an example of an addition or condensation polymerisation? Give a reason for your answer. (2) 5. A carboxylic acid monomer and an amine monomer has joined in an amide linkage as shown below. Name the type of polymerisation that occurs in this equation. Give a reason for your answer. (2) 6. Write an equation for polymerisation of ethane to form polyethene.(2) 7. Is (6) an example of addition or condensation polymerisation? Give a reason for your answer.  (2) 8. Name at least three industrial uses of polythene.  (3) [18] © Department of Basic Education 2015 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES 47 Mind the Gap Chemistry

1 Unit Solutions 1. A molecule that consists of a large number of atoms. 3(1) 2. A polymer is a large molecule that is made up of smaller units, covalently bonded to each other in a repeating pattern. 3 Each of these smaller units are called a momomere. 3(2) 3. Polymerisation 3(1) 4.1 prop-(1)-yne (1) 4.2 3 3 (2) 4.3. Addition polymerisation 3 monomers combined through an addition reaction. 3(2) 5. Condensation polymerisation. Two monomers with different functional groups are joined 3 and when they join it leads to the “loss” of a small molecule, in this case, water. 3(2) 6. 3 3 (2) 7. Addition polymerisation 3 monomers combinded through an addition reaction. 3(2) 8. Any of the following: sandwich bags, 3cling wrap, 3 car covers, 3 squeeze bottles, liners for tanks and ponds, moisture barriers in construction, freezer bags, water pipes, wire and cable insulation, extrusion coating. (3) [18] Keep going! 48 UNIT 1 ORGANIC COMPOUNDS AND MACROMOLECULES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 2 Rate and extent of reactions 2.1 Summary 2.2 Energy changes during chemical reactions During a chemical reaction, the reactants (written on the left hand side of the equation) react to produce products (written on the right hand side of the equation). Chemical reactions take place in steps Step 1: The reactant particles need energy so that the existing bonds (covalent, ionic or metallic bonds) between the particles in the reactants can break; Step 2: The separated particles that are present in the activated complex move and collide; Step 3: Energy is released by the particles as new bonds form when the products are produced. However, endothermic reactions, as we will later see, take in energy from the surroundings as they come to completion. © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 49 Mind the Gap Chemistry

2 Unit AL ATTENTION NB 2.3 Activation energy and the activated complex DEFINITIONS Activation energy is usually absorbed in the form of heat, light or mechanical Activation energy is the energy and is required to change reactants into an activated complex. It minimum kinetic energy is the energy required to break the bonds in the reactant particles to form required for a chemical the activated complex. reaction to take place. The activated complex is • Spontaneous reactions take place when the reactants have enough a temporary, unstable, high energy to start the reaction without additional heat. PAY SPECIAL ATTENTION • NB NBwhen the reactants do not ed examples energy intermediate state Non-spontaneous reactions take place between the reactants and the have enough energy themselves to start a reaction, and additional products. energy needs to be added. HINT! “Spontaneous” means “of its own accord” or “by itself” or “without hint PAY SPECIAL ATTENTION intervention”. NB NB A spontaneous reaction takes place without any addition of 2.4 e.g. Catalysts EG - worked examples energy. HINT! A positive catalyst increases the rate at which a reaction takes place while hint The mass ACTIVITIES boy of a catalyst a negative catalyst (also called an inhibitor) decreases the reaction rate. remains constant during a and girl-left and right of page chemical reaction. exams If you hear someone talk about a “catalyst”, they usually mean a positive catalyst. e.g. EG - worked examples Positive catalysts increase the rate of a chemical reaction... AL ATTENTION NB • by providing an alternative path for the reaction to take place comment exams DEFINITIONS • which requires less energy for the reactants to change into the p comment activity activated complex, A catalyst is a substance • by decreasing the activation energy. ACTIVITIES boy that changes the rate of a and girl-left and right of page chemical reaction without undergoing a permanent activity change during the reaction, 2.5 Energy changes during activity by decreasing the activation d examples energy of the reaction. chemical reactions: ∆H ACTIVITIES boy The heat of reaction or and girl-left and We know thatStep energy right of page comment enthalpy (∆H) is the net by step comment activity change in energy in a • is absorbed by the reactant particles when the intramolecular bonds chemical system when a break to form the activated complex, and is chemical reaction takes place. • is released when the particles in the activated complex form new bonds when the products form. Step by step comment comment In symbols: ∆H = Energy of products – Energy of reactants ACTIVITIES boy = Eproducts – Ereactants and girl-left and right of page = Efinal – Einitial OR ∆H = Hproducts – Hreactants = Hfinal – Hinitial p comment comment 50 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 2 2.6 Endothermic and exothermic reactions When we refer to a reversible reaction, the enthalpy (ΔH) is always stated for the forward reaction. It is important: • to know the differences between exothermic and endothermic reactions • to be able to identify whether a given reaction is exothermic or endothermic and • to be able to interpret graphs for exothermic and endothermic reactions. • Endo- means “into” or “inwards”, and “exo-” means “outwards”; so, “endothermic” means heat goes in (the reaction vessel will feel cold after the reaction completes), whereas “exothermic” means “heat goes out”, so the reaction vessel will feel hot when the reaction completes. 2.6.1 EXOTHERMIC REACTIONS 2.6.2 ENDOTHERMIC REACTIONS More energy is released when the new bonds form More energy is absorbed when the existing bonds than is absorbed when the existing bonds break. break than is released when the new bonds form. • Ereactants > Eproducts • Ereactants < Eproducts • ∆H is negative ∆H < 0 • ∆H is positive ∆H > 0 • ∆H = Eproducts – Ereactants < 0 • ∆H = Eproducts – Ereactants > 0 • Net energy is released to the surroundings in the • Net energy is absorbed from the surroundings in form of heat. the form of heat. • The temperature of the reaction mixture • The temperature of the reaction mixture increases. decreases. Example: fire: C (coal) + O2 (in air) → CO2 Example: dissolving ammonium nitrate in water. Writing conventions: Writing conventions: CO(g) + NO2(g) → CO2(g) + NO(g) ΔH = – 226 kJ·mol–1 H2O(ℓ) → H2O(g) ΔH = 41 kJ·mol–1 CO(g) + NO2(g) → CO2(g) + NO(g) ΔH < 0 H2O(ℓ) → H2O(g) ΔH > 0 CO(g) + NO2(g) → CO2(g) + NO(g) + Energy H2O(ℓ) + Energy → H2O(g) CO(g) + NO2(g) → CO2(g) + NO(g) + 226 kJ·mol–1 H2O(ℓ) + 41 kJ·mol–1 → H2O(g) Graphic representation: Graphic representation: Key to the symbols on the graphs: A: reactants (for the forward reaction) / products (for the E: activation energy for the forward reaction (without a reverse reaction) catalyst) B: products (for the forward reaction) / reactants (for the F: activation energy of the forward reaction (with a catalyst) reverse reaction) G: activation energy of the reverse reaction (without a C: activated complex (without a catalyst) catalyst) D: activated complex (with a catalyst) H: activation energy of the reverse reaction (with a catalyst) © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 51 Mind the Gap Chemistry

2 Unit 2.6.3 Law of conservation of energy The release of energy in chemical reactions occurs when the reactants have higher chemical energy than the products. The chemical energy in a substance is a type of potential energy stored within the substance. This stored chemical potential energy is the heat content or enthalpy of the substance. The collection of substances that is involved in a chemical reaction is referred to as a system and anything else around it is called the surroundings. If the enthalpy decreases during a chemical reaction, a corresponding amount of energy must be released to the surroundings. Conversely, if the enthalpy increases during a reaction, a corresponding amount of energy must be absorbed from the surroundings. This is simply the Law of Conservation of Energy. Now, to break a chemical bond, requires energy. Some substances need more and some substances need less energy to break the bond – depending on the bond energy (measure of the bond strength of the chemical bond) i.e. potential energy of the bonds of the initial reactants. Some reactants obtain the energy required to break the bondsPAYfrom the SPECIAL surroundings, which might include the energy ATTENTION NB NB that is released when the bonds in the products are formed. To form a bond releases energy; when the bond is formed in the product, the product releases energy, of which some energy is “given back” to break the bonds of the reactants – so they can break! For our purpose it is enough to simply talk about ‘the HINT! total energy released’ when the product is formed in an hint exothermic reaction is more (entalphy decreases) than the amount of energy required to break the initial reactants’ bonds. e.g. Worked example 1 EG - worked examples When there is a limited supply of oxygen, such as in a car which is not tuned properly, octane examsburns incompletely to produce carbon monoxide (amongst other emissions). The graph alongside represents the energy changes that take place during this reaction and the balanced chemical equation is given: activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 2C8H18(ℓ) + 17O2(g) ⇌ 16CO(g) + 18H2O(ℓ) ∆H < 0 1. What is the activation energy for the forward reaction? 2. What is the activation energy of the reverse reaction? 3. What is the heat of reaction for this reaction? 52 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 2 4. Is this reaction exothermic or endothermic? Motivate. 5. Use the chemical equation above and give a reason why vehicles with incorrectly tuned engines are a health hazard. 6. Part of the action of catalytic converters is to speed up the complete oxidation of carbon monoxide (CO) and petrol (C8H18) from incorrectly tuned engines to CO2. Why should people support legislation that makes catalytic converters a necessary component of exhaust systems of automobiles? 7. The graph below represents the relationship between potential energy and course of a reaction for a certain chemical reaction. The activation energy of this reaction is: A 1kJ B 2kJ C 3kJ D 4kJ Solutions 1. EActivated = EActivated Complex – Ereactants = 1 500 – 900 = 600 kJ·mol–1 2. EActivated = EActivated Complex – Ereactants = 1 500 – 300 = 1 200 kJ·mol–1 3. ∆H = Eproducts – Ereactants = 300 – 900 = – 600 kJ·mol–1 4. The reaction is exothermic because ∆H is negative. 5. CO is poisonous.  6. Catalytic converters ensure conversion of products to CO2 so that the danger of CO poisoning is minimised.  7. Eactivated = Ecomplex – Ereagents =5–3=2 (B): 2kJ  © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 53 Mind the Gap Chemistry

2 Unit 2.7 Rates of reactions In symbols: NB Where: PAY SPECIAL ATTENTION NB ∆[reactants] rr : reaction rate (mol·dm–3·s–1) rr = __________ ​  ∆t ​ DEFINITION ∆[reactants] : change in concentration of reactants OR (mol·dm–3) The rate of a reaction ∆[products] (reaction rate) is the change HINT! rr = _________ ​  ∆t ​ ∆[products] : change in concentration of products hint in concentration of reactants (mol·dm–3) or products per unit time. ∆t : change in time (s) e.g. EG - worked examples Activity 1 Consider the reaction exams H2 + I2 → 2HI (∆H < 0) 1. Is this reaction exothermic or endothermic? (1) 2. State a reason for your answer in 1.1. (1) 3. How does energy of the products compare to that of the reactants? (2) [4] activity Solutions ACTIVITIES boy and girl-left and 1. Exothermic (reaction). 3(1) right of page activity 2. ∆H < 0 3(1) 3. The energy of the products is less than 3 that of the reactants. 3(2) [4] Step by step comment comment 54 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 2 2.8 Factors which affect reaction rate The rate at which a chemical reaction takes place depends on the reaction conditions. In industry it is important to make sure that the chemical reactions take place at a rate as high as is safely and economically possible. The 5 factors that affect the rate of a chemical reaction are: FACTOR EFFECT ON REACTION RATE • The nature of the reactants. • The higher the reactivity of a reactant, the higher the reaction rate (E.g. reactive metals like Mg react faster than metals like Cu). • Inorganic reactions are faster than organic reactions. • The concentration of the reactants for gases (g) and • The higher the concentration of a reactant, the aqueous (aq) solutions. higher the reaction rate. • The pressure of gas (g) reactants. • The higher the pressure of a gas reactant, the higher its concentration and the higher the reaction rate. • The surface area (state of division) of solid • A powder has a large surface area. reactants (marked (s)). • Lumps and strips have a smaller surface area. • The larger the surface area, the higher the reaction rate. • The temperature of the reaction system. • The higher the temperature of the reaction system, the higher the reaction rate. However, if the temperature goes too high, a different reaction might take place (e.g. the reactants might burn up instead of reacting as expected). • The presence of a suitable catalyst. • A suitable positive catalyst increases the reaction rate. 2.9 The Collision Theory The collision theory is applied to explain how these factors affect the rates of chemical reactions. PAY SPECIAL ATTENTION • According to the Kinetic Molecular Theory, the particles in a reaction NB mixture are in a state of perpetual random motion and therefore collide with one another. • Only effective collisions between reactant particles lead to chemical reactions and the formation of new products. HINT! An effective collision between reactant particles takes place if: hint Remember • the colliding particles have enough kinetic energy to form an If a reaction is fast, it has activated complex i.e. if the energy of the colliding particles is equal to or more than the activation energy for the reaction AND a short reaction time but • the particles are correctly orientated when they collide. e.g. EG - worked examples a high reaction rate. exams © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 55 Mind the Gap Chemistry activity

2 Unit The rate of a chemical reaction depends on the number of effective collisions between the reactant particles per second; the number of PAY SPECIAL ATTENTION collisions for which the particles have energy more than or equal to NB NB energy and on the number of collisions that are correctly the activation orientated. Hence, smaller particles e.g. dust or gas, react more violently than larger particles, e.g. crystals, since the reactants can mix more thoroughly and therefore increase the number of effective collisions. HINT! Compare, for example, the explosive quality of gunpowder to say, lumps of hint Remember solid ingredients like sulphur, potassium nitrate, and coal. The more effective collisions In symbols: that occur between the reaction rate ∝ (number of effective collisions per unit time) e.g. reactant particlesEG -per second, worked the faster the reaction will examples ∝ (particles with correct orientation) occur. ∝ (particles with sufficient (enough) Ek) exams 2.10 The mechanism of reactions The Maxwell-Boltzmann distribution diagram shown below shows the distribution of the kinetic energy of the particles in a reaction system. The distribution of the kinetic energies of the reactant particles is used to activity explain the reaction rate in the system. ACTIVITIES boy • The graph shows the number of particles (on the vertical axis) versus and girl-left and right of page their kinetic energy (on the horizontal axis). activity Step by step comment comment • The curve always passes through the origin — this means that no molecules have zero kinetic energy since all molecules are in motion — unless they reach the temperature of 0 K or –273˚C. • The curve does not touch the x-axis at high kinetic energy i.e. there are always some molecules with very high kinetic energy. • The energy value that corresponds with the peak of the curve represents the most probable energy. • The area below the curve is equal to the total number of molecules in the system. 56 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 2 • The shaded area below the curve and to the right of the activation energy, equals the number of reactant particles that have enough energy to collide effectively and therefore can react. • The area below the curve and to the left of the activation energy, equals the number of reactant particles that can’t collide effectively and therefore can’t react. • If the shaded area is bigger, there are more particles that can collide effectively and the reaction rate is higher. a) The effect of increasing the temperature of the reaction PAY SPECIAL ATTENTION mixture NB NB If the temperature increases: DEFINITION • The particles in the sample have a higher average kinetic energy and the most probable energy is higher than at a low temperature. Temperature is a measure of • The graph is wider and the peak is lower. HINT! the average kinetic energy of • The total number of particles is the same, hint so he total area below the the particles in a sample. graph is the same. • More particles have kinetic energy > than the activation energy (the shaded area is bigger) • ∴ e.g. more effective collisions take place per second and EG - worked examples • the collisions are more energetic • ∴ the reaction rate increases. exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment b) The effect of increasing the concentration of the reactants If the concentration of the reactants increases: • The particles in the sample have the same average kinetic energy and the most probable energy is the same ∴ the graph has the same curve/shape • The graph is higher as there are now more particles per unit volume; • The total number of particles is greater, so the total area below the graph is bigger. • More particles have energy > than the activation energy (the shaded area is bigger) • ∴ more effective collisions take place per second • ∴ the reaction rate increases. © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 57 Mind the Gap Chemistry

2 Unit PAY SPECIAL ATTENTION NB NB HINT! hint Remember c) The effect of adding a catalyst Adding a suitable positive catalyst: When adding a suitable catalyst: • decreases the activation energy needed to change the reactants into an activated complex so e.g. EG - worked examples • the total number of • more particles have enough energy to form an activated complex reactant particles stays constant and • ∴ the number of possible effective collisions per second increases and • the average kinetic energy exams • ∴ the reaction rate increases. of the reactant particles remains constant as the temperature does not change. PAY SPECIAL ATTENTION activity NB NB ACTIVITIES boy and girl-left and A Maxwell-Boltzman right of page graph has a shallow gradient activity if the reaction rate is higher HINT! BUT in a graph of mass loss vs time, a shallow gradient hint means the reaction rate is lower. Step by step comment commentEG - worked examples e.g. Worked example 2 Consider the following Maxwell-Boltzman distribution diagram for a chemical reaction system examsand answer the questions that follow. activity ACTIVITIES boy and girl-left and right of page activity 58 UNIT 2 RATE AND EXTENT OF REACTIONS  Step by step comment comment © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 2 1. Did all the molecules have the same kinetic energy at low temperature? Motivate your answer. 2. What does the shaded area to the right of the vertical line represent? 3. If a catalyst is added to the reactants, will it cause the vertical dotted line to move to the left or right? Explain your answer. Solutions 1. No. Together they will have the same average kinetic energy, but each molecule’s kinetic energy is determined by its mass and the 1 square of its own average speed (remember, Ek = ​ __ 2 ​ mv ). 2 2. It indicates the % of molecules which will have more energy than the activation energy at that temperature. In other words, the % of molecules which are capable of colliding effectively and reacting. At the higher temperature, there is a higher % of molecules which are capable of colliding effectively (bigger area below the graph). 3. To the left. The activation energy of the reaction is decreased by adding a suitable catalyst. The line thus shifts to the left. It implies that at both the low and the high temperatures there will now be a greater % of molecules available for effective collisions.  2.11 Measuring rates of reactions We measure the rate of a reaction by studying the rate at which (how fast) the products appear or the rate at which the reactants are used up while the reaction takes place. Experimental techniques for measuring the rate of a given reaction include the following: • measuring of gas volumes produced, • turbidity (e.g. precipitate formation), • change of colour and • the change of the mass of the reaction vessel if a gaseous product is allowed to escape from it. Experimental Technique 1: Measuring turbidity Na2S2O3(aq) + 2HCℓ(aq) → 2NaCℓ(aq) + H2O(ℓ) + SO2(g) + S(s) Sodium thiosulphate reacts with diluted hydrochloric acid to form a yellow sulphur precipitate. The formation of the precipitate can be used to measure the reaction rate. © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 59 Mind the Gap Chemistry

2 Unit Apparatus: Method: PAY SPECIAL ATTENTION NB 1. Draw a black NBcross on a piece of paper. 2. Add an exact volume of Na2S2O3 to a conical flask and place it on the DEFINITIONS cross. Turbidity is the cloudiness 3. Quickly add HCℓ to the solution and start the stopwatch. or haziness of a fluid (liquid HINT! 4. Swirl (turn) the flask carefully and allow the reaction to continue until hintthat is caused by or solution) the black cross can no longer be seen. particles in the fluid which 5. Record the time. prevent light from passing through the liquid. In this experiment, the concentration of the HCℓ(aq) and Na2S2O3(aq) can e.g. Opaque: not transparent, EG - worked examples be changed, keeping the temperature constant OR In another experiment, does not allow light to pass the temperature can be changed, but keeping the concentrations constant. through. The time it takes for the solution to become opaque is a measure of the exams reaction rate. Experimental Technique 2: Measuring the volume of a gas product Consider the following reactions: Mg(s) + 2HCℓ(aq) → MgCℓ2(aq) + H2(g) OR activity CaCO 3(s) + ACTIVITIES 2HCℓ(aq) → CaCℓ2(aq) + H2O(ℓ) + CO2(g) OR boy and girl-left and right of page MnO2(s) 2KCℓO3(s) → 2 KCℓ(s) + 3O2(g) activity (The catalysed decomposition of potassium chlorate) 1. The rate of the reaction is determined by measuring the volume of the gas produced (the dependent variable) in specific time Step by step intervals comment comment (the independent variable). If you can’t recall what the terms “independent” and “dependent variables” mean, see the introduction section of this book. 2. The gas is collected in a graduated gas syringe or by the downward displacement of water in a burette. The volume of the gas produced can be determined accurately. 3. The stopwatch is started when the reactants are added and stopped when no new bubbles form – i.e. when the volume of gas produced stays constant. 60 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 2 Apparatus: Measuring volume with a gas syringe: Measuring volume by the downward displacement of water Graph of volume of gas produced vs. time 60 Experiment 1 50 Volume gas produced (cm3) Experiment 2 40 30 20 10 0 0 20 40 60 80 100 120 140 Time (s) Experimental Technique 3: Measuring the total mass of reactants and the mass loss: Consider the same reactions as used in Technique 2. © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 61 Mind the Gap Chemistry

2 Unit PAY SPECIAL ATTENTION NB NB Method: • The rate of the reaction is determined by measuring the mass of the gas produced (the dependent variable) in specific time intervals (the HINT! independent variable). hint • The reactants are added to the flask and the mouth of the flask is The mouth of the flask is plugged with some cotton wool. plugged with cotton wool • The mass of the reaction system (the reactants, flask and cotton wool) is recorded at the start of the experiment and the stop watch is e.g. because it is porous and examples EG - worked allows the gas to escape started. from the reaction system but • The gas that is produced passes through the cotton wool plug and prevents liquid spray that escapes from the system, so the mass (the reading on the scale) exams forms during the reaction, decreases. from escaping. • The mass of the system is recorded at regular time intervals e.g. every 30 seconds. • The stopwatch is stopped when no new bubbles form – i.e. when the mass reading on the scale stays constant. • The results are used to plot a graph of: mass of the flask and contents versus time OR mass loss (mass of gas escaped) versus activity time. ACTIVITIES boy and girl-left and right of page activity Step by step comment comment PAY SPECIAL ATTENTION NB NB HINT! hint • The gradients of the graphs indicate the rates of the reactions. • The steeper (more vertical) the gradient of the graph, the higher the reaction e.g. rate, because EG - worked examples more happens in less time. Limiting reactant: The reactant that is used up during a chemical reaction – all of the reactant reacts. Excess reactant: The reactant that is not all used up during a chemical reaction – exams there is always some of this reactant left in the reaction system once the reaction has stopped. activity ACTIVITIES boy and girl-left and right of page 62 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  activity Mind the Gap Chemistry

Unit 2 Interpreting the graphs that use Techniques 2 and 3: There are three time intervals shown in the graphs: Stage in the reaction Explanation Part 1 of both graphs • The concentration of the reactants (usually the • has the biggest slope (gradient) so acid) is the highest and • the reaction rate is the highest. • the reaction surface area (the solid), is the biggest therefore • the number of effective collisions taking place will be high • so the reaction rate is high and • the volume of gas produced per second is high. Part 2 shows • The reactants are being used to produce the • a gradual decrease in the slope so products so • the reaction rate decreases gradually. • the concentration of the acid is decreasing and • the reaction surface area of the solid reactant is decreasing therefore • the number of effective collisions taking place decreases and • the reaction rate decreases and • the volume of gas produced per second will decrease so that • fewer gas molecules escape per second and the rate at which the mass changes, decreases. Part 3 is where When the limiting reactant has been used up (all of it • each graph ends horizontally has reacted) • which means that the reaction is complete. • the reaction is complete and • no more products are produced so • no more gas escapes from the system and • the mass remains constant. © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 63 Mind the Gap Chemistry

HINT! 2 Unit hint e.g. Worked example 3 EG - worked examples A group of learners use the reaction between hydrochloric acid and magnesium powder to exams investigate one of the factors that influence the rate of a chemical reaction. The reaction that takes place is: Mg(s) + 2HCℓ(aq) → MgCℓ2(aq) + H2(g) The learners use the apparatus and follow the method shown below to conduct the investigation. activity ACTIVITIES boy and girl-left and right of page activity Method for Experiment 1: Step by step comment comment Step 1:  Place a spatula (spoon) of magnesium powder in a conical flask and add 50 cm3 HCℓ(aq) of known concentration. Step 2: Simultaneously start the stopwatch and close the flask with the rubber stopper containing the delivery tube. Step 3: Measure the volume of the H2(g) formed in time intervals of 20 seconds. Method for Experiment 2: Repeat steps 1 to 3 above, but use only 25 cm3 of the same HCℓ(aq) diluted to 50 cm3 with distilled water. 1. How does the concentration of the acid used in Experiment 2 differ from the concentration of the acid used in Experiment 1? Write down only greater than, lower than or equal to. 2. Write down a hypothesis for this investigation. 3. Why should the learners ensure that equal amounts of magnesium powder are used in each of the two experiments? 4. The learners use an excess of HCℓ(aq) for the two experiments. What meaning does the term ‘excess’ have in this situation? Give a reason why the excess HCℓ(aq) will not influence the results. 5. If the Mg powder used in Experiment 1 was impure with particles of dirt in between, describe and explain how the shape of the graph would be influenced. 64 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 2 e.g. Worked example 3 (continued) EG - worked examples 6. After completing the investigation, the learners represent the results obtained during each experiment on the graph below.  exams activity ACTIVITIES boy and girl-left and right of page activity Write down the volume of hydrogen gas formed during the first minute in: 6.1 ExperimentStep 1 by step comment comment 6.2 Experiment 2 7. Which one of the experiments took place at the faster rate? Refer to the shape of the curves to motivate your answer. PAY SPECIAL ATTENTION NB 8. Give a reason why the final volume of gas produced is the same in both experiments.  9. What conclusion can the learners draw from the results obtained?  10. How will an increase in the temperature influence the following? Write down HINT! increases, decreases or remains the same. hint 10.1 The final volume of gas obtained in each experiment The hypothesis must refer 10.2 The volume of gas obtained in each experiment after 40 s to the relationship between 11. Why do both graphs form horizontal lines respectively at t = 60 s and t = 120 s? e.g. dependent and EG - independent worked examples variables. It must be a 12. Give two reasons why the rate of formation of hydrogen gas is lower at statement that can be proved t = 50 s than at t = 20 s. to be correct or incorrect, it is 13. Determine the rate of formation of hydrogen gas in Experiment 2 at t = 20 s.  exams a prediction based on (prior) knowledge. 14. In a third experiment, Experiment 3, half the mass of magnesium was used in powder form. 14.1 How will this change influence the final volume of gas compared to Experiment 1 and 2? 14.2 Explain your answer. activity Solutions ACTIVITIES boy and girl-left and 1. Smaller than or lower than (not ‘less’ or ‘diluted’). right of page 2. Possible answer: activity If the concentration of the reactants in a chemical reaction increases, the reaction rate will increase. 3. The mass of the Mg is a controlled variable to ensure a fair test. There is only one independent variable which changes from the Step by step comment one experiment to the next i.e. the concentration of the acid. © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 65 Mind the Gap Chemistry

HINT! 2 Unit hint e.g. Worked example 3 (continued) EG - worked examples 4. ‘Excess’ means that there is more than enough hydrochloric acid for the reaction to take place. When the reaction is complete, exams all the Mg will have reacted, but some of the acid will be left over. The Mg is the limiting reactant and when it has all reacted, the reaction will stop. 5. The gradient would be less steep, indicating that the reaction rate is lower. Because the reaction surface area of the magnesium is smaller, fewer particles are available to collide, activity thus fewer effective collisions will occur. The final volume of hydrogen gas formed will beACTIVITIES less, boybecause the magnesium is the and girl-left and limiting reagent and a smaller mass right of page of pure magnesium was reacting, because of the dirt in between. 6.1 60 cmactivity 3 6.2 42 cm3 7. Experiment 1.The gradient or slope (of the tangent to the graph) is steeper. 8. The mass ofStep Mgbyused step in both experiments comment was the same. comment Mg is the limiting reagent and the limiting reagent determines the yield of the products. 9. If the concentration of the reactants in a chemical reaction increases, the reaction rate will increase. 10.1 Remains the same. 10.2 Increases. 11. The horizontal lines mean that the reactions are complete; all the magnesium has reacted with the excess HCℓ. Experiment 1 reaches completion first because of the higher concentration of HCℓ (more particles per unit volume to react), so more effective collisions will occur and the reaction rate is higher. 12. At 50 s the concentration of HCℓ has already decreased notably, and there is much less magnesium left, with a smaller exposed surface area. Thus the collisions are less effective and the reaction rate is lower. ∆y 13. Reaction rate = slope (gradient) = ​ ___ ∆x ​ ∆volume H2 = ​ _________ ∆time ​ (16–0) = ​ ______ (20–0) ​ 0,8 cm3 = ​ ______ s ​ 14.1 The final volume of hydrogen gas in Experiment 3 will be half the volume of hydrogen gas produced in Experiment 1 and 2. 14.2 Mg is the limiting reagent which determines the yield. From the balanced equation: 1 mol Mg(s) produces 1 1 1 mol H2(g), thus ​ __ __ 2 ​mol Mg(s) produces ​ 2 ​mol H2(g) 66 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 2 e.g. Worked example 4EG - worked examples A certain mass of calcium carbonate chunks is added to a hydrochloric acid solution in anexams open beaker on a scale as shown here. The equation for the reaction is as follows: CaCO3(s) + 2HCℓ(aq) → CaCℓ2(aq) + H2O(ℓ) + CO2(g) activity ACTIVITIES boy and girl-left and right of page activity CO2(g) is allowed to escape from the beaker. The data in the table below was obtained for a time interval of 8 minutes. Time (min) Mass of beaker and contents (g) Step by step comment comment 0 200,00 1 197,50 2 195,45 3 193,55 4 191,70 5 189,90 6 188,15 7 186,45 8 184,80 1. Explain the term ‘reaction rate’. 2. Calculate the change in the mass of the beaker and its contents during the 8 minutes. 3. Use your answer to Question 2 to show that the average reaction rate during the 8 minutes is 1,9 g•min–1. 4. Calculate the mass of calcium carbonate which reacted during the 8 minutes. 5. “The use of the beaker in this experiment influences the readings and therefore the calculation of the rate of the reaction and mass of the CO2 formed”. Explain this statement and suggest TWO ways to rectify this experimental problem and to make this a reasonable test. 6. One of the products formed in a chemical reaction is a gas. Which ONE of the following graphs of volume versus time best represents the formation of this gas until the reactants are used up? © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 67 Mind the Gap Chemistry

HINT! 2 Unit hint e.g. Worked example 4 (continued) EG - worked examples exams activity ACTIVITIES boy and girl-left and right of page Solutions activity 1. The change in amount/mass/volume/concentration of products formed per unit time. OR The change in amount/mass/volume/concentration of reactants used per unit time. Step by step comment comment 2. Change in mass = 200 – 184,8 = 15,2 g change in mass ____________ ____ 15,2 3. Rate of reaction = ​    time taken ​ = ​  8 ​ = 1,9 g × min–1 CO2 produced 4. mol CO2 formed: M(CO2) = 44 gmol–1 and M(CaCO3) = 100 g•mol–1 (for all the calculating molar masses) Mass CO2 produced = ∆m = 200 – 184,8 = 15,2 g ∴ mol CO2 produced: m 15,2 n = ​ __ ____ M ​ = ​  44 ​= 0,35 mol CO2 From the balanced equation: 1 mol CaCO3 : 1 mol CO2 ∴ 0,35 mol CaCO3 : 0,35 mol CO2 So the mass CaCO3 that reacted: m∴ m ∴ n = ​ __ ____ M ​ 0,35 = ​ 100 ​ m = (0,35)(100) = 35 g 5. Use a conical flask instead of a beaker and plug the mouth of the flask with cotton wool. The plug prevents the acid spray and water vapour from escaping from the flask. The CO2 gas can escape through the cotton wool. Escaping vapour from the beaker originally used will affect the change in mass readings and thus cause inaccurate measurement of the rate of reaction. The calcium carbonate must be placed in a small test tube that fits into the flask. When the flask is lightly shaken, the test tube Note: falls over and the acid reacts. This also prevents acid spray and In the calculation, n = 0,3454 water vapour from splashing out. Ensure that the stop watch is was rounded off to 0,35 so started when the two reactants come into contact and that it is that: stopped as soon as all the Mg has reacted. Mass = m(100) = 34,54 g 6. B. 68 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 2 Activity 2 1. Hydrogen gas may be prepared by the reaction of zinc metal with dilute hydrochloric acid. The chemical equation for this reaction is : Zn(s) + 2HCℓ(aq) → ZnCℓ2(aq) + H2(g) ∆H < 0 A learner determined the volume of the hydrogen produced with time at two temperatures and two grades of Zn; powder and solid pellets. He used the same mass of zinc and the same volume and concentration of hydrochloric acid for each experiment and plotted the following graphs. Which combination of temperature and surface area will be represented by A? (2) Temperature State of Zn A High Pellets B High Powder C Low Pellets C Low Powder 2. The graphs below represent the molecular distribution for a reaction at different temperatures. Which ONE of the graphs above represents the reaction at the highest temperature? (2) A P B Q C R D S [4] Solutions 1. B 33(2) 2. D 33(2) [4] © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 69 Mind the Gap Chemistry

2 Unit Activity 3 Give ONE term for each of the following descriptions by choosing a term from the list above. Write down only the term next to the question number. Surface Area; Catalyst; Elastic collision; Effective collision; Activated complex; Concentration; Temperature; Heat of reaction; Activation energy. 1. A chemical substance that speeds up the rate of a chemical reaction by lowering the net activation energy.  (1) 2. A collision in which the reacting particles have sufficient kinetic energy and correct orientation. (1) 3. The factor responsible for increasing the rate of a reaction when a solid is broken up into smaller pieces. (1) 4. The temporary unstable state that is formed during the course of a chemical reaction. (1) 5. A measure of the average kinetic energy of the particles in a gas.(1) 6. The net amount of energy released or absorbed during a chemical reaction. (1) [6] Solutions 1. Catalyst. 3 (1) 2. Effective collision. 3ATTENTION PAY SPECIAL (1) 3. NB Surface Area. 3 NB (1) 4. Activated complex. 3 (1) 5. Temperature. 3 (1) 6. Heat of reaction. 3 (1) HINT! [6] hint e.g. Worked example 5 EG - worked examples Learners use hydrochloric acid and a sodium thiosulphate (Na2S2O3) solution to investigate examsthe relationship between rate of reaction and temperature. The reaction that takes place is represented by the following equation: Na2S2O3(aq) + 2HCℓ(aq) → 2NaCℓ(aq) + S(s) + H2O(ℓ) + SO2(g) activity ACTIVITIES boy and girl-left and right of page activity 70 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015 Step by step comment comment  Mind the Gap Chemistry

HINT! hint Unit 2 e.g. Worked example 5 (continued) EG - worked examples They add 5 cm3 2 mol·dm–3 hydrochloric acid solution to 50 cm3 sodium thiosulphate solution in a flask placed examsover a cross drawn on a sheet of a white paper, as shown in the diagram below. The temperature of the mixture is 30°. They measure the time it takes for the cross to become invisible. The experiment is repeated with the temperature of the mixture at 40˚C, 50˚C and 60˚C. 1. Write down: 1.1 The possible hypothesis for this investigation. activity 1.2 NAME or FORMULA of the product that requires the need to work in a well-ventilated room. 1.3 NAME or FORMULA of the product that causes the cross to become invisible. ACTIVITIES boy 2. Apart from the volume of the reactants, stateand and girl-left ONE other variable that must be kept constant during this right of page investigation. 3. Why is itactivity advisable that the same learner observes the time that it takes for the cross to become invisible? 4. What experimental technique was used to measure the rate of reaction in this investigation? 5. The shown graph below is obtained from the results. Step by step comment comment 5.1 What is the relationship represented by ​  1 ___​on the vertical axis? time 5.2 What conclusion can be drawn from the results obtained? 6. The learners collected washed and dried sulphur. They weighed the dry sulphur and its mass was 0,12g. Calculate the percentage yield of sulphur. Solutions 1.1 E.g. The rate of reaction increase with increase in temperature. 1.2 Sulphur dioxide/SO2. 1.3 Sulphur/S. 2. Concentration of reactants (HCℓ and Na2S2O3). 3. Different people have different sight abilities/reaction times. (To ensure that the results are reliable). 4. Turbidity. 5.1 Reaction rate. 5.2 The reaction rate increases with increase in temperature. © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 71 Mind the Gap Chemistry

2 Unit 6. Na2S2O3(aq) + 2HCℓ(aq) → 2NaCℓ(aq) + S(s) + H2O(ℓ) + SO2(g) n(HCℓ) = CV n(HCℓ) = (2)(0,005) n(HCℓ) = 0,01 mol 2 n(HCℓ) = 1 n(S) n(S) = 0, 005 mol m PAY SPECIAL ATTENTION n = ​ __ NB M​ NB m 0,005 = ​ ___ 32 ​ m(s) = 0,16 g Percentage Yield = ​ ____ ( ) 0,12 ​  0,16 ​  ​ × 100 HINT! Percentage Yield = 75 (%) hint e.g. Worked example 6EG - worked examples A hydrogen peroxide solution dissociates slowly at room temperature according to the following exams equation: 2H2O2(aq) → 2H2O(ℓ) + O2(g) During an investigation, learners compare the effectiveness of three different catalysts on rate of decomposition of hydrogen peroxide. They place EQUAL AMOUNTS of sufficient hydrogen peroxide into three separate containers. They add EQUAL AMOUNTS of THREE catalyst, P, Q and R, to the hydrogen peroxide in the three containers respectively and measure the rate at which oxygen gas is produced. activitywrite down the: 1. For this investigation, ACTIVITIES boy 1.1 Independent variable. and girl-left and right of page 1.2 Dependent variable. activity Step by step comment comment 2. Which catalyst is the most effective? Give a reason for your answer. 72 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 2 e.g. Worked example 6 (continued) EG - worked examples 3. Fully explain, by referring to the collision theory, how the catalyst increases the rate of reaction. exams In another experiment, the learners obtain the following results for the decomposition of hydrogen peroxide: TIME(s) H2O2 CONCENTRATED (mol·dm–3) 0 0,0200 activity 200 0,0160 400 ACTIVITIES boy 0,0131 and girl-left and 600 right of page 0,0106 activity 800 0,0086 4. Calculate the AVERAGE rate of decomposition (in mol·dm–3) of H2O2(aq) in the first 400s. 5. Will the rate of decomposition at 600 s be GREATER THAN, LESS THAN or EQUAL TO the rate Step by step comment calculated in QUESTION comment 4.4? Give a reason for the answer. 6. Calculate the mass of oxygen produced in the first 600 s if the 50 cm3 of hydrogen peroxide decomposes in this time interval. Solutions 1.1 (Type of) catalyst. 1.2 Rate (of reaction). 2. R. It has the fastest reaction rate./ steepest gradient or slope./ Produces oxygen in a fastest or in shorter time. (Any of these reasons) 3. A catalyst provides an alternative reaction path with a lower activation energy. More molecules have sufficient/enough kinetic energy. OR More molecules have kinetic energy equal to or greater than activation energy. More effective collisions per unit time./Rate of effective collisions increases. ∆[H2O2] 4. Average rate = ​ ______ ∆t ​ ( 0,0131 – 0,020 ) ______________ = ​    400-0 ​ = 1,73 × 10–5 mol·dm–3·s–1 5. Less than. The concentration of hydrogen peroxide is decreasing as the reaction proceeds. © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 73 Mind the Gap Chemistry

2 Unit 6. OPTION 1 n c = __ ​ V ​ n (0,0200 – 0,0106) = _________ ​ (50 × 10–3) ​ 1 1 n(O2) = __ ​ 2 ​n(H2O2) = ​ __ 2 ​ (4,7 × 10 ) –4 n(O2) = 2,35 × 10–4 mol m n = __ ​ M ​ m 2,35 × 10–4 mol = ___ ​ 32 ​ m(O2) = 0,008g ≈ 0,01g OPTION 2 ∆c(H2O2) = (0,0200 – 0,0106) = 0,0094 1 1 ∆c(O2) = __ ​ 2 ​n(H2O2) = ​ __ 2 ​(0,0094) PAY SPECIAL ATTENTION NB = 4,7 × 10–4 NB m c = ___ ​ MV ​ m(O2) = cMV m(O2) = (0,0047)(3)(50 × 10–3) HINT! m(O2) = 0,008g ≈ 0,01g hint e.g. Worked example 7EG - worked examples The apparatus shown below is used to investigate the rate at which hydrogen gas examswhen a 6,5 g of zinc reacts with 30 ml of 1mol·dm HCℓ. produced –3 activity ACTIVITIES boy and girl-left and right of page The reactionactivity that takes place is represented by the following balanced equation: 2HCℓ(aq) + Zn(s) → ZnCℓ(aq) + H2(g) 1. What experimental method is used to measure the reaction rate in this experiment? 2. What other experimental Step by stepmethod can be comment used to measure comment the (reaction) rate for this reaction? 3. Determine the limiting reagent (using calculations). 4. Write down: 4.1 The name of the flask labelled Z. 4.2 ONE function of the item of apparatus labelled Y in THIS investigation. 74 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 2 e.g. Worked example 7 (continued) EG - worked examples Two experiments are conducted using apparatus above. The conditions for each experiment are given below. exams Experiment 1 Experiment 2 Zinc powdered powdered Hydrochloric acid 2 mol·dm–3 2 mol·dm–3 Temperature 25˚C 40˚C activity 5. FROM THE TABLE ABOVE, write down: ACTIVITIES boy and girl-left and 5.1 The independent variable for this investigation. right of page 5.2 ONE controlled variable. activity The volume of hydrogen gas produced is measured in each experiment. The graphs below show the results obtained. Step by step comment comment 6. Which graph, P or Q, represents Experiment 2? Refer to the data given in the table, as well as the shape of the graph, to explain how you arrived at the answer. 7. Give a reason why the rate of hydrogen production slows down towards the end in both experiments. 8. Calculate the mass of zinc used to prepare 0,024 dm3 of hydrogen of hydrogen at room temperature. Assume that 1 mole of hydrogen gas has a volume of 24,04 dm3 at room temperature. Solutions 1. Measuring the volume of a gas produced. 2. Measuring change in mass of reactants. n 3. n = ​ __ M​ 3,25 n = ​ ____ 65 ​ n = 0,05 mol Zn + 2HCℓ → ZnCℓ2 + 2H2 ( 30 ) n = CV = (2)​ _____ ​  1000 ​  ​ n = 0,06 mol Therefore the limiting agent is HCℓ 4.1 Conical/Erlenmeyer (flask). 4.2 Measure the volume of gas produced./ Collect the gas produced. 5.1 Temperature (of reactants). © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 75 Mind the Gap Chemistry

HINT! 2 Unit hint e.g. Worked example 7 (continued) EG - worked examples 5.2 Concentration (of HCℓ) 6.exams P. Experiment 2 has the highest temperature. The same volume of a gas was produced in a shortest period of time. OR Experiment 2 has the highest temperature. activity Graph has the steepest slope. 7. The concentration of reactants decreases as the reaction ACTIVITIES boy proceeds. and girl-left and right of page 8. activity Calculations Mark Allocation V n(H2) = ​ ___ Vm ​ • Substitute volume. 3 ( n(H2) = ​ _____ 0,24 ) ​  24,01 ​  ​ • Substitute molar volume. 3 n(H2) = 0,01 mol Step by step comment• comment 0,01mol 3 n n(H2) = ​ __ M​ • n(Zn) = n(H2) 3 m 0,01 = ​ ___ 65 ​ m = 0,65 g • Substitute 65 g•mol–1 3 • Answer. 3 OR 1 mol H2 gas = 24,04 dm3 0,01 mol H2 gas = 0,24 dm3 65 g Zn = 1 mol 0,65 g = 0,01 mol 76 UNIT 2 RATE AND EXTENT OF REACTIONS  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 2 Activity 4 81,1 g of nicotine consists of 60,07 g of carbon, 14,01 g of nitrogen and 7,02 g of hydrogen. Determine the: 1. The empirical formula of nicotine. (13) 2. Molecular formula of nicotine if its molar mass is 162,26g. (4) [17] Solutions 1. Determining the percentage composition by molar mass of each element: 60,07 C = ​ _____ 81,1 ​ × 100 = 74,04% 3 14,01 N = _____ ​  81,1 ​ × 100 = 17,27% 3 7,02 H = ​ ____ 81,1 ​ × 100 = 8,65% 3 m m Element g per 100g n = __ ​ M ​ n = __ ​ M ​ Simplest for 100g for 81,1g Ratio 74,04 n = _____ ​  12 ​ 3 60,07 _____ C 74,07 ​  12 ​= 5 3 5 = 6,17 3 17,27 n = _____ ​  14 ​ 3 14,01 _____ N 17,28 ​  14 ​= 1 3 1 = 1,23 3 8,65 n = ​ ____ 1 ​3 7,02 ____ H 8,65 ​  1 ​= 7 3 7 = 8,65 3 Empirical formula of nicotine is C5NH7 (13) 2. Now calculate the molar mass of the substance using empirical formula M(C5NH7) = 81 g/mol 3 x(81) = 162,2 3 x = 2 (meaning every atom in the empirical formula must be multiplied by 2) 3 Molecular formula of nicotine is C10N2H14 3 (4) [17]  © Department of Basic Education 2015 UNIT 2 RATE AND EXTENT OF REACTIONS 77 Mind the Gap Chemistry

3 Unit Chemical equilibrium When a chemical reaction takes place, the atoms or ions in the reactants form new bonds and new products with different chemical formulas and different chemical properties. 3.1 Summary 3.2 Key concepts In this section, we will be revising the basic concepts involved in chemical equilibria (plural of equilibrium). In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time. Usually, this state results when the forward reaction proceeds (continues) at the same rate as the reverse reaction. NB. This does not mean the reaction has stopped; instead it means that it is continuing to react, but the reactants and products are being produced (and re-produced) at a constant rate. 78 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 3 HOMOGENEOUS REACTIONS • are reactions where all the reactants and products are in the same phase, • i.e. all are solids, or all are liquids, or all are gases SO2(g) + O2(g) ⇌ SO3(g) HETEROGENEOUS REACTIONS • are reactions where not all species in the reaction vessel are in the same phase. CaCO3(s) + HCℓ(aq) → CaCℓ2(aq) + CO2(g) + H2O(ℓ) Most chemical reactions are non-reversible or irreversible – this means that the products cannot change back into the reactants. These reactions are essentially like baking. The ingredients, acting as the reactants, are mixed and baked together to form a cake, which acts as the product. This cake cannot be converted back to the reactants (the eggs, flour, etc.), just as the products in an irreversible reaction cannot convert back into the reactants (without e.g. human intervention). An example of an irreversible reaction is combustion. When carbon burns in oxygen to form carbon dioxide, the reaction is said to be non-reversible. C(s) + O2(g) → CO2(g) Pay special attention A non-reversible or irreversible reaction has a single arrow, indicating the products cannot form reactants during the chemical reaction. Combustion, for example, involves burning an organic compound—such as a hydrocarbon, in oxygen, to produce carbon dioxide and water. Because water and carbon dioxide are stable, they do not react with each other to form further reactants unless acted on by a strong external influence such as a lightning strike, an electrolytic cell set up by a human, etc. Some chemical reactions are reversible. The products react with each other to form the original reactants under suitable conditions. (E.g. change PAY SPECIAL ATTENTION in pressure, temperature, reactant concentration, etc.). NB N2(g) + 3 H2(g) ⇋ 2 NH3(g) HINT! hint A reversible reaction has a double arrow e.g. EG - worked examples exams © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 79 Mind the Gap Chemistry

3 Unit Therefore: Forward reaction Reverse reaction N2(g) + 3 H2(g) → 2 NH3(g) Reaction 2 NH3(g) → N2(g) + 3 H2(g) N2(g) + 3 H2(g) Reactants 2 NH3(g) 2 NH3(g) Products N2(g) + 3 H2(g) Learn these definitions. They are used to explain PAY SPECIAL ATTENTION what happens during a NB NB chemical reaction. DEFINITIONS • Open chemical system: reactants / products can escape from reaction vessel • Closed chemical system: reactants / products cannot escape from reaction vessel HINT! • Macroscopic changes: measurable or visible changes e.g. changes in colour, hint temperature, pressure, volume, concentration • Yield: the amount of product formed during a chemical reaction • An open system continuously interacts with its environment (conditions are not e.g. kept constant). EG - worked examples • A closed system is isolated from its surroundings (conditions are kept constant) • A reversible reaction occurs when the products can be converted back to reactants exams • Chemical equilibrium is dynamic when the rate of the forward reaction equals the rate of the reverse reaction • The concentration of a solution or of a gas is the number of moles of substance per unit volume (per dm3). • A dynamic chemical equilibrium is established in a closed system when the forward and reverse reactions of a reversible chemical reaction take place at the activity same time (simultaneously) and at the same rate. The conditions in the system PAY SPECIAL ATTENTION must also remain constant. ACTIVITIES boy and girl-left and NB NB right of page activity c = _​ n ​ m​ n = ​ __ ​ m ​ C = __ V where M OR MV HINT! hint c : concentration (mol·dm–3) Step by step comment comment You need to know the V : volume (dm3) formula, unit and symbols M : Molar mass (g·mol–1) e.g. used to calculateEGthe - worked examples concentration of a solution n : number of moles (mol) m : mass (g) exams activity ACTIVITIES boy and girl-left and right of page activity 80 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 3 AL ATTENTION NB You need to understand what a chemical equilibrium is and know the conditions needed for a chemical equilibrium. Chemical equilibrium • When a chemical system reaches dynamic chemical equilibrium, both the forward and the reverse reactions continue – they do NOT stop. • There are no macroscopic (visible with the eye) changes when the system is in ed examples equilibrium, but on a microscopic (atomic) level, the changes (reactions) continue. • Neither the reactants nor the products are ever used up in an equilibrium system (incomplete reaction – i.e. “incomplete” in the sense that there are still species of the original/starting species (reactants.)) ACTIVITIES boy and girl-left and right of page p comment comment Activity 1 Write the name of a reaction in which all reactants and products are in the same phase  (1) [1] Solution Homogeneous 3 [1] © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 81 Mind the Gap Chemistry

3 Unit 3.3 Factors that influence chemical equilibrium position Temperature (T): Increasing or decreasing the temperature of the equilibrium mixture Concentration (c): Adding a substance to or removing a substance from the system at constant volume. This only applies to solutions (aq) and to gases (g). Pressure (p): Increasing or decreasing the pressure on an equilibrium mixture (this ONLY applies to gases). 3.4 PAY SPECIAL ATTENTION Le Châtelier’s Principle NB NB If a closed system is in a state of dynamic chemical equilibrium, and one Le Châtelier’s Principle states of the factors that influences the equilibrium (temperature, concentration that: or pressure) is changed: When the equilibrium inHINT! a • the equilibrium position will be disturbed; hint closed system is disturbed, • either the forward or the reverse reaction will be favoured; the system will re-instate a • the applied change will be opposed; new equilibrium by favouring • the reaction rates will change and eventually the reaction that will oppose • a new equilibrium will be established e.g. the disturbance. EG - worked examples • so that the forward and the reverse reactions again take place at the PAY SPECIAL ATTENTION same rate. NB NB exams 3.4.1 Applying Le Châtelier’s Principle to HINT! changes in equilibrium conditions hint Changing the temperature of the reaction mixture (ΔT) If any factor changes, we can Increasing the temperature of the Decreasing the temperature of the reaction system (i.e. heating it) reaction system (i.e. cooling it) activity apply Le Châtelier’s Principle e.g. EG - worked examples to determine which changes • favours the reaction that ACTIVITIES boy • favours the reaction that increases the temperature of will happen to the system in decreases and girl-left and the temperature of equilibrium (what upsets the theof page right system (absorbs heat) the system (releases heat) balance). • ∴ favours the endothermic • ∴ favours the exothermic exams activity reaction. reaction. Changing the concentration of a solution or gas reactant (∆c) Increasing the concentrationof a Decreasing the concentration of a Step by step comment comment (g) or (aq) reactant (or product) (g) or (aq) reactant (or product) • favours the reaction that • favours the reaction that activity decreases the concentration of increases the concentration of that substance ACTIVITIES boy that substance and girl-left and ∴ right of•page the reaction that uses that • ∴ the reaction that makes that substance as a reactant. substance as a product. activity Step by step comment comment 82 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 3 Changing the pressure of a gaseous system (∆p) Increasing the pressure (or Decreasing the pressure (or decreasing the volume) of a increasing the volume) of a gaseous system gaseous system • favours the reaction that • favours the reaction that decreases the pressure of the increases the pressure of the PAY SPECIAL ATTENTION system system NB • by decreasing the total number • by increasing the total number of gaseous moles in the system. of gaseous moles in the system. Steps to follow HINT! hint By closed we do not mean Follow these steps to predict the change in a closed chemical system in a closed container. Rather dynamic equilibrium: e.g. we mean a closedEG - worked examples system where the conditions are kept Check that the reaction: constant. Step 1: Is reversible (⇌) (info given must indicate it is a dynamic exams equilibrium). Step 2: Equation is balanced. Step 3: Count the number of gaseous (g) moles on each side of the ⇌ . Determine: Step 4: If the forward reaction is exothermic or endothermic activity Step 5: If the reverse reaction is exothermic or endothermic ACTIVITIES boy and girl-left and Step 6: The change (which factor?) disturbs the equilibrium right of page Decide: activity Step 7: According to Le Châtelier’s Principle, which reaction is favoured (forward or reverse)? (Follow the reasoning in the table above.) Step 8: How the concentrations of the reactants and products change. Step by step comment BX2Y2 is a solid. Solids in equilibrium reactions have very little change in mass, hence they are ignored © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 83 Mind the Gap Chemistry

3 Unit Activity 2 Consider the following reversible reaction that is in equilibrium in a closed system (the symbols represent chemicals which are unnamed; if you see symbols like X or Y which aren’t on the Periodic Table, it often means you’ve got to work out what kinds of substances they are). Step 1, 2 & 3: Step 4/5: ∆H < 0 - the forward reaction is exothermic - the reverse reaction is endothermic In each of the following cases: • Explain how the equilibrium will be affected (which reaction will be favoured) • State how the concentration of the product A2(g) will change. CASE 1 The reaction system is heated.  (3) [3] Solution Case 1. PAY SPECIAL ATTENTION If the reaction system is heated, the temperature of the system increases. NB NB According to Le Châtelier’s Principle, increasing the temperature of the reaction system (i.e. heating it) favours the reaction that will decrease the temperature of the system HINT! ∴ favours the endothermic 3 reaction but ∆H < 0 hint ∴ the reverse reaction is endothermic You must be able to use ∴ the reverse reaction is favoured3 Le Châtelier’s Principle to ∴ the 3A2(g) + BX2Y2(s) reacts faster than it is produced e.g. EG - worked examples identify the factor that will increase the yield of the ∴ the concentration of the A2(g) decreases. 3 products and what influence (Remember: The opposite is true if the reaction vessel is cooled.) that factor will have on the exams reaction AT EQUILIBRIUM [3] activity ACTIVITIES boy and girl-left and 84 right of page UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry activity

Unit 3 CASE 2 Y2(g) is added to the vessel at constant pressure and temperature (3) [3] Solution Case 2. Adding Y2(g) increases the concentration of the Y2(g). 3 According to Le Châtelier’s Principle, increasing the concentrationof Y2(g) favours the reaction that decreases the concentration of Y2(g) ∴ the reaction that uses of Y2(g) as a reactant the forward reaction is favoured. 3 the rate at which A2(g) and BX2Y2(s) is produced, increases the concentration of the A2(g) increases. 3 [3] CASE 3  A2(g) is removed from the vessel at constant pressure and temperature.  (3) [3] Solution Case 3. Removing A2(g) decreases the concentration of the A2(g). 3 According to Le Châtelier’s Principle, decreasing theconcentration of the A2(g) favours the reaction that increases the concentration of the A2(g); the reaction that makes A2(g) as a product ∴ the forward reaction is favoured. 3 the rate at which A2(g) and BX2Y2(s) is produced, increases the concentration of the A2(g) increases. 3 [3] CASE 4  Some of the BX2Y2(s) is removed from the system at constant pressure and temperature. (3) [3] Solution Case 4. Removing BX2Y2(s) has no effect on the equilibrium as it is a solid. 3 Removing a solid from an equilibrium system does not disturb the concentration. 3 (Solids in equilibrium reactions have very little change in mass, hence they are ignored) The concentration of A2(g) remain the same. 3 [3] © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 85 Mind the Gap Chemistry

3 Unit CASE 5  The pressure on the system is increased (or the volume is decreased)  (3) PAY SPECIAL ATTENTION [3] NB NB Solution Case 5. HINT! According to Le Châtelier’s Principle, increasing the pressure (or hint decreasing the volume) of a gaseous system favours the reaction that decreases the pressure of the system by decreasing the total number of There are 4 gaseous moles gaseous moles in the system. 3 on the left, and 3 gaseous e.g. EG - worked examples moles on the right. the forward reaction is favoured. 3 the rate at which A2(g) and BX2Y2(s) is produced, increases the concentration of the A2(g) increases. 3 [3] exams Activity 3 Say whether the statement is TRUE or FALSE: If the equilibrium constant for the reaction A2(g) + B2(g) ⇌ 2AB(g) is equal to K, then the activity equilibrium constant for the reverse reaction 2AB(g) ⇌ A2(g) + B2(g) is also equal to K. ACTIVITIES boy (3) and girl-left and right of page [3] activity Solution 3 FALSE. ... for the reverse reaction 2AB(g) ⇌ A2(g) + B2(g) is equal 1 to ​ __ K ​ 3 3 OR ... less (smaller) than K [3] Step by step comment comment Activity 4 PAY SPECIAL ATTENTION The industrial preparation of hydrogen gas is represented by the equation NB below: NB CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) ∆H > 0 The reaction reaches equilibrium at 1 000 °C in a closed container. HINT! hint 1. State Le Châtelier’s principle. (2) 2. How will an increase in pressure at 1 000 °C (by decreasing There are various ways to the volume) affect the yield of hydrogen gas? Write down state Le Châtelier’s Principle only INCREASES, DECREASES OR NO EFFECT. Explain the e.g. correctly. EG - worked examples answer.(2) 3. Give TWO reasons why high temperatures are used for this reaction.(2) exams [6] activity 86 UNIT 3 CHEMICAL EQUILIBRIUM ACTIVITIES boy © Department of Basic Education 2015 and girl-left and  right of page Mind the Gap Chemistry

Unit 3 Solution 1. When the equilibrium in a closed system is disturbed the system will shift the equilibrium position (OR re-instate a new equilibrium)3 as to favour the reaction that will oppose OR cancel OR counteract the change OR disturbance. 3 OR When a stress / change is placed on a system in equilibrium the system shifts the equilibrium (position) OR re-instate a new equilibrium 3 so as to remove OR cancel OR oppose the stress / change. 3 (2) OR When the conditions affecting an equilibrium are changed, the equilibrium (position) shifts in such a way 3as to oppose the change OR cancel the change. 3 (any two) 2. Decreases 3 When the pressure is increased, the reverse reaction is favoured.3 OR The reaction that produced the smaller volume/amount of gas is favoured. 3 PAY SPECIAL ATTENTION OR NB 4 mol or volumes of gas produces 3 2 mol or volumes of (2) gas. 3 (any two) HINT! 3. Products form at faster rate. 3 Higher yield of products. 3 (2) hint [6] In physical chemistry, saturation is the point 3.4.2 Dynamic chemical equilibrium in e.g. EG - worked examples at which a solution of a substance can dissolve no more of that substance. solutions Additional amounts of the A dynamic chemical equilibrium in a solution: exams solute will appear as a • is only possible if the solution is saturated separate phase (usually as a • is only affected by changing the precipitate solid). This point of • concentration of the ions in the solution or maximum concentration, the • temperature of the solution. saturation point, depends on the temperature and pressure The common ion effect of the solution as well as activity the chemical nature of the ACTIVITIES boy substances involved. Steps to follow and girl-left and right of page activity When a solution is added to a reaction system: Step 1: Identify the common ion in the solution and the system Step 2: Explain the effect of the common ion (addition) on the Step by step comment chemical equilibrium. © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 87 Mind the Gap Chemistry

3 Unit Solution added Effect on equilibrium mixture: to the equilibrium H+(aq) + OH−(aq) ⇌ H2O(ℓ) mixture: • Increases the hydrogen ion concentration, [H+] or Hydrochloric acid, hydronium ion concentration [H3O+] HCℓ(aq) • Increases the chloride ion concentration, [Cℓ−] • Decreases the hydroxyl ion concentration [OH−] • Increases the hydrogen ion concentration, [H+] or hydronium ion concentration [H3O+] Sulphuric acid, • Increases the sulphate ion concentration, [SO42− ] PAY SPECIAL ATTENTION H2SO4(aq) • Decreases the hydroxyl ion concentration [OH−] NB NB because: Concentrated • Extracts water (H2O) from the system (this is also sulphuric acid, known as dehydration) PAY SPECIAL ATTENTION H2SO4(ℓ) HINT! NB NB hint A base like NaOH(aq) or KOH(aq) • Increases the hydroxyl ion concentration [OH−] When two different solutions A solution containing • Increases the chloride ion concentration, [Cℓ−] contain the same ion, this ion Cℓ(aq) ions e.g. HINT! • Decreases the silver ion concentration in a e.g. EG - worked examples is called a common ion NaCℓ(aq) hint solution because Ag+(aq) + Cℓ−(aq) → AgCℓ(s) exams e.g. Worked example 1 EG - worked examples Consider the following equilibrium: exams CoCℓ42–(aq) + 6H2O(ℓ) ⇋ Co(H2O)62+(aq) + 4Cℓ–(aq) ΔH < 0 blue purple-pink pink activity When cobalt chloride crystals (CoCℓ2⋅6H2O) are dissolved in ethanol (the solvent) in a ACTIVITIES boy and girl-left and flask, a blue solution is formed with CoCℓ42– ions. By carefully adding water to the blue right of page solution until it just turns purple-pink because of the formation of the pink Co(H2O)62+ activity ions, the above equilibrium is reached. activity Five test tubes each contain a small amount of the above purple-pink equilibrium ACTIVITIES boy solution. Describe and explain the colourandchange observed in each of the following girl-left and instances: right of page Step by step comment comment activity 1. The solution is cooled down. Becomes pink. According to Le Châtelier’s Principle, a decrease in temperature favours the exothermic, forward reaction therefore increasing the [Co(H2O)62+(aq)] which is pink. Step by step comment comment 2. Water is added to the solution. Becomes pink. According to Le Châtelier’s Principle, adding H2O will favour the production of more Co(H2O)62+(aq) ions so that the number of H2O molecules will decrease, therefore the forward reaction is favoured. The concentration of the Co(H2O)62+(aq) ions which are pink, increases. 88 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 3 e.g. Worked example 1 (continued) EG - worked examples 3. A few NaCℓ crystals are added to the solution. exams Becomes blue. Adding NaCℓ to the equilibrium mixture, increases the [Cℓ–] as a result of the common ion (Cℓ–) found in the reaction mixture and in the NaCℓ. The reverse reaction which decreases the [Cℓ–(aq)], is favoured, thereby increasing the concentration of the CoCℓ42–(aq) ions which are blue. 4. Concentrated sulphuric acid is added to the solution. Becomesactivity blue. Concentrated sulphuric acid (H2SO4)ACTIVITIES is a good boy dehydrating agent therefore it will and girl-left and extract water molecules (H2O) from the mixture. The reverse reaction that produces right of page H2O molecules is favoured and at the same time increasing the concentration of the activity CoCℓ42–(aq) ions which are blue. 5. A few drops of silver nitrate solution are added to the solution. Becomes pink. When the silver Step nitrate solution is added to by step the mixture, comment the silver ions, Ag+(aq) comment react with the Cℓ (aq) to form a AgCℓ(s) precipitate, thereby decreasing the [Cℓ–]. – According to Le Châtelier’s Principle, the forward reaction that increases the [Cℓ–] ions is favoured. More pink Co(H2O)62+(aq) ions form. Activity 5 1. The following hypothetical reaction reaches equilibrium in a closed container at a certain temperature: X2(g) + Y2(g) ⇋ 2XY(g) ∆H < 0 Which ONE of the following changes will increase the AMOUNT of XY(g)? A. Decrease in temperature B. Increase in temperature C. Increase in pressure D. Decrease in pressure (2) 2. The equation below represents a chemical reaction at equilibrium in a closed container. H2(g) + I2(g) ⇌ 2HI(g) ∆H < 0 Which ONE of the following changes will increase the yield of HI(g) in the above reaction? A. Increase the temperature B. Decrease the temperature C. Increase the pressure by decreasing the volume D. Decrease the pressure by increasing the volume (2) © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 89 Mind the Gap Chemistry

3 Unit 3. A chemical reaction reaches equilibrium. Which ONE of the following statements regarding this equilibrium is TRUE? A. The concentrations of the individual reactants and products are constant. B. The concentrations of the individual reactants and products are equal. C. The concentrations of the individual reactants are zero. C. The concentrations of the individual products increase until the reaction stops. (2) 4. The reaction represented by the equation below reaches equilibrium. 2CrO24−(aq) + 2H+(aq) ⇌ Cr2O27−(aq) + H2O(ℓ) yellow orange Which ONE of the following changes to the reaction mixture will change its colour from yellow to orange? A. Add a catalyst. B. Add water to the reaction mixture. C. Add a few drops of sodium hydroxide solution to the reaction mixture. D. Add a few drops of concentrated hydrochloric acid to the reaction mixture. (2) 5. Give one word for the following phrase: The stage reached in a reversible chemical reaction when the rate of the forward reaction is equal to the rate of the reverse reaction. (2) [10] Solution 1. A33 (2) AL ATTENTION 2. B33 (2) NB 3. A33 (2) Solids and pure liquids 4. D33 (2) are omitted from the 5. (Dynamic/Chemical) equilibrium 3 3 (2) Kc expression asPAY their SPECIAL ATTENTION [10] NB concentration is [1], as NB multiplying by 1 has no effect. 3.5 The Equilibrium Constant (Kc) ed examples HINT! (The Law of Mass Action) hint The concentrations of all the compounds (solutions and gases) in a closed system in dynamic chemical equilibrium are related by a mathematical Kc: equilibrium constant equation. The numerical value of this equation is called the equilibrium (no unit) constant (Kc). e.g. EG - worked examples [substance]: concentration of In the hypothetical equation below the equilibrium expression for this reactant or product (in mol·dm–3) reaction is: mol: number of moles 2P + 3R ⇋ 2S + 4T exams of each compound in the balanced reaction equation. [S]2 × [T]4 Kc = ​ ________ [P]2 × [R]3 ​ ACTIVITIES boy and girl-left and right of page 90 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  activity Mind the Gap Chemistry

Unit 3 PAY SPECIAL ATTENTION NB NB • This is the correct way of writing the Kc expression • The coefficients are the moles of each reactant and product in the balanced equation HINT! • The product of the concentration of reactants (not to be added, but multiplied!!), hint raised to the power of the number of moles is the numerator • The product of the concentration of products (not to be added, but multiplied!!), raised to the power of the number of moles is the denominator e.g. • The concentrations EG used in the - worked Law of Mass Action is the [reactant]equilibrium examples and [product]equilibrium (NOT initial concentrations!! i.e. the concentrations of reactants and products at equilibrium) exams Activity 6 Write Kc expressions for each of the following reactions: 1. N2(g) + 3H2(g) ⇋ 2NH3(g) (1) activity 2. CaCO3(s) ⇋ CaO(s) + CO2(g) (1) 3. P4(s) + 6Cℓ2(g) ⇋ 4PCℓ3(ℓ) ACTIVITIES boy and girl-left and (1) right of page [3] Solution activity [NH3]2 1. Kc = ​ ________ [N ]2·[H ]3 ​3 (1) 2 2 2. Kc = [CO2] 3 (1) 1 comment 3. Kc = ​ _____ Step by step comment [Cℓ2]6 ​ 3 (1) [3] 3.5.1 Factors that influence the equilibrium constant For a closed system in equilibrium: • Only the concentrations of • The value of Kc DOES change if aqueous solutions (aq) and gases the temperature of the system PAY SPECIAL ATTENTION NB(g) appear in the Kc expression; changes. NB • Solids (s) and pure liquids (ℓ) are NOT included in the Kc expression. The value of Kc DOES NOT change if: HINT! • the pressure in the system hint You need to know the: changes; • the concentration of a reactant or • conditions and product in the system changes; • factors • a catalyst is added to the system. e.g. that play EG - a role examples worked in HOW/WHEN Kc changes/does NOT change exams © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 91 Mind the Gap Chemistry

3 Unit a) The effect of a change in temperature on Kc If the temperature Kc for a specific reaction is constant at a specific temperature. changes, the Kc value for the reaction changes. If the change in the temperature of a closed equilibrium system: favours the forward reaction: favours the reverse reaction: • more products form; • more reactants form; • [products] increases; • [products] decreases; • [reactants] decreases and • [reactants] increases and • Kc increases • Kc decreases 3.5.2 The meaning of Kc-values The significance of the Kc value is important in industrial processes. The economic viability of an industrial process in the chemical industry depends on the manufacturing costs, the product yield (amount of product produced), and the retail cost. These factors determine the profit a company would make. Kc values are used to determine if the possible yield at a specific temperature is low or high. AL ATTENTION NB If Kc is small (Kc < 1) the: If Kc is large (Kc > 1) the: • equilibrium concentrations • equilibrium concentrations You must understand the of the products are relatively of the products are relatively meaning of Kc. low compared to those of the high compared to those of the reactants reactants The value of Kc is directly • product yield is low • product yield is high linked to the yield of the • industrial process is not • industrial process may be reaction. economically viable. economically viable. ed examples 3.5.3 Calculating the value of Kc for a closed system in dynamic equilibrium There are THREE types of calculations that you may be asked to perform. They will be covered in the following pages. Work through ACTIVITIES boy the given examples to and girl-left and make sure that you know right of page and understand the steps given to calculate the value of Kc for a specific equilibrium system. p comment comment 92 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 3 PAY SPECIAL ATTENTION HINT! NB i) Type 1: The initial amount (mol) of each reactant is hint given • UseHINT! a table e.g. Worked example 2 EG - worked examples hint (as indicated) to show each of the 1. Consider the following balanced reaction for the formation of sulphur trioxide, in the calculation steps when calculating second step of the Contact Process: exams 2SO2(g) + O2(g) ⇋ 2SO3(g) e.g. EG - worked examples the number of mol of each Exactly 8 mol SO2 and 10 mol O2 are sealed in a 500 cm3 reaction vessel which is compound at initially empty. The compounds react to produce SO3. At equilibrium, the vessel equilibrium. contains 6 mol SO3. exams • DO NOT use 1.1 Calculate the equilibrium concentrations of each of the compounds in the abbreviations such reaction vessel. activity as I, C and E in your table. Solution ACTIVITIES boy and girl-left and • Clearly indicate if right of page The Contact you work with mol activity or concentration activity Process is a three step process in the manufacturing of sulphuric acid. of each substance ACTIVITIES boy and girl-left and at equilibrium in right of page Step 1: the table. Always start by drawing a table with a column for the descriptions and activity a column for each of the reactants and products. Fill in the formulas of the reactantsStep and by step products andcomment comment the number of moles of each in the balanced reaction equation in the top line. The table should always consist of 5 rows with names as indicated: Step by step comment c Reactants Products balanced equation 2SO2(g) O2(g) 2 SO3(g) initial mol Always convert mol used or produced volume to dm3: final mol at equilibrium 500 cm3 = 500 ÷ 1000 final [ ] at equilibrium = 0,5 dm3 n c = ​ __ V​ © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 93 Mind the Gap Chemistry

HINT! 3 Unit hint e.g. Worked example 2 (continued) EG - worked examples Reactants Products exams balanced equation 2SO2(g) O2(g) 2 SO3(g) initial mol mol used or produced activity final mol at equilibrium ACTIVITIES boy final [ ] at equilibrium and girl-left and n c = ​ __ V​ right of page activity Step 2: Now fill in all the values given in the question. If only reactants were added at the beginning of the reaction, the initial mol of the products is zero. Step by step comment comment balanced equation 2SO2(g) O2(g) 2SO3(g) initial mol 8 10 0 mol used / produced final mol at equilibrium 6 n n n n n n final [ ] at equilibrium c = ​ __ ___ V ​ = ​  0,5 ​ c = ​ __ ___ V ​ = ​  0,5 ​ c = ​ __ ___ V ​ = ​  0,5 ​ Step 3: Find the column with 2 values. Complete the mol used or produced line for this compound. balanced equation 2SO2(g) O2(g) 2SO3(g) initial mol 8 10 0 mol used / produced 6 final mol at equilibrium 6 n n n n n n final [ ] at equilibrium c = __ ​ V ​ = ​ ___ 0,5 ​ c = ​ __ ___ V ​ = ​  0,5 ​ c = ​ __ ___ V ​ = ​  0,5 ​ Step 4: Compare the ratio of moles according to which the reactants react and to which the products are produced, according to the balanced reaction equation: SO2 : O2 : SO3 = 2 : 1 : 2 Consider the actual mol used or produced for the compound which is known. We have just calculated that 6 mol SO3 has been produced. This means that the ratio of SO2 : O2 : SO3 = 2 : 1 : 2 (balanced equation) =6 :3:6 94 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 3 e.g. Worked example 2 (continued) EG - worked examples Fill in these values. exams balanced equation 2SO2(g) 1 O2(g) 2SO3(g) initial mol 8 10 0 mol used / produced 6 3 6 final mol at equilibrium 6 n n n ___ n n ___ n activity final [ ] at equilibrium c = __ ​ V ​ = ​ ___ 0,5 ​ c = ​ __ V ​ = ​  0,5 ​ c = ​ __ V ​ = ​  0,5 ​ ACTIVITIES boy Step 5: and girl-left and right of page For the reactants: initial mol – mol used = final mol at equilibrium. activity For the products: initial mol + mol produced = final mol at equilibrium. balanced equation 2SO2(g) O2(g) 2SO3(g) initial mol 8 10 0 mol used / produced Step by step 6 comment 3 comment 6 final mol at equilibrium 8–6=2 10 – 3 = 7 6 n n n n 6 final [ ] at equilibrium c = __ ​ V ​ = ​ ___ 0,5 ​ c = ​ __ ___ V ​ = ​  0,5 ​ ___ ​  0,5 ​= 12 Step 6: Substitute the final mol values in the last line, and calculate the final concentrations. These are the answers to Question 1.1 above. balanced equation 2SO2(g) O2(g) 2SO3(g) initial mol 8 10 0 mol used / produced 6 PAY SPECIAL ATTENTION 3 6 NB NB final mol at equilibrium 2 7 6 2 ___ 7 ___ final [ ] at equilibrium ​  0,5 ​= 4 ​  0,5 ​= 14 12 ∴ at equilibrium: HINT! hint [SO2 (g)] = 4 mol·dm and [O2 (g)] = 14 mol·dm and [SO3 (g)] = 12 mol·dm −3 −3 −3 e.g. Worked example 3 EG - worked examples 1.2 Write an expression for the equilibrium constant for the forward reaction and calculate Kc for the forward reaction at this temperature. exams Solution [SO3]2 122 1. Kc = ​ ________ ​ = ​ _____​= 0,64 [SO ]2·[O ] 42.14 2 2 activity © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 95 ACTIVITIES boy Mind the Gap Chemistry and girl-left and right of page

NB NB 3 Unit HINT! ii) Type 2: The initial amount (mol) of one reactant and the hint concentration of a reactant or product are given e.g. Worked example 4 EG - worked examples 2 mol of NO2(g) and an unknown amount of N2O4(g) are sealed in a 2 dm3 container, examsthat is fitted with a plunger, at a certain temperature. The following reaction takes place: 2NO2(g) ⇋ N2O4(g) At equilibrium it is found that the NO2 concentration is 0,4 mol·dm–3. The equilibrium constant (Kc) at this temperature is 2. 1. Calculate the initial amount (in mol) of N2O4(g) that was sealed in the container. Solution activity ACTIVITIES boy and girl-left and Step 1: right of page Draw the table as before. Fill in the formulas of the reactants and products activity and the number of moles of each in the balanced reaction equation in the top line. Now fill in all the values given in the question. If only reactants were added at the beginning of the reaction, the initial mol of the products is zero. When the initial Step by step amount (mol) commentor product is NOT of a reactant comment zero and the value is not given (in this case the amount of N2O4) is not given, let this amount equal x. Write the statement regarding ‘x’ down, “Let …. be x”. Let the initial amount of N2O4 in the reaction vessel be x. A minus sign balanced equation 2 NO2 N2O4 is used for the reactants (in this case), because the initial mol 2 number of moles of the reactants decreases as they are mol used / produced used up. A plus sign is used for the final mol at equilibrium products (in this case), because final [ ] at equilibrium the number of moles of the n 0,4 products increases as c = ​ __ V​ they are formed. Step 2: n Use the concentration given and the formula for concentration (c = ​ __ V ​) to calculate the number of moles of NO2 in the equilibrium mixture. The volume of the container (2 dm3) was given. Show the calculation and substitute the calculated value in the table. n n c = __ ​ V ​ ∴ 0,4 = ​ __ ∴ 2 ​ n = (2)(0,4) = 0,8 mol Step 3: Find the column with Find the column with 2 values. Complete the mol used or produced line for this compound. 2 mol NO2 initially – ? mol NO2 used = 0,8 mol ∴ mol NO2 initial = 1,2 mol 96 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 3 e.g. Worked example 4 (continued) EG - worked examples Step 4: From the balanced equation ratio 2 : 1 determine the ratio of the exams moles used and produced. Therefore 2 : 1 becomes 1,2 (as calculated in Step 3): 0,6 Step 5: For the reactants: initial mol – mol used = final mol at equilibrium. For the products: initial mol + mol produced = final mol at equilibrium. Step 6: activity Substitute the final mol values into theboy ACTIVITIES last line, and calculate the final concentration of the N2O4. and girl-left and right of page activity 2NO2 N2O4 balanced equation 2 1,2 0,6 initial mol (Step 3) (Step 4) Step by step comment comment 0,8 + 0,6 mol used / produced (Step 2) (Step 5) final mol at equilibrium 0,4 ( x + 0,6 ) ​ ______ ​  2 ​  ​ (Step 6) Step 7: Write the expression for Kc and substitute all the known values – remember that the value of Kc was given. Solve for x. [N2O4] _____ ∴ ( x + 0,6 ) ​ ______ ​  2 ​  ​ ______ Kc = ​ [NO ]2 ​ 2 = ​  0,42 ​ 2 ∴ (2)(0,42) = ​ x + 0,6 ______ 2 ​ ∴ x = 0,4 mol The plunger is now pushed into the container and the volume of the container decreases. 2. How will this change influence the amount of nitrogen dioxide at equilibrium? Only write down increases, decreases or remains the same. Solution Decreases 3. Use Le Châtelier’s principle to explain your answer to Question 2.2. Solution When the volume of the container decreases, the pressure increases. According to Le Châtelier’s Principle the reaction that decreases the pressure will be favoured ∴ the (forward) reaction that produces fewer gas moles, 2 mol (g) ⇌ 1 mol (g), is favoured. © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 97 Mind the Gap Chemistry

3 Unit iii) Type 3: The initial mass of a reactant and the value of Kc are given and the mass of the reactant that remains unreacted, is asked Activity 7 The thermal decomposition of calcium carbonate (CaCO3) is an example of a heterogeneous equilibrium. The decomposition that takes place in a closed container can be represented by the following equation: CaCO3(s) ⇌ CaO(s) + CO2(g) Initially 5 g of CaCO3(s) is placed in a closed 500 cm3 container and then heated. Equilibrium is reached at 900°C. 1. Why is the above decomposition referred to as a heterogeneous equilibrium? (1) 2. Calculate the mass of unreacted CaCO3(s) that remains in the container at equilibrium if Kc for the reaction is 0,0108 at 900°C. (19) 3. It is found that the value of Kc increases when the container is heated. Is the forward reaction exothermic or endothermic? Use Le Châtelier’s principle to explain your answer. (5) 4. The volume of the container is now decreased to 250 cm3 while the temperature is kept constant.  How will each of the following be affected? Give a reason for your answer. 3.4.1 The value of Kc. 3.4.2 The number of mol CaCO3(s) present in the equilibrium mixture. (10) 3.4.3 The concentration of CO2(g) at the new equilibrium. 5. More CaCO3(s) is now added to the equilibrium mixture in the 500 cm3 container. PAY SPECIAL ATTENTION NB NBthis change influence the number of moles of CO2(g) How will in the equilibrium mixture? Give a reason for your answer. (2) [37] HINT! Solutions hint 1. Reactants and products are in different phases. 3 (1) Refer back to the definition of 2. This answer takes six steps. Answers for questions 3.3 to 3.5 a heterogeneous equilibrium appear below these steps. e.g. EG - worked examples Step 1: Convert the mass of the reactant to moles. ONLY moles may be used in exams the table. M(CaCO3) = 40 + 12 + 3(16) = 100 g·mol–1 3 m 5 n = ​ __ ____ M ​ = ​  100 ​= 0,05 mol 3 activity ACTIVITIES boy and girl-left and right of page 98 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015 activity  Mind the Gap Chemistry

PAY SPECIAL ATTENTION NB Unit 3 Solutions to Activity 7 (continued) HINT! Step 2: hint Write the expression for Kc and substitute all the known values – • Convert cm3 to dm3: remember that the value of Kc was given. • 500 cm3 = 500 ÷ 1000 Kc = [CO2] 3 e.g. = 0,5 dm3 . EG - worked examples Note: cm3 = mL, dm3 = L. 0,0108 = [CO2] It’s probably easier to ∴ [CO2] = 0,0108 mol·dm–3 3 remember e.g. that if exams 500 cm3 = 500 mL, Step 3: it’s the same as Use the calculated CO2(g) concentration and the formula for 0,5 L = 0.5 dm3 n concentration (c = __ ​ V )​ to calculate the number of moles of CO2 in the • Convert mass to mol equilibrium mixture. The volume of the container (500 cm3) was given. ​ m ​ using the formula: n = __ M Convert the volume to dm3. Show all working. n n activity c = ​ __ V​ ∴ 0,0108 = ​ ___ ∴ 0,5 ​ n = (0,5)(0,0108) = 0,0054 mol CO2 ACTIVITIES boy and girl-left and Pay special attention: right of page 3 3 Solids (s) are not included Step 4: activity in the Kc expression Find the column with 2 values. Complete the mol used or produced line ∴ only CO2 appears for this compound. (0 mol CO2 initially + produced mol CO2 = 0,0054 mol in Kc. 3 ∴ mol CO2 produced = 0,0054 mol) 3 Step 5: Step by step comment From the balanced equation ratio 1 : 1 : 1 determine the ratio of the moles used and produced. Therefore 1 : 1 : 1 becomes 0,0054 (from Step 3): 0,0054 : 0,0054 3 Step 6: 3 For the reactant CaCO3 ... initial mol – mol used = final mol at equilibrium ∴ 0,05 mol – 0,0054 = final mol at equilibrium = 0,0046 mol 3 3 balanced equation CaCO3 CaO CO2 initial mol 0,05 0 0 (Step 1) (Given) (Given) 3 mol used / 0,0054 0,0054 0,0054 produced (Step 5) (Step 5) (Step 4) 3 final mol at 0,05 – 0,0054 0,0054 0,0054 PAY SPECIAL ATTENTION equilibrium = 0,0046 (Step 3) (Step 3) 3 NB (Step 6) __ n n Equilibrium not asked not asked ​  V ​ = ​ ___ 0,5 ​ concentration = 0,0108 3 n (mol·dm–3) c = ​ __ V​ (Step 2) HINT! hint ∴ 0,0046 mol CaCO3 remains unreacted at equilibrium • Mass was asked, so M(CaCO3) = 40 + 12 + 3(16) = 100 g·mol–1 convert the mol calculated 3 m m e.g. __m to mass EG - worked examples n = ​ __ M​ ∴ 0,0046 = ​ ____ ∴ 100 ​ m = (0,0046)(10) = 0,46 g 3 • n = ​  ​ M ∴ 0,46g CaCO3 remains unreacted. 3 (19) exams © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 99 Mind the Gap Chemistry

3 Unit Solutions to Activity 7 (continued) 3. Endothermic. 3 Kc increases ∴ the forward reaction or reaction to the right is favoured. 3 3 According to Le Châtelier’s principle, an increase in temperature 3 will favour the endothermic reaction the forward reaction or reaction to the right must be endothermic. 3 (5) 3 3 4.1 Remains the same. Only temperature affects the value of Kc. 4.2 Increases. 3 3 Decreasing the volume increases the pressure. According to Le Châtelier’s Principle, when the pressure is increased, the reaction which decreases the pressure by producing fewer moles of gas products, is favoured. 3 As 0 mol (g) ∴ 1 mol (g), the reverse reaction is favoured. 3 4.3 Remains the same. 3 3 Kc = [CO2]. The temperature remains constant ∴ Kc remains constant ∴ the concentration of CO2 will remain constant. 3 (10) 3 5. Remains the same. Adding a solid does not affect the equilibrium, it does not change the concentration of a reagent. 3 3 (2) [37] Activity 8 PAY SPECIAL ATTENTION NB NB represents an equilibrium reaction in a sealed 1dm3 The equation below container. • When ∆H > 0 it means the reaction to the right NO2(g) + NO(g) ⇋ N2O(g) + O2(g) ΔH > 0 (the forward reaction)HINT! is Equilibrium was reached at a certain temperature and the value of Kc was hint PAY SPECIAL ATTENTION 3,93. The concentration (in mol·dm–3) of each reactant and product in the endothermic •NB When ∆H < 0 the forward NBequilibrium was: container at reaction is exothermic [NO2] = 0,06 mol·dm–3 e.g. EG - worked examples [NO] = 0,29 mol·dm–3 HINT! [N2O] = 0,18 mol·dm–3 hint [O2] = 0,38 mol·dm–3 These exams are the initial One of the conditions affecting the equilibrium is changed and a new concentrations – before equilibrium is established. At the new equilibrium, the concentration of e.g. the new equilibrium established. V = 1 dm3 EG - worked examples is the NO2(g) is 0,12mol·dm–3. (given) 1. Calculate the Kc value at the new equilibrium. (11) ∴ the values of the initial 2. Which condition, concentration or temperature, was changed? (1) exams activity concentrations are equal to 3. Give an explanation for the answer to Question 4.2. (1) the initial number of moles. ACTIVITIES boy [13] c = ​ _n ​and c = _​ n ​ and girl-left and V 1 right of page ∴ n = c(1) activity activity 100 UNIT 3 CHEMICAL EQUILIBRIUM  ACTIVITIES boy © Department of Basic Education 2015  and girl-left and Mind the Gap Chemistry Step by step right of page comment comment

Unit 3 Solutions 1. NO2(g) + NO(g) + Energy ⇌ N2O(g) + O2(g) 3 3 ∆ [NO2] = [NO2]final − [NO2]initial = (0,12 – 0,06) = 0,06 mol NO2 is formed. but NO2 is the product of the reverse reaction ∴ the reverse reaction was favoured. 3 From the balanced reaction equation, 1 NO2 : 1 NO : 1 N2O : 1 O2 Kc has NO ∴ 0,06 mol N2O + 0,06 mol O2 → 0,06 mol NO2 + 0,06 mol NO 3 UNIT! So at equilibrium: NO2 NO N2O O2 Initial number 0,06 0,29 0,18 0,38 of moles 3 (Given) (Given) (Given) (Given) (mol) Number of +0,06 +0,06 –0,06 –0,06 moles used / 3 (Step 4) (Step 5) (Step 5) (Step 5) formed (mol) Number of moles at 0,12 0,35 0,12 0,32 equilibrium (Step 3) (Step 6) (Step 6) (Step 6) 3 (mol) Equilibrium n 0,35 0,12 0,32 ​ __ 1 ​= 0,12 ​ ____ 1 ​= 0,35 ____ ​  1 ​= 0,12 ​ ____ 1 ​= 0,32 concentration 3 (mol·dm–3) (Step 2) (Step 7) (Step 7) (Step 7) [N2O][O2] (0,12)(0,32) 3 Kc = ​ ________ __________ [NO2][NO] ​ = ​ (0,12)(0,35) ​ = 0,91 3 (11) 3 2. Temperature. 3 (1) 3. Kc was initially 3,93 (given) now it is 0,91. If the Kc value changed, the temperature must have changed – only a change in temperature can change the value of Kc. 3 (1) [13] Activity 9 PAY SPECIAL ATTENTION NB NB Consider the following reaction equation for the production of ammonia in The Haber Process is the the Haber process. name of the chemical process N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H < 0 HINT! to manufacture NH3 In each of the following cases: hint a) State how Kc changes and b) Briefly explain your answer. 1. e.g. The concentration of the nitrogen gas is increased. EG - worked examples (1) 2. The pressure in the reaction system is decreased. (1) 3. FeO is added to the reaction system. (1) 4. The reaction system is heated. exams (5) [8] © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 101 Mind the Gap Chemistry activity ACTIVITIES boy

3 Unit Solutions 1. Remains constant because... 3 (1) 2. ... Kc is only affected by a change in temperature. 3 (1) 3. Kc Decreases 3 (1) 4. Heating the reaction system favours the endothermic, reverse reaction 3 which decreases the temperature 3 of the system ∴ the concentration of the forward product, [NH3] 3 decreases while the concentrations of the forward reactants, [N2] and [H2] increase 3 ∴ Kc decreases. 3 (5) [8] Activity 10 1. The reaction represented by the equation below reaches equilibrium. Co(H2O)62+(aq) + 4Cℓ–(aq) ⇋ CoCℓ42–(aq) + 6H2O(ℓ) ΔH > 0 pink blue Which ONE of the following changes to the reaction mixture will changes its colour from blue to pink? A. Add a catalyst B. Place the reaction mixture in a container with hot water. C. Add a few drops of concentrated hydrochloric acid to the reaction mixture. D. Add water to the reaction mixture. (2) [2] Solution 1. D 33 [2] 102 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 3 3.6 Interpretation of graphs for chemical systems in dynamic equilibrium Dynamic equilibrium in a chemical system can be represented graphically by a graph of: • reaction rate versus time • amount of each reagent (number of moles / mass / volume) • concentrationPAYofSPECIAL eachATTENTION reagent (in mol·dm–3) versus time. NB NB For each of these graphs: • time is the independent variable and is therefore placed on the horizontal axis and PAY SPECIAL ATTENTION • either the concentration, amount or the reaction rate is measured HINT! NB hint and is therefore the dependent variable which is placed on the vertical axis. HINT! e.g. Worked example 5EG - worked examples hint Graphs coincide: this means Equal amounts of SO2(g) and O2(g) react in a sealed container and reach equilibrium: the graphs lie on top of each exams 2SO2(g) + O2(g) ⇌ 2SO3(g) ∆H < 0 e.g. other EG - worked examples Use the explanation Pay special attention: exams below to interpret the We use square brackets [ ] graphs that follow. to indicate the concentration activity of a substance ACTIVITIES boy and girl-left and right of page activity activity Steps to follow ACTIVITIES boy and girl-left and right of page activity Step 1: We need to interpret each part of the graph for each reactant and product, step by step as Step by step comment the reaction comment proceeds and Step 2: identify what happens if there is a change in the rate/ concentration of each substance, which is visible by a change Step by step comment in the shape of the graph. (i) When the reaction initially starts: Initially • The forward reaction rate is high because the [reactants] is initially high and • The reverse reaction rate is zero as there are no products to react in the reverse reaction. © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 103 Mind the Gap Chemistry

HINT! 3 Unit hint e.g. Worked example 5 (continued) EG - worked examples As the reaction takes place the • exams concentrations of the reactants, [SO2] and [O2] decrease • concentration of the forward product, [SO3] increases as forward reaction rate > the reverse reaction rate. The forward reaction rate gradually decreases and the reverse reaction rate gradually increases. Reaction rate vs. time Concentration (or mol) vs. time activity ACTIVITIES boy and girl-left and right of page activity (ii) The reaction reaches a state of dynamic chemical equilibrium Step by step comment comment The forward and reverse reactions take place at the: • same time and • same rate. The concentrations of the reactants and of the products remain constant. Reaction rate vs. time Concentration (or mol) vs. time The lines coincide and are The lines on the graph are all horizontal. horizontal. The concentrations are not necessarily equal, so the lines do not coincide. (iii) O2(g) is added to the system. Adding O2 (g) to the system increases the concentration of the O2(g). O2(g) is a reactant in the forward reaction, therefore the rate of the forward reaction immediately shows a sharp increase. According to Le Châtelier’s Principle: • the reaction that opposes the change and decreases the concentration of the O2(g) is favoured so the • forward reaction is favoured • concentrations of the SO2(g) and O2(g) (forward reactants) decrease • concentration of the SO3(g) (forward product) increases. 104 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 3 e.g. Worked example 5 (continued) EG - worked examples The rate of the • exams forward reaction decreases as the [O2] and [SO2] decrease and • reverse reaction increases as the [SO3] increases • until a new dynamic equilibrium is established when the • forward and reverse reaction rates are again equal and • [O2] and [SO2] as well as the [SO3] again remain constant. activity Reaction rate vs. time Concentration (or mol) vs. time Increasing the concentration of ACTIVITIES boy Increasing the concentration of the forward reactant (O2) causes and a girl-left and the O (g) causes a sharp rise in its right of page 2 sharp rise in the forward reaction concentration. rate. activity Step by step comment comment (iv) SO3(g) is removed from the system. Removing SO3(g) to the system decreases the concentration of the SO3(g). SO3(g) is a reactant in the reverse reaction, therefore the rate of the reverse reaction immediately shows a sharp decrease. According to Le Châtelier’s Principle: • the reaction that opposes the change and increases the concentration of the SO3(g) is favoured so the • forward reaction is favoured • concentrations of the SO2(g) and O2(g) (forward reactants) decrease • concentration of the SO3(g) (forward product) increases. The rate of the • forward reaction decreases as the [O2] and [SO2] decrease • reverse reaction increases as the [SO3] increases • until a new dynamic equilibrium is established when the • forward and reverse reaction rates are again equal • [O2] and [SO2] as well as the [SO3] again remain constant. © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 105 Mind the Gap Chemistry

HINT! 3 Unit hint e.g. Worked example 5 (continued) EG - worked examples Reaction rate vs. time Concentration (or mol) vs. time exams the concentration of Decreasing Decreasing the concentration of the reverse reactant (SO3) causes a the reverse reactant (SO3) causes a sharp drop in the reverse reaction sharp drop in its concentration. rate. activity ACTIVITIES boy and girl-left and right of page activity (v) The temperature of the equilibrium mixture is increased. The heat of reaction is negative (∆H < 0) so the • forward reaction is exothermic • reverse reaction Step by is stependothermic. comment comment Increasing the temperature of the system increases both the forward and the reverse reaction rates. The rate of the endothermic reaction always increases the most if the system’s temperature increases, so for this reaction, the reverse reaction rate increases the most. According to Le Châtelier’s Principle: • The reaction that opposes the change and decreases the temperature of the equilibrium system, is favoured so • the endothermic reaction is favoured and therefore the • reverse reaction is favoured and the • concentration of the SO3(g) (reverse reactant) decreases and the • concentrations of the SO2(g) and O2(g) (reverse products) increase and 106 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 3 e.g. Worked example 5 (continued) EG - worked examples The rate of the • exams reverse reaction that was initially increased, decreases as the [SO3] decreases and • forward reaction increases as the [O2] and [SO2] increase • until a new dynamic equilibrium is established when the • forward and reverse reaction rates are again equal and • [O2] and [SO2] as well as the [SO3] again remain constant. activity Reaction rate vs. time Concentration (or mol) vs. time ACTIVITIES boy Both reaction rates initially A and change in temperature does and girl-left increase due to the increase in not cause any sharp increases or right of page temperature. decreases in concentration. (There activity are no vertical lines on the graph.) Step by step comment comment The rate of the endothermic reaction increases most when an equilibrium system is heated. In this case the endothermic reaction is the reverse reaction. (vi) The temperature of the equilibrium mixture is decreased. The heat of reaction is negative (∆H < 0) so the • forward reaction is exothermic • reverse reaction is endothermic. Decreasing the temperature of the system decreases both the forward and the reverse reaction rates. The rate of the endothermic reaction always decreases the most if the system’s temperature decreases, so for this reaction, the reverse reaction rate decreases the most. According to Le Châtelier’s Principle: • the reaction that opposes the change and increases the temperature of the equilibrium system, is favoured so the • exothermic reaction is favoured and therefore • forward reaction is favoured • concentrations of the SO2(g) and O2(g) (forward reactants) decrease and • concentration of the SO3(g) (forward product) increases and © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 107 Mind the Gap Chemistry

HINT! 3 Unit hint e.g. Worked example 5 (continued) EG - worked examples The rate of the • exams forward reaction decreases as the [O2] and [SO2] decrease • reverse reaction increases as the [SO3] increases • until a new dynamic equilibrium is established when • the forward and reverse reaction rates are again equal and • the [O2] and [SO2] as well as the [SO3] again remain constant. Reaction rate vs. time Concentration (or mol) vs. time activity Both reaction rates initially A change in temperature does ACTIVITIES boy decrease due to the decrease in and girl-left cause any sharp increases or notand temperature. decreases in concentration. (There right of page are no vertical lines on the graph.) activity PAY SPECIAL ATTENTION NB NB Step by step comment comment HINT! hint The rate of the endothermic reaction decreases most when an This applies to gases only. equilibrium system is cooled. In this case the endothermic reaction e.g. EG - worked examples is the reverse reaction. (vii) The pressure of the equilibrium mixture is increased exams From the balanced reaction equation, the number of gaseous moles is: 2SO2(g) + 1O2(g) ⇌ 2SO3(g) 2 + 1 = 3 mol(g) ⇌ 2 mol (g) According to Le Châtelier’s Principle: • the reaction that opposes the change and decreases the pressure activity in the equilibrium system, is favoured so ACTIVITIES boy• the reaction that produces the least moles of gas is favoured and girl-left and right of page • therefore the forward reaction is favoured and • the concentrations of the SO2(g) and O2(g) (forward reactants) activity decrease and • the concentration of the SO3(g) (forward product) increases and Step by step comment comment 108 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 3 e.g. Worked example 5 (continued) EG - worked examples The rate of the • exams forward reaction decreases as the [O2] and [SO2] decrease • reverse reaction increases as the [SO3] increases • until a new dynamic equilibrium is established when • the forward and reverse reaction rates are again equal and • the [O2] and [SO2] as well as the [SO3] again remain constant. Reaction rate vs. time Concentration vs. time activity Increasing the pressure in the When the pressure in the system ACTIVITIES boy system favours the forward increases, the volume decreases and girl-left and reaction and its rate increases right ofand page the concentration is inversely sharply. proportional to the volume activity n∴ (c = ​ __ V​ c ∝​  1 __​) so the V concentrations of all the gases in the system initially show a sharp increase. Step by step comment comment Thereafter the concentrations change as predicted by Le Châtelier’s Principle. Amount (mol) vs. time When the pressure in the system increases, there is no immediate change in the moles of each substance present, so there are no vertical lines on the mol vs. time graph. The amounts (moles) change gradually as predicted by Le Châtelier’s Principle. © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 109 Mind the Gap Chemistry

HINT! 3 Unit hint e.g. Worked example 5 (continued) EG - worked examples (viii) The pressure of the equilibrium mixture is decreased (the volume is exams increased). The same reasoning is applied as in the previous example and the graphs are as follows: activity ACTIVITIES boy and girl-left and right of page activity (ix) The addition of a suitable catalyst The addition of a The addition of a catalyst has no effect on the equilibrium of the system. catalyst has NO effect on the equilibrium. The rate of Reaction rate vs. time Concentration vs. time BOTH the forward and reverse reactions is increased due to the Adding a catalystStep by increases step the comment Adding a catalyst comment does not disturb presence of a catalyst, so there rate of the forward and the reverse the equilibrium, so the [O2], [SO2] is NO net change in the rate reactions equally. and [SO3] remain constant. of the reversible chemical reaction at equilibrium. ↑ ↑ Addition of the catalyst Addition of the catalyst 110 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 3 Activity 11 0,25 mol of A and 0,15 mol of B are introduced into a 1 dm3 vessel. A and B react and reach an equilibrium which can be represented by the following equation: A(g) + B(g) ⇋ AB(g) ΔH < 0 PAY SPECIAL ATTENTION NB HINT! hint Make use of the graph to read off the values e.g. EG - worked examples The graph shows the number of moles of A(g), B(g) and AB(g) vs time, under varying conditions. exams 1. What are the concentrations of A(g), B(g) and AB(g) at t = 20 s? (3) 2. How can we be sure that the system has reached equilibrium at t = 20 s? (1) 3. Give an explanation for the change that occurs at t = 25 s. (1) 4. Explain the change in concentrations of A(g) and B(g) between activity t = 25 s and t = 35 s. (3) ACTIVITIES boy and girl-left and 5. How does the equilibrium constant at t = 35 s compare with the right of page equilibrium constant at t = 25 s? (No calculation is required.) (1) 6. Assuming that the volume was kept constant, what was done at activity t = 40 s that produced the change as shown? Explain your answer. (5) 7. Do any reactions occur during the interval t = 30 s and t = 40 s? Briefly explain your answer. (4) Step by step comment [18] © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 111 Mind the Gap Chemistry

3 Unit Solutions 3 3 3 1. [A] = 0,15 mol·dm–3 [B] = 0,05 mol·dm–3 [AB] = 0,1 mol·dm–3 (3) 2. The number of mol of A, B and AB remains constant from 16 s to 25 s. 3 (1) 3. The amount of B is increased by adding B to the system, thus increasing the [B]. 3 (1) 4. With the addition of B, [B] increases and the forward reaction is favoured 3 according to Le Châtelier’s principle, so that some of B reacts with A to form more AB. Hence the [A] and [B] 3 decrease and [AB] increases. 3 (3) 5. Kc remains constant provided the temperature remains constant. 3 (1) 6. The temperature was lowered. 3 It was not a concentration change, because there are no peaks at 25 s 3. Therefore it must be a temperature change. From the graph, it can be seen that the forward reaction was favoured 3 because there is an increase in the number of moles of AB, and a decrease in the number of moles of A and B when establishing the new equilibrium. 3 According to the information given, the forward reaction is exothermic. 3 According to Le Châtelier’s Principle, a decrease in temperature favours the exothermic reaction. (5) 7. Yes, but since equilibrium has been reached, 3 the forward and reverse reactions take place at the same time 3 and at the same 3 rate, so that no changes occur in the concentrations of A, B or AB. 3 (4) [18] Activity 12 Nitrogen and oxygen gases react in a sealed container according to the following equation: O2(g) + N2(g) ⇋ 2NO(g) After the reaction reaches equilibrium, certain changes are made. The following graph of rate of reaction versus time represents the situation. 1. Write the equation for the reaction represented by the dashed line on this graph. (2) 112 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 3 2. What is represented by the section of the graph between the 10th and 15th minute? Explain your answer. (2) 3. A temperature change takes place at t = 15 minutes. 3.1 Was the temperature increased or decreased at t = 15 minutes? (1) 3.2 Explain whether the forward reaction is exothermic or endothermic. (3) 3.3 What effect does this temperature change have on the equilibrium constant (Kc)? Explain your answer. (3) 4. A pressure change is introduced at t = 20 minutes. 4.1 Was the pressure increased or decreased? (3) 4.2 Explain how this change in pressure affects the amounts of each gas at equilibrium. (2) [16] Solutions 1. 2NO(g) → N2(g) + O2(g) 3 3 (2) 2. The system is in equilibrium 3 because the rates of the forward and reverse reactions are equal. 3 (2) 3.1 Decreased. 3 (1) 3.2 The rates of both the forward and the reverse reactions decrease 3, but the rate of the reverse reaction decreases more. 3 The rate of the endothermic reaction will always decrease the most when a reaction mixture is cooled. Thus the reverse reaction is endothermic. That means that the forward reaction must be exothermic and was favoured by the decrease in temperature. 3 O2(g) + N2(g) ⇌ 2NO(g) + Energy (3) 3.3 Increases. 3 2 [NO] Since Kc = ​ ______ [N ][O ] ​ (3) 2 2 and the forward reaction is favoured 3 which causes an increase in [NO], the value of Kc increases. 3 (3) 4.1 Increased. 3 The pressure is probably caused by a volume n decrease. This results in the concentrations (c = __ ​ V ​) of the reactants, as well as the products, to increase 3, which in turn causes an equal increase in the forward and reverse reaction rates. 3 (3) 4.2 The amount (n) of reactants and products does not change, 3 since 2 moles of gaseous reactant molecules ⇌ 2 moles of gaseous product molecules. 3 (2) [16] © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 113 Mind the Gap Chemistry

3 Unit Activity 13 A fertiliser company produces ammonia on a large scale at a temperature of 450°C. The balanced equation below represents the reaction that takes place in a sealed container. N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H < 0 To meet an increased demand for fertiliser, the management of the company instructs their engineer to make the necessary adjustments to increase the yield of ammonia. In a trial run on a small scale in the laboratory, the engineer makes adjustments to the TEMPERATURE, PRESSURE and CONCENTRATION of the equilibrium mixture. The graphs below represent the results obtained. 1. Identify the changes made to the equilibrium mixture at each of the following times: 1.1 t1 (2) 1.2 t2 (2) 1.3 t3 (2) 2. At which of the above time(s) did the change made to the reaction mixture lead to a higher yield of ammonia? Write down only t1 and/or t2 and/or t3 (2) 3. The engineer now injects 5 mol N2 and 5 mol H2 into a 5 dm sealed 3 empty container. Equilibrium is reached at 450 °C. Upon analysis of the equilibrium mixture, he finds that the mass of NH3 is 20,4 g. Calculate the value of the equilibrium constant (Kc) at 450 °C (9) [17] 114 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 3 Solutions 1. 1.1 The concentration of nitrogen 3 is increased 3 / More Nitrogen 3 added 3 (any two) (2) 1.2 The pressure 3 on the system is increased 3 (2) 1.3 The temperature 3 is increased 3 (2) 2. t1 3and t2 3 (2) 3. Pay special attention: This is how marks are allocated to the answer of this Kc calculation: ​ m ​ 3 • Use of n = __ M • n(NH3) at equilibrium = 1,2 mol 3 • Using ratio n(N2) : n(H2) : n(NH3) = 1 : 3 : 2 3 • n(N2) at equilibrium (initial - change) 3 • n(H2) at equilibrium (initial - change) 3 • divide by volume (calculation of concentration) 3 • Kc expression 3 • Substitution into Kc expression 3 • Final answer 3 (9) m 20,4 n(NH3) = ​ __ ____ M ​ = ​  17 ​ 3= 1,2 mol 3 SO2 O2 SO3 Molar ratio 1 3 2 Initial quantity (mol) 5 5 0 Change (mol) 0,6 1,8 1,2 Quantity at equilibrium 4,4 3,2 1,2 Concentration (mol·dm–3) 0,88 0,64 0,24 [NH3] 2 (0,24) 2 Kc = ​ _______ [N ][H ]3 ​ = ​ __________ (0,88)(0,64)3 ​ = 0,25 2 2 [17] © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 115 Mind the Gap Chemistry

3 Unit Activity 14 2 mol of NO2(g) and an unknown amount of N2O4(g) are sealed in a 2 dm3 container, that is fitted with a plunger, at a certain temperature. The following reaction takes place: 2NO2(g) ⇌ N2O4(g) At equilibrium it is found that the NO2 concentration is 0,4 mol·dm–3. The equilibrium constant at this temperature is 2. 1. Calculate the initial amount (in mol) of N2O4(g) that was sealed in the container. (8) The plunger is now pushed into the container causing the pressure of the enclosed gas to increase by decreasing the volume 2. How will this change influence the amount of nitrogen dioxide at equilibrium? Only write down INCREASES, DECREASES or REMAINS THE SAME. (1) 3. Use Le Châtelier’s principle to explain your answer to QUESTION 2.1.2. (2) [11] Solutions 1. 2NO2 N2O4 Initial number of mole (mol) 2 x PAY SPECIAL ATTENTION NB NB of moles used/formed (mol) Number –1,23 +0,6 3 Number of moles at equilibrium(mol) 0,83 x + 0,6 3 (x + 0,6) _______ Equilibrium concentration (mol·dm–3) 0,4 ​  2 ​ 3 HINT! Remember, if you hint struggle to see how [N2O4] this calculation Kc = ​ _____ [NO2]2 ​3 worked, please revise the (x + 0,6) 2 “subject of formula” section at 2 3 = _________ ​  (0,4)2 ​ e.g. EG - worked examples the beginning of this book. x = 0,04 mol 3 (8) 2. Decreases 3 (1) exams 3. When the pressure is increased the system will try to decrease the pressure. 3 The forward reaction (2 mol to 1 mol) is favoured. 3 (2) [11] activity ACTIVITIES boy and girl-left and right of page activity 116 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 3 3.7 Applications of equilibrium principles in the chemical PAY SPECIAL ATTENTION NB industry • Many industrial processes involve reversible chemical reactions that reach equilibrium. HINT! • Two factors that are important from an economic point of view are hint the yield (amount of products produced relative to the amount of The success of chemical reactants) and the reaction rate (rate at which the products are processes in both the formed). e.g. EG - worked examples laboratory and the depends on: industry • the amount of product that is produced and exams • how cost effective this production is. PAY SPECIAL ATTENTION NB activity Remember ACTIVITIES boy Standard atmospheric and girl-left and right of page pressure (1 atm) is the HINT! hint activity atmospheric pressure at sea level. 1 atm = 101,3 kPa = 101 325 Pa (pascal) EG - worked examples e.g. Step by step comment exams You must understand: 1. the effect of changing the reaction conditions (this implies the change in the rate of a reaction) and 2. applying equilibrium activity principles to the Haber process and to the ACTIVITIE and girl-l Contact process right of p activity Step by step com © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 117 Mind the Gap Chemistry

3 Unit Haber Process Step 2 of the Contact Process Nitrogen, N2(g) Reactants Sulphur dioxide, SO2(g) Hydrogen, H2(g) Oxygen, O2(g) • N2(g) : fractional distillation of liquid air Source of • S (s) : sulphur • H2(g) : steam reforming of methane reactants • O2 (g) : fractional distillation of liquid air (CH4) from natural gas • S(s) + O2(g) → SO2(g) Ammonia, NH3(g) Products Sulphur trioxide, SO3(g) Reaction N2(g) + 3H2(g) ⇋ 2NH3(g) 2SO2(g) + O2(g) ⇋ 2SO3(g) equation 4 mol (g) : 2 mol (g) Gas mol ratio 3 mol (g) : 2 mol (g) ∆H < 0 Heat of reaction ∆H < 0 ∴ forward reaction is exothermic (Enthalpy) ∴ forward reaction is exothermic • Increasing the temperature Increase the • Increasing the temperature • Increasing the concentrations of the reaction rate by • Increasing the concentrations of the forward reactants N2 and H2 forward reactants SO2 • Adding a suitable catalyst and O2 • Adding a suitable catalyst • Decreasing the temperature Increase the • Decreasing the temperature • (favours the forward exothermic yield by • (favours the forward exothermic reaction) reaction) • Increasing the concentrations of the • Increasing the concentrations of the forward reactants SO2 forward reactants N2 and H2 and O2 • Increasing the pressure • Increasing the pressure • (favours the forward reaction that • (favours the forward reaction that produces less gas moles) produces less gas moles) Temperature: approx. 450 °C Optimal Reaction Temperature: approx. 450 °C Pressure: approx. 200 atm conditions Pressure: approx. 1 – 2 atm Fe (iron) OR Catalyst V2O5 (vanadium pentoxide) FeO (iron(II)oxide) CIAL ATTENTION NB • When the reaction system is in a state of dynamic equilibrium, high temperatures will increase the reaction rates BUT will decrease the yield of the products. Both the forward reactions are exothermic, therefore high temperature will, according to Le Châtelier’s Principle, favour the reverse endothermic reactions. This makes the reactions economically less viable! • Temperatures of ± 450 °C are maintained in both the Haber and Contact processes. • Industrial processes at high pressures require powerful compressors and strong reaction vessels which are very ked examples expensive. This makes the reactions economically less viable! Industrial plants that operate at high pressures also pose safety risks for the workers. • It is therefore necessary to find a compromise between the reaction rates, the yield, and the higher costs. • A pressure of 200 times that of atmospheric pressure is maintained in the Haber process to ensure that the yield is high enough. • The yield of SO3 in the Contact Process is however more than 90% at a pressure of 1 atm and therefore it is economically viable to keep the reaction at atmospheric pressure. It is unnecessary to maintain higher pressures and this saves unnecessary costs. 118 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  ACTIVITIES boy Mind the Gap Chemistry and girl-left and right of page

Unit 3 Activity 15 Name the industrial process for the production of ammonia (1) [1] Solution Haber (Bosch) process 3 [1] Activity 16 Sulphuric acid is an important substance used in the manufacture of fertilisers. The equation below represents one of the steps in the industrial preparation of sulphuric acid. 2SO2(g) + O2(g) ⇋ 2SO3(g) ∆H < 0 1. Write down the name of the process used to prepare sulphuric acid in industry. (1) 2. Write down the NAME or FORMULA of the catalyst used in the process in question (1) 3. Is the forward reaction exothermic or endothermic? Give a reason for the answer. The reaction reaches equilibrium at a certain temperature in a 2 dm3 closed container. On analysis of the equilibrium mixture, it is found that 0,6 mol of SO2(g), 0,5 mol of O2(g) and 0,4 mol of SO3(g) are present in the container. (1) 4. List THREE changes that can be made to this equilibrium to increase the yield of SO3(g) (3) 5. The temperature is NOW increased and the reaction is allowed to reach equilibrium for the second time at the new temperature. On analysis of this new equilibrium mixture, it is found that 0,2 mol of SO3(g) is present in the container. Calculate the equilibrium constant for this reaction at the new temperature. (8) [14] © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 119 Mind the Gap Chemistry

3 Unit Solutions 1. Contact process 3 (1) 2. V2O5 (vanadium pentoxide) 3 (1) 3. Exothermic ∆H < 0 3 (1) 4. Any three: 3 3 3 Decrease temperature Increase pressure Increase concentration of both/any one of the reactants Remove SO3 continuously (3) 5. SO2 O2 SO3 Molar ratio 2 1 2 Initial quantity (mol) 0,6 0,5 0,4 Change (mol) 0,2 0,1 0,2 Quantity at equilibrium 0,8 0,6 0,2 Concentration (mol·dm–3) 0,4 0,3 0,1 [SO3]2 (0,1)2 Kc = ​ ________ [SO ]2[O ] ​ = ​ ________ (0,4)2(0,3) ​ = 0,21 2 2 (8) PAY SPECIAL ATTENTION NB NB This how marks are allocated to the answer of this Kc calculation: • Change in n(SO3) = 0,2 mol 3 • Ratio n(SO2) : n(O2) :HINT! n(SO3) = 2 : 1 : 2 3 • n(SOhint 2) at equilibrium = initial + change 3 • n(O2) at equilibrium = initial + change 3 • divide three equilibrium amounts by 2 (calculation of concentration) 3 • Kc expression 3 EG - worked examples e.g. • Substitution into Kc expression 3 • Final answer 3 (8) exams [14] activity ACTIVITIES boy and girl-left and right of page activity 120 UNIT 3 CHEMICAL EQUILIBRIUM  © Department of Basic Education 2015  Step by step comment comment Mind the Gap Chemistry

Unit 3 Activity 17 The reaction below represents the catalysed step in the contact process: 2SO2(g) + O2(g) ⇋ 2SO3(g) ∆H <0 The reaction takes place in a closed container and reaches equilibrium at 427°C. How will a HIGHER temperature affect each of the following? Write down only INCREASES, DECREASES or REMAINS THE SAME. 1. The rate of production of SO3(g) (2) 2. The yield of SO3(g) (2) 3. The reaction is investigated on a small scale in the laboratory. Initially 4 mol of SO2(g) and an unknown mass, x, of O2(g) are sealed in a 2 dm–3 flask and allowed to reach equilibrium at a certain temperature. At equilibrium it is found that the concentration of SO3(g) present in the flask is 1,5 mol·dm–3. Calculate the mass of O2(g) initially present in the flask if the equilibrium constant (Kc) at this temperature is 4,5. (8) [12] PAY SPECIAL ATTENTION NB Solutions 1. Increases 3 3 (2) 2. Decreases 3 3 (2) HINT! 3. hint SO2 O2 SO3 Remember, if you struggle Molar ratio 2 1 2 to see how this calculation Initial quantity ___x e.g. EG - worked examples worked, please revise the “subject of formula” section 4 ​  32 ​ 3 3 0 (mol) at the beginning of this book. Change (mol) 3 1,5 3 ratio3 exams Quantity at x ___ 1 ​  32 ​– 1,5 3 33 equilibrium Concentration 3 n 1 (x – 48) ​ __ 2 ​= 1,5 (mol·dm–3) C = ​ __ __ V ​ = ​  2 ​= 0,5 ______ ​  64 ​ n (divide by 2) 3 (C = ​ __ V ​) [SO3]2 (1,5)2 activity Kc = ​ ________ [SO ]2[O ] ​ 3 ∴ 4,5 = ​ __________ (x – 48) ​ (0,5)2 ______ ACTIVITIES boy 2 2 ​  64 ​ and girl-left and ∴ x = 176 g 3 (8) right of page [12] activity Step by step comment Keep going! © Department of Basic Education 2015 UNIT 3 CHEMICAL EQUILIBRIUM 121 Mind the Gap Chemistry

4 Unit Acids and bases 4.1 Summary Acid (Arrhenius theory): is a substance that produces hydrogen ions (H+) / hydronium ions (H3O+) when it dissolves in water (Brønsted-Lowry theory): is a proton (H+ ion) donor. Strong acid Ionises completely in water to form a high concentration of H3O+ ions. Examples of strong acids are hydrochloric acid (HCℓ), sulphuric acid (H2SO4) and nitric acid (HNO3). Weak acid Ionises incompletely in water to form a low concentration of H3O+ ions. Examples of weak acids are ethanoic acid (CH3COOH) and oxalic acid (COOH)2. Concentrated Concentrated acids contain a large amount (number of acid moles) of acid in proportion to the volume of water. Diluted acid Dilute acids contain a small amount (number of moles) of acid in proportion to the volume of water. Base (Arrhenius theory): is a substance that produces hydroxide ions (OH–) when it dissolves in water. (Brønsted-Lowry theory): is a proton (H+ ion) acceptor. Strong base Dissociates (breaks up) completely in water to form a high concentration of OH– ions. Examples of strong bases are sodium hydroxide (NaOH) and potassium hydroxide (KOH). Weak base Dissociates/ionises incompletely in water to form a low concentration of OH– ions. Examples of weak bases are ammonia (NH3), calcium carbonate (CaCO3), potassium carbonate (K2CO3) , and sodium hydrogen carbonate (NaHCO3). Concentrated Concentrated bases contain a large amount (number of base moles) of base in proportion to the volume of water. Diluted base Dilute bases contain a small amount (number of moles) of base in proportion to the volume of water. Equivalence is the point at which the acid / base has completely reacted point with the base/acid. End point is the point where the indicator changes colour. Ionisation a process that takes place when a covalent compound reacts with water to form new ions OR Breaking up of a molecule into charged components (ions). In acid-base reactions this usually means dissolving in water. 122 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 Dissociation a process that take place when an ionic compound dissolves in water allowing the ions in the compound to separate. The same as ionisation. Hydrolysis ionisation of a salt in water, or, more generally, splitting a molecule by reacting it with water (e.g. in organic chemistry). Ampholyte is a substance that can act either as an acid or a base. (Amphiprotic) example: Water (H2O) Activity 1 1. An Arrhenius acid is a substance that A. Accept a proton B. Donates a proton C. Produces H+ in an aqueous solution D. Produces OH– in an aqueous solution (2) 2. Which of the following is an example of the strong base A. CaCO3 C. K2CO B. KOH D. NaHCO3(2) 3. An aqueous solution that contains more hydronium ions than hydroxyl ions is a(an) A. Acidic solution C. Basic solution B. Neutral solution D. Standardised solution (2) 4. A solution that has a large amount of dissolved substances in proportion to the volume of water A. Strong solution C. Concentrated solution B. Weak solution D. Diluted solution (2) [8] Solutions 1. D33 2. B33 3. A33 4. C33 [8] © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 123 Mind the Gap Chemistry

4 Unit 4.2 PAY SPECIAL ATTENTION Properties of acids and bases NB NB ACIDS BASES NOTE: • Have a sour taste; • Have a bitter taste and feel soapy; • You must know the • Influence the colour of indictors; • Influence the colour of indictors; Arrhenius and the HINT! PROPERTIES • Are electrolytes in solution; • Are electrolytes in solution; hint Brønsted-Lowry definitions • Increase the H+-ion • Decrease the H+-ion (or H3O+- for acids and bases, BUT concentration in a solution; ion) concentration in a solution; we will ALWAYS use the • Decrease the OH−-ion • Increase the OH−-ion Brønsted-Lowry theory concentration in a solution; concentration in a solution; e.g. EG - worked examples for all further discussions • Decrease the pH of a solution. • Increase the pH of a solution. of acids and bases. ARRHENIUS exams • An acid releases H+/ H3O+ • A base releases OH− -ions in ACID-BASE THEORIES -ions/ in aqueous solutions. aqueous solutions. • HCℓ → H+ + Cℓ− • NaOH → Na+ + OH− • HCℓ + H2O → H3O+ + Cℓ− BRØNSTED-LOWRY activity • An acid is a proton donor. • A base is a proton acceptor. • HCℓ + H2O → H3O+ + Cℓ− • OH− + H+ → H2O ACTIVITIES boy + and • NH 4 + H2O → H3O + NH3 • NH3 + H2O → OH− + NH4+ + girl-left and right of page • Ionise in water to produce • Ionic bases like NaOH REACTIONS WITH WATER activity hydronium ions dissociate in water to form hydroxyl ions. NaOH(s) → Na+(aq) + Cℓ−(aq) E.g. HCℓ + H2O → H3O+ + Cℓ− • Ammonia (NH3) ionises in water to form hydroxyl ions Step by step comment comment NH3 + H2O → OH− + NH4+ Practice the reaction equations of aqueous solutions of acids Activity 2 and bases. 1. Which of the following is the property of an acid A. Decreases H3O+ ion concentration in solution B. Decreases OH– ion concentration in solution C. Increases OH– ion concentration in solution D. Increases the pH of a solution (2) [2] Solution 1. B33 [2] 124 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 4.3 Common acids STRONG ACID FORMULA EXAMPLES & USES / WEAK Hydrochloric acid HCℓ Strong • Stomach acid (to digest food) Used • in swimming pools to control the Ionises almost completely in water STRONGEST pH (acidity) of the water; • to clean metals for soldering. Sulphuric acid H2SO4 Strong Used • to produce fertilisers; • to produce phosphoric acid; • as dehydrating agent; • as car battery acid and • in gold extraction. Nitric acid HNO3 Strong Used • to produce fertilisers and INCREASING STRENGTH • to produce explosives. Oxalic acid (COOH)2 Weak Used • to remove rust and stains and • as bleach. Phosphoric acid H3PO4 Weak Used • to add taste to gaseous cold drinks; Ionises partially in water • in dental cement and • to produce fertilisers like superphosphates. Ethanoic acid CH3COOH Weak Vinegar • Used WEAKEST • to flavour food; • to produce plastics; • to neutralise hair relaxers and • to treat jellyfish stings. Carbonic acid PAY SPECIAL ATTENTION H2CO3 Weak Used NB NB • to make gaseous cold drinks. StrongHINT! acid: HCℓ; H2SO4; HNO3 hint Weak acid: CH3COOH; (COOH)2 Strong base: NaOH; KOH; Mg(OH)2 Weak base: NH3; NaHCO3 ; CaCO3 e.g. EG - worked examples exams © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 125 Mind the Gap Chemistry activity

4 Unit 4.4 Common bases STRONG / BASE FORMULA EXAMPLES & USES WEAK Sodium hydroxide NaOH Strong Used (Caustic soda) • in the production of soap and • as drain cleaner and • to relax the curl in human hair. STRONGEST Potassium hydroxide KOH Strong Used (Caustic potash) • in the production of soap; Dissociates completely in water • in the production of biodiesels and • as electrolytes in cells. Magnesium Mg(OH)2 Strong Used hydroxide • in antacids (Milk of Magnesia); • in laxatives and • to neutralise acidic waste water. INCREASING STRENGTH Calcium hydroxide Ca(OH)2 Strong Used (Slaked lime) • to clarify water by removing particles from the water; • to neutralise acidic water so that water pipes are not damaged and • to neutralise acidic soil. Sodium carbonate Na2CO3 Weak Used (Washing soda) • to manufacture glass and • to soften water or laundry. Dissociates / Ionises partially in water Calcium carbonate CaCO3 Weak Found in marble and sea shells. WEAKEST (Limestone) Used • in the production of cement; • in chalk to write with; • as an antacid. Ammonia NH3 Weak Used • to produce nitric acid; • to produce fertilizers and • in cleaning materials. Sodium bicarbonate NaHCO3 Weak Used (Baking soda) • in baking to let dough rise 126 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 4.5 Mono- and polyprotic acids Acids can be classified according to the number of protons (H+) that they can donate. NB: Monoprotic acids have only one proton (H+) to donate. Polyprotic acids can donate two or three protons. The protons are donated in steps as shown in the examples in the table below. Monoprotic acids Polyprotic acids Can only donate only Can donate more than one proton (H+) one proton (H+) Diprotic acids Triprotic acids Can donate 2 protons Can donate 3 protons (H+). (H+). HCℓ, HNO3, CH3COOH H2SO4, H2CO3 H3PO4 E.g. E.g. E.g. HCℓ → H+ + Cℓ− H2SO4 → H+ + HSO4− H3PO4 → H+ + H2PO4− HSO4− → H+ + SO42− H2PO4− → H+ + HPO42− HPO42− → H+ + PO43− 4.6 NB Conjugate acid-base pairs PAY SPECIAL ATTENTION NB 4.6.1 Acid-base reactions PAY SPECIAL ATTENTION HINT! NB together” or to be a “Conjugate” is from Latin, it means literally “yoked NB hint couple. Acid-base reactions take place simultaneously. The acid donates a proton DEFINITION to the base, while the base accepts the proton from the acid. An acid-base (protolytic) HINT! reaction is a reaction in e.g. hint EG - worked examples which a proton (H+) is 4.6.2 Conjugate acids and bases transferred exams CONJUGATE ACIDS CONJUGATE BASES • When an acid donates a proton (H ), a + e.g. EG - worked examples • When a base receives a proton (H+), a conjugate conjugate base is produced. acid is produced. • The acid and its conjugate base are called a • The base and its conjugate acid are called a conjugate acid-base pair. exams conjugate acid-base pair. acid ⇌ H+ + conjugate base base + H+ ⇌ conjugate acid Examples: activity Examples: −ACTIVITIES boy HCℓ ⇌ H + + Cℓ and girl-left and OH− + H+ ⇌ H2 O right of page acid conjugate base base conjugate acid activity activity HSO4− + H+ ⇌ H2SO4 H2SO4 ⇌ H + + HSO4 − base conjugate acid ACTIVITIES boy acid conjugate base and girl-left and right of page Step by step comment comment activity Step by step comment comment © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 127 Mind the Gap Chemistry

4 Unit Activity 3 1. In the reaction: H2SO4(aq) + H2O(ℓ) ⇌ HSO4–(aq) + H3O+(aq), the Brønsted-Lowry bases are: A H2O and H3O+ C HSO4– and H3O+ B H2SO4 and H3O+ D H2O and HSO4– (2) [2] Solution 1. D 33 [2] Activity 4 Find the conjugate bases and conjugate acids. To find the conjugate base of an acid, remove one H+ (proton) from the acid. Acid Conjugate base Base Conjugate acid To find the conjugate acid HCℓ Cℓ− of a base, add one H+ (proton) to the base. HNO3 NO3− H2SO4 HSO4− HSO4− SO42− H3PO4 H2PO4− H2PO4− HPO42− HPO42− PO43− H2CO3 HCO3− HCO3− CO32− CH3COOH SO42− (COOH)2 HSO4− H2 O OH− NH4+ NH3 H3O+ H2 O [28] Note that some of these are marked in BOLD. That is because they are “amphiprotic”. We will see what this means later on. 128 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 Solution 1. Check your answers Acid Conjugate base Base Conjugate acid HCℓ Cℓ − 3 Cℓ − HCℓ 3 HNO3 NO3− 3 NO3 − HNO3 3 H2SO4 HSO4− 3 HSO4− H2SO4 3 HSO4 − SO4 32− SO42− HSO4− 3 H3PO4 H2PO4− 3 H2PO4− H3PO4 3 H2PO4 − HPO4 3 2– HPO4 2− H2PO4− 3 HPO42− PO43− 3 PO43− HPO42− 3 H2CO3 HCO3− 3 HCO3− H2CO3 3 HCO3 − CO3 32− CO3 2− HCO3− 3 CH3COOH CH3COO− 3 SO42− HSO− 3 (COOH)2 C2 O4 H− 3 HSO4− H2SO4 3 H2 O OH− 3 OH− H 2O 3 NH4 + NH3 3 NH3 NH4+ 3 H3O+ H 2O 3 H2 O H3O+ 3 [28] Did you know? Hydrogen carbonates are also known as bicarbonates. E.g. NaHCO3 is known as • sodium hydrogen carbonate OR as • sodium bicarbonate © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 129 Mind the Gap Chemistry

4 Unit Steps to follow when identifying conjugates 1. Find the acid on the left hand side of the arrow. Label it acid1. 2. Find the conjugate base of this acid on the right hand side of the arrow. Label it base1. 3. Draw a bracket to show that these two form an acid-base PAY SPECIAL ATTENTION NB conjugate pair. NB 4. Find the base on the left hand side of the arrow. Label it base2. 5. Find the conjugate acid of this base on the right hand side of the arrow. Label it acid2. 6. Draw a bracket HINT!to show that these two form an acid-base hint conjugate pair. e.g. Worked example 1 EG - worked examples For each of the following reactions, indicate the acid-base conjugate pairs. 1. exams HCℓ(g) + NH3(g) ⇌ Cℓ−(aq) + NH4+(aq) 2. H2SO4(g) + H2O(ℓ) ⇌ HSO4−(aq) + H3O+(aq) 3. H3PO4(g) + OH−(aq) ⇌ H2PO4−(aq) + H2O(ℓ) Solutions 1. activity ACTIVITIES boy and girl-left and right of page activity 2. Step by step comment comment 3. 130 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 4.7 Ampholyte (amphiprotic) PAY SPECIAL ATTENTION NB substance An ampholyte: Ampholyte is a substance • acts as a base in the presence of an acid and that can act as either a base • acts as an acid in the presence of a base. or acid. HINT! hint Example: Water (H2O) is an ampholyte: Water as an acid Water as a base e.g. EG - worked examples H2O(ℓ) + NH3(g) ⇌ OH−(aq) + NH4+(aq) HCℓ(g) + H2O (g) ⇌ Cℓ−(aq) + H3O+(aq) acid1 base2 conjugate conjugate acid1 base2 conjugate conjugate exams base1 acid2 base1 acid2 Example: Show that hydrogen sulphate ion (HSO4−) is an ampholyte. The hydrogen sulphate ion (HSO4−) donates a The hydrogen sulphate ion (HSO4−) accepts a proton when it acts as an acid proton when it acts as a base activity ACTIVIT HSO4−(aq) ⇌ H+(aq) + SO42−(aq) HSO4−(aq) + H+(aq) ⇌ H2SO4(aq) and girl right of acid conjugate base base conjugate acid activity Activity 5 In the acid-base equilibrium formed by adding HSO4– and OH– the acids Step by step com are: A HSO4– and H2SO4 C SO42– and H2SO– B HSO4– and H2O D SO42– and H2O (2) Which of the following is amphiprotic in water? A SO2 C HSO3– B SO3 2– D H2SO3 (2) [4] Solutions 1. B33 2. C33 [4] © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 131 Mind the Gap Chemistry

4 Unit 4.8 Salt hydrolysis A salt is formed in the reaction between an acid and a base. When an ionic salt dissolves in water, the ions in the salt dissociate. These ions react with the water and new ions form. PAY SPECIAL ATTENTION NB NB: Hydrolysis NB occurs when a salt (a compound made of a metal + non- metal portion) reacts with water. Hydrolysis, more generally, is the splitting of any compound by reacting it with water. In this chapter, we only deal with the hydrolysis of salts. (Salt is not just what you have on your table; that is HINT! just one type of salt, specifically NaCℓ. Any metal combined with any non- metal is a salt). hint e.g. Worked example 2 EG - worked examples To determine whether the solutions of each of the following salts are: 1. exams acidic or basic, and the pH of each solution is above or below 7? a. NH4NO3(aq) (2) b. K2SO4(aq) (2) Solutions 1a: 1b: activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 132 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 In summary: Remember that acids and bases react with each other to form new compounds: acid + base → salt + water We can determine whether the salt solution is basic or acidic by comparing the strengths of the reacting acids and bases. • A salt formed between a strong acid and a weak base is an acidic salt, for example NH4Cℓ. • When a salt reacts with water to form hydronium ions (H3O+), the solution is acidic (pH < 7). • A salt formed between a weak acid and a strong base is a basic salt, for example NaCH3COO. • When a salt reacts with water to form hydroxyl ions (OH−), the solution is basic (pH > 7). • Neutral salt is formed when a strong acid and a strong base are neutralized in the reaction. To determine the approximate pH of salts in salt hydrolysis H2O forms H3O+ H2O forms OH− • H2O acts as a base; • H2O acts as an acid; • it receives a proton (H+) and • it donates a proton (H+) and • H3O+ ions form so • OH− ions form so • the solution is acidic and • the solution is basic and • pH < 7 • pH > 7 Hydrolysis reaction: Hydrolysis reaction: ion + H2O ⇌ H3O+ + molecule / ion ion + H2O ⇌ OH− + molecule / ion REMEMBER: • When water reacts with a salt to form hydronium ions (H3O+), the solution is acidic (pH < 7). • When a salt reacts with water to form hydroxyl ions (OH–), the solution is basic (pH > 7). • The ions that don’t react are called spectator ions. • Positive spectator ions: cations from Groups I and II e.g. K+ and Mg2+ • Negative spectator ions e.g. SO42–; Cℓ–; NO3–. Salt of Nature of pH in an Example solution aqueous Acid Base solution Strong Strong Neutral pH = 7 NaCℓ(aq) (HCℓ + NaOH) Weak Weak Neutral pH = 7 CH3COONH4(aq) (CH3COOH + NH3) Strong Weak Acidic pH < 7 NH4Cℓ(aq) (HCℓ + NH3) Weak Strong Basic pH > 7 CH3COONa(aq) (CH3COOH + NaOH) © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 133 Mind the Gap Chemistry

4 Unit Step by step PAY SPECIAL ATTENTION NB NB Step 1: Write down the formula of the acid and of the base that reacted to form the salt. Step 2: Write down HINT! whether the acid is strong or weak. hint Step 3: Write down whether the base is strong or weak. Step 4: Is the salt solution acidic or basic? e.g. Worked example 3 EG - worked examples Determine whether each of the following salt solutions are acidic, basic orexams neutral. a) Na2SO4(aq) b) NH4NO3(aq) Solution a) Acid + Base Acid + Base activity H2SO4 + NaOH HNO3 + NH3 Strong Strong Strong + ACTIVITIES boy Weak and girl-left and ∴ neutral solution ∴ right of page acidic solution activity AL ATTENTION NB 4.9 Acid-base indicators Indicators: DEFINITION • can be used to determine whether a solution is acidic or basic; An indicator is a substance • are weak acids Stepthat by stepare in equilibrium comment with their conjugate bases (or comment that changes colour in the vice versa) and presence of an acid or a base. • have complex structures and formulae which will simply be represented as HIn (H followed by In for “indicator”). ed examples Indicator in equilibrium HIn ⇌ H+ + In− colour 1 colour 2 Indicator in acid Indicator in base • The high [H+] of the acid will • The low [H+] of the base will disturb the indicator equilibrium. disturb the indicator equilibrium. • According to Le Châtelier’s • According to Le Châtelier’s principle Principle, the reverse reaction, the forward reaction, which which lowers the [H+], will be increases the [H+], will be favoured. favoured. • More In− is produced and the • More HIn is produced and the indicator will exhibit (show) ACTIVITIES boy indicator will exhibit colour 1. colour 2. and girl-left and right of page p comment comment 134 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 4 e.g. Worked example 4EG - worked examples exams Bromothymol blue is an acid-base indicator. The colours it exhibits (shows) can be represented as follows: HIn ⇌ H+ + In− yellow blue A test tube contains a solution to which a drop of bromothymol blue has been added. The solution appears blue. activity a) What will you observe if a few drops of concentrated hydrochloric ACTIVITIES boy acid are added to the test tube? and girl-left and right of page b) Explain your answer. activity Solutions a) The colour of the solution changes from blue to yellow. b) Adding hydrochloric acid increases the concentration of the H+ ions in solution. Step byAccording step to Le Châtelier’s comment Principle the reverse comment reaction that opposes this change and decreases the [H+], will be favoured. Therefore the [HIn] increases and the colour changes to blue. The most common indicators that are used in laboratories are: INDICATOR pH RANGE COLOUR CHANGE methyl orange 3,1 – 4,4 red to orange to yellow methyl red 4,2 – 6,2 red to yellow litmus 4,5 – 8,3 red to blue bromothymol blue 6,0 – 7,8 yellow to blue phenolphthalein 8.3 – 10 pink-purple (magenta) in range, colourless outside range universal 3-11 red (3 and below); 3-6 orange/yellow; 7 green; 8-11 blue; 11 and above, violet. © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 135 Mind the Gap Chemistry

4 Unit Choose the most suitable indicator for a particular titration. If you do not have Universal Indicator available, you should follow these steps: A colour version of this diagram is available on the Step by step – universal indicator inside cover of this book. Step 1: Identify the strength of the acid and the base (pH range maximum and minimum) Step 2: Draw a bracket to show the strength of the identified acid and base (shown PAY SPECIAL as a line with arrowheads above) ATTENTION StepNB 3: Note or mark the mid point (centre) NB of the bracket (or line as shown above) Step 4: Use the indicator table to choose the indicator that shows the range of pH around the mid point (centre) of that bracket/line. HINT! hint RESULT: You now have selected the appropriate indicator to use. e.g. Worked example 5 EG - worked examples Which one of the indicators given below will be the most suitable to be exams used in the titration of ethanoic acid against sodium hydroxide? INDICATOR pH COLOUR CHANGE RANGE A bromothymol blue 6,0 – 7,8 B phenolphthalein 8,3 – 10 C methyl orange 3,1 – 4,4 D activity methyl red 4,2 – 6,2 ACTIVITIES boy and girl-left and Solution right of page activity Consult the table at Sections (3) and (4) above. We see Ethanoic Acid has a pH of about 2,4 (1M solution), and NaOH has a pH of about 14. The mid-point between 2,4 and 14 is about 5,8, so any indicator which changes from indicating acid to base around 5,8 would do. Answer: D; methyl red. Step by step comment comment 136 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 4.10 Acid-base titrations (volumetric analysis) Titrations are used to experimentally determine the concentration of an unknown acid or base. When the titration results are used to determine the concentration of the unknown solution it is called a volumetric analysis. a) Method and apparatus for an acid-base titration 4.11 Preparing a standard solution DEFINITION A standard solution has a known concentration which remains constant for a period of time. Standard solutions are often used in laboratories and it is important to know how to prepare a standard solution. n CONCENTRATION (c = ​ __ V ​) CONCENTRATED SOLUTION DILUTE SOLUTION Greater number of moles of Smaller number of moles of substance per unit volume. + H2O(ℓ) substance per unit volume. nsolute nsolute nsolute c = ​ ____ V1 ​ c = ​ __________ V +V ​= ​ ____​ V 1 water added 2 ndiluted = nconcentrated © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 137 Mind the Gap Chemistry

PAY SPECIAL ATTENTION HINT! NB 4 Unithint HINT! e.g. Worked example 6 EG - worked examples hint 1 cm3 = 1 ml ; 1 dm3 = 1 l Aim: To prepare a 200 cm NaOH standard solution with a concentration of 0,5 mol∙dm . 3 –3 exams Step 1: Calculate the mass of NaOH pellets required. M(NaOH) = 40 + 16 + 1 = 40 g·mol EG - –1 e.g. worked examples Step 2: Use an electronic scale (balance) to measure off ± 4,4 g NaOH. exams Note: A specific mass of NaOH(s) can’t always be activity measured because it consists of pellets (not a fine powder). ACTIVITIES boy and girl-left and right of page Step 3: Note the exact mass of NaOH on the scale. activity activity ACTIVITIES boy Step 4: Transfer the NaOH pellets to a volumetric flask. Add and girl-left and right of page ± 100 cm3 of distilled water to the flask and seal it with a stopper. Shake the flask carefully until all the solute activity (NaOH) has dissolved. Step by step comment comment Step 5: Slowly fill the flask to the calibration mark on the neck of the flask. Step 6: Now calculate the exact concentration of the prepared Step by step comment c solution – use the mass that was noted on the scale in Step 3. m 4,1 n 0,103 n = ​ __ ___ M ​ = ​  40 ​= 0,103 mol and c = __ ​ V ​ = ​ _____ 0,3 ​= 0,52 mol·dm –3 ∴ a standard NaOH(aq) solution of concentration 0,52 mol·dm−3 has been prepared. Solute: Solvent: Solution: The substance The substance The mixture of the (usually a solid) that (usually distilled solute and the is dissolved in the water) in which the solvent. solvent. solute is dissolved. 138 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 4.12 Dilution of solutions We sometimes need to dilute a solution so that we can use it in a laboratory. We do this by taking a small amount of the solution and adding distilled Remember: water to it. When a solution is diluted, the number of moles of When diluting a solution, we need to know: substance remains constant. • the exact amount of distilled PAY SPECIAL ATTENTION water that needs to added as well as Only the volume of the NB • the exact concentration of the dilute solution. NB solution changes (hence n is omitted in this equation.) In symbols: c1 : concentration 1 (mol·dm–3) c1V1 = c2V2 V1 : volume 1 (dm–3) HINT! c2 : concentration 2 (mol·dm–3) hint V2 : volume 2 (dm–3) e.g. Worked example 7EG - worked examples Solution 1 has a concentration of 0,2 mol∙dm–3. Exactly 150 cm3 of solution A is transferred examsto a beaker 2 and 250 cm of distilled water is added to the beaker. 3 Calculate the concentration of the diluted solution. Solution c1V1 = c2V2 (0,2)(0,15) = c2(0,4) activity (0,2)(0,15) c = ________ ​  ​ 2 0,4 ACTIVITIES boy = 0,075 mol·dm–3and girl-left and right of page • The volume of the final solution (V2) is: activity V2 = V1 + Vwater = 150 + 250 = 400 cm3 • 400 cm3 = 400 ÷ 1000 = 0,4 dm3 • convert to dm3 Step by step comment comment © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 139 Mind the Gap Chemistry

4 Unit 4.13 Acid-base titration calculations PAY SPECIAL ATTENTION NB In symbols: NB n : mol (mol) __ na caVa ​  nb ​= ​ ___ cbVb ​ c = __ n ​ V ​ c : concentration (mol·dm–3) m where OR c = ___ ​ MV ​ V : volume (dm–3) m = ​ __ nHINT!M​ a : acid hint b : base e.g. Worked example 8EG - worked examples During a titration 20 cm3 diluted H2SO4 precisely neutralises 25 cm3 of a NaOH solution. If the concentration of the H2SO4 solution is 0,5 mol∙dm–3, calculate the concentration of exams the NaOH. Solution First write down the balanced reaction. 1 H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + H2O(ℓ) activity Find the mol ratio of acid : base: H2SO4 : NaOH = 1 : 2 na caVa ​ __ ____ ACTIVITIES boy nb ​ = ​ c V ​ b b and girl-left and na = 1 (0,5)(20) right of page 1 ​ __ 2​ = ​ _______ cb (25) ​ nb = 2 activity 1∙ cb (25) = (0,5)(20)(2) Va = 20 cm3 Vb = 25 cm3 cb = 0,8 mol∙dm –3 ca = 0,5 mol∙dm–3 cb = ? Step by step comment comment 140 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 4 e.g. Worked example 9 EG - worked examples TITRATION CALCULATIONS exams Eight grams (8,0 g) of sodium hydroxide are dissolved in 350 cm3 of distilled water. 15 cm3 of this solution neutralises 20 cm3 of a sulphuric acid solution. The balanced equation for this reaction is: 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(ℓ) Calculate the concentration of the sulphuric acid solution. Solution STEP 1 activity ACTIVITIES boy m m and c = ​ ___ MV ​ OR __ n = ​  ​ and girl-left right of M page 8 8 c = ​ _________ (40 × 0,35) ​ n = ___ ​  40 ​= 0,2 mol NaOH activity n c = 0,57 mol·dm–3 c = ​ __ V​ 0,2 c = ​ ____ 0,35 ​ c = 0,5 mol·dm–3 Step by step comment comment Solution STEP 1 m m c = ​ ___ MV ​ OR n = ​ __ M ​ 8 8 c = ​ _________ (40 × 0,35) ​ n = ___ ​  40 ​= 0,2 mol NaOH n c = 0,57 mol·dm–3 c = ​ __ V​ 0,2 c = ​ ____ 0,35 ​ c = 0,5 mol·dm–3 STEP 2 __ na caVa ​  nb ​= ​ ___ cbVb ​ From the balanced equation 2 n(NaOH) : 1 n(H2SO4) 1 ​ __ 2 ​ = (ca × 0,02) (0,57 × 0,025) ca = 0,36 mol·dm–3 © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 141 Mind the Gap Chemistry

HINT! 4 Unit hint REMEMBER: To convert volume from cm3 to dm3 e.g. Worked example 10 EG - worked examples (by dividing by 1000). The titration of oxalic acid with sodium hydroxide exams Part 1: Aim: To prepare a 250 cm3 of oxalic acid standard solution with a concentration of 0,08 mol∙dm–3 Solutions Step 1 • Measure between 2,5 and 2,7 g of pure oxalic acid hydrate activity crystals (COOH)2 · 2H2O into the glass beaker and weigh again. • Add 30-60 cm3 of distilled water ACTIVITIEStoboythe glass beaker and dissolve the crystals. and girl-left and right of page • Transfer the solution into a clean 250 cm3 volumetric flask. • Rinseactivity the beaker with 15-20 cm3 of distilled water and pour this solution into the volumetric flask and repeat. This will ensure that all of the oxalic acid is transferred into the volumetric flask. • Fill the volumetric flask to within about 2 cm of the mark and allow it to sit for a minute. This will allow any water clinging to the edges of theStepneck to drain into the by step comment comment flask. Using an eyedropper, fill the flask to the calibrated mark with water. • Stopper (close) the flask and mix the solution by repeated inversion (turning upside-down) and swirling. This requires about 30 inversions and takes close to 1 minute. Step 2 • Calculate the exact concentration of the oxalic acid solution using the mass of acid used and the volume of the volumetric flask and record the concentration of your solution on the bottle. If you have used the following Mass oxalic acid crystals 2,6 g Final volume solution 250,0 cm3 Using the above information to calculate the number of moles and concentration of oxalic acid crystals. Part 2: Procedure to perform (to do) a titration of sodium hydroxide with oxalic acid. Aim: use a standard solution of oxalic acid to determine the concentration of sodium hydroxide H2C2O4(aq) + 2 NaOH(aq) → 2 H2O(ℓ) + Na2C2O4(aq) Solutions • Measure exact 25,00 cm3 of the oxalic acid standard solution (c = 0,084 mol·dm–3) into a flask and add few drops of phenolphthalein. • In this titration the oxalic acid solution is acidic and therefore phenolphthalein will be colourless. 142 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

HINT! hint Unit 4 e.g. Worked example 10 (continued) EG - worked examples • The sodium hydroxide solution will be added drop wise (drop by drop) from a burette into the flask containing the oxalic acid and exams indicator. • As the sodium hydroxide is added to the flask it will react with the oxalic acid and be neutralized. • At the point where all of the oxalic acid is reacted, the next drop of sodium hydroxide will make the entire solution basic and it will turn pink. At this point you have completed the titration. activity Results (If the following data and results ACTIVITIES boy were obtained after at least three trials) and girl-left and right of page Volume oxalic acid solution 25,00 cm3 Volume activity sodium hydroxide to titrate 19,43 cm3 Mole of sodium hydroxide reacted: __ na caVa ​  nb ​= ​ ___ cbVb ​ Step by step comment comment First write down the balanced reaction. 1 H2C2O4(aq) + 2 NaOH(aq) → 2 H2O(ℓ) + Na2C2O4(aq) Find the mol ratio of acid : base ... ratio of H2C2O4 : NaOH = 1 : 2 na caVa ​ __ ___ nb ​= ​ c V ​ na = 1 b b __ 1 (0,084)(25) nb = 2 ​  2 ​ = ​ _________ cb (25) ​ Va = 20 cm3 Vb = 25 cm3 1· cb (19,43) = (0,084)(25)(2) ca = 0,5 mol·dm–3 cb = 0,216 mol·dm–3 cb = ? Part 3: In order to determine the concentration of acetic acid in the vinegar solution you will titrate it with the standardized sodium hydroxide solution. The equation for this reaction is 1 CH3COOH(aq) + 1 NaOH(aq) → H2O(ℓ) + CH3COONa(aq) The following data was obtained during titration Data Volume vinegar solution 25,00 cm3 Volume sodium hydroxide titrated 22,84 cm3 Calculations for determining concentration of acetic acid in vinegar First write down the balanced reaction. 1 CH3COOH(aq) + 1 NaOH(aq) → H2O(ℓ) + CH3COONa(aq) Find the mol ratio of acid : base ... CH3COOH : NaOH = 1 : 1 © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 143 Mind the Gap Chemistry

HINT! 4 Unit hint e.g. Worked example 10 (continued) EG - worked examples From a balanced equation 2 n(NaOH) : 1 n(H2SO4) exams n caVa ​ __ ___ nb ​= ​ c V ​ na = 1 a b b (ca)(25) nb = 1 1 __ ​  1 ​ = ​ ___________  ​ (0,216)(22,84) Va = 25 cm3 Vb = 22,84 cm3 1· (0,216) (22,84) = ca (25)(2) cb = 0,216 mol·dm–3 ca = 0,197 mol·dm–3 ca = ? activity ACTIVITIES boy In order to get the best precision possible, you should repeat each and girl-left and titration until you get 3 trials that are right within 1% of each other. of page PAY SPECIAL ATTENTION activity NB 4.14 TheNBEquilibrium Constant (K ) c Solids and pure liquids (The Law of Mass Action) are omitted from the Kc expression asPAY their SPECIAL ATTENTION The concentrations Stepof by all step the compounds comment(solutions commentand gases) in a closed HINT! NB system in NB dynamic chemical equilibrium are related by a mathematical hint is [1], as concentration equation. The numerical value of this equation is called the equilibrium multiplying by 1 has no effect. constant (Kc). In the hypothetical equation below the equilibrium expression for this e.g. HINT! EG - worked examples reaction is: hint 2P + 3R ⇋ 2S + 4T Kc: equilibrium constant (no unit) exams [S]2 × [T]4 Kc = ​ ________ e.g. EG - worked examples [substance]: concentration [P]2 × [R]3 ​ of reactant or product (in mol·dm–3) PAY SPECIAL ATTENTION mol: number of moles of each NB NB exams compound in the balanced activity reaction equation. • This is the correct way of writing the Kc expression • The coefficients are the moles of each reactant and product in the balanced equation ACTIVITIES boy and girl-left and HINT! • right product of the concentration of reactants (not to be added, but multiplied!!), Theof page hint raised to the power of the number of moles is the numerator activity activity • The product of the concentration of products (not to be added, but multiplied!!), raised to the power of the number of moles is the denominator ACTIVITIES boy e.g. • The concentrations used in the Law of Mass Action is the [reactant]equilibrium EG - worked examples and girl-left and right of page and [product]equilibrium (NOT initial concentrations!! i.e. the concentrations of activity Step by step reactants comment andcomment products at equilibrium) exams Step by step comment comment activity ACTIVITIES boy and girl-left and 144 UNIT 4 ACIDS AND BASES rightof page © Department of Basic Education 2015  Mind the Gap Chemistry activity

Unit 4 4.15 Ka and Kb values PAY SPECIAL ATTENTION NB Since the ionisation of a weak acid is an equilibrium, a chemical equation and an equilibrium constant expression can be written as follows. An acid or base’s strength Remember that square brackets means “concentration”, or moles per refers to its degree of dm3. ionisation. A strong acidHINT! will [H3O+][A–] HA + H2O ⇋ H3O+ + A– Ka = ​ _______ [HA] ​ hint completely ionise in water while a weak acid will The equilibrium constant for the ionisation of an acid is called the acid only partially ionise. Since ionisation constant (Ka). there are different degrees A similar expression can be written for bases: e.g. of ionisation, there are EG - worked examples [OH–][HA+] different levels of weakness. A + H2O ⇋ HA+ + OH− Kb = ​ ________ [A–] ​ Fortunately, there is a simple The equilibrium constant for the ionisation of a base is called the base quantitative way of expressing ionisation constant (Kb) this. exams ACIDS BASES Ionisation of acid (HA): Ionisation of base (B): HA + H2O ⇋ H3O+ + A− B + H2O ⇋ OH− + BH+ (aq) [H3O+][A–] Ka = ​ _______ [HA] ​ [OH–][BH+] Kb = ​ ________ [B] ​ activity ACTIVIT (NB: here we substituted in values and girl from the line above!) right of Strong acid → high Ka-value (>1) Strong base → high Kb-value (>1) activity Weak acid → low Ka-value (<1) Weak base → low Kb-value (<1) Examples: HCℓ(g) + H2O(ℓ) ⇋ H3O+(aq) + NH3(g) + H2O(ℓ) ⇋ OH−(aq) + Step by step com Cℓ−(aq) NH4+(aq) Dissociation of base: H2O(ℓ) NaOH (s) → Na+ (aq) + OH− (aq) Kb = [Na+][OH–] Activity 6 Do you think a strong acid will have larger or smaller Ka value? Explain your answer. (3) [3] Solution The strong acid will have a larger Ka value. 3 A strong acid is a better proton donor, resulting in more products. 3 Since the concentration of the products is in the numerator of the Kaexpression, the stronger the acid, the larger the Ka. 3 [3] © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 145 Mind the Gap Chemistry

4 Unit Activity 7 Choose the strongest base in the list below by comparing their Kb values. Base Kb A Ammonia, NH3 1,8 × 10–5 B Hydroxylamine, HONH2 9,1 × 10–9 C Ethylamine, C2H5NH2 4,3 × 10–4(2) [2] Solids and pure liquids Solution are NEVER written in C 3 3 [The larger the Kb, the stronger the base] [2] the equations for the equilibrium constant Kc, or Ka or Kb as their concentration is [1] 4.16 The relationship between Ka and K b for a substance We know that the strength of a conjugate base is inversely proportional to the strength of the conjugate acid; i.e. weak acids produce strong conjugate bases, and vice versa. ACID CONJUGATE BASE HA(aq) + H2O(ℓ) ⇋ H3O (aq) + A (aq) + − A (aq) + H2O(ℓ) ⇋ HA(aq) + OH−(aq) − [H3O+][A–] [OH–][HA+] Ka = ​ _______ [HA] ​ Kb = ​ ________ [A–] ​ Then follows: [H3O+][A–] [HA][OH–] KaKb = ​ _______ _______ [HA] ​ × ​  [A–] ​= [H3O ][OH ] + – But: [H3O+][OH–] = Kw = 1 × 10–14 PAY SPECIAL ATTENTION NB NB You are aware that the pH scale runs from 1 to 14... now look at the values above again. HINT! hint e.g. EG - worked examples exams 146 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  activity Mind the Gap Chemistry

Unit 4 4.17 Auto-ionisation of water Water is an ampholyte and can react as an acid (donates a H+) or a base (accepts a H+). Two water molecules can undergo autoprotolysis or auto-ionisation. One water molecule can act as an acid and donate a H+ to the other water molecule, the base. • Water is a weak acid and a weak base. An acid : H2O ⇋ H+ + OH− • Low % auto-ionisation occurs. A base : H O + H ⇋ H O + + • Pure water is a weak electrical conductor. 2 3 Net reaction showing the conjugate acid-base pairs: 4.18 Equilibrium constant for water (Kw) From the acid-base reaction for water the equilibrium expression for water can be written. H2O(ℓ) + H2O(ℓ) ⇋ H3O+(aq) + OH−(aq) Kc = [H3O+][OH−] Kw : equilibrium constant At 25°C: Kw = [H3O ][OH ] + − for water Kw = 1 × 10–14 Water has a low percentage of ionisation with a low concentration of products. The Kw-value will be low. Due to the symbol H+ often being used instead of H3O+, the ionisation constant (ion product) for water can also be written as: Kw = [H+][OH−] Therefore: Ion product for water Kw = [H3O+][OH−] = [H+][OH−] = 1 × 10−14 at 25 °C 4.19 The pH scale We know Kw = [H3O+][OH−] = [H+][OH−] = 1 × 10−14 therefore [H3O+] = [OH−] = 10−7 or [H+] = [OH−] = 10−7 Acid solution Neutral solution Base solution [H3O ] > [OH ] + − [H3O ] = [OH ] + − [H3O+] < [OH−] OR OR OR [H ] > [OH−] + [H ] = [OH−] + [H ] < [OH−] + [H3O+] > 10−7 mol·dm–3 [H3O+] = 10−7 mol·dm–3 [H3O+] < 10−7 mol·dm–3 © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 147 Mind the Gap Chemistry

HINT! 4 Unit hint e.g. Worked example 11 EG - worked examples Determine whether the following solutions are acidic, basic or neutral: exams A solution with [H3O+] = 10−3 mol·dm–3 1. A solution with [H3O+] = 10−3 mol·dm–3 10−3 mol·dm–3 > 10−7 mol·dm−3 ∴ Solution is acidic [H3O+] = 10−12 2. A solution with [H3O+] = 10−12 mol·dm−3 mol·dm–3 10−12 mol·dm−3 < 10−7 mol·dm−3 activity ∴ Solution is basic 3. A solution with [OH−] = 10−7 mol·dm−3 ACTIVITIES boy and girl-left and− [OH ] = 10 mol·dm − −7 –3 right3O [H + ][OH of page ] = 10−14 activity ∴ [H3O+] (10−7) ∴ = 10−14 [H3O ] + = 10−14 ÷ 10−7 = 10−7 ∴ Solution is neutral PAY SPECIAL ATTENTION 4. A solution with [OH−] = 10−4 mol·dm−3 NB NB [OH ] = 10 mol·dm − −4 –3 [H3O+][OH−] = 10−14 Step by step comment comment−14 ∴ [H3O+] (10−4) = 10 DEFINITION ∴ [H3O+] = 10−14 ÷ 10−4 = 10−10 pH of a solution is the 10 mol·dm−3 −10 < 10−7 mol·dm–3 negative logarithm HINT! ∴ Solution is basic of the hint hydronium ion concentration in a solution. pH = log [H3O+] It is complicated to continually refer to the hydronium ion concentration. e.g. EG - worked examples The pH scale is therefore used as a simplified method of indicating the degree of acidity of a solution. exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 148 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 4.20 pH Calculations for strong acids and bases To calculate the pH of a strong acid or base, these steps should be followed: Acid Step Base 0,1 mol·dm–3 of HCℓ solution For 0,5 mol·dm–3 of NaOH solution example: Write down the ionisation reaction Write down the dissociation for the acid. 1 reaction for the base. HCℓ + H2O ⇋ 1 H3O+ + Cℓ− 1NaOH(s) ⇋ Na+(aq)+OH− (aq) Find the mol ratio of acid to Find the mol ratio of base to hydronium ions 2 hydroxyl ions 1 mol HCℓ : 1 mol H3O+ 1 mol NaOH : 1 mol OH− Determine the [H3O+] Determine the [OH–] [HCℓ] = 0,1 mol·dm–3 [NaOH] = 0,5 mol·dm–3 3 ∴ [H3O+] = 0,1 mol·dm–3 ∴ [OH−] = 1 (0,5) = 0,5 mol·dm–3 Determine the [H3O+] No step 4 for acids [H3O+][OH−] = 10−14 4 PAY SPECIAL ATTENTION NB NB∴ [H3O+] (0,5) = 10−14 ∴ [H3O+] = 10−14 ÷ 0,5 = 2 × 10−14 Calculate the pH Calculate the pH pH = − log [H3O+] pH = − log [H3O+] 5 = − log 0,1 HINT! = − log (2 × 10−14) hint =1 = 13,7 e.g. Worked example 12 EG - worked examples Calculate the pH of a H2SO4 solution of concentration 0,6 mol∙dm–3. exams Solutions Step 1: 1 H2SO4(g) + 2 H2O(ℓ) ⇋ 2 H3O+(aq) + SO42−(aq) Step 2: 1 mol HCℓ : 2 mol H3O+ Step 3: [H2SO4] = 0,6 mol·dm–3 ∴ [H3O+] = 2(0,6) = 1,2 mol·dm–3 Step 4: activity pH = − log [H3O+] = − log 1,2 ACTIVITIES boy = 0,08 and girl-left and right of page activity Step by step comment comment © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 149 Mind the Gap Chemistry

4 Unit 4.21 Summary: Strong and weak acids and bases Acids ionise in water to form ions: Examples: HCℓ(g) + H2O(ℓ) ⇋ H3O+(aq) + Cℓ−(aq) CH3COOH(g) + H2O(ℓ) ⇋ H3O+(aq) + CH3COO–(aq) STRONG AND WEAK ACIDS Strong acids Weak acids • High % ionisation • Low % ionisation • Donate many protons (H+) • Donate few protons (H+) Remember: • High [H3O+] or [H+] • Low [H3O+] or [H+] The electrical conductivity of • Low pH • Relatively low pH any electrolyte is determined • Good electric conductors • Poor electric conductors by the number of ions in • High reaction rates • Low reaction rates solution. The greater the ion concentration, the higher the Examples: Examples: conductivity. HCℓ (hydrochloric acid) CH3COOH (acetic acid) H2SO4 (sulphuric acid) (COOH)2 (oxalic acid) Reaction rate is determined by the concentration of a HNO3 (nitric acid) C6H8O7 (citric acid) reactant. The higher the H2CO3 (carbonic acid) concentration of the ions in a Strong acid ⇌ weak conjugate base Weak acid ⇌ strong conjugate solution, the higher the rate of + H+ base + H+ reaction. HCℓ ⇌ Cℓ− + H+ CH3COOH ⇌ CH3COO− + H+ Bases like: • NaOH are ionic and dissociate in water while • NH3 consists of covalent molecules which ionise in water forming NH4+. STRONG AND WEAK BASES Strong bases Weak bases • High % dissociation • Low % dissociation or ionisation • Accept many protons (H+) • Accept few protons (H+) • High [OH–] • Low [OH–] • Very high pH • Relatively high pH Remember: NaOH Sodium hydroxide NH3 Ammonia KOH Potassium hydroxide CaCO3 Calcium carbonate The pH of an acid is dependent on the relative strong base + H+ ⇌ weak conjugate weak base + H+ ⇌ concentrations of the H3O+ acid strong conjugate acid and OH− ions in the solution. OH− + H+ ⇌ H2O NH3 + H+ ⇌ NH4+ In general: pH 1 - 2 : strong acid pH 12 - 14 : strong base pH 3 - 6 : weak acid pH 8 - 11 : weak base pH 7 : neutral 150 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 4.22 Summary: Concentrated and dilute acids and bases Concentration of an acid or a base is an indication of the number of moles of solute per unit volume. Both strong and weak acids and bases can be either concentrated or diluted. Examples: 1 mol·dm–3 HCℓ(aq) : concentrated solution of a strong acid 0,01 mol·dm–3 HCℓ(aq) : diluted solution of a strong acid 1 mol·dm–3 CH3COOH(aq) : concentrated solution of a weak acid 0,01 mol·dm–3 CH3COOH(aq) : diluted solution of a weak acid 1 mol·dm–3 NaOH(aq) : concentrated solution of a strong base 0,01 mol·dm–3 NaOH(aq) : diluted solution of a strong base 1 mol·dm–3 NH3(aq) : concentrated solution of a weak base 0,01 mol·dm–3 NH3(aq) : diluted solution of a weak base Activity 8 1. A standard solution is a solution ... A. at 25˚C B of an acid or a base C of which the volume is known D of which the concentration is known (2) 2. Consider the following ionisation equilibrium: H2O(ℓ) ⇋ H+(aq) + OH−(aq) The ionisation constant of water (Kw) in the above reaction increases from 1 × 10–14 at 25˚C to 9,6 × 10–14 at 60˚C. Which one of the following statements is therefore correct A [H+] > [OH−] at 60˚C B The ionisation reaction is exothermic C The pH increases with an increase in temperature D The pH decreases with an increase in temperature (2) 3. Consider the reaction: CH3COOH(g) + H2O(ℓ) ⇋ H3O+(aq) + CH3COO–(aq) The ionisation constant for this reaction at 25°C is 1,8 × 10–5. The equilibrium constant for this reaction at 60°C, is 3,6 × 10–7. From this information we can deduce that forward reaction ... (Explain your choice). A is endothermic B is exothermic C is a redox reaction D is a precipitation reaction (2) © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 151 Mind the Gap Chemistry

4 Unit 4. Two beakers, A and B, contain solutions of the same concentration with a pH of 2 and 4,5 respectively. Which of the following combinations is correct? Beaker A Beaker B A Weak Acid Strong Acid B Strong Acid Weak Acid C Strong base Weak base D Weak base Strong base (2) 5. Andile rinses the apparatus before starting an acid-base titration experiment. Which rinsing method is likely to cause inaccurate results? A The Erlenmeyer (conical) flask is rinsed with distilled water B The burette is rinsed with the acids it is to be filled with C The pipette is rinsed with the base it is to be used for D The volumetric (measuring) flask, which is used to make up the standard solution of the base, is rinsed with distilled water. (2) 6. If base X is titrated against acid Y, the pH of the solution at the end point is 8. The base X and acid Y are respectively: X Y A NaOH CH3COOH B Na2CO3 HCℓ C NaOH H2SO4 D Na2CO3 CH3COOH (2) [12] Solutions 1. D 33 (2) 2. D __________ At 60˚C: [H+] = [OH–] = ​√9,6 × 10–14 ​= 3,1 × 10–7 mol·dm–3 3 pH = –log[H+] = –log(3,1 × 10–7) = 6,5 3 (2) 3. B3 3 Kc at 25°C is greater than Kc at 60°C Kc = [CH3COO–][ H3O+]. If the mixture is heated, the Kc decreases (2) 4. B33 Beaker A has a low pH which indicates that it contains more H+, therefore it has ionised completely. Whilst beaker B has a high pH which indicates few H+ ions, thus it has partially ionised. (2) 5. A33 The Erlenmeyer (conical) flask must be rinsed with distilled water. Only the number of moles of base which was measured in the pipette must be neutralised. (2) 6. A33 (2) [12] 152 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 4 Activity 9 1. Write down: 1.1 The meaning of the term diprotic acid. (2) 1.2 The formula of a diprotic acid. (1) 2. Magnesium hydroxide (Mg(OH)2) is often used as medicine to relieve an upset stomach. The pH of the HCℓ(aq) in a person’s stomach is 1. 2.1 Calculate the concentration of the hydrochloric acid in the person’s stomach. (3) 2.2 Will the pH in the stomach INCREASE, DECREASE or STAY THE SAME after taking in a dose of Mg(OH)2?(2) 2.3 A person takes in a dose of Mg(OH)2. Write down the balanced equation for the reaction that takes place in the stomach. (3) 3. A textbook states that calcium sulphate (CaSO4) is slightly soluble in water. Two learners decided to test the dam water from a local municipality for calcium sulphate. They took a 0,5 dm3 sample of the dam water and treated it with sodium carbonate solution to precipitate the calcium ions present according to the following equation: CaSO4(aq) + Na2CO3(aq) → Na2SO4(aq) + CaCO3(s) The precipitate is then dissolved in 30 cm3 of 0,1 mol·dm–3 HCℓ solution which converts the precipitate to aqueous calcium chloride, water and carbon dioxide according to the following equation: CaCO3 + 2HCℓ → CaCℓ2 + CO2 + H2O The HCℓ was in excess. They neutralised the excess HCℓ by adding 15,8 cm3 of a 0,1 mol·dm–3 NaOH solution. The equation for the reaction is: HCℓ + NaOH → NaCℓ + H2O Calculate the mass of calcium sulphate that was present in the sample of dam water. (10) [21] © Department of Basic Education 2015 UNIT 4 ACIDS AND BASES 153 Mind the Gap Chemistry

4 Unit Solutions 1. An acid that can release/donate3 two protons (H+)3(2) 2. H2SO43 (H2S ; H2SO3 ; H2CO3 ; (COOH)2)(1) 2.1 pH = –log [H3O+] = 13 ∴ [H3O+] = 10–1 = 0,1 3 ∴ [HCℓ] = 0,1 mol·dm–33(3) 2.2 INCREASE 33(2) 2.3 Mg(OH)2(s) + 2HCℓ(aq) → MgCℓ2(aq) + 2H2O(ℓ)  (3) 3 3 3 3. na ca Va VHCℓ (reacted with NaOH) = 15,8 cm3 ​ __ ___ nb ​= ​ cbVb ​ 3 na = nb = 1 VHCℓ (reacted with CaCO3) = 30 – 15,8 cm3 3 n(CaCO3) = 0,5 × n(HCℓ) 3 = 14,2 cm3 = 0,5 × 1,42 × 10–3 3 ∴nHCℓ (reacted with CaCO3) = 0,0142 × 0,1 = 7,1 × 10–4 mol 3 V = c V = 0,1 × 15,8 × 1 = 15,8 cm3 = 1,42 × 10–3mol 3 b b ∴ n(CaSO4) = 0,5 × n(HCℓ) 3 = 0,5 × 1,42 × 10–3 = 7,1 × 10–4 mol ∴ m(CaSO4) = nM 3 = 7,1 × 10–4 × 136 3 = 0,097 g 3(10) [21] Keep going! 154 UNIT 4 ACIDS AND BASES  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 Electrochemistry Just remember, Electrochemistry refers to chemical reactions during which chemical energy is the two half reactions converted to electric energy, or electric energy is converted to chemical energy. are added together to produce the overall or During these chemical reactions oxidation and reduction take place. These are net redox reaction. called redox reactions. 5.1 Summary Redox (reduction - oxidation) reactions: • Electron transfers take place. • can be represented by two half-reactions: • an oxidation half-reaction and • a reduction half-reaction. 5.2 Key concepts: Definitions and Another name terminology for a galvanic cell is a voltaic cell. A galvanic cell has self-sustaining L ATTENTION electrode reactions. NB Electrolysis is the chemical process in which electrical energy is converted to chemical energy. Electrolysis is the process where electrical energy is used to produce a chemical change. Remember You must remember these basic definitions for electrochemistry: d examples • Galvanic cells • Electrolytic cells • Oxidation • Reduction • Oxidising agent • Reducing agent • Anode • Cathode • Electrolyte • Electrolysis © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 155 Mind the Gap Chemistry

5 Unit A galvanic cell is a cell in which chemical energy is converted into electrical energy. An electrolytic cell is a cell in which electrical energy is converted into chemical energy. Oxidation Reduction PAY SPECIAL ATTENTION NB NBof electrons Oxidation is the loss Reduction is the gain of electrons by a substance (i.e. by an atom, a by a substance (i.e. by an atom, a Remember molecule or an ion). molecule or an ion). We define oxidation and Learn: LEO for Learn: GER for reduction in terms of electron HINT! Loss of Electrons is Oxidation. Gain of Electrons is Reduction. hint (e ) transfer – (because gaining electrons is gaining minuses, so reducing) RED CAT: A substance that is oxidised (i.e. A substance that is reduced e.g. REDuction – CAThode EG - worked examples loses electrons) is called a reducing agent. (i.e. gains electrons) is called an oxidising agent. AN OX : The oxidation number of a The oxidation number of a ANode exams - OXidation compound that is oxidised, compound that is reduced, increases (becomes less negative, decreases (becomes more negative, or becomes more positive). or becomes less positive). The anode (–) is the electrode where oxidation (+) takes place The cathode (+) is the electrode where reduction (–) takes place. activity The electrodes: • ACTIVITIES conduct and boy girl-left andelectricity • right areofplaced page in electrolytes which are solutions consisting of ions. activity The electrolyte is the solution/liquid/dissolved substance that conducts electricity through the movement of ions. Step by step comment comment Activity 1 The liquid inside Give one word for the following statements: the cell has a special name. 1. The chemical process when an electric current is passed through an ionic compound in solution or in molten state. (1) 2. An ionic solution that conducts electricity. (1) 3. The reactant that donates electrons during a redox reaction.  (1) 4. The electrode in an electrochemical cell where reduction takes place. (1) 5. TRUE OR FALSE? The reactions C(s) + O2(g) → CO2(g) and 2KCℓO3(s) → 2KCℓ(s) + 3O2(g) are examples of redox reactions. (1) 6. The electrode in an electrochemical cell where oxidation occurs. (1) 7. A substance that shows a decrease in oxidation number during chemical reactions. (1) 156 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 8. Which of the following substances can be used as an electrolyte? A. Mercury B. Molten copper C. Sugar dissolved in distilled water D. Table salt dissolved in distilled water.  (1) [8] Solutions 1. Electrolysis3 (1) 2. Electrolyte3 (1) Both types of cells 3. Reducing agent3 (1) contain electrodes 4. Cathode3 (1) where redox reactions 5. TRUE statement3 (1) occur 6. Anode3 (1) 7. Oxidising agent3 (1) 8. D3 (1) [8] 5.3 Electrochemical cells Electrochemical cells allow conversion between electrical and chemical energy. There are two types of electrochemical cells: • Galvanic (voltaic) cells • Electrolytic cells. Electrolytic cell Galvanic / voltaic cell Electrolytic cell Galvanic / voltaic cell voltmeter Power + electrode – source v electrode Cathode – Anode – where (where oxidation occurs. reduction occurs.) electrolyte electrolyte salt bridge electrolyte electrodes Electric energy → chemical energy Chemical energy → electric energy No salt bridge Salt bridge Non-spontaneous redox reactions Spontaneous redox reactions Endothermic reaction Exothermic reaction Positive electrode → anode → oxidation Negative electrode → anode → oxidation Negative electrode → cathode → reduction Positive electrode → cathode → reduction emf is always negative emf is always positive © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 157 Mind the Gap Chemistry

5 Unit Activity 2 Give ONE word for the following statements: 1. The electrode in a galvanic cell at which reduction occurs. (1) 2. Anode in an electrolytic cell. (1) 3. The type of electrochemical cell in which electrical energy is converted to chemical energy. (1) 4. TRUE or FALSE? An electrolytic cell converts mechanical energy to electrical energy.  (1) 5. Which ONE of the following half-reactions occurs at the cathode during the electrolysis of an aqueous CuCℓ2 solution? A. Cℓ2 + 2e– → 2Cℓ– B. Cu+ + e– → Cu C. 2Cℓ– → Cℓ2 + 2e– D. Cu2+ + 2e– → Cu (2) 6. The gain of electrons by a substance in a chemical reaction is known as … A. Oxidation B. Reduction C. Electrolysis D. Oxidation and reduction (2) [8] Solutions Multiple choice 1. Cathode3 (1) questions are given TWO marks in the 2. Positive electrode3 (1) exam 3. Electrolytic3 (1) 4. This statement is FALSE. This cell converts electrical energy to chemical energy.3 (1) 5. D 3 3 (2) 6. B 3 3 (2) [8] 158 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 5.4 Electrolytic cells Electrodes are made from unreactive substances (platinum, Pt or graphite, C) and are placed in the same ionic solution or molten ionic compound The electrons flow in the external circuit from Negative ions (anions) move Positive ions (cations) move the negative pole of the (migrate) to the anode (migrate to the cathode battery to the cathode (–) where they are oxidised where they are reduced a) Electrolysis of molten ionic compounds: • The negative ions migrate (move) to the anode where they are oxidised. • The positive ions migrate to the cathode where they are reduced. b) Electrolysis of ionic solutions: Remember these rules. The anions (negative ions) from the ionic compound and the hydroxyl They will help you ions (OH–) from the water migrate to the anode (positive electrode). understand the process The anions from the ionic compound compete with the hydroxide taking place inside the ions (OH–) from the solution, to be oxidised. electrolytic cell. Rules: If the ionic compound contains: • halide ions (Cℓ–; Br– or I–), the halide ion will be oxidised and NOT the hydroxide ion. The product will therefore be the corresponding halogen. • sulphate (SO42–) or nitrate (NO3–) ions, the hydroxide ion (OH–) will be oxidised to produce oxygen, O2(g). The cations (positive ions) from the ionic compound and the H+ ions from the water migrate to the cathode (negative electrode). The cations from the ionic compound compete with the H+ ions to be reduced. Rules: If the ionic compound contains: • cations of metals with a positive electrode potential, these PAY SPECIAL ATTENTION NB NB cations will be reduced to form the corresponding metal e.g. Cu. See the table on electrode potentials provided in introduction to this book. Remember to indicate the charge on the ion • cations of metals with a negative electrode potential (e.g. Zn2+), HINT! the H+ ions will be reduced to form hydrogen, H2(g). See the hint table on electrode potentials provided in his book. e.g. EG - worked examples exams © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 159 Mind the Gap Chemistry

HINT! 5 Unit hint e.g. Worked example 1 EG - worked examples 1. Explain how the electrolysis of a copper (II) chloride solution CuCℓ2(aq) takes place. exams Solution 1. The anions Cℓ– and OH– migrate to the positive anode. The Cℓ– ions are oxidised. oxidation half-reaction: 2Cℓ–(aq) → Cℓ2(g) + 2e– The cations (Cu2+ and H+) migrate to the cathode. The Cu2+ ions are reduced. activity PAY SPECIAL ATTENTION NB half-reaction: reduction Cu 2+ (aq)boy+ 2e–NB ACTIVITIES → Cu(s) and girl-left and right of page The net reaction: Cu2+(aq) + 2Cℓ–(aq) → Cu(s) + Cℓ2(g) activity The solution is initially blue due to the presence of the Cu2+ ions, but as they are reducedHINT! to Cu(s), the solution turns colourless and red-brown hint Cu(s) is deposited on the cathode. Step by step comment comment e.g. Worked example 2 EG - worked examples PAY SPECIAL ATTENTION 2. Predict the products of the electrolysis of a copper (II) sulphate solution CuSO4(aq) NB NB exams Solution 2. • At the anode:HINT!O2(g) •hint At the cathode: Cu(s) activity e.g. Worked example 3 EG - worked examples ACTIVITIES boy and girl-left and right of page activity 3. Predict the products of the electrolysis of a zinc chloride solution ZnCℓ2(aq) exams Solution 3. • At the anode: Cℓstep comment 2(g) Step by comment • At the cathode: Zn(s) activity c) Electrolysis of water: ACTIVITIES boy and girl-left and Water is a weak electrolyte right (weak electric conductor). A small amount of page of dilute sulphuric acid (H2SO4) is added to water to increase its activity conductivity. When an electric current is passed through the acidified water, oxidation and reduction reactions take place. Step by step comment comment 160 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 At the anode: oxidation half-reaction: 2H2O → O2(g) + 4H+ + 4e– E0 = +1,23 V At the cathode: reduction half-reaction: 2H+ + 2e– → H2(g) E0 = 0,00 V The net reaction: 2H2O(ℓ) → 2H2(g) + O2(g) 5.5 Application of electrolysis • Electroplating • Production of chemicals e.g. chlorine gas, hydrogen gas and sodium hydroxide (membrane cell, chlor-alkali industry) • Extraction of metals e.g. aluminium • Refining (purifying) of metals e.g. copper. a) Electroplating Electroplating is the process of putting a metallic coating on an object using electrolytic reactions. Electroplating is used to protect metals that oxidise easily, by covering them with a thin layer of a metal that does not oxidise easily e.g. chromium, silver or gold. A relatively cheap metal is covered by an expensive metal. Silver is used to cover cutlery — it is too expensive to make a spoon of pure silver, and it is too weak to use — while chromium can be used to cover car parts like bumpers. © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 161 Mind the Gap Chemistry

HINT! 5 Unit hint e.g. Worked example 4 EG - worked examples An attractive silver appearance can be created by electroplating artefacts (objects) made from cheaper metals, such as nickel, with silver. The simplified diagram here exams represents an arrangement that can be used to electroplate a nickel artefact with silver. activity ACTIVITIES boy and girl-left and right of page 1. Which electrode (cathode or anode) does the nickel artefact represent? activity 2. Name the metal represented by electrode Y. 3. Write down the half-reaction responsible for the change that occurs at the surface of the artefact. 4. Give a reason why the concentration of the electrolyte remains constant during electroplating. Step by step comment comment 5. In industry some plastic articles are sometimes electroplated. Explain why plastic must be coated with graphite before electroplating. 6. Give a reason why, from a business point of view, it is not advisable to plate platinum with silver. Solutions 1. Cathode 2. Silver 3. Ag+ + e– → Ag 4. The rate of oxidation of silver at the anode is equal to the rate of reduction of silver ions at the cathode. 5. Plastic is a non-conductor. It must be covered with a conducting layer so that it can act as the cathode. Graphite is a conductor. 6. Platinum is expensive and is more durable than other metals. You would normally electroplate a cheap metal with an expensive, durable metal and not the other way round. 162 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 Activity 3 1. TRUE or FALSE? During electroplating of a steel teaspoon with silver, the teaspoon is the cathode and the electrolyte is a solution of any soluble compound [2] Solution 1. FALSE. 3… the electrolyte is a solution of a soluble silver compound. 3 [2] b) Refining of copper Copper which is mined is impure and the copper ore can be refined as follows by means of electrolysis: Cu(s) → Cu2+(aq) + 2e– Cu2+(aq) + 2e– → Cu(s) © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 163 Mind the Gap Chemistry

HINT! 5 Unit hint e.g. Worked example 5 EG - worked examples Impure copper can be purified by the process of electrolysis. The simplified diagram represents examsan electrolytic cell used to purify copper. activity ACTIVITIES boy and girl-left and right of page activity 1. Define the term electrolysis. 2. Which electrode, P or Q, consists of the impure copper? Explain how you arrived at your Step answer. by step comment comment 3. Write down the half-reactions that take place at electrodes P and Q.  4. During purification, metals such as silver and platinum form sludge at the bottom of the container. Refer to the relative strengths of reducing agents to explain why these two metals do not form ions during the purification process. 5. Explain why the concentration of the copper (II) sulphate solution remains constant. Assume that the only impurities in the copper are silver and platinum. 6. Why is the sludge of economic importance? Solutions 1. Electrolysis is a process during which electrical energy is converted to chemical energy. It is the process in which electricity is used to bring about a chemical change / decompose / break compounds into components. 2. P: P is the positive electrode / anode. The impure Cu is oxidised at the positive electrode / anode. 3. P: Anode: Cu(s) → Cu2+(aq) + 2e– oxidation impure Q: Cathode: Cu2+(aq) + 2e– → Cu(s) reduction pure 4. Platinum and silver are both weaker reducing agents than copper and will not be oxidised to form ions. 5. The rate at which copper is oxidised (at the anode) is equal to the rate at which copper ions are reduced (at the cathode). 6. Silver and platinum are valuable and expensive metals and can therefore be sold at a profit. 164 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 Activity 4 1. Refer to the diagram in the worked example above. One of the electrodes consists of impure copper and the other one of pure copper. 1. What type of power source is used to drive the reaction in this cell? Write down only AC or DC. (1) 2. Give a reason why the copper(II) sulphate is dissolved in water before it is used in this cell. (1) When an electric current passes through the solution, electrode P becomes coated with copper. 3. Is electrode P the cathode or the anode? Support your answer by writing the half-reaction that takes place at electrode P.(2) 4. Write down the half-reaction that takes place at electrode Q.(2) It is found that the impure copper plate contains platinum. The platinum forms a residue at the bottom of the container during electrolysis. 5. Refer to the relative strengths of reducing agents to explain why platinum forms a residue at the bottom of the container. (2) 6. How will the concentration of the copper(II) sulphate solution change during electrolysis? Write down only INCREASES, DECREASES or REMAINS THE SAME. (3) [11] Solutions 1. DC 3 (1) 2. Free ions needed to conduct electricity 3 (1) 3. Cathode. 3 Cu2+ + 2e– → Cu 3 (2) 4. Cu → Cu2+ + 2e– 3 3 (2) 5. Pt is a weaker reducing agent than Cu 3and will not be oxidised 3 (2) OR Cu is a stronger reducing agent than Pt and will be oxidised 6. Remains the same.3 The rate at which Cu is oxidised at the anode 3 equals the rate at which Cu2+(aq) is reduced at the cathode 3 (3) [11] c) The recovery or extraction of aluminium PAY SPECIAL ATTENTION NB metal from bauxite Aluminium: A very important application • is one of the most abundant metals on earth, yet it is expensive – of electrolysis is the recovery largely because of the amount of electricity needed to extract it. of aluminium. South Africa HINT! • has the following properties: a low density; the ability to resist importshint bauxite, an aluminium corrosion; is very ductile; can be rolled out in thin layers; is ore. lightweight and is a good electrical conductor. © Department of Basic Education 2015 e.g. UNIT 5 ELECTROCHEMISTRY 165 EG - worked examples Mind the Gap Chemistry

5 Unit The process involves: • Aluminium oxide (Aℓ2O3) is extracted from the bauxite and it is then heated. • The melting point of Aℓ2O3 is higher than 2 000°C. Cryolite (Na3AℓF6) is added to the ore before it is heated. Cryolite does not lower the melting point of Aℓ2O3 but dissolves it. Aℓ2O3 dissolved in a molten cryolite can be electrolysed easily. This means that less electricity is required to extract the aluminium, thereby decreasing the extraction cost. • The ions, Aℓ3+ and O2–, are formed during the extraction. • The mass of the anode therefore gradually decreases. The cost of replacing the anode when it is depleted adds to the high cost of the aluminium extraction. At the high temperatures, the O2 produced at the anode reacts with the carbon electrodes to form CO2. C(s) + O2(g) → CO2(g) The O2– ions are oxidised at the anode to produce O2(g): 2O2– → O2 + 4e– • The ecological impact of Aℓ extraction: Loss of landscape due to the size of the chemical plant needed; disposal of red mud (iron (III) oxide) formed during extraction of aluminium oxide from bauxite, into rivers and lagoons and into ground water. • Environmental impact of Aℓ extraction: Carbon dioxide from the burning of the anodes contributes to the greenhouse effect, causing global warming. Fluorine and its compounds lost from the cryolite during the electrolysis process are poisonous. Chemicals in the red mud dams drain into the soil and contaminate groundwater. Pollution caused by power generation (for electrolytic process) using coal fired plants leads to acid rain and adds to the greenhouse effect. Noise pollution from the extraction plant. 166 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 Activity 5 Give ONE word for the following phrase: 1. The main ore from which aluminium is extracted. (1) 2. The name of the chemical substance in which Aℓ2O3 is dissolved to lower its melting point during the industrial extraction of aluminium.  (1) [2] Solutions 1. Bauxite 3 2 Cryolite 3 [2] Activity 6 1. In an aluminium smelter, aluminium metal is extracted from bauxite, a hydrated aluminium oxide, via an electrolytic process. 1.1 Write down the energy conversion that takes place in an electrolytic cell. (2) 1.2 Write down the equation for the half reaction responsible for the formation of aluminium metal in a smelter. (2) 1.3  Explain in terms of the relative strength of oxidizing agents why the electrolytic production of aluminum requires more electrical energy than that of iron or copper. (2) 1.4  Name TWO advantages that the use of aluminium has over that of iron.  (2) 2. A huge aluminium smelter is planned for Coega in the Eastern Cape. When operational, it will consume 1350 MW of electricity, or 4% of the nation’s total electrical energy. It is estimated that 5200 jobs will be created at the peak of construction. About 1000 workers will be employed on a full-time permanent basis, and between 200 and 300 full-time subcontractors will also be directly associated with the smelter. (Source: www.engineeringnews.co.za; www.groundwork.org.za). 2.1 Taking the present South African socio-economic realities into account, give ONE reason why the aluminium smelter should: (a) Not be built (1) (b) Be built  (1) 2.2 Give ONE reason why environmental activists oppose the construction of the smelter.  (1) [11] © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 167 Mind the Gap Chemistry

5 Unit Solutions 1.1 Electrical energy → chemical energy 3 3 (2) 1.2 Aℓ + 3e → Aℓ 3 3 3+ – (2) 1.3 Aluminium has a lower reduction potential (–1,66 V) 3/ Weaker oxidizing agent compared to that of iron (–0,44 V) [and copper (+0,34 V)]. The aluminium ions therefore require a large amount of energy to be reduced/ will reduce more difficultly than iron (and copper). 3 (2) 1.4 It is much lighter for the same strength (or stronger for the same mass). 3 It is corrosion free.3 (2) 2.1 (a) It will consume huge amount of electricity. 3/ (1) Will cause power failures 3 (any 2) (b) It will create jobs 3/Create foreign investment. 3 (1) Contribute to GDP 3 (any 3) 2.2 The production of the large amount of electricity used 3 enhances the greenhouse effect (or climate change) 3 OR The process is responsible for toxic fluoride waste OR (1) pollution. (any one) [11] 168 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 5.6 Voltaic (Galvanic) cells Remember: • A Galvanic or voltaic cell is a cell in which chemical energy is converted to electrical energy spontaneously. We therefore use a chemical reaction to produce electricity. E.g. standard AA / penlight batteries. Consider a copper-zinc cell: As the oxidation and A strip of zinc metal is placed in a zinc ion solution. A strip of copper is reduction reactions take placed in a separate beaker in an aqueous copper (II) ion solution. These place in two different solutions are the electrolytes. As they contain ions which dissociated when beakers, we refer to indirect or external the salts were dissolved in water, they are good electric conductors. electron transfer. • The half reactions that occur in each half cell: Zn(s) → Zn2+ + 2e– Cu2+ + 2e– → Cu(s) REMEMBER When redox half-reactions are to be written, the correct single arrow should be used. © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 169 Mind the Gap Chemistry

5 Unit Zn half-cell Cu half-cell Oxidation half-reaction at the Reduction half-reaction at the anode which is always the negative cathode which Is always the positive electrode (e– are given off). electrode. AL ATTENTION NB The stronger reducing agent is The stronger oxidising agent is oxidised (gives off electrons) reduced (receives electrons) • In redox reactions, Zn(s) → Zn2+ + 2e– Cu2+ + 2e– → Cu(s) most learners struggle to (read from right to left) (read from left to right) understand the difference between the name of the Zn is the stronger reducing agent Cu2+ is the stronger oxidising agent process, such as oxidation The anode always decreases in The cathode always increases in and the substance, for mass if the reducing agent is a solid mass if the reduced product is a example the reducing (The Zn rod is corroding). solid (copper deposits on rod). ed examples agent. Net cell reaction: • Be very careful not to • Add the two half reactions together, eliminating the electrons from both write reduction agent sides. instead write reducing • If necessary, multiply each half-reaction with suitable coefficients to ensure agent. that the number of electrons lost equals the number of electrons gained. • Be able to clearly Oxidation half-reaction: Zn(s) → Zn2+ + 2e– distinguish between the Reduction half-reaction: Cu + 2e → Cu(s) 2+ – oxidising and the reducing agent. Net cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) a) Salt bridge – part of electrochemical cell (usually a tube) containing ACTIVITIES boy and girl-left and electrolyte providing electrical contact between two solution. right of page The salt bridge: • Completes the circuit. • Keeps the two electrolytes in the two half-cells separate so that they do not mix. PAY SPECIAL ATTENTION NB • NB Allows movement of ions between the electrolytes, so as to ensure electrical neutrality i.e. it acts as an ion exchanger. p comment comment • Contains a saturated salt solution (not 1 mol⋅dm–3) of either Potassium chloride is not suitable for a silver half-cell, KNO3 or KCℓ. because AgCℓ is formedHINT! and • The electrolyte in the salt bridge must contain ions which are hint this compound is insoluble, weak reducing agents and weak oxidising agents. This will ensure that these ions are not oxidised or reduced, but act as thus a precipitate will form. spectator ions. e.g. EG - worked examples b) Electrolyte: • The salt of the compounds need to be soluble. • Suitable soluble salts of zinc and copper are zinc nitrate and copper AL ATTENTION exams nitrate. NB • The electrolyte solutions in this cell is Zn(NO3)2(aq) and Cu(NO3)2(aq). • As the Zn(s) is oxidised to Zn2+(aq), the concentration of the Zn2+ The nitrate salts of ionic ions increases in the diagram above. This means that the Zn half- compounds are very suitable cell will start building up a positive charge. As the electrons will only as an electrolyte, since the move away from a negative potential, the function of the cell would salts are solubleactivity in water. become impaired. • In the same way as the Cu2+ ions are reduced to Cu(s), the ACTIVITIES boy concentration and girl-left and of the Cu2+ ions decreases. This leads to the copper right of page d examples activity 170 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 half-cell becoming less positive. The electrons are therefore not attracted to the cathode as strongly and the function of the cell would become impaired (weaker). • The migration of the NO3– anions in the salt bridge to the anode (the Zn half-cell) and of the K+ ions in the salt bridge to the cathode (the Cu half-cell), cancel this build-up of undesired charge and maintain the cell’s electric neutrality. The cell therefore continues to function properly. c) Electron current: • Electrons always flow from the anode, through the external circuit, to the cathode. • In the Cu/Zn cell, electrons therefore flow from the zinc (Zn) to the copper (Cu). 5.7 Cell notation The structure of the galvanic cell may also be represented in symbols. Rules: • Always start with the anode on the left and end with the cathode on the right. • Use a / to separate the anode or cathode from its electrolyte. • Represent the salt bridge with the symbol //. • Inert electrodes (usually Pt or C) and phases are usually included: Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s) electrode / electrolyte // electrolyte / electrode anode (–) salt bridge cathode (+) e.g. Pt/reducing agent/oxidised species//oxidising agent/reduced species/Pt Pt / Cℓ–(aq) / Cℓ2(g) // F2(g) / F–(aq) / Pt © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 171 Mind the Gap Chemistry

5 Unit Activity 7 Give ONE word for the following phrase: 1. The electrode in a galvanic cell at which reduction occurs. (1) 2. The component of a galvanic cell that allows for the movement of ions between the half-cells. (1) 3. Which statement is CORRECT for a Zn-Cu galvanic cell that operates under standard conditions? (Standard Conditions are defined in the tables at the beginning of this book) A The concentration of the Zn2+ ions in the zinc half-cell gradually decreases. B The concentration of the Cu2+ ions in the copper half-cell gradually increases. C Negative ions migrate from the zinc half-cell to the copper half-cell. D The intensity of the colour of the electrolyte in the copper half-cell gradually decreases. (2) 4. The reactions below occur in two different electrochemical cells X and Y. Cell X: CuCℓ2(aq) → Cu(s) + Cℓ2(g) Cell Y: Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq) Which ONE of the following correctly describes the substance that forms at the CATHODE of each of these cells? Cell X Cell Y A Cℓ2(g) Cu(s) B Cu(s) Cu(s) C Cℓ2(g) ZnSO4(aq) D Cu(s) ZnSO4(aq) (2) 5. Which one of the following statements regarding a copper-silver galvanic cell is TRUE? A Silver is formed at the anode B Copper is formed at the anode C Silver is formed at the cathode D Copper is formed at the cathode (2) [8] Solutions 1. Cathode 3 (1) 2. Salt bridge 3 (1) 3. D 33 (2) 4. B 33 (2) 5. C 33 (2) [8] 172 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 5.8 Standard electrode potentials PAY SPECIAL ATTENTION The Reactivity Series is a list of substances which are arranged in order of NB their ability to act as reducing agents or as oxidising agents. You will notice that all ionic The Table of Standard Reduction Potentials lists the standard electrode compounds, for example, potentials (E0 values) for various compounds. There are two tables, they H2SO4, HNO3 and HCℓ HINT! are similar, but the entries are arranged in opposite directions. must behint dissociated into ions, before we can use • The tables of Standard Reduction Potentials can be used to: the information in the Table • identify oxidising and reducing agents of Standard Reduction • write balanced redox reaction equations • predict whether a redox reaction takes place spontaneously or e.g. Potentials. EG - worked examples not • calculate the emf of a voltaic cell by using one of the following The tables are given formulae (formulas): examsexaminations and in do NOT HAVE TO be memorised E0cell = E0cathode – E0anode OR E0cell = E0oxidising agent – E0reducing agent REMEMBER The formula must be written activity without any abbreviation in the ACTIVITI and girl- subscript. right of p activity PAY SPECIAL ATTENTION NB NB Step by step com Further information is discussed on the Table of Standard Electrode Potentials (shown here HINT! so you know which table to look for). This hint table is given at the beginning of this book. E.g. Stronger reducing agents displace weaker reducing agents, and remember to always start with the oxidation half-reaction. e.g. EG - worked examples Please read that information now. exams activity ACTIVITIES boy and girl-left and right of page activity © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 173 Mind the Gap Chemistry Step by step comment comment

HINT! 5 Unit hint e.g. Worked example 6 EG - worked examples A piece of zinc metal (Zn) is placed in a beaker containing a copper (II) sulphate solution and a piece of copper metal (Cu) is placed in a beaker containing a zinc (II) sulphate solution. exams 1. Predict which redox reaction will take place spontaneously. 2. Motivate your answer. 3. Write the oxidation half-reaction, the reduction half-reaction and the net cell reaction for the spontaneous redox reaction. Solutions activity 1. Zn is oxidised to Zn2+ and Cu 2+ is boy ACTIVITIESreduced to Cu. The redox reaction in the beaker containing the piece of zinc (Zn) and the copper (II) sulphate and girl-left and right of page solution with Cu2+ ions is spontaneous. activity 2. Zn is a stronger reducing agent than Cu (see reducing ability on table), i.e. Zn will be oxidised more readily than Cu and Cu2+ is a stronger oxidising agent than Zn PAY SPECIAL 2+ (see oxidising ability on table) i.e. Cu2+ will ATTENTION NBreduced more readily than Zn2+. Therefore be NB Zn will displace Cu from CuSO4(aq). Step by step comment comment 3. oxidation half-reaction : Zn(s) → Zn2+(aq) + 2e– reduction HINT! half-reaction: Cu2+(aq) + 2e– → Cu(s) hint Overall or net ionic reaction is: Zn(s) + Cu (aq) → Zn (aq) + Cu(s) 2+ 2+ e.g. Worked example 7 EG - worked examples 1. Use the Table of Standard Reduction Potentials to determine whether Ag will spontaneously displace Zn2+ ions from a zinc sulphate (ZnSO4) solution. Motivate your answer. exams 2. Write the oxidation half-reaction, the reduction half-reaction and the net ionic reaction which takes place when a piece of copper (Cu) is dipped in a silver nitrate (AgNO3) solution. Solutions 1. No, the reaction will not take place. activity Ag is a weaker reducing agent than Zn, therefore it will not be able to displace Zn from the zinc ACTIVITIES sulphate boy solution. and girl-left and 2. oxidation half-reaction: right of page Cu → Cu2+ + 2e– reduction half-reaction: Ag+ + e– → Ag+ activity Multiply this reaction by 2, so that the number of electrons released by the copper atom equals the number of electrons gained. 2Ag+ + 2e– → 2Ag+ net ionic reaction: Step by step comment Cu +comment 2Ag+ → Cu2+ + 2Ag+ 174 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 Activity 8 1. Which one of the following solutions can be stored in an aluminium container? (Use the Table of Standard Reduction Potentials.) A CuSO4(aq) B ZnSO4(aq) C NaCℓ(aq) D Pb(NO3)2(aq) (2) [2] Solution 1. C 33 [2] Activity 9 The galvanic cell represented in the diagram consists of a Mg electrode dipped into a Mg(NO3)2 solution, and a Pb electrode dipped into a Pb(NO3)2 solution. Assume that the cell operates under standard conditions. 1. State TWO standard conditions under which this cell operates. (2) 2. Write down the half-reaction that takes place in half-cell A. (2) 3. Write down the cell notation for this cell. (2) 4. Calculate the emf of this cell. (3) 5. How will each of the following changes influence the value of the cell’s emf calculated in QUESTION 2.4? Write down only INCREASES, DECREASES or REMAINS THE SAME 5.1 An increase in [Mg2+(aq)] (1) 5.2 An increase in [Pb2+(aq)] (1) 6. In which direction, from half-cell A to B or from half-cell B to A, do cations move within the salt bridge to maintain electrical neutrality? Explain how you arrived at your answer. (3) [14] © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 175 Mind the Gap Chemistry

5 Unit Solutions 1. Temperature: 298 K (25 °C) 33 (2) 2. Mg(s) → Mg2+(aq) +2e– 33 (2) 3. Mg(s)/Mg2+(1 mol·dm–3) 3// Pb2+(aq)(1 mol·dm–3)/Pb(s) 3 (2) 4. E0cell = E0cathode – E0anode 3 = - 0,13 – (–2,36) 3 = 2,23 V 3 (3) 5. 5.1 DECREASES 3 (1) 5.2 INCREASES 3 (1) 6. Half-cell A to half-cell B 3 Concentration of positive ions/cations 3/ Pb2+ ions decreases in half-cell B. 3 OR Concentration of positive ions/cations/Mg2+ ions increases in half- cell A To prevent a build-up of positive ions in half-cell A and negative ions in half-cell B /for electrical neutrality. Positive ions migrate from/through the salt bridge (3) [14] Activity 10 Learners conduct an investigation to determine which combination of two half-cells will provide the largest emf under standard conditions. Three half-cells, represented as A, B and C in the table below, are available. Half-cell A Half-cell B Half-cell C Mg/Mg2+ Pb/Pb2+ Aℓ/Aℓ3+ The learners set up galvanic cells using different combinations of the above half-cells. 1. Write down the standard conditions under which these cells operate.(2) 2. Write down the dependent variable in this investigation. If you can’t remember what the dependent (and independent) variables are, please see the introductory materials at the beginning of this book. (1) 3. Use the Table of Standard Reduction Potentials to determine which one of the three half-cells (A, B or C) contains the: 3.1 Strongest reducing agent. (1) 3.2 strogest oxidizing agent. (1) 176 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 4. Without any calculation, write down the combination of two half-cells which will produce the highest emf. Write down only AB, BC or AC.(1) 5. One group of learners set up a galvanic cell using half-cells A and B, as shown below. X is one of the components of the galvanic cell. 5.1 Write down the NAME or SYMBOL of the substance that will act as the anode in this cell. Give a reason for the answer.(2) 5.2 Calculate the initial emf of this cell. (3) 5.3 How will an increase in the concentration of the electrolyte in half-cell B affect the initial emf of this cell? Write down only INCREASES, DECREASES or REMAINS THE SAME. (1) 5.4 Briefly explain how component X ensures electrical neutrality while the cell is functioning. (1) [13] Solutions 1. Temperature 25 °C /298 K 3 Concentration of electrolytes = 1 mol·dm–3 3 (2) 2. emf/potential difference 3 (1) 3.1 Half-cell A 3 (1) 3.2 Half-cell B 3 (1) 4. Combination AB 3 (1) 5.1 Magnesium/Mg. 3Is oxidised/loses electrons/increase in oxidation number/stronger reducing agent 3 (2) 5.2 E cell = E cathode – E anode 3 0 0 0 = – 0,13 – (–2,36) 3 = 2,23 V 3 (3) 5.3 Increases 3 (1) 5.4 Allows for the migration of positive ions to the cathode half-cell 3 (1) OR Allows for the migration of negative ions to the anode half-cell [13] © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 177 Mind the Gap Chemistry

5 Unit 5.9 The standard hydrogen electrode The hydrogen gas/hydronium ion electrode has been chosen as standard half-cell. The potential of any electrode (E0) relative to the standard half-cell can be measured by connecting it to this Activity 11 standard half-cell as a voltaic cell. The reading on the voltmeter is 1. Which is the the reference electrode in the diagram above?  (1) assigned to the other 2. TRUE or FALSE? A standard cell that has a negative emf cannot half-cell. be used as a galvanic cell. (2) 3. TRUE or FALSE? The standard conditions used to measure standard electrode potentials are: A temperature of 273 K, a concentration of 1 mol·dm–3, and a pressure of 101,3 kPa. (2) 4. Which ONE of the following containers can be used to store an iron(II) sulphate solution? A Aℓ B Mg C Ni D Zn(2) [7] Solutions 1. Hydrogen half-cell 3 (1) 2. TRUE 33 (2) 3. FALSE. 3 A temperature of 298 K (25 °C) — the question gives the PAY SPECIAL ATTENTION NB temperatureNB incorrectly 3 (2) 4. C 33 (2) A half-cell potential of [7] 0,00 V has been assigned to the standard hydrogenHINT! hint electrode. The half-cell potential values are given on the right hand side of the table. e.g. EG - worked examples 178 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015 exams  Mind the Gap Chemistry

Unit 5 a) To determine the standard electrode potential of the copper half-cell b) To determine the standard electrode potential of the zinc half-cell © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 179 Mind the Gap Chemistry

5 Unit 5.10 The emf of an PAY SPECIAL ATTENTION NB NB electrochemical cell • A voltmeter connected to the zinc / copper cell has a reading of 1,1 V. It reads the difference between the electric potentials of the two separate HINT! electrodes. hint • This reading of 1,1 V can also be referred to as the potential difference or cell potential or emf of the cell. AL ATTENTION NB e.g. Worked example 8 EG - worked examples Always use the values exactly Using the Cu / Zn cell as an example: as they stand on the table; do not round off. exams E0cell = E0cathode – E0anode = +0,34 – (–0,76) = +0,34 + 0,76 = +1,1 V ed examples • The cell potential (voltage) does not depend on the surface area of Now you know why voltage is called the electrode. potential difference • activity The bigger the surface area of the electrode, the bigger the current produced by the cell. ACTIVITIES boy • In general the cell potential and increases girl-left andslightly with an increase in right of page temperature. • activity The reaction in the cell reaches equilibrium when the cell is completely discharged (flat or dead). This usually means that the concentration of the cations in the reduction half-cell (i.e. at the cathode) has decreased to zero. Using the Table of Standard Reduction Step by step comment commentPotentials to predict whether a ACTIVITIES boy and girl-left and redox reaction will be spontaneous or not: right of page • Assume that the reaction will take place as suggested in the question. • Identify the reducing agent and the oxidising agent according to the given equation. • Calculate the cell potential for this reaction. comment • If the cell potential is positive, the reaction will be spontaneous i.e. it p comment will take place without the addition of any other form of energy. If it is negative, the reaction will not occur spontaneously. • If the two half reactions are very close to each other on the Table of Standard Reduction Potentials, heating the reaction mixture may supply enough additional energy for the reaction to take place. Steps Step 1. Use the Reduction Potentials to determine whether a reaction will take place spontaneously. Step 2. If the calculated cell potential is a positive value the redox reaction will take place spontaneously. 180 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 PAY SPECIAL ATTENTION NB NB H2S → S + 2H+ + 2e– OR HINT! S + 2H+ + 2e– ← H2S hint is the only correct way of writing a half-reaction • Note the direction of the single arrow changes, depending on whether the reaction is an oxidation or reduction half reaction. e.g. EG - worked examples • Please do NOT include the double arrow in the half reaction. • Always remember to include the charges on the ions in the half-cell reactions. exams AL ATTENTION NB When you have to explain the relative strength of oxidizing and reducing agents, write the explanations as follows: • Cu is a stronger oxidising agent than Mg. Do NOT state the position of a substance activity in terms of its position in the Reduction Table (e.g. Cu is above Mg) ACTIVITIES boy • Do NOT state the relative strength in terms and of relative reactivity only, for example, girl-left and right of page Mg is more reactive than Cu. ed examples activity Your correct answer would read as follows: Mg is a stronger reducing agent than Cu and therefore Mg will be able to reduce Cu2+ ions to Cu. PAY SPECIAL ATTENTION NB OR NB Step by step Explain the relative strength of the oxidizingcomment comment and reducing agents in terms of their relative strength as electron acceptor and donors. Mg is a stronger reducing agent than Cu because Cu has a stronger tendency to accept electrons HINT!than Mg. OR: Mg has a stronger tendency to donate hintelectrons to Cu than Cu has to donate electrons to Mg. ACTIVITIES boy and girl-left and e.g. Worked example 9 right of page EG - worked examples Batteries consist of one or more galvanic cells. A galvanic cell is a combination of two half-cells. examsJohn wants to determine which one of Options A or B, shown below, can be used to assemble a galvanic cell with the highest potential difference. p comment comment Option Combination of half-cells A Ag(s) in AgNO3(aq) & Ni(s) in Ni(NO3)2(aq) B Mg(s) in Mg(NO3)2(aq) & Ag(s) in AgNO3(aq) activity 1. Draw a fully labelled diagram of theACTIVITIES galvanicboy cell that John can use to measure the potential difference forandthegirl-left cell inandOption B. Use a positive (+) right of page and negative (–) sign to indicate the positive and negative electrodes activity respectively. © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 181 Mind the Gap Chemistry Step by step comment comment

5 Unit 2. Write down the oxidation and reduction half reactions as well as a balanced chemical equation, excluding spectator ions, for the net (overall) cell reaction for the galvanic cell in Option B. 3. Calculate the initial potential difference that can be obtained under standard conditions for the galvanic cell in Option B. 4. State TWO standard conditions that John must adhere to during the experiment, to ensure that the measured potential difference is the same as the calculated potential difference. 5. Write down the cell notation for the galvanic cell in Option A. 6. Without doing any calculations, determine which one of Option A or Option B should result in the galvanic cell with the highest potential difference. Refer to the relative strengths of the two reducing agents, as well as the two oxidising agents involved, to explain your answer. 7. How will each of the following changes influence the value of the cell’s emf calculated in Question 3? 7.1 An increase in [Mg2+(aq)]. 7.2 An increase in [Ag+(aq)]. 8. In which direction, from half-cell Mg2+ / Mg to Ag + / Ag, or from half-cell Ag + / Ag to Mg2+ / Mg, do cations move within the salt bridge to maintain electrical neutrality? Explain how you arrived at your answer. 9. Give two functions of the salt bridge. 10. Name the cathode and the anode for the galvanic cell in option B. 11. Which electrode will show an increase in mass for the galvanic cell in option B? 12. Distilled water is added to the Ag+ solution for the galvanic cell in Option B. How will the emf of the cell be influenced? Explain your answer. 13. Consider the galvanic cell in Option B. If 0,6 moles of electrons flow from the anode to the cathode, what will the decrease in mass of the anode be? 14. A 2 V bulb is connected to the cell in option B instead of the voltmeter. Will the bulb light up? Justify your answer. 15. Consider the galvanic cell in Option B. It is observed that after a while, the reading on the voltmeter drops to zero. We say the cell is ‘flat’ or ‘dead’. Explain this observation in terms of the concentrations of the solutions in the cell. 16. Give two uses of galvanic cells. Solutions 1. Mg (s) Ag (s) Mg (–) KNO3(aq) Mg (+) – Anode + Cathode Mg(NO3)2(aq) Ag(NO3)(aq) 2. oxidation: Mg(s) → Mg2+(aq) + 2e– reduction: Ag+(aq) + e– → Ag(s) net reaction: Mg(s) + 2Ag+(aq) → Mg2+(aq) + 2 Ag(s) 182 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 3. E0cell = E0cathode – E0anode = 0,80 – (–2,36) = 3,16 V 4. Ensure a temperature of 25 °C and Mg2+, Ni2+ and Ag+ solutions of concentration 1 mol·dm3. 5. Ni(s) / Ni2+(aq) (1 mol·dm–3) // Ag(s)+(aq) (1 mol·dm–3) /Ag(s) 6. Option B. The reaction leading to the highest emf (or potential difference) will be between the strongest reducing agent (Mg) and the strongest oxidising agent (Ag+). 7. An increase in the rate of the forward reaction will increase the emf of a galvanic cell. 7.1 decreases; the reverse reaction is favoured 7.2 increases; the forward reaction is favoured 8. Half-cell Mg2+ / Mg to Ag + / Ag Concentration of positive ions or cations; [Ag+] ions decreases in half-cell Ag + / Ag. OR Concentration of positive ions or cations; [Mg2+] ions increases in half-cell Mg2+ / Mg. To prevent a build-up of positive ions in half-cell Mg2+ / Mg and negative ions in half-cell Ag+ / Ag. For electrical neutrality, positive ions migrate from/through the salt bridge to the cathode. 9. Any two: It completes the circuit. It ensures electrical neutrality by allowing the migration of ions through the salt bridge. It separates the electrolytes. 10. cathode = Ag anode = Mg 11. cathode = Ag 12. By adding water to the solution, the [Ag+] will decrease. Thus the reverse reaction is favoured and emf decreases. 13. Mg(s) → Mg2+(aq) + 2e– (anode) From the balanced equation 1 mol Mg → 2 mol e– thus 0,3 mol Mg → 0,6 mol e– m __ n = ​ M ​ m=n×M molar mass (M) of Mg = 24 g·mol–1 m = 0,3 × 24 = 7,2 g mass loss is 7,2 g 14. The bulb will light up, as the emf of the cell is 3,16 V which is more than the required 2 V. 15. While the cell is in operation, the concentration of the cations in the reduction half-cell Ag+(aq) decreases. At the same time, the concentration of the cations in the oxidation half-cell Mg2+(aq) increases. The result is a gradual decrease in the cell potential until there is no further change in the concentration and equilibrium is reached, and the cell potential will be zero. 16. Any two: Torch cells, car batteries, batteries in toys and small appliances like remote controls. © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 183 Mind the Gap Chemistry

HINT! 5 Unit hint e.g. Worked example 10 EG - worked examples Rusting is an unwanted redox reaction. Iron rusts when exposed to oxygen and moisture. exams The unbalanced ionic equation for one reaction that occurs during rusting is represented below. Fe(s) + O2(g) + H2O(ℓ) → Fe2+(aq) + OH–(aq) Use the Table of Standard Reduction Potentials to answer the following questions for this reaction: 1. Write down the oxidation and the reduction half-reactions. activity 2. Write down the NAME of the substance that is reduced. ACTIVITIES boy 3. Perform a calculation to verify that this reaction is spontaneous. and girl-left and right of page Solutions activity 1. oxidation: Fe → Fe2+ + 2e– reduction: 2H PAY2O + O2 SPECIAL + 4e– → 4OH– ATTENTION 2. NB oxygen NB 3. oxidation: Fe → Fe + 2e 2+ – E0anode = –0,44 V comment Step by step comment reduction: 2H2O + O2 + 4e– → 4OH– E0cathode = 0,40 V E0cell = E0cathode – E0anode = 0,4 – (–0,44) = 0,84 V Because E0cellHINT!is positive, the reaction is spontaneous. hint e.g. Worked example 11 EG - worked examples Magnesium is used to protect underground iron pipes against rusting (cathodic protection). exams The diagram here shows an iron pipe connected to a magnesium bar. activity ACTIVITIES boy and girl-left and 1. Use the Table of Standard Reductionright Potentials of page to explain why magnesium can be used to protect an iron pipe against rusting. activity 2. The iron pipe in contact with the magnesium bar forms an electrochemical cell. What serves as the salt bridge of this cell? 3. Give a reason why the magnesium bar must be replaced after some time. 4. Write down a half-reaction to support your answer to Question 3.3. 5. Name TWO otherStep by step methods comment that can be used comment to protect iron pipes against rust. 6. State ONE advantage and ONE disadvantage of using plastic pipes instead of iron pipes. 184 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 Solutions 1. Mg is a stronger reducing agent (than Fe) and will be oxidised more readily than Fe which is a weaker reducing agent. 2. The salts which are dissolved in the moist soil. 3. Mg is continuously oxidised to Mg2+ and therefore its mass decreases i.e. it is used up. 4. Mg → Mg2+ + 2e– 5. Any two: Paint; electroplating e.g. galvanising; oil or waterproofing; plastic coating. 6. Advantages: Plastic is cheaper; plastic does not rust. Disadvantages: Plastic not biodegradable; plastic is not as strong as iron. Activity 12 Give ONE word for the following phrase: 1. The process taking place in a cell when an electric current passing through its electrolyte, results in chemical reactions at its electrodes. (1) 2. TRUE or FALSE? A battery labelled as 3 000 mA·h can deliver a current of 500 mA for 6 hours. (2) [3] Solutions 1. Electrolysis 3 (1) 2. True. 33 (2) [3] Activity 13 1. The most common filling for tooth cavities is ‘dental amalgam’ – a solid solution of tin and silver in mercury. If you bite on a piece of aluminium foil that is in contact with a dental filling in your mouth, you may feel a painful sensation because A the aluminium foil is hard. B a temporary galvanic cell has been set up whilst the aluminium and filling are in contact. C electrons are being transferred to the aluminium. D a temporary electrolytic cell has been set up whilst the aluminium and filling are in contact. (2) © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 185 Mind the Gap Chemistry

5 Unit 2. Which one of the following can be classified as a redox reaction? A NH3(g)+ HNO3(g) →NH4NO3(s) B FeS(aq)+ 2HCℓ(aq) → FeCℓ2(aq) + H2S(g) C Aℓ(s) + Cℓ2(g) →AℓCℓ3(s) D Mg(NO3)2(s) → Mg2+(aq) + 2NO3–(aq) [with H2O catalyst] (2) 3. A silver (Ag) spoon is left in a beaker containing copper nitrate, Cu(NO3)2, solution. What will be observed after some time? A The spoon will be covered with a thin layer of copper. B The nitrate will form NO2 gas and copper will form in the beaker. C The spoon will start to erode and its mass will decrease. D There will be no reaction. (2) [6] Solutions 1. B. 33 (2) 2. C. 33 (2) 3. D. 33 (2) [6] Activity 14 1. A group of learners set up an electrochemical cell using lead and copper half cells. 1.1 Which ONE of copper or lead will be the negative electrode? Give a reason for your answer. (2) 1.2 Use the Table of Standard Reduction Potentials and write down the reduction half-reaction that will take place in this cell. (2) 1.3 A 2V bulb is connected to the cell. Will the bulb light up? Justify your answer with a calculation. (5) 1.4 A voltmeter is now connected to the cell instead of the bulb. It is observed that after a while the reading on the voltmeter drops to zero. We say the cell is ‘flat’ or ‘dead’. Explain this observation in terms of the concentrations of the solutions in the cell. (3) 2. A cell such as the one described above is not much useful. However, the principle is used in batteries for cars, torches, computers, et cetera. These batteries are called secondary cells. One such battery is the mercury cell. The half reactions occurring in this cell are shown below. Zn(s) + 2OH–(aq) → ZnO(s) + H2O + 2e–……… (i) HgO(s) + H2O + 2e– → Hg(ℓ) + 2OH–(aq)….….. (ii) 2.1 Write down the overall cell reaction. (3) 2.2 Why does the use of this battery pose an environmental hazard? (1) [16] 186 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 Solutions 1.1 Lead. 3 Stronger reducing agent 3 OR Is oxidised preferably (2) 1.2 Cu2+(aq) + 2e– → Cu(s) 33 (2) 1.3 E0cell = E0cathode − E0anode 3 = 0,34– (–0,13) 3 = 0,47 V 3 (5) Bulb will not light, 3 energy from cell not sufficient 3 OR emf of cell is much less than 2 V needed for the bulb 33 1.4. While the cell is in operation, the concentration of the reactants (Cu2+(aq)) decreases. 3 At the same time the concentration of the products (Pb2+(aq)) increases. 3 The result is a gradual decrease in the cell potential until there is no further change in concentration and equilibrium is reached where the cell potential will be zero. 3 (3) 2.1 Zn(s) + HgO(s) → ZnO(s) + Hg(ℓ) 333 (3) 2.2 Mercury is poisonous 3 (1) [16] Activity 15 The discovery of electrochemical cells has revolutionised our way of life. The diagram below represents an electrochemical cell. 1. Name the type of electrochemical cell that converts chemical energy to electrical energy. (1) 2. If the electrochemical cell is set up as illustrated, there will be no reading on the voltmeter. Give a reason for this observation. (1) 3. Write down the value of the standard emf of the electrochemical cell when it is functioning. (1) 4. Write down the voltmeter reading when the net cell reaction in the above electrochemical cell reaches equilibrium. (1) © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 187 Mind the Gap Chemistry

5 Unit 5. Write down the equation for the reaction that occurs at the anode. (1) 6. Another electrochemical cell is set up under standard conditions by replacing the standard hydrogen half-cell with a standard magnesium half-cell. 6.1 Which electrode will undergo a decrease in mass? Give a reason for your answer. (2) 6.2 Calculate the initial emf of this electrochemical cell under standard conditions. (3) 6.3 After a while the emf of this electrochemical cell decreases. Explain this observation by referring to the concentration of the electrolytes. (2) 7. Electrochemical cells such as motor car batteries with plastic casings can harm the environment if not disposed of safely. Suggest TWO ways how motor car batteries can be safely disposed of. (2) [14] Solutions 1. Galvanic/voltaic cell 3 (1) 2. Incomplete circuit/No salt bridge 3 (1) 3. 0,76 V 3 (1) 4. Zero 3 (1) 5. Zn(s) → Zn2+(aq) + 2e– 3 (1) 6.1 Mg. 3 Mg is oxidised 3 (2) 6.2 E cell = E cathode – E anode 3 = 0,76 – (–2,36) 3 = 1,6 V 3 0 0 0 (3) 6.3 As the cell functions, the concentration of zinc ions (reactants) decreases 3 relative to the standard conditions and the concentration of magnesium ions (products) increases relative to standard conditions. The reverse reaction starts opposing the forward reaction causing the emf to decrease relative to standard conditions. 3 (2) 7. Neutralise acid before disposal. 3 Recycle plastic casing and lead electrodes 3 (2) [14] Activity 16 In 1780, Luigi Galvani discovered that when copper and zinc metal were connected to each other and if each free end touched different parts of the same nerve of a frog leg at the same time, the frog’s leg contracted. He called this “animal electricity”. 1. Briefly explain what this “animal electricity” really was. (1) The diagram shows an electrochemical cell setup under standard conditions using aluminium (Aℓ) and nickel (Ni) electrodes. AℓCℓ3(aq) and NiCℓ2(aq) are used as the electrolytes, and a solution of sodium nitrate (NaNO3(aq)) is used in the salt bridge. 188 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 Answer each of the following questions on this electrochemical cell: 2. The diagram indicates that electrons flow from metal X to metal Y. Identify: 2.1 Metal X (1) 2.2 Electrolyte B (1) 3. What is the concentration of electrolyte B? (1) 4. Write down the FORMULA of the substance that moves towards metal Y in the salt bridge. (1) 5. Write down the half-reaction that occurs at the cathode of this cell. (2) 6. Calculate the reading on the voltmeter. (3) 7. State what happens to the concentration of metal ions in the solution containing electrolyte as time goes by? (1) 8. 8.1 Consider 5.7 again and recall your answer. What effect does this have on the voltmeter reading? (1) 8.2 Briefly explain your answer to 5.8.1. (2) [14] Solutions 1. The chemical reaction between the zinc and the copper r eleased electrical energy 3 (1) 2.1 Metal X: Aℓ 3 (1) 2.2 Electrolyte B: NiCℓ2 3 (1) 3. 1mol·dm 3 –3 (1) 4. NO3 3– (1) 5. Ni2++ 2e– → Ni 3 3 (2) 6. E0cell = E0oxidising agent − E0reducing agent3 = –0,25V–(–1,66V) 3 = 1,41 V 3 (3) 7. Increase 3 (1) 8.1 The voltmeter reading decreases 3 (1) 8.2 As the [Aℓ ] increases, 3 the reverse reaction in the 3+ AℓAℓ3+ half-cell: Aℓ ⇌ Aℓ3+ +3e– is favoured.3 (2) OR The tendency of the net reaction Aℓ + Ni2+ → Aℓ3+ + Ni to proceed from left to right is reduced. This lowers the electrode potential of this half-cell, resulting in a lower cell potential. [14] © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 189 Mind the Gap Chemistry

5 Unit Activity 17 Tina wants to investigate the effect of the area of the metal plates used as electrodes in a galvanic cell on the emf of the cell. She sets up the following Zn/Pb cell under standard conditions and measures the emf. 1. Which electrode will show an increase in mass when this cell is functioning? (1) 2. Write down the equation for the half-reaction occurring at the anode. (2) 3. Calculate the emf that you would expect Tina to read on the voltmeter. (3) 4. Name TWO variables that should be controlled for during this investigation. (2) 5. Tina now replaces the two metal plates with ones of larger surface area, and takes the readings again. 5.1 How would you expect the new emf to compare with the one calculated in Question 6.3? (Only write SMALLER THAN, LARGER THAN or EQUAL TO) (1) 5.2 Explain your answer to Question 5.1 (1) 6. Tina now connects a resistor of low resistance across terminals A and B. She notes that the reading on the voltmeter immediately drops. 6.1 Give a reason for this observation. (1) 6.2 After some time she observes a further drop in the reading on the voltmeter. Give a reason for this observation. (2) [13] Solutions 1. Pb 3 (1) 2. Zn → Zn2+ + 2e– 33 (2) 3. E0cell = E0oxidising agent − E0reducing agent3 = –0,13 – (–0,76) 3 = 0,63 V 3 (3) 190 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 5 4. Temperature, 3 (initial) concentration of electrolytes 3 (2) 5.1 Equal to 3 (1) 5.2 Area/size of electrodes has no effect on the emf of a cell. It is still a standard cell 3 (1) 6.1 the cell has internal resistance 3 (1) 3 6.2 The emf decreases as the concentration of Pb (aq)2+ decreases. 3/The cell is running flat as the electrolyte concentration in the Pb cell decreases. 3 3 (any 2) (2) [13] Activity 18 Electrolysis is an important industrial process used to decompose compounds, extract metals from their ores and to purify metals like gold or copper. The simplified diagram below represents an electrolytic cell used to purify copper. 1. Define the term electrolysis. (2) 2. Which electrode, P or Q, consists of the impure copper? Explain how you arrived at your answer. (2) 3. Write down the half-reaction that takes place at electrode Q. (2) 4. During purification, metals such as silver and platinum form sludge at the bottom of the container. Refer to the relative strengths of reducing agents to explain why these two metals do not form ions during the purification process. (2) 5. Explain why the concentration of the copper(II) sulphate solution remains constant. Assume that the only impurities in the copper are silver and platinum.  (2) 6. Why is the sludge of economic importance? (1) [11] © Department of Basic Education 2015 UNIT 5 ELECTROCHEMISTRY 191 Mind the Gap Chemistry

5 Unit Solutions 1. The process in which electricity is used to bring about a chemical change / decompose / break compounds into components 33 (2) OR A process in which electrical energy is converted to chemical energy. 2. P. 3 P is the positive electrode /anode 3 (2) OR Oxidation takes place at the positive electrode/anode 3. Cu2+(aq) + 2e– → Cu(s) 33 (2) 4. Pt and Ag are both weaker reducing agents 3 (than copper) and will not be oxidised 3 (2) OR Cu is a stronger reducing agent (than Pt and Ag) and will be oxidised 5. The rate at which copper is oxidised at the anode, 3 is equal to the rate at which copper ions are reduced at the cathode. 3 (2) 6. Contains platinum and silver that are valuable/expensive metals 3 (1) [11] Keep going! 192 UNIT 5 ELE CTROCHEMISTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 6 The chlor-alkali industry The chlor-alkali industry is a large worldwide multimillion industry which produces compounds that are essential to human existence. The industry is based on the electrolysis of sodium chloride solutions. 6.1 Summary We only have to understand the electrolysis of a concentrated solution of sodium chloride and its use in the chlor-alkali industry. Let’s briefly discuss the membrane cell. Chlor-Alkali industry involves the electrolysis of a sodium chloride solution which produces: • chlorine gas; • hydrogen gas and • sodium hydroxide (aq). © Department of Basic Education 2015 UNIT 6 THE CHLOR-ALKALI INDUSTRY 193 Mind the Gap Chemistry

6 Unit 6.2 Definitions and terminology The processes involved are electrolysis, oxidation and reduction. Electrolysis: • is an endothermic reaction • involves a process during which electric energy is converted to chemical energy or brings about a chemical change • involves a non-spontaneous redox reaction. Oxidation: • takes place when electrons are given off • at the anode. Reduction: • takes place when electrons are received • at the cathode. You must know how to: • Identify the Terminology: processes involved Corrosive: Substance that damages and destroys substances (including • Give a description of the concept or term living matter) it comes into contact with (touches). • Give the correct definition Leaching: Movement of a substance (like NaOH (aq), pesticides or • Describe the process fertilisers) that are carried by water, down through soil (ground) to the and chemicals involved/ underground water table. The water table is the level below which the produced ground is saturated with water. • State uses of the products Carbon footprint: The amount of carbon dioxide and other greenhouse • State the gases released into the atmosphere due to the activities of a person, environmental issues organisation or process. Brine: A concentrated solution of NaCℓ(aq) 194 UNIT 6 THE CHLOR-ALKALI INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 6 6.3 The Chlor-alkali industry – Learn this table. reactants and products The table summarises the products of REACTANTS the electrolysis of a concentrated solution of • Sodium chloride solution, NaCℓ(aq) sodium chloride and • Produced by evaporation from sea water, salt lakes the uses of these products PRODUCTS Chlorine gas Hydrogen gas Sodium hydroxide (caustic soda) Cℓ2(g) H2(g) NaOH(aq) • Yellow-green gas • Colourless gas • Colourless solution PROPERTIES & • Poisonous: Do not • Explosive: No inhale. open flames or • Highly corrosive electronic sparks • Prevent leaching PRECAUTIONS nearby. into the ground • Prevent skin and eye contact. Reactant for Reactant for Reactant for production of: production of: production of: • hydrochloric acid, • hydrochloric acid, • soap and HCℓ(aq); HCℓ(aq); bleaching agents; • extraction of Cu, • ammonia, NH3(g), • paper; Au, Ti; • H2O2(g) (hydrogen • textiles; • paint; peroxide); • oven cleaners. USES • PVC plastics; • nylon; • insecticides and • rocket fuel; Used to: weed killers; • margarine • treat waste water • bleaching agents. from plant oils (sewage water). (hydrogenation of unsaturated Used to: hydrocarbons). • disinfect water. PAY SPECIAL ATTENTION NB NB REMEMBER: Due to the properties The membrane cell is preferred of the products, chlor-alkali plants (factories) are built far from living because the: areas as a safety precaution. HINT! • components of the cell are hint non-toxic; • products are more pure and • production costs are relatively low. e.g. EG - worked examples • Sodium chloride (table sale) NaCℓ(s) is an ionic compound and is very soluble in water (a polar solvent). • NaCℓ (s) dissociates in an aqueous exams solution – the ions in the ionic crystal lattice separate. • NaCℓ(s) H2O (ℓ) Na+ (aq) + Cℓ– (aq) • A sodium chloride solution, NaCℓ (aq), is known as brine. activity ACTIVITIES boy © Department of Basic Education 2015 UNIT 6 THE CHLOR-ALKALI and girl-left right of page and INDUSTRY 195 Mind the Gap Chemistry activity

6 Unit 6.4 The chlor-alkali industry – industrial process The chlor-alkali industry products can be produced by using a: • membrane cell • mercury cell • diaphragm cell Membrane Cell Diaphragm Cell Mercury Cell Purity and • Produces • Produces • Low yield of yield of Cℓ2 (g) relatively pure Cℓ2(g) that is Cℓ2(g) and of produced Cℓ2(g) that is mixed with NaOH(aq). mixed with O2(g) so the very small product is less amount of pure. O2(g). Purity of NaOH • High • Medium • Low (aq) produced • NaOH (aq) is mixed with NaCℓ (aq) Energy • Lowest • Relatively low • Very high consumption (Affects costs) Environmental • Lowest • High • Highest impact • Non-toxic, • Diaphragm • Mercury reusable is made from reacts with components asbestos. the NaCℓ (aq) - low setup and leaches costs. into the • Lowest energy groundwater. consumption therefore small carbon footprint. Health risks • No potential • Diaphragm • Mercury is health risks / is made from toxic. environmental asbestos – • Risks due to risks due to health risk contamination components. for workers of • H2(g) is – asbestosis groundwater. explosive and cancer. (Fish, animals and Cℓ2(g) is • H2(g) is and humans poisonous. explosive use the water. and Cℓ2(g) is • H2(g) is poisonous. explosive and Cℓ2(g) is poisonous. PAY SPECIAL ATTENTION NB NB • Very few companies still use mercury cells but many still use diaphragm cells. When new industrial sites are set up today, membrane cells are mostly used. HINT! hint 196 UNIT 6 THE CHLOR-ALKALI INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

' ' “(be)HORN ‘uoRnjos H epixospAy wnipos e WO) + ' epoue ay} 0} eueJquoW e6ueYyoxo ' H uones oy) YySnosyy eyes6yu ' youued YoIyA suojue (be) HO SY - | { Pue jusWpedWwoo spoue oy] Woy « | H ' euequew i aquoneo ayy YBnosy payes6iw eney yoiuMm suoned (be),eN OUL + “@poyjeo ay) SPseMo] BAOW ' (be),eN pue (be),y suoneo ey, + | jUaUedWOD apouzed 9y} Uj t ' H H : H “pasoys pue juawyeduioo H H oy) Woy panouas s} (6) “HOUL + | ' (a¢e'0-=,3) H ' (be) Hoz + (6)"H +_9z+(ao7HZ + | 1 “poonpes Ajipeas sJ0W Ss! pue EN { | ue quabe Busippro seBuonse sO'H | + se (6)'H O peonpes si (OH Woy) H+ | H Peonpes eq oO} H | Byeduioo (QF Woy) SuopeD (be),H Oy + | ' pue juewedwos epoue ey Woy ! 1 auesquiow sJuoNeS ay YBnosy payes6nu i ' ‘eney yoy suoged (be) en eyL «+ | H 1{-) powzes ous iy H (be)HOeNZ + (6)7H + (6)-20<— (2)07HZ + (be)2DeNZ 2U01}9BAd [89 (j]249A0) JON “paws Bureq (5)"0 pue (6)475 jo yonpoud pexjw e Ul JINSeJ Pinom siLQ se ‘epoUe Oy 0} BugesBjW ‘wWo4j jUaWEdWOD epoYjeD ey} UI (SUC! _HO ~SUO} -) sUOIUE sjuaNeg ‘weUEdWOD epoyjed ey} 0} UELIUEdWOO epouR ey) WO eyes 0} pue y YGnosy) ssed 0} (be) eN escjas0y) ‘(SuUd) +) suoQed smoyE AIUD + ‘(euesquiew d}U0HeD) euRIqQuew eBueYoxe UOReD “quoupedwos oy; oyu! podwnd ujeSe si uoANjos payejusouce (©) © © (-) epouzeo (+) epoue aut‘! 0} pappe si (s)OeN e0w Jeyy ‘uawyedwco pour Ou} JO No pedwnd sj uoANjos: (be) .8N (Parnp) pere|dep oy, epoue 8U) SPIEMO} BAOW SUOjUR (be) 39D eUL ‘opoujyeo 9y} SpeMO] SUBIqQUIDWY 2]UOeD 24) yGnosy) saw SuoHed (be) .eN ey quewpedwoo epoue ey) oyu! peduund s} (be)2DeN peyequesueD squowpedwos apour o43 Uy (be) HOeN (6) 7H (6) 49. (Q) 07H “(be) ;0e8N :S{ONPOld ssyuepeoy Ayoupeje jo sjunowe a6ue; sasn — 199 on Ajo09/3 ‘ ' Afiuaua jeojwayo + ABsaua oyj9a/3 ‘ ‘ 1149 ANVYEWAW AHL -9v +(b2) Hy + (6)?O=- (a)07HZ “se ouYO|Y> O43 JO And oy) seseaslap IU — PeSIPIXO S} LONN|OS (be) JOeN eu Ul JUeAjos OF} Ey Woy (be) HO ‘suo! \AxoupAy ay; uaym paonpaid aq Aew (6)%¢ ‘s26 ueBAxo jo yunowe jews y “peuols pue JuaWLedWCo 6u] Woy perowes S| (6)775 euL (A €8'0-=,3) 9% +(6'20— (be) 292 (8)Fa9 ©} posip)xo oue sucjue (be) 39 pour ou; 1¥ AYLSNOGNI IWH1V-YOTHD SHL ‘SNOLLOWSY IWOINSHOOYLOANA ‘AZONVHO IWOINSHO UNIT 6 THE CHLOR-ALKALI INDUSTRY 197 © Department of Basic Education 2015 Mind the Gap Chemistry

HINT! 6 Unit hint e.g. Worked example 1 EG - worked examples 1. Chlorine is produced industrially by electrolysis in an electrolytic cell and can be represented as follows: exams 2NaCℓ(aq) + 2H2O(ℓ) → H2(g) + Cℓ2(g) + 2NaOH(aq) I 2H2O + 2e– → H2 + 2OH– II activity Cℓ 2 + 2e– → 2Cℓ– ACTIVITIES boy III 2Cℓ– and girl-left and → Cℓ2 + 2e– right of page IV H2 + 2OH– → 2H2O + 2e– activity The correct statement(s) is/are: A I and III B I only C II and IV D III only Step by step comment comment 2. Chlorine is a poisonous gas commonly used as a bleaching agent. Chlorine is produced in industry by _____________. Solutions 1. A 2. Electrolysis of brine 198 UNIT 6 THE CHLOR-ALKALI INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 6 Activity 1 The electrolysis of saturated sodium chloride can be illustrated as follows: Hydrogen and chlorine bubble off at the electrodes. Cℓ2(g) H2(g) brine product NaCℓ(ℓ) electrode porous electrode diaphragm 1. Give an equation to show how chlorine bubbles are formed at the electrode. (2) 2. At which electrode are the chlorine bubbles formed? (1) 3. At which electrode does hydrogen gas form? (1) 4. What is the name of the product which leaves the cell? (1) 5. Give TWO applications of the product formed in 2.4. (2) 6. What purpose does the porous diaphragm serve? (2) 7. The chlorine gas produced, dissolves in water to form chlorine water. Write down a balanced equation for the reaction that takes place. (2) [11] Solutions 3 3 This question is based on 1. 2Cℓ–(aq) → Cℓ2(g) + 2e– (2) the electrolysis of NaCℓ in 2. Anode 3 (1) the diaphragm cell 3. Cathode 3 (1) 4. Sodium hydroxide 3 (1) 3 3 3 5. Making soap and detergents; paper; rayon and other fibres; dyeing textiles 3 (any 2) (2) 6. Stops chlorine passing through; helps to separate sodium hydroxide from NaCℓ(aq). 3 3 (2) 7. Cℓ2 + H2O → HCℓ + HOCℓ (2) 3 3 [11] © Department of Basic Education 2015 UNIT 6 THE CHLOR-ALKALI INDUSTRY 199 Mind the Gap Chemistry

HINT! 6 Unit hint e.g. Worked example 2 EG - worked examples The diagram below shows a type of membrane cell used in the chlor-alkali industry. exams activity ACTIVITIES boy and girl-left and right of page activity 1. Name the gases A and B 2. Why is the membrane called a cationic membrane? 3. Write down the half-reaction that takes place at electrode N. Step by step 4. Apart from its use comment in household products, name ONE comment industrial use of chlorine. 5. Explain why this electrolytic process cannot be done in one large container without a membrane. Solutions 1. A: Chlorine B: Hydrogen 2. Allows only the positive ions (cations) to pass through it. 3. 2H2O + 2e– → H2 + 2OH– 4. Manufacture of PVC, paper, drugs, etc Disinfectant for water 5. In a single pot the chlorine will react with water to form chlorine water/or the chlorine will react with the OH– ions to form bleach OR the products formed will be contaminated 200 UNIT 6 THE CHLOR-ALKALI INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 6 Activity 2 The simplified diagram of a cell used in the chlor-alkali industry is shown below. 1. Write down the CHEMICAL FORMULA of brine. (1) 2. At which electrode, X or Y is chlorine gas formed? (1) 3. Write down a half-reaction that explains the formation of hydrogen gas at one of the electrodes. (4) 4. The purity of the sodium hydroxide produced in the chlor-alkali industry depends on the extent to which it is separated from the chlorine gas produced by this cell. Briefly describe how chlorine gas and sodium hydroxide are prevented from mixing in this cell. (2) 5. Apart from the advantages and disadvantages of products produced, write down for this process: 5.1 ONE positive impact on humans. (1) 5.2 ONE negative impact on humans (1) [10] Solutions 1. NaCℓ (aq) 3 (1) 2. Y3 (1) 3 –3 3 3 3. 2H2O + 2e → H 2 + 2OH – (4) 4. The membrane prevents chloride ions from moving to the cathode3 , only allows positive ions through. 3 (2) 5.1 Job creation resulting in more people having a better life. 3 (1) 5.2 Uses huge amounts of electricity resulting in load shedding. 3 OR  Chemical plant uses a lot of space that could have been used for housing/gardens, etc. 3 (any one) (1) [10] Keep going! © Department of Basic Education 2015 UNIT 6 THE CHLOR-ALKALI INDUSTRY 201 Mind the Gap Chemistry

7 Unit The fertiliser industry The need for fertilisers is increasing due to: • the growth in the world population – more people means more food is necessary and • the decrease in available agricultural land – less land means the available land must produce a lot of high quality food at a high rate (fast). Nutrients in the soil are used by the plants that grow and these nutrients must be replaced. Inorganic fertilisers are produced industrially in the Haber, Contact & Ostwald processes at chemical plants like SASOL. 7.1 Nutrients required by plants MACRO NUTRIENTS MICRONUTRIENTS Primary non- Primary mineral Secondary mineral Trace elements nutrients nutrients nutrients • C carbon • N nitrogen • Ca calcium • Cu copper Elements • H hydrogen • P phosphorus • Mg magnesium • Fe iron • O oxygen • K potassium • S sulphur • Mn manganese • Zn zinc • I iodine • From the air • Absorbed from • Absorbed from • Absorbed from as CO2(g) and the soil the soil the soil Sources • in rainwater • by the roots • by the roots • by the roots as H2O(ℓ) • as water- • as water-soluble • as water-soluble IAL ATTENTION NB soluble salts. salts. salts. • Sufficient • Used up • Sufficient • Sufficient Availability DEFINITIONS available in and must be available in the available in the the air replaced by soil soil Promote: to help or using fertilisers encourage Essential: absolutely necessary Inorganic: not produced Did you know? by a living organism; not • Bone meal is produced when the bones of dead animals are ground into powder. ked examples containing carbon • Kelp meal is dried seaweed. Mineral: an inorganic • Granite meal is finely ground granite rock. solid substance that occurs naturally (not man-made) Inert: unreactive Sufficient: enough 202 UNIT 7 THE FERTILISER INDUSTRY  © Department of Basic Education 2015  ACTIVITIES boy Mind the Gap Chemistry and girl-left and

Unit 7 Activity 1 1. Which of the following is a primary mineral nutrient that is needed by plants? A) N B) C C) Mg D) Na (2) [2] Solution 1. A 3 3 [2] 7.2 Functions and sources of the primary mineral nutrients PRIMARY MINERAL NUTRIENTS Nutrient (N) Phosphorous (P) Potassium (K) • Promotes growth of • Promotes development • Ensures high quality and leaves; of roots and stems; colourful flowers, fruit • gives a healthy green • is essential for good and vegetables; colour to the plant; growth; • promotes resistance • is part of chlorophyll • is necessary for against diseases. which is needed for photosynthesis; photosynthesis. • is active during Function production of oils, sugar and starch. Sources • Guano • Bone meal • Wood ash Pre- (Sea bird excrement) • Mined in Germany World • Plant Compost War II • Animal manure • Bone meal • Kelp meal • Fish emulsion (ground seaweed) Organic • Dried animal blood • Guano (bat excrement) • Compost Sources Post World War II • Nitrate (NO3−) salts and • Superphosphates • KCℓ, potassium chloride • ammonium (NH4+) salts • CaH4(PO4)2·H2O • is imported to South • from ammonia (NH3) • Monoammonium Africa. produced in the Haber phosphate (MAP) process NH4H2PO4 • Also: Inorganic • E.g. • Diammonium phosphate • KNO3 • NH4NO3 ammonium (DAP): (NH4)2HPO4 potassium nitrate nitrate • K2SO4 • KNO3 potassium potassium sulphate nitrate • (NH4)2SO4 ammonium sulphate © Department of Basic Education 2015 UNIT 7 THE FERTILISER INDUSTRY 203 Mind the Gap Chemistry

PAY SPECIAL ATTENTION NB NB 7 Unit HINT! Atmospheric nitrogen is fixed into compounds containing nitrates (NO3−) or ammonium ions (NH4+) so that the nitrogen can be taken in by plants and hint used as nutrient. “Fixed” means “attached”. • Atmospheric fixation: fixation by lightning; • Biological fixation: fixation by bacteria in the ground and by the roots e.g. EG - worked examples of legumes; • Industrial fixation: fixation by industrial processes like the Haber process. exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 204 UNIT 7 THE FERTILISER INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 7 1. In the Haber process: Steam reforming of methane (natural gas) in the presence of a platinum catalyst to form a mixture of carbon monoxide (CO) and hydrogen (H2) gases (synthesis gas). 2. In the Contact process: 2.1 If SO2(g) is not dried before step 2, it reacts with water to form sulphurous acid, H2SO3(aq) which forms acid rain, a type of rain which is produced as a side-effect of human industrial activities, which is corrosive to structures and harmful to living things. SO2(g) + H2O(ℓ) → H2SO3(aq) 2.2 SO3(g) reacts with water to form gaseous sulphuric acid, H2SO4(g) which escapes into the atmosphere and forms acid rain. SO3(g) + H2O(ℓ) → H2SO4(g) 3. In the Contact process: 3.1 The 1st step is known as the catalytic oxidation of ammonia. Pt 4NH3(g) + 5O2(ℓ) → 4NO(g) + 6H2O(g) 3.2 NO2 (nitrogen dioxide) is a brown gas which reacts with water to form HNO3(g) which escapes into the atmosphere and forms acid rain. Fluorapatite, Ca5(PO4)3F 1. is mined at Phalaborwa; 2. is insoluble in water and can’t be absorbed by plant roots; 3. reacts with sulphuric acid, H2SO4(ℓ) to produce phosphoric acid, H3PO4(aq). Superphosphate 1. is produced when fluorapatite reacts with sulphuric acid H2SO4 and is a mixture of Ca(H2PO4)2·H2O and CaSO4 (gypsum) Triple Superphosphate 2. is produced when fluorapatite reacts with phosphoric acid H3PO4 to form Ca(H2PO4)2·H2O (no CaSO4 is formed). © Department of Basic Education 2015 UNIT 7 THE FERTILISER INDUSTRY 205 Mind the Gap Chemistry

7 Unit 7.3 The industrial production of fertilisers Study this table in conjunction with the flowchart on the next page ORGANIC Ammonium Ammonium Monoammonium Superphosphate Urea sulphate nitrate phosphate (MAP) Ca(H2PO4)2·H2O + (NH2)2CO (NH4)2SO4 NH4H2PO4 NH4NO3 CaSO4 Fertiliser Diammonium Triple phosphate (DAP) Superphosphate (NH4)2HPO4 Ca(H2PO4)2·H2O NH3(g) NH3(g) NH3(g) Ca5(PO4)3F NH3(g) Ammonia Ammonia Ammonia fluorapatite Ammonia (Haber process) (Haber process) (Haber process) (Haber process) Reactants H2SO4 H2SO4(ℓ) HNO3(aq) H3PO4(aq) sulphuric acid CO2(g) Sulphuric acid Nitric acid Phosphoric acid (Contact process) Carbon dioxide (Contact process) (Ostwald process) OR (by-product of the Haber process) H3PO4 Phosphoric acid 2NH3(g) + NH3(g) + HNO3(aq) NH3(g) + H3PO4(aq) Not required 2NH3(g) + CO2(g) H2SO4(ℓ) → → NH4NO3(aq) → → Reaction equation (NH4)2SO4(aq) NH4H2PO4(aq) (NH2)2CO(g) + memorise! H2O(g) memorise! 2NH3(g) + H3PO4(aq) → (NH4)2HPO4(aq) • Very soluble in • Very soluble in • Very soluble in • Soluble in • Very soluble in water; water; water; water; water; • easily absorbed • easily absorbed • easily absorbed • superphosphate • easily absorbed by plant roots; by plant roots; by plant roots; also supplies by plant roots; Properties • provides plants • provides plants • provides plants the secondary • provides plants with both with nitrogen with both nutrient S, with nitrogen nitrogen (N) and (N). nitrogen (N) sulphur; (N). sulphur (S). and • widely used. phosphorous (P). 206 UNIT 7 THE FERTILISER INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 7 7.4 Flowchart: The industrial production of fertilisers © Department of Basic Education 2015 UNIT 7 THE FERTILISER INDUSTRY 207 Mind the Gap Chemistry

HINT! 7 Unit hint e.g. Worked example 1 EG - worked examples 1. Study the diagram below that illustrates the industrial preparation of nitric acid and answer the questions that follow. exams 1.1 Write down a balanced chemical equation for the formation of product A. 1.2 activity Name the catalyst used during the reaction in Question 2.1. 1.3 What name is given to the reaction ACTIVITIESinboyQuestion 2.1? Give a reason for this name. and girl-left and right of page 1.4 Write down a balanced chemical equation for the formation of product B. 1.5 activity Name the products A and B. 1.6 Write down a balanced chemical equation for the formation of product C. 1.7 Name product C. 2. Ammonium nitrateStepisbyimportant step for its use in fertilisers comment and explosives. It can be comment prepared by the reaction of nitric acid and ammonia. 2.1 Write down a balanced equation for the preparation of ammonium nitrate. 2.2 What kind of fertiliser is ammonium nitrate? 2.3 Which characteristic of ammonium nitrate makes it suitable for use as a fertiliser? 2.4 Which characteristic of ammonium nitrate makes it suitable to be used in explosives? 3. Ammonium sulphate is often used as a fertiliser to supplement nitrogen and sulphur shortages in plants. 3.1 Compile a flow chart that indicates all the industrial steps for the preparation of ammonium sulphate from air, natural gas and sulphur. 3.2 Give balanced equations for all the reactions that take place during the preparation of ammonium sulphate from air, methane and sulphur. Solutions Pt 1.1 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) 1.2 Platinum or Rhodium 1.3 Catalytic oxidation of ammonia 1.4 2NO(g) + O2(g) → 2NO2(g) 1.5 A: Nitrogen monoxide B: Nitrogen dioxide 1.6 3NO2(g) + H2O(ℓ) → 2HNO3(aq) + NO(g) 1.7 Nitric Acid 2.1 HNO3 + NH3 → NH4NO3 2.2 Primary mineral nutrient 2.3 High nitrogen content per mass 2.4 Nitrogen bonds are easily broken and therefore it decomposes rapidly and easily (explodes). 208 UNIT 7 THE FERTILISER INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 7 Answers for 3.1 and 3.2 • O2 + N2 (air) • CH4 (methane) • S • To produce: (NH4)2SO4 (ammonium sulphate) Step 1: Produce ammonia NH3 in the Haber process: Fe N2 (distilling liquid air) + 3H2 (steam reforming of CH4) → 2NH3 Step 2: Produce sulphuric acid in the Contact process. a) S + O2 (from air) + heat → SO2 (g) b) 2SO2 + O2 V2O5 ⇌ 2SO3 (g) c) H2SO4 (ℓ) + SO3 (g) → H2S2O7 (ℓ) d) H2S2O7 (aq) + H2O → 2H2SO4 (ℓ) Step 3: Produce ammonium sulphate: 2NH3 (g) + H2SO4 (ℓ) → (NH4)2SO4 (aq) © Department of Basic Education 2015 UNIT 7 THE FERTILISER INDUSTRY 209 Mind the Gap Chemistry

7 Unit 7.5 N:P:K fertilisers As plants need a large amount of the primary mineral nutrients (nitrogen, phosphorus and potassium) and these need to be replenished (replaced) in the soil to make sure that crops grow well to provide enough food, NPK fertilisers that contain a mix of these nutrients, are usually used. They are produced by mixing a nitrogen fertiliser, a phosphorus fertiliser and potassium chloride (KCℓ). Step by step Step 1. Find the mass of the bag. Step 2. Find the percentage (%) of fertiliser in the bag (the number in brackets). Step 3. Find the mass of fertiliser in the bag. percentage fertiliser mass of fertiliser = _______________ ​    100 ​ × mass of bag percentage fertilizer mass of fertilizer = ​ _______________    100 ​ × mass of bag 26 = ​ ____ 100 ​ × 10 kg = 2,6 kg 210 UNIT 7 THE FERTILISER INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 7 Step by step (continued) Refer to the steps on the previous page Step 4. Find the total number of parts of nutrients in the bag. (Add the numbers representing the ratios of each of the elements, N, P and K.) Ratio total = 3 + 1 + 5 = 9 from step two 3/ 1/ 5/ 9 parts N 9 parts P 9 parts K Step 5. For each element (N, P and K) find the percentage of the element in the fertiliser: ratio element % element = ​ __________ total ratio ​ × % fertiliser ratio N 3 % N = ​ ________ __ total ratio ​ × % fertiliser = ​ 9 ​ × 26 = 8,67 ratio P 1 % P = ​ ________ __ total ratio ​ × % fertiliser = ​ 9 ​ × 26 = 2,89 ratio K 5 % P = ​ ________ __ total ratio ​ × % fertiliser = ​ 9 ​ × 26 = 14,44 Step 6. For each element (N, P and K) find the mass of that element in the bag. ratio element % fertiliser mass element = __________ ​  total ratio ​ × ________ ​  100 ​ × mass of bag ratio N % fertiliser 3 26 mass N = ________ ​ total ratio ​ × ________ ​  100 ​ × mass of bag = __ ​ 9 ​ × ​ ____ 100 ​ × 10 kg = 0,87 kg ratio P % fertiliser 1 26 mass P = ________ ​ total ratio ​ × ________ ​  100 ​ × mass of bag = ​ __ ____ 9 ​ × ​  100 ​ × 10 kg = 0,29 kg ratio K % fertiliser 5 26 mass K = ________ ​ total ratio ​ × ________ ​  100 ​ × mass of bag = ​ __ ____ 9 ​ × ​  100 ​ × 10 kg = 1,44 kg The per component totals (0,87; 0,29; 1,44) add up to 2.6kg. © Department of Basic Education 2015 UNIT 7 THE FERTILISER INDUSTRY 211 Mind the Gap Chemistry

HINT! 7 Unit hint e.g. Worked example 2 EG - worked examples A bag of fertiliser has the following information on it: 3 : 2 : 3 (26). 1. exams What information can you deduce from these numbers? 2. Calculate the percentage composition of the fertiliser in the bag. Solutions 1. It shows the ratio N : P : K. 3+2+3=8 therefore the fertiliser mixture consists of: 3 activity 2 3 ​  8 ​parts N; ​ 8 ​parts P and __ __ __ ​ 8 ​parts K These nutrients make up 26% of the ACTIVITIES boy mass of the bag content. and girl-left and The other 74% is made up of gypsum, right of page lime and sand. activityratio N 2. % N = ​ ________ __ 3 total ratio ​ × % fertiliser = ​ 8 ​ × 26 = 9,75 ratio P 2 % P = ​ ________ __ total ratio ​ × % fertiliser = ​ 8 ​ × 26 = 6,5 ratio K 3 ​ total ratio ​ × % fertiliser = ​ __ % K = ________ 8 ​ × 26 = 9,75 Step by step comment comment 212 UNIT 7 THE FERTILISER INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 7 7.6 Excessive use of fertiliser and the environment The correct application of fertiliser to crops is essential for high quality, fast growing crops but using too much or unnecessary fertiliser has a negative effect on the environment. • Groundwater is contaminated by fertiliser leaching (spreading by water) into the ground. • Soil becomes acidic (pH decreases). Many plants do not grow in acidic soil. • Invasive plants grow excessively while indigenous plants die. Invasive plants are undesirable non-indigenous (foreign) plants that grow too fast and out-compete local plants. • Fertiliser in dams and rivers leads to “eutrophication”, defined below. • High nitrate concentrations in drinking water decreases the ability of haemoglobin in the blood to carry oxygen and leads to ‘blue baby syndrome’. 7.6.1 Eutrophication In water (dams and rivers) On land PAY SPECIAL ATTENTION NB • Poor (bad) control of waste products. • Excessive / unnecessary application of fertilisers. REMEMBER • Waste products (urine and • Poor (bad) farming practises: An overabundance (excess, Causes faeces) and • Excessive / unnecessary use too much) of nutrients (NHINT! and • excessive fertiliser nutrients of fertilisers hint P) in water leads to: • leach into ground water and • Animal waste in feeding and • the fast and excessive milking stalls contains urea growth of • run into rivers and dams. and ammonia and leaches algae and other water into ground water. e.g. EG - worked examples plants (this is called ‘algal • Excessive growth of algae; • Invasive plants grow excessively, bloom’) • Algae die; indigenous plants die; and bacteria • Bacteria feed on the algae and • Bird and insect species move exams • causing a drop in the Effects use oxygen; away disturbing the ecosystems; oxygen concentration in • Oxygen levels in the water drop; • Soil erosion increases due to the water and absence of indigenous plants • Fish and other living organisms • the death of fish and other with strong root systems. die. living organisms in the • Strict control over all waste and • Test soil to determine need for water. sewage must be enforced. fertiliser – apply correct amount and type; activity • Waste products: ACTIVITI Possible solutions • must be treated and • Plant indigenous plants along and girl- rivers to prevent erosion and right of p • not allowed to leach into prevent inflow of contaminated groundwater or water; activity • run into rivers and dams. • Conserve natural swamps which break down excess nitrogen nutrients naturally; • Control animal waste in feeding Step by step com and milking stalls. © Department of Basic Education 2015 UNIT 7 THE FERTILISER INDUSTRY 213 Mind the Gap Chemistry

PAY SPECIAL ATTENTION NB NB 7 Unit HINT! hint VOCABULARY Overabundance / excess: more than is needed e.g. EG - worked examples Shortage: not enough Invasive plants: plants that do not naturally grow in a certain area Indigenous plants: plants that grow naturally in (are native to) a certain area Swamps: ground that is uncultivated but is usually covered with exams water Erosion: soil and rock is removed from the Earth’s surface by wind, rain or water Stunted growth: reduced growth leading to smaller, underdeveloped plants activity ACTIVITIES boy Activity 2 and girl-left and right of page activity Which ONE of the following is NOT associated with eutrophication in water? A Dead zones B Algal bloom C Depletion of oxygen D Increased aquatic life Step by step comment comment (2) [2] Solution D 33 [2] 7.6.2 The effect of a shortage or excess of the primary nutrients (N, P and K) on plants N (nitrogen) P (phosphorus) K (potassium) Shortage • Little growth • Stunted growth • Poor quality • Weak stems • Slow growth flowers and fruit (colour and • Yellow leaves • Purple leaves taste) • Leaves have brown or yellow edges Excess • Dark green • Causes shortage • Causes shortage leaves of micro of micro • Poor production nutrients Fe and nutrients Mg and of seed and fruit Zn Ca • Stunted growth • Poor quality flowers and fruit 214 UNIT 7 THE FERTILISER INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 7 7.7 Alternatives to inorganic fertilisers Although fertilisers are essential for the fast growth of high quality crops, the negative effects of inorganic compounds on the environment must be taken into account. Alternative sources of organic nutrients that can be used to ensure good crops are: • Bone meal; • Animal manure; • Natural plant compost; • Bat guano (faeces); • Fish emulsions; • Kelp meal Learn the advantages and disadvantages of organic Advantages of organic fertilisers: fertilisers. • Break down and release nutrients more slowly than inorganic fertilisers, so there is less chance of the fertiliser leaching into the soil and causing contamination of groundwater; • Usually cost less and • Are often available for free. Disadvantages of organic fertilisers: • Not enough is available for large scale usage; • Provide less nutrients – more has to be used; • Slow release of nutrients sometimes harms plants; • Slow release may cause nutrients to be available too late in the plant’s growth cycle. Activity 3 Fertilisers allow farmers to grow crops in the same soil year after year. However, environmental problems, such as eutrophication, are associated with the application of fertilisers. 1. State ONE PRECAUTION that a maize farmer can take to prevent eutrophication.(1) Nitric acid is an important reactant in the production of ammonium nitrate, a nitrogen-based fertiliser. 2. Write down the name of the industrial process for the production of nitric acid. (1) 3. Write down a balanced equation for the preparation of ammonium nitrate from nitric acid. (3) [5] © Department of Basic Education 2015 UNIT 7 THE FERTILISER INDUSTRY 215 Mind the Gap Chemistry

7 Unit Solutions 1. Use fertilisers sparingly 3 / Do not over-fertilise 3 Make use of precision (computerised) application of fertilisers 3 Ensure that water from fields does not run into rivers/dams 3 Redirect water from fields into reservoirs/away from rivers/dams 3 (any one) (1) 2. Ostwald process 3 (1) 3. HNO3 3 + NH3 3 → NH4NO3 3 (3) [5] Activity 4 The rapidly increasing human population is resulting in an ever-increasing demand for food. To meet this demand, farmers apply fertiliser to the same cultivated land EACH YEAR. 1. Explain why farmers have to apply fertilisers to their land each year. (1) 2. Write down one negative impact that over-fertilisation can have on humans. (1) 3. Sulphuric acid is an important substance used in the manufacture of fertilisers. The equation below represents one of the steps in the industrial preparation of sulphuric acid. 2SO2(g) + O2(g) ⇋ 2SO3(g) ∆H < 0 3.1 Write down the name of the process used to prepare sulphuric acid in industry (1) 3.2 Write down the name or the formula if the catalyst used in 2.3.1 (1) 3.3 Is the forward reaction endothermic or exothermic? Give a reason for your answer. (1) 3.4 Write down the name or formula of the fertiliser formed when sulphuric acid reacts with ammonia. (2) [7] 216 UNIT 7 THE FERTILISER INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

Unit 7 Solutions 1. Fertilisers replenish nutrients depleted by the growing of crops 3 (1) 2. Damage to crops/soil 3 resulting in small or no harvest/ less income. Excessive fertiliser seeps into groundwater and contaminates drinking water 3 or runs into rivers and/or dams and causes eutrophication 3 which may result in less income 3/famine/starvation 3/poor quality drinking water 3/fewer recreation areas/3 environmental damage/3 death of wild animals 3 (any one) (1) 3.1 Contact process 3 (1) 3.2 V2O5 3 vanadium pentoxide 3 (any one) (1) 3.3 Exothermic as ∆H < 0 3 (1) 3.4 (NH4)2SO4 33 OR / Ammonium 3 sulphate 3 (2) [7] Activity 5 Ammonia, ammonium nitrate and ammonium sulphate are three important nitrogen-containing fertilisers. The flow diagram below shows how these fertilisers are produced in industry. 1. Use the information in the flow diagram above and write down the following: 1.1 Name Process 1 (1) 1.2 Balanced equation for Process 2 (3) 1.3 Name or Formula for compound X (1) 1.4 Balanced equation for the preparation of ammonium sulphate using sulphuric acid and compound Y (3) 1.5 Name or Symbol of the primary nutrient in ammonium sulphate (1) © Department of Basic Education 2015 UNIT 7 THE FERTILISER INDUSTRY 217 Mind the Gap Chemistry

7 Unit 2. Write down one positive impact of fertilisers on humans (1) 3. Write down two negative impacts of the use of ammonium nitrate as fertiliser, on humans. (2) [12] Solutions 1.1 Fractional distillation of liquid air 3 (1) 1.2 N2 + 3H2 → 2NH3 (reactants 3 products 3 balance 3) (3) 1.3 Nitric acid / HNO3 3 (1) 1.4 H2SO4 + 2NH3 → (NH4)2SO4 (reactants 3 products 3 balance 3) (3) 1.5 Nitrogen / N 3 (1) 2. Enhance growth of crops/plants 3 to produce more food for humans 3 food security for humans 3 production / application of fertiliser 3 results in job creation selling of fertilisers 3 stimulates the economy 3 (any one) (1) 3. (Excessive) nitrates in water (eutrophication) 3 can result in blue baby syndrome or cancer 3 (Excessive) nitrates/ammonium ions in water can result in poor quality drinking water 3 or death of fish 3 or less food 3 or fewer recreational facilities 3 or famine due to killing plants / crops from the excess 3 or excessively changing the pH of the soil and thereby reducing the food production 3 (any two) (2) [12] Well done! 218 UNIT 7 THE FERTILISER INDUSTRY  © Department of Basic Education 2015  Mind the Gap Chemistry

The Mind the Gap study guide series assists you to make the leap by studying hard to achieve success in the Grade 12 exam. This publication is not for sale. © Copyright Department of Basic Education www.education.gov.za This publication has a Creative Commons Attribution NonCommercial Sharealike license. Call Centre 0800 202 933