Success Chemistry SPM 9789834707569

Table of contents :
Form 4
1 Introduction to Chemistry
1.1 Chemistry and Its Importance
1.2 Scientific Method
1.3 Scientific Attitudes and Values in Conducting Scientific Investigations
SPM Exam Practice 1
2 The Structure of the Atom
2.1 Matter
2.2 The Atomic Structure
2.3 Isotopes and Their Importance
2.4 The Electronic Structure of an Atom
2.5 Appreciating the Orderliness and Uniqueness of the Atomic Structure
SPM Exam Practice 2
3 Chemical Formulae and Equations
3.1 Relative Atomic Mass and Relative Molecular Mass
3.2 Relationship between the Number of Moles and the Number of Particles
3.3 Relationship between the Number of Moles of a Substances and Its Mass
3.4 Relationship between the Number of Moles of a Gas and Its Volume
3.5 Chemical Formulae
3.6 Chemical Equations
3.7 Scientific Attitudes and Values in Investigating Matter
SPM Exam Practice 3
4 Periodic Table of Elements
4.1 Periodic Table of Elements
4.2 Group 18 Elements
4.3 Group 1 Elements
4.4 Group 17 Elements
4.5 Elements in a Period
4.6 Transition Elements
4.7 Appreciating the Existence of Elements and Their Compounds
SPM Exam Practice 4
5 Chemical Bonds
5.1 Formation of Compounds
5.2 Formation fo Ionic Bonds
5.3 Formation of Covalent Bonds
5.4 The Properties of Ionic Compounds and Covalent Compounds
SPM Exam Practice 5
6 Electrochemistry
6.1 Electrolytes and Non-electrolytes
6.2 Electrolysis of Molten Compounds
6.3 Electrolysis of Aqueous Solutions
6.4 Electrolysis in Industries
6.5 Voltaic Cells
6.6 The Electrochemical Series
6.7 Developing Awareness and Responsible Practices when Handling Chemicals used in the Electrochemical Industries
SPM Exam Practice 6
7 Acids and Bases
7.1 Characteristics and Properties of Acids and Bases
7.2 The Strength of Acids and Alkalis
7.3 Concentration of Acids and Alkalis
7.4 Neutralisation
SPM Exam Practice 7
8 Salts
8.1 Salts
8.2 Qualitatives Analysis of Salts
8.3 Practising Systematic and Meticulous Methods when Carrying Out Activities
SPM Exam Practice 8
9 Manufactured Substances in Industry
9.1 Sulphuric Acid
9.2 Ammonia and Its Salts
9.3 Alloys
9.4 Synthetic Polymers
9.5 Glass and Ceramics
9.6 Composite Materials
9.7 Appreciating Various Synthetic Industrial Materials
SPM Exam Practice 9
Form 5
1 Rate of Reaction
1.1 Rate of Reaction
1.2 Factors that Affect the Rate of Reaction
1.3 The Collision Theory
1.4 Practising Scientific Knowledge to Enhance Quality of Life
SPM Exam Practice 1
2 Carbon Compounds
2.1 Carbon Compounds
2.2 Alkanes
2.3 Alkenes
2.4 Isomerism
2.5 Alcohols
2.6 Carboxylic Acids
2.7 Esters
2.8 Oils and Fats
2.9 Natural Rubber
2.10 Order in Homologous Series
2.11 The Variety of Organic Materials in Nature
SPM Exam Practice 2
3 Oxidation and Reduction
3.1 Redox Reactions
3.2 Rusting as a Redox Reaction
3.3 The Reactivity Series of Metals and Its Applications
3.4 Redox Reactions in Electrolytic Cell and Chemical Cell
3.5 Appreciating the Ability of the Elements to Change their Oxidation Numbers
SPM Exam Practice 3
4 Thermochemistry
4.1 Energy Changes in Chemical Reactions
4.2 Heat of Precipitation
4.3 Heat of Displacement
4.4 Heat of Neutralisation
4.5 Heat of Combustion
4.6 Appreciating the Existence of Various Energy Sources
SPM Exam Practice 4
5 Chemicals for Consumers
5.1 Soaps and Detergents
5.2 Uses of Food Additives
5.3 Medicine
5.4 Appreciating the Existence of Chemicals
SPM Exam Practice 5
SPM Model Test
Answers
Glossary

Citation preview

OXFORD FAJAR ADVISORY BOARD The board consists of a team of experienced teachers who review our titles to ensure that the contents are in line with the current syllabus and examination requirements as set by the Examination Syndicate, Ministry of Education Malaysia. Success Chemistry SPM e-book was reviewed by • Ilani bte Ibrahim • Tan Sze Chuan • Tay Geok It

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Oxford Fajar Sdn. Bhd. (008974-T) (Formerly known as Penerbit Fajar Bakti Sdn. Bhd.) 4 Jalan Pemaju U1/15, Seksyen U1 Hicom-Glenmarie Industrial Park 40150 Shah Alam Selangor Darul Ehsan © Oxford Fajar Sdn. Bhd. (008974-T) 2013 First published 2013 ISBN 978 983 47 0756 9 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of Oxford Fajar Sdn. Bhd. (008974-T) Text set in 10 point ITC Giovanni by Leo & Libra Creative, Kuala Lumpur

Word Power

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Contents Keywords v

4.7 Appreciating the Existence of Elements and Their Compounds 98 SPM Exam Practice 4 100

FORM 4 CHAPTER 1

Introduction to Chemistry

1

CHAPTER 5 2

1.1 Chemistry and Its Importance 1.2 Scientific Method 1.3 Scientific Attitudes and Values in Conducting Scientific Investigations SPM Exam Practice 1

2 5

Chemical Bonds

107

5.1 5.2 5.3 5.4

108 109 114

Formation of Compounds Formation of Ionic Bonds Formation of Covalent Bonds The Properties of Ionic Compounds and Covalent Compounds SPM Exam Practice 5

7 9

CHAPTER 2

The Structure of the Atom

14

2.1 2.2 2.3 2.4 2.5

15 23 26 29

Matter The Atomic Structure Isotopes and Their Importance The Electronic Structure of an Atom Appreciating the Orderliness and Uniqueness of the Atomic Structure SPM Exam Practice 2

CHAPTER 6 2

Electrochemistry 134 6.1 6.2 6.3 6.4 6.5 6.6 6.7

Electrolytes and Non-electrolytes 135 Electrolysis of Molten Compounds 137 Electrolysis of Aqueous Solutions 142 Electrolysis in Industries 152 Voltaic Cells 158 The Electrochemical Series 165 Developing Awareness and Responsible Practices when Handling Chemicals used in the Electrochemical Industries 172 SPM Exam Practice 6 173

32 33

2 CHAPTER 3

Chemical Formulae and Equations 3.1 Relative Atomic Mass and Relative Molecular Mass 3.2 Relationship between the Number of Moles and the Number of Particles 3.3 Relationship between the Number of Moles of a Substance and Its Mass 3.4 Relationship between the Number of Moles of a Gas and Its Volume 3.5 Chemical Formulae 3.6 Chemical Equations 3.7 Scientific Attitudes and Values in Investigating Matter SPM Exam Practice 3

121 128

39 40 43

CHAPTER 7 2

45

Acids and Bases 181

48 51 57 60 62

7.1 Characteristics and Properties of Acids and Bases 182 7.2 The Strength of Acids and Alkalis 192 7.3 Concentration of Acids and Alkalis 196 7.4 Neutralisation 203 SPM Exam Practice 7 211

Periodic Table of Elements

67

CHAPTER 8 2

4.1 4.2 4.3 4.4 4.5 4.6

68 74 75 82 89 93

Salts 218

2 CHAPTER 4

Periodic Table of Elements Group 18 Elements Group 1 Elements Group 17 Elements Elements in a Period Transition Elements

8.1 Salts 8.2 Qualitative Analysis of Salts 8.3 Practising Systematic and Meticulous Methods when Carrying Out Activities SPM Exam Practice 8

iii

219 240 255 256

2 CHAPTER 9

2 CHAPTER 3

Manufactured Substances in Industry

262

Oxidation and Reduction

9.1 9.2 9.3 9.4 9.5 9.6 9.7

Sulphuric Acid 263 Ammonia and Its Salts 266 Alloys 272 Synthetic Polymers 278 Glass and Ceramics 281 Composite Materials 284 Appreciating Various Synthetic Industrial Materials 288 SPM Exam Practice 9 289

Redox Reactions 385 Rusting as a Redox Reaction 412 The Reactivity Series of Metals and Its Applications 418 Redox Reactions in Electrolytic Cell and Chemical Cell 430 3.5 Appreciating the Ability of the Elements to Change their Oxidation Numbers 441 SPM Exam Practice 3 444

FORM 5

2 CHAPTER 4

Thermochemistry 452

CHAPTER 1

4.1 4.2 4.3 4.4 4.5 4.6

Energy Changes in Chemical Reactions Heat of Precipitation Heat of Displacement Heat of Neutralisation Heat of Combustion Appreciating the Existence of Various Energy Sources SPM Exam Practice 4

Rate of Reaction 295 1.1 1.2 1.3 1.4

Rate of Reaction Factors that Affect the Rate of Reaction The Collision Theory Practising Scientific Knowledge to Enhance Quality of Life SPM Exam Practice 1

296 305 322 328 329

CHAPTER 2

Carbon Compounds

384

3.1 3.2 3.3 3.4

453 463 469 474 482 489 492

2 CHAPTER 5

340

Chemicals for Consumers 499

2.1 Carbon Compounds 341 2.2 Alkanes 343 2.3 Alkenes 346 2.4 Isomerism 352 2.5 Alcohols 357 2.6 Carboxylic Acids 362 2.7 Esters 366 2.8 Oils and Fats 370 2.9 Natural Rubber 371 2.10 Order in Homologous Series 376 2.11 The Variety of Organic Materials in Nature 376 SPM Exam Practice 2 378

5.1 Soaps and Detergents 5.2 Uses of Food Additives 5.3 Medicine 5.4 Appreciating the Existence of Chemicals SPM Exam Practice 5

500 510 515 519 521

SPM Model Test 526 Answers 537 Glossary 592

iv

Key Words 1 Introduction to Chemistry conclusion – kesimpulan constant variable – pembolehubah yang dimalarkan manipulated variable – pembolehubah yang dimanipulasikan procedure – kaedah/prosedur responding variable – pembolehubah yang bergerakbalas scientific attitudes – sikap saintifik variable – pembolehubah

denominator – penyebut element – unsur empirical formula – formula empirik ionic compound – sebatian ion mass – jisim molar volume – isipadu molar molecular formula – formula molekul numerator – pengangka product – hasil tindak balas reactant – bahan tindak balas reduced – diturunkan relative atomic mass – jisim atom relatif 4 Periodic Table of Elements

2 The Structure of the Atom charged particles – zarah bercas chemical reaction – tindakbalas kimia collision – perlanggaran compressibility – kemampatan condensation – kondensasi diffusion – peresapan duplet – duplet electronic configuration –susunan elektron electron shells – petala elektron forces of attraction – daya tarikan freezing point – takat beku half-life – setengah hayat isotope – isotop matter – jirim melting point – takat lebur non-renewable – tidak boleh diperbaharui nucleon number – nombor nukleon octet – oktet

3 Chemical Formulae and Equations anion (negatively-charged ion) – anion (ion bercas negatif) cation (positively-charged ion) – kation (ion bercas positif) compound – sebatian crucible – mangkuk pijar covalent compound –sebatian kovalen

boiling point – takat didih chemical bonding – ikatan kimia covalent bond – ikatan kovalen double bond – ikatan ganda dua electrical conductivity – kekonduksian elektrik electrostatic force of attraction – daya tarikan elektrostatik inert gas – gas adi ionic bond – ikatan ion Lewis structure – struktur Lewis non-polar – tidak berkutub organic solvent – pelarut organik polar – berkutub shell – petala single bond – ikatan tunggal solubility – kelarutan triple bond – ikatan ganda tiga valence electron – electron valens volatility – kemeruapan 5 Chemical Bonds aqueous solution – larutan akueus giant molecules – molekul raksasa intermolecular force – daya tarikan antara molekul noble gas – gas adi 6 Electrochemistry alkaline cell – sel alkali anode – anod cathode – katod concentration – kepekatan

v

decomposition – penguraian discharge – nyahcas displacement reaction – tindak balas penyesaran dry cell – sel kering electrochemical series – siri elektrokimia electrochemistry – elektrokimia electrode – elektrod electrolysis – elektrolisis electrolyte – elektrolit electroplating – saduran elektrik half-reaction – tindak balas setengah non-electrolyte – bukan elektrolit non-rechargeable cell – sel yang tidak boleh dicas semula potential difference – beza keupayaan primary cell – sel primer rechargeable cell – sel yang boleh dicas semula secondary cell – sel sekunder 7 Acids and Bases basicity – kebesan degree of dissociation – darjah penceraian dilution – pencairan diprotic acid – asid dwibes end point – takat akhir molarity – kemolaran monoprotic acid – asid monobes hydroxide ion – ion hidroksida hydroxonium ion –ion hidroksonium neutralisation – peneutralan partial dissociation –penceraian separa standard solution – larutan piawai titration – pentitratan triprotic acid – asid tribes universal indicator –penunjuk semesta 8 Salts brown ring test – ujian cincin perang confirmatory test –ujian pengesahan continuous variation –perubahan berterusan crystal – hablur Key Words

KEY WORDS

FORM 4

decolourised – nyahwarna double decomposition –penguraian ganda dua evaporation – sejatan filtrate – hasil turasan impurities – benda asing insoluble salts –garam tak terlarutan precipitate – mendakan qualitative analysis – analisa kualitatif recrystallisation – penghabluran semula residue – baki soluble salts – garam terlarutan solution – larutan

KEY WORDS

9 Manufactured Substances in Industry alloy – aloi biodegradable – terbiodegradasikan borosilicate glass – kaca borosilikat brass – loyang bronze – gangsa catalyst – mangkin ceramic – seramik coagulation – penggumpalan Contact process – proses sentuh corrosion – kakisan density – ketumpatan ductility – kemuluran explosive – bahan letupan fibre optic – gentian optik fused glass – kaca silika terlakur lead glass – kaca plumbum malleability – kebolehtempaan photochromic glass –kaca fotokromik polymerisation – pempolimeran refrigerant – bahan penyejuk rust – karat soda glass – kaca soda kapur solder – pateri stainless steel – keluli nirkarat superconductor – superkonduktor synthetic fibre – gentian sintetik FORM 5

energy profile diagram – rajah profil tenaga observable change – perubahan yang dapat diperhatikan rate of reaction – kadar tindak balas 2 Carbon Compounds addition – penambahan alkanes – alkana alkenes – alkena alkynes – alkuna combustion – pembakaran fractionating column –turus pemeringkat functional group – kumpulan berfungsi general formula – formula am homologous series – siri homolog hydration – penghidratan hydrogenation – penghidrogenan IUPAC nomenclature – sistem penamaan IUPAC saturated – tepu sootiness – kejelagaan straight chain – rantai lurus structural formula – formula struktur substitution – penukargantian unsaturated – tak tepu carboxylic acid – asid karboksilik coagulation – penggumpalan dehydration – pendehidratan distillation – penyulingan drying agent – agen pengontangan elasticity – kekenyalan esterification – pengesteran extraction – pengekstrakan fatty acid – asid lemak fermentation – penapaian hydroxonium ion – ion hidroksonium hydroxyl group – kumpulan hidroksil polyunsaturated fats –lemak poli tak tepu volatility – kemeruapan vulcanised – tervulkan

1 Rate of Reaction activation energy –tenaga pengaktifan average rate – kadar purata catalyst – mangkin collision frequency – frekuensi perlanggaran collision theory – teori perlanggaran effective collision –perlanggaran berkesan energy barrier – rintangan tenaga Key Words

3 Oxidation and Reduction blast furnace – relau bagas cast iron – besi tuangan chemical cell – sel kimia displacement reaction – tindak balas penyesaran electrolytic cell – sel elektrolisis extraction – pengekstrakan impurity – bendasing

vi

metal displacement –penyesaran logam oxidation state –keadaan pengoksidaan oxidising agent – agen pengoksidaan reactivity series – siri kereaktifan reducing agent – agen penurunan sacrificial metal – logam korban

4 Thermochemistry bond energy – tenaga ikatan endothermic reaction –tindak balas endotermik energy content –kandungan tenaga energy level diagram –gambar rajah aras tenaga exothermic reaction –tindak balas eksotermik fuel value – nilai haba bahan api heat of combustion – haba pembakaran heat of displacement –haba penyesaran heat of formation – haba pembentukan heat of neutralisation –haba peneutralan law of conservation of energy – hukum keabadian tenaga precipitation – pemendakan reversible reaction – tindak balas berbalik specific heat capacity –muatan haba tentu thermal dissociation –penceraian terma thermochemical equation – persamaan termokimia

5 Chemicals for Consumers additive – bahan tambahan analgesic – analgesik antioxidant – pengantioksida/ antipengoksida antipsychotic – antipsikotik biodegradable – terbiodegradasikan biological enzyme – enzim biologi codeine – kodeina detergent – detergen flavouring agent – agen perisa preservative – pengawet saponification – saponifikasi soap – sabun stabiliser – pengstabil thickening agent – agen pemekat

FORM 4 THEME: Introducing Chemistry

CHAPTER

1

Introduction to Chemistry

SPM Topical Analysis 2008

Year Paper

1

Section Number of questions

–

2009

2 A

B

C

–

–

–

3

1

–

–

2010

2 A

B

C

–

–

–

3

1

–

–

2011

2 A

B

C

–

–

–

3

1

–

–

3

2 A

B

C

–

–

–

ONCEPT MAP

INTRODUCTION TO CHEMISTRY

History of chemistry

Importance of chemistry

Meaning of chemistry

The scientific method

Science that studies the properties, composition and structure of substances and the changes they undergo

In the fields of: • Food processing • Medicine • Agriculture • Transportation • Telecommunications • Daily usage of chemical products

Methodology in chemistry

Careers that need knowledge of chemistry: • Medicine and Dentistry • Pharmacy • Geology • Biochemistry • Engineering

• Observe a situation • Identify all variables • Suggest a problem statement • Form a hypothesis • Select suitable apparatus • Carry out an experiment • Collect and tabulate data • Interpret the data • Write a report

Scientific attitudes and values

• Attitudes • Avoid wastage • Maintain cleanliness • Avoid accidents

–

1

1.1

(a) Medicine: to fight diseases and prolong life.

Chemistry and Its Importance

1 Human beings used chemical processes before 500 BC to extract metals such as copper and iron for making ornaments. They also found ways to make ceramics from clay. However, they could not explain the chemical processes that took place. 2 The next 1700 years of chemical history were dominated by a pseudo-science called alchemy (pseudo means not genuine or false). The word alchemy originated from the Arabic word ‘alkimiya’ (al === the; kimiya === art of changing). The alchemists in Egypt believed they could change cheap metals like lead into gold. Their efforts were unsuccessful but along the way they (a) discovered other substances like mercury, sulphur, antimony and phosphorus, (b) developed some reliable techniques of chemical manipulation, (c) learned to prepare some mineral acids such as sulphuric, hydrochloric and nitric acids.

Medicine

(b) Fertilisers and pesticides: increase crop yields.

(c) Preservatives: prolong the storage of food.

Food preservative

(d) Materials used for making clothing such as cotton, silk and nylon. A chemist carrying out research

3 Modern chemistry originated from an Englishman named Robert Boyle. In 1661, he wrote a book called The Sceptical Chymist which introduced the modern concept of chemical elements. An element is a substance that cannot be broken down into simpler substances by chemical means. In the centuries that followed, many elements were discovered.

Nylon

(e) Building materials concrete and glass.

such

as

cement,

The Importance of Chemistry 1 Chemistry is the science concerned with the composition of substances, the basic forms of matter and the interactions between them. 2 Chemical substances or chemicals are very important in our lives. The following are a few examples of chemicals. Introduction to Chemistry

Building

2

(b) how chemicals interact among each other, and (c) how to use the knowledge of the properties of these chemicals to produce new substances.

(f) Components of automobiles and computers.

1 Many careers require knowledge of chemistry. For example, a dentist uses hydrogen peroxide gel (H2O2) to bleach teeth (making it whiter). For extraction of a tooth, the dentist will administer a local anaesthetic (procaine) before extracting the tooth of a patient. To fill a tooth, he/she will use amalgam which is an alloy of mercury and silver. Hydrogen peroxide, procaine and amalgam are all chemicals.

(g) Consumer products such as soap and detergents.

Consumer products

3 Table 1.1 shows the uses of some chemical substances in our daily life. 4 In chemistry, we study (a) the basic units that make up these materials,

A dentist at work

Table 1.1

Name of substance

Uses

Chemical formula

Oxygen

O2

Respiration and combustion

Nitrogen

N2

Manufacture of ammonia

Carbon dioxide

CO2

In photosynthesis and in making carbonated drinks

Sodium chloride

NaCl

Food preservation, for example, salted fish

Iron(II) sulphate

FeSO4

Iron pills to treat anaemia

Aspirin Calcium sulphate hemihydrate Copper-nickel alloy Urea Sulphuric acid Ethanol Sodium stearate Ethanoic acid (acetic acid) Calcium carbonate

CH3COOC6H4COOH

An analgesic drug to treat pain and fever

2CaSO4. H2O

Used as a cast to support broken bones of accident victims

25% Nickel + 75% Copper CO(NH2)2

To make coins A nitrogenous fertiliser

H2SO4

As an electrolyte in a lead-acid accumulator

C2H5OH

As a solvent and manufacture of industrial chemicals

C17H35COONa

Soap

CH3COOH

Preservation of fruits and manufacture of food flavourings

CaCO3

Calcium supplement 3

Introduction to Chemistry

1

Chemistry Related Careers

2 A medical doctor needs a knowledge of chemistry to administer the correct amount of medicine to a patient. Categories of medicine include antibiotics, hormones, psychiatric medicine, analgesics, alkaloids and fungal creams. All these medicine are chemicals.

Food processing

6 A farmer uses fertilisers to increase the yield of his crops. Pesticides, herbicides and fungicides are used to control pests. Therefore, even the farmer is required to have a knowledge of chemistry.

1

A doctor administering an injection

3 Pharmacy is a branch of science which deals with the interaction of medicine with the human body. It also finds ways to synthesise new drugs. Most medicine are organic compounds. Therefore a pharmacist must have an understanding of organic chemistry. A farmer spraying pesticides

Chemical-based Industries in Malaysia and their Contributions Local chemical industries have contributed greatly to Malaysia’s economy. These industries not only provide job opportunities but also earn foreign exchange for the country when the chemicals produced are exported. Some notable chemical industries in Malaysia are: 1 Plants in Pasir Gudang, Johor and Gebeng, Pahang produce chemicals such as polyethylene. Polyethylene and polypropylene are used to make many household items such as chairs, raincoats, pails and basins. Table 1.2 shows the chemicals produced by the petrochemical plants.

Pharmacist

4 The expertise of forensic chemists can help the police to solve crimes. The analyses and identification of samples of blood, drugs, semen, poison, weapons and a host of other items collected from the crime scene are used as evidence to convict criminals.

Table 1.2 Chemicals produced by petrochemical plants in Malaysia

Petrochemical plant Analysis of DNA

5 Many types of chemicals, namely, preservatives, colourings, antioxidants, flavour enhancers, food stabilisers and artificial flavourings are used in the food processing industry. Thus, food technologists require knowledge of chemistry to ensure the correct mixture of these chemicals. Introduction to Chemistry

4

Product

BASF Petronas Chemicals Sdn Bhd

Acrylic polymers

Titan Petrochemicals (M) Sdn Bhd

Polyethylene

Petrochemicals (M) Sdn Bhd

Expandable polystyrene





1.1 1 Name two examples of chemicals used in each of the following fields. Field

Chemicals used

Agriculture Medicine Food processing



1.2

Scientific Method

1 Chemistry is an experimental science similar to Biology and Physics and requires scientific research. 2 There are some basic guidelines in approaching any scientific research. These guidelines are known as the scientific method. 3 The scientific method is a systematic approach to research. It consists of the following steps: (a) Making an observation about a situation A scientific research starts with an observation. For example, a student would have observed a situation as follows:

Does the solubility of sugar increase proportionally with the increase in the temperature of water?

When he adds 20 g of sugar to 100 cm3 of hot water and stirred, all the sugar dissolved. However, when 20 g of sugar is added to 100 cm3 of water at room temperature and stirred, some sugar remains undissolved in the water.



of the water affect the solubility of sugar in water. (ii) A constant variable is the factor which is kept the same throughout the experiment. To study the effect of temperature on the solubility of sugar in water, the volume of the water used in the experiment must be kept constant. The volume of water is called the constant variable. (iii) A variable which is changed during the experiment is called the manipulated variable. An experiment can be carried out by heating the water to temperatures of 30 °C, 40 °C, 50 °C, 60 °C and 70 °C. The mass of sugar that dissolves at different temperatures of water is then measured. The temperature of water is called the manipulated variable. (iv) A responding variable is the variable that responds to the change made by the manipulated variable. The amount of sugar that dissolves in water at different temperatures is called the responding variable. Thus the variables are: Manipulated variable: Temperature of the water Responding variable: Amount of sugar that dissolves in water at different temperatures Constant variable: Volume of water (c) Suggesting a problem statement This is a question which identifies the problem related to the observation. For example,

This will lead to the forming of a hypothesis. (d) Forming a hypothesis A hypothesis is a proposition, idea, theory or any other statement used as a starting point for discussion, investigation or study. For example, to study the effect of the temperature of water on the amount of sugar that dissolves, a probable hypothesis would be:

(b) Identifying variables (i) A variable is a factor which affects the results of the experiment. The factors that affect the solubility of sugar in water are called variables. It is found that the temperature and volume 5

Introduction to Chemistry

1

2 The Asean Bintulu Fertiliser (ABF) plant in Sarawak produces urea. This is a project undertaken by some Asean countries. Urea is a nitrogenous fertiliser. Lack of nitrogen in plants will cause chlorosis whereby the leaves of the plants turn yellowish. 3 Composite Technology Research of Malaysia (CTRM) in Malacca produces fibreglass used in the making of aircraft and boats.

the scientist needs to draw a conclusion based on the experimental results. (j) Writing a report Lastly, the scientist has to write a report of his/her work. This will enable him/her to communicate with other scientists. The general format of a report:

1

The higher the temperature of the water, the greater the amount of sugar that can dissolve in it. (e) Apparatus and materials When planning an experiment, suitable apparatus and materials that are required to carry out the experiment are selected. (f) Listing a work procedure The procedure is the list of steps that needs to be taken to carry out an experiment. It is advisable to list the steps in point form. (g) Carrying out the experiment After planning the experiment, a scientist will carry out the experiment according to the procedure. (h) Data collection The scientist will then record the results of the experiment accurately. He or she should not change the results of the experiment and must be honest. (i) Data interpretation and conclusion After collecting the data, the scientist will analyse the results of his/her experiment. The results can be presented in various forms, such as a table, graph or calculation. Then,

Title: Aim: Problem statement: Hypothesis: Variables: (a) Manipulated variable: (b) Responding variable: (c) Constant variable(s): Materials: Apparatus: Procedure: Data and observation: Interpreting data: Discussion: Conclusion: 4 The following is an example of an experimental report.

Experiment 1.1

1.1 To investigate the effect of the temperature of water on the solubility of sugar Procedure Problem statement 1 100 cm3 of water is Does the amount of sugar that dissolves in water increase measured using a when the temperature of the water increases? measuring cylinder and Hypothesis is poured into a 250 The higher the temperature of the water, the greater cm3 beaker. the mass of sugar that dissolves in it. 2 The temperature of the Variables water is recorded using a thermometer. • Manipulated variable: Temperature of water 3 A 100 cm3 beaker is • Responding variable: Amount of sugar that filled with sugar. The dissolves at different temperatures Figure 1.1 beaker and its contents • Constant variable: Volume of water and size of are then weighed and recorded as a gram. sugar 4 The sugar is added a little at a time to the water Sugar and water. Materials in the beaker using a spatula. The mixture is then Apparatus stirred using a glass rod. 3 3 5 The process is continued until no sugar can 100 cm measuring cylinder, 250 cm beaker, 100 3 further dissolve in the water. cm beaker, electronic balance, Bunsen burner, 6 The beaker and its contents (sugar) are weighed tripod stand, wire gauze, spatula, thermometer and again and recorded as b gram. glass rod. Introduction to Chemistry

6

7 The amount of sugar that dissolved in the water at room temperature is (a – b) gram. 8 The experiment is repeated by heating the water to temperatures of 40 °C, 50 °C, 60 °C and 70 °C respectively. 9 The results are recorded in Table 1.3. Results

Interpreting data

Room temperature

40

50

60

70

Figure 1.2 Graph of mass of sugar dissolved against temperature

Initial mass of beaker and its contents (g)

a

b

c

d

e

Final mass of beaker and its contents (g)

b

c

d

e

f

A graph of the mass of sugar dissolved against temperature is plotted as shown in Figure 1.2. (Note: Both axes must be labelled with their units and the title of the graph must be stated)

Mass of sugar dissolved (g)

(a – b)

Temperature (°C)

Conclusion

(b – c) (c – d) (d – e) (e – f)

The amount of sugar that dissolves in the water increases when the temperature of the water increases. The hypothesis is accepted.

(Note: The unit of each reading must be stated:temperature in °C and mass in gram)

1.2 (iii) the constant variable of the experiment. (c) List the materials and apparatus needed to carry out the experiment. (d) Give a brief procedure of the experiment. (e) Tabulate your results.

1 You are required to investigate whether table salt dissolves in water and kerosene. (a) State a hypothesis for the experiment. (b) State (i) the manipulated variable, (ii) the responding variable,

1.3

1 A student must develop the following good laboratory practices. (a) Positive attitudes A student should (i) have an enquiring outlook, (ii) cooperate with other students while carrying out an experiment, (iii) be honest and not alter the results of an experiment. (b) Safety (i) Do not carry out an experiment without the supervision of the teacher. (ii) Do not taste any chemicals. (iii) Do not use burning paper to light a Bunsen burner. (iv) Always check the label of the chemical before using it.

Scientific Attitudes and Values in Conducting Scientific Investigations

Scientific Attitudes and Values

Students carrying out an experiment

7

Introduction to Chemistry

1

Table 1.3

1



(v) Dispose of all toxic waste in a proper container. (vi) Do not play with electrical appliances. (c) Wastage (i) Do not waste chemicals. Take only whatever is necessary. (ii) Switch off the gas supply or electricity when it is not required. (d) Cleanliness After carrying out an experiment, (i) the apparatus must be cleaned and returned to the same place, (ii) the table must be wiped dry with a towel or rag, (iii) all solid waste must be thrown into the dustbin and not into the sink.



1.3 1 What are the safety precautions that must be taken when carrying out the following experiments? (a) Diluting concentrated acid. (b) Heating a solution in a test tube. (c) Carrying out an experiment that involves the release of a poisonous gas.

(b) Responding variable: A variable that responds to the change of the manipulated variable. (c) Constant variable: The factor that is kept constant throughout the experiment. 5 After carrying out the experiment, you have to write a report which includes the following: (a) Write the aim or problem statement (b) State the hypothesis (c) List all the variables (d) List the chemicals and apparatus used in the experiment (e) Tabulation of your data (f) Interpret your result (g) Make a conclusion

1 The scientific method is a systematic approach to research. 2 The scientific approach begins with a hypothesis. A hypothesis is an intelligent guess relating a manipulated variable with a responding variable. 3 A variable is a factor that affects the result of a reaction. For example, the mass of salt that can dissolve in water depends on the volume and the temperature of the water. Volume and temperature are called variables. 4 There are three types of variables: (a) Manipulated variable: A variable that is changed during the experiment.

Introduction to Chemistry

(e) Accidents (i) Any chemical spilled on the body, clothing or eyes must be washed immediately with plenty of water. (ii) Any chemical unintentionally ingeste­d must be spat out immediately and the mouth must be washed with plenty of water.

8

1 1.1

Chemistry and Its Importance

1 The chemical used to neutralise acidity in soil is A potassium nitrate B calcium hydroxide C copper(II) oxide D sodium carbonate 2 The chemical used in raising flour is A calcium carbonate B sodium nitrate C magnesium sulphate D sodium bicarbonate 3 Which of the following careers below do not need a knowledge of chemistry? A Geologist B Forensic scientist C Meteorologist D Pharmacist 4 The branch of chemistry that studies carbon compounds is A organic chemistry B polymer chemistry C inorganic chemistry D industrial chemistry 5 Chloroform has the formula of CHCl3. Which of the following statements are true about the chloroform molecule? I It is made up of three elements. II It is made up of five elements. III The molecule consists of five atoms. IV The molecule consists of four atoms. A I and III only B I and IV only C II and III only D II and IV only 6 The substance that cannot be broken down into simpler form is called A compound C molecule B element D particle

7 Which of the following chemicals is synthetic? A Neon B Protein C Sodium hydroxide D Citric acid 8 DDT is a chemical used as pesticide. It is made up of carbon, chlorine and hydrogen atoms. The molecular formula of DDT is CCl3CH(C6H4Cl)2. What is the total number of atoms in a DDT molecule? A 3 C 18 B 17 D 28 9 Chemical X is used as electrolyte in the accumulator. Chemical Y is used in soap making. What is chemical X and Y? X

Y

A

Sulphuric acid

Sodium hydroxide

B

Sodium hydroxide

Sulphuric acid

C

Hydrochloric acid

Sodium hydroxide

D

Sodium hydroxide

Hydrochloric acid

1.2

Scientific Method

10 The factor that affects the result of an experiment is called a A solute C result B solution D variable For questions 11 – 13, use the information given below: Hydrogen peroxide decomposes as represented by the equation: 2H2O2 → 2H2O + O2 A student is required to study the effect of magnesium oxide and manganese(IV) oxide on the rate of decomposition of hydrogen peroxide.

9

11 What is the manipulated variable of the experiment? A Magnesium oxide and manganese(IV) oxide B Mass of magnesium oxide and manganese(IV) oxide C Temperature of hydrogen peroxide solution D Concentration of hydrogen peroxide solution 12 What is the constant variable of the experiment? I Volume of oxygen released II Mass of magnesium oxide and manganese(IV) oxide III Temperature of hydrogen peroxide solution IV Concentration of hydrogen peroxide solution A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 13 The responding variable for the experiment is A the rate of release of oxygen gas. B the decreasing rate in volume of hydrogen peroxide. C the rate of increase in concentration of hydrogen peroxide. D the decreasing rate in mass of the metal oxide. 14 Magnesium ribbon reacts with hydrochloric acid as shown. Mg + 2HCl → MgCl2 + H2 If you are required to study the effect of concentration of hydrochloric acid on the rate of reaction above, what variables must be constant? I Time of reaction II Temperature of hydrochloric acid III Size of beaker IV Length of magnesium ribbon A IV only B II and IV only C I, II and IV only D II, III and IV only Introduction to Chemistry

1

Multiple-choice Questions

15 “The greater the quantity of sodium chloride added to ice, the lower its melting point”. If you are required to study the above hypothesis, what is the manipulated variable? A Mass of ice B Types of salt added C Mass of salt added D Temperature of ice 16 “Without water iron will not rust”. Which of the following is correct in carrying out the experiment to prove the statement above?

1

Manipulated variable

Constant variable

A

Presence or absence of water

Rusting of iron

B

Presence or absence of water

Presence or absence of air

C

Presence or absence of water

Presence of air

D

Presence or absence of air

Presence of water

17 A student wants to find out the effect of temperature on the solubility of sugar in water. Which of the following is correct? Manipulated variable

Responding variable

Constant variable

A

Temperature of water

Mass of sugar dissolved

Volume of water

B

Volume of water

Mass of sugar dissolved

Temperature of water

C

Mass of sugar dissolved

Temperature of water

Volume of water

D

Temperature of water

Mass of sugar dissolved

Humidity of the air

18 An experiment is carried out to study the solubility of sodium chloride in water and in benzene. What is the (i) manipulated variable (ii) constant variable of the experiment? Manipulated variable

Constant variable

A

Solvent

Rate of stirring

B

Solvent

Mass of sodium chloride

C

Solute

Volume of solvent

D

Solute

Rate of stirring

19 A hypothesis I is a law of science. II can be a true or false statement. III is a conclusion derived from the result of the experiment. IV is a statement that relates the manipulated variable and the responding variable. A I and III only B II and IV only C I, II and III only D II, III and IV only Introduction to Chemistry

10

20

Q

Analyse data

R

Observe a situation

S

Make a hypothesis

T

Carry out the experiment

U

Collect data

The steps above are the steps taken to carry out a scientific investigation. The correct order in carrying out the investigation is A S, R, T, U, Q B R, S, T, U, Q C R, T, U, Q, S D R, T, U, S, Q 21 During electrolysis, the mass of metal deposited at the cathode is dependent on the time and the amount of current passed through the electrolyte. If you are required to show that the mass of metal deposited is proportional to the current passed through the electrolyte, what are the variables in this experiment? Manipulated variable

Constant variable

A

Time

Types of electrodes used

B

Time

Amount of current

C

Amount of current

Types of electrodes used

D

Amount of current

Time

1.3

Scientific Attitudes and Values

22 What precautions must you take when storing concentrated nitric acid? A Store it in a dark place B Store it in a fume cupboard C Store it in a locked cupboard D Store it away from any Bunsen burner 23 What precaution must you take when diluting concentrated sulphuric acid? A Add the concentrated sulphuric acid to water

25

A bottle of chemical has a label shown in the diagram. What does this label represents? A Flammable chemical B Corrosive chemical C Radioactive chemical D Oxidising chemical

24

A bottle of chemical has a label as shown in the diagram. What precaution must be taken when storing this chemical? A Store it in a dark place B Store it in a fume cupboard C Store it in a locked cupboard D Store it away from any Bunsen burner

26 When heating a solution in a boiling tube, what precaution must you take? A Never heat the solution too strongly B Never hold the boiling tube vertically C Never use a Pyrex boiling tube to heat the solution D Never direct the mouth of the boiling tube at your classmates

27 An experiment should be carried out in the fume cupboard if it involves A the release of poisonous gas. B the release of flammable gas. C the use of corrosive chemicals. D the use of oxidising chemicals. 28 Why is it important to understand the experimental procedures before carrying out the experiment? I To prevent wastage of chemicals II To prevent accidents from happening III To prevent repetition of the experiment IV To know what apparatus is needed to carry out the experiments A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV

Structured Questions seconds. From the graph in (i), determine the concentration of the hydrochloric acid solution. [1 mark]

1 Table 1 shows the time taken for a 5 cm length of magnesium ribbon to dissolve in 50 cm3 of dilute hydrochloric acid of different concentrations.

2 Table 2 shows the mass of two salts P and Q that dissolved in 100 cm3 of water at different temperatures.

Concentration of 0.1 0.2 0.3 0.4 0.5 hydrochloric acid (mol dm–3) Time taken for a 5 cm magnesium ribbon to dissolve (s)

Temperature (°C)

30 26 22 18 14

Table 1

(a) State (i) the manipulated variable, (ii) the responding variable and (iii) the constant variable of the experiment above. [3 marks] (b) State a hypothesis for this experiment.

[1 mark]

Solubility of salt (mass of salt soluble in 100 cm3 of water) Salt P (g)

Salt Q (g)

30

5

7

40

10

14

50

15

21

60

20

28

70

25

35

Table 2

(c) (i) Plot a graph of concentration of hydrochloric acid against time taken for the magnesium to dissolve. [5 marks] (ii) If a 5 cm length of magnesium ribbon is added to a hydrochloric acid solution of unknown concentration, the time taken for the magnesium ribbon to dissolve is 17

(a) State one other variable, besides temperature, that affects the solubility of salt. [1 mark] (b) In the experiment above, state (i) the manipulated variable (ii) the constant variable (iii) the responding variable

11

[3 marks]

Introduction to Chemistry

1

B Add water to the concentrated sulphuric acid C Mix equal volumes of the concentrated sulphuric acid and water together D Mix one volume of concentrated sulphuric acid to three volumes of water together

(c) Plot a graph of the solubility of salts P and Q against temperature on the same axis. [4 marks]

(a) State a hypothesis for the experiment above. [1 mark]

(d) What can you conclude from the graphs in (c)?

(b) What is the (i) manipulated variable (ii) responding variable (iii) constant variable when carrying out the experiment? [3 marks]

[2 marks]

1

3 The procedure below shows the sequence in carrying out an experiment to study the effect of the temperature of water on the mass of sugar that can dissolve.

(c) Calculate the mass of sugar that dissolved in water at various temperatures and write your answer in the right column of the table. [2 marks]

Procedure: • Initial mass of beaker P and sugar is taken (a gram). • 50 cm3 of water is poured into a separate 100 cm3 beaker. The water is heated to 30 °C. • Sugar is added to the 50 cm3 of water at 30 °C a little at a time while stirring the mixture until no more sugar can further dissolve. • The final mass of beaker P and sugar is taken (b gram). The mass of sugar that dissolved is (a – b) gram. • The experiment is repeated by dissolving the sugar in water heated to 40 °C, 50 °C, 60 °C and 70 °C.

(d) Plot a graph of the mass of sugar that dissolved against the temperature of water. [3 marks] (e) From your graph, estimate the mass of sugar that can dissolve in water at 45 °C. [1 mark] 4 (a) Name four chemicals used in food processing. [4 marks]

(b) Name six careers that need a knowledge of chemistry. [6 marks]

The results are tabulated in Table 3. Initial mass Final mass Temperature of beaker P of beaker P of water (°C) and contents and contents (g) (g) 30

92.50

87.50

40

87.50

77.50

50

77.50

62.50

60

62.50

42.50

70

42.50

17.50

(c) Name three contributions of chemical industries to the country. [3 marks]

Mass of sugar dissolved (g)

(d) Name five scientific values that must be observed when carrying out scientific research. [5 marks] (e) Name two types of chemicals that can increase the yield of crops. [2 marks]

Table 3

Essay Questions (b) The list below shows the steps involved in carrying out scientific research:

1 ‘Without air, an iron nail will not rust’. You are required to plan an experiment to verify the statement. (a) List the apparatus and materials needed to carry out the experiment. [3 marks]

Making a hypothesis, making a conclusion, collecting data, making an inference, making an observation, carrying out an experiment, interpreting data, identifying variables and planning the procedure of the experiment.

(b) State (i) the manipulated variable, (ii) the responding variable and (iii) the constant variable of the experiment above. [3 marks]

(c) Briefly write the procedure for the experiment. [10 marks]

[8 marks]



(d) Tabulate your results. [4 marks] 2 (a) Explain the meaning of the scientific method. [2 marks]

Introduction to Chemistry

(i) Arrange the steps in the correct order.



12

(ii) Explain the difference between inference and hypothesis. [4 marks] (iii) State three ways of presenting the experimental results. [6 marks]

Experiments 1 ‘The greater the volume of water, the higher the solubility of salt’. Plan an experiment to prove the statement. Your answer should include the following items: (a) Aim of experiment (b) Statement of hypothesis (c) All variables (d) List of materials and apparatus (e) Procedure (f) Tabulation of data

[3 marks] [3 marks] [3 marks] [3 marks] [3 marks] [3 marks]

Concentration of sodium hydroxide solution (mol dm–3)

0.1

0.2

0.3

0.4

0.5

pH value

13.0

13.3

13.5

13.6

13.7

1

2 The table shows the pH values of 25 cm3 sodium hydroxide solutions of different concentrations measured by a student using a pH meter.

(a) State the variables of this experiment. [3 marks] (b) Suggest a hypothesis for the experiment. [3 marks] (c) Plot a graph of pH value against concentration of the NaOH solution.

[3 marks]

(d) Using the graph that you have plotted, determine (i) the pH value of a sodium hydroxide solution with a concentration of 0.35 mol dm–3. (ii) the concentration of NaOH solution with a pH value of 13.4.

[3 marks]

3 Manganese(IV) oxide is a catalyst that speeds up the decomposition of hydrogen peroxide (H2O2) to form water and oxygen gas as represented by the equation: 2H2O2(l) → 2H2O(l) + O2(g) A student carried out an experiment by adding different amounts of manganese(IV) oxide to 50 cm3 of 0.2 mol dm–3 hydrogen peroxide solution. The table shows the results obtained by the student. Quantity of manganese(IV) oxide (g)

0.2

0.4

0.6

0.8

1.0

Time taken to collect 50 cm of oxygen (s)

30

25

20

15

10

3

(a) State the (i) manipulated variable, (ii) responding variable, (iii) constant variable of the experiment.

[3 marks]

(b) What can you conclude from the results of the experiment?

13

[3 marks]

Introduction to Chemistry

FORM 4 THEME: Matter Around Us

CHAPTER

2

The Structure of the Atom

SPM Topical Analysis 2008

Year 1

Paper

3

2

Section

A

Number of questions

1 — 2

5

2009

B –

1

2010 3

2

C

A

B

C

–

1 — 2

–

–

–

6

–

1

2

2011

2

3

A

B

C

1

–

–

1

–

2

3

A

B

C

1

–

–

3

–

ONCEPT MAP

MATTER

Kinetic theory of matter

Changes in states of matter

Diffusion in a solid, liquid and gas

Atomic structure

Particles in matter: atom, molecule and ion

Subatomic particles: proton, electron and neutron Determination of the melting and freezing points of naphthalene

Electron arrangement in atoms and valence electrons

Symbols of elements A Z

X

Isotopes

2.1

Matter

(b) When the gas tap in the laboratory is turned on, the smell of the gas is immediately detected. This shows that the gas is also made up of particles in motion. 7 An element is a substance that cannot be made into anything simpler by means of a chemical reaction. 8 The particles in some elements are made up of atoms. For example, metals like gold, copper, iron, zinc are all made up of atoms.

SPM

’08/P1

1 Substance

2

1 Chemistry is the study of matter, its composition and the changes it undergoes. 2 Matter is anything that occupies space and has mass. In other words, matter is anything that has volume and mass. 3 Examples of matter are books, pens, chairs, water, air and plants. Examples of non-matter are electricity and light. 4 The particle theory of matter states that matter is made of very tiny discrete particles. The particulate nature of matter is investigated in Activity 2.1. 5 Elementary particles that make up matter may be atoms, molecules or ions.

’04

Chemical formula

Naphthalene

C10H8

Iron

Fe

Sodium chloride

NaCl

Figure 2.1 Copper foil is made up of atoms

9 A compound is a substance that can be made into something smaller by means of a chemical reaction. 10 Compounds contain more than one element. The elements in a compound are not just mixed together. They are joined by strong forces called chemical bonds. Compounds do not have the same properties as the elements they contain. Compounds are (a) formed by chemical reactions, and (b) they have different properties from the elements they contain. 11 The particles in compounds may be molecules or ions. Molecules are made up of two or more atoms held together by chemical bonds. Molecules are particles that are not charged. 12 A molecule may consists of atoms of the same element, for example, oxygen molecules (O2), nitrogen molecules (N2), hydrogen molecules (H2) and sulphur molecules (S8) (Figure 2.2(a)). 13 A molecule may also consist of dissimilar atoms of two or more elements. For example, a water molecule (H2O) consists of one oxygen and two hydrogen atoms, and a carbon dioxide molecule (CO2) consists of one carbon and two oxygen atoms (Figure 2.2(b)). 14 Some molecules can be very large. For example, quinine which is a drug used to treat malaria patients has the formula C20H24N2O2.

State the particles present in each of the above substances. Solution Naphthalene – molecules, iron – atoms, sodium chloride – ions.

All metals and noble gases are made up of atoms. A compound formed between non-metallic elements (example: naphthalene, C10H8), is made up of molecules. A compound formed between a metal and a non-metal (example: sodium chloride, NaCl) is made up of ions.

6 The existence of these particles is supported by some observations. Some examples are: (a) When a drop of ink falls into a glass of water, the colour of the ink spreads throughout the water. This shows that ink is made up of particles in Dropping ink into a motion. glass of water 15

The Structure of the Atom

(a) The particles (atoms, molecules or ions) possess kinetic energy. They are in constant motion and constantly collide with each other. (b) The velocities of the particles in the three physical states of matter—solid, liquid and gas—are different. (c) The higher the temperature, the higher the kinetic energy, as the velocity of the particles increases. (d) At a given temperature, the lighter particles move faster than the heavier ones. 2 In 1827, Robert Brown (a botanist) made an observation through a microscope. He found that pollen grains on the surface of water are in constant motion. He explained that the pollen grains are moving because the moving water molecules are constantly colliding with the pollen grains. The visible motion of these pollen grains is called the Brownian motion. 3 The Brownian motion gives the evidence that a liquid consists of particles in constant movement.

(a) Model of nitrogen, oxygen and sulphur molecules

(b) Model of carbon dioxide and water molecules

2

Figure 2.2

15 However, some compounds consist of atoms or a group of atoms that carry positive or negative charges. These charged particles are called ions. For example, table salt, NaCl, consists of sodium ions (Na+) and chloride ions (Cl–) (Figure 2.3). The rust on an iron nail consists of iron(III) ions (Fe3+) and oxide ions (O2–).

Figure 2.3 Model of sodium chloride crystal

16 Ions which are positively-charged are called cations. For example, sodium ions (Na+) and iron(III) ions (Fe3+) are cations. 17 Ions which are negatively-charged are called anions. For example, chloride ions (Cl–) and oxide ions (O2–) are anions. 18 Generally, metals form positive ions and nonmetals form negative ions. Some examples of cations and anions are given in Table 2.1.

Figure 2.4 Pollen grain being bombarded by water molecules

4 Another evidence of the movement of particles is diffusion. Diffusion is the random movement of particles from a region of high concentration to a region of low concentration.

Table 2.1 Examples of positive ions (cations)

You can smell perfume while you walk past cosmetic counters.The perfume particles have left the open perfume bottles and spread out through the air by diffusion.

H+, K+, Cu2+, Al3+, NH4+, Mg2+, Ca2+, Zn2+, Pb2+ and Ag+

Examples of negative Br–, I–, OH–, NO3–, SO42–, CO32–, PO43–, O2–, S2– ions (anions) and S2O32–

The Kinetic Theory of Matter 1 The kinetic theory is an extension of the particle theory of matter. According to the kinetic theory: The Structure of the Atom

5 There are three states of matter, namely, solid, liquid and gas. Table 2.2 shows the comparison between the three states of matter. 16

SPM

Table 2.2 Comparison between the three states of matter

solid

liquid

The particles are very closely packed.

The particles are closely packed but there are more empty spaces between them compared to the solid state.

The particles are very far apart from each other.

Forces of The very strong forces attraction of attraction restrict the between particles movement of the particles. The particles in a solid are held in fixed positions.

The forces of attraction are weaker than in the solid state. The particles are no longer held in fixed positions.

The forces of attraction are very weak. The particles move randomly in all directions at great speed.

Volume and shape

Solids have fixed volumes and shapes.

Liquids have fixed volumes. However, they do not have fixed shapes but take the shapes of the containers.

Gases do not have fixed shapes or volumes.

Types of movement

Vibration and rotation

Vibration, rotation and translation

Vibration, rotation and translation

Kinetic energy of particles

The kinetic energy of the particles are low.

The kinetic energy of the particles are high, on average.

The kinetic energy of the particles are very high and they move at high speed.

Compressibility

Very difficult to be compressed because the particles are packed closely

Not easily compressed because the particles are packed quite closely

Easily compressed because the particles are very far apart

Arrangement of particles

Rate of diffusion

Very low

Average Diffusion

gas

Very high SPM

’08/P2

1 Diffusion refers to the process by which particles intermingle as a result of their kinetic energy of random motion. 2 Figure 2.5(a) shows a container that consists of gases A and B. The two gases are separated by a partition. The particles of both gases are in constant motion and make numerous collisions with the partition. 3 If the partition is removed as in Figure 2.5(b), the gases will mix because of the random motion of their particles. In time, a uniform mixture of gases A and B particles will be produced in the container. 4 The rate of diffusion depends on the temperature and the molecular mass of the particles. The higher the molecular mass, the lower the rate of diffusion. 17

The Structure of the Atom

2

States of matter

’11/P2

To investigate the diffusion of particles in a gas, liquid and solid

SPM

’09/P1

Apparatus

Two gas jars with plastic covers, beaker, teat pipette, boiling tube, spatula and rubber stopper.

Materials

Liquid bromine, potassium manganate(VII), KMnO4 crystals, water and hot jelly solution.

Procedure (A) Diffusion in a gas

2

1 A few drops of liquid bromine are dropped into a gas jar using a teat pipette. 2 The gas jar is covered with a gas jar cover. 3 An empty gas jar is placed upside down on top of the first jar. 4 The cover is removed and any colour change is recorded. The time taken for the brown bromine vapour to spread into the second gas jar is recorded. Figure 2.6

(B) Diffusion in a liquid 2 filled with water. 1 A beaker is — 3 2 A few potassium manganate(VII) crystals are placed at the bottom of the water using a spatula. 3 Any colour change is recorded. The time taken for the purple manganate(VII) ions to spread throughout the water is recorded. Figure 2.7

(C) Diffusion in a solid 1 Some freshly cooked jelly solution is poured into a boiling tube until it is almost full. 2 The jelly is allowed to set. 3 A small potassium manganate(VII) crystal is placed on top of the jelly. 4 The boiling tube is then stoppered using a rubber stopper. 5 Any colour change is recorded. The time taken for the purple manganate(VII) ions to spread throughout the solid jelly is recorded. Figure 2.8

Results

Activity 2.1

Experiment

Observation

A

The brown bromine vapour spreads out into the upper gas jar. The time taken is very short.

B

After about 10 minutes, the purple colour of the manganate(VII) ions had spread throughout the water.

C

After a week, the purple colour of the manganate(VII) ions had spread throughout the solid jelly.

The Structure of the Atom

18

Discussion

2

1 Diffusion has taken place in the gas (air in experiment A), liquid (water in experiment B) and solid (jelly in experiment C). 2 The rates of diffusion of the particles in the solid, liquid and gaseous states are different. It is highest in gases, lower in liquids and lowest in solids. 3 This shows that there are more and bigger spaces between particles in the gas. The spaces between liquid particles are smaller. The particles in the solid state are very close with little space between them. 4 The occurence of diffusion proves that matter (bromine and potassium manganate(VII)) consist of particles in constant motion. 5 The diffusion experiments show that because particles possess kinetic energy, they are in constant motion.

SPM

The Changes in the States of Matter

’10/P2

1 A substance can be changed from one state into another when it is heated or cooled. 2 The changes in the state of the substance can be explained using the kinetic theory model.

Heating

Heating

Gas

Solid

Liquid

1 The particles in a solid are packed closely in a fixed pattern. 2 When the solid is heated, the particles receive heat energy. The kinetic energy of the particles increases and the particles vibrate faster. 3 At the melting point, the particles vibrate so much that they break away from their fixed positions. The solid becomes a liquid. 4 The temperature at which the solid changes into the liquid state is called the melting point.

1 When a liquid is continuously heated, the par­­ti­­­cles receive more energy and move even faster. They collide with each other more often. 2 At the boiling point, the particles receive enough energy to overcome the forces of attraction holding them together. The particles in the liquid state break loose to become the gaseous state. 3 When the liquid is cooled, the movement of the particles slows down. Stronger forces of attraction between the particles are formed. 4 The particles are arranged in an orderly ma­n­­ner in the solid state. The process whereby the liquid changes into a solid is called solidification. The temperature at which this process occurs is called the freezing point. 5 The melting point and the freezing point of a substance have the same value.

Cooling

1 When a gas is cooled, the particles lose kinetic energy. The movement of the particles slows down. 2 The forces of attraction between the particles are formed which hold the particles together in the liquid state. 3 The process whereby the gas changes into a liquid is called condensation. 4 The temperature at which the gas condenses to the liquid state is the same as the boiling point.

Cooling

19

The Structure of the Atom

3 Examples of substances that undergo sublima­ tion are iodine, ammonium chloride and solid carbon dioxide (dry ice). The process in which substances change directly from the gaseous to the solid state is also called sublimation.

2

Melting Point and Boiling Point 1 No two substances have the same melting and boiling points. We can thus identify a substance by its melting and boiling points. 2 The melting and boiling points of a substance will change when there is a small amount of impurity in it. For example, the melting point of pure water is 0 °C and its boiling point is 100 °C. A small amount of salt added to the water will decrease its melting point to –2 °C and increase its boiling point to 102 °C. 3 As the melting and boiling points of an impure substance will deviate slightly from its standard values, we can determine the purity of a substance by the melting and boiling points of the substance.

When a state of matter gains or loses heat, it undergoes a change. A gain in heat is called an endothermic change. A loss in heat is called an exothermic change.

Sublimation 1 Certain substances do not melt when heated. They change directly from the solid to the ’11/P1 gaseous state. 2 This process is called sublimation. SPM

To determine the melting and freezing points of naphthalene (A) Heating of naphthalene 1 3 spatulas of naphthalene powder are placed in a boiling tube. 2 A 500 cm3 beaker is filled with water until it is 3 about — full. It is then placed on a tripod stand. 4 3 The boiling tube containing naphthalene is clamped in the beaker of water, making sure the naphthalene powder is below the water level of the water bath. 4 The water bath is heated until it reaches a temperature of about 65 °C as shown in Figure 2.9. The water is then heated with a low flame. 5 A stopwatch is started and the temperature of the naphthalene is recorded at 30-second intervals until the temperature reaches 90 °C. The naphthalene is stirred continuously during the experiment. 6 The results are recorded in a table.

Apparatus Boiling tube, retort stand and clamp, tripod stand, Bunsen burner, wire gauze, thermometer (0 – 110 °C), 500 cm3 beaker, 250 cm3 conical flask, test tube holder and stopwatch. Materials

Naphthalene and water.

Procedure Figure 2.9 Heating of naphthalene

Activity 2.2

Figure 2.10 Cooling of naphthalene

The Structure of the Atom

(B) Cooling of naphthalene 1 The boiling tube containing the molten naphthalene is removed from the hot water bath using a test tube holder. 2 It is immediately transferred into a conical flask to be cooled slowly as shown in Figure 2.10. 20

Discussion 1 In the heating of naphthalene, a water bath is used instead of direct heating. This is to ensure that an even heating process is carried out. 2 In the cooling of naphthalene, the boiling tube containing the liquid naphthalene is cooled inside a conical flask. This is to ensure that an even cooling process is carried out. 3 Stirring the naphthalene continuously also ensures even heating or cooling. 4 A water bath is suitable in this experiment because the melting point of naphthalene is below 100 °C, the maximum temperature that can be attained by the water bath. 5 If the melting point of the substance is above 100 °C, the water bath will have to be replaced by an oil bath or a sand bath. 6 Besides naphthalene, the other substance that is suitable for heating by water bath is acetamide. 7 The heating curve of naphthalene consists of three regions: AB, BC and CD as in Figure 2.11.

Results (A) Heating of naphthalene Time (s)

Temperature (°C)

0 30 60 90 120 150 180 210 (B) Cooling of naphthalene Time (s)

Temperature (°C)

0 30 60 90 120 150 180 210

Region in the graph

State of substance and the energy change

Region AB Naphthalene is in the solid state. As napthalene is heated, heat energy is SPM ’04/07 converted to kinetic energy. Kinetic P2 energy increases and the molecules vibrate faster about their fixed positions. Temperature increases as the molecules receive more heat energy.

Analysis of data Point B 1 A graph of temperature SPM against time is plotted ’10/P1 for the heating of naphthalene. The graph is shown in Figure 2.11. 2 A graph of temperature against time is plotted Figure 2.11 Heating curve Region BC for the cooling of of naphthalene SPM ’11/P1 naphthalene. The graph is shown in Figure 2.12. 3 When plotting a graph, make sure that: (a) The axes are labelled with their units. (b) The points are transferred correctly. Figure 2.12 Cooling curve (c) The curve is smooth. of naphthalene 21

As the kinetic energy of the molecules increases, the molecules vibrate faster. At point B, some molecules vibrate so much that they break away from their fixed positions. The solid naphthalene begins to melt. Naphthalene now consists of a mixture of solid and liquid. At this region the temperature remains constant because the heat energy supplied by the water bath is the same amount as the heat energy absorbed. Heat energy is absorbed to overcome the forces of attraction holding the naphthalene molecules together in the solid state. The heat absorbed to overcome the forces of attraction is called the latent heat of fusion. Latent heat of fusion of The Structure of the Atom

2

3 The stopwatch is started and the temperature of the naphthalene is recorded at 30-second intervals until it drops to about 70 °C. The naphthalene is stirred continuously during the experiment. 4 The results are recorded in a table.

Region in the graph

2

Point C

State of substance and the energy change

Region in the graph

State of substance and the energy change

a substance is the heat required to convert a solid into a liquid without a change in temperature.

moving except for small vibrations. At point Q, the liquid naphthalene begins to solidify or freeze.

All the naphthalene has completely melted.

Region QR Naphthalene now consists of a mixture of liquid and solid. At this region the temperature remains constant because the heat energy lost to the environment is the same amount as the heat energy released. Latent heat of fusion is released when forces of attraction are formed between the molecules as the liquid naphthalene solidify (or freezes).

Region CD Naphthalene is in the liquid state. As the liquid naphthalene is heated, the molecules gain more heat energy. The temperature continues to increase. 8 The cooling curve of naphthalene consists of three regions: PQ, QR and RS as in Figure 2.12.

Point R

All the naphthalene has completely solidified.

Region PQ Naphthalene is in the liquid state. The liquid naphthalene loses heat to the environment. The kinetic energy of the molecules decreases as the temperature decreases.

Region RS

Naphthalene is in the solid state. The solid naphthalene continues to lose heat to the environment and hence the temperature drops down to room temperature.

Point Q

Conclusion The melting point and the freezing point of naphthalene is 80 °C.

Region in the graph

State of substance and the energy change

As the kinetic energy of the molecules decreases, the molecules move slower. At point Q, some molecules stop

2

’09

A I and III only B II and IV only

C I, II and III only D I, II and IV only

Comment From time 0 to t1 the substance loses heat to the surroundings. Hence the temperature decreases. (Statement I is incorrect) From time t1 to t2 condensation takes place and heat energy is released. The kinetic energy of particles becomes lower and the forces of attraction become stronger. (Statement II is correct) The graph shows the cooling curve for gas X. Which of the following statements are true? I From time 0 to t1 heat energy is absorbed. II From time t1 to t2 forces of attraction between particles become stronger. III From time t2 to t3 the kinetic energy of particles increases. IV From time t3 to t4 heat energy released is equal to the heat lost to the surroundings. The Structure of the Atom

From time t2 to t3 the particles continue to lose heat to the surroundings. Hence the kinetic energy of particles decreases. (Statement III is incorrect) From time t3 to t4 freezing takes place. The temperature of the substance remains constant because during freezing, heat energy released is equal to the heat lost to the surroundings. (Statement IV is correct) Answer B 22

2.1 1 State the type of particles (atoms, molecules or ions) that make up the substances below. (a) Ammonia gas (d) Potassium iodide (b) Sodium chloride (e) Copper wires (c) Iron nail (f) Cooking oil the

heating

curve

for Figure 2.13 John Dalton and his model

3 However, Dalton’s atomic model had its weakness. It was found that: (a) The atom is not the smallest particle in an element. There are subatomic particles (proton, electron and neutron) in an atom. (b) A radioactive atom decomposes spon­ taneously, which means that an atom can be destroyed. A new atom can also be created by a process called transmutation. (c) Not all atoms of an element are alike. They may differ in atomic mass. For example, hydrogen has three isotopes 11H, 21H and 31H. 4 In 1897 J. J. Thomson discovered negativelycharged particles which he called electrons. Thomson then suggested that an atom is a positively-charged sphere with electrons embedded in it like a raisin pudding.

(a) State the melting point of naphthalene. (b) What is the physical state of naphthalene at time t second? (c) Why does the temperature remain constant at region BC although heating is carried on? (d) Draw the cooling curve obtained when the molten naphthalene is cooled from T3 to room temperature.

2.2

The Atomic Structure

The Historical Development of the Atomic Model 1 The concept of the atom originated from Democritus, a Greek philosopher. He proposed that if a piece of gold is divided repeatedly, it will reach a state whereby the smallest particle, which is indivisible, is obtained. He called the smallest indivisible particle atomos, which means ‘indivisible’ in Greek. 2 In 1808, John Dalton proposed the atomic theory. In this theory, Dalton proposed that: (a) All elements are made up of small indivisible particles called atoms. (b) Atoms are neither created nor destroyed in chemical reactions. (c) The atoms of an element are alike, but differ from the atoms of other elements. (d) When atoms combine, they do so in a simple ratio. (e) All chemical reactions result from the combination or separation of atoms.

Figure 2.14 J. J. Thomson and his model

5 In 1911, Ernest Rutherford bombarded a thin gold foil with alpha particles (helium nuclei, He2+). (a) It was found that most of the alpha particles passed directly through the gold foil without deflection. Rutherford then suggested that most of the atom must be empty space.

Figure 2.15(a) Rutherford’s experiment

23

The Structure of the Atom

2

2 The graph shows naphthalene.

7 In 1932, James Chadwick discovered rays of electrically neutral subatomic particles which he called neutrons. The neutron has a mass almost the same as that of a proton. Chadwick suggested that the nucleus of the atom contains protons and neutrons, and the nucleus is surrounded by a cloud of electrons.

2

Figure 2.15(b) Magnified view showing alpha particles deflected by the nuclei of gold atoms

(b) However, some of the alpha particles were deflected at very acute angles. To explain the deflection of the alpha particles, Rutherford proposed that all the positive charge of an atom is concentrated in the nucleus, which repelled the positivelycharged alpha particles in the opposite direction. Further experimental studies led to the discovery of positive particles in the nucleus. Rutherford called the positively-charged particles protons. (c) Rutherford proposed that an atom consists of a positively-charged nucleus with a cloud of electrons surrounding the nucleus.

Figure 2.18 James Chadwick and his model

SPM

8 The atomic model in the present day is based on the contributions of the above scientists. In this atomic model: (a) The nucleus of an atom consists of protons and neutrons occupying a small space in the centre of the atom. (b) Electrons are moving around the nucleus in permissible orbits or electron shells (also known as quantum shells).

’08/P1

Figure 2.16 Ernest Rutherford and his model

Subatomic Particles of an Atom

6 In 1913, Niels Bohr proposed that the electrons in the atom are arranged in permitted orbits called electron shells surrounding the nucleus.

An atom is made up of three smaller particles which are called protons, neutrons and electrons. These particles are called subatomic particles. Table 2.3 shows the relative masses and charges of these particles. Table 2.3 The symbols, relative masses and the charges of subatomic particles

Subatomic particle

Figure 2.17 Niels Bohr and his model The Structure of the Atom

24

Symbol

Relative mass

Charge

Proton

p

1

+1

Electron

e

1 — — — — 1840

–1

Neutron

n

1

0

3

that the number of neutrons in phosphorus is 31 – 15 = 16. 6 The relative masses of the proton and neutron are almost similar. However, the relative mass of the electron is very small. So the mass of an atom is determined by the number of protons and neutrons in the atom. 7 The nucleon number and proton number of SPM an element is written in the following way: ’09/P2

’05

The diagram shows a model of an atom.

2

Who introduced this model? A Niels Bohr B J. J. Thomson C John Dalton D Rutherford Answer A Niels Bohr. He proposed that electrons are arranged in shells surrounding the nucleus.

Proton Number and Nucleon Number 1 Protons and neutrons are located in the nucleus and the electrons are arranged in electron shells surrounding the nucleus. 2 The nucleus is positively-charged because it contains protons, each of which carry a positive charge. 3 The proton number of an element is the number of protons in its atom. The proton number is also known as the atomic number. Each element has its own proton number. No two different elements can have the same proton number. For example, sodium, with a proton number of 11 means that it has 11 protons in its nucleus and an element with 11 protons in its nucleus must be sodium. 4 In a neutral atom, the proton number also tells us the number of electrons. For example, the proton number of magnesium is 12. Therefore, a magnesium atom has 12 protons and 12 electrons. The proton number of nitrogen is 7 and hence a nitrogen atom has 7 protons and 7 electrons. 5 The nucleon number (also known as the SPM mass number) of an element is the sum of the ’11/P2 number of protons and neutrons in its atom.

A student need not memorise the proton number and nucleon number. It will be given in the examination. The proton number is smaller than the nucleon number.

4

’03

State the number of protons, electrons and neutrons 37 in a chlorine atom, 17 Cl. Solution 17 protons, 17 electrons and 20 neutrons (37 – 17 = 20)

Symbols of Elements 1 Each element is represented by a symbol, consisting of either one letter or two letters of the alphabet. 2 Some elements are represented by the first letter of its name. Examples are in the following table. Name of element Hydrogen Nitrogen Oxygen Fluorine Sulphur

Nucleon Number of Number of = + number protons neutrons OR Nucleon Proton Number of = + number number neutrons

Symbol H N O F S

3 The names of some elements start with the same letter. For example, the names of the elements Nitrogen, Neon, Nickel and Nobium start with the letter ‘N’. Therefore, a second letter is added to differentiate between these elements. The second letter used is always a small letter. Examples are in the following table.

For example, a sodium atom has 11 protons and 12 neutrons; hence the nucleon number of sodium is 23. The proton number of phosphorus is 15 while its nucleon number is 31. This means 25

The Structure of the Atom

and 53 stand for? How many protons, electrons and neutrons are there in an iodine atom?

Symbol Si Ne Cl Ca Br Mg

Name of element Silicon Neon Chlorine Calcium Bromine Magnesium

2 (a) A list of elements are represented by the letters given below: 11 5

Which two letters represent the same element? Explain your answer. (b) State four facts that you can derive from the nuclear symbol, 27 Al. 13

2

4 Some elements are represented by the letters of their Latin names. For example, Name of element Silver Copper Iron Gold Lead Tin Potassium Sodium Mercury

Latin Name

Symbol

Argentum Cuprum Ferrum Aurum Plumbum Stannum Kalium Natrium Hydrargyrum

Ag Cu Fe Au Pb Sn K Na Hg

A, 126 B, 24 C, 23 D, 146 E and 147 F 12 11

2.3

Isotopes and Their Importance

Isotopes 1 Isotopes are atoms of the same element with SPM the same proton number but different nucleon ’10/P1 numbers. Alternatively, isotopes can be defined as atoms of an element with the same number of protons but different numbers of neutrons. 2 Many elements exhibit the phenomenon of isotropy, whereby an element can have more than one type of isotope. 3 The isotopes of an element have the same chemical properties because they have the same electron arrangement but their physical properties such as densities and melting points differ. 4 Table 2.4 shows examples of isotopes of some elements.

2.2 1 (a) An atom of uranium (U) has 92 protons and 143 neutrons. What is the proton number and nucleon number? Write its atomic symbol. (b) Seaweed is rich in the element iodine, represented by 127 l. Lack of iodine in our diet 53 can cause goiter. What do the numbers 127

Table 2.4 Examples of isotopes of some elements

SPM

’07/P2

Hydrogen, 11H Deuterium, 12H Tritium, 13H

Proton number 1 1 1

Nucleon number 1 2 3

Number of protons 1 1 1

Number of neutrons 0 1 2

Percentage abundance 99.985% 0.015% Man-made isotope

Carbon-12, 126C Carbon-13, 136C Carbon-14, 146C

6 6 6

12 13 14

6 6 6

6 7 8

98.1% 1.1% Trace amount

17 17

35 37

17 17

18 20

75.5% 24.5%

8 8 8

16 17 18

8 8 8

8 9 10

99.757% 0.038% 0.205%

Element

35 Cl Chlorine-35, 17 37 Chlorine-37, 17 Cl

Oxygen-16, 168O Oxygen-17, 178O Oxygen-18, 188O

Same Different Same Different Nucleon number = Number of protons + Number of neutrons The Structure of the Atom

26

5 Some elements, such as fluorine, F, have only one isotope. However, most elements have more than one isotope. 6 The relative atomic mass of an element is based on the average mass of all the isotopes of the element. For example, the relative atomic mass of chlorine is 35.5 because chlorine has 75% of 35 Cl and 25% of 37 Cl. 17 17 7 In an element, some isotopes are stable while the rest are unstable isotopes. Unstable isotopes are radioactive isotopes. 8 Radioactive isotopes will undergo spontaneous decay to emit radioactive rays: alpha, beta and gamma. After radioactive decay, the proton

Uses of Isotopes in Daily Life

Isotopes are atoms of an element with the same number of protons but different numbers of neutrons. Alternatively, isotopes can be defined as atoms of an element with the same proton number but different nucleon numbers.

SPM

’05/P2

Medicine 1 Cobalt-60 is a radioactive isotope of cobalt. It decays by giving out gamma radiation. In radiotherapy, malignant cancer cells are killed by directing a beam of gamma rays towards the cancer cells.

to the thyroid gland. The radiation given out by the radioactive iodide ions will kill the malignant cancer cells without affecting the other parts of the body.

SPM

’09/P1

Radiotherapy is used to kill cancer cells

Patient suffering from thyroid cancer

2 Patients suffering from thyroid cancer are given a drink containing sodium iodide, (NaI) containing radioactive iodide ions. The radioactive iodide ions move preferen­ tially

3 Some medicine, surgical gloves, bandages, plastic hypodermic syringes are sterilised by using gamma radiation. These materials cannot be sterilised by boiling.

Agriculture 1 Using the radioactive carbon-14 (14C) in carbon dioxide, the path of carbon during the photosynthesis process can be determined. The rate of absorption of phosphorus by the plant can be determined by adding radioactive phosphate ions (32PO3– ) to the ground. 4 2 Male pests can be attracted into traps using female hormones (pheromone). The male

pests are then exposed to gamma radiation which can cause genetic mutation to the gametes (sperms). The male pests are then released to be allowed to mate with the females. The offsprings produced will have physical defects such as undeveloped digestive organs and wings. This will terminate the survival of the following generation.

27

The Structure of the Atom

2

number and nucleon number of the isotope may change. 9 There are many uses of radioisotopes, namely, in the field of medicine, agriculture, industry, archaeology, food preservation and electricity generation.

Industry 1 Beta radiation is used to control the thickness of paper, plastic, metals and rubber made in industry. A radioactive source is located at the bottom of the material being produced. A detector is located on top of the material. Any change in the reading of the recorder signifies a change in thickness of the material.

partially filled in which case a higher reading will be recorded.

2

Figure 2.20 Using radiation to detect if a container is fully filled

3 Radioisotopes are used to detect leaks in pipes carrying gas. A radioisotope (for example, sodium-24) is added to the gas so that it will be carried along by the gas flowing through the pipe. A detector is then moved along the external wall of the pipe. The detection of a high radioactive reading will signify the location of the leakage. Figure 2.19 Using radiation to control the thickness of materials

2 Gamma radiation is used to detect whether canned food or bottled drink is completely filled or only partially filled. A radioactive source emitting gamma radiation is directed to the bottled or canned food. More radiation will pass through if the container is only Figure 2.21 Using radiation to detect a leak

Archaeology Carbon-14 is used to determine the age of archaeological artifacts. Plants take in carbon-14 in the form of carbon dioxide (14CO2) during photosynthesis. Carbon-14 is incorpo­ rated into animals or human beings when the plants are eaten. As long as the organism is alive, the amount of carbon-14 in it remains constant. This is because the intake of carbon-14 through food is offset by its spontaneous decay. However, when the organism dies, the intake of carbon-14 is stopped. The amount of C-14 ‘locked’ in the body will continue to decay. The amount of C-14 remaining (measured by its activity) is inversely proportional to the age of the artifacts.

The age of bones dug out from a historical site can be estimated using carbon-14 dating. For very old bones, much of the C-14 would have decayed. The minute amount of C-14 left will show little radio­activity. A recent archaeological sample will have a high reading of C-14.

Source: Jabatan Muzium Malaysia

The Structure of the Atom

SPM

’08/P1

28

Food preservation 1 Food such as vegetables, fruits and meat rot due to the activity of fungus and bacteria. These microorganisms can be kill­ed by irradiating the food with gamma radiation of cobalt-60. The shelf-life of the food can be extended using this method. Irradia­tion is better than chemical preserva­ tives because it does not have adverse effects on health. 2 Irradiation can also slow down budding in potatoes and onions, thus extending their shelf-life. Gamma radiation can also slow the ripening of fruits to be exported.

Radiation can be used to delay rotting of fruits and vegetables

Nuclear energy is an alternative source of energy to replace fossil fuels such as petroleum, natural gas or coal. The nuclear fuel used is uranium-235. The uranium atoms become un­stable when bombarded with fast neutrons.

5

This causes the uranium nuclei to split, producing heat energy. The heat energy released is used to produce steam from water. The steam drives the turbine of the generator, producing electricity.

2.4

’05

Name an isotope and state its purpose for each of the following fields: (a) Medicine (c) Archaeology (b) Industry (d) Food preservation Solution (a) Cobalt-60 (gamma radiation from decay of Co-60 is used to kill cancer cells) (b) Sodium-24 (beta radiation from decay of Na-24 is used to detect leakages in pipes) (c) Carbon-14 (it is used to estimate the age of archaeological artifacts) (d) Cobalt-60 (gamma radiation from decay of Co-60 is used to kill fungus or bacteria that can cause food to rot)

U,

234 92

U and

235 92

The Electronic Structure of an Atom

Electron Arrangement in an Atom 1 Niels Bohr suggested that the electrons in an atom occupy orbits with definite energy levels. Each of these orbits or energy levels can hold a certain number of electrons. The electrons are not static but are moving around. 2 The electron orbits are also known as quantum shells. The shells are labelled first shell, second shell, third shell and so on, away from the nucleus. 3 The first shell is the one nearest to the nucleus and is filled first. It can hold a maximum of two electrons. 4 After the first shell is full, the remaining electrons are filled into the second shell. The second shell can hold a maximum of eight electrons.

2.3 1 Uranium has three isotopes:

2

Generation of electricity

U.

238 92

What do you understand by the term isotopes? State the differences between these isotopes. 2 A list of elements are represented by letters of the alphabet as given below. Choose a pair of isotopes from these elements. Explain your answer.

Figure 2.22 The electron shells of an atom is labelled away from the nucleus

A, 127 B, 79 C, 131 D, 131 E, 55 F 53 34 55 53 25

23 11

5 After the first and second shells are full, the remaining electrons are filled into the third shell. The third shell can take a maximum of eight or 18 electrons. If the number of electrons of an

3 Give an example of a radioactive isotope of carbon. What is meant by radioisotope? Give a use of the isotope given in your example.

29

The Structure of the Atom

atom is more than 20, the third shell will hold 18 electrons. If the number of electrons is 20 or less, the third shell will hold 8 electrons. 6 Table 2.5 shows the maximum number of electrons permitted in each shell. 7 The way in which the electrons are distributed in the shells of an atom is called the electron arrange­­­­­­ment or electronic configuration of the atom. SPM

’11/P2

9 The electrons in the outermost occupied shell are called the valence electrons. Therefore, carbon atom has four valence electrons, chlorine atom has seven valence electrons and calcium atom has two valence electrons. 10 Elements with the same number of valence electrons have the same chemical properties. For example, lithium, sodium and potassium of Group 1 of the Periodic Table have the same chemical properties because each atom has one valence electron.

Table 2.5 Maximum number of electrons permitted in each shell of an atom

Maximum number of electrons

Shell 2

Eight electrons are filled in the second shell. Eight electrons are filled in the third shell. Two electrons are filled in the fourth shell. The electron arrangement of calcium is 2.8.8.2.

First

2

Second

8

Third

8 or 18

Fourth

32

8 The examples below show the electron arrangement of some elements:

1 The carbon atom, 126C has six protons. In a neutral atom, the number of electrons = the number of protons. Hence there are six electrons and are arranged as follows: Two electrons are filled in the first shell. Four electrons are filled in the second shell. The electron arrangement of carbon is 2.4.

Electron arrangement

Number of valence electrons

Lithium Sodium Potassium

2.1 2.8.1 2.8.8.1

1 1 1

11 Group 17 elements have the same chemical properties because each element has seven valence electrons.

2 The chlorine atom, 3517Cl has 17 protons. In a neutral atom, the number of electrons = the number of protons. The 17 electrons are arranged as follows: Two electrons are filled in the first shell. Eight electrons are filled in the second shell. Seven electrons are filled in the third shell. The electron arrangement of chloride is 2.8.7.

Group 17 element

Electron arrangement

Number of valence electrons

Fluorine Chlorine Bromine Iodine

2.7 2.8.7 2.8.18.7 2.8.18.18.7

7 7 7 7

12 The inert or noble gases of Group 18 of the Periodic Table are very stable. They have filled outer shells of electrons.

3

Group 18 element

Electron arrangement

Number of valence electrons

Helium Neon Argon

2 2.8 2.8.8

2 8 8

13 Helium has exactly two electrons in the first shell. It has attained the duplet electron arrange­ment which is stable. Neon and argon each has eight electrons in the outermost shell. It has attained the octet electron arrangement which is stable.

The calcium atom, 40 Ca has 20 protons. 20 In a neutral atom, the number of electrons = the number of protons. The 20 electrons are arranged as follows: Two electrons are filled in the first shell. The Structure of the Atom

Group 1 element

30

14 Table 2.6 shows the diagrammatic electronic structures and the electron arrangements of elements with proton numbers 1 to 20. SPM

’04,05 06/P1 ’08/P1

2

Table 2.6 The diagrammatic electronic structure of elements with proton numbers 1 to 20

6

X, 3517Y, 126Z. Write the electronic configuration of each of these elements. Solution

23 11

’04

The atomic symbol of element X is 199X. Which of the following is true about the subatomic particles of element X?

Proton number

Number Number Electronic of of configuration protons electrons

Proton number

Nucleon number

Electronic configuration

X

11

11

11

2.8.1

A

9

19

2.7

Y

17

17

17

2.8.7

B

9

19

2.8.8.1

Z

6

6

6

2.4

C

19

9

2.7

D

19

9

2.8.8.1

Comment The proton number of X is 9. Hence it has 9 protons and 9 electrons. The 9 electrons are arranged as follows: Two electrons in the first shell and the remaining seven electrons are arranged in the second shell. Its electronic configuration is 2.7. Answer A

7

2.4 1 Write the electron arrangement and draw the atomic structures of carbon and magnesium atoms. [Proton number: C, 6; Mg, 12] 2 The diagram shows the atomic structure of an element X. (a) In an atom of X, how many of the following are there? (i) Valence electrons (ii) Protons (b) What is the nucleon number of X if it has 16 neutrons? (c) Write the atomic symbol of element X.

’04

The chemical symbols of three elements X, Y and Z are shown as follows: 31

The Structure of the Atom

2

2.5

Rutherford discovered the proton in 1911 and James Chadwick discovered the neutron in 1932. Niels Bohr explained the arrangement of the electrons in an atom. 3 We now know that the protons and neutrons are located at the center of the atom called the nucleus. The electrons are arranged in orbits around the nucleus. 4 The atomic structure of an atom can help us understand the chemical properties of the elements better and how they are bonded together to form compounds.

Appreciating the Orderliness and Uniqueness of the Atomic Structure

1 John Dalton proposed the atomic theory about 200 years ago in 1807. Before that scientists thought that atoms were solid particles like marbles. 2 About 100 years later, other scientists discovered the subatomic particles. J.J. Thomson discovered the electron in 1897. Ernest

5 During freezing, the temperature remains constant because heat energy is released and the energy released is equals to the heat lost to the surrounding during cooling. 6 The proton number is the number of protons in the nucleus of an atom. 7 The nucleon number is the total number of protons and neutrons in the nucleus of an atom. 8 Isotopes are atoms of the same element which contain the same number of protons but different numbers of neutrons. 9 The protons and neutrons are enclosed in the nucleus whereas the electrons are arranged in shells surrounding the nucleus. (a) The first shell can hold a maximum of two electrons. (b) The second shell can hold a maximum of eight or 18 electrons. (c) The third shell can hold a maximum of 18 electrons. However for atoms with proton numbers 1 – 20, the atom attains stability when its third shell has eight electrons. (d) The valence electron is the electron in the outermost shell of the atom. For example, the electronic configuration of the calcium atom, 40 Ca is 2.8.8.2. 20 The calcium atom has two valence electrons.

1 There are three states of matter: solid, liquid and gas. 2 When a substance is heated or cooled it will change state. 3 The table shows the energy involved during the change in state: Change of state

Process

Change in energy

Solid to liquid

Melting

Heat energy is absorbed

Liquid to gas

Boiling/ evaporation

Heat energy is absorbed

Solid to gas

Sublimation

Heat energy is absorbed

Liquid to solid

Freezing

Heat energy is released

Gas to liquid

Condensation

Heat energy is released

Gas to solid

Sublimation

Heat energy is released

4 During melting the temperature remains constant because heat energy absorbed is used to overcome the forces of attraction between the molecules.

The Structure of the Atom

32

2 Multiple-choice Questions Matter

1 What process and change in heat energy takes place when iodine ’11 crystals are heated at room temperature and pressure? Process A

Melting

B

Melting

C

Sublimation

D

Sublimation

Heat energy absorbed Heat energy released Heat energy absorbed Heat energy released

3 The diagrams show the spacing of the molecules of a substance at two different temperatures.

at –110 °C

at 85 °C

What is the likely melting point and boiling point of the substance? Melting point Boiling point (°C) (°C) A

–125

90

B

–117

78

C

–102

75

D

–98

105

D The air particles diffuse out of the balloon at a faster rate at higher temperature.

5 Carbon dioxide(CO2), sulphur dioxide(SO2) and nitrogen dioxide(NO2) are three gases that cause acid rain. Which of the following lists the molecules in order of increasing average speed? [Relative atomic mass: C, 12; N, 14; O, 16; S, 32]

Change in heat energy

2 Which statements below are true about a gas? I They move at low speed. II They are easily compressed. III They have a higher rate of diffusion compared to a liquid. IV They spread throughout the vessel in which they are contained. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV



4 An inflated balloon will shrink faster at higher temperature than at lower temperature.

Which of the following is the best explanation for this observation? A The air particles liquefy at lower temperature. B The air particles react to form other compounds at higher temperature. C The air particles come closer together at lower temperature.

Slowest

Fastest

A Sulphur Nitrogen Carbon dioxide dioxide dioxide B Sulphur Carbon dioxide dioxide

Nitrogen dioxide

C Nitrogen Sulphur Carbon dioxide dioxide dioxide D Carbon dioxide

Sulphur Nitrogen dioxide dioxide

6 The table shows the changes in physical states and energies of four substances. Process

Name of process

Change of physical state

Change of energy

I

Freezing

Solid to liquid

Heat is released

II

Melting

Solid to liquid

Heat is absorbed

III

Boiling

Solid to gas

Heat is absorbed

IV

Condensation

Gas to liquid

Heat is released

Which of the following processes above are correct? A I and III only C I, II and IV only B II and IV only D II, III and IV only 7 Which of the following statements is true about pentane molecules when it is cooled to a temperature of –129 °C? [Melting point of pentane is –135 °C and its boiling point is 36 °C]. A The pentane molecules remain static. B The pentane molecules move randomly. C The pentane molecules are arranged closely together. D The distance between the pentane molecules increases.

33

The Structure of the Atom

2

2.1

8 The graph shows the temperature against time of a substance X when it is heated.

Magnesium oxide

’07

Sodium

Ammonia

A

Ions

Atoms

Molecules

B

Ions

Molecules

Molecules

C

Molecules

Molecules

Atoms

D

Ions

Ions

Ions

2.2

The Atomic Structure

11 Which of the following sets is correct? The scientists who discovered the electron, proton and neutron are

2

’08

Which of the statements below are true about X ? I X starts to melt at Q. II The melting point of X is T1 °C. III X exists in gaseous state at region TU. IV At region RS, there is a mixture of solid and liquid X. A I and III only B II and IV only C I, II and III only D I, III and IV only 9 The diagram shows the graph of temperature against time for the heating of substance X. temperature (°C)

65

t1

t2

time (s)

Which statements below are true about substance X? I It is a gas at room temperature. II It undergoes physical change at 65 °C. III It absorbs heat at time intervals t1 and t2. IV It exists as a mixture of liquid and solid at time intervals t1 and t2. A I and II only B I and III only C II and III only D II and IV only 10 State the particles in magnesium oxide, sodium and ammonia. The Structure of the Atom

Electron

Proton

Neutron

A

Ernest Rutherford

J.J. Thomson

James Chadwick

B

J.J. Thomson

Ernest Rutherford

James Chadwick

C

J.J. Thomson

Ernest Rutherford

Niels Bohr

D

J.J. Thomson

James Chadwick

Ernest Rutherford

12 What can be deduced from the symbol 31 P? 15 I Phosphorus atom has five valence electrons. II Phosphorus atom has 15 protons and 31 neutrons. III Phosphorus atom has 16 neutrons. IV Phosphorus atom has proton number of 15 and nucleon number of 31. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV

I It belongs to Group 15 in the Periodic Table. II It belongs to Period 3 of the Periodic Table. III It forms an ion with a charge of –3. IV It is a metal. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 15 The diagrams show three models of the atom.

13 Two particles P and Q have the following compositions: Particle Electron Neutron Proton P

10

10

9

Q

10

12

11

It follows that A P and Q are both negativelycharged B P and Q have the same nucleon number. C P and Q are particles of the same element. D P is negatively-charged and Q is positively-charged. 14 An atom X has an electron arrangement of 2.8.5. Which of the following statements about X are correct?

34

Name the scientists who proposed these models? I

II

III

A

James J.J. Chadwick Thomson

Ernest Rutherford

B

Niels Bohr John Dalton

Ernest Rutherford

C

Niels Bohr J.J. Thomson

Ernest Rutherford

J.J. D Niels Bohr Ernest Rutherford Thomson

17 The atoms 126 C and 115 B have the same A number of protons B number of neutrons C physical properties D chemical properties

2.3

Isotopes and Their Importance

18 Two uranium isotopes are 235 U 92 and 238 U. Which of the following 92 statements below is true? A The 235 U atom has fewer 92 electrons than 238 U atom. 92 B The 235 U atom has 92 92 protons and 235 neutrons. C The 238 U atom has 92 92 protons and 146 neutrons. D The 235 U atom and 238 U atom 92 92 have the same number of neutrons. 19 Isotopes are different atoms with the same number of A protons but different number of neutrons. B electrons but different number of protons. C protons, electrons and neutrons. D protons but different number of electrons and neutrons. 20 Which of the following pairs are correct? Isotope

Use

I Uranium-235 To generate electricity II Iodine-131

To kill cancerous thyroid cells

Isotope

Use

III Carbon-14

To estimate the age of archaeological artifacts

IV Sodium-24 A B C D

To detect leakages in pipes

I, II and III only I, II and IV only II, III and IV only I, II, III and IV

21 Which of the following statements are true about isotopes? I They have the same chemical properties. II They have different physical properties. III The have a different number of neutrons. IV They have the same number of valence electrons. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 22 Oxygen has the isotope 16 O, 17 O and 18 O. Which of the following oxygen gas has the lowest rate of diffusion? A 16 O = 16 O B 17 O = 17 O C 18 O = 18 O D 17 O = 18 O 23

X

Y

Which term describes the particles X and Y shown above? A Isotopes C Anions B Isomers D Cations 24 An element has two isotopes, which are represented by 127 X and 131 X. How does 127 X differ from 131 X ? A It has four less neutrons and three less electrons. B It has four less neutrons. C It has four less protons and three less electrons. D It has four less protons.

35

25 Which of the following comparisons between 79 Br and 35 81 Br are correct? 35 Bromine-79

Bromine-81

I

Has 35 protons

Has 35 protons

II

Has 35 electrons

Has 35 electrons

III

Has 35 neutrons

Has 35 neutrons

IV

Has 44 neutrons

Has 46 neutrons

A B C D

I and IV only II and III only I, II and IV only I, III and IV only

26 The radioisotope that is used to kill cancerous cells is A uranium-235 B cobalt-60 C carbon-14 D phosphorus-32

2.4

The Electronic Structure of an Atom

27 Which of the following particles have eight valence electrons? 40 I 168 W III Y 18 23 + 35 – II 11 X IV Z 17 A I and III only B II and IV only C I, II and III only D II, III and IV only 28 Which of the following particles contains 18 electrons, 19 protons and 20 neutrons? A 39 X + C 39 X– 19 18 B

40 20

D

X 2+

39 19

X

29 How many protons and neutrons are there in one tin atom with nucleon number 119? Protons

Neutrons

A

50

68

B

50

69

C

50

71

D

50

119

The Structure of the Atom

2

16 An atom has the symbol 11 X. Which of the following 5 statements about X are correct? I It has 5 valence electrons in its atom. II It has 6 neutrons in its atom. III It belongs to Group 13 of the Periodic Table. IV It belongs to Period 2 of the Periodic Table. A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV

2

30 The symbol of an element X is 40 X. We can deduce that an atom 18 of element X I has eight valence electrons. II has 22 neutrons in its nucleus. III has three electron shells. IV has a total of 18 electrons in its atom. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 31 Two particles X and Y have the following composition: Particle Electrons Neutrons Protons X

10

12

11

Y

11

12

11

Which of the following statements are true about X and Y? I Both X and Y are negativelycharged. II Both X and Y are positivelycharged. III Both X and Y have the same nucleon number. IV Both X and Y are particles of the same element. A I and III only B II and IV only C III and IV only D II, III and IV only 32 An element 39 X 19 A has one valence electron. B forms a positively-charged ion of charge +2. C is located in Group 17 of the Periodic Table. D has 19 protons and 39 neutrons.

33 Which of the following elements given below have the same number of valence electrons? 19 9

A B C D

V ; 27 W; 35 X; 39 Y; 40 Z 13 17 19 20

Element

W

X

Y

Z

Proton number

6

7

12

15

A B C D

W and Y only X and Z only V and X only V and Y only

34 Which of the following list are the electron arrangements of all nonmetals? A 2.6 2.7 2.8.5 B 2 2.5 2.8.3 C 2.1 2.7 2.8.6 D 2.1 2.8.2 2.8.3 35 An atom 39 Y 19 I has 20 neutrons. II has 19 protons. III has one valence electron. IV has four electron shells. A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV 36 What is the number of subatomic particles in 60 Co2+ ion? 27 Protons Neutrons Electrons A

27

33

27

B

27

33

25

C

33

27

27

D

33

27

25

37 The table shows the proton numbers of four elements. Which of the following pairs of elements has the same number of valence electrons?

Y and Z only X and Z only X and Y only W and Z only

38 The electronic configuration of arsenic is 2.8.18.5. Which of the following statements is true? A Arsenic has three valence electrons. B Arsenic is in Group 14 of the Periodic Table. C The nucleon number of the arsenic atom is 33. D Arsenic is in the same group of the Periodic Table as an element with proton number 7. 39 The electronic configuration of the ion X– is 2.8.18.18.8. The ion X– has 74 neutrons. Determine the nucleon number of element X. A 127 B 128 C 129 D 130 40 The electronic configuration of the strontium ion, Sr2+ is 2.8.18.8. The Sr2+ ion has 49 neutrons. Determine the nucleon number of strontium. A 85 B 86 C 87 D 88

Structured Questions

1 Carbon has two isotopes as shown in Table 1 below. Atom

Proton number

Nucleon number

(b) Draw the atomic structure of represent an electron.

C

12 6

C

14 6

’08

Table 1

(a) (i) Complete Table 1 with the proton numbers and nucleon numbers of the two different carbon isotopes. [2 marks] The Structure of the Atom

(ii) What is the difference between the two isotopes 126C and 146C? [1 mark]

36

(c) Give one use of 146C.

C using, x, to

14 6

[2 marks] [1 mark]

(d) What is the number of valence electrons in both of the carbon atoms above? [1 mark]

(g) (i) Explain the meaning of the term isotope. [2 marks] (ii) State a pair of isotopes from the particles in Table 2. [1 mark]

2 Diagram 1 shows a graph of temperature against time of substance M when it is heated until it boils.

4 Table 3 shows four substances and their respective formulae. ’04

Chemical formula

Bromine

Br2

Iron

Fe

Diagram 1

Naphthalene

C10H8

(a) State the physical state of M at the region (i) PQ (iii) RS (ii) QR (iv) ST [3 marks]

Sodium chloride

NaCl

(b) When does M begin to boil?

[1 mark]

(c) What is the melting point of M?

[1 mark]

2

Substance

Table 3

(a) State two substances that consist of molecules. [1 mark]

(b) Which of the following substances has the highest melting point: bromine, iron or naphthalene?

(d) Explain why the temperature of M remains constant from time t1 to t2. [1 mark]

[1 mark]

(e) Sketch the graph obtained when molten M is cooled from 450 °C to room temperature. [2 marks]

(c) (i) State the substance that can conduct electricity in the solid state. [1 mark] (ii) Draw the arrangement of the particles of this substance. [1 mark]

3 Table 2 shows the proton numbers and nucleon numbers of five particles represented by the letters V, W, X, Y and Z. Particle

Proton number

Nucleon number

V

6

12

W

8

16

X

8

18

Y

11

23

Z

16

32

Electron arrangement

(d) Name the particles present in sodium chloride. [1 mark]

(e) Diagram 2 shows the graph of temperature against time obtained when solid naphthalene is heated.

Table 2

(a) Write the electron arrangements of all the particles in Table 2. [2 marks] (b) What is the number of valence electrons in particle V ? [1 mark]

Diagram 2

(c) Draw the atomic structure of particle Y. [2 marks] (d) State the number of electron shells in particle Z. [1 mark] (e) Explain the meaning of nucleon number. [1 mark] (f) What is the number of neutrons in particle Y ? [1 mark]



(i) State the melting point of naphthalene. [1 mark] (ii) Explain why there is no change in temperature from Q to R. [2 marks] (iii) State how the movement of naphthalene particles changes between R and S during heating. [1 mark]

Essay Questions You are given two substances X and Y. They are either naphthol or naphthalene. You are required to carry out an experiment to identify X and Y. Design an experiment to determine X and Y.

1 (a) Compare the three physical states of matter in terms of particle arrangements, forces of attraction between the particles, kinetic energy of the particles and compressibility. [8 marks] (b) Table 1 shows the melting points of naphthol and naphthalene.

37

The Structure of the Atom

Chemical Naphthol

65

Naphthalene

80



2 (a) Define the following terms: (i) Proton number (ii) Nucleon number (iii) Valence electron

Melting point (°C)

Table 1

[3 marks]

(b) (i) What are isotopes? [3 marks] (ii) Give an example of a pair of isotopes. [2 marks] (iii) Discuss six uses of isotopes. [12 marks]

[12 marks]

2

Experiments 1 An experiment is carried out to determine the melting point of naphthalene. Solid naphthalene is heated and its temperature is recorded every 30 seconds. ’05 Diagram 1 shows the recorded temperature readings at 30-second intervals.

Diagram 1

(a) Record the temperatures in the spaces provided in Diagram 1.

[3 marks]

(b) Draw a labelled diagram of the apparatus used to carry out the experiment.

[3 marks]

(c) Plot a graph of temperature against time for the heating of naphthalene.

[3 marks]

(d) State the melting point of naphthalene.

[3 marks]

(e) What is the physical state of naphthalene at time 90 seconds?

[3 marks]

(f) Explain why the temperature between time 60 s to 120 s remained constant.

[3 marks]

(g) Sketch a graph you expect to obtain if the molten naphthalene is cooled to room temperature.

[3 marks]

2 ‘The melting point of a substance is lowered by the presence of impurities’. Using naphthalene and a mixture of naphthalene with some acetamide, describe an experiment to prove the statement above. Your answer should include the following items: (a) Aim of experiment

[3 marks]

(b) All variables involved

[3 marks]

(c) List of apparatus and materials used

[3 marks]

(d) Procedure of experiment

[3 marks]

(e) Tabulation of results

[3 marks]

The Structure of the Atom

38

FORM 4 THEME: Matter Around Us

CHAPTER

3

Chemical Formulae and Equations SPM Topical Analysis 2008

Year 1

Paper

2009 3

2

Section

A

B

C

Number of questions

1 — 2

–

–

4

–

1

9

2010

2

3

A

B

C

1

–

–

–

1

2011

2

4

3

A

B

C

–

–

–

1

–

4

2

3

A

B

C

–

–

–

–

ONCEPT MAP FORMULAE AND CHEMICAL EQUATIONS

Atom

Molecule

Relative atomic mass

Relative molecular mass

Relative atomic mass, Ar in gram

Relative molecular mass, Mr in gram

Molar mass

Mass of matter

3 Molar mass

4 Molar mass

4 Molar volume Volume of gas

3 Avogadro constant Number of moles

3 Molar volume

Number of particles 4 Avogadro constant

Reactants

Chemical equation

Products of reaction

Empirical formula

Chemical formula

Molecular formula

3.1

Relative Atomic Mass and Relative Molecular Mass

1 It is impossible to weigh an atom in gram. So chemists compared how heavy one atom is to another atom which is taken as the standard. The comparison of the mass of an atom to another is called the relative atomic mass (r.a.m.).

Relative Mass of one atom of the element atomic mass = — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — 1 of an element —  mass of one carbon-12 atom 12

3

For example, a sodium atom, Na is 23 times heavier than one-twelfth of the mass of one carbon-12 atom. Thus the relative atomic mass of Na is 23. Mass of one Na atom — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — = 23 1 —  mass of one carbon-12 atom 12 (Note: the mass of one carbon-12 atom is 12 units) 5 A molecule is a small group of atoms joined together. The simplest being diatomic molecules like O2, N2 and Cl2. Examples of triatomic molecules are CO2 and H2O. Some examples of larger molecules are ammonia (NH3), methane (CH4), sulphur (S8), phosphorus (P4) and ethanol (C2H5OH). 6 The relative molecular mass (Mr) of a compound is defined as the number of times one molecule of the compound is heavier than one-twelfth of the mass of a carbon-12 atom.

The relative atomic mass of atoms or relative molecular mass of molecules can be determined using the mass spectrometer with carbon-12 as the standard

2 In 1961, scientists agreed to use carbon-12 as the standard. The mass of a carbon-12 atom is assigned a value of exactly 12 units. 3 Carbon is chosen as the standard because (a) the abundance of carbon-12 isotope is almost 99%. Carbon-13 and carbon-14 isotopes make up about only 1%. Thus the mass of a carbon atom using carbon-12 isotope or using the average mass of the three isotopes of carbon is still 12.00 units. (b) carbon is a solid at room temperature. Unlike hydrogen and oxygen which are gases, it does not require a container with a lid to contain it. (c) carbon is present in many organic substances, namely, wood, natural gas and petroleum. Thus carbon is easily available. Carbon can be obtained by burning these organic substances in a limited supply of oxygen. 4 The relative atomic mass (Ar) of an element SPM is defined as the number of times one atom ’11/P1 of the element is heavier than one-twelfth of the mass of a carbon-12 atom, that is:

Chemical Formulae and Equations

Relative Mass of one molecule molecular of the compound — — — — — — — — — — — — — — — — — — — — — — — — mass of a = — 1 compound —­  mass of one 12 carbon-12 atom For example, a molecule of methane, CH4, is 16 times heavier than one-twelfth of the mass of a carbon-12 atom. Thus the relative molecular mass of CH4 is 16.

A student need not memorise the relative atomic mass of elements. They will be given in the examination. However, one must know how to calculate the relative molecular mass of compounds from the Ar given. To determine the Mr of a molecule, we sum up the relative atomic mass of every atom present in the molecule.

40

Concept of relative atomic mass and relative molecular mass using analogy Apparatus

Twin-pan balance

3

Ball bearings, iron nails, screws and nuts.

Figure 3.1(a)

Figure 3.1(b)

1 An iron nail is put into the pan on the right of the balance (Figure 3.1(a)). 2 Ball bearings are added to the pan on the left of the balance until it is balanced. 3 The number of ball bearings needed to balance the small nail is counted. 4 The procedure is repeated for a screw, a nut, a screw and a nut (Figure 3.1(b)) and a screw and two nuts. For each sample the number of ball bearings required to balance the object is counted and recorded in the table as follows:

OR Mass of one iron nail ——————————————————— = 5 Mass of one ball bearing Hence, relative mass of one iron nail = 5 2 If we assume that a ball bearing represents onetwelfth of the mass of carbon-12 atom and the nail, screw and nut represent the atoms of other elements, then the relative atomic mass of these elements will be 5, 12 and 7 respectively. 3 If we assume that the screw and the nut form a molecule, relative molecular mass of the molecule = relative atomic mass of a screw + relative atomic mass of a nut = 12 + 7 = 19 4 A total of 26 ball bearings are required to balance the mass of one screw and two nuts. Hence, Mr of a screw and two nuts = r.a.m. of a screw + r.a.m. of two nuts = 12 + 7 + 7 = 26 Therefore to determine the relative molecular mass of a molecule, we sum up the relative atomic mass of each atom present in the molecule. 5 In this experiment, we do not need to know the actual mass of a ball bearing to determine the relative mass of other objects. Similarly, we do not need the actual mass of a carbon atom to determine the relative atomic mass of an element or the relative molecular mass of a molecule.

Number of Relative mass ball bearings of object needed to (compared to balance object a ball bearing)

Iron nail

5

5

Screw

12

12

Nut

7

7

Screw + nut

19

19

Screw + 2 nuts

26

26

(Assuming the relative mass of a ball bearing is 1 unit) 1 If we assume that the ball bearing has a mass of 1 unit, then the iron nail which is equivalent to five ball bearings has a relative mass of five units.

41

Chemical Formulae and Equations

Activity 3.1

Object

Mass of 1 iron nail = mass of five ball bearings

Solution 24 (a) Z is heavier than Y by — — times = 1.5 times. 16 (b) Assume that n atoms of X has the same mass as the sum of 3 atoms of Y and 2 atoms of Z. 12n = 3(16) + 2(24) 12n = 96 96 n = — — 12 =8

1

3

Iridium is a very dense metal and was discovered in 1804 by Smithson Tennant. Determine how many carbon atoms will have the same mass as one iridium atom. [Relative atomic mass: C, 12; Ir, 192] Solution Assuming n carbon atoms has the same mass as one iridium atom. 12n = 192 192 n = — — — 12 = 16

2

5 Adrenaline is produced by the adrenal gland. Adrenaline has the formula C9H13NOx. If its r.m.m. is 183, determine the value of x. Then write the molecular formula of adrenaline. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16]

SPM

’08/P1

Three cobalt atoms have the same mass as fifteen carbon atoms. Determine the relative atomic mass of cobalt. [Relative atomic mass: C, 12]

Solution Relative molecular C9H13NOx= 183

Solution Assume that the relative atomic mass of Co = a

mass

of

adrenaline

is

9(12) + 13(1) + 14 + x(16) = 183 108 + 13 + 14 + 16x = 183 16x = 183 – 135 48 x =— — 16 x = 3

Co + Co + Co = 15 C 3a = 15  12 15  12 a = — — — — — — — 3 = 60

The formula of adrenaline is C9H13NO3.

3 1

The mass of a rutherfordium (Rf) atom is equal to the sum of three sodium atoms and six sulphur atoms. What is the relative atomic mass of rutherfordium? [Relative atomic mass: Na, 23; S, 32]

The relative formula mass of X3(PO4)2 is 310. Determine the relative atomic mass of element X. [Relative atomic mass: O, 16; P, 31]

Solution Relative atomic mass of Rf = 3(23) + 6(32) = 69 + 192 = 261

Solution Assume that the relative atomic mass of the element X is p. Relative formula mass of X3(PO4)2 = 310 3(p) + 2[31 + 4(16)] = 310 3p + 2[31 + 64] = 310 3p + 190 = 310 3p = 310 – 190 120 p =— — 3 = 40

4 The relative atomic mass of elements X, Y and Z are 12, 16 and 24 respectively. (a) How much is an atom of Z heavier than an atom of Y? (b) How many atoms of X will have the same mass as the sum of 3 atoms of Y and 2 atoms of Z? Chemical Formulae and Equations

’09

42

1 (a) A platinum atom is five times heavier than a potassium atom. What is the relative atomic mass of platinum? [Relative atomic mass: K, 39] (b) Calculate the number of carbon atoms that has the same mass as one molybdenum atom. [Relative atomic mass: C, 12; Mo, 96] (c) Five aluminium atoms have the same mass as the sum of six lithium atoms and three phosphorus atoms. Determine the r.a.m. of phosphorus. [Relative atomic mass: Li, 7; Al, 27] (d) The relative atomic mass of elements W, X, Y and Z are 7, 39, 56 and 195 respectively. (i) One atom of thorium (Th) has the same mass as the sum of six W atoms, two X atoms and two Y atoms. What is the r.a.m. of thorium? (ii) How many W atoms will have the same mass as the sum of two X atoms, one Y atom and one Z atom? 2 Determine the relative molecular mass (or relative formula mass) of the following compounds: (a) Sodium stearate, C17H35COONa (Soap molecule) (b) Complex ion Cu(NH3)4SO4 [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Na, 23; S, 32; Cu, 64] 3 Borax is a compound used to kill cockroaches. Its molecular formula is X2B4O7. If the relative molecular mass of borax is 202, determine the relative atomic mass of the element X. Identify the element X from the list of elements given below. [Relative atomic mass: B, 11; C, 12; O, 16; F, 19; Na, 23; Mg, 24]

3.2

Amedeo Avogadro Amedeo Avogadro was a professor of physics at the University of Turin, Italy. In 1811, he proposed the hypothesis which states that under the same temperature and pressure, equal volumes of different gases contain equal numbers of molecules. He showed that 22.4 dm3 of any gas at a temperature of 0 °C and a pressure of 1 atmosphere contains 6.02  1023 molecules. Therefore the value of 6.02  1023 is called Avogadro’s number or the Avogadro constant in honour of him.

Relationship between the Number of Moles and the Number of Particles

Definition of the Mole

Concept of the Mole

SPM

’08/P1

1 One mole is the amount of substance which contains the same number of particles as there are in 12 grams of carbon-12. 2 The number of atoms in 12 grams of carbon-12 is 6.02  1023. 3 This number, 6.02  1023, is called Avogadro’s number or the Avogadro constant (NA). 4 The particles in matter can be atoms, ions or molecules. 5 For elements, the particles are atoms. For example, 1 mol of gold contains 6.02  1023 gold atoms.

1 The relative atomic mass of carbon atom is 12 and the relative atomic mass of helium atom is 4. This means that a carbon atom is three times heavier than a helium atom. 2 Thus, a sample containing 12 grams of carbon and four grams of helium will contain the same number of atoms, that is, number of atoms in 12 grams of carbon = number of atoms in 4 grams of helium

43

Chemical Formulae and Equations

3

3 Let us extend the reasoning to other elements. The number of atoms in a sample of any element with its relative atomic mass in grams is equal to the number of atoms in 12 g of carbon-12. For example, 1 g of hydrogen, 14 g of nitrogen, 23 g of sodium, 56 g of iron will all contain the same number of atoms as in 12 g of carbon-12. 4 Now the question that arises is: how many atoms are there in 4 g of helium, 1 g of hydrogen, 14 g of nitrogen, 23 g of sodium, 56 g of iron and 12 g of carbon-12? Through several experiments scientists have found that this number is 602 000 000 000 000 000 000 000 or 6.02  1023. 5 In Chemistry, the number 6.02  1023 is called one mole (or mol in short).

3.1

3

Conversion of the Number of Moles to the Number of Particles

6 For ionic compounds, the particles are ions. For example, 1 mol of magnesium ions contains 6.02  1023 Mg2+ ions. 1 mol of potassium iodide, KI contains 6.02  1023 K+ ions and 6.02  1023 I– ions. 1 mol of magnesium chloride, MgCl2 contains 6.02  1023 Mg2+ ions and 2  6.02  1023 Cl– ions. 7 For covalent compounds, the particles are molecules. For example, 1 mol of carbon dioxide contains 6.02  1023 CO2 molecules.

6

Calculate the number of particles in: (a) 0.75 mol of aluminium atoms, A1, (b) 1.2 mol of chloride ions, Cl–, (c) 0.07 mol of carbon dioxide molecules, CO2. [Assume NA = 6  1023 mol–1]

Conversion of the Number of Moles to the Number of Particles and Vice Versa 1 Since one mole of any substance contains 6.02  1023 particles, n moles of the substance will contain n  6.02  1023 particles. 2 Hence,



number of particles = number of mole 3 NA (where NA = 6.02 3 1023)

3 If 6.02  1023 particles are found in 1 mol, then one particle is found in 1 particle — — — — — — — — — — — — — — — — — —  1 mol 6.021023 particles 1 =— — — — — — ­­— — — — mol 6.021023

Solution (a) 0.2 mol of SO2 contains 0.2  6  1023 molecules = 1.2  1023 molecules. 1 sulphur dioxide molecule (SO2) has 3 atoms (one sulphur and two oxygen atoms). Therefore the number of atoms = 3  1.2  1023 = 3.6  1023 atoms

5 Generally,  NA mol

(b) 0.125 mol of CH4 contains 0.125  6  1023 molecules = 7.5  1022 molecules. 1 methane molecule (CH4) has 5 atoms (one carbon and four hydrogen atoms). Therefore the number of atoms = 5  7.5  1022 = 3.75  1023 atoms

A student need not memorise that the Avogadro constant is 6.02  1023. It will be given in the examination. However in most cases, the value of NA given is 6  1023 for easy calculation.

Chemical Formulae and Equations

SPM

’11/P1

Calculate the number of atoms in: (a) 0.2 mol of sulphur dioxide gas, SO2, (b) 0.125 mol of methane gas, CH4. [NA = 6  1023 mol–1]

number of moles = number of particles  NA (where NA = 6.02  1023)

 NA

Solution (a) 1 mol of aluminium contains 6  1023 Al atoms. 0.75 mol of aluminium contains Make sure that the 0.75 mol ———————  6  1023 Al atoms numerator and the 1 mol denominator have the same unit. = 4.5  1023 Al atoms (b) 1 mol of chloride ions contains 6  1023 Cl– ions. 1.2 mol of chloride ions contain 1.2 mol — — — — — —  6  1023 Cl– ions. 1 mol = 7.2  1023 Cl– ions (c) 1 mol of carbon dioxide contains 6  1023 CO2 molecules. 0.07 mol of carbon dioxide contains 0.07 mol — — — — — — —  6  1023 CO2 molecules 1 mol = 4.2  1022 CO2 molecules

7

Therefore, x particles are found in x ————————— mol 6.02  1023 4 Thus, the

number of particles

SPM

’09/P1

44

Conversion of the Number of Particles to the Number of Moles

3 The mole-atom is the relative atomic mass of SPM an atom expressed in gram. ’10/P1 [Relative atomic mass: C, 12; Al, 27; S, 32] 1 = 1 = 1 =

Calculate the number of moles of the following substances: (a) 6  1021 iron atoms, (b) 7.5  1023 carbon monoxide molecules. [NA = 6  1023 mol–1]

mole-atom of carbon 12 g mole-atom of aluminium 27 g mole-atom of sulphur 32 g

Each sample contains 6.02  1023 atoms

4 The mole-molecule is the relative molecular mass of a compound expressed in gram. [Relative molecular mass: H2O, 18; CO2, 44; NH3, 17; C2H5OH, 46; CH4, 16]

Solution (a) 1 mol of iron contains 6  1023 atoms. Therefore 6  1021 iron atoms 6  1021 atom =— — — — — — — — — — —  1 mol 6  1023 atom = 0.01 mol (b) 1 mol contains 6  1023 molecules. Therefore 7.5  1023 CO molecules contain 7.5  1023 molecules — — — — — — — — — — — — — — — — —  1 mol 6  1023 molecules = 1.25 mol

1 mole-molecule of water, H2O = 18 g 1 mole-molecule of carbon dioxide, CO2 = 44 g 1 mole-molecule of ethanol, C2H5OH = 46 g

Each sample contains 6.02  1023 molecules

Conversion of the Number of Moles of a Substance to Its Mass

3.2

1 Since 1 mol of an element is the relative atomic mass in gram, x mol of the element has x  relative atomic mass in gram.

1 Calculate the number of particles in 1 (a) — mol of copper, 6 (b) 0.0625 mol of water molecule, H2O, (c) 1.3 mol of sodium ions, Na+. [NA = 6  1023 mol–1]



2 Calculate the number of atoms in (a) 0.012 mol of ethane gas, C2H6, (b) 1.1 mol of sulphur trioxide, SO3. [NA = 6  1023 mol–1]

Number of mole-atom mass in gram =— — — — — — — — — — — — — — — — — — — — — relative atomic mass

2 Similarly, since 1 mol of a compound is the relative molecular mass in gram, x mol of the compound has x  relative molecular mass in gram.

3 Calculate the number of moles of the following substances: (a) 6  1022 sodium ions, (b) 1.8  1024 H2S molecules. [NA = 6  1023 mol–1]

3.3

SPM

’07/P2

Number of mole-molecule mass in gram =— — — — — — — — — — — — — — — — — — — — — — — — — relative molecular mass

Relationship between the Number of Moles of a Substance and Its Mass

3 Generally, mole

1 The mass of a substance that contains one mole of the substance is called the molar mass. 2 One mole of substance contains 6.02  1023 particles. Therefore the molar mass of any substance contains 6.02  1023 particles.

 Ar or Mr 4 Ar or Mr

mass in gram

number of moles

45

Chemical Formulae and Equations

3

8

Conversion of the Number of Moles of a Substance to Its Mass

Conversion of the Number of Particles of a Substance to Its Mass and Vice Versa 1 Two steps are involved in the conversion of the mass of substance to the number of particles. Step 1: Mass in gram is converted to number of moles by dividing the mass by the relative atomic mass or relative molecular mass. Step 2: Number of moles is converted to number of particles by multiplying the number of moles by the Avogadro constant. 2 Two steps are involved in the conversion of the number of particles to mass. Step 1: Number of particles is converted to the number of moles by dividing the number of particles by the Avogadro constant. Step 2: Number of moles is converted to mass in gram by multiplying the number of moles by the relative atomic mass or relative molecular mass. 3 In general,  Ar  NA or Mr mass in number of mole particles gram  NA  Ar or Mr

3

9 Determine the mass for each of the following substances: 2 (a) — mol of aluminium atoms, 3 (b) 0.08 mol of ascorbic acid, C6H8O6, (c) 0.125 mol of magnesium hydroxide, Mg(OH)2. [Relative atomic mass: H, 1; C, 12; O, 16; Mg, 24; Al, 27; Cl, 35.5] Solution (a) 1 mol of Al = 27 g 2 2 — mol of Al = — 27 g = 18 g 3 3 (b) 1 mol of C6H8O6 = 6(12) + 8(1) + 6(16) g = 176 g 0.08 mol C6H8O6 = 0.08  176 g = 14.08 g (c) 1 mol of Mg(OH)2 = 24 + 2(16 + 1) g = 58 g 0.125 mol of Mg(OH)2 = 0.125  58 g = 7.25 g

Conversion of the Mass of a Substance to the Number of Moles

10

Conversion of the Mass of a Substance to the Number of Particles

SPM

’08/P1

11

Calculate the number of moles of the following substances: (a) 23.5 g of copper(II) nitrate, Cu(NO3)2, (b) 0.97 g of caffeine, C8H10N4O2 (a stimulant). [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Si, 28; S, 32; Cu, 64]

Calculate the number of particles in: (a) 12.8 g of copper, (b) 8.5 g of ammonia, NH3. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Fe, 56; Cu, 64; NA = 6  1023 mol–1] Divide mass

Solution (a) 1 mol of Cu(NO3)2 = 64 + 2[14 + 3(16)] g = 188 g 23.5 23.5 g of Cu(NO3)2 = — — — —  1 mol 188 = 0.125 mol (b) 1 mol of C8H10N4O2 = 8(12) + 10 + 4(14) + 2(16) g = 194 g

in gram by the Solution relative atomic 12.8 to find the (a) 12.8 g of Cu = — — — mol = 0.2 mol mass number of moles 64 1 mol contains 6  1023 atoms. 0.2 mol contains 0.2  6  1023 atoms = 1.2  1023 atoms Number of moles is converted to (b) 1 mol of NH3 = 17 g number of particles by multiplying the 8.5 8.5 g of NH3 = — — mol = 0.5 mol number of moles 17 by the Avogadro 1 mol contains 6  1023 molecules. constant 0.5 mol contains 0.5  6  1023 molecules = 3  1023 molecules

0.97 g of C8H10N4 O2 0.97 = ———  1 mol 194 = 0.005 mol Chemical Formulae and Equations

46

Conversion of the Number of Particles of a Substance to Its Mass

3.3

12 Calculate the mass of the following substances: (a) 1.2  1022 zinc atoms, (b) 3  1023 ethanol (C2H5OH) molecules. [Relative atomic mass: H, 1; C, 12; O, 16; Zn, 65; NA = 6  1023 mol–1] Solution (a) 1 mol contains 6  1023 atoms. 1.2  1022 atoms Number of particles is converted to the number of 1.2  1022 moles by dividing the number =— — — — — — — — — mol 23 6  10 of particles by the Avogadro constant = 0.02 mol 1 mol of Zn = 65 g 0.02 mol of Zn = 0.02  65 g = 1.3 g Number of moles is (b) 1 mol contains converted to mass in 23 6  10 molecules. gram by multiplying 3  1023 C2H5OH molecules the number of moles by the relative atomic 3  1023 mass is contained in — — — — — — mol 6  1023 = 0.5 mol 1 mol of C2H5OH = 46 g 0.5 mol of C2H5OH = 0.5  46 g = 23 g

2

2 Calculate the number of moles of the following substances: (a) 2.8 g of iron, (b) 4.05 g of nicotine, C10H14N2 (an addictive substance in cigarette), (c) 1.49 g of ammonium phosphate, (NH4)3PO4 (a fertiliser), (d) 2.3 g of ethanol, C2H5OH. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; P, 31; Fe, 56] 3 Calculate the mass of the following substances: (a) 3  1023 titanium atoms, (b) 1.2  1024 argon atoms, (c) 7.5  1022 citric acid (C12H16O14) molecules. [Relative atomic mass: H, 1; C, 12; O, 16; Ar, 40; Ti, 48; NA = 6  1023 mol–1] 4 Calculate the number of particles in the following substances: (a) 4 g of sulphur, (b) 2.24 g of cadmium, (c) 36 g of glucose, C6H12O6. [Relative atomic mass: H, 1; C, 12; O, 16; S, 32; Cd, 112; NA = 6  1023 mol–1]

’04

The relative atomic mass of X and Y is 64 and 16 respectively. Which of the following is about the atoms of X and Y? The mass of one atom of Y is 16 g. The number of protons in X is 64. 4 mol of Y have the same mass as 1 mol of X. The density of one atom of X is 4 times that of an atom of Y.

5 Geranial is a compound found in lemon grass (daun serai). Its molecular structure is shown below. CH3 H H H CH3 H H ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ C == C –– C –– C –– C == C –– C == O ⎮ ⎮ CH3 H H

Solution 16 The mass of one atom of Y is — — — — — — — g. 6  1023 (A is incorrect) The nucleon number of X is 64. (B is incorrect) Mass of 4 mol of Y = 4  16 g = 64 g (C is correct) Mass of 1 mol of X is 64 g.

(a) Determine the mass of 1 mol of geranial. (b) Determine the mass of 0.02 mol of geranial. (c) Determine the number of molecules present in 7.6 g of geranial. (d) Determine the mass of 7.5  1022 geranial molecules. [Relative atomic mass: H, 1; C, 12; O, 16; NA = 6  1023 mol–1]

Densities cannot be determined because the volume of the atom is not given. (D is incorrect)

47

Chemical Formulae and Equations

3

1 Calculate the mass of each of the following substances: (a) 1.25 mol of helium gas, 2 (b) — mol of cobalt, Co, 5 (c) 0.15 mol of hydrated copper(II) sulphate, CuSO4.5H2O, (d) 0.05 mol of potassium manganate(VII), KMnO4. [Relative atomic mass: H, 1; He, 4; O, 16; S, 32; K, 39; Mn, 55; Co, 59; Cu, 64]

3.4

Conversion of the Number of Moles of a Gas to Its Volume and Vice Versa

Relationship between the Number of Moles of a Gas and Its Volume

1 Since 1 mol of any gas occupies 22.4 dm3 at s.t.p. (or 24 dm3 at r.t.p.), n mol of the gas will occupy n  22.4 dm3 at s.t.p. (or n  24 dm3 at r.t.p.) 2 Hence, volume of gas = number of moles of gas  molar volume where the molar volume is 22.4 dm3 at s.t.p. or 24 dm3 at room temperature. 3 If a volume of 22.4 dm3 (or 22 400 cm3) is occupied by 1 mol of gas, 1 1 cm3 of gas is occupied by — — — — — — ‑ mol. 22 400 4 Conversely, volume of gas number of moles of gas = — — — — — — — — — — — — — — — — — — — ‑ molar volume

3

Conversion of the Number of Particles of a Substance to Its Volume 1 The volume occupied by a gas depends on the temperature. As the temperature increases, the gas expands and occupies a larger volume. If the gas is cooled, the gas contracts and occupies a smaller volume. 2 The volume occupied by a gas also depends on the pressure. If a gas is compressed (with increased pressure), the volume of the gas decreases. If the pressure is decreased, the volume of the gas increases. 3 At the same temperature and pressure, equal volumes of all gases contain the same number of particles. Accordingly, one mole of any gas (which contains 6.02  1023 particles) will occupy the same volume at a particular temperature and pressure. 4 It is found that one mole of any gas at room temperature (25°C) and pressure of 1 atmosphere occupies a volume of 24 dm3 (or 24 000 cm3). 5 At standard temperature and pressure (s.t.p.), that is, at a temperature of 0 °C and pressure of 1 atmosphere, one mole of any gas occupies a volume of 22.4 dm3 (or 22 400 cm3). 6 The volume occupied by one mole of any gas SPM is called the molar volume. ’11/P1 Example • 1 mol of oxygen gas, O2(32 g) • 1 mol of carbon dioxide gas, CO2(44 g)

A student need not memorise that molar volume is 22.4 dm3 at s.t.p. or 24 dm3 at r.t.p. It will be given in the examination.

Conversion of the Number of Moles to Volume of Gas

13 Calculate the volume of 0.75 mol of nitrogen gas at s.t.p. [1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.] Solution 1 mol of gas occupies a volume 22.4 dm3 at s.t.p. 0.75 mol of N2 gas occupies Make sure that the numerator and 0.75 mol 3 3 — — — — — — — —  22.4 dm = 16.8 dm denominator have the 1 mol same unit

occupies 24 dm3 at room temperature or occupies 22.4 dm3 at s.t.p.

[Relative molecular mass: O2, 32; CO2, 44]

Conversion of Volume of Gas to Number of Moles

7 One mole of gas contains 6.02  10 molecules and therefore at s.t.p.

23

22.4 dm3 of oxygen gas, O2(32 g) 22.4 dm3 of carbon dioxide gas, CO2(44 g)

Chemical Formulae and Equations

SPM

’04/P2

14 Calculate the number of moles of the following gases at room temperature and pressure. (a) 4.8 dm3 of chlorine gas, (b) 1200 cm3 of methane gas. [1 mol of gas occupies a volume of 24 dm3 at room temperature]

each contains 6.02  1023 molecules

48





Solution (a) 1 mol of gas occupies a volume of 24 dm3 at room temperature. 4.8 dm3 4.8 dm3 Cl2 gas contain — — — — — —  1 mol 24 dm3 = 0.2 mol (b) 1 mol of gas occupies a volume of 24 000 cm3 at room temperature. 1200 cm3 of CH4 gas contain 1200 cm3 — — — — — — — — —  1 mol 24 000 cm3 = 0.05 mol

Solution 1 mol of NH3 gas = 17 g 3.4 g 3.4 g of NH3 = — — — — 1 mol = 0.2 mol 17 g 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p. 0.2 mol of ammonia gas occupies a volume of 0.2  22.4 dm3 = 4.48 dm3. Conversion of Volume of Gas to Mass

Conversion of the Volume of Gases to Mass and Vice Versa

SPM

Calculate the mass of the following gases at room temperature and pressure: (a) 7.2 dm3 of sulphur dioxide gas, SO2, (b) 600 cm3 of methane gas, CH4. [Relative atomic mass: H, 1; C, 12; O, 16; S, 32; 1 mol of gas occupies a volume of 24 dm3 at room temperature]

’06/P2 ’08/P2

1 Two steps are involved in the conversion of volume of gas to mass. Step 1: Volume of gas is converted to number of moles (by dividing the volume of gas by the molar volume). Step 2: Number of moles is converted to mass in gram (by multiplying the number of moles by the relative atomic mass or relative molecular mass of the gas). 2 Two steps are involved in the conversion of mass to volume of gas. Step 1: Mass in gram is converted to number of moles (by dividing the mass by the relative atomic mass or relative molecular mass). Step 2: Number of moles is converted to volume of gas (by multiplying the number of moles by the molar volume). 3 In general:  22.4 dm3

volume  22.4 dm3 (at s.t.p.)

number of moles

 Ar or Mr

3

16

Solution (a) 1 mol of gas occupies a volume of 24 dm3 at room temperature. 7.2 dm3 7.2 dm3 of SO2 = — — — — — — — — — — 1 mol = 0.3 mol 24 dm3 1 mol of SO2 = (32 + 32) g = 64 g 0.3 mol of SO2 = 0.3  64 g = 19.2 g (b) 1 mol of gas occupies a volume of 24 000 cm3 at room temperature. 600 cm3 600 cm3 of CH4 gas = — — — — — — — — —  1 mol 24 000 cm3 = 0.025 mol 1 mol of CH4 = 16 g 0.025 mol of CH4 = 0.025  16 g = 0.4 g Conversion of the Volume of Gases to the Number of Particles and Vice Versa

mass in gram

 Ar or Mr

SPM

’04,06 /P2

1 Two steps are involved in the conversion of the volume of gas to the number of particles. Step 1: Volume of gas is converted to number of moles (by dividing the volume of gas by the molar volume). Step 2: Number of moles is converted to number of particles (by multiplying the number of moles by the Avogadro constant). 2 Two steps are involved in the conversion of the number of particles to volume of gas.

Conversion of Mass to Volume of Gas

15 Calculate the volume occupied by 3.4 g of ammonia gas, NH3 at standard temperature and pressure. [Relative atomic mass: H, 1; N, 14; 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.] 49

Chemical Formulae and Equations

Solution 1 mol contains 6  1023 molecules. 1.5  1023 molecules 1.5  1023 molecules =— — — — — — — — — — — — — — — — — —  1 mol = 0.25 mol 6  1023 molecules

Step 1: Number of particles is converted to number of moles (by dividing the number of particles by the Avogadro constant). Step 2: Number of moles is converted to volume of gas (by multiplying the number of moles by the molar volume). 3 In general: volume of gas

 22.4 dm3

number of moles

3

 22.4 dm3 (at s.t.p.)

 NA  NA

1 mol of gas occupies a volume of 24 dm3 at room temperature. 0.25 mol of gas occupies a volume of 0.25  24 dm3 = 6 dm3

number of particles

volume

Conversion of Volume of Gas to Number of Particles

 Mr  Mr

mass in gram

 (61023)

number of particles

3

’03

Which of the following gases contain 6  1022 molecules? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Avogadro constant = 6  1023 mol–1] I 1.0 g of hydrogen gas II 2.8 g of nitrogen gas III 4.4 g of carbon dioxide IV 1.8 g of water vapour I, II and III only II, III and IV only I, III and IV only I, II, III and IV

Solution (a) 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p. 0.28 dm3 0.28 dm3 of nitrogen gas = — — — — — — —  1 mol 22.4 dm3 = 0.0125 mol 1 mol of N2 contains 6  1023 molecules. 0.0125 mol of N2 contains 0.0125  6  1023 molecules = 7.5  1021 molecules (b) 1 mol of gas occupies a volume of 22 400 cm3 at s.t.p. 448 cm3 448 cm3 of CO gas = — — — — — — — — —  1 mol 22 400 cm3 = 0.02 mol 1 mol of CO contains 6  1023 molecules. 0.02 mol of CO contains 0.02  6  1023 molecules = 1.2  1022 molecules

Solution 6  1022 6  1022 molecules = — — — — — — — ‑  1 mol = 0.1 mol 6  1023 I 1 mol of H2 = 2 g 1 1 g of H2 = — mol = 0.5 mol (I is incorrect) 2 II 1 mol of N2 = 28 g 2.8 2.8 g of N2 = — — mol = 0.1 mol (II is correct) 28 III 1 mol of CO2 = 44 g 4.4 4.4 g of CO2 = — — mol = 0.1 mol (III is correct) 44 IV 1 mol of H2O = 18 g 1.8 1.8 g of H2O = — — mol = 0.1 mol (IV is correct) 18 Answer

Conversion of Number of Particles to Volume of Gas

18 Calculate the volume of 1.5  1023 molecules of ethane, C2H6 gas at room temperature and pressure. [1 mol of gas occupies a volume of 24 dm3 at r.t.p., NA = 6  1023 mol–1] Chemical Formulae and Equations

number of moles

 (61023)

Calculate the number of molecules present at s.t.p. in (a) 0.28 dm3 of N2 gas, (b) 448 cm3 of carbon monoxide, CO gas. [1 mol of gas occupies a volume of 22.4 dm3 at s.t.p., NA = 6  1023 mol–1]



 22.4 dm (or 24 dm3)

3

17



 22.4 dm3 (or 24 dm3)

50

3.5

’05

The activity of microorganisms on waste products at dump sites produces methane gas. If 180 dm3 of methane gas is collected, calculate the mass of methane obtained. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies a volume of 24 dm3 at room temperature and pressure]

Chemical Formulae

SPM

’05/P2 ’08/P1

1 A chemical formula is used to represent a chemical compound. The chemical formula shows (a) the elements, denoted by their symbols, present in the compound. (b) the relative numbers, indicated by subscripts written after the symbols, of each element present in the compound. 2 For example, the chemical formula of sulphuric acid is H2SO4. The chemical formula indicates that (a) the elements present in sulphuric acid are hydrogen, sulphur and oxygen. (b) two hydrogen atoms, one sulphur atom and four oxygen atoms combine to form the compound. 3 Table 3.1 shows the chemical formulae of some covalent compounds.

Solution 1 mol of gas occupies a volume of 24 dm3 at room temperature. 180 dm3 180 dm3 of CH4 gas = — — — — —  1 mol 24 dm3 = 7.5 mol 1 mol of CH4 = 16 g 7.5 mol of CH4 = 7.5  16 g = 120 g

Table 3.1 The chemical formulae of some covalent compounds

3.4

Name of compound

1 Calculate the volume of 0.55 mol of oxygen gas at room temperature and pressure. [1 mol of gas occupies a volume of 24 dm3 at room temperature]

Oxygen

2 Calculate the number of moles of 672 cm3 of carbon dioxide gas at s.t.p. [1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.] 3 Calculate the volume occupied by 1.4 g of ethene gas, C2H4 at room temperature and pressure. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies a volume of 24 dm3 at room temperature]

Chemical formula O2

Number of each element in the compound 2 oxygen atoms

Water

H2O

2 hydrogen atoms and 1 oxygen atom

Ammonia

NH3

1 nitrogen atom and 3 hydrogen atoms

Sulphuric acid

H2SO4

2 hydrogen atoms, 1 sulphur atom and 4 oxygen atoms

4 The chemical formula of an ionic compound can be written if the charge of the cation (positively-charged ion) and the anion (negatively-charged ion) forming the ionic compound are known. 5 Table 3.2 shows the charges of some ions.

4 Calculate the mass of each of the following gases at standard temperature and pressure: (a) 16.8 dm3 of methane, CH4, (b) 6720 cm3 of carbon monoxide gas, CO. [Relative atomic mass: H, 1; C, 12; O, 16; 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.]

Table 3.2 Charges of some cations and anions

5 Calculate the number of molecules present in: (a) 3.6 dm3 of N2 gas, (b) 1200 cm3 of ammonia gas at room temperature and pressure. [1 mol of gas occupies a volume of 24 dm3 at room temperature, NA = 6  1023 mol–1]

Charge

+1

6 Calculate the volume of 9  1021 molecules of CO at standard temperature and pressure. [1 mol of gas occupies a volume of 22.4 dm3 at s.t.p., NA = 6  1023 mol–1]

51

Cation Sodium ion Potassium ion Lithium ion Silver ion Copper(I) ion Hydrogen ion Ammonium ion Nickel(I) ion

Symbol Na+ K+ Li+ Ag+ Cu+ H+ NH4+ Ni+

Chemical Formulae and Equations

3

4

Charge

+2

3

+3

Cation Magnesium ion Calcium ion Zinc ion Iron(II) ion Copper(II) ion Manganese(II) ion Lead(II) ion Nickel(II) ion

Mg2+ Ca2+ Zn2+ Fe2+ Cu2+ Mn2+ Pb2+ Ni2+

Iron(III) ion Aluminium ion Chromium(III) ion

Fe3+ Al3+ Cr3+

Charge

must be equal to the total negative charge of the anion. 7 To write the chemical formula of an ionic compound, the following steps can be used: (a) Write the formulae of the ions involved in forming the compound and their charges. (b) Then, balance the positive and negative charges. This can be done by writing the numerical charge of the cation next to the anion as a subscript and the numerical charge of the anion next to the cation as a subscript. (c) Finally write the chemical formula of the ionic compound without the charges. 8 Generally, if the ionic compound is formed by the ions Xm+ and Yn–, then the chemical formula of the compound is XnYm. If m = n, then the chemical formula is XY. Examples, (a) Chemical formula of sodium sulphate Charge of ion +1 –2 Formula of ion Na+ SO42– Ratio 2 1 Chemical formula is Na2SO4. (b) Chemical formula of iron(III) chloride Charge of ion +3 –1 Formula of ion Fe3+ Cl– Ratio 1 3 Chemical formula is FeCl3. (c) Chemical formula of magnesium oxide Charge of ion +2 –2 2+ Formula of ion Mg O2– Ratio 2 2 Chemical formula is MgO. Note that if the charges of the ions are the same, the ratio of the ions that combine is 1 : 1. 9 In general, the chemical formulae of compounds formed between the ions are summarised in Table 3.3.

Symbol

Anion

Symbol

–1

Fluoride ion Chloride ion Bromide ion Iodide ion Hydroxide ion Nitrate ion Nitrite ion Bicarbonate ion Permanganate ion Hydride ion

F– Cl– Br– I– OH– NO3– NO2– HCO3– MnO4– H–

–2

Oxide ion Sulphide ion Sulphate ion Sulphite ion Carbonate ion Thiosulphate ion Chromate(VI) ion Dichromate(VI) ion

O2– S2– SO42– SO32– CO32– S2O32– CrO42– Cr2O72–

–3

Phosphide ion Phosphate ion Nitride ion

P3– PO43– N3–

6 Since a chemical compound is always electrically neutral, the total positive charge of the cation

Table 3.3 Formulae of ionic compounds

Cation

Anion

Chemical formula

Examples

X+ X+ X+

Y– Y 2– Y 3–

XY X2Y X3Y

Sodium chloride, NaCl Potassium dichromate(VI), K2Cr2O7 Ammonium phosphate, (NH4)3PO4

X 2+ X 2+ X 2+

Y– Y 2– Y 3–

XY2 XY X3Y2

Calcium chloride, CaCl2 Copper(II) sulphate, CuSO4 Magnesium nitride, Mg3N2

X 3+ X 3+ X 3+

Y– Y 2– Y 3–

XY3 X2Y3 XY

Iron(III) chloride, FeCl3 Chromium(III) sulphate, Cr2(SO4)3 Aluminium nitride, AlN

Chemical Formulae and Equations

52

Name of compound

Cu+ Cu2+

O2– O2–

Copper(I) oxide Copper(II) oxide

Cu2O CuO

Fe2+ Fe3+

Cl– Cl–

Iron(II) chloride Iron(III) chloride

FeCl2 FeCl3

Mn2+ Mn4+

O2– O2–

Manganese(II) oxide Manganese(IV) oxide

MnO MnO2

All transition elements are metals. Many of them like iron, manganese, copper, silver, gold, nickel and titanium are of major technological importance. A large number of the transition elements combine with each other to form useful alloys.

Compound

Molecular formula

C:H:O =1:2:1

CH2O

C20H24N2O2 C : H : N : O C10H12NO = 10: 12 : 1 : 1

’09

The table shows the mass of elements M and O in an oxide, and relative atomic mass of M and O.

1 The empirical formula of a compound shows SPM the simplest ratio of the atoms of the elements ’07/P2 that combine to form the compound. 2 The molecular formula of the compound shows the actual numbers of the atoms of the elements that combine to form the compound. 3 Table 3.5 shows the molecular and empirical formulae of some compounds.

Simplest ratio of the elements

C6H12O6

Empirical formula

5

Empirical and Molecular Formulae

Table 3.5 The molecular and empirical formulae of some compounds

Glucose

Simplest ratio of the elements

4 The following steps can be used to determine the empirical formula of a compound: Step 1: Write the mass or percentage of each element in the compound. Step 2: Calculate the number of moles of each element in the compound by dividing the mass or percentage of the element by the relative atomic mass of the element. Step 3: Next, divide each number by the smallest number to obtain the simplest ratio. Step 4: Finally write the empirical formula of the compound based on the ratio of the elements.

Chemical formula

Anion

Molecular formula

Quinine

Table 3.4 Formulae of some compounds of transition elements

Cation

Compound

Element

M

O

Mass (g)

2.4

1.6

Relative atomic mass

48

16

What is the empirical formula of this compound? Solution Element

SPM

Step 1

Mass

Step 2

Number of moles

Step 3

Simplest ratio

M

O

2.4 g

1.6 g

’09/P1

Empirical formula

Water

H2O

H:O=2:1

H2O

Ethene

C2H4

C:H=1:2

CH2

Butane

C4H10

C:H=2:5

C2H5

Ethane

C2H6

C:H=1:3

CH3

2.4 —— mol 48 = 0.05 0.05 — — — 0.05 =1

1.6 — — mol 16 = 0.1 0.10 — — — 0.05 =2

Empirical formula of compound is MO2. 53

Chemical Formulae and Equations

3

10 Transition elements can form ions with different charges. In the IUPAC (International Union of Pure and Applied Chemistry) system, the standardised chemical nomenclature is denoted by Roman numerals in brackets to indicate the charge of the ion. For example, iron(II) represents Fe2+ ion and iron(III) represents Fe3+ ion. Table 3.4 shows examples of some compounds of transition elements.

6

Simplest 4.8 —— ratio 1.6 =3

’05

2.5 g of X combined with 4 g of Y to form a compound with the formula XY2. If the relative atomic mass of Y is 80, determine the relative atomic mass of X.

1.6 —— — 1.6 =1

4.84 ——— 1.6 =3

Empirical formula of boric acid is H3BO3.

Solution Assume the relative atomic mass of X is a. Element

3

Mass Number of moles

X

Y

2.5

4

20

A gaseous hydrocarbon X contains 85.7% of carbon by weight. 4.2 g of the gas X occupies a volume of 3.36 dm3 at standard temperature and pressure. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.] (a) Determine the empirical formula of X. (b) Determine the relative molecular mass of X. (c) What is the molecular formula of X?

2.5 4 —— mol — — mol a 80 = 0.05

Simplest ratio

1

2

Since the empirical formula is XY2, the ratio of X : Y=1:2 2.5 — — a

Solution

1 — — —= — 0.05 2 2.5 0.05 — —= — — — a 2 2.5  2 a=— — — — — — 0.05 = 100

Element

C

H

Percentage

85.7%

100 – 85.7% = 14.3%

Number of moles

Boric acid is used to preserve prawns and fish. Chemical analysis of the compound shows that it contains 4.8% hydrogen, 17.7% boron and the rest is oxygen. Determine the empirical formula of boric acid. [Relative atomic mass: H, 1; B, 11; O, 16]

H

B

O

Percentage

4.8%

17.7%

100 – 4.8 – 17.7% = 77.5%

14.3 — — — — 7.14 =2

Therefore the relative molecular mass of X is 28. (c) Assume the molecular formula of X is (CH2)n. The relative molecular mass of X is (12 + 2)n = 28 28 n=— — 14 =2

4.8 17.7 77.5 —— — 100 g —— —100 g ———  100 g 100 100 100 = 4.8 g = 17.7 g = 77.5 g

Number 4.8 17.7 77.5 —— mol ——— mol ——— mol of moles 1 11 16 = 4.8 = 1.6 = 4.84

Chemical Formulae and Equations

14.3 — — — mol 1 = 14.3

(a) Empirical formula of X is CH2. (b) 4.2 g of the gas has a volume of 3.36 dm3 at s.t.p. 1 mol of the gas (22.4 dm3) has a mass of 22.4 dm3 — — — — — — —  4.2 g 3.36 dm3 = 28 g

Solution Element

85.7 — — — — mol 12 = 7.14

Simplest ratio 7.14 — — — 7.14 =1

19

Mass in 100 g of compound

SPM

’04/P2

Thus the molecular formula of X is C2H4.

54

To determine the empirical formula of magnesium oxide by experiment combustion of the magnesium. Care is taken to prevent white smoke (magnesium oxide powder formed) from escaping by closing the lid immediately after it is lifted. 6 When the magnesium is completely burnt, the crucible is cooled and is weighed again with its lid. The weight is recorded. 7 The heating, cooling and weighing process is repeated until a constant mass is obtained.

Procedure

Results Mass of crucible + lid = a gram Mass of crucible + lid + magnesium = b gram Mass of crucible + lid + magnesium oxide = c gram Calculations Mass of magnesium used Mass of oxygen that combined with magnesium

1 The 20 cm length of magnesium ribbon is polished with sandpaper to remove the oxide layer on its surface. 2 The magnesium ribbon is rolled into a loose coil. 3 An empty crucible with its lid is first weighed and its weight is recorded. 4 The rolled magnesium ribbon is placed in the crucible. The crucible is weighed again and its weight recorded. 5 The crucible is placed on a clay pipe triangle and then heated strongly. The lid is lifted at intervals to allow the oxygen from the air to enter for the

= (b – a) gram = (c – b) gram

Element

Mg

O

Mass (g)

(b – a)

(c – b)

Number of moles Simplest ratio

(b – a) — — — — mol 24

(c – b) — — — — mol 16

x

y

Conclusion The empirical formula is MgxOy.

To determine the empirical formula of copper oxide

Materials

SPM

’09/P2

1 The combustion tube with a porcelain dish is weighed. The weight is recorded. 2 A spatula of copper oxide powder is placed in the porcelain dish. The combustion tube with its contents is weighed again. The weight is recorded. 3 Dry hydrogen gas is passed through the tube for a few minutes to expel all the air. 4 The copper oxide is then heated strongly and the hydrogen gas passing through the end of the combustion tube is lit. 5 When all the copper oxide is reduced to copper metal (the black copper oxide has all become brown), heating is stopped. 6 A continuous stream of hydrogen gas is allowed to pass through the tube until it is cooled.

The empirical formula of copper oxide is determined by reducing the copper oxide using hydrogen gas. Apparatus

3

Materials

Crucible with lid, tripod stand, sandpaper, Bunsen burner, clay pipe triangle, electronic balance and tongs. A magnesium ribbon of about 20 cm in length and oxygen.

Porcelain dish, combustion tube and electronic balance. Copper oxide powder and dry hydrogen gas.

Procedure

55

Chemical Formulae and Equations

Activity 3.2 & 3.3

Apparatus

7 The combustion tube with its contents is weighed and the weight recorded. 8 The heating, cooling and weighing process is repeated until a constant weight is obtained.

3

Results Mass of combustion tube + empty porcelain dish = 54.31 g Mass of combustion tube + porcelain dish + copper oxide = 62.32 g Mass of combustion tube + porcelain dish + copper = 60.71 g Calculations Mass of copper obtained = (60.71 – 54.31)g = 6.40 g Mass of oxygen that combined with the copper = (62.32 – 60.71)g = 1.61 g Element

Cu

O

Mass (g)

6.40

1.61

6.4 — — — mol = 0.1 64

1.61 — — — mol = 0.1 16

Number of moles

Simplest ratio 0.1 — — —= 1 0.1

Precautions taken 1 Hydrogen gas must be allowed to pass through SPM the combustion tube for a few minutes before the ’11/P1 copper oxide is heated. This is to remove the air in the tube. A mixture of air and hydrogen can explode when lit. 2 The flow of hydrogen gas must be continued throughout heating. This is to ensure that air does not enter the combustion tube. Hence the hydrogen gas must be seen to be burning continuosly at the end of the combustion tube. 3 The hot copper metal is allowed to cool in a stream of hydrogen gas. This is to ensure that oxygen from the air does not oxidise the hot copper to copper oxide again. 4 The heating, cooling and weighing process is repeated until a constant weight is obtained. This is to ensure that all the copper oxide has been reduced. Conclusion The empirical formula of copper oxide is CuO.

0.1 — — —= 1 0.1

3.5 1 Write down the chemical formulae of the following ionic compounds: (a) Silver nitrate (b) Sodium thiosulphate (c) Ammonium phosphate (d) Calcium hydroxide (e) Magnesium carbonate (f) Zinc phosphide (g) Iron(III) hydroxide (h) Aluminium oxide (i) Chromium(III) chloride (j) Copper(II) sulphate (k) Nickel(I) chloride (l) Magnesium nitride

4 10.2 g of vanadium metal combined with 8 g of oxygen to form a compound with empirical formula V2O5. Determine the relative atomic mass of vanadium. [Relative atomic mass: O, 16] 5 Hydrazine is used as a rocket fuel. It contains 87.5% nitrogen and 12.5% hydrogen. [Relative atomic mass: H, 1; N, 14] (a) Determine the empirical formula of the compound. (b) If the relative molecular mass of hydrazine is 32, determine its molecular formula. 6 x g of iron combined with 5.04 g of oxygen to form an oxide with empirical formula Fe2O3. Determine the value of x. [Relative atomic mass: O, 16; Fe, 56]

2 Silicon hydride contains 87.5% silicon by mass. Determine the empirical formula of silicon hydride. [Relative atomic mass: H, 1; Si, 28]

7 A gaseous hydrocarbon Q contains 20% hydrogen. 6 g of this hydrocarbon occupies a volume of 4.48 dm3 at s.t.p. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies 22.4 dm3 at s.t.p.]. Determine: (a) The empirical formula of Q (b) Relative molecular mass of Q (c) Molecular formula of Q

3 Reduction of 7.55 g of tin oxide using hydrogen gas yields 5.95 g of tin metal. Determine the empirical formula of tin oxide. [Relative atomic mass: Sn, 119; O, 16]

Chemical Formulae and Equations

56

3.6

Chemical Equations Remember that a chemical equation cannot be balanced by altering the formulae of the reactants or products. You can only balance the equation by putting a number in front of each of the formulae.

Reactants and Products of Chemical Equations 1 A chemical reaction can be represented by a chemical equation. For example, when sodium hydroxide reacts with sulphuric acid, sodium sulphate and water are produced. The reaction can be represented symbolically by the equation:

1 Lets’s look at a chemical reaction to practice these rules. During combustion, magnesium reacts with the oxygen in air to produce a white powder, magnesium oxide. You might start by writing an equation for this reaction by writing the symbols for the reactants and products.

2NaOH + H2SO4 → Na2SO4 + 2H2O (a) The reactants are the chemicals that are reacting and they are written on the left-hand side of the equation. Hence, sodium hydroxide and sulphuric acid are the reactants. (b) The products are the chemicals formed in a reaction and they are written on the righthand side of the equation. Hence, sodium sulphate and water are the products. (c) 2 mol of NaOH react with 1 mol of H2SO4 to produce 1 mol of Na2SO4 and 2 mol of H2O. 2 Some symbols which appear in a chemical equation are: Symbol

Meaning

∆

Heating of substance

↑ or (g)

Gas evolved

↓ or (s)

Precipitate formed



reactants

Mg

+

O2

yield product

⎯→

MgO

2 On the left-hand side of the equation there is 1 atom of magnesium and there is also 1 atom of Mg on the product side of the equation. However, there are 2 oxygen atoms on the left and only 1 on the right. To balance the number of atoms of each type you will need to add coefficients. If you multiply Mg on the left-hand side of the equation by 2 then you must do the same to the Mg atom on the right which will mean doubling the number of oxygen atoms on the right-hand side of the equation too since you may not change the ratio of atoms within a molecule because it will change the identity of the substance. The new equation will look like this:

Reversible reaction

2Mg + O2

3 When writing a chemical equation, the following steps are followed: (a) Write the correct formulae of all reactants on the left-hand side of the equation. (b) Write the correct formulae of all products on the right-hand side of the equation. (c) The equation is then balanced. This involves making sure that the number of atoms of each element before and after the reaction are the same. (d) Finally the physical state of each of the reactants or products is written as: (s) – represents solid state (l) – represents liquid state (g) – represents gaseous state (aq) – represents aqueous state, that is, the substance is dissolved in water.

⎯→ 2MgO

3 Now the equation is balanced. There are the same number of Mg and O atoms on both sides of the equation. A balanced equation is the source of a great deal of information about both products and reactants during a chemical change. Examples of Chemical Equations 1 (a) When green copper(II) carbonate is heated, it decomposes to form black copper(II) oxide and carbon dioxide gas which turns limewater cloudy. CuCO3(s) ⎯→ CuO(s) + CO2(g)

reactant

57

products

Chemical Formulae and Equations

3

Writing Chemical Equations

If 0.12 g of magnesium is added to excess hydrochloric acid, calculate (a) the mass of magnesium chloride salt formed. (b) the volume of hydrogen gas evolved at room temperature and pressure. [Relative atomic mass: Mg, 24; Cl, 35.5; 1 mol of gas occupies a volume of 24 dm3 at r.t.p.]

(b) When calcium carbonate is reacted with hydrochloric acid, the products formed are calcium chloride, carbon dioxide and water. CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) (c) When an aqueous solution of potassium iodide is added to an aqueous solution of lead(II) nitrate, a yellow precipitate of lead(II) iodide is formed.

Solution (a) Referring to the equation, 1 mol of Mg (24 g) produces 1 mol of MgCl2(95 g). Hence 0.12 g of Mg will produce 0.12 g Mg — — — — — — — — — — — — —  95 g MgCl2 24 g Mg

3

2KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2KNO3(aq)

Interpreting Chemical Equations Qualitatively and Quantitatively

= 0.475 g of MgCl2

SPM

’11/P1

(b) 1 mol of Mg (24 g) produces 1 mol of H2 gas (24 dm3) at r.t.p. Hence 0.12 g of Mg will produce 0.12 — — —  24 dm3 of H2 24 = 0.12 dm3 of H2 = 120 cm3 of H2

1 From a chemical equation, we can obtain information on (a) the reactants taking part in the reaction (on the left-hand side of the equation). (b) the products formed in the reaction (on the right-hand side of the equation). (c) the number of moles of each substance taking part in the reaction and the number of moles of the products formed (from the coefficients/numbers before the substances). (d) the physical states of all the reactants and products (from the symbols in bracket after the substance). For example: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) The chemical equation shows that: 2 mol of solid sodium react with 2 mol of liquid water to form 2 mol of aqueous sodium hydroxide and 1 mol of hydrogen gas.

Solving Numerical Problems Using Chemical Equations

8

3.2 g of copper(II) oxide powder is reacted with excess dilute nitric acid. (a) Write a chemical equation for the reaction. (b) Calculate the mass of copper(II) nitrate salt formed in the reaction. [Relative atomic mass: N, 14; O, 16; Cu, 64] Solution (a) CuO(s) + 2HNO3(aq) → Cu(NO3)2(aq) + H2O(l) (b) 1 mol of CuO produces 1 mol of Cu(NO3)2. 80 g of CuO produces 188 g of Cu(NO3)2. Hence 3.2 g of CuO will produce 3.2 g CuO ——————  188 g of Cu(NO3)2 80 g CuO = 7.52 g of Cu(NO3)2

SPM

’08/P1

The quantity of reactants and products can be calculated using stoichiometric equations when numerical information is given (stoichiometry).

7

21 Ethanol burns in air as represented by the equation C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(l) Calculate the mass of ethanol burnt if 2.4 dm3 of carbon dioxide is produced at room temperature. [Relative atomic mass: H, 1; C, 12; O, 16; 1 mol of gas occupies a volume of 24 dm3 at room temperature]

’09

Magnesium reacts with hydrochloric acid as represented by the equation Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Chemical Formulae and Equations

’04

58

Solution Referring to the equation, 1 mol of C2H5OH (46 g) produces 2 mol (2  24 dm3) of CO2. Therefore 2.4 dm3 of CO2 is produced from 2.4 dm3 — — — — — — — — —  46 g of C2H5OH 2  24 dm3 = 2.3 g of ethanol, C2H5OH

3 Iron metal reacts with excess hydrochloric acid to produce iron(II) chloride and hydrogen gas. (a) Write a balanced chemical equation for the reaction. (b) If 2.8 g of iron metal is used in the reaction, calculate (i) the maximum mass of iron(II) chloride formed. (ii) the volume of hydrogen gas produced at room conditions. [Relative atomic mass: Cl, 35.5; Fe, 56; 1 mol of gas occupies a volume of 24 dm3 at r.t.p.]

22 The chemical name of baking powder is sodium bicarbonate (NaHCO3). When sodium bicarbonate is heated, it decomposes to sodium carbonate, carbon dioxide and water. (a) Write a balanced chemical equation for the decomposition of sodium bicarbonate when it is heated. (b) If 2.1 g of sodium bicarbonate is heated, calculate the volume in cm3 of CO2 produced at room temperature. [Relative atomic mass: H, 1; C, 12; O, 16; Na, 23; 1 mol of gas occupies a volume of 24 dm3 at room temperature]

4 Acetylene gas (C2H2) is used in metal welding. This gas can be prepared by reacting calcium carbide with excess water as represented by the equation CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g) If 4.8 g of calcium carbide is reacted with excess water, calculate (a) the volume of acetylene gas evolved at room temperature. (b) the mass of calcium hydroxide formed. [Relative atomic mass: H, 1; C, 12; O, 16; Ca, 40; 1 mol of gas occupies a volume of 24 dm3 at r.t.p.]

Solution (a) 2NaHCO3(s) ⎯→ Na2CO3(s) + CO2(g) + H2O(l) ∆ (b) 2 mol of NaHCO3 (2  84 g) produces 1 mol of CO2 (24 dm3) at r.t.p. 2.1 g of NaHCO3 will produce 2.1 — — — — — —  24 dm3 of CO2 = 0.3 dm3 of CO2 2  84 = 300 cm3 of CO2

5 Hydrogen gas is prepared by reacting methane gas with steam using platinum as catalyst. The reaction is represented by the equation CH4(g) + H2O(g) → CO(g) + 3H2(g) If 60 dm3 of hydrogen gas is produced at room temperature and pressure, calculate (a) the mass of methane that is used in the reaction. (b) the number of carbon monoxide molecules released. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies a volume of 24 dm3 at r.t.p; NA = 6 1023 mol–1]

3.6 1 Balance the following chemical equations: (a) Na(s) + H2O(l) → NaOH(aq) + H2(g) (b) Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g) (c) CuCO3(s) + HNO3(aq) → Cu(NO3)2(aq) + CO2(g) + H2O(l) (d) HCl(aq) + Na2S2O3(aq) → NaCl(aq) + SO2(g) + S(s) + H2O(l) (e) Pb(NO3)2(s) ⎯⎯→ PbO(s) + NO2(g) + O2(g) ∆

6 Zinc phosphide (Zn3P2) is used as a rat poison. This substance can be prepared by reacting excess zinc powder with phosphorus. (a) Write a balance chemical equation for the reaction. (b) Calculate the mass of phosphorus needed to react with the excess zinc to produce 51.4 kg of zinc phosphide. [Relative atomic mass: P, 31; Zn, 65]

2 Write balanced equations for the following reactions: (a) Reaction between sulphur dioxide and oxygen to form sulphur trioxide. (b) Neutralisation reaction between potassium hydroxide and sulphuric acid to form potassium sulphate and water. (c) Burning of ethane (C2H6) in air to form carbon dioxide and water.

59

Chemical Formulae and Equations

3

(d) Displacement reaction between zinc metal and copper(II) sulphate solution to form copper metal and zinc sulphate salt. (e) Decomposition of zinc carbonate to zinc oxide and carbon dioxide gas when heated. (f) Reduction of solid lead(IV) oxide, PbO2 by hydrogen gas to form lead metal and water.

3.7

Scientific Attitudes and Values in Investigating Matter

3

1 In the 19th century, scientists introduced a simple way to represent an element. Each element is represented by a letter or two letters of the alphabet; the first letter is a capital letter and the second letter is a small letter. For example, the reaction between carbon and oxygen to form carbon dioxide can be represented by the following equation:

to produce 2 mol of magnesium oxide, we need to burn 2 mol of magnesium in air. If we need to produce 1 mol of MgO (40 g), then we have to burn 1 mol of Mg (24 g) in air. Similarly, to produce 40 kg of MgO, the required amount of Mg needed is 24 kg. Any excess amount of magnesium used is considered a wastage. 5 The use of relative mass by scientists enabled the mass of atoms and molecules to be determined although they are too small to be measured by any weighing machines. The development of the standard to be used in the determination of relative atomic mass and relative molecular mass started with the use of the hydrogen atom, followed by the oxygen atom, and finally carbon-12 which is now used as the standard. 6 The Italian scientist, Amedeo Avogadro proposed the hypothesis that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. With perseverance and ingenuity, he calculated that 22.4 dm3 of any gas contains 6.02  1023 particles at standard temperature and pressure. The number of 6.02  1023 is now known as Avogadro’s number or the Avogadro constant in honour of him. 7 Some products can be produced by different processes. For example, calcium oxide can be produced by (a) burning calcium in air: 2Ca(s) + O2(g) → 2CaO(s) or (b) heating calcium carbonate: CaCO3(s) → CaO(s) + CO2(g)

C(s) + O2(g) → CO2(g) 2 The symbols of elements using letters of the alphabet are universal in that they are agreed upon by all scientists, regardless of the countries of origin. They become a tool of communication between chemists although the spoken languages may be different. 3 Many chemical reactions involve the formation of compounds. However, the names of some compounds are long. This makes the writing of equations in words cumbersome. For example, Sulphuric acid + sodium hydroxide → sodium sulphate + water To simplify the writing of the chemical equations, scientists used chemical formulae to represent compounds. The above reaction can then be represented by H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) Thus the writing of chemical equations is made easy. 4 In the formation of a desired product through a chemical reaction, it is important to calculate the correct amount of reactants required so as to prevent wastage. This is made possible by the mole concept used by chemists in the quantitative calculations of chemical reactions. For example, in the reaction between magnesium and oxygen,

8 The ideal process is one in which (a) a high percentage yield is obtained. For example, a process which produces 80% yield is superior to one which produces only 45% yield. The mole concept makes it possible to calculate the percentage yield. (b) the cost of reactants used is cheap. In general, the cheaper the reactants, the better the process is because the cost of producing the product is lower. Thus, the product can be priced competitively.

2Mg + O2 → 2MgO

Chemical Formulae and Equations

60

3.7 2 Explain the meaning of the following chemical equations: (a) CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l) (b) 2Mg(NO3)2(s) ⎯→ 2MgO(s) + 4NO2(g) + ∆ O2(g) (c) H2(g) + PbO(s) → Pb(s) + H2O(l) (d) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

1 The relative atomic mass of an element is the number of times one atom of the element is heavier than one-twelfth the mass of one carbon-12 atom. 2 The relative molecular mass of a compound is the number of times one molecule of the compound is heavier than one-twelfth the mass of one carbon-12 atom. 3 One mole is the quantity of substance which contains the same number of particles as there are in 12.00 grams of carbon-12. 4 One mole of element is the relative atomic mass of the element expressed in gram. For example,

6 The molar volume of a gas is the volume occupied by one mole of the gas. At standard temperature and pressure, one mole of gas occupies a volume of 22.4 dm3. 7 The interconversion between mole, mass, volume and number of particles can be summarised below:

3

1 Write the chemical formulae for the following compounds: (a) Hydrobromic acid (b) Zinc hydroxide (c) Potassium nitrate (d) Sodium sulphate (e) Silver nitrate (f) Magnesium chloride

1 mol of C = 12 g 1 mol of Na = 23 g 1 mol of S = 32 g

 22.4 dm3 Volume 3 22.4 dm3

3 (6 3 1023)

All contain 6.02 3 1023 atoms

Mass in gram

 RAM or RMM

 (6 3 1023)

Number of particles

5 One mole of compound is the relative molecular mass of the compound expressed in grams. For example, 1 mol of H2O = 18 g 1 mol of CO2 = 44 g 1 mol of C2H6 = 30 g

Number of moles

3 RAM or RMM

All contain 6.02 3 1023 molecules

61

Chemical Formulae and Equations

3 Multiple-choice Questions

3

3.1

Relative Atomic Mass and Relative Molecular Mass

1 The relative formula mass of Mg(XO3)2 is 148. Determine the relative atomic mass of element X. [Relative atomic mass: O, 16; Mg, 24] A 10 C 14 B 12 D 18 2 A compound with formula M2S2O3.5H2O has a relative formula mass of 248. What is the relative atomic mass of M? [Relative atomic mass: H, 1; O, 16; S, 32] A 23 C 27 B 24 D 39 3 Veronal is a barbiturate used to induce sleep in psychiatric patients. The molecular formula of veronal is C4H2N2O3(C2H5)2. Determine the relative molecular mass of veronal. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16] A 160 C 186 B 184 D 196

[Relative atomic mass: H, 1; C, 12; O, 16] A 136 B 140 C 151 D 152 5 Three elements are represented by the letters X, Y and Z. One ’08 atom of Z is two times heavier than one atom of Y. One atom of Y is three times heavier than one atom of X. If the relative atomic mass of X is 39, what is the relative atomic mass of Z? A 78 C 195 B 117 D 234 6 Which compound in the table below is correctly matched with its relative formula mass? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Na, 23; P, 31; S, 32; Cl, 35.5; Ca, 40; Co, 59]

9 Calculate the number of molecules in 0.88 g of vitamin C, C6H8O6. [Relative atomic mass: H, 1; C, 12; O, 16; NA = 6 3 1023 mol–1] A 3.0 3 1021 B 3.0 3 1022 C 7.5 3 1021 D 7.5 3 1022

Compound

Relative formula mass

I Ca3(PO4)2

310

II C14H18N2O5

294

III C17H35COONa

306

H O

IV CoCl2.6H2O

170



C C H | C C–O–C–H | H C H | O–H Determine the relative molecular mass of vanillin. Chemical Formulae and Equations

Relationship between the Number of Moles and the Number of Particles

8 Which of the following samples contain 3.0 3 1022 molecules? [Relative atomic mass: H,1; C, 12; N, 14; O,16; S, 32; 1 mol contains 6 3 1023 molecules] I 0.9 g of H2O II 0.85 g of NH3 III 1.4 g of C2H4 IV 1.6 g of SO2 A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV

4 The diagram shows the molecular structure of vanillin molecule which gives vanilla its taste.

C | H C H

3.2

A I and III only B II and IV only C I, II and III only D I, III and IV only 7 Calculate how many beryllium atoms that will have the same mass as one quinine molecule, C20H24N2O2 (an anti-malarial drug). [Relative atomic mass: H,1; Be,9; C,12; N,14; O,16] A 35 C 37 B 36 D 38

62

10 Calculate the number of atoms in 0.96 g of titanium. [Relative atomic mass: Ti, 48; NA = 6 3 1023 mol–1] A 1.2 3 1021 B 1.5 3 1021 C 1.2 3 1022 D 1.5 3 1022 11 Which of the following has the most number of molecules? [Relative atomic mass: H, 1; C, 12; O, 16; Br, 80] A 3.6 g of water, H2O B 4.0 g of methane, CH4 C 19.6 g of sulphuric acid, H2SO4 D 13.2 g of ethanoic acid, CH3COOH 12 The relative atomic mass of oxygen and sulphur are 8 and 32 respectively. Which of the following statements is/are true about oxygen and sulphur?

13 What is the total number of atoms present in 6.05 g dichlorodifluoromethane, CCl2F2. [Relative atomic mass: C, 12; F, 19; Cl, 35.5; NA = 6 3 1023 mol–1] A 1.2 3 1022 C 1.2 3 1023 B 3.0 3 1022 D 1.5 3 1023 14 Pyrethrin is an insecticide with molecular formula of C19H26O3. Calculate the number of pyrethrin molecules contained in a spray with 15.1 g of the compound. [Relative atomic mass: H, 1; C, 12; O, 16; NA = 6 3 1023 mol–1] A 1.5 3 1022 C 1.5 3 1023 B 3.0 3 1022 D 3.0 3 1023 15 If m is the number of atoms in 3 g of carbon, the number of atoms in 3 g of magnesium in terms of m is [Relative atomic mass: C, 12; Mg, 24] 1 A —m C m 2 1 B —m D 2m 3

3.3

Relationship between the Number of Moles of a Substance and Its Mass

16 Which of the following substances contain the same ’09 number of atoms as in 36 gram of carbon? [Relative atomic mass: H, 1; C, 12; O, 16; S, 32] I 3 g of hydrogen II 64 g of sulphur III 18 g of water IV 22 g of carbon dioxide

A I and III only B I and IV only C II and III only D II and IV only 17 Calculate the mass of 7.5 3 1021 aspirin, C9H8O4 molecules. [Relative atomic mass: H, 1; C, 12; O, 16; NA = 6 3 1023 mol–1] A 1.25 g C 2.25 g B 1.44 g D 3.00 g 18 Caffeine is found in coffee beans. Its molecular formula is C4H5N2O. A pill contains 0.05 mol of caffeine. Determine the mass of the compound in the pill. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16] A 1.94 g C 4.85 g B 2.42 g D 9.70 g 19 A cockroach repellent has the formula CH3(CH2)5CHCHCHO. Determine the mass of 0.02 mol of this substance. [Relative atomic mass: H, 1; C, 12; O, 16] A 1.40 g C 3.50 g B 2.80 g D 5.60 g 20 Acetaminophen is a medicine used to relieve pain. 0.0002 mol of acetaminophen has a mass of 0.0302 g. Which of the following is the molecular formula of acetaminophen? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16] C C8H9NO A C8H9NO2 D C8H9N2O B C8H8NO2 21 Ethyl ethanoate is a liquid used as nail varnish remover. If 0.025 mol of ethyl ethanoate has a mass of 2.20 g, determine the relative molecular mass of ethyl ethanoate. A 44 C 77 B 55 D 88 22 The diagram shows the molecular structure of allicin which is a compound obtained from garlic. Allicin have powerful antibiotic and antifungal properties. O i C S C C H H2 H2 C C S C H2 H2 H allicin

63

Calculate the number of moles in 4.86 g allicin. [Relative atomic mass: H, 1; C, 12; O, 16; S, 32] A 0.02 mol C 0.03 mol B 0.025 mol D 0.05 mol 23 Cocaine, C17H21O4N is a drug. Calculate the mass of 1.2 3 1021 cocaine molecules. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; NA = 6 3 1023 mol–1] A 0.202 g C 0.404 g B 0.303 g D 0.606 g

3.4

Relationship between the Number of Moles of a Gas and Its Volume

24 1.0 g of calcium carbonate is added into excess hydrochloric acid. CaCO3 + 2HCl → CaCl2 + H2O + CO2 Determine the volume of carbon dioxide gas evolved at room temperature. [Relative atomic mass: C, 12; O, 16; Ca, 40; 1 mol of gas occupies a volume of 24 dm3 at r.t.p.] C 180 cm3 A 120 cm3 B 150 cm3 D 240 cm3 25 Which of the following gases will occupy the same volume as 2.42 g dichlorodifluoromethane, CCl2F2? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; F, 19; S, 32; Cl, 35.5] I 0.34 g of ammonia, NH3 II 0.88 g of carbon dioxide, CO2 III 1.28 g of sulphur dioxide, SO2 IV 0.40 g of methane, CH4 A I, II and III only B I, II and IV only C I, III and IV only D II, III and IV only 26 Which of the following gases occupy a volume of 600 cm3 at room temperature and pressure? [Relative atomic mass: H, 1; C, 12; O, 16; S, 32; 1 mol of gas occupies a volume of 24 dm3 r.t.p.] I 0.64 g of oxygen, O2 II 0.75 g ethane, C2H6 III 0.56 g of carbon monoxide, CO Chemical Formulae and Equations

3

I 4 g of oxygen and 8 g of sulphur contain the same number of atoms. II The sulphur atom has four times more neutrons than an oxygen atom. III The sulphur atom is four times denser than the oxygen atom. IV 8 g of oxygen contain two times more atoms than 8 g of sulphur. A I and III only B II and IV only C II and III only D IV only

3

IV 1.15 g nitrogen dioxide, NO2 A I and III only B II and IV only C I, II and III only D II, III and IV only 27 Which of the following gases has the heaviest mass at room temperature and pressure? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; S, 32; 1 mol of gas occupies a volume of 24 dm3 r.t.p.] A 3 dm3 of sulphur dioxide, SO2 B 6 dm3 of nitrogen dioxide, NO2 C 15 dm3 of methane, CH4 D 96 dm3 of hydrogen, H2 28 Which of the following gases occupies the greatest volume at room temperature and pressure? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; 1 mol of gas occupies a volume of 24 dm3 r.t.p.] A 0.64 g oxygen, O2 B 0.70 g nitrogen, N2 C 1.10 g carbon dioxide, CO2 D 0.51 g ammonia, NH3 29 What is the number of atoms in 0.05 mol of ammonia gas, NH3? [Avogadro number : 6 3 1023 mol–1] A 1.2 3 1022 B 3.0 3 1022 C 1.2 3 1023 D 3.0 3 1023 30 Which of the statements below are true? I 1.10 g of CO2 and 1.25 g of SO2 gases occupy the same volume at s.t.p. II 0.42 g of N2 and 0.66 g of CO2 gases occupy the same volume at s.t.p. III 2.24 dm3 of C2H6 and 1.12 dm3 of SO2 gases at s.t.p. have the same mass. IV 4.48 dm3 of O2 and 8.96 dm3 of CH4 gases at s.t.p. have the same mass. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; S, 32; 1 mol of gas occupies 22.4 dm3 at s.t.p.] A I and III only B II and IV only C I, II and IV only D II, III and IV only

Chemical Formulae and Equations

3.5

Chemical Formulae

31 1.04 g of element X reacted with 0.48 g of oxygen to form an oxide with the empirical formula X2O3. Determine the relative atomic mass of element X. [Relative atomic mass: O, 16] A 24 C 52 B 48 D 56 32 When 3.64 g of a metal oxide of M is reduced, 2.04 g of the metal is obtained. Determine the empirical formula of the metal oxide. [Relative atomic mass: O, 16; M, 51] A MO2 C M3O2 B M2O3 D M2O5 33 2.75 g of metal M combines with 1.6 g of oxygen to form an ’10 oxide with the empirical formula of MO2. Determine the relative atomic mass of M. [Relative atomic mass: O, 16] A 48 C 55 B 52 D 56 34 x gram of antimony (Sb) combines with 0.48 g of oxygen to form an oxide with the empirical formula of Sb2O3. Determine the value of x. [Relative atomic mass: O, 16; Sb, 122] A 1.22 g B 2.44 g C 3.66 g D 4.88 g 35 An element E forms a fluoride compound with the formula EF3, which contains 16.2 % of E by mass. What is the relative atomic mass of E? [Relative atomic mass; F, 19] A 11 B 24 C 27 D 31

3.6

Chemical Equations

36 Calculate the mass of zinc required to react with excess ’09 nitric acid to produce 360 cm3 of hydrogen at room conditions.

64

[Molar volume: 24 dm3 mol–1 at room conditions; Relative atomic mass: Zn, 65] A 0.325 g B 0.650 g C 0.975 g D 4.333 g 37 0.31 gram of copper(II) carbonate is heated. Determine the volume of carbon dioxide gas released at room conditions. [Relative atomic mass: C, 12; O, 16; Cu, 64; Molar volume : 24 dm3 at room conditions] A 60 cm3 B 120 cm3 C 240 cm3 D 360 cm3 38 Calculate the mass of aluminium oxide, Al2O3 formed if 5.4 g of aluminium is heated in air. [Relative atomic mass: O, 16; K, 39] A 8.4 g B 8.6 g C 10.2 g D 11.8 g 39 Magnesium oxide reacts with nitric acid to form magnesium nitrate and water. If 8.0 g of magnesium oxide is reacted with excess nitric acid, calculate the mass of salt formed. [Relative atomic mass: N, 14; O, 16; Cu, 64] A 14.8 g B 17.2 g C 20.4 g D 29.6 g 40 Iron reacts with chlorine according to the equation ’09 below: 2Fe + 3Cl2 → 2FeCl3 If 1.68 gram of iron burns completely in chlorine, calculate the mass of product formed. [Relative atomic mass: Cl, 35.5, Fe, 56] A 2.438 g B 4.875 g C 6.555 g D 9.750 g

Structured Questions 1 A hydrocarbon X contains 82.76% carbon by mass. 2.9 g of hydrocarbon X occupies a volume of 1.2 ’10 dm3 at room temperature and pressure. [Relative atomic mass: H, 1; C, 12; O, 16; 1 mol of gas occupies a volume of 24 dm3 at r.t.p.; NA = 6  1023 mol–1] (a) What is a hydrocarbon? [1 mark]

mass



(ii) Write a chemical equation for the reduction of M oxide to metal M using hydrogen gas. [2 marks]

(e) Can the empirical formula of magnesium oxide be determined using the same arrangement of apparatus as above? Explain your answer. [2 marks] 3 Table 1 shows the positive and negative ions in three salt solutions.

of

[2 marks]

’05

Name of salt

(d) Determine the molecular formula of hydrocarbon X. [2 marks]

Copper(II) sulphate

(e) Combustion of X in air produces carbon dioxide and water. Write a chemical equation for the reaction. [2 marks]

(f) If 11.6 g of X is burnt, calculate (i) the mass of water formed. [1 mark] (ii) the number of carbon dioxide molecules produced at room temperature. [1 mark]

Negative ion

Cu

SO42–

2+

Potassium iodide

K+

I–

Lead(II) nitrate

Pb2+

NO3–

Table 1

(a) What is another name for a positively-charged ion? [1 mark] (b) Write the formula of lead(II) nitrate.

2 The apparatus shown is used to determine the ’07

Positive ion

[1 mark]

(c) When aqueous lead(II) nitrate solution is added to aqueous potassium iodide solution, a yellow precipitate is formed. (i) Write a chemical equation for the reaction.

empirical formula of the oxide of metal M by reducing the metal oxide with dry hydrogen gas. [Relative atomic mass: O, 16; M, 55]

[2 marks]



(ii) Describe the chemical equation in (i).



(iii) Name the yellow precipitate. [1 mark] (iv) If 0.04 mol of aqueous potassium iodide solution is added to excess lead(II) nitrate solution, calculate the maximum mass of the yellow precipitate formed. [Relative atomic mass: I, 127; Pb, 207] [2 marks]

[1 mark]

Diagram 1

(a) State one precaution that must be taken when carrying out the experiment. [1 mark]

4 Table 2 shows the descriptions and observations of two experiments, I and II.

(b) How can you ensure that all the oxide of metal M has been reduced? [1 mark]



(c) (i) Name two chemicals used to prepare hydrogen gas in the laboratory. [1 mark] (ii) Write an equation for the reaction in (i).



(iii) Name a chemical used to dry hydrogen gas.



’05

Experiment

Combustion of 1.2 g of magnesium powder in excess oxygen

Magnesium burns brightly and a white powder is formed

II

Heating copper(II) carbonate strongly in a test tube

Black solid X is formed and a gas P which turns limewater cloudy is evolved

[1 mark]

Mass of combustion tube + asbestos paper = 39.25 g Mass of combustion tube + asbestos paper + oxide of metal M before heating = 47.95 g Mass of combustion tube + asbestos paper + metal M after heating = 44.75 g

Observation

I

[1 mark]

(d) The information below shows the results of the experiment:

Description

Table 2

[Relative atomic mass: C, 12; O, 16; Mg, 24; Cu, 64; 1 mol of gas occupies a volume of 24 dm3 at room temperature and pressure]

65

Chemical Formulae and Equations

3

molecular

(i) Determine the empirical formula of M oxide. [2 marks]

(b) Determine the empirical formula of hydrocarbon X. [2 marks] (c) Calculate the relative hydrocarbon X.



3

(a) Based on Experiment I: (i) The white powder formed is magnesium oxide. Write the chemical equation for the [2 marks] reaction that takes place. (ii) Calculate the mass of magnesium oxide formed if 3 g of magnesium is completely [2 marks] burnt in excess oxygen. (iii) State one use of magnesium oxide. [1 mark] (iv) The magnesium oxide is basic and reacts with nitric acid (HNO3) to form magnesium nitrate and water. Write a chemical equation [2 marks] for this reaction.

(a) Write a balanced chemical equation for the decomposition of sodium bicarbonate on heating. [2 marks] (b) State a chemical test for carbon dioxide gas. [2 marks]



[3 marks]

[Relative atomic mass: H, 1; C, 12; O, 16; Na, 23; 1 mol of gas occupies a volume of 24 dm3 at room temperature and pressure]

(b) Based on Experiment II: (i) Name the black solid X and gas P formed when copper(II) carbonate is heated strongly.

(d) Sodium bicarbonate reacts with nitric acid (HNO3) to form sodium nitrate, carbon dioxide and water. (i) Write a balanced chemical equation for this reaction. [2 marks] (ii) Calculate the mass of sodium bicarbonate that reacts with the excess acid to produce 17 g of sodium nitrate. [3 marks] (iii) State one use of sodium nitrate. [2 marks] [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Na, 23]

[2 marks]



(c) If 8.4 g of sodium bicarbonate decomposes, calculate (i) the volume of carbon dioxide gas envolved at room temperature and pressure. [3 marks] (ii) the mass of sodium carbonate formed.

(ii) Write the chemical equation for the reaction [1 mark] that takes place. (iii) If 6.2 g of copper(II) carbonate had reacted, calculate [1 mark] (a) the mass of solid X formed. (b) the volume of gas P formed at room [1 mark] temperature and pressure.

5 When sodium bicarbonate (NaHCO3) is heated, it decomposes to sodium carbonate, carbon dioxide and water.



(e) (i) Name another chemical that reacts with nitric acid to form sodium nitrate. [1 mark] (ii) Write an equation for this reaction. [2 marks]

Essay Questions (d) Describe a laboratory method of determining the empirical formula of lead oxide. Your answer should include (i) the procedure of the experiment [4 marks] (ii) tabulation of result [3 marks] (iii) calculation of the results obtained. [4 marks] [Relative atomic mass: O, 16; Pb, 207]

1 (a) Using a suitable example, explain the meaning of the following terms: (i) Empirical formula [2 marks] (ii) Molecular formula [2 marks] (b) State a suitable method that can be used to determine the empirical formula of lead(II) oxide. [2 marks] (c) Can the same method in (b) be used to determine the empirical formula of magnesium oxide? Explain your answer. [3 marks]

Experiment 1 You are required to plan an experiment to determine the empirical formula of magnesium oxide. Your explanation should include the following: (a) Statement of the problem (b) All the variables (c) List of materials and apparatus (d) Procedure (e) Tabulation of data

Chemical Formulae and Equations

66

[17 marks]

FORM 4 THEME: Matter Around Us

CHAPTER

4

Periodic Table of Elements

SPM Topical Analysis 2008

Year 1

Paper

3

2

Section

A

Number of questions

1 — 2

4

2009

B –

1

2010 3

2

C

A

B

C

–

1 — 4

–

–

1

3

–

1

3

2011

2

3

A

B

C

1

–

–

1

1

6

2

3

A

B

C

1

–

–

–

ONCEPT MAP

PERIODIC TABLE

Historical development of the Periodic Table

Transition elements

J. W. Dobereiner John Newlands Lothar Meyer Dmitri Mendeleev Henry G. J. Moseley

Metallic properties of transition elements

Group

Physical and chemical properties of elements in: Group 1 Group 17 Inert property and uses of Group 18 elements

Special characteristics of transition elements

Period

Identifying the group and period of an element based on the electron arrangement of the element

• Changes in the properties of the elements and their oxides across Period 3 • Uses of semi-metals such as silicon and germanium in the microelectronics industry

4.1

3 This systematic method of classification of the elements will enable us to (a) study and generalise the chemical and physical properties of elements in the same group. (b) predict the position of an element in the Periodic Table from its properties. (c) identify and compare elements from different groups. (d) predict the chemical and physical pro­per­ ties of new elements in the same group. 4 Chemists such as Lavoisier, Dobereiner, New­lands, Meyer, Mendeleev and Moseley contributed to the development of the Periodic Table in use today.

Periodic Table of Elements

4

Historical Development of the Periodic Table 1 Many of the elements known today were disco­ vered from the years 1800 to 1900. Chemists noted that certain elements have similar chemi­ cal properties. For example: chlorine, bromine and iodine; potassium and sodium; magnesium and calcium have similar chemical properties. 2 Chemists then tried to group the elements with the same chemical properties together. This led to the development of the Periodic Table.

Contribution of Scientists to the Development of the Periodic Table Antoine Lavoisier 1 Antoine Lavoisier, a Frenchman, was the first chemist who attempted to classify the elements into four groups as in Table 4.1. 2 The four groups consisted of gases, metals, non-metals and metal oxides. 3 His classification was not accurate as light and heat, which are not elements, were inclu­­ded. Furthermore, some elements in each group did not have the same chemical properties. Table 4.1 Lavoisier’s Periodic Table

Group I

Group II

Group III

Group IV

Oxygen Nitrogen Hydrogen Light Heat

Sulphur Phosphorus Carbon Chlorine Fluorine

Arsenic Bismuth Cobalt Lead Zinc, Nickel, Tin, Silver

Calcium oxide Barium oxide Silicon(IV) oxide Magnesium oxide Aluminium oxide

Antoine Lavoisier (1743–1794)

Johann W. Dobereiner 1 Dobereiner classified the elements with the same chemical properties into groups of three called triads. Example: 2 Dobereiner discovered the relationship between the relative Element in the Relative atomic atomic mass (r.a.m.) of elements in triad mass (r.a.m.) each triad. He found that the r.a.m. Lithium (Li) 7 of the element in the middle of each Sodium (Na) 23 triad is approximately equal to the average of the total r.a.m. of the Potassium (K) 39 Johann W. Dobereiner other two elements. Average of the total r.a.m. of Li and K is (1780–1849) 3 However, this relationship did 7 + 39 not apply to most of the other ⎯⎯⎯ = 23 2 elements. Periodic Table of Elements

68

John Newlands 1 John Newlands arranged the elements in order of increasing nucleon number (mass number) in horizontal rows. Each row consisted of seven elements. 2 He found that the chemical properties of every eighth element are similar. This was known as the law of octaves. 3 Table 4.2 shows the arrangement of elements by John Newlands. Table 4.2 The classification of elements by Newlands (law of octaves)

Be

B

C

N

O

F

Na

Mg

Al

Si

P

S

Cl

K

Ca

John Newlands (1837–1898)

4

Li

4 The classification of elements by Newlands was not successful because: (a) The law of octaves was only accurate for the first 16 elements (from Li to Ca). (b) There were no positions allocated for elements yet to be discovered. 5 Nevertheless, Newlands was the first scientist who discovered the existence of periodicity in the elements.

Lothar Meyer

SPM

’09/P1

1 Meyer, a German chemist, Volume of an atom calculated the volume of mass of one mole-atom of the element an atom of an element = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ density of the element using the formula: 2 He plotted a graph of volume of atoms of elements against their relative atomic masses. The graph obtained is shown in Figure 4.1. Lothar Meyer (1830–1895)

Figure 4.1 Lothar Meyer’s atomic volume curve

3 From the shape of the atomic volume curve, Meyer showed that elements occupying the corresponding positions of the curve exhibit similar chemical properties. For example: (a) Li, Na, K and Rb which are located at the peak of the curve show similar chemical properties. (b) Be, Mg, Ca and Sr which are located at positions after the maximum points also show similar chemical properties. 4 Just like Newlands, Meyer showed that the properties of the elements recur periodically. 69

Periodic Table of Elements

Dmitri Mendeleev 1 Dmitri Mendeleev, a Russian chemist, arranged the elements in order of increasing atomic mass. 2 Table 4.3 below shows the Periodic Table suggested by Mendeleev.

4

Table 4.3 The partial Periodic Table by Mendeleev

1 2 3 4 5 6

I

II

H Li Na K Cu Rb

Be Mg Ca Zn Sr

III

IV

V

VI

B C N O Al Si P S () Ti V Cr () () As Se Y Zr Nb Mo ( ) represents unknown elements yet to be discovered

VII

VIII

F Cl Mn Br ()

Fe, Co, Ni

Dmitri Mendeleev (1834–1907)

Ru, Rh, Pd

3 Dmitri Mendeleev was more successful for several reasons. (a) First, he left gaps for elements yet to be discovered. He even used the table to predict the existence and properties of undiscovered elements. Mendeleev correctly predicted the properties of the elements gallium, scandium and germanium which were only discovered much later. (b) Secondly, although the elements were arranged in order of increasing atomic mass, he changed the order of the elements if the chemical properties are not similar. Henry G. J. Moseley 1 Moseley was a British physicist. He bom­barded different elements with high energy electrons and measured the frequency (f) of the X-ray emitted by the element. 2 He then plotted the square-root of the frequency of the X-ray ( ⎯ f ) against the proton number (atomic number) of the element and obtained a straight line graph (Figure 4.2). 3 Thus from the square-root of the frequency of the X-ray emitted by an unknown element, he could determine its proton number. 4 After obtaining the proton number of the elements, Moseley went on to arrange the elements in the Periodic Table in order of increasing proton number (atomic number). 5 Just like Mendeleev did, Moseley left gaps ( ) for elements yet to be discovered. He reasoned that there should be an element corresponding to each proton number. 6 Moseley successfully predicted the existence of four undiscovered elements from the atomic numbers. These elements were later determined to be technetium, promethium, hafnium and rhenium. With this prediction, Moseley was able to prove the separate existence of each element in the lanthanide series.

Henry G. J. Moseley (1887–1915)

Figure 4.2

Modern Periodic Table 2 In the modern Periodic Table, the elements are arranged in order of increasing proton number. This order is also related to the electron arrangement of the elements.

1 There are 117 discovered elements now. Most of these elements are naturally occurring elements. However, a few of these elements are made artificially in nuclear reactors. Periodic Table of Elements

70

(b) Periods 2 and 3 have eight elements each. The first three periods are called the short periods. (c) Periods 4 and 5 have 18 elements each. They are called the long periods. (d) Period 6 has 32 elements. Not all the elements can be listed on the same horizontal row. The elements with proton numbers 58 to 71 are separated and are grouped below the Periodic Table. These elements are called the Lanthanide Series. (e) Period 7 has 31 elements and not all can be listed on the same horizontal row. The elements with proton numbers 90 to 103 are grouped below the Periodic Table. They are called the Actinide Series. The Electron Arrangement of Elements in the Periodic Table 1 Table 4.4 shows the electron arrangement of the elements with proton numbers 1–20. 2 All members of the same group have the same number of valence electrons. Valence electrons are electrons in the outermost shell. For example: (a) Group 1 elements (Li, Na, K) each has one valence electron. (b) Group 2 elements (Be, Mg, Ca) each has two valence electrons. (c) Group 17 elements (F, Cl, Br, I, At) each has seven valence electrons. 3 (a) The number of valence electrons of Group 1 and Group 2 elements is the same as its group number. SPM

’11/P1, P2

Table 4.4 The electron arrangement of the first 20 elements in the Periodic Table

Group Period

1

2

13

14

15

16

17

18 He 2

1

H 1

2

Li 2.1

Be 2.2

B 2.3

C 2.4

N 2.5

O 2.6

F 2.7

Ne 2.8

3

Na 2.8.1

Mg 2.8.2

Al 2.8.3

Si 2.8.4

P 2.8.5

S 2.8.6

Cl 2.8.7

Ar 2.8.8

4

K 2.8.8.1

Ca 2.8.8.2 71

Periodic Table of Elements

4

3 The Periodic Table is a classification of elements whereby elements with the same chemical properties are placed in the same group. This makes the study of the chemistry of these elements easier and more systematic. 4 The modern Periodic Table is shown on page 78. 5 The vertical columns of the Periodic Table are called groups. There are 18 groups in the Periodic Table. 6 Each member of a group shows similar chemi­ cal properties although their physical properties such as density, melting point and colour may show a gradual change when descending the group. • Group 1 elements (Li, Na, K, Rb, Cs, Fr) are called alkali metals. • Group 2 elements (Be, Mg, Ca, Sr, Ba, Ra) are called alkaline earth metals. • Group 17 elements (F, Cl, Br, I, At) are called halogens. • Group 18 elements (He, Ne, Ar, Kr, Xe, Rn) are called noble gases. 7 A block of elements called the transition elements separates Group 2 and Group 13. (a) The elements in Groups 1, 2, 13 and the transition elements are metals. (b) The elements in Groups 15, 16 and 17 are non-metals. (c) Group 14 consists of non-metals at the top of the group, followed by semi-metals (or metalloids) and metals lower down in the group. 8 The horizontal rows are called periods. There are seven periods. (a) Period 1 has two elements only: hydrogen and helium.

4

3

(b) Except for helium, the elements with more than 2 valence electrons (Groups 13 to 18), the group number = 10 + (number of valence electrons). 4 The period number is indicated by the number SPM of filled electron shells. For example: ’05/P2 (a) Elements in Period 1 (H and He) each has only one electron shell filled with electrons. (b) Elements in Period 2 (Li, Be, B, C, N, O, F, Ne) each has two electron shells filled with electrons. (c) Elements in Period 3 (Na, Mg, Al, Si, P, S, Cl, Ar) each has three electron shells filled with electrons. (d) Elements in Period 4 (K and Ca) each has four electron shells filled with electrons. 5 All elements in the same period have the same number of filled electron shells.

1

The proton number of element X is 20. Which of the following statements are true concerning the element X? I X can conduct electricity. II X belongs to Group 2 of the Periodic Table. III X belongs to Period 4 of the Periodic Table. IV X belongs to Period 3 of the Periodic Table. A II and III only B II and IV only C I, II and III only D I, II and IV only Comment X has 20 electrons. The electronic configuration of X is 2.8.8.2. X belongs to Group 2 of the Periodic Table because it has two valence electrons. (II is correct) Group 2 elements are metals and can conduct electricity. (I is correct)

’05

X belongs to Period 4 of the Periodic Table because it has four electron shells filled with electrons. (III is correct, IV is incorrect) Answer C

Which of the following represents the electron arrangement of an element in Group 17? A

B

C

D

4 Element

Answer D (The element has seven valence electrons)

2

Electron arrangement ’03

R

S

2.8.6

2.8.8

2

Determine the group in which Q, R and S belong to in the Periodic Table.

In the Periodic Table, Y is below Z in the same group. If the proton number of atom Z is 11, what is the electron arrangement of atom Y ? A 2.2 C 2.8.3 B 2.7 D 2.8.8.1

Solution For elements with one or two valence electrons, Group number = Number of valence electrons

Comment In an atom, the number of protons is equal to the number of electrons. Thus Z has 11 electrons. The electron arrangement of Z is 2.8.1. The elements in the same group have the same number of valence electrons. Therefore the electron arrangement of Y is 2.8.8.1 because each atom of Y and Z has one valence electron. Answer D

Periodic Table of Elements

Q

For elements with more than two valence electrons, Group number = 10 + Number of valence electrons Therefore element Q belongs to Group 16 and element R belongs to Group 18 of the Periodic Table. S has two electrons in the first electron shell. It is helium and belongs to Group 18 of the Periodic Table.

72

1 Q

R

T

X

Z

3

15

18

19

35

Element Electron arrangement

The proton numbers of elements Q, R, T, X and Z are given in the table above. Which of the following statements are true? I Elements Q and X belong to the same group in the Periodic Table. II Elements T and X belong to the same period. III Elements Q and X are metals. IV Elements R and Z are non-metals. A I and III only B II and IV only C I, III and IV only D II, III and IV only

Q

R

2.1

T

X

Z

2.8.5 2.8.8 2.8.8.1 2.8.18.7

Comment Elements Q and X belong to the same group in the Periodic Table because they have the same number of valence electrons. (Statement I is correct) Element T belongs to Period 3 (it has 3 electron shells), whereas element X belongs to Period 4 (it has four electron shells). (Statement II is wrong) Elements in Groups 1, 2 and 13 have one, two and three valence electrons and they are metals. (Statement III is correct) Elements with five, six, seven or eight valence electrons are non-metals. (Statement IV is correct) Answer C

4.1 3 Arsenic is represented by the symbol 75 As. In which 33 group and period does arsenic belong to in the Periodic Table? Name two elements that have the same chemical properties as arsenic. Explain your answer.

1 Explain the term valence electrons. State the number of valence electrons of the elements in the following groups in the Periodic Table. (a) Group 1 (b) Group 2 (c) Group 15 (d) Group 17 2

Element

W

X

Y

4 The following table shows the proton numbers of 10 elements represented by the letters A to K: Element

Z

A

Proton number 2

Electron 2.8.5 2.1 2.8.8.3 2.8.18.32.18.7 arrangement

B

C D

E

F

G H

J

K

9 13 19 18 16 7 20 17 6

Electron arrangement

The table shows the electron arrangement of four elements. (a) State the group of each element: (i) W (ii) X (iii) Y (iv) Z

Group number Period number (a) Write the electron arrangement of each element. Then state the group number and period number of each element.

(b) State the period of each element: (i) W (ii) X (iii) Y (iv) Z

(b) Pick a pair of elements which have the same chemical properties.

73

Periodic Table of Elements

4

Element Proton number

’04

4.2

Group 18 Elements

’05

General He Ne Ar Kr Xe Rn

4

Figure 4.3 The elements of Group 18 in the Periodic Table

• The noble gases are also known as the inert gases. They are elements of Group 18, at the far right of the Periodic Table (Figure 4.3). • The group consists of six elements: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). • The atomic radius of the noble gases increases down the group. This is because as the number of filled electron shells increases down the group, the valence electron is further from the nucleus.

Physical Properties This is because as the size of the atoms increases Physical properties of noble gases: down the group, the van der Waals forces of • All noble gases are insoluble in water and do attraction become stronger. not conduct electricity or heat. • All noble gases have low melting points and • The noble gases have very low densities. This is low boiling points. This is because the noble because the atoms are very far apart. The density of the noble gases increases when descending gases exist as monatoms which are attracted by the group. This is due to the increase in the very weak van der Waals forces of attraction. The melting points and boiling points of the relative atomic mass of the element. noble gases increase when descending the group. Chemical properties stable duplet electron arrangement. Chemical properties of noble gases: – The other noble gases have eight electrons in • Noble gases are chemically inert which means their outermost electron shells. They are said that they are unreactive in nature. They do not to have attained the stable octet electron react with any other elements. arrangement. • Noble gases are the only elements which exists • Therefore the noble gases do not need to as monatoms (as single atoms). accept, donate or share electrons with other • Noble gases are unreactive because they all elements. have filled outer shells of electrons which are a stable electron arrangement. (All chemical reactions involve either gaining, – Helium has only one filled electron shell and losing or sharing electrons.) it has exactly two electrons in this outermost electron shell. It is said to have attained the Table 4.5 The proton numbers, relative atomic mass, electron arrangement and physical properties of inert gases

Element He Ne Ar Kr Xe Rn

Proton Relative atomic Electron number arrangement mass 2 10 18 36 54 86

Periodic Table of Elements

4 20 40 84 131 222

2 2.8 2.8.8 2.8.18.8 2.8.18.18.8 2.8.18.32.18.8 74

Melting Atomic radius (nm) point (°C) 0.06 0.07 0.094 0.109 0.130 –

–270 –248 –189 –156 –112 –71

Boiling point (°C)

Density (g dm–3 )

–269 –246 –186 –152 –107 –62

0.17 0.84 1.66 3.54 5.45 –

4.2

’04

1 2P, 11Q, 13R, 18S, 20T. The above is a set of elements with their proton numbers. Choose two elements that exist as monatomic gases. Explain your answer.

A car distributor wants to use colourful electric lamps to attract customers. Which of the following substances A, B, C or D in the Periodic Table is suitable for use in the lamps?

2 Inert gases are the elements of Group 18 of the Periodic Table. Name two inert gases and state one use of each. 3 All the inert gases have low melting and boiling points. Explain why as we go down Group 18, the melting points and boiling points increase.

Comment B is neon gas. Neon-filled electric bulbs produce an attractive bright red light which is used in advertising. Answer B

4.3

4 One of the characteristics of Group 18 gases is that they exist as monatoms. Explain why neon does not form compounds with other elements.

Group 1 Elements General

Figure 4.4 The elements of Group 1 in the Periodic Table

• Group 1 elements are also known as the alkali metals. • The elements in Group 1 are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs) and francium (Fr). • All Group 1 elements are soft metals. The metals are grey in colour and are silvery when the surface is freshly cut, before being exposed to air.

Physical Properties Physical properties of Group 1 elements: between the atoms becomes weaker down the • All Group 1 elements are metals; hence they group as the atomic radius increases. are good conductors of heat and electricity. • The electropositivity of the metals increases • The atomic radius increases down the group. down the group. Electropositivity is a measurement The reason is that as the number of filled of the ability of an atom to lose an electron electron shells increases when descending and form a positive ion. the group, the distance between the outermost M → M+ + e– (M = Li, Na, K, Rb, Cs, Fr) electron shell and the nucleus increases. • The density increases down the group. The densities of Li, Na and K is lower than that As the atomic radius becomes larger down of water, and hence these metals can float on the group, the force of attraction between water. the nucleus and the single valence electron • The melting point decreases when descending becomes weaker. Hence the elements lose the group. This is because the metallic bond the single valence electron more easily when descending the group.

75

Periodic Table of Elements

4

2

SPM

4

Reactivity

’07/P2, ’11/P2

Reactivity of Group 1 elements M(s) → M+ + e– (M = Li, Na, K, Rb, Cs, Fr) • All Group 1 elements are very reactive. The Li(s) → Li+ + e– reactivity increases down the group. 2.1 2 • The elements in Group 1 have one valence Na(s) → Na+ + e– electron each. During a chemical reaction, reactivity increases 2.8.1 2.8 an atom of a Group 1 element will donate a valence electron to form a univalent positive K(s) → K+ + e– ion to attain the stable duplet or octet in its 2.8.8.1 2.8.8 electron arrangement. • The reactivity of Group 1 elements depends on how easily it can donate its valence electron. The atomic radius of Group 1 elements increases down the group. This causes the force of attraction between the nucleus and the valence electron to become weaker, making it easier for a metal lower in the group to donate its valence electron. Therefore, the reactivity of Group 1 elements increases when descending the group. Chemical Properties Chemical properties of Group 1 elements • The elements in Group 1 have the same chemical properties because each has one valence electron. • Group 1 elements react with

– cold water to produce hydrogen gas and alkalis, – oxygen to produce metal oxides, – halogen to produce metal halides.

Table 4.6 Some physical properties of Group 1 elements

SPM

’06/P2, ’08/P1

Element

Colour

Electron arrangement

Atomic radius (nm)

Density (g dm–3)

Li

Silvery

2.1

0.15

0.53

Conductor

181

Na

Silvery

2.8.1

0.16

0.97

Conductor

98

K

Silvery

2.8.8.1

0.23

0.86

Conductor

63

Ru

Silvery

2.8.18.8.1

0.25

1.53

Conductor

39

Cs

Silvery

2.8.18.18.8.1

0.26

1.87

Conductor

29

Fr

Silvery

2.8.18.32.18.8.1

0.29

–

Conductor

27

Electrical Melting Electro­positivity conductivity point (°C)

⏐ ⏐ ⏐ Increases ⏐ ⏐ ⏐ ↓

4.1

Experiment 4.1

To study the reaction of alkali metals with oxygen Problem statement How do lithium, sodium and potassium differ in reactivity with oxygen?

Variables (a) Manipulated variable : The alkali metals used (b) Responding variable : Reactivity of the metals with oxygen (c) Constant variable : Excess supply of oxygen and the size of the metal piece used Materials Small pieces of Li, Na and K metals, oxygen gas, filter paper and phenolphthalein indicator.

Hypothesis The alkali metals show similar chemical properties in their reactions with oxygen but the reactivity of the alkali metals with oxygen increases down the group (in the order from lithium, sodium to potassium).

Periodic Table of Elements

76

3 The lithium metal is heated until its starts to burn. The spoon is then put into a gas jar containing oxygen gas. 4 The observation is recorded. 5 After the reaction has stopped, about 20 cm3 of water is added to the gas jar. The gas jar is shaken so that the product of the combustion is dissolved in the water. The solution is tested with a few drops of phenolphthalein indicator. The observation is recorded. 6 The experiment is repeated using sodium and potassium metals.

Apparatus Pen knife, tongs, gas jar with cover, gas jar spoon and Bunsen burner. Safety precautions Alkali metals, especially sodium and potassium, are very reactive. Therefore we have to (a) use small pieces of each metal. (b) use goggles while carrying out the experiment. (c) ensure that we do not handle the metal with our bare hands.

4

Procedure 1 A piece of lithium metal is removed from the bottle with tongs. A small piece of the metal is cut using a pen knife. A piece of filter paper is used to absorb the paraffin oil from the piece of metal. 2 The lithium metal is then transferred onto a gas jar spoon using tongs.

Figure 4.5 Reaction of Group 1 metals with oxygen

Results Observation

Metal

Lithium

The lithium metal burns with a red flame forming a white metal oxide. The metal oxide dissolves in water producing a solution which turns phenolphthalein indicator red.

Sodium

The sodium metal burns with a bright yellow flame forming a white metal oxide. The metal oxide dissolves in water producing a solution which turns phenolphthalein indicator red.

Potassium

The potassium metal burns with a very bright purplish flame forming a white metal oxide. The metal oxide dissolves in water producing a solution which turns phenolphthalein indicator red.

Discussion 1 Alkali metals burn in oxygen to form white metal oxides 4Li(s) + O2(g) → 2Li2O(s) 4Na(s) + O2(g) → 2Na2O(s) 4K(s) + O2(g) → 2K2O(s)

Reactivity increases down the group

⎯⎯→

SPM

’11/P1

The presence of hydroxide ions, OH– causes the solution to be alkaline. Conclusion All alkali metals burn in oxygen gas to produce white metal oxides. Group 1 elements show similar chemical properties. Their reactivity increases down the group from lithium, sodium to potassium.

2 These metal oxides dissolve in water to form alkaline solutions which turn phenolphthalein indicator red.

4M(s) + O2(g) → 2M2O(s),

Li2O(s) + 2H2O(l) → 2LiOH(aq) Na2O(s) + 2H2O(l) → 2NaOH(aq) K2O(s) + 2H2O(l) → 2KOH(aq)

where M = Group 1 element.

77

Periodic Table of Elements

Periodic Table of Elements Periodic Table of Elements

4

Periodic Table of Elements

Periodic Table of Elements

78

SPM

4.2

’05/P2 ’08/P3 ’09/P1 ’10/P3

To study the reaction of alkali metals with water Problem statement How do lithium, sodium and potassium differ in reactivity with water?

Figure 4.8 Reaction of potassium with water

Hypothesis The alkali metals show similar chemical properties in their reactions with water but the reactivity of alkali metals with water increases down the group (in the order from lithium, sodium to potassium).

Results Metal

Observation

The lithium metal moves slowly on the surface of the water (Figure 4.6). A colourless solution is produced. This solution turns red litmus paper blue. Sodium The sodium metal moves at a fast speed on the surface of the water with a ‘hissing’ sound. It is ignited during the reaction and burns with a yellow flame (Figure 4.7). A colourless solution is obtained. This solution turns red litmus paper blue. Potassium The potassium metal moves at a very fast speed on the surface of the water. It is ignited during the reaction and burns with a purple flame with a ‘pop’ sound (Figure 4.8). A colourless solution is obtained. This solution turns red litmus paper blue.

Variables (a) Manipulated variable : The alkali metals used (b) Responding variable : Reactivity of the metal with water (c) Constant variable : Size of alkali metal and the temperature of water Materials Small pieces of lithium, sodium and potassium metals, basin filled with water, filter paper and red litmus paper. Apparatus Pen knife and tongs. Procedure 1 A piece of lithium metal is removed from the bottle. A small piece of the metal is cut using a pen knife. A piece of filter paper is used to absorb the paraffin oil from the piece of metal. 2 The lithium metal is then dropped into a basin of water carefully using a pair of tongs. 3 The observation is recorded. 4 The solution formed in the basin is tested with a piece of red litmus paper. 5 The experiment is repeated using small pieces of sodium and potassium metals.

4

Lithium

2 The metal hydroxides are alkaline and turn the colour of red litmus paper blue. Conclusion 1 Elements of Group 1 have similar chemical properties. All of them react with cold water to produce an alkaline solution and hydrogen gas. 2M(s) + 2H2O(l) → 2MOH(aq) + H2(g), where M = Group 1 element. 2 The reactivity increases down the group from lithium, sodium to potassium.

Figure 4.6 Reaction of lithium with water

Figure 4.7 Reaction of sodium with water

79

Periodic Table of Elements

Experiment 4.2

Discussion 1 Alkali metals are reactive with water and can SPM displace hydrogen from water. They react with ’11/P2 water to produce hydrogen gas and aqueous solutions of metal hydroxides. 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) Reactivity 2Na(s) + 2H2O(l) → increases down 2NaOH(aq) + H2(g) the group 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)

react vigorous­ly with water forming hydrogen gas and alkali solutions. If the size of the metal is quite large, it will ignite and start to burn giving out ‘pop’ sounds. When carrying out an experiment on the reaction of alkali metals with water, we must ensure that no flam­mable organic solvents are nearby because the fire from the burning alkali metals can spread to the organic solvents. Therefore alkali metals must be kept in paraffin oil in bottles to prevent them from reacting with water.

When the Group 1 elements react with water they give off hydrogen gas. The heat generated by the chemical reaction sets the hydrogen gas alight and it burns with a coloured flame. Sodium burns with a yellow flame.

Safety Precaution to be Taken When Handling Alkali Metals

4

Alkali metals are extremely reactive. With the exception of lithium, the rest of the alkali metals

4.3

SPM

’09/P1

To study the reaction of alkali metals with halogens Results

Problem statement How do lithium, sodium and potassium differ in reactivity with chlorine gas?

Metal

Hypothesis The alkali metals show similar chemical properties in their reactions with chlorine but the reactivity of alkali metals with chlorine increases down the group (in the order from lithium, sodium and potassium). Variables (a) Manipulated variable : The alkali metals used (b) Responding variable : Reactivity of the metal with chlorine (c) Constant variable : Supply of chlorine gas and size of metal pieces used

Lithium burns slowly with a reddish flame. A white solid is obtained.

Sodium

Sodium burns brightly with a yellowish flame. A white solid is obtained.

Potassium

Potassium burns very brightly with a purplish flame. A white solid is obtained.

2Li(s) + Cl2(g) → 2LiCl(s) 2Na(s) + Cl2(g) → 2NaCl(s) 2K(s) + Cl2(g) → 2KCl(s)

Procedure 1 A small piece of lithium is cut and the paraffin oil on it is blotted using filter paper. 2 The lithium metal is then transferred onto a gas jar spoon using tongs. 3 The lithium metal is heated until it starts to burn. The spoon is then put into a gas jar containing chlorine gas. 4 The observation is recorded. 5 The experiment is repeated using sodium and potassium metals.

⎯⎯⎯→

Apparatus Pen knife, tongs, gas jar with cover, gas jar spoon and Bunsen burner.

Reactivity increases down the group

2 If the experiments are repeated using bromine gas, the brown colour of bromine gas will disappear and white metal bromides will be formed. 2Li(s) + Br2(g) → 2LiBr(s) 2Na(s) + Br2(g) → 2NaBr(s) 2K(s) + Br2(g) → 2KBr(s)

⎯⎯→

Experiment 4.3

Lithium

Discussion 1 All alkali metals react with chlorine gas to form white metal chlorides The reactivity of the metal increases down the group from Li, Na to K.

Materials Small pieces of lithium, sodium and potassium metals chlorine gas and filter paper.

Periodic Table of Elements

Observation

Reactivity increases down the group

3 Similarly, Group 1 elements will react with iodine vapour to produce white metal iodides. The purple colour of iodine vapour will disappear in the reactions. 80

4 Generally, all alkali metals react with halogens to produce metal halides. 2M(s) + X2(g) → 2MX(s) where M = Group 1 elements, X = halogen.

2M(s) + Cl2(g) → 2MCl(s) where M = Group 1 elements 2 The reactivity increases down the group from lithium, sodium to potassium.

Conclusion 1 Elements of Group 1 have similar chemical properties. All react with chlorine to produce white metal chlorides.

’05

An atom of element X has three electron filled shells. When element X reacts with chlorine, it forms a compound with the formula XCl. Which of the following is element X? [Proton number of Li, 3; Na, 11; Mg, 12; K, 19] A Lithium C Magnesium B Sodium D Potassium

Element

Electron arrangement

X

2.8.1

Y

2.8.8.1

The atomic radius of Y is larger than X. Hence Y can donate its valence electron more easily than X. Thus Y is more reactive. (I is correct) Both X and Y have one valence electron. Both are alkali metals which can conduct electricity. (II and III correct) Alkali metals react with water to form metal hydroxides and hydrogen gas. (IV is incorrect) Answer A

Comment Alkali metals react with chlorine to form metal chloride with formula XCl. X is sodium because sodium has three electron shells. Answer B

4.3

4

’03

1 Imagine that a new element is discovered. It is named Pentium (Pn) and is below sodium metal in Group 1 of the Periodic Table. (a) Predict three physical and three chemical properties of pentium. (b) Which is more reactive, pentium or sodium? Explain your answer.

The table below shows the proton numbers of two elements: Element

Proton number

X

11

Y

19

2 List the alkali metals in order of decreasing reactivity. 3 How do each of the following properties of alkali metals change as we go down the group? (a) Melting point (b) Density (c) Hardness (d) Chemical reactivity

Which of the following statements are true? I Y is more reactive than X. II Both X and Y can conduct electricity. III Both X and Y are in the same group of the Periodic Table. IV Both X and Y react with water to form metal oxides and hydrogen gas. A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV

4 When a piece of burning sodium metal is placed in a gas jar containing bromine gas, the brown colour of bromine gas will disappear and white powder will be formed. Explain the above observation with a suitable equation.

81

Periodic Table of Elements

4

Comment

3

4

4.4

SPM

Group 17 Elements

’08/P2

General

Physical properties

• The elements in Group 17 are also known as the halogens. • The elements in Group 17 are fluorine (F), chlorine (Cl), bromine (Br), iodine (I) and astatine (At). • Halogens are very reactive elements and most of them exist naturally as halide salts. • The halogen molecules exist as diatomic molecules: F2, Cl2, Br2, I2 and At2.

Physical properties of the halogens • All Group 17 elements are non-metals. Hence, they are non-conductors of heat and electricity. • The atomic radius increases down the group. The reason is that as the number of filled electron shells increases down the group, the distance between the outermost electron shell and the nucleus increases. • The density increases down the group. This is due to the increase in relative molecular mass. • Halogens have low boiling points. The forces of attraction between the molecules are weak. • The melting points and boiling points of the halogens increase down the group. This is because the molecular size increases down the group. As the size increases, the van der Waals forces of attraction between the molecules become stronger. More heat is required to overcome the forces of attraction and therefore the melting points and boiling points increase. The first two elements (fluorine and chlorine) are gases at room temperature. Bromine is a liquid whereas iodine and astatine are solids at room temperature. • The colour of the halogen becomes darker down the group. Fluorine is a colourless gas; chlorine is a yellowish-green gas; bromine is a dark brown liquid and iodine is a black solid. • All halogens have high electro­negativities. They are electronegative non-metals. Electronegativity is a measurement of the tendency of an element to attract electrons. The electronegativity decreases down the group from fluorine to iodine. As the atomic radius becomes larger down the group, the force of attrac­tion between the nucleus and the electrons becomes weaker and thus electronegativity decreases.

Figure 4.9 The elements of Group 17 in the Periodic Table

Halothane (CHClBrCF3) is used as a general anaesthetic during a major operation. Can you name the halogens present in this compound? Answer Bromine, chlorine and fluorine

Chemical properties Chemical properties of Group 17 elements • The elements in Group 17 have the same chemical properties because each has seven valence electrons. • Group 17 elements react with – water to produce acids, – metals such as iron to produce metal halides, – sodium hydroxide to produce salts and water. Periodic Table of Elements

82

SPM

Reactivity

Reactivity of Group 17 elements X2 + 2e– → 2X–, where X = F, Cl, Br or I • All Group 17 elements are very reactive. However, the Cl2 + 2e– → 2Cl– reactivity decreases down the group. 2.8.7 2.8.8 • The elements in Group 17 have seven valence electrons – each. During a chemical reaction, the atom of a Br2 + 2e → 2Br– Group 17 element will accept a valence electron to 2.8.18.7 2.8.18.8 form univalent negative ion to attain the stable octet – I2 + 2e → 2I– in its electron arrangement. 2.8.18.18.7 2.8.18.18.8 • The reactivity of Group 17 elements depends on its ability to gain an electron. The atomic radius of Group 17 elements increases down the group. Thus the force of attraction between the nucleus and the valence electrons become weaker. As a result, the halogen lower in the group has a lower tendency to attract an electron to form a negative ion. Therefore, the reactivity of halogens decreases down the group. Table 4.7 Some physical properties of three halogens

2.8.7

Atomic radius (nm) 0.099

Melting point (°C) –101

Boiling point (°C) –35

Physical state at room temperature Gas

2.8.18.7 2.8.18.18.7

0.114 0.133

–7 114

58 183

Liquid Solid

Halogen

Proton number

Electron arrangement

Chlorine

17

Bromine Iodine

35 53

4

’06/P2, ’08/P2

SPM

’08/P2

Electro­ negativity 3.0 2.8 2.5

Colour Yellowishgreen gas Brown liquid Black solid

4.4 To study the reactions of chlorine, bromine and iodine with water Materials Chlorine gas, liquid bromine, iodine crystals and blue litmus paper.

Problem statement How do chlorine, bromine and iodine react with water? Hypothesis The halogens show similar chemical properties when they react with water but the reactivity decreases down the group from chlorine to iodine.

Apparatus Test tube, rubber stopper, test tube holder, delivery tube and teat pipette.

Variables (a) Manipulated variable : Types of halogens used (b) Responding variable : The rate at which the halogen dissolves in water and products of reactions (c) Constant variable : Temperature of water

Figure 4.10 Reaction of halogen with water

(a)

(b)

(c)

83

Periodic Table of Elements

Experiment 4.4

Safety precautions (a) Chlorine gas and bromine vapour are poisonous. The experiments should be carried out in a fume cupboard. (b) The chlorine gas and bromine vapour irritate the eyes. So goggles should be worn while carrying out the experiments.

4

Procedure (A) Reaction of chlorine with water 1 Chlorine gas is passed into a test tube containing water. 2 The solution produced is tested with blue litmus paper. (B) Reaction of bromine with water 1 A few drops of liquid bromine are added to some water in a test tube. Results

2 The test tube is tightly closed with a rubber stopper and then shaken. 3 The solution produced is tested with blue litmus paper. (C) Reaction of iodine with water 1 Some iodine crystals are added to some water in a test tube. 2 The test tube is tightly closed with a rubber stopper and then shaken. 3 The solution produced is tested with blue litmus paper.

SPM

’11/P1

Observation

Halogen

Solubility

Effect on litmus paper

Chlorine

Dissolves quickly in the water to form a light yellowish solution

The solution first turns the blue litmus paper red, then it quickly decolourises it.

Bromine

Dissolves slowly in water to form a brown solution

The solution first turns the blue litmus paper red. The red colour of the litmus takes a longer time to be decolourised.

Iodine

A little of the iodine crystals dissolves slightly in water to form a pale brown solution

The solution turns the litmus paper from blue to red. The red litmus paper is not decolourised.

Discussion 1 Chlorine, bromine and iodine dissolve in water to form acidic solutions which turn blue litmus paper red. The solubility of halogens decreases from chlorine to iodine. 2 Chlorine dissolves in water to form hydrochloric acid and hypochlorous(I) acid.

Hypobromous(I) acid is a weak bleaching agent and takes a longer time to decolourise the red colour of litmus paper. 4 Iodine is only very slightly soluble in water. It forms hydroiodic acid and hypoiodous(I) acid. I2(s) + H2O(l) → HI(aq) + HOI(aq) hydroiodic hypoiodous(I) acid acid

Cl2(g) + H2O(l) → HCl(aq) + HOCl(aq) hydrochloric hypochlorous(I) acid acid

Hydroiodic acid, HI is an acid and it turns blue litmus red. Hypoiodous(I) acid has a very weak bleaching property.

Hydrochloric acid, HCl is an acid and it turns blue litmus red. Hypochlorous(I) acid, HOCl is a strong bleaching agent. It decolourises the red colour of litmus paper quickly. 3 Bromine dissolves slowly in water to form hydrobromic acid and hypobromous(I) acid.

Conclusion 1 Chlorine, bromine and iodine show similar chemical properties. They dissolve in water to form acidic solutions. 2 The solubility of halogens in water decreases down the group. 3 Aqueous chlorine and bromine solutions have bleaching properties. Aqueous iodine solution does not bleach the colour of litmus paper.

Br2(l) + H2O(l) → HBr(aq) + HOBr(aq) hydrobromic hypobromous(I) acid acid Hydrobromic acid, HBr is an acid and it turns blue litmus red.

Periodic Table of Elements

84

4.5 SPM

To study the reactions of halogens with aqueous sodium hydroxide solution

’06/P2

Hypothesis The halogens show similar chemical properties when they react with sodium hydroxide solution but the reac­tivity decreases down the group from chlorine to iodine. Variables (a) Manipulated variable : Types of halogens used (b) Responding variable : The products of the reactions (c) Constant variable : Concentration of sodium hydroxide solution

Halogen

Observation

Chlorine

The greenish chlorine gas dissolves quickly in NaOH solution to form a colourless solution.

Bromine

The brownish liquid bromine dissolves steadily in NaOH solution to form a colourless solution.

Iodine

The dark iodine crystal dissolves slowly in NaOH solution to form a colourless solution.

4

Results

Problem statement How do chlorine, bromine and iodine react with aqueous sodium hydroxide solution?

Discussion 1 Chlorine gas reacts rapidly with sodium hydroxide solution to produce sodium chloride salt, sodium chlorate(I) salt and water. Cl2(g) + 2NaOH(aq) → NaOCl(aq) + NaCl(aq) + H2O(l)

Materials Chlorine gas, liquid bromine, iodine crystals and sodium hydroxide solution.

sodium sodium chlorate(I) chloride

Apparatus Test tube, rubber stopper, test tube holder and teat pipette.

2 Bromine reacts moderately fast with sodium hydroxide solution to produce sodium bromide salt, sodium bromate(I) salt and water. Br2(l) + 2NaOH(aq) → NaOBr(aq) + NaBr(aq) + H2O(l)

Procedure (A) Reaction of chlorine with aqueous sodium hydroxide solution 1 Chlorine gas is bubbled into aqueous sodium hydroxide solution. 2 The colour change of chlorine is recorded.

sodium bromate(I) sodium bromide

3 Solid iodine reacts slowly with sodium hydroxide solution to produce the salts sodium iodide, sodium iodate(I) and water. I2(s) + 2NaOH(aq) → NaOI(aq) + NaI(aq) + H2O(l)

(B) Reaction of bromine with aqueous sodium hydroxide solution 1 Two drops of liquid bromine are added to aqueous sodium hydroxide solution using a teat pipette. 2 The test tube is tightly closed with a rubber stopper and the mixture is shaken. 3 The colour change of bromine is recorded.

sodium iodate(I) sodium iodide

Conclusion 1 Chlorine, bromine and iodine react with sodium hydroxide solution to form two types of salts and water.

(C) Reaction of iodine with aqueous sodium hydroxide solution 1 Some iodine crystals are added to aqueous sodium hydroxide solution. 2 The test tube is tightly closed with a rubber stopper and the mixture is shaken. 3 The colour change of iodine crystal is recorded.

X2(g) + 2NaOH(aq) → NaX(aq) + NaOX(aq) + H2O(l), where X = Cl, Br, I 2 The reactivity of halogens with sodium hydroxide solution decreases down the group from chlorine to iodine. 85

Periodic Table of Elements

Experiment 4.5

[Sodium chlorate(I), sodium bromate(I), sodium iodate(I) are also called sodium hypochlorite, sodium hypobromite and sodium hypoiodite respectively]

4.6 SPM

To study the reactions of halogens with iron ’06/P2 Problem statement How do chlorine, bromine and iodine react with iron?

(B) Reaction of bromine gas with iron wool

4

Hypothesis The halogens show similar chemical properties when they react with iron but the reactivity decreases down the group from chlorine to iodine.

Figure 4.12 Reaction of bromine with iron wool

Variables (a) Manipulated variable : Types of halogen used (b) Responding variable : Products of reactions and rate of the reactions (c) Constant variable : Iron wool

1 A small roll of iron wool is placed in the middle of a combustion tube and is heated strongly. 2 The liquid bromine is warmed up by using a Bunsen burner. 3 The bromine is vaporised and bromine gas passed through the heated iron wool. 4 The excess bromine gas is absorbed by the soda lime.

Materials Chlorine gas, liquid bromine, iodine crystals, soda lime, potassium manganate(VII), concentrated hydro­chloric acid and iron wool.

(C) Reaction of iodine with iron wool

Apparatus Combustion tubes, Bunsen burner, retort stand and clamp, conical flask and thistle funnel. Procedure (A) Reaction of chlorine gas with iron wool 1 A small roll of iron wool is placed in the middle of a combustion tube. The iron wool is then heated strongly. 2 Chlorine gas is prepared in the laboratory by adding concentrated hydrochloric acid to potassium manganate(VII). 3 The chlorine gas produced is allowed to pass through the heated iron wool. 4 The excess chlorine gas is absorbed by the soda lime.

Figure 4.13 Reaction of iodine with iron wool

1 A few crystals of iodine are placed in a boiling tube. 2 A small roll of iron wool is then placed in the middle of a combustion tube. 3 The iron wool is heated strongly first, followed by the iodine crystals (sublimation will take place). 4 The iodine vapour produced is allowed to pass through the hot iron wool. Results

Experiment 4.6

Halogen

Chlorine Hot iron wool glows brightly when chlorine gas is passed over it. A brown solid is formed. Bromine Hot iron wool glows moderately bright when bromine gas is passed over it. A brown solid is formed. Iodine Hot iron wool glows dimly when iodine vapour is passed over it. A brown solid is formed.

Figure 4.11 Reaction of chlorine with iron wool

Periodic Table of Elements

Observation

86

Discussion 1 The halogens react with hot iron wool to form iron(III) halides which are brown in colour but the reactivity of the halogen decreases down the group. reactivity decreases down the group

Conclusion 1 Chlorine, bromine and iodine show the same chemical properties when they react with iron wool, producing brown iron(III) halides. 2 The reactivity of the halogen decreases down the group from chlorine to iodine.

2 The reaction between concentrated hydrochloric acid and potassium manganate(VII) produces chlorine gas:

X and Y reacts with water to form two kinds of acids. (III is correct)

Safety Precaution to be Taken When Handling Group 17 Elements

X2 + H2O → HX + HOX

1 Fluorine, chlorine and bromine gases are very poisonous. In fact chlorine was used in the First World War to kill people. 2 Hence the experiments which involve the use of these gases should be carried out in a fume cupboard.

5

Answer C

4.4 1 Explain why a solution of chlorine is (a) acidic (b) able to bleach things.

’05

2 Aqueous bromine and iodine solutions are both brown. (a) How do you differentiate between the two solutions? (b) Carry out an experiment to show that bromine is more reactive than iodine.

X Y

Which of the following statements below are true concerning the elements X and Y in the above Periodic Table? I Y is more reactive than X. II X is more electronegative than Y. III They react with water to produce two kinds of acids. IV Both form univalent negative ions of charge –1 when reacted with sodium. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV

3 Iodine is an element below chlorine in Group 17 of the Periodic Table. (a) Does iodine show similar chemical properties as chlorine? Explain your answer. [Proton number of iodine is 53 and proton number of chlorine is 17] (b) How does the (i) density (ii) melting point of iodine compare to chlorine? (c) Write equations for the reaction of iodine with (i) aqueous sodium hydroxide solution (ii) iron wool

Comment X has a smaller atomic radius. Thus it has a higher tendency to accept an electron to form a univalent negative ion of charge –1. Thus X is more reactive and more electronegative. (I is incorrect)

4 Explain why the reactivity of Group 17 elements decreases down the group. 5 Name five compounds containing halogens and state their uses.

87

Periodic Table of Elements

4

2Fe(s) + 3Br2(g) → 2FeBr3(s) iron(III) bromide 2Fe(s) + 3I2(g) → 2FeI3(s) iron(III) iodide

3 Soda lime is a mixture of calcium hydroxide and sodium hydroxide. It is used to absorb the excess halogen gas. The excess chlorine and bromine gas have to be absorbed because they are poisonous.

⎯⎯⎯⎯⎯⎯⎯→

2Fe(s) + 3Cl2(g) → 2FeCl3(s) iron(III) chloride

2KMnO4(s) + 16HCl(aq) → 2KCl(aq) + 2MnCl2(aq) + 5Cl2(g) + 8H2O(l)

4

Modern Periodic Table (a) Elements are arranged in order of increasing proton numbers. (b) The vertical column is known as Group whereas the horizontal rows are called Periods. (c) The number of valence electrons in the element corresponds to the group the element is in. (d) The number of filled electron shells of an element corresponds to the period of the element.

Group 18 elements (a) Group 18 elements are inert. (b) They have attained the duplet or octet electronic configuration. (c) Thus they do not need to share, donate or receive electrons from other elements.

Group 1 elements (Alkali metals) (a) As we go down Group 1, (i) melting point decreases, (ii) density increases, (iii) reactivity and electropositivity increases. (b) Chemical properties of Group 1 elements: (i) Alkali metals react with chlorine to form metal halide salts. 2M(s) + Cl2(g) → 2MCl(s) (M = Li, Na, K, …) (ii) Alkali metals react with water to form metal hydroxides and hydrogen gas. 2M(s) + 2H2O(l) → 2MOH(aq) + H2(g) (M = Li, Na, K, …) (iii) Alkali metals burn in air to form metal oxides which are soluble in water. 4M(s) + O2(g) → 2M2O(s) (M = Li, Na, K, …) M2O(s) + H2O(l) → 2MOH(aq)

Group 17 elements (Halogen) (a) As we go down group 17, (i) melting point and density increases, (ii) reactivity and electronegativity decreases. (b) Chemical properties of Group 17 elements: (i) Halogens dissolve in water to form two types of acids. X2(g) + H2O(l) → HX(aq) + HOX(aq)

(X = F, Cl, Br, …)

(ii) Halogens react with iron to form brown iron(III) halide salts. 2Fe(s) + 3X2(g) → 2FeX3(s)

(X = F, Cl, Br, …)

(iii) Halogens react with sodium hydroxide to form two types of salts and water. 2NaOH(aq) + X2(g) → NaX(aq) + NaOX(aq) + H2O(l) (X = F, Cl, Br, …)

Periodic Table of Elements

88

period, other than Period 3, based on the changes in the properties of elements in Period 3. 4 Table 4.8 shows the trends across Period 3.

Elements in a Period

Elements in Period 3 1 The elements in Period 3 are sodium (Na), magnesium (Mg), aluminium (Al), silicon (Si), phosphorus (P), sulphur (S), chlorine (Cl), and argon (Ar). 2 The study of the elements in Period 3 will show a gradual change of physical and chemical properties across the period from left to right. 3 Period 3 is a typical period. Thus we can predict the trend of changes in properties across a

Figure 4.14 The elements in Period 3 of the Periodic Table SPM

Table 4.8 The trends across Period 3

Group

’11/P1

1

2

13

14

15

16

17

18

Element

Na

Mg

Al

Si

P

S

Cl

Ar

Proton number

11

12

13

14

15

16

17

18

2.8.1

2.8.2

2.8.3

2.8.4

2.8.5

2.8.6

2.8.7

2.8.8

1

2

3

4

5

6

7

8

0.156

0.136

0.125

0.117

0.111

0.104

0.099

0.094

Electro­negativity

0.9

1.2

1.5

1.8

2.1

2.5

3.0

—

Melting point (°C)

98

649

660

1410

590

119

–101

–189

Boiling point (°C)

883

1107

2467

2355

Ignites

445

–35

–186

Nature of elements Metal

Metal

Metal

Formula of oxide

Na2O

MgO

Al2O3

SiO2

P4O10

SO2

Cl2O7

None

Character of oxide

Basic

Basic

Ampho­teric

Acidic

Acidic

Acidic

Acidic

—

Electron arrangement Number of valence electrons Atomic radius (nm)

Metalloid Non-metal Non-metal Non-metal Non-metal

Trends of Changes across Period 3 SPM

’05/P2 ’09/P2

Atomic Radius

Valence Electrons

The atomic radius decreases across the period. • All the elements in Period 3 have three filled electron shells but the proton number increases by one unit across the period. • As a result, the increase in the number of protons increases the electrostatic force between the nucleus and the valence electrons. • The valence electrons are pulled closer to the nucleus, causing the atomic radius to decrease.

The number of valence electrons increases across the period. • As the proton number increases, the number of electrons increases. • The number of valence electrons increases by 1 from one element to the next across the period.

89

Periodic Table of Elements

4

4.5

4

Electronegativity

Melting & Boiling Points

The electronegativity increases across the period. Electronegativity is a measurement of the tendency of an atom to attract electrons. • The atomic radius decreases across the period. • The proton number increases across the period. • The increase in the number of protons (positive charge in the nucleus) and the decrease in the distance between the nucleus and the outermost electron shell across the period cause an increase in the force of attraction of the nucleus. The atoms will have a higher tendency to attract electrons. Therefore electronega­tivity increases. • Elements on the left side of the period tend to lose electrons to form positive ions. Elements on the right side of the period tend to gain electrons to form negative ions. • Thus the elements on the left side of the period (such as Na) are electropositive while the elements on the right side of the period (such as Cl) are electronegative.

The melting points and boiling points of the elements increase from the left of the period to the middle of the period and then decrease again. • Sodium, magnesium and aluminium are metals with strong metallic bonds between the metal atoms. Hence they have high melting and boiling points. The strength of the metallic bonds increase with the increase in the number of valence electrons in the order: Na < Mg < Al. • Silicon has very high melting and boiling points. It has strong covalent bonds between atoms forming a 3-dimensional gigantic network. • Phosphorus, sulphur, chlorine and argon are non-metals with weak van der Waals forces of attraction between molecules. They are lowest at the right with chlorine and argon existing as gases at room temperature.

Nature of Metals As the electronegativity of the elements increases, the elements change from metals to metalloid and finally to non-metals across the period. • The elements on the left of the period are metals (Na, Mg and Al). • Silicon has some metallic and some non-metallic properties. It is called a metalloid or semi-metal. • The elements on the right of the period are non-metals (P, S, Cl and Ar).

Nature of Oxides

SPM

’08/P1, ’10/P1

The oxides of the elements change from basic to • Acidic oxides react with alkali to form salts and amphoteric and then to acidic across the period. water. • Elements on the left of the period, which are For example, sulphur trioxide reacts with metals, form metal oxides. The metal oxides sodium hydroxide to form sodium sulphate are usually basic oxides. salt and water: • Basic oxides react with acids to form salts and water. SO3(g) + 2NaOH(aq) → Na2SO4(aq) + H2O(l) For example, magnesium oxide reacts with sulphuric acid to form a salt and water: • An amphoteric oxide can react with both acids and bases to form salt and water. Aluminium MgO(s) + H2SO4(aq) → MgSO4(aq) + H2O(l) oxide is an example of an amphoteric oxide. Aluminium oxide can react with both acids and • Elements on the right of the period are nonalkalis to form salts and water. metals. Non-metallic oxides are acidic oxides. • Argon as an inert gas, does not form oxide.

Periodic Table of Elements

90

4.7 To determine the properties of the oxides of elements in Period 3 Problem statement How do the properties of the oxides of elements in Period 3 change across the period? Hypothesis The oxides change from basic to amphoteric and then to acidic across Period 3. 4

Variables (a) Manipulated variable : Oxides of Period 3 (b) Responding variable : Reaction with acid or alkali (c) Constant variable : Concentrations of sodium hydroxide and nitric acid solutions

Figure 4.15 Reaction of oxides of Period 3 with (a) an acid, (b) an alkali

3 Two drops of universal indicator are added and the pH of the solution is noted. 4 The experiment is repeated with magnesium oxide, aluminium oxide, silicon(IV) oxide, phosphorus(V) oxide, sulphur dioxide and dichlorine heptoxide respectively in place of sodium oxide. (B) Reaction of the oxides of Period 3 elements with 2 mol dm–3 nitric acid and 2 mol dm–3 sodium hydroxide solutions 1 A little sodium oxide powder is put into two separate test tubes. 2 5 cm3 of nitric acid and 5 cm3 of sodium hydroxide are added separately to the contents in each test tube. 3 The contents in each test tube are heated slowly while being stirred with glass rods. 4 The solubility of sodium oxide in the two solutions is recorded. 5 The experiment is repeated with magnesium oxide, aluminium oxide, silicon(IV) oxide, phosphorus(V) oxide, sulphur dioxide and dichlorine heptoxide respectively in place of sodium oxide.

Materials Sodium oxide, magnesium oxide, aluminium oxide, silicon(IV) oxide, phosphorus(V) oxide, sulphur dioxide, dichlorine heptoxide, nitric acid solution and sodium hydroxide solution of 2 mol dm–3, distilled water and universal indicator. Apparatus Test tube, test tube holder, Bunsen burner, rubber stopper and glass rod. Procedure (A) Reaction of the oxides of Period 3 elements with water 1 A little sodium oxide powder is added to some distilled water in a test tube. 2 The test tube is tightly closed with a rubber stopper and the contents are shaken. Results Experiment A

Observation Solubility in water

pH value of solution 13 – 14

Al2O3

Dissolves in water to form a colourless solution Slightly soluble in water to form a colourless solution Insoluble in water

SiO2

Insoluble in water

Na2O MgO

Inference Solution obtained is a strong alkali. Sodium oxide is basic. Sodium shows metallic properties. Solution obtained is a weak alkali. Magnesium oxide is basic. Magnesium shows metallic properties. pH measured is pH of water as the oxide is insoluble in water. pH measured is pH of water as the oxide is insoluble in water.

8–9

7 7

91

Periodic Table of Elements

Experiment 4.7

Oxide

Oxide P4O10 SO2 Cl2O7

Observation Solubility in water

pH value of solution

Dissolves in water to form a colourless solution Dissolves in water to form a colourless solution Dissolves in water to form a colourless solution

2–3

Inference Solution obtained is acidic. P4O10 is acidic. Phosphorus shows non-metallic properties. Solution obtained is acidic. SO2 is acidic. Sulphur shows non-metallic properties. Solution obtained is a strong acid. Cl2O7 is acidic. Chlorine shows non-metallic properties.

2–3 1

4

Experiment B Oxide

Reaction with 2 mol dm–3 NaOH

Reaction with 2 mol dm–3 HNO3

Inference

Na2O

Does not react

Reacts with nitric acid to form a colourless solution

Sodium oxide is basic because it reacts with the acid. Sodium shows metallic properties.

MgO

Magnesium oxide does not dissolve

Dissolves in nitric acid forming a colourless solution

Magnesium oxide is basic because it reacts with the acid. Magnesium shows metallic properties.

Al2O3

Dissolves in sodium Dissolves in nitric acid hydroxide forming a forming a colourless solution colourless solution

SiO2

Reacts with sodium hydroxide solution

Does not dissolve in nitric acid

Silicon(IV) oxide is acidic because it reacts with an alkali. Silicon shows non-metallic properties.

P4O10

Reacts with sodium hydroxide solution

Does not react with nitric acid

Phosphorus(V) oxide is acidic because it reacts with an alkali. Phosphorus shows non-metallic properties.

SO2

Reacts with sodium hydroxide solution

Does not react with nitric acid

Sulphur dioxide is acidic because it reacts with an alkali. Sulphur shows non-metallic properties.

Cl2O7

Reacts with sodium hydroxide solution

Does not react with nitric acid

Dichlorine heptoxide is acidic because it reacts with an alkali. Chlorine shows non-metallic properties. SiO2(s) + 2NaOH(aq) → Na2SiO3(aq) + H2O(l)

Discussion 1 (a) Metallic oxides are basic. Metal oxides react with acids to form salts and water.

P4O10(s) + 12NaOH(aq) → 4Na3PO4(aq) + 6H2O(l) SO2(g) + 2NaOH(aq) → Na2SO3(aq) + H2O(l)

Na2O(s) + 2HNO3(aq) → 2NaNO3(aq) + H2O(l)

Conclusion 1 Across Period 3 from left to right, (a) the element changes from being a metal, to a metalloid and a non-metal. (b) the oxides change from being basic to amphoteric and acidic. 2 Silicon is classified as a metalloid because it is a very weak conductor of electricity. However, its oxide is acidic. 3 Aluminium is classified as a metal because it is a very good conductor of electricity. However, its oxide is amphoteric. Aluminium oxide shows both metallic and non-metallic properties.

MgO(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2O(l) (b) Aluminium oxide is amphoteric because it can react with both acids and alkalis.

SPM

’05/P2

Al2O3(s) + 6HNO3(aq) → 2Al(NO3)3(aq) + 3H2O(l) Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4(aq) (sodium aluminate)

Other examples of amphoteric oxides are lead(II) oxide and tin(II) oxide. (c) Non-metallic oxides are acidic. Non-metallic oxides react with alkalis to form salts and water. Periodic Table of Elements

Aluminium oxide is amphoteric because it reacts with both acid and alkali.

92

Uses of Semi-metals (or Metalloids) 1 A semi-metal or metalloid is an element with properties intermediate between those of metals and non-metals. 2 For example, silicon is a non-metal and is a very poor conductor of electricity. However, the conductivity increases with temperature. It becomes a good conductor of electricity at high temperatures. 3 This type of substance is known as a semi-metal. Examples of semi-metals are silicon, germanium, boron, antimony, and arsenic. These elements are important industrial materials and are used to make semiconductors. 4 Adding of foreign elements (called doping) can increase the conductivity of semi-metals. (a) If silicon is doped with Group 13 elements such as boron, a p-type semiconductor is produced. (b) If it is doped with Group 15 elements such as arsenic or antimony, a n-type semiconductor is produced. 5 Semiconductors are very important in the microelectronic industry and are used to make transistors, diodes, rectifiers, thermistors and microprocessors. Hundreds of these electronic components can be built onto a crystal of silicon to make a microchip.

Microchip wafer

4

Do you know that a silicon wafer can contain hundreds of microchips? Each microchip itself contains hundreds of electronic components. Our earth contains about 27.7% silicon and a large portion is found in sand.

4.6

Transition Elements

Figure 4.16 The transition elements in the Periodic Table

1 The transition elements are elements in a block located in-between Group 2 and Group 13 in the Periodic Table. 2 There are 10 elements in each series. The first series is in Period 4 and consists of the elements: scandium(Sc), titanium(Ti), vanadium(V), chromium(Cr), manganese (Mn), iron(Fe), cobalt(Co), nickel(Ni), copper(Cu) and zinc(Zn). 3 Table 4.9 shows some physical properties of the transition elements in the first series in Period 4. 4 The transition elements are all metals with the following physical properties: (a) High density (b) High hardness (c) High electrical conductivity (d) High tensile strength (e) Silvery surface (f) Ductile and malleable (g) High melting point (h) High boiling point 5 The atomic radius and electronegativity of the transition elements are almost the same.

4.5 1 Predict the changes in the properties of the elements across Period 3 in the Periodic Table. (a) Atomic radius (b) Electronegativity (c) Acidic-base property of the oxides 2 The proton numbers of element X and Y are 3 and 9 respectively. (a) To which period of the Periodic Table do X and Y belong? (b) Which is more electronegative? Explain your answer. 3 Write the formula of the oxides of the elements in Period 3. State whether the oxide is basic, acidic or amphoteric. 4 Silicon is a semi-metal. State one difference between (a) silicon and sulphur, (b) silicon and iron.

93

Periodic Table of Elements

Table 4.9 Some physical properties of the transition elements in the first series

4

Element

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

Atomic radius (nm)

0.162

0.147

0.134

0.130

0.135

0.126

0.125

0.124

0.128

0.138

Melting point (°C)

1539

1668

1900

1875

1245

1536

1495

1453

1083

419

Boiling point (°C)

2730

3260

3450

2665

2150

3000

2900

2730

2959

906

Density (g cm–3)

3.0

4.51

6.1

7.19

7.43

7.86

8.9

8.9

8.96

7.14

Electronegativity

1.3

1.5

1.6

1.6

1.5

1.8

1.8

1.8

1.9

1.6

Hardness (Mohs’ scale)

soft

6.0

7.0

8.5

6.0

4.0

5.0

4.0

3.0

2.5

(Note: Hardness is measured using Mohs’ scale. Diamond, which is the hardest substance known, has a hardness of 10 on Mohs’ scale.)

Titanium alloy is corrosion resistant and lightweight. It is used in orthopaedic and dental implants.

Vanadium steel alloy is used in making gears and crankshafts of vehicles.

Titanium alloy is light and have very high tensile strength. It is used to make aircraft engines. An engine of a A380 Airbus uses 11 tons of titanium.

Uses of transition metals Cobalt is alloyed with iron, nickel and other metals to make Alnico, an alloy of unusual magnetic strength.

Nickel is used in many industrial and consumer products, including stainless steel, magnets, coinage and special alloys.

Chromium is used to make stainless steel and for the electroplating of iron.

Periodic Table of Elements

Iron is alloyed with carbon to make steel which is used in making cars, ships and in building industries.

94

Copper is used as electrical conductors and in piping.

Special Properties of Transition Elements Transition elements have variable oxidation numbers 1 Unlike elements in the main group of the Periodic Table in which each has only one oxidation number, a transition element has more than one oxidation number in its compounds. 2 Oxidation number is the charge on the ion. In other words, a transition element can form ions with different charges. For example, magnesium (in Group 2), can form only Mg2+ ion, with an oxidation number of +2. Iron, however, as a transition element, can form Fe2+ ion (oxidation number of +2) and Fe3+ ion (oxidation number of +3). 3 Table 4.10 shows the different oxidation numbers of transition elements in their compounds.

Table 4.10 The oxidation numbers of transition elements in their compounds

Chromium(III) chloride Potassium dichromate(VI)

CrCl3 K2Cr2O7

+3 +6

Manganese(II) chloride Manganese(IV) oxide Potassium manganate(VI) Potassium manganate(VII)

MnCl2 MnO2 K2MnO4 KMnO4

+2 +4 +6 +7

Iron(II) sulphate Iron(III) chloride

FeSO4 FeCl3

+2 +3

Nickel(II) sulphate Nickel(III) chloride

NiSO4 NiCl3

+2 +3

Copper(I) oxide Copper(II) oxide

Cu2O CuO

+1 +2

Transition elements form coloured com­pounds 1 Unlike main group metal compounds which are usually white, transition elements can form compounds of different colours. 2 Unlike aqueous solutions of main group compounds or ions which are usually colourless, aqueous solutions of transition element compounds or their ions are coloured. 3 Table 4.11 shows the colours of some aqueous solutions of ions of transition elements.

Formula of ion of transition element

Colour of aqueous solution

Manganese(II) ion

Mn2+

Pink

Chromium(III) ion

3+

Cr

Green

Nickel(II) ion

Ni2+

Green

4 Precious stones are coloured due to the presence of com­pounds of transition elements. Table 4.12 shows the transition elements present which are responsible for the colours of some precious stones.

Table 4.11 The colours of some aqueous solutions of ions of transition elements

Name of ion of transition element

Name of ion of transition element

Formula of ion of transition element

Colour of aqueous solution

Chromate ion

CrO

Yellow

Dichromate ion

Cr2O

Orange

Permanganate ion

MnO4–

Purple

Precious stone

Colour

Transition element present

Iron(II) ion

Fe2+

Green

Ruby

Red

Chromium

Iron(III) ion

Fe

Brown

Sapphire

Blue

Iron and Titanium

Copper(II) ion

Cu2+

Blue

Emerald

Green

Chromium

Cobalt(II) ion

Co2+

Pink

Amethyst

Purple

Manganese

2– 4 2– 7

3+

Table 4.12 Examples of some precious stones and the transition elements which give them their distinctive colours

95

Periodic Table of Elements

4

Compounds of transition Chemical Oxidation elements formula number

Transition metals or their compounds have catalytic properties 1 A catalyst is a substance that speeds up the rate of a reaction. A catalyst does not change chemically after a reaction. Catalysts are used in almost all chemical manufacturing plants. 2 Many catalysts are transition elements or their compounds.

SPM

’08/P1/P2

3 Table 4.13 shows the uses of transition elements or their compounds as catalysts in industries.

Table 4.13 The uses of transition elements or their compounds as catalysts in industries

4

Transition element or its compound Fine iron powder, Fe

Industrial process which use the catalyst Haber process in the manufacture of ammonia. Iron catalyses the reaction between nitrogen and hydrogen gas to produce ammonia. Fe

N2(g) + 3H2(g) ⎯⎯→ 2NH3(g) Vanadium(V) oxide, V2O5 SPM

’10/P1

Contact process in the manufacture of sulphuric acid. V2O5 catalyses the oxidation of sulphur dioxide to sulphur trioxide. V2O5

2SO2(g) + O2(g) ⎯⎯→ 2SO3(g) SO3 is used to manufacture sulphuric acid. Nickel

Manufacture of margarine. Nickel catalyses the hydrogenation of unsaturated vegetable oil into saturated oil in the production of margarine.

Platinum

Ostwald process in the manufacture of nitric acid.

Transition elements can form complex ions 1 A complex ion is a polyatomic cation or anion consisting of a central metal ion with other groups bonded to it. 2 Table 4.14 shows some examples of complex ions formed by transition elements.

Table 4.14

Complex ions

Reaction of Aqueous Solutions of Transition Element Compounds with Sodium Hydroxide and Ammonia Solutions

Tetraamminecopper(II)

Cu(NH3)2+ 4

Hexaamminechromium(III)

Cr(NH3)3+ 6

Hexaaquocobalt(II)

Co(H2O)2+ 6

Hexacyanoferrate(II)

Fe(CN)64–

Hexacyanoferrate(III)

Fe(CN)63–

2 The ions of transition elements react with the hydroxide ions to form coloured metal hydroxide precipitates. 3 Table 4.15 gives some examples of the reactions of aqueous solutions of ions of transition elements with sodium hydroxide solution and ammonia solution.

1 The presence of ions of transition elements in a solution can be confirmed by using sodium hydroxide solution or ammonia solution.

Periodic Table of Elements

Formula

96

Table 4.15

Aqueous sodium hydroxide solution, NaOH(aq)

Fe2+

Green precipitate of iron(II) hydroxide is formed. Precipitate is insoluble in excess NaOH solution.

Green precipitate of iron(II) hydroxide is formed. Precipitate is insoluble in excess aqueous NH3 solution.

Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) (from NaOH) (green precipitate)

Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) (from NH3) (green precipitate)

Brown precipitate of iron(III) hydroxide is formed. Precipitate is insoluble in excess NaOH solution.

Brown precipitate of iron(III) hydroxide is formed. Precipitate is insoluble in excess aqueous NH3 solution.

Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) (from NaOH) (brown precipitate)

Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) (from NH3) (brown precipitate)

Blue precipitate of copper(II) hydroxide is formed. Precipitate is insoluble in excess NaOH solution.

Blue precipitate of copper(II) hydroxide is formed.

Fe3+

Cu2+

Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s) (from NH3) (blue precipitate)

Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s) (from NaOH) (blue precipitate)

6

Aqueous ammonia solution, NH3(aq)

4

Ions

Precipitate is soluble in excess NH3 solution to form a complex ion which is dark blue in colour. Cu(OH)2(s) + 4NH3(aq) → Cu(NH3)2+ (aq) + 2OH–(aq) 4 (dark blue solution) Cu(NH3)2+ is a dark blue complex ion. Thus, Q is 4 copper(II) oxide. Answer A

’03

Q is a compound. Study the flowchart below and identify Q. Solid Q + HNO3(aq)

+NaOH(aq)

Blue solution is formed + NaOH(aq)

Blue precipitate

+ excess ammonia solution

A Copper(II) oxide B Iron(III) oxide

Aqueous ammonia contains hydroxide ions: NH3 + H2O NH4+ + OH–

Not soluble Dark blue solution

C Copper(II) sulphate D Iron(II) sulphate

Have high melting, boiling points and densities

Comment Q is a base, since it dissolves in nitric acid. Metal oxides are bases. Since it forms a blue precipitate, Q must contain copper(II) ions. The above reactions can be represented by the equations CuO(s) + 2HNO3(aq) → Cu(NO3)2(aq) + H2O(l) Cu(NO3)2(aq) + 2NaOH(aq) → 2NaNO3(aq) + Cu(OH)2(s) (blue precipitate) Cu(OH)2(s) + 4NH3(aq) → Cu(NH3)2+ (aq) + 2OH–(aq) 4

Form coloured compounds

Have more than one oxidation state

Transition elements

The presence of their ions can be confirmed using NaOH(aq) or NH3(aq)

97

Used as catalysts in chemical reactions

Form complex ions

Periodic Table of Elements

in a table for easy studying. Scientists who played prominent roles in the development of the Periodic Table were J. W. Dobereiner, John Newlands, Lothar Meyer, Dmitri Mendeleev and Henry Moseley.

4.6 1 You are given two blue aqueous solutions; one containing Cu2+ ions and another containing a blue food colouring. Explain how you can differentiate between the two solutions. 2 The body of U-2 spy plane is made of titanium alloy. Give three reasons why titanium alloy is used?

4

4.7

Uses of the Elements and Compounds in Our Daily Life 1 An element is a substance that cannot be broken down into simpler substances. Table 4.17 shows the uses of some of these elements in daily life. 2 A compound is a substance made by chemically combining atoms of two or more elements. 3 Many compounds have been synthesised by chemists to improve our standard of living. Many items in our home are made up of elements. The shirt you are wearing may be made of Terylene, your sofa may be made of polyvinyl chloride, the cooking utensils in your kitchen are made of stainless steel, glass or ceramics, part of the motor of your ceiling fan is made of copper and the microprocessors in the computer are made of silicon and germanium. Elements are used to make vehicles, medicine and communication tools like the handphone. Imagine our life without these elements. Coloured compounds of transition elements are used as paint pigments and they make our homes more colourful. Table 4.18 shows the uses of some of these compounds.

Appreciating the Existence of Elements and Their Compounds

1 In ancient times, gold, mercury, copper, iron, sulphur, tin, antimony, lead, diamond, and graphite were already discovered. Credit should be given to the scientists who, through their persistent efforts, discovered and isolated other elements. Table 4.16 below shows the scientists who isolated some of these elements. 2 Then about a century (100 years) later, scientists discovered the subatomic particles of atoms. Credit is given to J. J. Thomson, Ernest Rutherford, James Chadwick and Niels Bohr in helping us understand the atomic structure of atoms in elements. The atomic structure of atoms helps us understand how elements react to form compounds. 3 As more elements are isolated, there is a need to classify the elements. Elements with the same chemical properties are grouped together

Table 4.16

Element

Scientist

Symbol

Isolated in

Hydrogen

H

Henry Cavendish (British)

1766

Nitrogen

N

Daniel Rutherford (British)

1772

Oxygen

O

Carl Wilhelm Scheele (Swedish)

1774

Sodium

Na

Sir William Ramsay (British)

1807

Potassium

K

Sir Humphry Davy (British)

1807

Magnesium

Mg

Sir Humphry Davy (British)

1808

Aluminium

Al

Hans Christian Orsted

1825

Uranium

U

Eugene Melchior Peligot (French)

1841

Helium

He

Sir William Ramsay (British)

1895

Radium

Ra

Pierre Curie and Marie Curie (French)

1898

Periodic Table of Elements

98

Table 4.17 Uses of some elements

Formula

Uses

Hydrogen

H2

Used in the hydrogenation of unsaturated oil to produce margarine. Used in the manufacture of ammonia.

Aluminium

Al

Used in the production of duralumin alloy for use in the construction of aeroplane bodies.

Silicon

Si

To make microchips

Sulphur

S

To make matches, fireworks and for the manufacture of sulphuric acid

Chlorine

Cl2

To kill germs in drinking water

Iron

Fe

Production of steel

Copper

Cu

To make electrical wire, electrical motor, dynamo and coins (cupronickel alloy)

Cobalt-60

Co

The gamma rays emitted from this radioactive element is used to kill cancer cells. Table 4.18 Uses of some chemical compounds

Chemical Formula

Uses

Magnesium oxide

MgO

In antacid drugs to treat gastric patients. Also used to treat acid poisonings.

Tin fluoride

SnF2

Is added to toothpaste. Fluoride ions can strengthen the teeth.

Compound

Sodium bicarbonate

NaHCO3

Silver bromide

AgBr

Vinyl chloride

CH2CHCl

Ammonia Sodium hydroxide

NH3 NaOH

Used as a baking powder. Also used to treat acid burns. Used in the making of photographic films. To make PVC pipes, toys, raincoats and cushions To manufacture fertiliser, nitric acid and explosives To make soap

Preventing Wastage

pollution because many of these items are non-biodegradable. 3 We can minimise wastage in the school laboratory by practising the following: (a) Weigh or use the correct amount of chemicals that is required to carry out an experiment. (b) Read the label of the chemicals carefully, so that we do not take the wrong substance. (c) Read the procedure of the experiment and plan the experiment carefully before carrying them out, so as to avoid the need to repeat the experiment.

1 Most of the elements needed to keep life going are obtained from the Earth’s crust. A number of these elements are obtained from the oceans and the air. After extraction, these chemicals cannot be replaced and will be depleted one day. So it is important that we avoid wastage in using these elements. 2 Recycling of used materials is an alternative method. Used glass bottles, plastic bottles and aluminium cans should be separated and thrown into different bins provided by the government. This will also help to reduce

99

Periodic Table of Elements

4

Element

4

Cl2(g) + H2O(l) → HCl(aq) + HOCl(aq) (b) Halogens react with alkali metals to form halide salts, e.g. 2Na(s) + Cl2(g) → 2NaCl(s) (c) Halogens react with hot iron wool to form brown iron(III) halide salt, e.g. 2Fe(s) + 3Cl2(g) → 2FeCl3(s) 5 The reactivity of Group 17 element increases up the group because the atomic radius decreases. The smaller the atomic radius, the stronger the electrostatic force of attraction between the nucleus and the electron. Thus the elements higher in the group can accept electrons more easily. 6 Group 18 elements exist as monatoms because they have attained the duplet or octet electronic configuration. They do not need to donate, accept or share electrons with other elements. 7 Transition elements are a block of elements between Group 2 and Group 13 in the Periodic Table. The characteristics of transition elements are (a) they form coloured compounds (b) they have more than one oxidation number (c) they catalyse chemical reactions (d) they form complex ions

1 Chemical elements are classed into groups in the Periodic Table, with elements in the same group having the same chemical properties. 2 Group 1 elements are called alkali metals. (a) They react with cold water to produce an alkaline solution and hydrogen gas, e.g. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) (b) They react with halogen gas to form halide salts, e.g. 2Na(s) + Cl2(g) → 2NaCl(s) (c) They burn in air to form metal oxides, e.g. 4Na(s) + O2(l) → 2Na2O(s) All oxides of Group 1 metals can dissolve in water to form alkaline solutions. Na2O(s) + H2O(l) → 2NaOH(aq) 3 The reactivity of Group 1 elements increases down the group because the valence electron is further from the nucleus. The electrostatic force of attraction between the nucleus and the valence electron becomes weaker. Thus the elements lower in the group can release their valence electrons more easily. 4 Group 17 elements are called halogens. (a) Halogens dissolve in water to form acidic solutions, e.g.

4 Multiple-choice Questions 4.1

Periodic Table of Elements

1 The diagram below shows the electron arrangement of atom Y. x xx x x xx

Y

x x x

x xx

Which of the following is the position of atom Y in the Periodic Table? A B C D

Group Period TC 52 2 3 3 3 2 13 3 13

Periodic Table of Elements

2 The structure of a compound containing element Q and hydrogen is shown as follows. H H

xo xo x x x H Q Q H o x o xo xo

H H To which group of the Periodic Table does element Q belong? A 14 C 16 B 15 D 17 3 Which of the following elements belong to the same group in the Periodic Table?

7 3

P

9 4

Q

27 13

A P and R B R and T

R

35 17

S

40 20

C P and S D Q and T

100

T

4 Atom of element X has proton number of 19. Which of the following statements are true concerning X ? I X has one valence electron. II The oxide of element X is soluble in water. III X is in Period 4 of the Periodic Table. IV X has the same chemical properties as fluorine. A I and III only B II and IV only C I, II and III only D I, III and IV only 5 An atom of element Y has a nucleon number of 31 with 16 neutrons.

6 An element X has a proton number of 9. In which group and period does X belong to in the Periodic Table? Group

Period

A

16

2

B

17

2

C

16

3

D

17

3

4.2

Group 18 Elements

7 The symbols for four elements are shown below.

4 2

W

16 8

X

24 12

Y

40 18

Z

Which of the following statements is true? A Both elements W and Z are monatomic. B Element Z is more reactive than element X. C Elements W and Y react to form a compound with the formula YW. D Elements X and Y react to form a compound with the formula YX2. 8 Which of the inert gases below is used in a diver’s oxygen tank? A Neon C Helium B Argon D Krypton

9 Which element, as represented by the symbols below, exists as monatoms?

A W

B X

C Y

D Z

10 The element which does not form compound with other elements is likely to have a proton number of ’06 A 4 B 6 C 8 D 10 11 As we go down Group 18, I the density increases. II the reactivity decreases. A I and III only B II and IV only

III the boiling point increases. IV the atomic size decreases. C I, II and III only D I, III and IV only

12 Argon does not form compound with chlorine because A it has three filled electron shells. B it has low melting and boiling points. C it has eight electrons in the outermost electron shell. D the atoms have the same number of protons and neutrons in the nucleus.

4.3

Group 1 Elements

13 Rubidium is below sodium in Group 1. Which statements below about rubidium are correct? ’08 I It is more reactive than sodium. II It burns in air to form rubidium oxide which is insoluble in water. III It is produced from the electrolysis of molten rubidium chloride. IV It reacts with cold water to form rubidium hydroxide and hydrogen. A I and III only C I, III and IV only B II and IV only D I, II, III and IV

101

14 Which of the following explains why sodium is more reactive ’06 than lithium? A Sodium has more protons than lithium. B Sodium has less valence electrons than lithium. C Sodium has a lower melting point than lithium. D Sodium can release its valence electron more easily. 15 An element X is burned in air. The product of combustion is then dissolved in water. The solution gives a pH value of 14. The element X is A potassium B phosphorus C aluminium D sulphur 16 Which statement below is true about sodium? A It burns in air to form sodium oxide of formula NaO which is soluble in water. B It burns in air to form sodium oxide of formula Na2O which is soluble in water. C It burns in air to form sodium oxide of formula NaO which is insoluble in water. D It burns in air to form sodium oxide of formula Na2O which is insoluble in water. 17 Which statement concerning the ions of Group 1 elements is correct? A Each contains more protons than electrons. B Each contains more electrons than protons. C Each has one electron in its outer electron shell. D Each contains the same number of protons and electrons. 18 As we go down Group 1, I the density increases II the melting point increases III the reactivity increases IV the electropositivity decreases A I and II only B I and III only C II and III only D II and IV only Periodic Table of Elements

4

Which of the following statements is true concerning element Y? A Atom of element Y has 16 protons. B Atom of element Y has two filled electron shells. C Atom of element Y has six valence electrons. D Element Y is in Group 15 of the Periodic Table.

4

19 The element X has a proton number of 19. Element X A forms a basic oxide that is insoluble in water. B reacts with cold water to form an acidic solution. C reacts with chlorine to form a compound with formula XCl2. D is more reactive than an element with a proton number of 11.

4.4

Group 17 Elements

20 Which of the following explains why chlorine is more reactive than bromine? A Chlorine is less dense than bromine. B Chlorine has a lower boiling point than bromine. C Chlorine atom accepts an electron more easily than bromine atom. D Chlorine atom contains less protons than bromine atom. 21 Which of the following halogens dissolve in water to produce an acidic solution which can decolourise the colour of litmus paper? I Chlorine II Bromine III Iodine IV Astatine A I and II only B I and III only C II and IV only D III and IV only 22 An element from Group X can dissolve in water to form an acidic solution. This resulting solution reacts with silver nitrate reagent to form a white precipitate. In which group of the Periodic Table does the element belong to? A 1 B 2 C 16 D 17 Periodic Table of Elements

23 Which of the statements below is not true concerning Group 17 elements? A Bromine is a gas at room temperature. B Chlorine is more reactive than bromine. C As we go down Group 17, the density increases. D Chlorine is more electronegative than bromine. 24 As we go down Group 17, I the reactivity decreases II electronegativity increases III the melting point increases IV the solubility of the element in water decreases A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 25 Which statement about ions of Group 17 elements is correct? A Each has eight valence electrons. B Each contains more protons than electrons. C Each contains more protons than neutrons. D Each has an odd number of electrons. 26 Which of the following statements are true about chlorine and bromine? I Both are gases at room temperature. II They react with sodium to form soluble salts. III They react with heated iron wool to form iron(II) halides IV They dissolve in water to form solutions with pH values of less than 7. A I and III only B II and IV only C I, II and IV only D II, III and IV only 27 An element X has a proton number of 35. Which statement about X is true? A X forms a positive ion during chemical reactions. B It belongs to Group 15 of the Periodic Table.

102

C Reacts with iron wool to form a compound with the formula FeX3. D Reacts with sodium to form a compound with formula NaX2. 28 Which of the following statements about the Periodic Table is true? A All inert gases have eight valence electrons. B All Group 1 elements are equally reactive. C The atomic radius increases in size across a period from left to right. D The electronegativity of Group 17 elements increases up the group.

4.5

Elements in a Period

29 The table shows the proton numbers of four elements. Element

P

Q

R

S

T

Proton 10 11 14 17 19 number Arrange the atomic radii of the elements in increasing order. A P, Q, R, S, T B P, S, R, Q, T C T, S, R, Q, P D T, Q, R, S, P 30 Which of the following metallic oxides can react with both acid ’06 and alkaline solutions? I Aluminium oxide II Tin(II) oxide III Lead(II) oxide IV Copper(II) oxide A I and III only B II and IV only C I, II and III only D I, III and IV only 31 X, Y and Z are elements in the same period of the Periodic Table. The oxide of X is acidic, the oxide of Y is basic and the oxide of Z is amphoteric. Arrange the elements X, Y and Z in increasing proton number. A Z, X, Y C Y, X, Z B X, Z, Y D Y, Z, X

Metal oxides

X oxide

Y oxide

Z oxide

Observation Sodium hydroxide solution

Nitric acid solution

Dissolves to form colourless solution Dissolves to form colourless solution No change

Dissolves to form colourless solution No change

Dissolves to form colourless solution

What is the correct arrangement in increasing proton number of the elements? A X, Y, Z B Y, X, Z C Z, Y, X D Z, X, Y

Element

X

Y

Z

Proton number

11 13 17

Which of the following statements is true? A X, Y and Z are all conductors of electricity. B All the elements above are made up of atoms. C The atomic radius decreases in the order Z, Y, X. D The electronegativity increases in the order X, Y, Z.

4.6

Transition Elements

35 Which of the following elements will form coloured compounds? I Cobalt III Aluminium II Nickel IV Manganese A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV 36 The diagram shows part of a Periodic Table.

Which of the following statements below are true about the elements in the Periodic Table? I Element W is inert. II Element Y has more than one oxidation state. III Element X reacts with cold water to form X oxide and hydrogen. IV Element Z reacts with iron to form a compound with formula FeZ3. A II and IV only B I, II and III only C I, II and IV only D I, II, III and IV 37 The properties of the transition metals include ’08 I they form white compounds. II they act as catalysts in chemical reactions. III they have more than one oxidation state. IV they form complex ions. A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV

38 The table shows the properties of four elements. Which element is most likely to be a transition element?

33 The graph shows the change in a property as we go across Period 3 of the Periodic Table.

Element

Electrical conductivity

Melting point (°C) Density (g cm–3) 360

6.8

Good

B

620

3.1

Good

C

3500

3.5

Poor

D

3400

8.1

Good

A

39 An element X forms two oxides with the formula XO and X2O. Which of the following statements is true about the element X ? A It is a transition metal. B It is a Group 1 element. C It is a Group 17 element. D The oxides of element X are amphoteric. Which of the properties below corresponds to the change as shown in the graph above? A Atomic radius B Electropositivity C Electrical conductivity D Density 34 The table shows the proton numbers of three elements X, Y and Z.

40 Which of the following shows correctly the colour of the ions of the transition elements? Colour Fe2+

Cr2O72–

MnO4–

Co2+

Brown

Orange

Purple

Pink

B

Green

Orange

Purple

Pink

C

Brown

Purple

Orange

Blue

D

Green

Purple

Orange

Blue

A

103

Periodic Table of Elements

4

32 X, Y and Z are elements in Period 3 of the Periodic Table. The table shows the properties of the oxides of X, Y and Z when reacted with sodium hydroxide and nitric acid solutions.

Structured Questions 4 A list of elements represented by the letters with the nucleon numbers and proton numbers are given below:

1 Bromine is an element of Group 17 in the Periodic Table. (a) What is the physical state of bromine at room temperature? [1 mark]

P, 126Q, 199R, 27 S, 35 T, 39 U 13 17 19

1 1

(b) Write an equation for the reaction between bromine and water. [1 mark]

(a) Choose two elements from the list above that belong to the same group in the Periodic Table.

4

(c) Draw a labelled diagram for the apparatus that can be used to carry out a reaction between bromine and iron wool. [2 marks]

[1 mark]

(b) State the (i) group and (ii) period of element U. [2 marks]

(d) What is the number of valence electrons in bromine? [1 mark]

(c) Give one use of the element P.

[1 mark]

(d) Draw the atomic structure of element R. [2 marks]

(e) Iodine is below bromine in Group 17. Which of the two elements, iodine or bromine, is more reactive? Explain your answer. [3 marks]

(e) (i) Which of the elements in the list reacts with cold water to produce hydrogen gas?

2 Rubidium is placed below potassium in Group 1 in the Periodic Table.

[1 mark]



(a) Give two physical properties of rubidium. [2 marks]

(ii) Write a balanced chemical equation for the reaction in (i). [1 mark] (f) Write the formula of the ion formed by the element U. [1 mark]

(b) How is rubidium stored in the laboratory? [1 mark]

(g) (i) What is electronegativity? (ii) Which is the more electronegative element between R and T? Explain your answer.

(c) Write the equations for the reactions of rubidium (Rb) with (i) water and (ii) chlorine gas. [2 marks]

[3 marks]

(d) Explain why rubidium is more reactive than potassium. [2 marks]

5 A list of the symbols of the transition elements in Period 4 is given below:

(e) Write the formula of (i) rubidium nitrate and (ii) rubidium sulphate. [2 marks]

Ti, V, Cr, Mn, Fe, Co, Ni, Cu

(f) What is the colour of rubidium nitrate? [1 mark]

(a) Give two physical properties of the transition [2 marks] elements.

3 Diagram 1 shows a portion of the Periodic Table. The letters in the Periodic Table do not represent the ’05 actual symbols of the elements.

(b) What is the colour of the aqueous Cu2+ ion solution? [1 mark] (c) Name a transition element in the list which is used as a catalyst in the (i) Haber process to manufacture ammonia. [1 mark]



Diagram 1

(a) Choose two elements which are metals in the Periodic Table above. [2 marks]

(ii) hydrogenation of unsaturated oil to make margarine. [1 mark]

(d) Other than forming coloured compounds and having catalytic properties, name two other properties of transition elements. [2 marks]

(b) Write the formula of the ions formed by the elements (i) E and (ii) Q. [2 marks] (c) Choose an element from the Periodic Table above that can form a coloured compound. [1 mark]

(e) Name a reagent that can be used to differentiate between the ions of transition elements in the list above. [1 mark]

(d) Choose an element that exists as monatoms. Give a use of this element. [2 marks]

6 The symbols of the elements in Period 3 of the Periodic Table are given below:

(e) Which element from the Periodic Table above can form an acidic oxide? [1 mark]

Na, Mg, Al, Si, P, S, Cl, Ar

(f) Which is the more reactive element between E and R? Explain your answer. [2 marks]

(a) Name an element that can conduct electricity in Period 3. [1 mark]

Periodic Table of Elements

104

(b) How does the atomic radius change across the period? Explain your answer. [3 marks]

(f) (i) Name an acidic oxide of Period 3 that can dissolve in water. [1 mark] (ii) Write an equation for the reaction that takes place when the acidic oxide in (i) is dissolved in water. [1 mark]

(c) Name an element that exists as monatoms in Period 3. Explain your answer. [2 marks] (d) Write the formulae of the oxides of the elements in Period 3. [3 marks]

(g) The oxide of aluminium can react with both acid and alkali solutions. What is the term given to such an oxide? [1 mark]

(e) (i) Name a basic oxide of Period 3 that can dissolve in water. [1 mark] (ii) Write an equation for the reaction that takes place when its basic oxide in (i) is dissolved in water. [1 mark]

[1 mark]

4

(h) Give a use of the element silicon.

Essay Questions 1 (a)

(b) Explain why Group 18 elements exist as monatoms. [5 marks] II (c) Iron(III) chloride Chlorine

The reactivity of alkali metals increases down the group in the order: lithium < sodium < potassium Explain the statement given above.

I

[7 marks]

Sodium chloride

(b) Rubidium (Rb) is placed below potassium in Group 1 of the Periodic Table. Predict three physical properties and three chemical properties of rubidium. [10 marks] (c) The nucleon number of sodium is 23 and its atom has 12 neutrons. The nucleon number of chlorine atom is 35 and it has 18 neutrons. Prove that sodium and chlorine belong to the same period in the Periodic Table. [3 marks]



The flowchart shows the conversion of chlorine to sodium chloride and iron(III) chloride. Explain a method to carry out an experiment (i) to obtain sodium chloride from chlorine in conversion I. [5 marks] (ii) to obtain iron(III) chloride from chlorine in conversion II. [5 marks]

4 (a) Using a suitable period in the Periodic Table as an example, explain the trend in the properties of the elements in terms of metals, non-metals and semi-metals. [5 marks]

2 (a) Explain the following statements: (i) The reactivity of Group 17 elements ’06 decreases down the group. [7 marks] (ii) The atomic radius of Period 3 elements decreases across Period 3 from left to right.

(b) Using a suitable period in the Periodic Table as an example, explain why the electronegativity of the elements increases across a period from left to right. [5 marks]

[5 marks]

(b) With suitable examples, discuss the four properties of transition elements in the Periodic Table.

(c) Write the formulae of all the oxides of the elements in Period 3. Describe a suitable experiment to show that the oxides change from basic to amphoteric and then to acidic across Period 3. [10 marks]

[8 marks]

3 (a) Name three elements in Group 18 and state their uses. [5 marks]

Experiments 1 The reactivity of a halogen in the reaction with iron wool depends on its position in Group 17. Diagram 1 shows the set-up of apparatus for an experiment to determine the reactivity of halogens in Group 17.

Diagram 1

105

Periodic Table of Elements

The experiment is carried out using bromine gas, chlorine gas and iodine vapour to react with heated iron wool respectively. Observation of the experiment is shown in Table 1 below. Halogen

Observation

Bromine

Hot iron wool glows moderately bright when bromine gas is passed over it.

Chlorine

Hot iron wool glows brightly when chlorine gas is passed over it.

Iodine

Hot iron wool glows dimly when iodine vapour is passed over it. Table 1

(a) Complete Table 2 below based on the experiment.

4

Variables

Action to be taken

(i) Manipulated variable :

(i) The way to manipulate variable :

(ii) Responding variable :

(ii) What to observe in the responding variable :

(iii) Constant variable :

(iii) The way to maintain the constant variable :

Table 2

2 ’08

[6 marks]

(b) State one hypothesis for the experiment.

[1 mark]

(c) Based on the observation, arrange bromine, chlorine and iodine in descending order of reactivity of halogens with iron wool.

[1 mark]

(d) The proton numbers of the halogens increase in the order: chlorine < bromine < iodine. Make a conclusion regarding the positions of halogens in Group 17 in relation to their reactivities.

[1 mark]

(e) Astatine is a halogen below iodine in Group 17. Predict the reactivity of astatine with iron wool.

[1 mark]

Lithium, sodium and potassium are in Group 1 of the Periodic Table. The reactivity of Group 1 elements increases down the group from lithium to potassium. You are required to design a laboratory experiment to prove the statements above. Your explanations should include the following: (a) Problem statement

[3 marks]

(b) Hypothesis

[3 marks]

(c) List of materials and apparatus

[3 marks]

(d) Procedure

[3 marks]

(e) Tabulation of data

[3 marks]

Periodic Table of Elements

106

FORM 4 THEME: Interaction between Chemicals

CHAPTER

5

Chemical Bonds SPM Topical Analysis 2008

Year 1

Paper Section Number of questions

2009 3

2

5

A

B

C

1 — 3

–

–

–

1

3

2010 3

2 A

B

C

1 — 4

–

–

1

–

6

2011

2

3

A

B

C

1

–

–

1

–

5

2

3

A

B

C

–

–

1 — 2

–

ONCEPT MAP CHEMICAL BONDS

To attain the stable electron arrangement of the noble gases

Ionic bonds

Covalent bonds

Formed by transfer of electrons from metal atoms to non-metal atoms

Formed by sharing of electrons between non-metal atoms

Metal atoms donate electrons

Non-metal atoms share electrons

Non-metal atoms accept electrons

Example: CH4 Example: NaCl

/ / 

¶ /

5H

*S

5H

*

*S

/

/

* /

/

Differences in physical properties

Melting point and boiling point

Solubility

Electrical conductivity

/

5.1

9 Helium has only one electron shell filled with 2 electrons (a duplet electron arrangement). All other noble gases have 8 electrons in the valence shell. This is known as an octet electron arrangement. 10 A duplet electron arrangement (as in helium) or an octet electron arrangment (as in the other noble gases) is very stable. As such, atoms of noble gases do not donate, accept or share electrons with other elements. Thus, atoms of noble gases do not combine with other elements or with itself. They exist as monatoms.

Formation of Compounds

5

Stability of Noble Gases 1 A compound is a chemical substance that is formed by combining two or more elements chemically in fixed proportions. 2 Almost all chemical substances exist as compounds in nature. Examples of compounds are water (H2O), carbon dioxide (CO2), table salt sodium chloride (NaCl) and minerals such as metal silicates, metal oxides, metal carbonates and metal sulphides. 3 Only noble gases and a few minerals such as gold, diamond and platinum exist as pure elements. 4 The tendency of elements to combine with other elements to form compounds shows that compounds are more stable than elements. 5 The formation of compounds proves that chemical bonds hold atoms of elements together. 6 Noble gases are elements in Group 18 of the Periodic Table. They are also known as inert gases which consists of helium, neon, argon, krypton, xenon and radon. 7 Noble gases exist as elements. They are very stable and are inert, which means they are non–reactive chemically. 8 The stability of noble gases is due to their electron arrangements (or electronic configurations) as shown in Table 5.1.

Conditions for the Formation of Chemical Bonds 1 Noble gases do not form chemical bonds because they have the stable duplet or octet electron arrangement. 2 Atoms of elements from Group 1 to Group 17: (a) Have less than 8 valence electrons (b) Each atom will tend to donate, accept or share electrons to achieve the stable duplet or octet electron arrangement as that of a noble gas 3 In the process of attaining the stable duplet or octet electron arrangement, chemical bonds will form between atoms of these elements. 4 The two types of chemical bonds are (a) ionic bonds (formed by transfers of electrons), and (b) covalent bonds (formed by sharing of electrons). 5 In the formation of chemical bonds, only valence electrons are involved in the donation, acceptance or sharing of electrons. Electrons in the inner shells are not involved. 6 The valence shell will then achieve an octet electron arrangement or a duplet electron arrangement (in the case where there is only one electron shell).

Table 5.1 Electron arrangements of noble gases

Noble gas

Symbol

Electron arrangement

Helium

He

2

Neon

Ne

2.8

Argon

Ar

2.8.8

Krypton

Kr

2.8.18.8

Xenon

Xe

2.8.18.18.8

Radon

Rn

2.8.18.32.18.8

Chemical Bonds

Neon is inert not just because it is a Group 18 element but because it has a stable octet electron arrangement with eight electrons in the outermost shell.

108

electrons respectively. In chemical reactions, these metal atoms tend to donate all their valence electrons to achieve the stable duplet or octet electron arrangement. 6 A negative ion (or anion) is formed when an atom accepts one or more electrons. The ion formed is negatively-charged because there are more electrons than protons. For instance,

5.1 1 The proton numbers of neon and argon are 10 and 18 respectively. Write the electron arrangements of neon and argon. Explain why these two elements exist as monatoms. 2 The electron arrangements of atoms P, Q and R are given in the table below.

Electron arrangement

P

2.8.2

Q

2.8.7

R

2.8.8

2– accepts 2 electrons 8p

Oxygen atom (O) (8p, 8e)

3 State two types of chemical bonds.

Formation of Ionic Bonds

SPM

’08/P1

Formation of Ions 1 Atoms are neutral because the number of protons is the same as the number of electrons. 2 An ion is formed when an atom donates or receives one or more electrons. 3 A positive ion or cation is formed when an atom donates one or more electrons. The ion formed has less electrons than protons and is positively-charged. For example,

11p

Sodium atom (Na) (11p,11e)

donates an electron

Oxide ion (O2–) (8p, 10e)

7 Generally, (a) non–metals usually form negative ions, (b) charge of negative ion = number of electrons received by an atom X + ne– → Xn– 8 Non–metal atoms from Groups 15, 16 and 17 in the Periodic Table have 5, 6 and 7 valence electrons respectively. In chemical reactions, non-metal atoms will accept electrons so that the ion formed achieves the stable octet electron arrangement.

• The term electron arrangement is used interchangeably with electronic configuration. • Donate electrons can also be explained as lose electrons or release electrons. • Accept electrons can also be explained as gain electrons or receive electrons. • The name of a metal ion is the same as the metal atom. Examples: sodium ion (Na+), magnesium ion (Mg2+), aluminium ion (Al3+). • The name of the non-metal ion ends with -ide or -ate (when oxygen is attached to the non-metal). Examples: chloride ion (Cl–), sulphide ion (S2–), sulphate ion (SO42–), carbonate ion (CO32–).

+ 11e

8p

8e

(a) Which atom is chemically inert? Explain your answer. (b) Which atom will take part in chemical bonding? Explain your answer.

5.2

10e

5

Atom

10e 11p

Sodium ion (Na+) (11p,10e)

4 Generally, (a) metal atoms usually form positive ions, (b) charge of positive ion = number of electrons released by an atom M → Mn+ + ne– 5 Metal atoms from Groups 1, 2 and 13 in the Periodic Table have 1, 2 and 3 valence

• The duplet or octet electron arrangement of noble gases are very stable. • Atoms form positive ions or negative ions so as to attain the electron arrangement as that of the noble gases.

109

Chemical Bonds

5

To prepare ionic compounds Materials

Magnesium ribbon, sodium, chlorine gas, iron wool and sodium hydroxide solution.

Apparatus

Tripod stand, clay pipe triangle, Bunsen burner, crucible and lid, sandpaper, gas jar, gas jar spoon, combustion tube, filter funnel, retort stand, clamp and beaker.

2 The ignited sodium is placed in a gas jar filled with chlorine gas. Any changes that occur are recorded.

(A) Preparation of magnesium oxide Procedure 1 A 5 cm length of magnesium ribbon is cleaned with a piece of sandpaper. 2 The magnesium ribbon is placed in the crucible. 3 The magnesium ribbon is heated strongly. Any changes that occur are recorded.

Figure 5.2 Preparation of sodium chloride

(C) Preparation of iron(III) chloride Procedure 1 A little iron wool is placed inside a combustion tube. 2 The end of the combustion tube is connected to a filter funnel inverted into a beaker with some sodium hydroxide solution. 3 The iron wool is heated strongly until it glows. 4 Chlorine gas is passed through the iron wool while being heated. Any changes that occur are recorded.

Figure 5.1 Preparation of magnesium oxide

(B) Preparation of sodium chloride Procedure 1 A small piece of sodium metal is placed in a gas jar spoon and is heated carefully until it begins to ignite. Figure 5.3 Preparation of iron(III) chloride

Results

Activity 5.1

Method

Observation

Inference

Heating of magnesium in air

• The magnesium ribbon burns with a bright flame • White powder is formed

The white powder formed is magnesium oxide

Burning of sodium in chlorine gas

• Sodium burns with a bright yellow flame • The yellowish-green colour of chlorine gas is decolourised. • White fumes are produced and deposited as white powder

The white powder formed is sodium chloride

Heating of iron in chlorine gas

• The iron wool continues to glow brightly in the chlorine gas • A brown powder is formed

The brown powder formed is iron (III) chloride

Chemical Bonds

110

2Mg(s) + O2(g) → 2MgO(s) 2 Magnesium oxide is an ionic compound that contains Mg2+ ions and O2– ions. 3 When sodium burns in chlorine gas, the white powder formed is sodium chloride.

Conclusion 1 Generally, the reaction between metals and nonmetals produces ionic compounds. 2 Ionic compounds such as magnesium oxide, sodium chloride and iron(III) chloride can be prepared by direct combination of the metal and non-metal elements.

2Na(s) + Cl2(g) → 2NaCl(s) 4 Sodium chloride is an ionic compound that contains Na+ ions and Cl– ions. 5 When iron burns in chlorine gas, the brown powder formed is iron(III) chloride.

Metal Non-metal Magnesium + oxygen → Sodium + chlorine → Iron + chlorine →

2Fe(s) + 3Cl2(g) → 2FeCl3(s) 6 Iron(III) chloride is an ionic compound that contains Fe3+ ions and Cl– ions.

Formation of Ionic Bonds

Ionic compound magnesium oxide sodium chloride iron(III) chloride

SPM

’09/P1

(c) non–metal atoms receive electrons to form negative ions (anions). (d) positive ions and negative ions are then attracted to each other by the strong electrostatic force of attraction. The bond formed between ions of opposite charges is known as ionic bond or electrovalent bond.

1 Metals from Groups 1, 2 or 13 react with non–metals from Groups 15, 16 or 17 in the Periodic Table to form ionic compounds. 2 In the formation of an ionic bond, (a) electrons are transferred from a metal atom to a non–metal atom. (b) metal atoms donate valence electrons to form positive ions (cations).

Formation of ionic bond in sodium chloride 1 A sodium atom with an electron arrangement of 2.8.1 achieves stability after it donates one valence electron to form a sodium ion, Na+. The electron arrangement of the sodium ion, Na+, is 2.8, with an octet of valence electrons.

SPM

’06/P2

Cl + e– ⎯→ Cl– 2.8.7 2.8.8 3 Sodium ions, Na+ and chloride ions, Cl– with opposite charges are attracted to each other by the electrostatic force of attraction. This force of attraction is called the ionic bond.

Na ⎯→ Na+ + e– 2.8.1 2.8 2 A chlorine atom with an electron arrangement of 2.8.7 achieves stability after it accepts one electron from a sodium atom – to form a chloride ion, Cl­ . The electron arrangement of the chloride ion, Cl–, is 2.8.8, with an octet of valence electrons. 111

Chemical Bonds

5

7 Sodium hydroxide solution is used to absorb the excess chlorine gas. Besides sodium hydroxide, soda lime can also be used. 8 A filter funnel is used to prevent the sodium hydroxide solution from being suctioned back into the combustion tube as chlorine gas is very soluble in water.

Discussion 1 When magnesium is heated, magnesium atom combines with oxygen in the air to form magnesium oxide.

Formation of ionic bond in magnesium oxide 1 A magnesium atom with the electron arrangement of 2.8.2 achieves the stable octet electron arrangement when it donates two valence electrons to form a magnesium ion, Mg2+.

O + 2e– ⎯→ O2– 2.6 2.8 3 The electrostatic force of attraction that exists between the oppositely-charged magnesium ions, Mg2+ and oxide ions, O2– forms the ionic bond.

Mg ⎯→ Mg2+ + 2e– 2.8.2 2.8

5

SPM

’09/P1

2 An oxygen atom with an electron arrangement of 2.6 achieves the stable octet electron arrangement when it accepts two electrons to form an oxide ion, O2–. Formation of ionic bond in potassium oxide electrons from two potassium atoms to form an oxide ion, O2–. 3 Hence each of the two potassium atoms donates one valence electron to be accepted by one oxygen atom. Ionic bonds are formed between the two potassium ions and the oxide ion.

1 A potassium atom with an electron arrangement of 2.8.8.1 achieves the stable electron arrangement of an octet when it donates one valence electron to form a potassium ion, K+. K ⎯→ K+ + e– 2.8.8.1 2.8.8 2 An oxygen atom with an electron arrangement of 2.6 achieves the stable octet electron arrangement when it accepts two valence

How to Predict the Formula of an Ionic Compound When drawing the electron arrangements to show the formation of ionic bonds, • do not overlap the outermost electron shells of atoms. • the outermost shells of all ions must have eight electrons except Li+ and H+. • use ‘dots‘ or ‘crosses‘ to represent the electrons from different atoms. • show the charge of the ions clearly outside the brackets of the ions.

The chemical formula of sodium chloride, NaCl, tells us that 1 mol of sodium ions combines with 1 mol of chloride ions and it is not 1 molecule of sodium chloride.

Chemical Bonds

1 Metal atoms will donate their valence electrons to achieve the stable duplet or octet electron arrangement of the noble gases. 2 Non-metal atoms will accept electrons in order to achieve the stable octet electron arrangement of the inert gases. 3 For cations Mb+ and anion Xa–, the formula of an ionic compound formed between them is written as MaXb Number of electrons that will be received by X (or charge of X ion)

Number of electrons that will be donated by M (or charge of M ion)

4 The overall positive charge of the cation must be equal to the overall negative charge of the anion in an ionic compound. Hence the formula of an ionic compound formed between them can also be derived as aMb++ bXa– → MaXb

112

1

2

Atoms in element J and element Q have proton numbers 12 and 17 respectively. Explain what type of bond will form between J and Q. Predict the formula of the compound formed.

Element M is an element from Group 13 and element X is an element from Group 16 in the Periodic Table. What is the formula of the compound formed between element M and element X?

Solution An atom of J, with a proton number of 12 and an electron arrangement of 2.8.2, will achieve the stable octet electron arrangement by donating two electrons to form a J 2+ ion.

Solution M from Group 13 has three valence electrons and will form M 3+ ion. X from Group 16 will receive two electrons to form X 2– ion. The formula of the compound formed is

M2X3 Charge of X 2– ion

Charge of M 3+ ion

An atom of element Q with proton number 17 and an electron arrangement of 2.8.7 will achieve the stable octet electron arrangement by accepting one electron to form a Q– ion.

5.2

Q + e ⎯⎯→ Q 2.8.7 2.8.8 –

5

J ⎯⎯→ J 2+ + 2e– 2.8.2 2.8

–

1 Give the formulae of the ions formed by the following elements: (a) Calcium (b) Phosphorus (c) Sulphur (d) Potassium (e) Nitrogen

Two Q atoms will accept one electron each from the two electrons donated by every J atom. Ionic bonds are formed between J 2+ ions and Q – ions to produce a compound with formula JQ2.

[Proton number: N, 7; P, 15; S, 16; K, 19; Ca, 20]

1

2 The proton numbers of element P and element Q are 9 and 20 respectively. (a) Write the equations that show the transfer of electrons and the formulae of the ions formed by atom P and atom Q respectively. (b) Draw the valence electron arrangement to show the formation of a compound formed between P and Q.

’02

R reacts with S to form an ionic compound with the formula R2S3. Which of the following electron arrangements of atoms R and atoms S is true? Electron arrangement of atom R

Electron arrangement of atom S

A

2.8.2

2.8.3

B

2.8.3

2.8.2

Group

1

2

16

17

C

2.8.2

2.5

Element

E

F

G

H

D

2.8.3

2.6

3 The table shows the groups of four elements in the Periodic Table represented by the letters E, F, G and H.

(a) Write the formulae of the ions formed by the elements E, F, G and H. (b) What is the formula of the compound formed from (i) E and G ? (ii) E and H ? (iii) F and G ? (iv) F and H ?

Comments Since an ionic compound is formed, R must be a metal and S must be a non-metal. The formula of R2S3 shows that R will donate three electrons whereas S will accept two electrons to achieve the octet electron arrangement. Hence, R has three valence electrons and S has six valence electrons. Answer D

113

Chemical Bonds

3 A Lewis structure is a diagram which shows only the valence electrons of the atoms represented by dots. The Lewis structure of the hydrogen molecule is

Formation of Covalent Bonds

1 A covalent bond is a bond that is formed from the sharing of valence electrons between ’08/P1 non-metal atoms to achieve the stable octet or a duplet electron arrangement. 2 Non-metals from Groups 15, 16 or 17 in the Periodic Table react with other non-metals of the same group or different groups to form covalent compounds. 3 Hydrogen is a non-metal element. It can form covalent bonds with other non-metal atoms from Groups 14, 15, 16 and 17. Examples are CH4, NH3, H2S and HCl. 4 Molecules with covalent bonds can be formed from: (a) Atoms of the same element Examples: Hydrogen (H2), chlorine (Cl2), oxygen (O2) and nitrogen (N2). (b) Atoms of different elements Examples: Water (H2O), ammonia (NH3), carbon dioxide (CO2), methane (CH4), tetrachloromethane (CCl4). 5 The types of covalent bond formed depend on the number of pairs of electrons shared between two atoms. There are three types of covalent bonds. (a) Single bond: One pair of electrons shared between two atoms. (b) Double bond: Two pairs of electrons shared between two atoms. (c) Triple bond: Three pairs of electrons shared between two atoms.

H• + H• → H •• H (or H — H) Formation of a chlorine molecule, Cl2 1 A chlorine atom with an electron arrangement of 2.8.7, needs to share one electron to achieve the stable octet electron arrangement of 2.8.8.

• •

• •

• •

• •

• •

• •

• •

2 Each chlorine atom contributes one valence electron to be shared. The sharing of one pair of electrons results in the formation of one single covalent bond between two chlorine atoms. 3 The Lewis structure showing the formation of chlorine molecule is • •

5

SPM

• •

5.3

Cl • + • Cl •• → •• Cl •• Cl ••

Formation of methane molecule, CH4

Formation of Single Covalent Bonds

1 A carbon atom with an electron arrangement of 2.4 needs to share four electrons to achieve the stable octet electron arrangement of 2.8. 2 A hydrogen atom needs to share one electron to achieve the stable duplet electron arrangement. 3 One carbon atom contributes four valence electrons to be shared with four hydrogen atoms respectively to form four single covalent bonds in the methane molecule, CH4.

Formation of a hydrogen molecule, H2 1 A hydrogen atom with an electron arrangement of 1, needs to share one electron to achieve the stable duplet electron arrangement. 2 Each hydrogen atom will contribute one valence electron to be shared between two hydrogen atoms. One pair of electrons shared between two hydrogen atoms form a single covalent bond.

Chemical Bonds

114

4 The Lewis structure for the formation of methane molecule is ⏐

4H • + • C • → H •• C •• H or H ⎯ C ⎯H ⏐

H

Formation of tetrachloromethane (carbon tetrachloride), CCl4

SPM

’10/P1, ’11/P1

1 A carbon atom with an electron arrangement of 2.4 needs to share four electrons to achieve the stable octet electron arrangement of 2.8. 2 A chlorine atom with an electron arrangement of 2.8.7 needs to share one electron to achieve the stable octet electron arrangement of 2.8.8. 3 Four chlorine atoms contribute one valence electron each to be shared with one carbon atom to form four single covalent bonds in the tetrachloromethane molecule, CCl4.

4 The Lewis structure showing the formation of ammonia molecule, NH3 is •

•

• •

• ⎯ ⎯ 3H • + N → H •• N • H or H N H • • ⏐ H H • •

• •

H

• •

•

Formation of water molecule, H2O 1 An oxygen atom with an electron arrangement of 2.6 needs to share two electrons to achieve the stable octet electron arrangement of 2.8. 2 A hydrogen atom needs to share one electron to achieve the stable duplet electron arrangement. 3 One oxygen atom contributes two valence electrons to be shared with two hydrogen atoms (which contributes one electron respectively) to form two single covalent bonds in the water molecule, H2O.

4 The Lewis structure showing the formation of water molecule, H2O is

• •

• •

4 The Lewis structure of the formation of tetrachloromethane molecule is • • • •

• • • •

H H ⏐ • • • • • • ⎯O • H O H or 2H • + •• O → • • • • • • • • •

• •

• •

• • • •

• •

• •

• • • •

• •

• •

• •

• • Cl •• • Cl • ⏐ 4 •• Cl • + • C • → •• Cl •• C •• Cl •• or •• Cl ⎯ C⏐ ⎯ Cl •• • • • • • Cl • • Cl • • •

115

Chemical Bonds

5

1 A nitrogen atom with an electron arrangement of 2.5 needs to share three electrons to achieve the stable octet electron arrangement of 2.8. 2 A hydrogen atom needs to share one electron to achieve the stable duplet electron arrangement. 3 One nitrogen atom contributes three valence electrons to be shared with three hydrogen atoms (which contributes one electron respectively) to form three single covalent bonds in the ammonia molecule, NH3.

H

• •

•

H

•

Formation of ammonia molecule, NH3

Formation of Double Covalent Bonds

Formation of carbon dioxide molecule, CO2

1 A carbon atom with an electron arrangement of 2.4 needs to share four electrons to achieve the stable octet electron arrangement of 2.8. 2 An oxygen atom with an electron arrangement of 2.6 needs to share two electrons to achieve the stable octet electron arrangement of 2.8. 3 One carbon atom contributes four valence electrons to be shared with two oxygen atoms (which contribute two electrons each) to form two double bonds in the carbon dioxide molecule, CO2 as shown in the diagram below.

1 An oxygen atom with an electron arrangement of 2.6 needs to share two electrons to achieve the stable octet electron arrangement of 2.8. 2 Each oxygen atom will contribute two valence electrons to be shared between two oxygen atoms. 3 The sharing of two pairs of electrons between two atoms results in the formation of a double covalent bond in an oxygen molecule, O2.

4 The Lewis structure below shows the formation of a carbon dioxide molecule, CO2. • •

• •

• •

• •

• •

• • • •• •• 2O • + • C • → O • • C • • O or O == C == O • • • • • • • • • •

• •

Formation of Triple Covalent Bonds Formation of nitrogen molecule, N2

• •

• •

• •

1 Each nitrogen atom with an electron arrangement of 2.5 will contribute three valence electrons to be shared between two nitrogen atoms so as to achieve the stable octet electron arrangement of 2.8. 2 The sharing of three pairs of electrons results in the formation of a triple covalent bond in the nitrogen molecule, N2.

In the formation of a covalent bond • an atom that requires one more electron to achieve the stable electron arrangement as in the noble gases will form one single covalent bond. For example: hydrogen and chlorine form single covalent bonds in H – H; Cl – Cl; H – Cl. • an atom (such as oxygen and sulphur) that requires two electrons to achieve the stable electron arrangement will form two single covalent bonds or one double covalent bond. For example:

3 The Lewis structure for nitrogen molecule, N2 is shown below.

O=O 1 double covalent bond per oxygen atom

•

Chemical Bonds

116

••

N • + • N • → ••N •• •• N •• or •• N ≡ N •• •

2 single covalent bonds per oxygen atom

;

• •

H – O – H

• •



SPM

’10/P1

• •

• • •• • == O •• O • + • O → O • • O or • O • • • •

•

• •

• •

4 The Lewis structure showing the formation of oxygen molecule is • •

5

Formation of oxygen molecule, O2

SPM

’06/P2 ’08/P1

How to Predict the Formula of a Covalent Compound

2

Comments P, with 6 valence electrons, will need to share 2 electrons in order to achieve the stable octet electron arrangement. Q, with 4 valence electrons, will need to share 4 electrons to achieve the octet electron arrangement. Hence the formula of the compound is

Table 5.2

Number of valence electrons (x)

Valency (8–x)

Group 14

4

4

Group 15

5

3

Group 16

6

2

Group 17

7

1



2 electrons to be shared with Q by P

3 Element X is from Group 14, element Y is from Group 16 and element Z is from Group 17 of the Periodic Table. What is the formula of the compound formed between (a) element X and element Y? (b) element X and element Z?

MyNx valency of N

valency of M

Solution (a) An atom of element X from Group 14 with 4 valence electrons needs to share 4 electrons in order to achieve the octet electron arrangement. An atom of element Y from Group 16 has 6 valence electrons and needs to share 2 electrons in order to achieve the octet electron arrangement. The formula of the compound formed is

5 Table 5.3 below shows the formulae of covalent compounds formed between elements from different groups. Table 5.3

Non-metal element, M

Non-metal element, N

Formula of covalent Examples compound MN4

CH4, CCl4, SiCl4

M2N4 or MN2

CO2, SiO2

Group 15 H or Group 17 (valency 3) (valency 1)

MN3

NH3, PH3, PCl3

Group 16 H or Group 17 (valency 2) (valency 1)

MN2

H2O, H2S, SCl2

Group 14 H or Group 17 (valency 4) (valency 1) Group 14 (valency 4)

Group 16 (valency 2)

4 electrons to be shared with P by Q

Answer A

4 If a non-metal element M, with a valency of SPM x, combines with another non-metal element, ’09/P1 N, with a valency of y, the formula of the covalent compound formed will be MyNx.

Q2P4 or QP2

X2Y4 or XY2 2 electrons required by Y to be shared to achieve the octet electron arrangement

4 electrons required by X to be shared to achieve the octet electron arrangement

(b) An atom of element Z from Group 17 requires 1 electron to be shared, hence the formula of the compound is X1Z4 or XZ4 Z shares 1 electron with X

117

X shares 4 electrons with Z

Chemical Bonds

5

The electron arrangement of atom P is 2.8.6 and atom Q has four valence electrons. What is the formula of the compound formed between P and Q? A QP2 B QP4 C Q2P D Q4P

1 Covalent compounds are formed from the non-metal elements (from Group 14 to Group 17). 2 The number of electrons required by a nonmetal atom to achieve the stable octet electron arrangement as that of a noble gas is known as the valency. 3 The valency of a non-metal is (8–x) where x is the number of valence electrons. Table 5.2 shows the valency of non-metal elements.

Element

’03

4 The proton numbers of element J and element Q are 16 and 17 respectively. Explain the type of bond and the formula of the compound formed between J and Q.

The electron arrangement of an atom of Q is 2.8.7. Both J and Q are non-metals and they will form a covalent compound. An atom J needs to share 2 electrons and an atom Q needs to share 1 electron in order to achieve the stable octet electron arrangement. The formula of the compound formed is JQ2.

Solution The electron arrangement of an atom of J is 2.8.6.

5

Guidelines in drawing the electron arrangement of a covalent compound formed by atom M and atom N. Step 1

Step 2

Determine the number of valence electrons and the valency of atom M and atom N, then determine the formula of the compound. Example Atom M has 5 valence electrons and valency is 3. Atom N has 7 electrons and valency is 1. Formula of compound is M1N3. [Proton number: M, 7; N, 9]

Draw the positions of atom M and atoms N in the formula with atoms N (more number of atoms) surrounding atom M. Example:

Step 4

Step 3

Determine the number of electrons to be shared (= 8 – x) where x = number of valence electrons. Draw the electrons to be shared as dot or crosses at the overlapped area. Example:

Draw the valence electron shells overlapping between the two types of atoms.

Step 5

Step 6

Draw the balance of the valence electrons not shared (x – number of electrons shared) on the shell outside the overlapped area.

Draw the inner electron shells for atom M and atoms N and the electrons as dots and crosses.

N has 7 valence electrons: 1 shared, remainder 6 unshared.

Chemical Bonds

118

How to predict the type of chemical bonds formed and the formula of the compound formed from two elements, X and Y. Step 1

Step 2

Write the electron arrangement to determine the number of valence electrons of elements X and Y.

5

• If one of the elements (say X) has 1, 2 or 3 valence electrons and the other element (say Y) has 5, 6 or 7 valence electrons then X and Y form an ionic bond. • If both elements have 4, 5, 6 or 7 valence electrons, or one of the elements is hydrogen, X and Y form a covalent bond.

Step 3 Determine the number of electrons X or Y needs to donate/ accept/share to achieve the stable octet electron arrangement.

Ionic Bonding

Covalent Bonding

Formula is Xa Yb.

Number of electrons that atom Y needs to accept to achieve an octet electron arrangement

Formula is Xa Yb.

Number of electrons that atom X needs to donate to achieve an octet electron arrangement

Number of electrons that atom Y needs to share to achieve an octet electron arrangement

Number of electrons that atom X needs to share to achieve an octet electron arrangement

1 Elements with 1, 2 or 3 valence electrons can only form ionic bonds with non-metals (elements with 5, 6 or 7 valence electrons). 2 Elements with 4 valence electrons can only form covalent bonds. 3 Elements with 5, 6 or 7 valence electrons can form both ionic and covalent bonds.

119

Chemical Bonds

3

5.3

’04

1 Write the Lewis structures for the following compounds: (a) Hydrogen chloride, HCl (b) Water, H2O (c) Hydrogen cyanide, HCN (d) Phosphorus trichloride, PCl3 [Proton number: H, 1; C, 6; N, 7; P, 15; Cl, 17]

5

The diagram shows the electron arrangement of a compound formed between atoms X and Y. Which of the following statements is true about the compound? A The compound is formed by ionic bonds. B The compound is formed by electron transfer. C Atom X is a metal and atom Y is a non-metal. D It is a covalent compound.

2 The table shows four elements in different groups of the Periodic Table represented by the letters V, X, Y and Z.

Comments The diagram shows the overlap of the outermost electron shells of atoms X and Y where the sharing of electrons is represented by dots and crosses. The diagram represents a covalent compound formed by the sharing of electrons between non-metal atoms. Answer D

Similarity

15

16

17

Element

V

X

Y

Z

3 The proton numbers of elements P and Q are 6 and 9 respectively. Draw the electron diagrams for the formation of compounds between (a) Q and hydrogen (b) Q and P (c) Q and Q

Covalent bonding

Atoms achieve the stable (duplet or octet) electron arrangement after the formation of bonds.

Difference Involves the transfer of electrons from metal atoms to non-metal atoms.

Chemical Bonds

14

(a) Write the formula of the compound formed between (i) V and Z (ii) V and Y (iii) X and Z (b) Draw the Lewis structures for the compounds formed in (a). (c) What type of compounds are formed in (a)?

Comparison between Ionic Bonding and Covalent Bonding Ionic bonding

Group

Involves the sharing of electrons between non-metal atoms.

Positivelycharged ions and negativelycharged ions are formed.

No charged ions are formed. Molecules are formed.

Strong electrostatic force of attraction holds the oppositelycharged ions together.

Van der Waals forces of attraction exist between the covalent molecules.

CFCs (chlorofluorocarbon) are covalent compounds that consist of chlorine, fluorine and carbon atoms bonded by covalent bonds. Two examples of CFCs are CF2Cl2 and CFCl3. CFCs are colourless, odourless and non-toxic gases with low boiling points. CFCs are used in aerosol and as refrigerants in air-conditioners. However, since 1992, the usage of CFC was banned because it was believed to deplete the ozone layer which protects the earth from getting too much harmful ultraviolet rays from the sun.

• Metals only form ionic bonding. • Non-metals form both ionic and covalent bondings.

120

8 The differences between the physical properties of ionic compounds and covalent compounds ’09/P1 are shown in Table 5.4.

SPM

The Properties of Ionic Compounds and Covalent Compounds

Properties of Ionic Compounds and Covalent Compounds

’08/P1

SPM

Table 5.4 Comparison of properties of ionic and covalent compounds

Ionic compound

Covalent compound (simple molecules)

High

Low

(b) Volatility

Non-volatile

Volatile (can change to vapour when heated)

(c) Solubility

Usually soluble in water and polar solvents but insoluble in organic solvents

Usually soluble in organic solvents such as benzene but insoluble in water

(d) Electrical conductivity

Conducts electricity in the molten state or aqueous solution

Does not conduct electricity in any state

SPM

’10/P2

Properties

1 The physical properties of ionic and covalent compounds are different because of the different types of bonds formed and the difference in the structures of the compounds. 2 An ionic compound is formed when ions of opposite charges are held by the strong electrostatic force of attraction. 3 The ions of ionic compounds are arranged in an orderly and compact manner and form large ionic structures. 4 Figure 5.4 shows the arrangement of ions in a three-dimensional network forming a large structure in an ionic compound.

(a) Melting point and boiling point

Figure 5.4 Three-dimensional network of ions

5 Most covalent compounds consist of simple molecules. 6 The covalent bond within a molecule is strong but the intermolecular forces (van der Waals forces of attraction) are weak. 7 Figure 5.5 shows the intermolecular forces (van der Waals forces of attraction) between methane molecules and the intramolecular bonds (covalent bonds) in the methane molecules. The dotted line represents the weak intermolecular forces (van der Waals forces).

5

5.4

Most ionic compounds are soluble in water. However there are some ionic compounds that are insoluble in water. For example, lead(II) bromide, lead(II) chloride and calcium carbonate are insoluble in water. There are also some covalent compounds that are polar and are soluble in water. For example, ethanol, ethanoic acid and sugar are soluble in water.

• The low melting and boiling points of covalent compounds are due to the weak intermolecular forces between covalent molecules. • They are not due to the strength of covalent bonds inside the molecules.

Figure 5.5 Types of bonds in methane molecules

121

Chemical Bonds

To study the physical properties of ionic and covalent compounds (A) To investigate the melting point and volatility of compounds Apparatus

5

Materials

SPM

’06/P2

3 Steps 1 and 2 are repeated with naphthalene and hexane respectively to replace magnesium chloride.

Crucible, spatula, tripod stand, wire gauze and Bunsen burner. Magnesium chloride, naphthalene and hexane.

Procedure 1 A spatula of magnesium chloride solid is placed in a crucible. 2 Magnesium chloride solid is heated slowly at first and then strongly. The change in the physical state of the compound is recorded.

Figure 5.6 To investigate the melting point of a compound

Results Compound

Observation Physical state

Magnesium chloride

Solid

No change

High melting point

Naphthalene

Solid

Melts rapidly

Low melting point

Hexane

Liquid

Vaporise rapidly

Low melting point and very volatile

Conclusion 1 Magnesium chloride is an ionic compound and has high melting and boiling points. Naphthalene and hexane are covalent compounds and they have low melting and boiling points. 2 Ionic compounds are non-volatile whereas covalent compounds are volatile. (B) To investigate the solubility of compounds

Activity 4.2

Inference

Action of heat

Apparatus

Boiling tubes and spatula

Materials

Magnesium chloride, naphthalene, hexane, water and cyclohexane.

Procedure 1 A boiling tube is filled with 5 cm3 of distilled water. 2 Half a spatula of magnesium chloride solid is added to the distilled water and shaken. The solubility of magnesium chloride in water is noted. 3 Steps 1 to 2 are repeated with naphthalene and hexane to replace magnesium chloride. 4 Another boiling tube is filled with 5 cm3 of cyclohexane (an organic solvent). 5 Half a spatula of magnesium chloride is added to the cyclohexane and shaken. The solubility of magnesium chloride in cyclohexane is noted. 6 Steps 4 and 5 are repeated with naphthalene and hexane in turn to replace magnesium chloride. Chemical Bonds

Results Compound

Solubility In water

In organic solvent

Magnesium chloride

Soluble

Insoluble

Naphthalene

Insoluble

Soluble

Hexane

Insoluble

Soluble

Conclusion 1 Magnesium chloride is an ionic compound and is soluble in water but insoluble in organic solvents. 2 Naphthalene and hexane are covalent compounds and are insoluble in water but soluble in organic solvents. 3 Hexane and water form two layers of liquid in a test tube. This shows hexane is insoluble in water. (C) To investigate the electrical conductivity of compounds Apparatus

Crucible, spatula, graphite rods, batteries, light bulb, switch, connecting wires, tripod stand, clay triangle and Bunsen burner.

Materials

Lead(II) bromide, magnesium chloride, sugar and naphthalene.

122

Figure 5.7 To investigate the electrical conductivity of compounds

Results Compound Lead(II) bromide

Naphthalene

Magnesium chloride

Sugar

State of compound

Observation

Inference

Solid

Light bulb does not light up

Conducts electricity in the liquid but not in the solid state

Molten

Light bulb lights up

Solid

Light bulb does not light up

Molten

Light bulb does not light up

Solid

Light bulb does not light up

Aqueous solution

Light bulb lights up

Solid

Light bulb does not light up

Aqueous solution

Light bulb does not light up

Conclusion 1 Ionic compounds such as lead(II) bromide and magnesium chloride do not conduct electricity in the solid state but conduct electricity in the molten state or in aqueous solution due to the presence of mobile ions. 2 Covalent compounds such as naphthalene and sugar do not conduct electricity in any state due to the absence of mobile ions. Discussion 1 Besides light bulbs, galvanometers and ammeters can be used to determine the electrical conductivity. The needle of a galvanometer will deflect if

Does not conduct electricity in any state

Conducts electricity in aqueous state but not in the solid state Does not conduct electricity in any state

an electric current flows through the liquid or aqueous solution. 2 Lead(II) bromide has a lower melting point than magnesium chloride. Therefore, lead(II) bromide is more suitable for use in the investigation of the electrical conductivity of ionic compounds in the liquid state compared to magnesium chloride. 3 Lead(II) bromide is insoluble in water whereas magnesium chloride is soluble in water. Therefore, magnesium chloride is more suitable for the investigation of the electrical conductivity of ionic compound in an aqueous solution. 4 Naphthalene does not dissolve in water. Some covalent compounds such as sugar are soluble in water but they still do not conduct electricity. 123

Chemical Bonds

5

SPM

Procedure ’04/P2 1 Three spatulas of lead(II) bromide solid is placed in a crucible. 2 Two graphite rods are dipped in the lead(II) bromide solid and the circuit is completed by connect­ing to the batteries and switch. 3 The switch is turned on and the bulb is checked if it lights up. 4 Lead(II) bromide is heated strongly until it melts. The switch is turned on again to check if the bulb lights up. 5 Steps 1 to 4 are repeated using naphthalene to replace lead(II) bromide. 6 Three spatulas of magnesium chloride solid is placed in a crucible. 7 The switch is turned on to check if the bulb lights up. 8 Water is added to the magnesium chloride. The mixture is stirred with a glass rod until all the magnesium solid dissolves in water. 9 The switch is turned on again to check if the bulb lights up. 10 Steps 6 to 9 are repeated using sugar to replace magnesium chloride.

Explanation for the Differences between the Physical Properties of Ionic and Covalent Compounds

consist of a covalent network of molecules forming a giant molecular structure. Some examples are carbohydrates, proteins and silicon dioxide. Strong covalent bonds hold the atoms together in these giant molecules. A lot of heat energy is required to overcome the strong covalent bonds of these giant molecules.

5

Melting points, boiling points and volatility 1 Ionic compounds have higher melting and boiling points than covalent compounds. This is because the electrostatic force of attraction between oppositely charged ions is very strong. A lot of heat energy is required to overcome this strong force of attraction. 2 There are two types of covalent compounds: (i) Simple molecules (ii) Giant molecules 3 Below are some examples of simple molecules and giant molecules. Most simple molecules exist as gases or liquids while giant molecules exist as solids. Simple covalent molecules

Solubility 1 Ionic compounds are soluble in water because they can form bonds with water molecules but are insoluble in organic solvents. 2 Covalent compounds are soluble in organic solvents because they can form bonds with organic solvent molecules but are insoluble in water. 3 Some covalent molecules can dissolve in water because they form hydrogen bonds with water. Some examples of these covalent compounds are sugar, ethanol, acetone and carboxylic acids. 4 Most giant covalent molecules are in­soluble in both water and organic solvents.

Giant covalent molecules

Carbon dioxide, CO2

Graphite

Tetrachloromethane, CCl4

Diamond

Ammonia, NH3

Silicon

Hydrogen gas, H2

Silicon dioxide

Oxygen gas, O2

Carbohydrates

Water, H2O

Protein

Electrical conductivity 1 All ionic compounds can conduct electricity in the molten state and in aqueous solution because the charged ions can move freely. Ionic compounds in the solid state do not conduct electricity because the ions are held by the strong electrostatic force of attraction in the lattice structure and are not free to move. 2 Covalent compounds do not conduct electricity in any state because they consist of molecules. There are no freely moving charged particles. 3 A few covalent compounds can dissociate into ions when dissolved in water. So, they can conduct electricity in aqueous solution. Some examples are ammonia (NH3), hydrogen chloride (HCl) and sulphur dioxide (SO2). 4 Electrical conductivity in the molten or liquid state is the best physical property to differentiate between ionic compounds and covalent compounds.

4 Melting and boiling points of covalent compounds of simple molecules such as tetrachloromethane, naphthalene, chlorine, bromine and iodine are low because the intermolecular forces of attraction between molecules (van der Waals forces of attraction) are very weak. Little heat energy is required to overcome the weak intermolecular forces. 5 The low boiling points of the covalent compounds cause the covalent compounds to be volatile (easily changed to the vapour state when heated). Hence covalent compounds should be kept far away from a heat source. 6 Some covalent compounds have high melting and boiling points because they Chemical Bonds

124

1 The flowchart below helps us to determine the type of chemicals. Chemical Does it conduct electricity in the solid state? No

Yes

Metal

Ionic compound or covalent compound Does it conduct electricity in the liquid state? No

5

Yes

Ionic compound

Covalent compound Does it have high melting and boiling points? No

Yes

Giant covalent compound

Simple covalent compound Does it dissolve in water? No

Yes

Polar covalent compound

Non-polar covalent compound

4

’07

The table shows electrical condutivity and melting points of three substances X, Y, and Z. Substance

Electrical conductivity

Melting point (°C)

Solid

Molten

Aqueous

X

No

No

No

–117

Y

No

No

No

80

Z

No

Yes

Yes

801

(a) State the types of structure and bonding of substances X, Y and Z. (b) Explain why substance X has a low melting point. (c) Show how the bonds are formed in substance Y and substance Z. (d) Based on the information in the table, predict the solubility of substances X, Y and Z in water. Solution (a) Substances X and Y are covalent molecules with covalent bonds. Substance Z is an ionic compound with a giant

network of ions held by ionic bonds. (b) Substance X consists of simple covalent molecules held by weak intermolecular forces of attraction that requires little energy to overcome them. (c) The bonds in substance Y are formed by the sharing of valence electrons to achieve the stable octet electron arrangement. The bonds in substance Z are formed by the transfer of electrons to achieve the stable octet electron arrangement. (d) Substances X and Y are insoluble in water. Substance Z is soluble in water.

125

Chemical Bonds

5

Uses of Covalent Compounds as Organic Solvents

• Covalent bonds formed between atoms are very strong. When a covalent compound is heated, heat energy is used to overcome the weak intermolecular forces of attraction which is the van der Waals forces of attraction. The covalent bonds are not broken. • Some covalent compounds are polar (examples: ethanol, hydrogen chloride, ammonia) and can dissolve in water. • Some ionic compounds are insoluble in water (examples: Al2O3, PbCl2).

1 Organic solvents are liquids consisting of simple covalent compounds. 2 Examples of organic solvents are ethanol, ether, acetone, benzene, turpentine, petrol and tetrachloromethane. 3 Organic solvents are used (a) to clean or remove stains that cannot be cleaned by water. (b) to extract organic compounds. (c) as solvents for drugs in medicine or compounds used in cosmetics. 4 Some common useful organic solvents and their uses are given in Table 5.5.

Property

Table 5.5 Uses of some organic solvents

Organic solvents

Uses

Petrol, kerosene and turpentine

To remove grease and paint stains

Ethanol, acetone

To prepare medicated solution, perfume and ink

Acetone, turpentine

To dissolve varnish compounds such as shellacs, lacquer and paints

Ether

Ionic compound

Simple molecule

Giant molecule

High

Low

High

(b) Volatility

Non-volatile

Very volatile

Non-volatile

(c) Solubility

Soluble in water, insoluble in organic solvents

Soluble in organic solvents, insoluble in water

Insoluble in both water and organic solvents

(d) Electrical conductivity

Conducts electricity in liquid and in aqueous solution

Does not conduct electricity in any state

(a) Melting and boiling points

To extract organic compounds

Covalent compound

5.4 2 The table shows the number of valence electrons of three elements represented by the letters Q, R and S.

Aluminium oxide, silicon dioxide, bromine, ethanol, ammonium nitrate, barium sulphate, naphthalene, tetrachloromethane, hydrogen iodide, copper(II) chloride 1 Based on the chemicals in the list above, choose (a) two compounds with high melting points. (b) two compounds which can conduct electricity in the aqueous state. (c) one compound which dissolves in water but cannot conduct electricity. (d) one covalent compound with a high melting point. (e) two compounds which can be used as organic solvents. (f) one compound which cannot dissolve in water but can conduct electricity in the molten state. Chemical Bonds

Element

Q

R

S

No. of valence electrons

2

4

7

Choose two elements which can combine to form a compound that (a) has low melting and boiling points, (b) conducts electricity in the liquid state, (c) is soluble in water. 3 Explain clearly why carbon dioxide is a gas whereas magnesium chloride is a solid at room temperature. Which compound can conduct electricity in the liquid state? Explain your answer.

126

(b) Double bond: Sharing of two pairs of electrons between two atoms. (c) Triple bond: Sharing of three pairs of electrons between two atoms. 14 For a covalent compound formed by a non-metal element P combined with another non-metal element, Q, the formula of the covalent compound formed will be PYQX. where

1 Inert gases are non-reactive because they have either a stable duplet or octet electron arrangement. 2 Inert gases do not release, accept or share electrons with other elements. 3 Atoms of elements from Group 1 to Group 17 have less than eight electrons. These elements will form chemical bonds. 4 Ionic bonds are formed between atoms of metals and atoms of non-metals. 5 Metal atoms from Groups 1, 2 and 13 will release their valence electrons to achieve the stable duplet or octet electron arrangement. 6 A positive ion (cation) is formed when an atom releases electrons. 7 Non-metal atoms from Groups 15, 16 and 17 will accept electrons to achieve the stable octet electron arrangement. 8 A negative ion (anion) is formed when an atom receives electrons. 9 The positive ions and negative ions are attracted to each other by strong electrostatic force of attraction in ionic bonds. 10 The total positive charge of the cation must be equal to the total negative charge of the anion in an ionic compound. 11 For an ionic compound consisting of cations Mb+ and anions Xa–, the formula of the ionic compound formed between them is written as

Number of electrons required by Q to achieve duplet or octet electron arrangement

Number of electrons required by P to achieve duplet or octet electron arrangement

15 The differences in physical properties between ionic compounds and covalent compounds:

Property

Melting point and boiling point

MaXb Number of electrons that will be received by X (or charge of X ion)

5

PYQX

Number of electrons that will be released by M (or charge of M ion)

12 A covalent bond is formed between atoms of nonmetals. 13 Non-metal atoms share valence electrons to achieve the stable duplet or octet electron arrangement in covalent compounds. There are three types of covalent bond. (a) Single bond: Sharing of one pair of electrons between two atoms.

127

Ionic compound

Covalent compound (simple molecule)

High

Low

Volatility

Non-volatile

Volatile

Solubility

Soluble in water, insoluble in organic solvents

Soluble in organic solvents, insoluble in water

Electrical conductivity

Conducts electricity in molten form and in aqueous solution

Does not conduct electricity in any state

Chemical Bonds

5 Multiple-choice Questions

5

5.1

Formation of Compounds

1 Which of the statements below best explains why elements of Group 18 of the Periodic Table such as neon, argon and krypton are chemically inert? A Exists as monatoms B Have 8 valence electrons C Have 8 electrons in every electron shell D Have weak covalent bonds between atoms 2 Which of the following proton numbers belong to an element that does not form chemical bonds? I 8 II 10 III 18 IV 20 A I and II only B I and III only C II and III only D II and IV only 3 Which of the following statements best explains the formation of chemical bonds between elements? A The elements have high reactivities. B The atoms of the elements have less than eight valence electrons. C The atoms of the elements have too many electrons. D The elements are either electropositive or electronegative.

5.2

Formation of Ionic Bonds

4 Oxide and fluoride ions have the same number of ’03 [Proton number: O, 8; F, 9] A charge C neutrons B electrons D protons Chemical Bonds

5 Which of the following substances has particles bonded with very strong electrostatic forces? A Carbohydrate B Graphite C Napthalene D Magnesium sulphide 6 The proton numbers of three elements represented by the letters P, Q and R are given as follows: P: 4

Q: 13

R: 16

What will be the charges of the most stable ions of P, Q and R? P ion

Q ion

R ion

A

+4

+3

+6

B

+2

+3

+6

C

+2

+3

–2

D

+4

–3

+2

7 The table shows the proton numbers for four elements R, S, T and U. Element

R

S

Proton number

8

11 13 20

T

U

Which of the following ions have the same number of electrons as the sulphide ion? [Proton number of S = 16] C T 3+ A U 2+ + B S D R 2– 8 The proton number of sodium and oxygen are 11 and 8 respectively. ’07 Which of the following occurs when sodium metal is burned in air to produce sodium oxide? A One oxygen atom receives one electron from one sodium atom. B One oxygen atom receives two electrons from one sodium atom. C One oxygen atom receives two electrons, one from each sodium atom. D One sodium atom and one oxygen atom share two electrons.

128

9 Element P and element Q are located in Group 2 and ’10 Group 17 in the Periodic Table respectively. Element P reacts with element Q to form a compound. What is the chemical formula of the compound? A PQ B P2Q C PQ2 D P2Q3 10 Element P reacts with element Q to form an ionic compound with formula P2Q3. If the electronic configuration for Q is 2.8.6, what is the possible electron arrangement of P? A 2.8.2 C 2.8.5 B 2.8.3 D 2.8.6 11 An ionic nitride compound has formula X3N2. What is the possible proton number of atom X if nitrogen is in Group 15 of the Periodic Table? A 2 C 13 B 12 D 15 12 Which of the following represents the valence electron arrangement ’06 for the compound sodium oxide? [Proton number: O, 8; Na, 11] A

B

C

D

What is the nucleon number of element J ? A 18 C 38 B 20 D 40 14 G is an element in Group 2 and X is an element in Group 16. Which of the following is the formula of the compound formed by G and X? A GX C G3X D GX2 B G2 X6

5.3

Formation of Covalent Bonds

15 The diagram shows the electron arrangement of an atom of element X.

18 The following diagram shows the electron arrangements of atoms ’11 X and Y. X and Y are not the actual symbols of the elements.

X Y Which pair of formula and the type of compounds is correct? Formula

Type of compound

A

XY3

Ionic

B

XY3

Covalent

C

X3Y

Ionic

D

X3Y

Covalent

19 The electronic configuration of the atoms of four elements represented by the letters P, Q, R and S are shown in the table below. Element

P

Q

R

S

Electronic 1 2.7 2.8 2.8.8.2 configuration

The atom of element X can form a covalent bond with another atom through the A acceptance of two electrons. B donation of two electrons. C sharing of two pairs of electrons. D sharing of six electrons. 16 An element forms a covalent compound with hydrogen. Atoms of this element also form covalent molecules themselves. This element may be I sodium III chlorine II oxygen IV nitrogen A I and II only B III and IV only C I, II and IV only D II, III and IV only 17 How many pairs of electrons are shared by the oxygen atoms in a molecule of oxygen gas? [Proton number: O, 8] A 1 C 3 B 2 D 4

Which of the following pairs of elements will form a covalent compound? A P and Q C Q and S B Q and R D P and R 20 Which of the following compounds have covalent double bonds? I O2 III CCl4 II N2 IV CO2 A I and II only B III and IV only C I and IV only D I, II and IV only 21 The diagram shows the electron arrangement in a compound ’05 formed by elements X and Y.

X

Y

A

2

14

B

4

6

C

14

16

D

18

18

22 Element R reacts with element S to produce a covalent compound with the formula R2S3. Which of the following electron arrangements of atoms R and S are true? Electron Electron arrangement arrangement of atom R of atom S A

2.8.2

2.3

B

2.8.2

2.5

C

2.8.3

2.6

D

2.8.5

2.6

23 The electronic configuration of atom E is 2.8.6 and atom G has ’03 four valence electrons. What is the formula of the compound formed between E and G? A GE2 C G2E B G4E2 D G2E4 24 The diagram shows the symbols of atoms P and Q. 28

16 P

14

Q 8

Atom P reacts with atom Q to form a compound. Calculate the relative molecular mass for the compound formed. A 21 C 60 B 43 D 113

5.4

Properties of Ionic and Covalent Compounds

25 Element W and element Y have proton numbers 17 and 19 as shown below.

Which of the following is the correct groups of X and Y in the Periodic Table of Elements?

129

Chemical Bonds

5

13 The following diagram shows the electron arrangement for the J 2+ ’07 ion. An atom of element J has 20 neutrons.

5

Which of the following are true of the compound formed between element W and element Y? I Has high melting point II Dissolves in organic solvents III Dissolves in water IV Conducts electricity in the solid state A I and III only B II and IV only C III and IV only D I, III and IV only 26 Which of the following elements will form an oxide that conducts electricity when dissolved in water? A Potassium B Silicon C Hydrogen D Aluminium 27 The electronic configuration of element X is 2.5. Which of the following statements are true of element X? I X forms a hydride with formula XH3. II Molecule X has a triple bond. III X forms an oxide with high melting point. IV Chlorine and X form a covalent compound. A I and III only B II and IV only C III and IV only D I, II and IV only 28 Which of the following statements best explains why bromine liquid is very volatile? A Bromine consists of diatomic covalent molecules. B Intermolecular forces of attraction between bromine molecules are weak. C Size of bromine molecules is small. D The covalent bond in bromine molecules is weak. 29 Covalent compounds do not conduct electricity because A they do not have free moving ions. B they have strong covalent bonds.

Chemical Bonds

C they are insoluble in water. D they can become vapour easily when heated. 30 Covalent compounds have low melting and boiling points because A they have weak covalent bonds. B they have weak intermolecular forces of attraction. C they are very volatile. D they cannot withstand strong heating. 31 Two elements represented by the letters X and Y have proton numbers as given in the table. Element

X

Y

Proton number

16

20

From the information given in the table, it can be deduced that I X and Y will form a compound with formula XY2. II Y can form ionic compounds but X can form both ionic and covalent compounds. III element X is a non-conductor of electricity whereas element Y is a conductor of electricity. IV the compound formed between element X and element Y has a high melting point. A I and II only B III and IV only C I, II and III only D II, III and IV only 32 Both carbon and magnesium form compounds with oxygen. Which of the following is the difference between magnesium oxide and carbon dioxide? A Magnesium oxide is soluble in water whereas carbon dioxide is insoluble in water. B Magnesium oxide solid conducts electricity whereas carbon dioxide solid does not. C Magnesium oxide is basic whereas carbon dioxide is neutral.

130

D Magnesium oxide has a high melting point whereas carbon dioxide has a low melting point. 33 Element X and element Y form a compound that has a low melting point and does not conduct electricity. Element X and Y maybe I hydrogen and sulphur II chlorine and oxygen III sodium and sulphur IV potassium and oxygen A I and II only B I and III only C II and IV only D III and IV only 34 Which of the following compounds have a high melting point and are soluble in water? I Calcium chloride II Sulphur dichloride III Copper(II) chloride IV Tetrachloromethane A I and II only B III and IV only C I and III only D II and IV only 35 Element X has an electronic configuration of 2.8.2. Which of the following are true of element X? I It can form negative ions. II It is a metallic element. III It can form a covalent compound with chlorine. IV It can form a basic oxide. A I and II only B II and IV only C I and III only D III and IV only 36 Which of the following compounds can be used as solvents for covalent compounds? I Water II Ethanol III Benzene IV Tetrachloromethane A I and II only B II and IV only C III and IV only D II, III and IV only

Structured Questions 1 Diagram 1 shows the Periodic Table of elements. The letters P, Q, R, S, T, U, V and W represent elements and are not the symbols of the actual elements. 1

2

13

14

15

16

17

18

1 2

P

3

R

4

V

Q

U

S

T

W

5



5

6

Diagram 1

Answer the following questions with reference only to the letters given above. [1 mark] (a) (i) State the type of chemical bond found in the compound formed between R and T. (ii) Draw a diagram showing the electron arrangement of the compound formed in (i). [2 marks] (iii) State a physical property of the compound formed in (i). [1 mark] (b) (i) Write the electronic arrangement of an ion formed by element Q. [1 mark] (ii) Q exists as diatomic molecules. Draw a diagram that shows the electron arrangement of the diatomic molecule. [2 marks] (c) Can element U form a compound? Explain your answer. [2 marks] (d) Write the equation for the reaction between V and water. [1 mark] 2 Diagram 2 shows two elements represented by the letters P and Q. (a) Write the electron arrangement of atom P and atom Q. (b) State the groups of elements P and Q in the Periodic Table.

16

23

P

[2 marks]

Q

11

8

[2 marks]

(c) Write an equation for the formation of (i) ion P. (ii) ion Q.

Diagram 2 [2 marks]

(d) P can combine with Q to form a compound. (i) Write the formula of the compound formed. [1 mark] (ii) What type of bond is formed in this compound? [1 mark] (iii) Draw the electron arrangement of this compound. [1 mark] (e) Q can combine with itself to form a compound. State a physical property of the compound formed. [1 mark] 3 Table 1 shows a few properties of substances P, Q, R and S. Conductivity

Substance

Melting point (°C)

Solubility in water

Solubility in organic solvent

Solid state

Liquid state

P

114

Insoluble

Soluble

Non-conducting

Non-conducting

Q

782

Soluble

Insoluble

Non-conducting

Conducting

R

840

Insoluble

Insoluble

Conducting

Conducting

S

2750

Insoluble

Insoluble

Non-conducting

(Does not exist in the liquid state)

Table 1

Based on the information given in Table 1, answer the questions below. (a) What type of bond is found in the particles of (i) substance P? (ii) substance Q? (b) Explain why P has a low melting point.

[1 mark] [1 mark] [2 marks]

131

Chemical Bonds

(c) Name a substance that has the same physical property as Q. [1 mark]



(d) Which of the above substances is a metal? [1 mark]

(e) Z can react with heated iron to form a compound. (i) Write the formula of this compound.

(e) Give reasons why Q conducts electricity in the liquid state but does not do so in the solid state. [2 marks]

[1 mark]

(ii) State one compound.

5

(f) Suggest one difference between the types of particles in substance R and in substance S. Give an example each for substance R and substance S. [2 marks]

physical

property

of

this

[1 mark]

5 Diagram 3 shows the diagram of the valence electron arrangement of a compound formed by two types of atoms represented by the letters P and Q.

4 Table 2 shows the number of protons and neutrons for a few atoms represented by the letters V, W, X, Y and Z. Atoms

(d) (i) State two atoms in Table 2 that are elements in the same group in the Periodic Table. [1 mark] (ii) Explain your answer in (i). [1 mark]

Number of protons Number of neutrons

V

3

4

W

6

6

X

6

7

Y

9

10

Z

17

18

Diagram 3

Table 2

(a) What is the formula of this compound?

[1 mark]

Use the information in Table 2 to answer the questions below.

(b) Predict the groups to which the atoms P and Q belong to in the Periodic Table. [2 marks]

(a) What is the nucleon number of element Y? [1 mark]

(c) Name the type of bond formed in the compound shown in Diagram 3. [1 mark]

(b) W can combine with Y to form a compound. (i) Name the type of chemical bond found in this compound. [1 mark] (ii) Write the formula of this compound. [1 mark] (iii) Predict the relative molecular mass of this compound. [1 mark]

(d) State three physical properties of this compound. [3 marks]

(e) Q is an element that exists as diatomic molecules. (i) Draw a diagram to show the valence electron arrangement of the molecule. [2 marks] (ii) Explain why element Q has a low boiling point. [1 mark]

(c) Name two atoms that are isotopes. Explain your answer. [2 marks]

Essay Questions 1 Diagram 1 shows the chemical symbols for three elements: carbon, sodium and chlorine:

Diagram 1

(a) Construct a table to compare the three elements in terms of the number of protons, the number of neutrons, the electron arrangement and the number of valence electrons. [4 marks] (b) Carbon reacts with chlorine to form a covalent compound whereas sodium reacts with chlorine to form an ionic compound. Explain how these ionic and covalent compounds are formed.

[8 marks]

(c) State four differences in physical properties between the two types of compounds formed in (b). [4 marks] Chemical Bonds

132

2

Group

Period

W

1

3

Y

14

2

Z

17

3

Element

Table 1

[14 marks]

(b) State the differences between an ionic compound and a covalent compound.

[6 marks]

3 (a) Using suitable examples, explain the meaning of single covalent bond, double covalent bond and triple covalent bond.

[15 marks]

(b) Explain why ionic compounds can conduct electricity in the liquid state and in aqueous solution whereas covalent compounds cannot.

[5 marks]

Experiments 1 A group of students carried out an experiment to determine the types of particles in three compounds X, Y and Z. The results obtained is recorded in Table 1.

Chemical compound

Physical state

X

Observation when shaken with Water

Acetone

Solid

Forms a colourless solution

X does not dissolve in acetone

Y

Liquid

Forms two immiscible layers

Y and acetone mix completely

Z

Liquid

Z and water mix completely

Z and acetone mix completely

Potassium chloride Table 1

(a) If the experiment is repeated using potassium chloride, predict what will be observed and complete Table 1.

[3 marks]

(b) Deduce the solubility of X, Y and Z in water and in acetone.

[3 marks]

(c) Classify the four chemical compounds used in Table 1 into (i) ionic compound, (ii) covalent compound.

[3 marks]

(d) Potassium nitrate is used as a nitrogenous fertiliser in agriculture. Which of the compounds X, Y and Z in the experiment may be potassium nitrate? Explain your answer.

[3 marks]

(e) Compound Z is used as a solvent in medicine and perfume. Name a substance that may be compound Z. Explain your answer. [3 marks] 2

Covalent compounds and ionic compounds differ in their ability to conduct electricity in the liquid state. By referring to the statement above, design a laboratory experiment relating to the difference in electrical conductivity of ethanol and sodium nitrate. In designing your experiment, the following aspects must be included: (a) Problem statement (b) Hypothesis (c) All the variables (d) List of substances and apparatus (e) Experimental procedures (f) Tabulation of data [17 marks]

133

Chemical Bonds

5



(a) With the help of an electron arrangement diagram, explain how two elements from Table 1 can combine to form (i) an ionic compound, and (ii) a covalent compound.

FORM 4 THEME: Interaction between Chemicals

CHAPTER

6

Electrochemistry

SPM Topical Analysis 2008

Year 1

Paper Section Number of questions

5

2009

2 A

B

C

–

1

–

3

1

–

6

2010

2 A

B

C

–

1

–

3

1

1

4

2011

2 A

B

C

1

–

1

3

1

–

5

3

2 A

B

C

1

–

–

–

ONCEPT MAP ELECTROCHEMISTRY types of substances

Electrolytes and non-electrolytes conversion of electrical energy into chemical energy

conversion of chemical energy into electrical energy

Electrolysis

Electrolysis of molten compounds

Electrolysis of aqueous solution

products

• Metals are formed at the cathode • Non-metals are formed at the anode

Voltaic cells

products

Selective discharge of ions determined by: • Position of ions in the electrochemical series • Concentration of ions • Types of electrodes

Uses of electrolysis: • Extraction of metals • Purification of metals • Electroplating of metals

can be used to determine

Various types of voltaic cells: • non-rechargeable cell and • rechargeable cell

To construct the electrochemical series: • Potential difference between two different metals • Displacement reaction of metals

Electrochemical series

Uses of the electrochemical series: • Predict the products of electrolysis • Determine the terminals of cells • Prediction of displacement reactions

6 Metals which are conductors are not regarded as electrolytes because they are not de­composed by the passage of an electric current. 7 The electrical conductivity of any substance is based on the movement of charged particles. (a) In metals, electricity is conducted by freely moving electrons.

Introduction 1 Electrochemistry is the study of the inter­ conversion of chemical energy and electrical energy. 2 The energy change in electrochemistry consists of the (a) conversion of electrical energy into chemical energy (in electrolysis), (b) conversion of chemical energy into electrical energy (in voltaic cells).

8 Most covalent substances are non-electrolytes because they consist of molecules, which do not form ions in aqueous solution. 9 Some covalent substances such as hydrogen chloride and ammonia are electrolytes because they react with water to produce freely moving ions. Examples:

Electrolytes and NonSPM electrolytes

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1 An electrolyte is a chemical compound which conducts electricity in the molten state or in an aqueous solution and undergoes chemical changes. 2 When an electrolyte conducts electricity, chemical changes occur and the electrolyte decomposes into its component elements at the electrodes. Electrolysis is said to have taken place. 3 Examples of electrolytes are acid solutions, alkali solutions, molten salts or aqueous salt solutions. 4 A non-electrolyte is a chemical compound which does not conduct electricity in any state. 5 Examples of non-electrolytes are metals and covalent substances such as naphthalene, latex and sugar solution.

6

6.1

(b) In electrolytes, electricity is conducted by freely moving ions: positive ions and negative ions.

HCl(g) + H2O(l) → H3O+(aq) + Cl–(aq) NH3(g) + H2O(l) → NH4+(aq) + OH–(aq) 10 Ions are charged (positive or negative) particles. All metal ions and hydrogen ions are positive ions (also known as cations). All non-metal ions are negative ions (also known as anions). All electrolytes will dissociate into cations and anions in the molten states or aqueous solutions.

6.1

Problem statement How to identify electrolytes and non-electrolytes? Hypothesis Substances that, in the molten state or in aqueous solution, conduct electricity and then undergo chemical reactions are electrolytes. Substances that do not conduct electricity in any state are non-electrolytes.

Apparatus

Crucible, spatula, carbon (graphite) electrodes, batteries, light bulb, switch, rheostat, connecting wires, tripod stand, clay pipe triangle and Bunsen burner.

Materials

Glucose, naphthalene, lead(II) bromide and potassium iodide.

Procedure (A) To investigate the electrical conductivity of substances in the solid state and in the molten state 1 A crucible is half-filled with lead(II) bromide solid.

Variables (a) Manipulated variable : Types of compounds (b) Responding variable : Electrical conductivity (c) Constant variable : Numbers of batteries, type of light bulb and amount of substance used 135

Electrochemistry

Experiment 6.1

To identify electrolytes and non-electrolytes

6

2 The crucible with its contents is placed on a clay triangle on a tripod stand. 3 Two carbon electrodes are dipped in the lead(II) bromide solid and are connected to the batteries, rheostat, switch and a light bulb with connecting wires (Figure 6.1). 4 The switch is turned on and the light bulb is checked if it lights up. 5 The lead(II) bromide solid in the crucible is heated up until it melts. The switch is turned on again to check if the light bulb lights up. 6 Steps 1 to 5 of the experiment are repeated using naphthalene in place of lead(II) bromide.

(B) To investigate the electrical conductivity of substances in the solid state and in aqueous solutions 1 Three spatulas of potassium iodide solid are put in a beaker. 2 Two carbon electrodes are dipped in the potassium iodide solid and then connected to the batteries, rheostat, switch and a light bulb with connecting wires (Figure 6.2). 3 The switch is turned on and the light bulb is checked if it lights up. 4 Distilled water is added to the beaker and the mixture is stirred until all the potassium iodide has dissolved. 5 The switch is turned on again and the light bulb is checked if it lights up. 6 Steps 1 to 5 of the experiment is repeated using glucose in place of potassium iodide.

Figure 6.1 To investigate the conductivity of substances in the solid state and in the molten state

Figure 6.2 To investigate the conductivity of substances in the solid state and in aqueous solution

Results Chemical substance Lead(II) bromide Naphthalene Potassium iodide

Glucose

Physical state

Does the light bulb light up?

Inference

Solid

No

No noticeable change

Non-electrolyte

Liquid (molten)

Yes

Brown gas is evolved

Electrolyte

Solid

No

No noticeable change

Non-electrolyte

Liquid (molten)

No

No noticeable change

Non-electrolyte

Solid

No

No noticeable change

Non-electrolyte

Aqueous solution

Yes

Solution turns to a brown colour

Electrolyte

Solid

No

No noticeable change

Non-electrolyte

Aqueous solution

No

No noticeable change

Non-electrolyte

Disscussion 1 Electrolytes can conduct electricity because they have freely moving charged ions. 2 In the solid state, an ionic compound is not an electrolyte because the ions are held together by strong ionic bonds and are not free to move. 3 When an ionic compound is heated to its melting point, the heat energy supplied overcomes Electrochemistry

Observation: Does reaction occur?

the ionic bond. The ions are free to move in the molten state. Hence ionic compounds are electrolytes in the molten state. 4 When ionic compounds are dissolved in water, the water molecules separate the ions into freely moving ions. Hence aqueous solutions of ionic compounds are electrolytes. 136

5 Covalent compounds such as glucose and naphthalene do not conduct electricity because they consist of molecules which are uncharged particles. 6 When electricity passes through molten lead(II) bromide, the brown gas evolved is bromine gas. 7 When electricity passes through aqueous potassium iodide solution, the brown iodine solution is formed. 8 The presence of freely moving ions enable a molten compound or aqueous solution to conduct electricity.

6

Conclusion 1 Lead(II) bromide is an electrolyte in the liquid state but not in the solid state. 2 Potassium iodide is an electrolyte in aqueous solution but not in the solid state. 3 Lead(II) bromide and potassium iodide are ionic compounds. Ionic compounds are electrolytes in the molten state or aqueous solution but are non-electrolytes in the solid state. 4 Naphthalene and glucose are covalent compounds and are non-electrolytes in any state.

6.2 All liquid or aqueous solutions of ionic compounds are electrolytes. All covalent compounds (except those that can dissociate to form ions when dissolved in water) are non-electrolytes. An electrolyte can conduct electricity because it has freely moving ions.

Electrolysis of Molten Compounds

Meaning of Electrolysis and Electrolytic Cell • Electrolysis is a process of decomposition of an electrolyte by an electric current. • An electrolytic cell consists of batteries, a cathode, an anode and an electrolyte consisting of cations and anions. • During electrolysis, anions (negatively-charged ions) move towards the anode and cations (positively-charged ions) move towards the cathode. • Graphite or platinum is usually used as electrodes because they are inert; they do not react with the electrolyte or the products of electrolysis.

6.1 1 (a) (i) What is meant by the term electrolyte? Give two examples of electrolytes. (ii) What is meant by the term non-electrolyte? Give two examples of non-electrolytes. (b) State the difference between a conductor and an electrolyte. 2 Classify the ten substances below into electrolytes and non-electrolytes. Molten sulphur, molten zinc oxide, zinc oxide solid, aqueous zinc chloride solution, zinc metal, molten zinc, acetone, aqueous glucose solution, aqueous ethanoic acid solution, molten sodium chloride. 3 Explain why magnesium chloride solid cannot conduct electricity but becomes an electrolyte when it is in the molten state.

Electrons flow from the anode to the cathode through the connecting wire in the external circuit.

A rheostat can be used to control the quantity of electric current that flows through the electrolyte.

An ammeter or a bulb can be used to indicate the flow of electric current.

• Anode is the electrode connected to the positive terminal of the batteries. • At the anode, anions discharge by donating electrons. Example: 2Cl– → Cl2 + 2e–

• Cathode is the electrode that is connected to the negative terminal of the batteries. • At the cathode, cations discharge by accepting electrons, Example: Na+ + e– → Na

An electrolyte conducts electricity in aqueous solution or in the molten state as a result of the presence of freely moving cations (positive ions) and anions (negative ions).

Figure 6.3 Electrolytic cell

137

Electrochemistry

Electrolysis Process 1 In the solid state, the cations and anions of an electrolyte are unable to move freely because they are held together by strong ionic bonds in fixed positions in a lattice.

6

3 Generally, a molten compound produces Am+ cations and Bn– anions. SPM

’09/P1



2 When the solid is heated until it melts (in the molten form), the heat energy supplied is used to overcome the electrostatic force of attraction between the ions. Hence, the cations and the anions are free to move.

AnBm

AnBm(s) → nAm+(l) + mBn–(l) cation anion 4 Two steps occur during electrolysis. (a) Movement of ions to the electrodes Cations (positive ions) move towards the cathode (negative electrode) whereas anions (negative ions) move towards the anode (positive electrode). (b) Discharge of ions at the electrodes Cations are discharged by accepting (gaining) electrons. Generally: An+ + ne– → A Anions are discharged by donating (losing) electrons. Generally: Bn– → B + ne–

Example: PbBr2(s) → Pb2+(l) + 2Br–(l) lead(II) ion bromide ion • Examples of cations are hydrogen ions, H+, ammonium ions, NH4+ and metal ions such as K+, Mg2+, Pb2+ and Al3+. • Examples of anions are Cl–, Br–, I–, OH–, SO42– and NO3–.

5 The electrons donated by anions at the anode are accepted by the cations at the cathode. The discharge of ions at the anode and cathode results in the (a) conduction of electricity by the electrolyte. (b) decomposition of the electrolyte into its component elements.

Writing Half-equations for the Discharge of Ions at the Anode and the Cathode 4 Addition of the two half-equations at the anode and cathode gives the overall chemical equation for the reaction. Half-equation at the anode: Bn– → B + ne– ………… equation 1 Half-equation at the cathode: An+ + ne– → A ………… equation 2 Overall chemical reaction equation: An+ + Bn– → A + B …… equation 1 + equation 2

1 The reactions that take place at the anodes or the cathodes involve ions and electrons. The equations representing these reactions are known as half-equations. 2 The half-equation of the anode shows the discharge of the anions by the loss of electrons to produce neutral atoms. Half-equation at the anode:

Bn– → B + ne–

3 The half-equation of the cathode shows the discharge of the cations by the gain of electrons to produce neutral atoms. Half-equation at the cathode: Electrochemistry

An+ + ne– → A 138

1 Step 2: Balance the number of atoms on both sides of the equation.

Write the half-equation for the discharge of oxide ions, O2– at the anode.

2O2– → O2

Solution Step 1: Write the formulae of the reactant and the product of the reaction.

Step 3: Balance the charge by adding in the right number of electrons.

O2– → O2

6

2O2– → O2 + 4e–

2 When molten aluminium oxide, Al2O3 is electrolysed, aluminium metal and oxygen gas are produced. (a) Write half-equations for the reactions at the anode and cathode. (b) Write the overall chemical equation of electrolysis.

At the cathode: Cations receive electrons Al3+ + 3e– → Al ……… (2) (Number of electrons donated at the anode = number of electrons accepted at the cathode)

Solution At the anode: Anions lose electrons

Equation (1)  3: 6O2– → 3O2 + 12e– Equation (2)  4: 4Al3+ + 12e– → 4Al Overall chemical equation: 4Al3+(l) + 6O2–(l) → 4Al(l) + 3O2(g)

2O2– → O2 + 4e– ……… (1)

To investigate the electrolysis of molten lead(II) bromide

SPM

’06/P2

Apparatus Crucible, spatula, graphite electrodes, batteries, light bulb, ammeter, switch, rheostat, connecting wires, tripod stand, clay pipe triangle and Bunsen burner.

Procedure 1 A crucible is half-filled with lead(II) bromide solid. 2 The solid lead(II) bromide is heated strongly until it melts to a molten state. 3 Two carbon electrodes are dipped in the molten lead­­ (II) bromide and are then connected to batteries, rheo­stat, switch and light bulb by the connecting wires (Figure 6.4). 4 Electric current is allowed to flow through for 15 minutes and the changes that occur at the light bulb, ammeter, cathode and anode are recorded.

Figure 6.4 To investigate the electrolysis of lead(II) bromide

139

Electrochemistry

Activity 6.1

Materials Lead(II) bromide.

Results

6

Apparatus

Observation

Inference

Light bulb

Light bulb lights up

Molten lead(II) bromide conducts electricity

Ammeter

Ammeter needle is deflected

Anode

Pungent brown gas that changes damp blue litmus paper to red is evolved

Bromine gas is evolved

Cathode

Shiny grey metal is deposited

Lead metal is formed 5 At the cathode, a lead(II) ion discharges by accepting 2 electrons to form a lead metal atom.

Discussion 1 Solid lead(II) bromide consists of lead(II) ions, Pb2+ and bromide ions, Br– that are held by strong ionic bonds. 2 When heated until it melts to form the molten state, lead(II) ions and bromide ions are free to move.

Pb2+ + 2e– → Pb 6 A larger quantity of the products would be formed at the anode and the cathode if the electroysis is carried out (a) for a longer time period, (b) with a larger current.

PbBr2(s) → Pb2+(l) + 2Br –(l) 3 During electrolysis, bromide ions are attracted to the anode while lead(II) ions are attracted to the cathode. 4 At the anode, a bromide ion discharges by releasing one electron to form a bromine atom. Two bromine atoms combine to form a bromine molecule that exists as a brown gas.

Conclusion 1 The lighting up of the bulb and the deflection of the ammeter needle shows that molten lead(II) bro­ mide is an electrolyte and can conduct electricity. 2 Electrolysis of molten lead(II) bromide produces bromine gas at the anode and lead metal at the cathode.

2Br – → Br2 + 2e–

formed at the cathode while the non-metal component is formed at the anode. 2 Products of electrolysis of molten electrolytes can be predicted by the following steps.

Prediction of the Products of Electrolysis of Molten Electrolytes 1 When molten compound is electrolysed, the metal component of the compound is Step 1

Step 2

Step 3

Identifying the cations and anions that are present in the molten electrolyte. Generally, a salt with formula AxBy will produce ions as below.

Identify the cathode and anode and the movement of ions in the electrolyte to the electrodes. • Cations move to cathode (electrode connected to the negative terminal of the battery). • Anions move to anode (electrode connected to the positive terminal of the battery).

Write the half-equations for the discharge of the ions at the electrode. • At the cathode: Cations discharge (losing the positive charge) by accepting electrons.

AxBy → xAy+ + yBx– cation anion

Electrochemistry

140

Ay+ + ye– → A • At the anode: Anions discharge (losing the negative charge) by donating electrons. Bx– → B + xe–

Electrolysis of molten sodium chloride, NaCl

Electrolysis of molten lead(II) oxide, PbO 1 Cations and anions produced from the molten lead(II) oxide are lead(II) ions, Pb2+ and oxide ions, O2–.

1 The cations and anions produced from molten sodium chloride are sodium ions, Na+ and chloride ions, Cl–.

PbO → Pb2+ + O2–

NaCl → Na+ + Cl–

2 Lead(II) ions are attracted to the cathode while the oxide ions are attracted to the anode. 3 At the cathode, a lead(II) ion accepts 2 electrons to form a lead metal atom (grey metal).

2 The sodium ions are attracted to the cathode while the chloride ions are attracted to the anode. 3 At the cathode, a sodium ion accepts an electron to form a sodium metal atom (grey metal).

6

Pb2+ + 2e– → Pb 4 At the anode, an oxide ion donates 2 electrons to form an oxygen atom. Two oxygen atoms combine to form an oxygen gas molecule.

Na+ + e– → Na

2O2– → O2 + 4e– 4 At the anode, a chloride ion donates an electron to form a chlorine atom. Two chlorine atoms combine to form a chlorine molecule (greenish-yellow gas). The term electrolysis was introduced by Michael Faraday. ‘Lysis’ means loosening in Greek. Electrolysis means loosening by electric current.

2Cl– → Cl2 + 2e–

6.2 1 Write the formulae of the ions produced in the molten chemical compounds in the table below. Identify the ions that move to the anode and to the cathode respectively during electrolysis by completing the table below. Name of compound

Ions that move to the

Ions produced

anode

cathode

(a) Zinc chloride (b) Magnesium oxide 2 For every compound below, write the half-equations that occur at the anode and the cathode, as well as the products formed during electrolysis. Name of compound

Half-equation at the

Products formed at the

anode

anode

cathode

cathode

(a) Calcium oxide (b) Aluminium iodide 3 The figure shows the arrangement of apparatus used in an experiment. The switch is turned on for 20 minutes. (a) Predict the observations at (i) electrode P, (ii) electrode Q. (b) What are the products formed at (i) electrode P, (ii) electrode Q? (c) Write the half-equations for the reactions that take place at (i) electrode P, (ii) electrode Q.

141

Electrochemistry

6.3

Electrolysis of Aqueous Solutions

(A) Positions of ions in the electrochemical series 1 The tendency of ions to be selectively discharged at electrodes depends on their positions in a series known as the electrochemical series. 2 The tendency of the ions to be discharged is shown below in ascending order.

6

Identifying Cations and Anions in Aqueous Solutions 1 A molten salt consists of one type of cation and one type of anion only. During electrolysis, the products formed are the component elements of the compound. 2 An aqueous solution is produced when a solute is dissolved in water. An aqueous solution of a salt consists of two types of cations (cations of the salt and hydrogen ions, H+) and two types of anions (anions of the salt and hydroxide ions, OH–). 3 H+ ions and OH– ions are always present together with the ions produced from the dissociation of salts in aqueous solutions. This is because water dissociates partially to form hydrogen ion and hydroxide ion. H2O

Cation K+ Na+ Mg2+ Al3+ Zn2+ Fe2+ Sn2+ Pb2+ H+ Cu2+ Hg+ Ag+

3 The lower the position of the ion in the electrochemical series, the easier the ion will be discharged. 4 For example, in the electrolysis of aqueous sodium sulphate (Na2SO4) solution, with carbon electrodes,

H+ + OH–

4 For example, in an aqueous sodium chloride solution, 4 types of ions are present in the solution from the dissociation of NaCl and H2O.

Na2SO4 → 2Na+ + SO42– H2O H+ + OH–

cation anion NaCl ⎯→ Na+ + Cl– + H2O + OH– H

The cations present are Na+ ions and H+ ions. The anions present are SO42– ions and OH– ions. 5 Both the Na+ ions and the H+ ions are attracted to the cathode. However, only the H+ ions are discharged at the cathode because the H+ ion is lower in position than the Na+ ion in the electrochemical series. Half-equation at the cathode:

The four types of ions present are Na+, Cl–, H+ and OH–. 5 During electrolysis of an aqueous salt solution, 2 types of cations will move to the cathode while 2 types of anions will move to the anode. For example, during electrolysis of aqueous sodium chloride solution, the cations that move to the cathode are Na+ ions and H+ ions. The anions that move to the anode are Cl– ions and OH– ions. However, only one type of cation and anion will be discharged at each electrode.

2H+ + 2e– → H2 Hence, hydrogen gas is produced at the cathode. 6 Both the SO42– ions and the OH– ions are attracted to the anode. However, only the OH– ions are discharged at the anode because the OH– ion is lower in position than the SO42– ion in the electrochemical series. Half-equation at the anode:

Factors that Determine the Selective Discharge of Ions at the Electrodes The factors that determine the types of ions to be discharged at the electrodes are: (a) Positions of ions in the electrochemical series (b) Concentration of ions in the solution (c) Types of electrodes used Electrochemistry

tendency to discharge increases

Anion F– SO42– NO3– Cl– Br– I– OH–

4OH– → 2H2O + O2 + 4e– Hence, oxygen gas is produced at the anode. 142

(C) Effect of types of electrodes used

(B) Effect of concentration of ions in the solution

SPM

’04/05 P2

1 The common materials used as electrodes are carbon and platinum because they are inert. Both of these materials do not react with the electrolytes or the products of electrolysis. 2 The types of electrodes used can determine the types of ions discharged in electrolysis. 3 For example, in the electrolysis of aqueous copper(II) sulphate (CuSO4) solution,

1 When the concentration of a particular type of ion is high, that ion will more likely to be discharged in electrolysis irrespective of its position in the electrochemical series. 2 Usually, in the electrolysis of concentrated halide (Cl–/Br–/I– ions) solutions, the concentration of the halide ion is always higher than the hydroxide ion, OH–. Hence, halide ions will be selectively discharged at the anode. 3 For example, in the electrolysis of aqueous copper(II) chloride solution, CuCl2 with carbon electrodes,

6

CuSO4 → Cu2+ + SO42– H2O H+ + OH– both the anions, sulphate ions, SO42– and hydroxide ions, OH– are attracted to the anode. (a) If carbon is used as the electrodes, OH– ions are discharged at the anode because of the position of OH– ion in the elec­trochemical series. Half-equation at the anode:

CuCl2 → Cu2+ + 2Cl– H2O H+ + OH–

4OH– → 2H2O + O2 + 4e– Hence, oxygen gas is produced at the anode. (b) If copper is used as the anode, both SO42– ions and OH– ions are not discharged. Instead the copper anode dissolves by releasing electrons to form copper(II) ions, Cu2+. Half-equation at the anode:

Both types of anions, chloride ion, Cl– and hydroxide ions, OH– are attracted to the anode. (a) If a concentrated solution is electrolysed, chloride ions, Cl–, will be selectively discharged at the anode because the concentration of the chloride ions is higher than that of the hydroxide ions. Half-equation at the anode:

Cu → Cu2+ + 2e– Hence, the mass of anode decreases. Copper acts as an active electrode here because it takes part in the chemical reaction during electrolysis.

2Cl– → Cl2 + 2e– Hence chlorine gas is evolved at the anode. (b) If a dilute solution is electrolysed, hydroxide ions, OH–, will be discharged at the anode because the concentration of the chloride ions is low and OH– ion is more easily discharged because of its position in the electrochemical series. Half-equation at the anode:

Generally, in the electrolysis of a halide solution using carbon electrodes: • A concentration of more than 0.5 mol dm–3 halide solution is concentrated solution whereby the halide ions will be selectively discharged at the anode. • A concentration of less than 0.005 mol dm–3 halide solution is considered a dilute solution whereby the hydroxide ions will be selectively discharged at the anode. • Electrolysis of a solution with a concentration of between 0.005 mol dm–3 and 0.5 mol dm–3 may result in two types of products: halogen and oxygen to be produced at the anode.

4OH– → 2H2O + O2 + 4e– Hence, oxygen gas is evolved at the anode.

143

Electrochemistry

6.2 To investigate the effect of the positions of ions in the electrochemical series on the selective discharge of ions and the products of electrolysis of aqueous solutions

6

Problem statement How do the positions of ions in the electrochemical series determine the types of ions selectively discharged during electrolysis?

Procedure 1 Aqueous copper(II) sulphate solution is put into an electrolytic cell with carbon electrodes. 2 Two test tubes, filled with copper(II) sulphate solution are inverted over the carbon anode and cathode respectively (Figure 6.5).

Hypothesis If an aqueous solution consists of more than one type of ion, the lower the position of the ion in the electrochemical series, the higher the tendency it is for the ion to be discharged. Variables (a) Manipulated variable : Position of ions in the electrochemical series (b) Responding variable : Types of ions discharged at the anode and the cathode (c) Constant variable : Concentration of electro­ lytes, types of electrodes, duration of electrolysis Apparatus Batteries, electrolytic cell, carbon electrodes, ammeter, switch, connecting wires with crocodile clips and test tubes.

Figure 6.5 Electrolysis of copper(II) sulphate solution

3 The switch is turned on and electric current is allowed to flow for 15 minutes. 4 Steps 1 to 3 of the experiment are repeated by replacing copper(II) sulphate solution with dilute sulphuric acid and sodium nitrate solution in turn.

Materials Aqueous 0.5 mol dm–3 copper(II) sulphate, CuSO4 solution, 0.5 mol dm–3 dilute sulphuric acid, H2SO4 and 0.5 mol dm–3 sodium nitrate, NaNO3 solution. Results Electrolyte

Observation

Copper(II) sulphate At the cathode solution • Brown deposit is formed

Inference Copper metal is deposited

At the anode • Gas bubbles are formed Oxygen gas is produced • Gas produced lights up a glowing wooden splint

Experiment 6.2

Colour of electrolyte • The blue colour of the solution becomes paler Dilute sulphuric acid and sodium nitrate solution

At the cathode Hydrogen gas is produced • Gas bubbles are formed • When a lighted wooden splint is placed near the mouth of the test tube, a ‘pop’ sound is produced At the anode • Gas bubbles are formed • Gas lights up a glowing wooden splint

Electrochemistry

Concentration of Cu2+ ion decreases

144

Oxygen gas is produced

8 The ions present in the sodium nitrate solution are Na+ ions, NO3– ions, H+ ions and OH– ions.

Discussion 1 The ions present in the aqueous copper(II) sulphate solution are Cu2+ ions, SO42– ions, H+ ions and OH– ions. CuSO4 → Cu2+ + SO42– H2O H+ + OH–

9 Both types of cations, H+ ions and Na+ ions are attracted to the cathode. H+ ions are selectively discharged at the cathode because the position of the H+ ion is lower than that of the Na+ ion in the electrochemical series. Hydrogen ions are discharged to form hydrogen gas. Half-equation at the cathode:

2 Both types of cations, Cu2+ ions and H+ ions are attracted to the cathode. Cu2+ ions are selectively discharged at the cathode because the position of the Cu2+ ion is lower than that of the H+ ion in the electro­che­mical series. A Cu2+ ion is discharged by accepting 2 electrons to form a copper atom at the cathode. Half-equation at the cathode:

2H+ + 2e– → H2 10 Both types of anions, NO3– and OH– ions are attracted to the anode. OH– ions are selectively discharged at the anode because of the position of the OH– ion in the electrochemical series. Hydro­xide ions are discharged to form water and oxygen. Half-equation at the anode:

Cu2+ + 2e– → Cu 3 Both types of anions, SO42– ions and OH– ions are attracted to the anode. OH– ions are discharged at the anode because the position of the OH– ion is lower than that of the SO42– ion in the electroche­ mical series. 4OH– ions are dischar­ged by donating 4 electrons to form water and oxygen. Half-equation at the anode:

4OH– → 2H2O + O2 + 4e– 11 Electrolysis of dilute sulphuric acid and sodium nitrate solution is actually electrolysis of water because H+ ions and OH– ions are discharged. The decrease in the concentrations of H+ ions and OH– ions in the solution results in the increase of sulphuric acid and sodium nitrate concentration during electrolysis. 12 The ratio of H2 gas to O2 gas produced is 2 : 1. This is because the release of 4 electrons in the for­ ma­ tion of 1 molecule of O2 results in the formation of 2 molecules of H2 when these 4 electrons are accepted by 4 H+ ions.

4OH– → 2H2O + O2 + 4e– 4 The blue colour of the copper sulphate solution is due to the presence of copper(II) ion, Cu2+. The blue colour of the electrolyte becomes paler during electrolysis because the Cu2+ ion concen­ tra­tion decreases when Cu2+ ions are discharged. 5 The ions present in the dilute sulphuric acid are SO42– ions, H+ ions and OH– ions. H2SO4 → 2H+ + SO42– H2O H+ + OH–

Conclusion 1 Electrolysis of aqueous copper(II) sulphate solution using carbon electrodes produces copper at the cathode and oxygen gas at the anode. 2 Cu2+ ions and OH– ions are selectively discharged because of their lower positions in the electrochemical series. 3 Electrolysis of dilute sulphuric acid and sodium nitrate solution using carbon electrodes produces hydrogen gas at the cathode and oxygen gas at the anode. 4 H+ ions and OH– ions are selectively discharged because their positions are lower in the electrochemical series. The hypothesis is accepted.

6 Hydrogen ions are attracted to the cathode. H+ ion is discharged by receiving one electron to form a hydrogen atom. Two hydrogen atoms will combine to form a hydrogen gas molecule. Half-equation at the cathode: 2H+ + 2e– → H2 7 Both types of anions, SO42– ions and OH– ions are attracted to the anode. OH– ions are selectively discharged at the anode because of the position of the OH– ion in the electrochemical series. Four OH– ions are discharged to form water and oxygen. Half-equation at the anode: 4OH– → 2H2O + O2 + 4e–

145

Electrochemistry

6

NaNO3 → Na+ + NO3– H2O H+ + OH–

6.3

SPM

’09/P2

To investigate the effect of the concentration of ions on the selective discharge of ions and the products of electrolysis of aqueous solutions

6

Problem statement How does the concentration of ions determine the types of ions discharged during electrolysis? Hypothesis Ions of higher concentration will be selectively discharged during electrolysis. Variables (a) Manipulated variable: Concentration of ions in the solution (b) Responding variable: Types of ions to be discharged at the anode and cathode (c) Constant variables: Types of ions in the electrolyte, types of electrodes, duration of electrolysis.

Figure 6.6 Electrolysis of copper(II) chloride solution

Apparatus Batteries, electrolytic cell, carbon electrodes, ammeter, switch, connecting wires with crocodile clips and test tubes. Materials Aqueous 2.0 mol dm–3 copper(II) chloride, CuC12 solution and aqueous 0.001 mol dm–3 copper(II) chloride solution. Procedure 1 Concentrated aqueous copper(II) chloride solution of 2.0 mol dm–3 is put into an electrolytic cell with carbon electrodes.

2 Two test tubes, filled with copper(II) chloride solution are inverted over the carbon anode and cathode respectively (Figure 6.6). 3 The switch is turned on and electric current is allowed to flow for 15 minutes. 4 Any change in colour of the electrolyte and any other changes that occur around the carbon electrodes are recorded. 5 Steps 1 to 4 of the experiment are repeated using the dilute copper(II) chloride solution of 0.001 mol dm–3 to replace the concentrated copper(II) chloride solution.

Results Electrolyte

Experiment 6.3

Concentrated copper(II) chloride solution of 2.0 mol dm–3

Dilute copper(II) chloride solution of 0.001 mol dm–3

Electrochemistry

Observation

Inference

At the cathode: Brown deposit is formed

Copper metal is produced

At the anode: Bubbles of pungent greenish-yellow gas are produced. The gas turns the damp blue litmus paper to red and then bleaches it

Chlorine gas is produced

Colour of electrolyte: The blue colour of the solution becomes paler

Concentration of Cu2+ ions in copper(II) chloride solution decreases

At the cathode: Brown deposit is formed

Copper metal is produced

At the anode: Bubbles of colourless gas are produced. The gas lights up a glowing wooden splint

Oxygen gas is produced

Colour of electrolyte: The blue colour of the solution becomes paler

Concentration of Cu2+ ions in copper(II) chloride solution decreases

146

Cu2+ ions are discharged because the position of Cu2+ ions is lower than that of H+ ions in the electrochemical series. Hence copper metal is deposited.

Discussion 1 Aqueous copper(II) chloride consists of Cu2+ ions, H+ ions, OH– ions and Cl– ions. CuCl2 → Cu2+ + 2Cl– H2O H+ + OH–



(b) The blue colour of the solution becomes paler because the concentration of the Cu2+ ions decreases when Cu2+ ions are discharged at the cathode during electrolysis. Conclusion 1 In the electrolysis of concentrated aqueous copper­(II) chloride solution, copper metal is pro­duced at the cathode and chlorine gas is produced at the anode. At the anode, the Cl– ions are selective­ly dis­­char­ged, producing chlorine gas because the con­ cen­ tration of Cl– ions is – higher than that of OH ions. 2 In the electrolysis of dilute aqueous copper(II) chloride solution, copper metal is produced at the cathode and oxygen gas is produced at the anode. At the anode, OH– ions are selectively discharged, producing oxygen gas because the concentration of Cl– ions is low. 3 The type of ions that is selectively discharged at the electrode is determined by the concentration of the ions. The hypothesis is accepted.

Half-equation at the anode: 2Cl– → Cl2 + 2e–

Hence chlorine gas is produced. 3 Electrolysis of dilute aqueous copper(II) chloride solution (0.001 mol dm–3): At the anode: 2 types of anions, Cl– ions and OH– ions are attracted to the anode. OH– ions are discharged because the concentration of Cl– ions is lower than that of OH– ions.

Half-equation at the anode: 4OH– → 2H2O + O2 + 4e–

Hence oxygen gas is produced. 4 (a) At the cathodes of both concentrated and dilute aqueous copper(II) chloride solution:

6.4

6

Cu2+ + 2e– → Cu

2 Electrolysis of concentrated aqueous copper(II) chloride solution (2 mol dm–3): At the anode: 2 types of anions, Cl– ions and OH– ions are attracted to the anode. Cl– ions are discharged because the concentration of Cl– ions is higher than that of OH– ions.

Half-equation at the anode:

SPM

’09/P3

Hypothesis The products of electrolysis of copper(II) sulphate solution with copper electrodes are different from that with carbon electrodes.

Apparatus Batteries, electrolytic cell, carbon electrodes, copper electrodes, ammeter, switch, rheostat, connecting wires with crocodile clips and test tubes.

Variables (a) Manipulated variable : Types of electrodes (b) Responding variable : Products of electrolysis

Materials Aqueous 1.0 mol dm–3 copper(II) sulphate, CuSO4 solution 147

Electrochemistry

Experiment 6.4

To investigate the effect of the types of electrodes on the selective discharge of ions and the products of electrolysis of aqueous solutions (c) Constant variables : Types of ions in the Problem statement electrolyte and the concen­ How do the types of electrodes determine the types tration of ions of ions discharged during electrolysis?

6

Procedure 1 Aqueous 1.0 mol dm–3 copper(II) sulphate solution is put into an electrolytic cell with carbon electrodes. 2 A test tube filled with copper(II) sulphate solution is inverted over the carbon anode (Figure 6.7). 3 The switch is turned on and the electric current is allowed to flow for 15 minutes. 4 Any change in colour of the electrolyte and any other changes that occur around the carbon electrodes are recorded. 5 Steps 1 to 4 of the experiment are repeated using copper electrodes to replace carbon electrodes.

Figure 6.7 Electrolysis of copper(II) sulphate solution

Results Type of electrodes Carbon

Copper

Observation At the cathode: Brown deposit is formed

Copper metal is produced

At the anode: Bubbles of colourless gas are produced The gas lights up a glowing wooden splint

Oxygen gas is produced

Colour of electrolyte: The blue colour of the solution becomes paler

Concentration of Cu2+ ions decreases

At the cathode: Formation of brown deposit makes the cathode thicker

Copper metal is produced

At the anode: Anode corrodes and becomes thinner

Copper anode dissolves to form Cu2+ ions

Colour of electrolyte: The blue colour of the solution remains unchanged

Concentration of Cu2+ ions in copper(II) sulphate remains constant

4 The colour intensity of the blue solution decreases because the concentration of Cu2+ ions in the copper(II) sulphate decreases. 5 During electrolysis, the concentration of OH– ions decreases, leaving H+ ions behind. As a result, the solution becomes acidic. 6 If copper is used as the anode, both SO42– ions and OH– ions are not discharged. Instead, the copper anode dissolves by releasing electrons to form Cu2+ ions. Hence, the mass of the anode decreases and the anode becomes thinner. The types of electrodes (copper electrode is an active electrode) determine the product formed at the anode during electrolysis. Half-equation at the anode:

Discussion 1 Aqueous copper(II) sulphate solution consists of Cu2+ ions, H+ ions, SO42– ions and OH– ions. 2 During electrolysis, OH– ions and SO42– ions move to the anode. If a carbon electrode is used as the anode, OH– ion is selectively discharged due to its position in the electrochemical series. Oxygen gas is produced. Half-equation at the anode: 4OH– → 2H2O + O2 + 4e– 3 Cu2+ ions and H+ ions move to the cathode. Cu2+ ion which is at a lower position than the H+ ion in the electrochemical series will be discharged. Copper metal is produced. Half-equation at the cathode:

Cu → Cu2+ + 2e– 7 Cu2+ ions are still discharged at the cathode, producing copper metal. This causes the mass

Cu2+ + 2e– → Cu Electrochemistry

Inference

148

Conclusion 1 In the electrolysis of aqueous copper(II) sulphate solution: (a) If a carbon electrode is used as the anode, OH– ions are discharged and oxygen gas is produced. (b) If a copper electrode is used as the anode, both OH– ions and SO42– ions are not discharged. Instead the copper anode dissolves to produce Cu2+ ions. (c) Cu2+ ions are discharged at the cathode producing copper metal whether the cathode used is a carbon electrode or a copper electrode. 2 The types of electrodes used during electrolysis determine the types of ions discharged and the products of electrolysis. The hypothesis is accepted.

Cu2+ + 2e– → Cu 8 The colour intensity of the blue solution does not change because the concentration of Cu2+ ions is constant. The number of moles of Cu2+ ions discharged at the cathode is the same as the number of moles of Cu2+ ions produced at the anode. 9 The mass of the electrodes can be weighed before and after electrolysis using an electronic balance. 10 The decrease in the mass of the copper anode is the same as the increase in the mass of the copper cathode.

How to Predict the Products of Electrolysis of Aqueous Solutions

1

’08

The diagram shows the arrangement of apparatus for the electrolysis of silver nitrate solution.

Write the half-equation representing the reaction at electrode Q. Comments The ions present in the electrolyte are Ag+, NO3–, H+ and OH–. Electrode Q is the anode as it is connected to the posi­tive terminal of the battery. OH– ions are discharged at the anode producing oxygen gas. (Factor: positions of ions in the electrochemical series) 4OH– → 2H2O + O2 + 4e–

The cations at a higher position in the electrochemical series are very stable. These ions are unlikely to accept electrons to form neutral atoms. Hence K+ ions and Na+ ions are never discharged in an aqueous solution in electrolysis. The cations at the lower position of the electrochemical series are less stable. They are more likely to accept electrons to form neutral atoms. Similarly with anions. Anions at a higher position in the electrochemical series are never discharged in an aqueous solution in electrolysis. Hence F –, SO42– and NO3– ions are stable compared to the lower anions at a lower position in the electrochemical series. Cl–, Br –, I– or OH– ions will be discharged instead depending on their ionic concentration in the aqueous solution.

149

Electrochemistry

6

of the cathode to increase and the cathode becomes thicker. Half-equation at the cathode:

How to predict the products of electrolysis of aqueous solutions. Step 1

Step 2

Identify the cations and anions that are present in the aqueous solution. Generally, an aqueous solution of MaXb will produce ions as follows:

Identify the anode and cathode • Anode is the electrode connected to the positive terminal of the battery. Positive electrode attracts negative ions (anions). • Cathode is the electrode connected to the negative terminal of the battery. Negative electrode attracts positive ions (cations).

Cation Anion From MaXb : M b+ , X a– From H2O : H+ , OH–

6



SPM

’10/P2, ’11/P2

Step 3

Step 4

Identify the movements of ions • M b+ ions and H+ ions move to the cathode. • X a– ions and OH– ions move to the anode.

Identify the ions to be discharged at the cathode H+ ions are discharged at the cathode (producing H2 gas) except if Mb+ is Cu2+ or Ag+ (because these ions are lower than H+ ions in the electrochemical series). • H+ ions discharge by accepting electrons to form H2 gas. 2H+ + 2e– → H2 • Cu2+ ions or Ag+ ions discharge by accepting electrons to form metal atom. Cu2+ + 2e– → Cu Ag+ + e – → Ag

Inert electrode

Step 5 Identify the types of electrode as anode Inert electrode: carbon or platinum. Active electrode: Ag or Cu.

SPM

’09/P1

OH – ions are discharged (producing O2 gas) except if X b– ions are Cl–/Br –/I – of high concentration. • OH– ions discharge by donating electrons to form water and O2 gas. 4OH– → O2 + 2H2O + 4e– • Cl–/Br –/I– of high concentration discharge by donating electrons to form halogen. Example: 2Cl– → Cl2 + 2e–.

Electrochemistry

150

Active electrode Anions are not discharged. Instead anode dissolves to form ions. Examples: Ag → Ag+ + e– Cu → Cu2+ + 2e–

Electrolysis of dilute acids and alkalis

Electrolysis of aqueous copper(II) nitrate solution, Cu(NO3)2 with copper electrodes

1 Electrolysis of all dilute acids (HCl/H2SO4/ HNO3) and dilute alkali solutions (NaOH/ KOH) is actually the electrolysis of water. 2 At the cathode: H+ ions are discharged producing hydrogen gas.

1 At the cathode: Cu2+ ions are discharged. Copper metal is deposited. Mass of cathode will increase. (Factor: positions of ions in the electrochemical series).

2H+ + 2e– → H2 3 At the anode: OH– ions are discharged producing oxygen gas.

2 At the anode: Copper dissolves from the anode forming Cu2+ ions. Mass of the anode will decrease. (Factor: types of electrodes).

4OH– → 2H2O + O2 + 4e–

Cu → Cu2+ + 2e–

4 The removal of H+ ions and OH– ions from the solution during electrolysis causes the concentration of the acid or alkali to increase.

Electrolysis of aqueous silver nitrate solution, AgNO3 with carbon electrodes

Electrolysis of concentrated aqueous sodium chloride solution, NaCl with carbon electrodes

1 At the cathode: Ag+ ions are discharged. Silver metal is deposited. (Factor: positions of ions in the electrochemical series).

SPM 1 Cations present : Na+ ions, H+ ions ’08/P2 Anions present : Cl– ions, OH– ions 2 At the cathode: H+ ions are discharged producing hydrogen gas. (Factor: positions of ions in the electrochemical series).

Ag+ + e– → Ag 2 At the anode: OH– ions are discharged producing oxygen gas. (Factor: positions of ions in the electrochemical series).

2H+ + 2e– → H2 3 At the anode: Cl– ions are discharged producing chlorine gas. (Factor: concentration of ions).

4OH– → 2H2O + O2 + 4e–

2Cl– → Cl2 + 2e–

Electrolysis of aqueous silver nitrate solution, AgNO3 with silver electrodes

Electrolysis of aqueous copper(II) nitrate solution, Cu(NO3)2 with carbon electrodes 1 Cations present : Cu2+ ions, H+ ions Anions present : NO3– ions, OH– ions 2 At the cathode: Cu2+ ions are discharged. Copper metal is deposited. (Factor: positions of ions in the electrochemical series).

1 At the cathode: Ag+ ions are discharged. Silver metal is deposited. Mass of silver cathode will increase. (Factor: positions of ions in the electrochemical series). Ag+ + e– → Ag

Cu2+ + 2e– → Cu

2 At the anode: silver anode dissolves forming Ag+ ions. Mass of silver anode electrode will decrease. (Factor: types of electrodes).

3 At the anode: OH– ions are discharged producing oxygen gas. (Factor: positions of ions in the electrochemical series).

Ag → Ag+ + e–

4OH– → 2H2O + O2 + 4e–

151

Electrochemistry

6

Cu2+ + 2e– → Cu

such as sodium, calcium, magnesium and aluminium are extracted from their compounds using electrolysis. 2 Electrolysis is used because these reactive metals cannot be extracted from the minerals by reduction using carbon. 3 Examples are: (a) Extraction of aluminium from aluminium oxide, Al2O3. (b) Extraction of sodium from sodium chloride, NaCl.

6.3 1 Write the formulae of the ions present in the aqueous solutions below. Identify the ions that will be discharged at the anode and the cathode during electrolysis using carbon electrodes in every case. Ions present in the solution

Aqueous solution

Ions discharged at the cathode

Ions discharged at the anode

6

(a) Aqueous nitric acid (b) Silver nitrate (c) Very dilute copper(II) chloride

Extraction of aluminium metal from the mineral bauxite (Hall-Heroult process) 1 Bauxite is the major ore of aluminium consisting of aluminium oxide, Al2O3. 2 Cryolite (Na3AlF6) is added to aluminium oxide to lower its melting point from 2000°C to about 950°C. 3 Molten aluminium oxide is electrolysed using carbon as electrodes in the electrolytic cell as shown in Figure 6.8.

2



The figure above shows the arrangement of apparatus in an electrolysis experiment. (a) Name all the ions present in Cell I and Cell II. (b) Electrolysis is carried out for 20 minutes. (i) Predict the observations at electrodes P and Q. (ii) What will be the change in colour in Cell I? Explain your answer. (iii) Predict the observations at electrodes R and S. (iv) Write the half-equations at electrodes P and R. (c) What is the factor that determines the formation of the products in (i) electrode Q? (ii) electrode R?

6.4

Figure 6.8 Extraction of aluminium metal from aluminium oxide.

4 Molten aluminium oxide dissociates into aluminium and oxide ions as follows: Al2O3(l) → 2Al3+(l) + 3O2–(l) (a) At the cathode: Al3+ ions discharge to form aluminium metal

Electrolysis in Industries

Uses of Electrolysis in Industries

Al3+ + 3e– → Al

SPM

(b) At the anode: O2– ions discharge to form oxygen gas

’09/P1

1 Electrolysis process is used widely in industries. 2 Some common industrial applications of electrolysis are (a) extraction of reactive metals (b) purification of metals (c) electroplating of metals

2O2– → O2 + 4e– Overall equation of electrolysis: 2Al2O3 → 4Al + 3O2 5 The carbon anode is required to be replaced from time to time because the oxygen gas generated oxidises the carbon anode to form carbon dioxide.

(A) Extraction of Reactive Metals Using Electrolysis 1 Metals that are very reactive (placed at the top positions of the electrochemical series) Electrochemistry

152

Extraction of sodium metal from sodium chloride (Downs process) 4 Molten sodium chloride dissociates into ions as follows:

1 Sodium chloride is the most abundant and cheapest sodium compound. 2 Electrolysis of molten sodium chloride is carried out using iron as cathode and carbon as anode as in Figure 6.9.

NaCl(l) → Na+(l) + Cl–(l) (a) At the cathode: Na+ ions discharge to form sodium metal. Sodium metal is less dense and floats on top of the electrolyte to be collected.

chlorine gas

sodium is produced here

Na+ + e– → Na

6

molten sodium chloride

cathode (iron)

(b) At the anode: Cl– ions discharge to form chlorine gas. Chlorine gas is a useful by-product.

anode (carbon)

Figure 6.9 Extraction of sodium metal from molten sodium chloride

2Cl– → Cl2 + 2e– Overall chemical electrolysis:

3 Calcium chloride is added to lower the melting point of sodium chloride.

equation

of

the

2NaCl → 2Na + Cl2

(B) Purification of Metals Using Electrolysis

the anode to the cathode. After electrolysis, the mass of anode is reduced while that of the cathode is increased. 3 For example, in the purification of impure copper metal: (a) Anode: impure copper (b) Cathode: pure copper (c) Electrolyte: copper(II) ion solutions such as copper(II) sulphate

To investigate the purification of copper metal using electrolysis Apparatus Batteries, electrolytic cell, beaker, connecting wires with crocodile clips, ammeter and rheostat. Materials 1 mol dm–3 copper(II) sulphate solution, pure copper plate and impure copper plate. Procedure 1 About 200 cm3 of 1 mol dm–3 copper(II) sulphate solution is poured into a beaker.

SPM

’10/P1

2 A piece of impure copper plate is connected to the positive terminal of the batteries. This plate acts as the anode. 3 A piece of pure copper metal is connected to the negative terminal of the batteries. This plate acts as the cathode. 4 The circuit is completed using the connecting wires, rheostat and ammeter. The two copper plates are immersed in the copper(II) sulphate solution. The solution is electrolysed for 30 minutes (Figure 6.10). 153

Electrochemistry

Activity 6.2

1 Impure metals containing impurities can be purified using electrolysis as outlined below. (a) The impure metal is used as the anode. (b) A piece of pure metal is used as the cathode. (c) The electrolyte is a solution containing the ions of the metal to be purified. 2 In the process of purification of a metal using electrolysis, metal is transferred from

Discussion 1 At the anode: Copper dissolves from the anode by releasing electrons to form Cu2+ ions. The mass of the anode decreases. Half-equation:

6

Figure 6.10 Purification of copper metal

Results Observation At the anode: • The copper anode becomes thinner • Impurities are deposited below the anode

2 At the cathode: Cu2+ ions are discharged by receiving electrons to form copper atoms. Copper metal is deposited on the surface of the cathode. As a result, the copper cathode becomes thicker. The mass of the cathode increases. Half-equation:

Inference Copper anode dissolves to form Cu2+ ions

At the cathode: • The copper cathode becomes thicker

Copper metal is deposited at the cathode

Colour of electrolyte: • Colour intensity of the blue solution does not change

Concentration of Cu ions in the electrolyte does not change

2+

Cu → Cu2+ + 2e–

Cu2+ + 2e– → Cu

3 In this process, Cu2+ ions are transferred from the anode to the cathode and are deposited as pure copper metal. Impurities that are collected below the anode are known as anode mud. 4 The colour intensity of the blue solution does not change because the concentration of Cu2+ ions remains constant throughout electrolysis. The rate of formation of Cu2+ ions at the anode is the same as the rate of discharge of Cu2+ ions at the cathode. Conclusion When copper(II) sulphate solution is electrolysed using pure copper as the cathode and impure copper as the anode, purification of copper takes place. Pure copper is deposited at the cathode.

(C) Electroplating of Metals Using Electrolysis (a) The object to be electroplated is used as the cathode. (b) The anode is the electroplating metal. (c) The electrolyte is a solution that contains the electroplating metal ions. 4 An even and lasting layer of metal can be produced if (a) the surface of the object to be electroplated is first polished using sandpaper. (b) a low electric current is used so that electroplating is carried out slowly during the electroplating process. (c) the object to be plated is rotated steadily during electrolysis.

1 Electroplating is a process carried out to coat the surface of metal objects with a thin and even layer of another metal. 2 Two main aims of electroplating metals are (a) to prevent corrosion. For example, iron objects are plated with a thin layer of chromium or nickel metal to protect the iron from rusting. (b) to improve the appearance. For example, electroplating with gold, platinum and silver makes the surface of the objects appear shiny and more attractive. 3 In general, there are three conditions in the SPM electroplating of metal.

’08/P1

Electrochemistry

154

2

Comments At the anode, the silver foil dissolves by releasing electrons, thus becoming thinner:

SPM

’10/P1

The diagram shows the set-up of the apparatus used to electroplate an iron key with silver.

Ag → Ag+ + e– At the cathode, silver ions discharge by receiving electrons to form silver atoms, forming a shiny grey deposit on the iron key: Ag+ + e– → Ag Answer B

A B C D

Anode

Cathode

Shiny grey deposits are formed Silver foil becomes thinner Shiny grey deposits are formed Silver foil becomes thinner

Silver foil becomes thicker Shiny grey deposits are formed Gas bubbles are released Gas bubbles are released

6

What is observed at the anode and cathode after 30 minutes?

Presently, plastic electroplating is carried out to coat a thin layer of metal onto the surface of plastic objects. The object produced will have the advantanges of plastic: light, cheap, resistant to corrosion as well as having a shiny surface like a metal. As plastic is an electric insulator, a layer of graphite powder is coated onto the surface of the plastic, so that it can conduct electricity, before electroplating is carried out.

6.5 To investigate the electroplating of an iron spoon with copper using electrolysis Problem statement How is electrolysis used to electroplate an iron spoon with copper metal? Hypothesis Electroplating of an iron spoon with copper occurs if the iron spoon is used as the cathode, copper metal is used as the anode and aqueous copper(II) sulphate solution as the electrolyte. Variables (a) Manipulated variable : The position of the iron spoon as an electrode (b) Responding variable : The deposition of copper on the iron spoon (c) Constant variable : Type of electrolyte and arrangement of apparatus

Materials 0.5 mol dm–3 copper(II) sulphate solution, copper plate and iron spoon. Procedure 1 About 200 cm3 of 0.5 mol dm–3 copper(II) sulphate solution is poured into a beaker. 2 An iron spoon is polished using sandpaper and is connected to the negative terminal of the batteries. The spoon acts as the cathode. 3 A piece of copper metal, as the anode, is connected to the positive terminal of the batteries. 155

Electrochemistry

Experiment 6.5

Apparatus Batteries, electrolytic cell, beaker, connecting wires with crocodile clips, ammeter and rheostat.

4 The circuit is completed using the connecting wires, rheostat and ammeter. The iron spoon and the copper metal are immersed in the copper(II) sulphate solution. The solution is electrolysed for 30 minutes using a small current (0.5 A). 5 Steps 1 to 4 of the experiment are repeated by interchanging the positions of the iron spoon and copper metal, whereby the iron spoon is made the anode and the copper metal is made the cathode.

6

Set Set 1: Iron spoon as the cathode, copper metal as the anode

Set 2: Copper metal as the cathode, iron spoon as the anode

Figure 6.11 Electroplating of an iron spoon with copper

Observation

Inference

At the cathode: A brown metal is deposited on the surface of the iron spoon

The iron spoon is plated with copper metal

At the anode: The copper anode becomes thinner

The copper anode dissolves to form Cu2+ ions

Colour of the electrolyte: Colour intensity of the blue solution does not change

Concentration of Cu2+ ions in the electrolyte remains constant

At the cathode: The copper plate becomes thicker

Copper metal is deposited on the copper electrode

At the anode: No noticeable change in the appearance of the iron spoon

Electroplating of copper on the iron spoon does not take place

Colour of the electrolyte: The blue colour of the solution becomes paler

Concentration of Cu2+ ions in the electrolyte decreases

3 The colour intensity of the blue solution does not change because the concentration of Cu2+ ions remains constant throughout the electroplating process. 4 A slow electrolysis process using a small current will ensure that the layer of copper sticks firmly to the surface of the iron spoon.

Discussion 1 The brown metal deposited on the iron spoon is copper metal. (a) At the anode: copper dissolves from the anode by releasing electrons to form Cu2+ ions. Half-equation:

Cu → Cu2+ + 2e– Conclusion 1 In electroplating an iron spoon with copper using electrolysis, the iron spoon is made the cathode and a piece of copper metal is made the anode. 2 Copper metal is transferred from the copper anode to the iron spoon and is deposited as a thin layer of copper metal. 3 Electroplating does not take place if the iron spoon is made the anode. The hypothesis is accepted.

(b) At the cathode: Cu2+ ions are discharged by accepting electrons to form copper atoms. Copper metal is deposited on the surface of the iron spoon. Half-equation:

Cu2+ + 2e– → Cu

2 In this process, Cu2+ ions are transferred from the anode to the cathode (iron spoon) and are deposited as a thin and even layer of copper metal.

Electrochemistry

156

Benefits and Harmful Effects of Electrolysis in Industries 1 Table 6.1 shows the advantages and disadvantages of using electrolysis. Table 6.1 Advantages and disadvantages of using electrolysis.

1 Electrolysis is an effective method of extracting 1 Electrolysis is a process that uses a large quantity of electricity. For example, the recycling reactive metals from their compounds. Some of aluminium requires only 9% of the electrical chemical substances such as chlorine and sodium energy used to produce the same quantity of can be manufactured in large quantities using aluminium in electrolysis. electrolysis. 2 In electroplating, the whole surface area of a 2 The problem of environmental pollution especially in the electroplating process. metal such as iron is coated with a thin, even (a) In the electroplating of iron by chromium and valuable metal (such as gold, platinum and nickel, the waste chemicals contain and silver). This layer of metal protects iron chromium ions and nickel ions that can from being exposed to air and water so as endanger human health as well as pollute to prevent corrosion. Besides that, the layer water sources. of metal also gives an attractive appearance. (b) In silver electroplating, potassium silver Electroplating can also be performed using cyanide, KAg(CN)2 solution is sometimes polymers as the coating material. This has used as an electrolyte. The waste chemical been used to coat new cars with paint. The of the electrolyte contains cyanide ions advantage of this process is that it can be done which are toxic. in water thereby eliminating the use of volatile (c) Metal objects to be electroplated are organic solvents used in spray-paints. cleaned by acids to remove the layer 3 Electrolysis process is used to purify metals of metal oxide on the surface before such as zinc, silver, nickel, copper, lead and electroplating. The used acid wastes will aluminium. This process is also known as electropollute water in the drains, rivers and refining. lakes, thus destroying aquatic life. 2 Steps taken to overcome the problems of electrolysis in industries (a) Recycling such as the recycling of aluminium cans is encouraged to reduce the use of electrolysis in the extraction of aluminium.

(b) Waste chemicals in electrolyte from electroplating are treated to remove the toxic substances before being drained out as effluent.

6.4 1 State three main uses of the electrolysis process in industries. 2



A student carried out an experiment to electroplate an iron key with silver using the apparatus as shown in the above figure.

157

(a) What is a suitable metal that can be used as metal M? (b) State two observations that will be obtained in this experiment. (c) Write the half-equations for the reactions that take place at (i) the iron key (ii) the metal M (d) How will the concentration of silver nitrate solution change after electrolysis? Explain your answer. (e) Explain how the student can ensure that an even and lasting layer of silver metal stays on the surface of the iron key.

Electrochemistry

6

Disadvantages of using electrolysis

Advantages of using electrolysis

6.5

Simple Voltaic Cells (Chemical Cells)

6

series) will accept electrons and acts as the positive terminal (cathode). 5 A continuous flow of electrons from the negative terminal to the positive terminal of the cell through the external circuit produces an electric current. 6 The flow of electric current can be detected by the lighting up of a light bulb or the deflection of a galvanometer needle. 7 Voltaic cells are also known as galvanic cells or chemical cells. 8 The potential difference (voltage) of the cell is the electromotive force (e.m.f.) that moves electrons and can be measured by a voltmeter. 9 The further the distance between the positions of two metals in the electrochemical series, the bigger the voltage of the cell. For example, a magnesium/copper cell will produce a higher voltage than a zinc/copper cell.

Voltaic Cells SPM

’09/P2

1 A simple voltaic cell can be made by immersing two different types of metals in an electrolyte and connecting the two metals by wires in the external circuit. 2 In a simple voltaic cell, electrons flow from one metal to another metal through the connecting wires in the external circuit. 3 The more electropositive metal (metal that is at a higher position in the electrochemical series) will release electrons and thus acts as the negative terminal (anode) of the voltaic cell. 4 The less electropositive metal (metal that is at a lower position in the electrochemical

6.6 To investigate the production of electricity from chemical reactions in a simple voltaic cell Problem statement How does a chemical reaction produce electrical energy in a simple voltaic cell?

Procedure 1 A piece of magnesium plate and a piece of copper plate are polished with sandpaper. 2 Both pieces of the magnesium and copper plates are immersed in 200 cm3 of aqueous sodium chloride solution in a beaker as shown in Figure 6.12. 3 Both plates are connected by the connecting wire to a voltmeter. 4 The experiment is repeated using two pieces of copper plates as electrodes.

Hypothesis Electric current is produced when two different metals connected by wires are immersed in an electrolyte. Variables (a) Manipulated variable : Pairs of different metals (b) Responding variable : Deflection of a voltmeter needle by the electric current produced (c) Constant variable : Types of electrolyte and arrangement of apparatus

Experiment 6.6

Apparatus Voltmeter, beaker, connecting wires with crocodile clips and sandpaper. Materials 1 mol dm–3 sodium chloride solution, copper plates and magnesium plate. Electrochemistry

SPM

’05/P2

Figure 6.12

158

Results

Magnesium metal and copper metal

Observation

Inference

• Voltmeter needle deflects but the deflection decreases after awhile • Magnesium metal corrodes • Bubbles of colourless gas are evolved around the copper metal

Two pieces of copper metal

• Voltmeter needle does not show a deflection • No noticeable change occurs at the copper electrode

Discussion 1 The deflection of the voltmeter needle shows that an electric current is produced. The decreasing deflection indicates that the electric current decreases rapidly. 2 Magnesium metal is more electropositive than copper (at a higher position in the electrochemical series). Hence, it has a higher tendency to donate electrons than copper. 3 Magnesium atoms will donate electrons to form magnesium ions, Mg2+ in the solution, hence magnesium metal corrodes. Half-equation:

• Electric current is not produced • No reaction occurs

Figure 6.13 Movement of electrons and ions in a simple Mg/Cu cell using sodium chloride solution as the electrolyte

Mg → Mg2+ + 2e–

8 If copper(II) solution is used as the electrolyte, Cu2+ ions will receive electrons and are discharged because its position is lower than H+ ions and Mg2+ ions in the electrochemical series. Copper metal is produced. The overall equation of the cell will be

4 Electrons accumulate at the surface of the magnesium metal. This makes magnesium act as the negative terminal (also known as the anode) of the cell. 5 The electrons flow through the external circuit from the magnesium metal (negative terminal) to the copper metal (positive terminal or cathode of the cell) producing electricity. 6 When sodium chloride solution is used as the electrolyte, H+ ions (from water), Na+ ions and Mg2+ ions move towards the copper metal. H+ ions will accept electrons from the copper metal and be discharged because its position is lower than Na+ ions and Mg2+ ions in the electro­ chemical series. Hydrogen gas is produced. Half-equation:

• Electric current is produced. The voltage produced is not constant and decreases rapidly • Magnesium dissolves to form Mg2+ ions • Hydrogen gas is produced

Mg + Cu2+ → Mg2+ + Cu Conclusion 1 An electric current is produced when a chemical reaction occurs in a simple voltaic cell consisting of two different metals, connected by wires externally and immersed in an electrolyte. 2 In a simple voltaic cell, chemical energy released from chemical reactions is converted into electrical energy. 3 No electric current will be produced if both electrodes are of the same material because there is no potential difference between them. The hypothesis is accepted.

2H+ + 2e– → H2

7 The overall chemical equation in the cell is: Mg + 2H+ → H2 + Mg2+

159

Electrochemistry

6

Type of metal used as electrodes

3

’04

The diagram shows the set-up of the apparatus of a simple chemical cell.

Metal P

Metal Q

Iron Copper Zinc Lead

Aluminium Magnesium Iron Magnesium

A B C D

What are metals P and Q?

6

Comments The diagram shows that electrons flow from metal P to metal Q. Metal P must be more electropositive (higher in position in the electrochemical series) than metal Q. This is because the more electropositive metal will release electrons to become the negative terminal of the chemical cell. Answer C

SPM

Different Types of Voltaic Cells

’10/P2

oltaic cells can be divided into two categories as V shown in Table 6.2. Table 6.2 Two categories voltaic cells

Primary cells

Secondary cells

• Non-rechargeable cells (cells that cannot be charged again)

• Rechargeable cells (cells that can be charged again)

• Examples (a) Daniell cell (b) Dry cell (c) Mercury cell (d) Alkaline cell

• Examples (a) Lead-acid accumulator (b) Nickel/cadmium cell

Figure 6.14 Daniell cell using a salt bridge

(b) A porous pot as shown in Figure 6.15.

Figure 6.15 Daniell cell using a porous pot

Daniell Cell

4 A salt bridge contains inert ions or salt that does not react with the electrolyte. Examples are sodium chloride, potassium chloride, potassium nitrate, ammonium chloride and dilute sulphuric acid. 5 A simple salt bridge can be made by immersing a piece of filter paper in sulphuric acid or in a salt solution. 6 A porous pot has fine pores that allow ions to flow through but can prevent the two different aqueous solutions from mixing.

1 A Daniell cell has copper metal as the positive terminal and zinc metal as the negative terminal. 2 The zinc metal is immersed in zinc sulphate solution and the copper metal is immersed in copper(II) sulphate solution. 3 The two solutions of the Daniell cell are connected using either of the following: (a) A salt bridge as shown in Figure 6.14.

Electrochemistry

160

2 The electrolyte is ammonium chloride in the form of a paste. 3 The cross-section of a dry cell is shown in Figure 6.16.

7 The functions of salt bridges and porous pots are SPM (a) to allow the flow of ions so that the ’11/P1 circuit is completed. (b) to prevent the two aqueous solutions from mixing. This will prevent displace­ ment reaction between a more electro­ positive metal and the salt solution of the less electropositive metal from taking place. 8 In a simple voltaic cell made by immersing both the zinc metal and copper metal in copper(II) sulphate solution, (a) zinc metal reacts directly with copper(II) sulphate solution in a displacement reaction.

6

Figure 6.16 Dry cell

4 When the dry cell is in use, the zinc metal releases electrons and dissolves to form Zn2+ ions.

Zn + CuSO4 → ZnSO4 + Cu As a result, the zinc metal will be coated by a layer of copper metal. (b) the electric current decreases rapidly. 9 The salt bridge or porous pot prevents the zinc from reacting directly with the copper(II) sulphate solution. 10 When the negative terminal (zinc) is connec­ ted to the positive terminal (copper), the highest voltage produced is 1.10 V if both zinc sulphate solution and copper(II) sulphate solution have a concentration of 1.0 mol dm–3. 11 The reactions that take place in a Daniell cell are as follows: At the negative terminal:

Zn → Zn2+ + 2e–

At the positive terminal:

Cu2+ + 2e– → Cu

At the negative terminal: Zn → Zn2+ + 2e– 5 Electrons flow from the zinc metal casing through the external circuit to the carbon rod, where NH4+ ions receive electrons to produce ammonia gas and hydrogen gas. At the positive terminal: 2NH4+ + 2e– → 2NH3 + H2 6 When the cell produces an electric current, zinc metal dissolves. When the zinc metal casing is perforated and the electrolyte starts to leak out, the dry cell can no longer be used. 7 Usually a dry cell produces quite a stable voltage of about 1.5 V.

Overall chemical equation of cell: Zn + Cu2+ → Zn2+ + Cu 12 When the Daniell cell is in use, (a) the copper metal becomes thicker (mass increases), (b) the zinc metal becomes thinner (mass decreases), (c) the concentration of copper(II) sulphate solution decreases, hence the blue colour of the solution becomes paler, (d) the concentration of zinc sulphate solution increases.

Dry cells of different sizes

Dry Cell

Dry cells that are out of electricity (old) need to be removed from the electrical appliance. This is because when the container is corroded, the electrolyte will leak out to damage the electrical appliance.

1 A dry cell consists of a carbon rod (positive terminal) and a metal casing made of zinc (negative terminal). 161

Electrochemistry

Lead-acid Accumulator

6 In the recharging process, (a) reverse reactions occur at both electrodes. (b) lead(II) sulphate is converted back into lead(IV) oxide and hence lead(II) sulphate dissolves. (c) sulphuric acid is formed. 7 An accumulator normally consists of 6 pairs of plates and produces a voltage of 12 V.

1 A lead-acid accumulator is made of pieces of lead plates immersed in moderately concentrated sulphuric acid as shown in Figure 6.17.

Other Types of Voltaic Cells

6

Mercury cell 1 A mercury cell consists of zinc (negative terminal), mercury(II) oxide, HgO (positive terminal) and a mixture of potassium hydroxide, KOH and zinc oxide, ZnO as electrolyte.

Figure 6.17 Lead-acid accumulator is used as car batteries.

2 When the accumulator is used to produce current, the following changes occur. (a) At the negative terminal, a lead atom donates 2 electrons to form a Pb2+ ion. Pb → Pb2+ + 2e– (b) At the positive terminal, lead(IV) oxide accepts electrons and reacts with H+ ions in dilute sulphuric acid to form Pb2+ ions.

Mercury cells are small

PbO2 + 4H+ + 2e– → Pb2+ + 2H2O

2 Mercury cells are small and long-lasting, producing a constant voltage of 1.3 V. 3 Mercury cells are used in hearing aids, digital watches and heart pacemakers.

(c) Lead(II) ions from both electrodes combine with SO42– ions in sulphuric acid to produce lead(II) sulphate, PbSO4. Pb2+ + SO42– → PbSO4

Alkaline cell

The overall chemical reaction is represented by the equation below: Pb + PbO2 + 4H+ + 2SO42– → 2PbSO4 + 2H2O 3 In this reaction, sulphuric acid is used up and water is produced. Hence, sulphuric acid becomes more dilute and its density decreases. 4 Lead(II) sulphate is insoluble and exists as a white precipitate. When the precipitate covers the surface of both electrodes, further reaction is prevented and no electric current will be produced. 5 The accumulator can be recharged by passing an electric current in the opposite direction to renew the cell. Electrochemistry

1 Alkaline cells are non-rechargeable cells. 2 An alkaline cell consists of zinc (negative terminal), carbon rod (positive terminal) surrounded by manganese(IV) oxide, MnO2 and alkali (potassium hydroxide and sodium hydroxide) as the electrolyte. 162

Nickel-cadmium cell

Nickel-cadmium cells are rechargeable cells

A nickel-cadmium cell consists of cadmium (negative terminal), nickel(IV) oxide, NiO2 (positive terminal) and alkali, potassium hydroxide, KOH as the electrolyte.

Other new types of cells include lithium ion, nickel hydride and polymeric cells. These cells are rechargeable cells. Unlike the lithium ion and nickel hydride cells which require battery casings, the polymeric cell is flexible and can be specifically shaped to fit the device it will power.

6

Advantages and Disadvantages of Various Types of Voltaic Cells Table 6.3 Advantages and disadvantages of various types of voltaic cells

Type of cell

Advantages

Disadvantages

1 Daniell cell

• Can be prepared easily in the laboratory

• A type of wet cell, the electrolyte spills easily • Voltage is not constant

2 Dry cell

• Cheap • No spillage as it is a dry cell • Produces a moderately constant current and voltage • Portable, can be carried around easily • Available in different sizes

• Does not last long • Cannot be recharged • Zinc metal casing dissolves and the electrolyte that leaks out may corrode electrical instruments • Current and voltage produced is low

3 Alkaline cell

• Lasts longer than a dry cell • Produces a higher and more stable voltage • Portable

• Cannot be recharged • Cost more than a dry cell • If leakage occurs, electrolyte is corrosive

4 Mercury cell

• Can be made into very small sizes • Produce a constant voltage for a long period • Can last for a long period of time

• Expensive • Cannot be recharged • Mercury which is produced is toxic

5 Lead-acid accumulator

• Can be recharged repeatedly • Produces a high current (175 A), suitable for heavy work such as starting a car engine • Produce a high voltage (12 V) for a long period

• Acid may spill • Heavy and is difficult to be carried around • Loss of charge occurs if not used for a long time • Lead plates are easily corroded after a long period of usage

6 Nickel-cadmium cell

• Can be charged repeatedly • No spillage occurs because it is a dry cell • The size is smaller that an accumulator

• Expensive • Requires a transformer for the recharging process

163

Electrochemistry

Comparison of Voltaic Cells and Electrolytic Cells

SPM

’10/P2

1 Table 6.4 below shows several similarities and differences between an electrolytic cell and a voltaic cell.

6

Table 6.4

Electrolytic cell

Voltaic cell

Figure 6.18 Electrolytic cell

Figure 6.19 Voltaic cell

Similarities • Contains an electrolyte • Consists of an anode and a cathode • Electrons move from the anode to the cathode in the external circuit (connecting wires) Electrolytic cell

• Positive ions and negative ions move in the electrolyte • Chemical reactions involve the donation (at the anode) or acceptance (at the cathode) of electrons

Differences

• A battery is required to supply electrical energy • Graphite (carbon) is usually used as electrodes

• A battery is not required to supply electrical energy Basic structure

• Electrodes are not made up of different metals • Electrical energy is converted into chemical energy • Anode (positive electrode): Anions (negative ions) lose electrons at the anode

Energy conversion

Transfer of electrons at the positive electrode

Electrochemistry

• Chemical energy is converted into electrical energy • Cathode (positive electrode): Oxidising agent accepts electrons from the cathode Cu+(aq) + 2e– → Cu(s) … reduction

Transfer of electrons at the negative electrode

Y 2+ + 2e– → Y ... reduction • Electrons flow from the anode (positive electrode) to the cathode (negative electrode)

• Graphite (carbon) is not used as electrodes • Electrodes are made up of two different metals

2X – → X2 + 2e– ... oxidation • Cathode (negative electrode): Cations (positive ions) accept electrons from the cathode

Voltaic cell

Transfer of electrons in the external circuit

164

• Anode (negative electrode): Reducing agent releases electrons Zn(s) → Zn2+(aq) + 2e– … oxidation • Electrons flow from the anode (negative electrode) to the cathode (positive electrode)

The table shows the differences between the terminals in voltaic and electrolytic cells as a result of the transfer of electrons. Transfer of electrons Type of cells

Electrons are donated

Electrons are accepted

Voltaic cells

At the negative terminal (anode) At the positive terminal (cathode)

Electrolytic cells

At the positive terminal (anode) At the negative terminal (cathode)

(d) Predict the observations obtained after the voltaic cell is used for some time.

1 (a) Draw the circuit diagram for a simple voltaic cell consisting of iron metal, copper metal and copper(II) sulphate solution. Show the direction of the flow of electrons in the circuit diagram. (b) Which metal serves as the negative terminal? (c) Write the half-equations for the reactions that occur at both electrodes.

6.6

2 Compare the advantages and disadvantages of dry cells and alkaline cells. 3 Give an example of a voltaic cell that can be recharged. Explain how the reactions occur at the positive terminal and the negative terminal to produce an electric current.

4 The electrochemical series can be constructed by two methods. (a) The potential difference (voltage difference) between pairs of metals (b) The ability of a metal to displace another metal from its salt solution.

The Electrochemical Series

1 The electrochemical series is an arrangement of elements according to their tendencies to form ions. 2 In the electrochemical series, a metal that has a higher tendency to ionise and form positive ions (by releasing electrons) is placed at a higher position in the series. Hence, metal ions at the upper positions of the electrochemical series are less likely to receive electrons to form metal atoms. 3 Part of the electrochemical series (for metal elements) is shown below.

(A) To Construct the Electrochemical Series Based on the Potential SPM ’08/P1 Difference (Voltage Difference) ’09/P2 1 Metals are arranged in the electrochemical series according to their tendencies to donate electrons to form cations. 2 The electrochemical series can be constructed based on the measurement of the potential diffe­rence between two metals in voltaic cells. 3 When two different metals (immersed in their respective salt solutions) are connected in the external circuit through a voltmeter and a salt bridge: (a) The metal that serves as the negative terminal of the voltaic cell has a higher tendency to release electrons. Hence, that metal is placed at a higher position in the electrochemical series. Conversely, the metal that serves as the positive terminal is placed at a lower position in the electrochemical series. (b) The further apart the positions of two metals in the electrochemical series, the greater the potential difference (voltage).

Metals Positive ions (cations) K ⎯⎯⎯→ K+ + e– Na ⎯⎯⎯→ Na+ + e– Ca ⎯⎯⎯→ Ca2+ + 2e– Tendency Tendency Mg ⎯⎯⎯→ Mg2+ + 2e– of cations of metal Al ⎯⎯⎯→ Al3+ + 3e– to accept atoms to Zn ⎯⎯⎯→ Zn2+ + 2e– electrons donate Fe ⎯⎯⎯→ Fe2+ + 2e– to form electrons Sn ⎯⎯⎯→ Sn2+ + 2e– metals to form Pb ⎯⎯⎯→ Pb2+ + 2e– increases ions increases H ⎯⎯⎯→ H+ + e– Cu ⎯⎯⎯→ Cu2+ + 2e– Ag ⎯⎯⎯→ Ag+ + e– 165

Electrochemistry

6

6.5

6.7 To construct the electrochemical series through the potential difference (voltage) of pairs of metals Procedure

Problem statement How to construct the electrochemical series based on the measurement of the potential differences between pairs of metals in simple voltaic cells?

6

SPM

’06/07 P3

Hypothesis Two principles are used in the construction of the electrochemical series: (a) A metal that serves as the negative terminal of a cell is placed at a higher position in the electrochemical series. (b) The bigger the voltage differences of the voltaic cells, the further apart the positions of the two metals in the electrochemical series.

Figure 6.20

1 Pieces of zinc, magnesium, iron, aluminium and silver metals are polished with sandpaper. 2 A piece of zinc metal and a piece of copper metal are connected to a voltmeter by the connecting wires with crocodile clips. 3 The two metals are then dipped in the sodium chloride solution in a beaker as shown in Figure 6.20. 4 The highest cell voltage obtained is recorded. 5 The direction of the flow of electrons is also noted to determine the terminals of the voltaic cell. Electrons flow from the negative terminal to the positive terminal. If the voltmeter reading shows a negative value, the metal pairs connected to the terminals of the voltmeter should be reversed. 6 Zinc metal is then replaced by other metals in turn: magnesium, iron, aluminium and silver. The highest cell voltage of every pair of metals is recorded.

Variables (a) Manipulated variable : Pairs of metals as electrodes (b) Responding variable : Voltage values of voltaic cells (c) Constant variables : Type and concentra­ tion of electrolytes Apparatus Voltmeter, beaker, connecting wires with crocodile clips and sandpaper. Materials Sodium chloride solution of 1.0 mol dm–3, pieces of copper, zinc, magnesium, iron, aluminium and silver metals.

Experiment 6.7

Results Pairs of metals

Positive terminal

Negative terminal

Potential difference (V)

Zn/Cu

Copper

Zinc

1.1

Mg/Cu

Copper

Magnesium

2.7

Fe/Cu

Copper

Iron

0.8

Al/Cu

Copper

Aluminium

2.0

Ag/Cu

Silver

Copper

1.1

2 Silver serves as the positive terminal when it is connected to copper. Hence, silver is placed at a lower position than copper in the electrochemical series. 3 The further apart the distance between the metals in the electrochemical series, the greater the potential difference (voltage).

Conclusion 1 Copper metal serves as the positive terminal of the voltaic cells when paired with zinc, magne­ sium, iron and aluminium metal. Hence, copper is at a lower position than zinc, magnesium, iron and aluminium in the electrochemical series. Electrochemistry

166

4 The arrangement of the metals in the electrochemical series based on the voltage (potential difference) of the cell is as follows: Magnesium Aluminium Higher tendency to release electrons

2.0 V

Zinc 1.1 V

Iron 0.8 V

Copper

2.7 V

The bigger the voltage reading, the further the distance between the metals

1.1 V

6

Silver

4

SPM

’10/P1

The table shows information about three simple cells. Metal pairs

Potential difference (V)

Positive terminal

P and Q

1.7

P

Q and S

2.1

S

R and S

0.6

R

What is the potential difference of the metal pair P and R? Q Higher tendency to release electrons

Comments • In the metal pair of P and Q, P is the positive terminal. Hence P is placed below Q in the electrochemi­cal series. Similarly, S is placed below Q in the electrochemical series. • The potential difference between Q and S is bigger than that between Q and P. Thus S is placed below P. • In the metal pair R and S, R is the positive terminal. Hence R is placed below S. • The arrangement of the metals according to their increasing tendencies to form metal ions is as follows: 1.7 V

2.1 V

P S

0.6 V

The bigger the voltage reading, the further the distance between the metals.

R

Answer The potential difference between P and R = (2.1 – 1.7) + 0.6 = 1.0 V

(a) metal M is more likely to release electrons than metal N. (b) metal M is more electropositive than metal N. (c) metal M is placed at a higher position than metal N in the electrochemical series. 3 Alternatively, if metal P is immersed in an aqueous Q2+ ion solution and no reaction takes place, then metal P is at a lower position than metal Q in the electroche­mical series.

(B) To Construct the Electrochemical Series from the Displacement Reactions of Metals 1 The electrochemical series can also be constructed based on the ability of a metal to displace another metal from its salt solution. 2 If metal M can displace metal N from an aqueous N salt solution, then

167

Electrochemistry

6.8 SPM

To construct the electrochemical series from displacement reactions

6

Problem statement How to construct the electrochemical series based on the ability of a metal to displace another metal from its salt solution?

’08/P2, ’07/P1, ’04/P2

3 A piece of magnesium metal is placed in the solution of every test tube except that of its salt solution (Figure 6.21).

Hypothesis A metal that can displace another metal from its salt solution is placed at a higher position in the electrochemical series. The greater the number of metals that can be displaced by a metal from their solutions, the higher its position in the electrochemical series. Variables (a) Manipulated variable : Different types of metal and their salt solutions (b) Responding variable : Deposition of metals or colour change in the salt solutions (c) Constant variable : Concentration of nitrate salt solutions Figure 6.21

Apparatus Test tubes, test-tube rack and sandpaper.

4 Observations are made after awhile to check if (a) there is any colour change in the solution, (b) there are any solid deposits on the magnesium metal, (c) magnesium metal dissolves. 5 If any of the above occurrences (a), (b) or (c) is observed, displacement reaction has taken place: a tick symbol, (✓) is marked in the table of results. 6 If there is no noticeable observation, a cross symbol, (✗) is marked at the table to indicate that displacement reaction did not take place. 7 The experiment is repeated using different metals and fresh solutions of ions. The results of the experiment are shown in the table below.

Materials Pieces of magnesium, zinc, iron, tin, lead and copper metals, solutions of copper(II) nitrate, lead(II) nitrate, tin(II) nitrate, iron(II) nitrate, zinc nitrate and magnesium nitrate (concentration and volume of all salt solutions are 0.5 mol dm–3 and 10 cm3 respectively). Procedure 1 Pieces of magnesium, zinc, copper, tin, lead and iron metals are polished with sandpaper. 2 10 cm3 of 0.5 mol dm–3 solutions of copper(II) nitrate, lead(II) nitrate, tin(II) nitrate, iron(II) nitrate, zinc nitrate and magnesium nitrate are placed into separate test tubes. Results Solution

Cu(NO3)2

Pb(NO3)2

Sn(NO3)2

Fe(NO3)2

Zn(NO3)2

Mg(NO3)2

Magnesium, Mg

✓

✓

✓

✓

✓

–

Zinc, Zn

✓

✓

✓

✓

–

✗

Iron, Fe

✓

✓

✓

–

✗

✗

Tin, Sn

✓

✓

–

✗

✗

✗

Lead, Pb

✓

–

✗

✗

✗

✗

Copper, Cu

–

✗

✗

✗

✗

✗

Experiment 6.8

Metal

Electrochemistry

168

4 The result of the experiment shows that the order of the positions of the metals in the electrochemical series is:

Conclusion 1 Metals can be arranged according to the number of tick symbols (3) recorded (or the number of metals displaced in reactions). The more (3) symbols, the more reactive the metal is and the position of the metal is placed higher in the electrochemical series. 2 Magnesium is placed at the highest position in the electrochemical series because it can displace all the other metals from their solutions. 3 Copper is placed at the lowest position in the electrochemical series because copper cannot displace any other metals in this experiment.

Mg

Zn

Fe

Sn

Pb

Cu

Electropositivity of metal decreases

6

5 The electrochemical series can be constructed from displacement reactions. The hypothesis is accepted.

The Uses of the Electrochemical Series To determine the terminals of voltaic cells

To predict the ability of a metal to displace another metal from its salt solution

1 When two different metals are connected by wires and then immersed in an electrolyte, a simple voltaic cell is formed. The metal that is placed at a higher position in the electrochemical series will become the negative terminal of the cell. The metal that is placed lower in the electrochemical series will become the positive terminal of the cell. 2 The metal that is placed higher in the electrochemical series is more electropositive and has a higher tendency to release electrons. Electrons will flow from the negative terminal to the positive terminal. 3 For example, in the zinc/copper simple voltaic cell, zinc metal will become the negative terminal of the cell because zinc is above copper in the electrochemical series. Copper metal will become the positive terminal of the cell.

To compare the standard voltage of the voltaic cell

1 A metal that is at a higher position in the electrochemical series can displace another metal that is lower than itself in the electro­ chemical series from its salt solution. 2 For example, aluminium is above iron in the electrochemical series, hence aluminium can displace iron from an iron(II) salt solution (such as iron(II) sulphate solution).

To predict whether a metal can displace hydrogen from an acid 1 Hydrogen ion is placed between lead(II) ion and copper(II) ion in the electrochemical series. 2 All other metals that are placed at higher positions than hydrogen ion in the electrochemical series can displace hydrogen from acids. For example, zinc is above hydrogen in the electrochemical series; hence zinc can displace hydrogen from hydrochloric acid.

SPM

’09/P1

1 The further the distance between two metals in the electrochemical series, the greater the cell voltage will be. 2 For example, the distance between magnesium and copper is further than that between zinc and copper in the electrochemical series. Hence the cell voltage produced by a magnesium/copper voltaic cell is greater than that from a zinc/copper voltaic cell.

Zn + 2HCl → ZnCl2 + H2 3 Metals below hydrogen ion in the electro­ chemical series cannot displace hydrogen from acids. Examples are copper, mercury and silver. 169

Electrochemistry

6.9 To confirm the predictions of displacement reactions

6

Problem statement How can the prediction of the displacement reaction of a metal from its salt solution by another metal be confirmed?

Materials Pieces of magnesium, iron and copper, solutions of copper(II) sulphate, iron(II) sulphate and magnesium sulphate (concentration and volume of all salt solutions are 0.5 mol dm–3 and 10 cm3 respectively).

Hypothesis If metal X is at a higher position than metal Y in the electrochemical series, then metal X can displace metal Y from a salt solution of metal Y.

Procedure 1 Pieces of magnesium, copper and iron metals are polished with sandpaper. 2 10 cm3 of 0.5 mol dm–3 solutions of copper(II) sulphate, iron(II) sulphate and magnesium sulphate are put into different test tubes. 3 Magnesium, copper and iron metals are placed in different salt solutions in the test tubes. 4 Observations are made after 20 minutes to check if there is (a) any change in colour of the solution, (b) any deposits of metal, (c) any corrosion of metal. 5 The result of the experiment is tabulated in the table below.

Variables (a) Manipulated variable : Different types of metals and salt solutions (b) Responding variable : Deposits of metal or colour change in the salt solution (c) Constant variable : Concentration of salt solu­tions Apparatus Test tubes, test-tube rack and sandpaper.

Experiment 6.9

Results Metal + salt solution

Prediction

Observation

Magnesium + iron(II) sulphate solution

Displacement occurs because magnesium is more electropositive than iron

• Grey deposit is formed • The colour of the green solution becomes paler • Magnesium dissolves

Magnesium + copper(II) sulphate solution

Displacement occurs because magnesium is more electropositive than copper

• Brown deposit is formed • The colour of the blue solution becomes paler • Magnesium dissolves

Iron + magnesium sulphate solution

Displacement does not occur because iron is less electropositive than magnesium

No noticeable change

Iron + copper(II) sulphate solution

Displacement occurs because iron is more electropositive than copper

• Brown deposit is formed • The blue coloured solution changes to green • Iron dissolves

Copper + magnesium sulphate solution

Displacement does not occur because copper is less electropositive than magnesium

No noticeable change

Copper + iron(II) sulphate solution

Displacement does not occur because copper is less electropositive than iron

No noticeable change

Electrochemistry

170

Discussion 1 A more electropositive metal can displace a less electropositive metal from its salt solution. 2 Magnesium is at a higher position than iron and copper in the electrochemical series. Hence magnesium can displace iron from iron(II) sulphate solution and copper from copper(II) sulphate solution.

Fe + Cu2+ → Cu + Fe2+

Conclusion The prediction that a metal at a higher position in the electrochemical series can displace a metal which is at a lower position from its salt solution is confirmed. The hypothesis is accepted.

Mg + Fe2+ → Mg2+ + Fe Mg + Cu2+ → Mg2+ + Cu 3 Iron is at a higher position than copper in the electrochemical series. Hence iron can displace copper from copper(II) sulphate solution.

Electrochemical series:

A series of metals based on their tendencies to form metal ions

Potassium, K Sodium, Na Calcium, Ca Magnesium, Mg Aluminium, Al Zinc, Zn Iron, Fe Tin, Sn Lead, Pb Hydrogen, H Copper, Cu Silver, Ag Gold, Au

A metal placed higher in the series has a higher tendency to form positive ions (it is more electropositive). For example: Zn is more electropositive than Fe. A metal placed at a lower position in the series can be displaced by a metal above it.

This series is used to: (a) predict the tendency of a metal to form ions. (b) predict the ability of a metal to displace another metal from its ionic solution. (c) determine the terminals and voltage of a voltaic cell. In a voltaic cell, the metal that becomes the negative terminal is the metal that is higher in position in the electrochemical series as it has a higher tendency to form ions. The further the distance between the metal pairs in a voltaic cell, the greater the cell voltage will be.

6.6

1 The table below shows the voltages obtained from three voltaic cells using different pairs of metals. Voltaic cell Metal pairs Voltage (V)

Positive electrode

1

X and Y

1.2

X

2

X and Z

0.9

X

3

Y and W

0.4

Y



(i) Which metal will become the terminal of the cell? (ii) Predict the voltage of the cell. (c) Predict what will happen if (i) metal Y is immersed in a solution (ii) metal X is immersed in a solution (iii) metal W is immersed in a solution

negative

of Z salt. of Y salt. of X salt.

2 Silver is placed at a position lower than copper and magnesium in the electrochemical series. Predict the observation and reaction that will occur in the following experiment: (a) Silver in copper(II) sulphate solution, (b) Copper in silver nitrate solution, (c) Magnesium in silver nitrate solution.

(a) Based on the observation above, arrange the metals W, X, Y and Z in an ascending order according to their electropositivity. (b) A voltaic cell is made from metal Z and metal W.

171

Electrochemistry

6

4 Copper cannot displace either magnesium or iron from their salt solutions because copper is below mag­ ne­ sium and iron in the electrochemical series.

6

6.7

2 However, a safe and systematic method of disposal of used batteries and industrial by-products in electrochemical industries is important to prevent environmental pollution. (a) Used batteries should be separated from other household disposal. They are required to be disposed off separately to prevent the chemicals of the batteries from leaking and polluting water sources. Parts of batteries that are useful should be recycled. (b) Chemical wastes from electrolytic industries should be treated to remove poisonous chemicals before being disposed as industrial waste. For example, (i) acids that are used to clean metals before electroplating should be diluted and neutralised before draining off as waste water. (ii) metal ions that are toxic and hazardous to human health such as cadmium ion, chromium ion and nickel ion need to be treated and removed from industrial effluent.

Developing Awareness and Responsible Practices when Handling Chemicals used in the Electrochemical Industries

1 Electrochemical industries play an important role in our daily life by improving our quality of life. (a) For example, useful metals such as aluminium, sodium and magnesium are extracted from their minerals or compounds using electrolysis. (b) Useful chemical substances such as chlorine and sodium hydroxide are manufactured on a large scale using electrolysis. (c) Electroplating of iron with chromium protects the iron components of machinery from corrosion. Silver-plating is commonly used in the making of fine cutleries. (d) Various voltaic cells are used in different devices such as radio, torchlight, quartz watch, handphone and others.

1 An electrolyte is a chemical compound which conducts electricity in the molten state or in an aqueous solution and undergoes chemical changes. 2 A non-electrolyte is a chemical compound which does not conduct electricity in any state. 3 Electrolysis is the decomposition of an electrolyte (molten or in aqueous solution) by the passage of an electric current. 4 Graphite or platinum is usually used as electrodes because they are inert. 5 The anode is the electrode connected to the positive terminal of the batteries. 6 The cathode is the electrode connected to the negative terminal of the batteries. 7 Two steps occur during electrolysis. (a) Movement of ions to the electrodes: Cations (positive ions) move towards the cathode (negative electrode) whereas anions (negative ions) move towards the anode (positive electrode). (b) Discharge of ions at the electrodes: Cations discharge by receiving electrons. Generally: An+ + ne– → A Anions discharge by releasing electrons. Generally: Bn– → B + ne–

Electrochemistry

8 The factors that determine the types of ions to be discharged at the electrodes are (a) positions of ions in the electrochemical series: The lower positioned ion will be discharged (b) concentration of ions in the solution: The more concentrated ion will be discharged (c) types of electrodes used 9 Uses of electrolysis in industries (a) Extraction of reactive metals. For example: Extraction of aluminium from molten bauxite (Al2O3) using carbon electrodes. (b) Refining of metals. For example: Purification of copper. (c) Electroplating of metals. For example: Copper plating or silver plating. 10 There are two types of voltaic cells: (a) Primary cells: Non-rechargeable cells (cells that cannot be charged again). (b) Secondary cells: Rechargeable cells (cells that can be charged again). 11 The electrochemical series is an arrangement of elements based on their tendencies to form ions.

172

6 6.1

Electrolytes and Nonelectrolytes

1 Which of the following can conduct electricity? A Ethanol B Solid lead(II) nitrate C Magnesium chloride solution D Liquid tetrachloromethane 2 Calcium carbonate powder does not conduct electricity because A it does not contain ions. B it contains covalent molecules. C it contains calcium ions and carbonate ions that are not free to move. D all the atoms in calcium carbonate are bonded by strong covalent bonds. 3 The diagram shows the set-up of apparatus to test the conductivity ’06 of the chemical in the beaker. It was found that there is no deflection on the ammeter needle.

4 Which of the following statements are true about an ’06 electrolyte? I It has ions that conduct electricity in the solid state. II It can conduct electricity in the molten state or in aqueous solution. III It is a compound with ionic bonds only. IV It can be decomposed by electric current. A I and II only B III and IV only C II and IV only D I, II, III and IV 5 Which of the following statements are true about electrolysis? I The cathode is the positive electrode. II Molten covalent compounds can be used as electrolytes. III Platinum can be used as inert electrodes. IV A compound is decomposed by electric current. A I and II only B III and IV only C II, III and IV only D I, III and IV only

6.2

Electrolysis of Molten Compounds

6 A

Which of the following action will cause a deflection of the ammeter’s needle? A Add more dry cells in the circuit B Add water to glacial ethanoic acid C Add ethanol to glacial ethanoic acid D Substitute the platinum electrodes with carbon electrodes

?

@

heating

Which of the following occurs when molten zinc chloride is 6/9 electrolysed in the apparatus as shown in the diagram?

173

A Zinc metal is formed at electrode X. B Chlorine gas is formed at electrode Y. C Zinc ions are attracted to the anode. D Chloride ions are discharged at the positive electrode. 7 When molten lead(II) iodide solution is electrolysed using carbon electrodes, which of the following represents the half-equation that occurs at the cathode? A Pb2+ + 2e– → Pb B Pb → Pb2+ + 2e– C I2 → 2l– + 2e– D 2I– + 2e– → I2 8 Which of the following substances will produce aluminium metal when electrolysis is carried out using carbon electrodes? A Aqueous aluminium sulphate solution B Aqueous aluminium chloride solution C Solid aluminium oxide D Molten aluminium oxide

6.3

Electrolysis of Aqueous Solutions

9 Which of the following compounds produces oxygen gas and hydrogen gas during electrolysis? A Aqueous potassium hydroxide solution B Saturated sodium chloride solution C Aqueous copper(II) nitrate solution D Concentrated hydrochloric acid 10 The diagram below shows the apparatus set-up for the ’09 electrolysis of potassium nitrate solution, KNO3. Electrochemistry

6

Multiple-choice Questions

carbon electrode Y potassium nitrate solution carbon electrode X

What are the products formed at electrodes X and Y ? TC 54

6

X Nitrogen gas

Hydrogen gas

B

Nitrogen dioxide gas

Potassium

C

Oxygen gas

Hydrogen gas

D

Hydrogen gas

Oxygen gas

11 The products formed at the electrodes during the electrolysis of aqueous sodium sulphate solution using carbon electrodes are Anode

A

Sodium

B

Hydrogen

Sulphur dioxide

C

Hydrogen

Oxygen

D

Sodium

Oxygen

Sulphur

12 Electrolysis of dilute sodium chloride solution using carbon electrodes produces oxygen and hydrogen at the anode and cathode respectively. The products formed at the electrodes will change if A platinum is used as the cathode. B a bigger current flows through the circuit. C a concentrated sodium chloride solution is used. D the distance between the electrodes is reduced. 13 Which of the following is true about the electrolysis of aqueous copper(II) chloride solution using copper electrodes? A The mass of cathode decreases. Electrochemistry

14 The diagram shows the set-up of apparatus for the electrolysis ’03 of iron(II) nitrate solution.

Y

A

Cathode

B Chlorine gas is evolved at the cathode. C Copper metal deposits at the anode. D The intensity of the blue colour of the solution remains constant.

What is formed at electrode X ? A Iron B Oxygen C Hydrogen gas D Nitrogen dioxide gas 15 When aqueous magnesium sulphate solution is electrolysed using graphite electrodes, A the mass of cathode increases. B the mass of anode decreases. C magnesium metal deposits at the cathode. D the concentration of magnesium sulphate solution increases. 16 Electrolysis of aqueous sodium iodide solution is carried out ’09 using carbon electrodes. Which half-equation shows the reaction at the cathode? A 2I– → I2 + 2e– B 4OH– → 2H2O + O2 + 4e– C Na+ + e– → Na D 2H+ + 2e– → H2

17 In an experiment, dilute aqueous potassium iodide solution is electrolysed using carbon electrodes. Which of the following statements are true about this experiment? I A gas that produces a small ‘pop’ sound when tested with a lighted wooden splint is produced at the cathode. II A gas that rekindles a glowing wooden splint is produced at the anode. III The solution around the anode changes to brown colour. IV The concentration of potassium iodide solution increases. A I and II only B II and IV only C I and III only D I, II and IV only 18 Metal Y is placed at a high position in the electrochemical series. When a dilute Y chloride solution is electrolysed using carbon electrodes, the product formed at the cathode is A hydrogen B oxygen C chlorine D metal Y

6.4

Electrolysis in Industries

19 Electrolysis is used to extract aluminium metal from molten aluminium oxide. Which chemical is used to lower the melting point of aluminium oxide to 900°C? A Cryolite B Bauxite C Silicon dioxide D Calcium carbonate

20 What are the suitable chemicals and apparatus used to electroplate a spoon with silver metal by electrolysis? Anode

Cathode

Electrolyte

A

Silver

Spoon

Silver chloride solution

B

Spoon

Silver

Silver nitrate solution

C

Carbon

Spoon

Silver nitrate solution

D

Silver

Spoon

Silver nitrate solution

174

22 The diagram shows the arrangement of apparatus to ’06 electroplate a metal key with chromium. It is found that electroplating does not occur. How would you change the arrangement of the apparatus in order to plate a layer of chromium on the surface of the key?

A Replace chromium nitrate solution with chromium chloride solution B Change the supply of direct current to alternating current C Reverse the terminals of the batteries D Replace the chromium electrode with carbon

6.5

Voltaic Cells

24 The diagram below shows a simple chemical cell. Two ’11 different metals are used as electrodes. 0

1

2

3

4

5

26 Voltaic cells that are used in watches and calculators are A dry cells B alkaline cells C mercury cells D nickel-cadmium cells 27 The diagram shows the set-up of apparatus of a chemical cell. ’05

zinc plate

copper plate

sodium chloride solution

6

21 The presence of foreign metals in copper metal can reduce the conductivity of copper wire. Which of the following is suitable to be used as the cathode in the purification of copper by electrolysis? A Pure copper B Impure copper C Carbon D Platinum

Which of the following metals can be used to replace the zinc plate to obtain the brightest light in the light bulb and the highest voltage reading? A Magnesium B Iron C Aluminum D Lead 25 When magnesium metal and copper metal are connected by wire and then immersed in copper(II) sulphate solution, which of the following does not happen? A Electron flows from copper metal to magnesium metal. B Mass of copper increases. C Mass of magnesium decreases. D The colour intensity of blue copper(II) sulphate solution decreases.

Which of the following occurs in the chemical cell? A The magnesium rod becomes thicker. B The iron rod becomes thinner. C Electrons flow from iron to magnesium. D The green colour of iron(II) sulphate becomes paler. 28 The diagram shows the set-up of apparatus for an electrochemical cell.

23 The diagram shows the set-up of the apparatus used for the purification of a metal through electrolysis.

Which of the following combinations is suitably used for the purification of copper metal? Electrode X

Electrode Y

Electrode Z

A

Pure copper

Impure copper

Copper(II) sulphate

B

Impure copper

Pure copper

Copper(II) nitrate

C

Pure copper

Impure copper

Sulphuric acid

D

Impure copper

Pure copper

Copper(II) carbonate

175

Which of the following observations are true for this experiment? I Zinc electrode becomes thinner. II Brown colour is formed around electrode X. III Gray deposit is formed at electrode Y. IV Intensity of blue colour in beaker M becomes paler. A I and III only B II and III only C II and IV only D I, II and IV only Electrochemistry

29 The diagram shows the set-up of apparatus for a simple cell.

Which of the following pairs of metals gives the highest voltmeter reading?

6

Metal X

Metal Y

A

Magnesium

Iron

B

Zinc

Copper

C

Aluminium

Silver

D

Silver

Copper

30 A voltaic cell is made using metal X and Y as the electrodes. If electrons flow from metal X to metal Y, metal X and metal Y may be Metal X

Metal Y

A

Iron

Silver

B

Silver

Copper

C

Iron

Magnesium

D

Copper

Zinc

31 Which of the following statements is not true about lead-acid accumulator? A The electrolyte used is sulphuric acid. B Lead plate is the negative terminal. C Carbon is the positive terminal. D Lead(II) sulphate is formed when the cell is being used.

6.6

The Electrochemical Series

32 An experiment is carried out to measure the potential ’06 differences produced in voltaic cells made from metal electrode pairs Q-P, R-P, S-T or S-P metals. The results of the experiment is recorded in the table below. Electrochemistry

Metal electrode pairs

Negative terminal

Potential difference (V)

Q-P

Q

2.7

R-P

R

1.1

S-T

S

1.3

S-P

S

2.1

What is the potential difference of a voltaic cell made of metal electrode pair Q-T ? A 0.8 V B 1.4 V C 1.9 V D 3.5 V 33 If a piece of metal X is immersed in copper(II) sulphate solution, a brown deposit is formed. Metal X may be A copper B platinum C aluminium D silver 34 Two voltaic cells are constructed as shown in the diagram. The voltmeter reading of cell I is 1.1 V while that of cell II is 2.5 V.

Which of the following is true of a voltaic cell constructed using metal Q and metal R? A Metal Q will be the negative terminal. B Electrons will flow from metal R to metal Q. C The cell will produce a reading of 3.6 V. D R ions and Q ions are formed. 35 A piece of zinc metal is immersed in a beaker containing a mixture of copper(II) sulphate and magnesium nitrate solution. Which of the following does not happen? A Zinc metal dissolves.

176

B Concentration of magnesium ion in the solution decreases. C A brown deposit is formed. D Zinc ions are formed. 36 The table below shows information about three voltaic ’10 cells. Metals P, Q, R and S are used as electrodes in the cells. Voltaic Negative Positive Voltage cell terminal terminal (V) I

P

Q

0.9

II

R

Q

1.3

III

R

S

2.1

What is the order of the metals from the most electropositive to the least electropositive? A P, Q, R, S B P, R, Q, S C R, P, Q, S D S, Q, P, R 37 Which of the following pairs can undergo a displacement reaction? A Magnesium and potassium chloride solution. B Calcium and zinc sulphate solution. C Iron and calcium nitrate solution. D Copper and magnesium nitrate solution. 38 When an iron nail is immersed in X solution, Fe2+ ions are produced. Solution X may be A magnesium sulphate B zinc nitrate C copper(II) nitrate D sodium chloride 39 Excess metal X powder is added to copper(II) sulphate solution and is stirred. After half an hour, the solution becomes colourless and brown deposit is formed. Metal X may be I calcium II aluminium III magnesium IV silver A I and II only B III and IV only C I, II and III only D II, III and IV only

40 The diagram shows four simple chemical cells. In each cell, copper is one of the electrodes.

6

’05

In which cell does copper act as the negative terminal? A Cell I B Cell II C Cell III D Cell IV

Structured Questions 1 In an experiment, different chemical substances are tested using the set-up of apparatus as shown in Diagram 1.

2 In an experiment, electrolysis of 0.001 mol dm–3 hydrochloric acid is carried out using a electrolytic cell as shown in Diagram 2. Gases are collected at both the electrodes.

Diagram 1 Diagram 2

(a) When naphthalene is used as the chemical in the experiment, the light bulb does not light up. Explain this observation. [1 mark]

(a) Write the formulae of all the ions present in hydrochloric acid. [1 mark]

(b) When lead(II) bromide solid is used as the chemical in the experiment, the light bulb does not light up but lights up when lead(II) bromide is heated to the molten form. Explain this observation. [2 marks]

(b) Name a suitable material that can be used as the electrodes in this experiment. [1 mark] (c) (i) Name gas X and gas Y. [2 marks] (ii) Write the half-equation for the reaction that occurs at the anode. [1 mark]

(c) Predict the observation that will take place at the anode and the cathode when molten lead(II) bromide is used as the chemical in this experiment. [2 marks]

(d) After the electrolysis is carried out for 50 minutes, the concentration of hydrochloric acid increases and a different gas is collected at the anode. (i) Explain why the concentration of hydrochloric acid increases. [2 marks] (ii) Name the new gas collected at the anode and explain why this gas is produced. [2 marks] (iii) Write the half-equation for the reaction that occurs at the anode in (ii). [1 mark]

(d) Write the half-equations for the reactions that occur at the anode and the cathode in (c). [2 marks]

(e) Predict the products that will be formed if molten zinc chloride is used instead of lead(II) bromide in this experiment. [2 marks]

177

Electrochemistry

(a) Write the formula of all the cations present in the copper(II) sulphate solution. [1 mark]

3 Diagram 3 shows the arrangement of apparatus in an electrochemistry experiment.

(b) State the direction of the flow of electrons in Cell Q. [1 mark] (c) (i) State the observation at the cathode of Cell P. [1 mark] (ii) Write a half-equation for the reaction that occurred at the cathode of Cell P. [1 mark] (iii) Predict the change of colour intensity of the copper(II) sulphate solution of cell P. [1 mark] (iv) Name the product formed at the anode if copper electrodes in Cell P are replaced by carbon electrodes. [1 mark]

6

Diagram 3

(a) What is the difference between the energy change in Cell A and Cell B? [2 marks] (b) Write half-equations for the reactions that occur at the (i) magnesium electrode in Cell A. [1 mark] (ii) copper electrode in Cell A. [1 mark]

(d) Based on cell Q: (i) State the observation on the zinc plate. [1 mark]

(ii) Write the half-equations for the reaction that occurs at the zinc plate. [1 mark] (iii) Write an overall ionic equation for the reaction that has taken place. [1 mark] (iv) What happens to the cell voltage if the copper plate is replaced with a silver plate?

(c) (i) Name the electrode that serves as the negative terminal in Cell B. [1 mark] (ii) State the reason for your answer in (i). [1 mark]



(iii) State the direction of the flow of electrons in Diagram 3. [1 mark]

[1 mark]

(d) In Cell B, (i) name the product formed at the carbon electrode Q and write an equation for the reaction that occurs. [2 marks] (ii) name the product formed at the carbon electrode P and write an equation for the reaction that occurs. [2 marks]

5 Diagram 5 shows a voltaic cell that is formed from copper metal and lead metal.

(e) What would happen if the magnesium electrode in Cell A is replaced with a silver electrode? [1 mark]

(f) What would happen if carbon electrodes P and Q are replaced with copper electrodes? [1 mark]

Diagram 5

(a) State the positive terminal and the negative terminal of the voltaic cell. [2 marks]

4 Diagram 4 shows two types of cell.

(b) Write ionic equations showing the reactions that occur at (i) the negative terminal of the cell. [1 mark] (ii) the positive terminal of the cell. [1 mark] (c) Write the overall ionic equation of the cell. [1 mark] (d) What is the function of the salt bridge? [1 mark] (e) The voltage of the above cell is 0.57 V. If magnesium is above lead in the electrochemical series, what would be the expected voltage produced from a magnesium/copper voltaic cell? [1 mark] 6 An electrolysis process is carried out using the arrangement of apparatus as shown in Diagram 6. Diagram 4 Electrochemistry

178



(c) Write the ionic equation that occurs at (i) electrode L (ii) electrode M [2 marks]



(d) What is the product of electrolysis formed at (i) electrode R? (ii) electrode S? [2 marks] (e) Predict any colour change of the solution that may occur in beakers I and II after electrolysis has been carried out for an hour. [2 marks]

Diagram 6

(a) Name the electrodes that serve as the anode. [2 marks]

(f) (i) Name instrument Q in the diagram. (ii) What is the function of instrument Q?

(b) Write the formulae of all the ions present in beaker I. [2 marks]

6

[2 marks]

Essay Questions (b) Using a labelled diagram, describe an experiment to show how you can electroplate an iron spoon with another metal. In your description, give the observation and equations for the reactions that occur. [8 marks]

1 (a) What is meant by the term electrolysis? [2 marks] (b) Discuss in terms of ionic theory, the reasons why solid magnesium chloride (crystals) does not conduct electricity whereas molten magnesium chloride does. [4 marks] (c) You are supplied with magnesium chloride crystals and all the necessary apparatus. Describe an experiment to extract magnesium metal from magnesium chloride crystals using electrolysis. What would you observe in this experiment? Using ionic theory, explain how the products are formed at the cathode and the anode.

3 (a) What is the difference between an electrolytic cell and a voltaic cell? [4 marks] (b) You are supplied with metal P, metal Q, their nitrate salt solutions and all the necessary apparatus. Metal P is higher than metal Q in the electrochemical series and both metals have a valency of 2. Describe an experiment to show how you can produce an electric current from chemical reactions. Include a circuit diagram and show how you can detect the flow of electric current in your description. [12 marks]

[14 marks]

2 (a) The products of electrolysis may be different even though the same type of electrolyte is used. Using a suitable electrolyte, explain how (i) the types of electrodes, (ii) the concentration of ions can determine the products of electrolysis of an aqueous solution. [12 marks]

(c) Predict what will happen when a piece of metal P is placed in Q nitrate solution. Explain your answer. [4 marks]

Experiments 1 A group of students carried out three experiments to determine the products of electrolysis of sodium hydroxide solution, potassium iodide solution and aqueous X solution using carbon electrodes. The results of the experiment obtained is tabulated in Table 1. Experiment Chemical substance I

0.1 mol dm–3 sodium hydroxide solution

Observation at the cathode Colourless gas is evolved which produces a ‘pop’ sound when a lighted wooden splint is placed near the mouth of the test tube.

179

Observation at the anode Colourless gas is evolved which lights up a glowing wooden splint.

Electrochemistry

Experiment Chemical substance

Observation at the cathode

II

0.5 mol dm–3 aqueous potassium iodide solution

Colourless gas is evolved which produces a ‘pop’ sound when a lighted wooden splint is placed near the mouth of the test tube.

A brown solution is formed.

III

0.5 mol dm–3 aqueous X solution

Brown deposit is formed.

A brown gas is evolved which changes blue litmus paper to red and decolourises the litmus paper subsequently.



6

Observation at the anode

Table 1

(a) (i) In experiment I, name the products formed at the anode and cathode. (ii) What factor determines the type of ions discharged at the cathode?

[3 marks]

(b) (i) What is the product formed at the anode in experiment II? (ii) Suggest a test to identify the product of (i). (iii) What factor determines the type of ions discharged at the anode in experiment II?

[3 marks]

(c) Write ionic equations for the formation of the product(s) (i) at the anode and the cathode in experiment II. (ii) at the anode in experiment I.

[3 marks]

(d) In experiment III, aqueous X solution is blue in colour. (i) What is the brown deposit formed at the cathode? (ii) Name the brown gas produced at the anode. (iii) Suggest a chemical substance that may be X.

[3 marks]

2 Diagram 1 and Diagram 2 show the set-ups of two electrolytic cells using copper(II) chloride solutions of different concentrations. Plan an experiment to investigate the factor that affects the products of electrolysis of aqueous solutions as shown in Diagrams 1 and 2.









Diagram 1

Diagram 2

Your planning should include the following aspects: (a) Aim of experiment (b) Statement of hypothesis (c) All the variables (d) List of substances and apparatus (e) Procedure of the experiment (f) Tabulation of data [17 marks]

Electrochemistry

180

FORM 4 THEME: Interaction between Chemicals

CHAPTER

7

Acids and Bases

SPM Topical Analysis 2008

Year Paper

1

Number of questions

2 A

Section 5

2009

–

3

B

C

–

1 — 2

1

2 A

–

5

2010

–

3

B

C

–

2 — 3

1

1

3

2011

2

3

A

B

C

1

–

–

1

1

4

2

3

A

B

C

1

–

1 — 2

–

ONCEPT MAP

pH scale: measurement of the H+ ion concentration • Acids: pH < 7 • Alkalis: pH > 7 • pH value changes with the concentration and strength of acids/bases

ACIDS AND BASES

Acids: compounds that produce H+ ions in water • Strong acids: complete ionisation to form H+ ions in water • Weak acids: partial ionisation to form H+ ions in water

Properties of Acids: • Colour change with indicators • React with bases • React with reactive metals • React with metal carbonates

Concentration: units in • g dm–3 • mol dm–3 Relationship between pH values and concentration

Bases: compounds that react with acids to form salts and water • Alkalis: soluble bases that produce OH– ions in water • Strong alkalis: complete ionisation to form OH– ions in water • Weak alkalis: partial ionisation to form OH– ions in water

Neutralisation: • Reactions between acids and bases • Uses of acids/bases and neutralisation • Determination of end point in titration using acid/base indicators or a computer interface

Properties of Bases: • Colour change with indicators • React with acids • React with ammonium salt on heating • React with metal ions to form metal hydroxide

7.1

5 Without the presence of hydrogen ions, a substance does not show any acidic property. Dry hydrogen chloride gas, HCl(g) dissolved in an organic solvent (such as methylbenzene), glacial ethanoic acid and solid ethanedioic acid do not show any acidic property. 6 Acids can be divided into two types: mineral acids and organic acids. Mineral acids are obtained from minerals and most do not contain the element carbon. Organic acids are extracted from living things and contain the element carbon.

Characteristics and Properties of Acids and Bases

7

The Meaning of Acids 1 The definition of acids according to Arrhenius Theory: an acid is a chemical compound that produces hydrogen ions, H+ or hydroxonium ions, H3O+ when it dissolves in water. 2 A substance has acidic properties because of the formation of hydrogen ions or hydroxonium ions in water. 3 Dissociation of acids in water produces SPM hydrogen ions and anions. Examples: ’09/P1 (a)

Table 7.1 Examples of mineral acids and organic acids

H2O

HCl(g) ⎯⎯→ H+(aq) + Cl–(aq) hydrogen hydrogen chloride chloride ion ion

H2O HNO3(l) ⎯⎯→ H+(aq) + NO3–(aq) nitric acid hydrogen nitrate ion ion

(b)

Type of acid

Examples

Mineral acid

Hydrochloric acid, HCl, sulphuric acid, H2SO4 and nitric acid, HNO3

Organic acid

Ethanoic acid (CH3COOH), methanoic acid (HCOOH), ethanedioic acid (H2C2O4), citric acid, tartaric acid, malic acid and ascorbic acid.

H2O

(c)

H2SO4(l) ⎯⎯→ 2H+(aq) + SO42–(aq) sulphuric acid

H2O

(d) CH3COOH(l) ethanoic acid

hydrogen sulphate ion ion

Malic acid is found in apples. Citric acid is found in citrus fruits such as oranges. Tartaric acid is found in grapes. Ascorbic acid is vitamin C. Ethanoic acid is found in vinegar. Lactic acid is found in sour milk. Tannic acid is found in tea leaves.

H+(aq) + CH3COO–(aq)

hydrogen ethanoate ion ion

4 In actual fact, the hydrogen ion, H+ does not exist individually but is combined with a water molecule (hydrated) to form a hydroxonium ion. H+ + H2O → H3O+

An acid is a substance that produces hydrogen ions in the presence of water.

However, H3O+ is usually written as H+(aq) in the simplified way.

Hydrochloric acid, nitric acid and sulphuric acid are acids that are usually used in the school laboratories. Our stomachs contain hydrochloric acid that is required for digestion of food. Aspirin, which is used as an analgesic (a type of medicine for reducing pain), is also a type of acid.

Figure 7.1 The dissociation (ionisation) of a hydrogen chloride molecule to produce hydroxonium ion in water. Acids and Bases

182

6 A chemical substance has alkaline properties because of the formation of freely moving hydroxide ions, OH– in water. 7 In the presence of water, an alkali dissociates to hydroxide ions and cations. Examples:

The Meaning of Bases and Alkalis 1 A base is defined as a chemical substance that can neutralise an acid to produce salt and water only. For example, HCl + NaOH → NaCl + H2O salt

(a) NH3(g) + H2O(l) → NH4+(aq) + OH–(aq)

water

ammonia

2 Examples of bases are metal oxides and metal hydroxides that contain oxide ions, O2– and hydroxide ions, OH– respectively. Examples: copper(II) oxide, magnesium hydroxide. 3 The reaction between an acid and a base is known as neutralisation. In neutralisation the O2– ions or the OH– ions of a base react with the H+ ions of an acid to form water.







potassium hydroxide ion ion

(c) NaOH(s)

⎯→ Na+(aq) + OH–(aq)

H2O

H2O

(d) Ca(OH)2(s) ⎯→ Ca2+(aq) + 2OH–(aq) calcium hydroxide ion ion

8 A compound does not show any alkaline property in the absence of freely moving hydroxide ions. Examples: dry ammonia gas, ammonia gas dissolved in organic solvent (such as propanone), solid sodium hydroxide and solid potassium hydroxide do not show alkaline properties.

Figure 7.2 Venn diagram for bases and alkalis

Figure 7.4 The association (ionisation) of an ammonia molecule to produce a hydroxide ion

Bases

Zinc oxide, zinc hydroxide, copper(II) oxide, copper(II) hydroxide

potassium hydroxide

calcium hydroxide

4 Most bases are not soluble in water. Bases that are soluble in water are known as alkalis. 5 An alkali is defined as a chemical compound that dissolves in water to produce freely moving hydroxide ions, OH–.

examples

H2O

⎯→ K+(aq) + OH–(aq)

sodium sodium hydroxide hydroxide ion ion

O2– + 2H+ → H2O OH– + H+ → H2O

Bases that are insoluble in water

(b) KOH(s)

ammonium hydroxide ion ion

• An alkali is a compound that produces hydroxide ions in the presence of water. • A base is a compound that neutralises an acid and produces salt and water only.

Bases that are soluble in water (alkalis) examples Sodium oxide, sodium hydroxide, potassium oxide, potassium hydroxide, calcium hydroxide, ammonia

Theories on acids and alkalis: (a) Arrhenius theory: An acid is a compound that produces hydrogen ions when it dissolves in water. An alkali is a compound that produces hydroxide ions when it dissolves in water. (b) Brnsted–Lowry theory: An acid is a proton (hydrogen ion) donor. An alkali is a proton acceptor.

Figure 7.3 Flowchart showing types and examples of bases

183

Acids and Bases

7

acid base

1

’07

7

Which of the following statements is true about all bases? A React with acids B Dissolve in water C Produces hydroxide ions D Change red litmus paper to blue

Uses of Acids, Bases and Alkalis in Our Daily Life

Comments All bases react with acids to form salts. Only soluble bases (alkalis) dissolve in water to produce hydroxide ions that change red litmus paper to blue. Answer A

3 Examples of bases and their uses are given in Table 7.3.

SPM

’08/P2, ’09/P1

Table 7.3 Uses of bases

1 Acids and bases are widely used in our everyday life in agriculture, medicine, industry and in the preparation of food. 2 Examples of acids and their uses are given in Table 7.2.

Base

Sodium hydroxide To make soaps, detergents, bleaching agents and fertilisers

Table 7.2 Uses of acids

Acid

Uses

Ammonia

Uses

To make fertilisers, nitric acid, grease remover and to maintain latex in liquid form

To make paints, detergents, polymers, fertilisers, as an electrolyte in lead-acid accumulators

Calcium hydroxide To make cement, limewater and to neutralise the acidity of soil

Hydrochloric acid

To clean metals before electroplating

Magnesium hydroxide

To make toothpaste, gastric medicine (antacid)

Nitric acid

To make fertilisers, explosive substances (such as T.N.T.), dyes and plastics

Aluminium hydroxide

To make gastric medicine (antacid)

Benzoic acid

To preserve food

Carbonic acid

To make gassy (carbonated) drinks

Ethanoic acid

A component of vinegar

Tartaric acid

To make baking powder

Sulphuric acid

Cleaning agent contains ammonia

Methanoic acid is used in the coagulation of rubber latex

Fertilisers are made from acids and alkalis

Acids and Bases

Soaps and detergents are made from sodium hydroxide

184

7.1

SPM

’09/P2, ’10/P3, ’11/P2

Conclusion 1 Aqueous ethanoic acid turns blue litmus paper to red, indicating its acidic property. 2 Ethanoic acid in a dry condition or dissolved in organic solvents does not show any acidic property. 3 Ionisation of acids will only occur in the presence of water to produce hydrogen ions which are responsible for the acidic properties. 4 Water is essential for the formation of hydrogen ions which gives the acidic properties in an acid. The hypothesis is accepted.

Problem statement Is water needed for an acid to show its acidic properties? Hypothesis An acid will only show its acidic properties when dissolved in water. Variables (a) Manipulated variable : Types of solvents-water and propanone (b) Responding variable : Change in the colour of blue litmus (c) Constant variable : Type of acid and blue litmus paper Apparatus Test tube and droppers. Materials Glacial (dry) ethanoic acid, aqueous ethanoic acid, ethanoic acid dissolved in dry propanone and blue litmus paper. Procedure 1 A piece of dry blue litmus paper is placed in a test tube. 2 A few drops of glacial ethanoic acid are placed onto the blue litmus paper using a dropper. 3 The effect of the glacial ethanoic acid on the blue litmus paper is recorded. 4 Steps 1 to 3 of the experiment are repeated using aqueous ethanoic acid and ethanoic acid dissolved in propanone to replace glacial ethanoic acid. 5 The observations are then tabulated.

7

To investigate the role of water in showing the properties of acids

Discussion 1 In the presence of water, an acid dissociates into hydrogen ions that cause acidity in an acid. 2 Dry acids do not show any acidic properties in the absence of water because dry acids exist as covalent molecules. Hydrogen ions are not produced. 3 Solvents such as methylbenzene, propanone and trichloromethane cannot replace water for an acid to show its acidic properties. This is because an acid exists as covalent molecules in these organic solvents; H+ ions are not produced in these solutions. 4 Glacial ethanoic acid (CH3COOH) consists of acid molecules only. CH3COOH molecule is a covalent compound. 5 Figure 7.5 shows the types of particles that are present in ethanoic acid dissolved in propanone and in water.

Condition of ethanoic acid Glacial (dry)

Aqueous (dissolved in water) Dissolved in propanone

Observation

Inference

No noticeable colour change in the litmus paper Blue litmus paper has changed to red No noticeable colour change in the litmus paper

Does not show any acidic properties Shows acidic properties Does not show any acidic properties

Figure 7.5 Particles in ethanoic acid dissolved in (a) propanone (b) water

185

Acids and Bases

Experiment 7.1

Results

7.2 To investigate the role of water in showing the alkaline properties of alkali

7

Problem statement Is water essential for an alkali to show its alkaline properties?

2 The test tube must be stoppered immediately after the red litmus paper is put in.

Hypothesis An alkali will only show its alkaline properties when dissolved in water.

Results

Variables (a) Manipulated variable : Types of solvents–water and propanone (b) Responding variable : Change in the colour of red litmus paper (c) Constant variable : Type of alkali and red litmus paper Apparatus

Test tubes and droppers.

Materials

Dry ammonia gas stoppered in a test tube, ammonia gas dissolved in propanone, aqueous ammonia solution and red litmus paper.

Condition of ammonia

Experiment 7.2

Inference

Dry

No colour change in the red litmus paper

Does not show alkaline property

Aqueous (dissolved in water)

Red litmus has changed to blue

Shows alkaline properties

Dissolved in propanone

No colour change in the red litmus paper

Does not show alkaline property

Conclusion 1 Aqueous ammonia solution turns the red litmus paper to blue, indicating its alkaline property. 2 Dry ammonia gas or ammonia gas dissolved in organic solvents does not show any alkaline property. 3 An alkali shows its alkaline properties only in the presence of water. When water is present, ammonia ionises to produce OH– ions that are responsible for its alkaline properties. 4 Water is essential for the formation of hydroxide ions that cause alkalinity in an alkali. The hypothesis is accepted.

Procedure 1 A piece of dry red litmus paper is put into a stoppered test tube of dry ammonia gas and the test tube is then stoppered back immediately (Figure 7.6). 2 The effect of the dry ammonia gas on the red litmus paper is recorded. 3 Another piece of dry red litmus paper is put in 5 cm3 of aqueous ammonia solution in a separate test tube. 4 Step 3 of the experiment is repeated using ammonia dissolved in propanone to replace aqueous ammonia solution.

Discussion 1 In the presence of water, an alkali ionises to form hydroxide ions, OH– that change red litmus paper to blue. 2 Aqueous ammonia solution (ammonia dissolved in water) consists of NH4+ ions, OH– ions and NH3 molecules. An aqueous ammonia solution is alkaline due to the presence of hydroxide ions. NH3 + H2O

Figure 7.6 Testing for the alkaline properties of ammonia gas

NH4+ + OH–

3 Dry alkalis, solid alkalis (such as solid calcium hydroxide and barium hydroxide) and alkalis dissolved in organic solvents (such as propanone) do not show any alkaline properties. This is because the alkalis do not dissociate into hydroxide ions.

Safety precautions 1 Ammonia gas is poisonous. This experiment involving dry ammonia gas should be carried out in a fume cupboard.

Acids and Bases

Observation

186

Glacial ethanoic acid is the pure and dry form of ethanoic acid. It is named ‘glacial’ because it appears as ice when it solidifies below its melting point.

1 Acids are sour in taste. 2 Acid solutions have pH values of less than 7. 3 Acids change colours of indicators as shown in Table 7.4. 4 Acids can react with (a) bases to produce salts and water, (b) metals to produce salts and hydrogen gas, (c) carbonates to produce salts, carbon dioxide gas and water.

1 If the electrical conductivity of ethanoic acid in pro­ panone and aqueous ethanoic acid is tested in turn, only the aqueous solution of acid conducts electri­ city (light bulb is lighted up or ammeter needle is deflected). 2 This shows the presence of freely moving ions in an aqueous solution of acid. CH3COOH(l) ethanoic acid

CH3COO –(aq) + H+(aq) ethanoate ion hydrogen ion

H+ ions and CH3COO– ions conduct electricity

Table 7.4 Effects of acids on indicators

H+ ions change blue litmus to red

Colour of indicator in acidic solution

Indicator

3 Dry acids do not conduct electricity. This is because there are no freely moving ions. Dry acid exists as covalent molecules. 4 Similarly, ammonia dissolved in propanone does not conduct electricity. It exists as covalent molecules. 5 An aqueous ammonia solution can conduct electricity, showing the presence of freely moving ions.

Blue litmus paper

Red

Universal indicator

Orange and red

Methyl orange

Red

To investigate the chemical properties of acids Apparatus

Test tube, test tube holder, spatula, Bunsen burner, delivery tubes with stopper and wooden splint.

Materials

1.0 mol dm–3 sulphuric acid, copper(II) oxide, zinc powder, sodium carbonate powder and limewater.

7

Chemical Properties of Acids

Procedure 1 A little copper(II) oxide is added to 5 cm3 of sulphuric acid in a test tube. The mixture is heated slowly (Figure 7.7) and any changes that occur are recorded. 2 A little zinc powder is added to 5 cm3 of dilute sulphuric acid in a test tube. The gas evolved is tested by placing a lighted wooden splint near the mouth of the test tube (Figure 7.8). 3 A little sodium carbonate powder is added to 5 cm3 of dilute sulphuric acid in a test tube. The gas evolved is tested with limewater (Figure 7.9).

Figure 7.8 An acid with a metal

Activity 7.1

Figure 7.7 An acid with a base

SPM

’10/P2

Figure 7.9 An acid with a metal carbonate

187

Acids and Bases

Results Test on acid

Observation

Inference

Heating with copper(II) oxide

• Black powder dissolved • Blue solution is formed

Test with zinc powder

• Effervescence occurred • Gas produced a ‘pop’ sound when it is tested with a lighted wooden splint • Zinc powder dissolved

7

Test with sodium carbonate

• Copper(II) salt solution is formed

• Effervescence occurred • Gas evolved turned limewater milky • White solid of sodium carbonate dissolved

3 A dilute acid will react with a metal carbonate to produce a salt, carbon dioxide gas and water. acid + metal carbonate → salt + carbon dioxide + water

CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)

Examples: (a) Sodium carbonate reacts with dilute sulphuric acid to produce a salt, sodium sulphate, carbon dioxide gas and water.

(b) Copper(II) oxide dissolves in ethanoic acid to form a salt, copper(II) ethanoate and water. CuO(s) + 2CH3COOH(aq) → Cu(CH3COO)2(aq) + H2O(l)

Na2CO3(s) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(l)

(c) Nitric acid reacts with sodium hydroxide (an alkali) to produce a salt, sodium nitrate and water.

(b) Calcium carbonate reacts with dilute hydrochloric acid to produce a salt, calcium chloride, carbon dioxide gas and water.

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)

acid + reactive metal → salt + hydrogen Examples: (a) Zinc dissolves in sulphuric acid to form a salt, zinc sulphate and hydrogen gas. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

Acids and Bases

• Carbon dioxide gas is produced • A salt solution is formed

Mg(s) + 2CH3COOH(aq) → Mg(CH3COO)2(aq) + H2(g)

Acid + base → salt + water

2 A dilute acid will react with a reactive metal to produce a salt and hydrogen gas.

• A salt solution is formed

(b) Magnesium dissolves in ethanoic acid to form a salt, magnesium ethanoate and hydrogen gas.

Discussion 1 A dilute acid reacts with a base to produce salt and water only.

If the salt solution is evaporated until saturated, salt crystals will form upon cooling. Examples: (a) Black copper(II) oxide powder (a base) dissolves in dilute sulphuric acid to produce a salt, copper(II) sulphate (blue colour) and water.

• Hydrogen gas is produced

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) Conclusion 1 Sulphuric acid reacts with a base (copper(II) oxide) to produce salt and water. 2 Sulphuric acid reacts with a reactive metal (zinc) to produce a salt and hydrogen gas. 3 Sulphuric acid reacts with a metal carbonate (sodium carbonate) to produce a salt, water and carbon dioxide gas. 188

Chemical Properties of Alkalis 1 Alkalis are bitter in taste and feel soapy. 2 Alkaline solutions have pH values of more than 7. 3 Alkalis change the colours of indicators as shown in Table 7.5 below.

Non-reactive metals such as copper and silver do not react with dilute acid. Very reactive metals such as sodium and potassium will react with dilute acid vigorously and may produce an explosion.

2

Table 7.5 Effects of alkalis on indicators

’01

Colour of indicator in alkaline solution

Indicator

Which of the following compounds reacts with limestone powder to produce a gas that turns limewater milky? A Nitrogen dioxide gas B Hydrogen chloride gas dissolved in tetra­ chloromethane C Sulphur dioxide gas dissolved in propanone D Sulphur dioxide gas dissolved in water

Blue 7

Red litmus paper

Blue or purple

Universal indicator

Yellow

Methyl orange

4 An alkali reacts with an acid to produce salt and water. For example:

Comments An acidic gas must first dissolve in water before reacting with calcium carbonate (limestone) powder to produce carbon dioxide gas which turns limewater milky.

KOH(aq) + HCl(aq) → KCl(aq) + H2O(l) 5 When an alkali is heated with an ammonium salt, ammonia gas is produced. For example:

Answer D

NH4+(aq) + OH–(aq) → NH3(g) + H2O(l) from ammonium salt

Generally, (a) metal oxides and metal hydroxides are basic. For example: MgO + H2O → Mg(OH)2 magnesium oxide magnesium hydroxide

from alkali

6 An aqueous alkali forms metal hydroxide as precipitate when added to an aqueous salt solution. For example: Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s)

(b) non-metal oxides are acidic. For example, SO2, NO2 or CO2. SO2 + H2O → H2SO3 sulphur dioxide sulphurous acid

from copper(II) salt solution

from alkali

copper(II) hydroxide as blue precipitate

Apparatus Test tubes, test tube holder, spatula, Bunsen burner, delivery tubes with stopper and red litmus paper. Materials 2.0 mol dm–3 sodium hydroxide solution, benzoic acid powder, ammonium chloride powder, 1.0 mol dm–3 iron(III) sulphate solution. Procedure 1 A little benzoic acid powder is added to 5 cm3 of sodium hydroxide solution in a test tube. Any

changes that occur are recorded. 2 A little ammonium chloride powder is added to 5 cm3 of sodium hydroxide solution in a test tube. The mixture is heated gently. The gas evolved is tested with a piece of damp red litmus paper. 3 5 cm3 of sodium hydroxide solution is added to 5 cm3 of iron(III) sulphate solution in a test tube. Any changes that occur are recorded.

189

Acids and Bases

Activity 7.2

To investigate the chemical properties of alkalis

Results

7

Test on sodium hydroxide

Observation

Inference A salt solution is formed

With benzoic acid powder added

White powder dissolves and a colourless solution is formed

Heating with ammonium chloride powder

Ammonia gas A pungent gas that turns damp is produced red litmus paper blue is evolved

With iron(III) A brown precipitate is sulphate solution added formed

NaOH(aq) + C6H5COOH(s) → C6H5COONa(aq) + H2O(l) 2 If the salt solution is evaporated in an evaporating dish until a saturated solution is formed, white crystals of sodium benzoate will be crystallised upon cooling. 3 When sodium hydroxide is heated with ammonium chloride (an ammonium salt), ammonia gas is produced. NH4Cl(s) + NaOH(aq) → NaCl(aq) + H2O(l) + NH3(g)

Iron(III) hydroxide is formed

Conclusion 1 Sodium hydroxide reacts with benzoic acid to produce salt and water. 2 When sodium hydroxide is heated with ammonium chloride, ammonia gas which turns red litmus to blue is produced. 3 Sodium hydroxide solution reacts with an aqueous iron(III) solution to produce a brown precipitate, iron(III) hydroxide.

In this reaction, ammonium ions react with hydroxide ions to produce ammonia gas. The ionic equation for this reaction is NH4+(aq) + OH–(aq) → NH3(g) + H2O(l) 4 Sodium hydroxide solution hydroxide ions in water.

to

NaOH → Na+ + OH– Hydroxide ions combine with iron(III) ions from iron(III) sulphate solution to form insoluble iron(III) hydroxide as a brown precipitate. Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s)

Discussion 1 Sodium hydroxide as an alkali reacts with benzoic acid, C6H5COOH to produce a salt, sodium benzoate and water in a neutralisation reaction.

from iron(III) sulphate

from sodium hydroxide

CH3COOH(aq)

Basicity of Acids 1 Basicity of an acid is the number of moles of OH– ions that are required to react with one mole of the acid. 2 Since one mole of OH– ions reacts with one mole of H+ ion, the basicity of an acid is also the number of moles of H+ ion that can be produced by one mole of the acid when it dissolves in water. 3 A monoprotic acid (or monobasic acid) is an acid that will produce one mole of H+ ion when one mole of the acid dissolves in water. For example, although ethanoic acid has four hydrogen atoms in the molecule, only one of the hydrogen dissociates to form H+ ion in water.

Acids and Bases

dissociates

iron(III) hydroxide as brown precipitate

CH3COO–(aq) + H+(aq)

Three H atoms bonded to carbon do not dissociate

Only one H atom dissociates to form H+ ion

4 A diprotic acid (or dibasic acid) is an acid that will produce two moles of H+ ions from one mole of the acid in water. For example: H2SO4(aq) → 2H+ (aq) + SO42–(aq) 1 mol sulphuric acid

2 mol hydrogen ions

sulphate ion

5 A triprotic acid (or tribasic acid) is an acid that will produce three moles of H+ ions from one mole of the acid in water. For example: H3PO4(aq) 3H+(aq) + PO43–(aq) 1 mol phosphoric acid

190

3 mol hydrogen ions

phosphate ion

Table 7.6 Examples of monoprotic acid and diprotic acid

Examples of monoprotic acid

Hydrochloric acid (HCl), nitric acid (HNO3), ethanoic acid (CH3COOH) and methanoic acid (HCOOH)

Basicity of an acid is not the same as the number of H atoms in the formula of the acid.

Examples of diprotic acid

Sulphuric acid (H2SO4), ethanedioic acid (H2C2O4), carbonic acid (H2CO3) and chromic acid (H2CrO4)

Basicity is the number of moles of H+ ions produced by one mole of acid in water.

1 (a) Explain what you understand by the term (i) an acid (ii) a base (iii) an alkali (b) What is the effect of an acid and an alkali on moist litmus paper?

3 Identify the chemicals Q, R, X, Y and gas Z in the following reactions: (a) H2SO4 + Q → MgSO4 + H2O + CO2 (b) Ca(OH)2 + 2R → Ca(NO3)2 + 2H2O (c) 2Al + 6X → 2AlCl3 + 3H2 heat (d) Y + NH4NO3 ⎯⎯→ KNO3 + H2O + Z

2 Identify the correct uses of the following acids and bases.

4 Write equations to show the reactions between (a) sulphuric acid and magnesium oxide (b) nitric acid and aluminium metal (c) hydrochloric acid and calcium carbonate (d) ethanoic acid and sodium hydroxide (e) potassium hydroxide and ammonium chloride when heated

Acids or bases: H2SO4, HNO3, CaO, Ca(OH)2, Mg(OH)2, NH3, NaOH Acids or bases

Uses To make antacid

5 Effervescence occurs when magnesium powder is added to aqueous hydrochloric acid. However, no noticeable change takes place when magnesium powder is added to hydrogen chloride dissolved in methylbenzene. Explain why.

To make fertiliser To make soap To neutralise acidity in soil

Acid

Alkali

dissolves in water

dissolves in water

produces H+ ions

produces OH– ions

reacts with

carbonate

metal

reacts with

base

acid

ammonium salt

metal ions

heat salt + carbon dioxide + water

salt + hydrogen

salt + water

191

ammonia gas

metal hydroxides

Acids and Bases

7

7.1

1 The pH scale is a set of numbers used to indicate the degree of acidity or alkalinity of a solution. 2 The values of the pH scale range from 0 to 14. • pH < 7 ⇒ acidic solution • pH = 7 ⇒ neutral solution

• pH > 7 ⇒ alkaline solution 3 pH is actually a measurement of the con­cen­ tration of hydrogen, H+ ions in a solution. 4 The higher the concentration of the H+ ions, the lower the pH value and the more acidic the solution. 5 The higher the concentration of the OH– ions, the higher the pH value and the more alkaline the solution. 6 The relationship between the pH scale, acidity or alkalinity and concentration of H+ ions is shown below.

• All acids have pH < 7. • The lower the pH value, the higher the H+ ion concentration.

• All alkalis have pH > 7. • The higher the pH value, the higher the OH– ion concentration.

7.2

The Strength of Acids and Alkalis

7

The pH Scale

4 A pH meter is an electric meter that is used to measure the pH value of a solution accurately. A pH meter will show the pH value when its probe is immersed in a solution pH meter to be tested. 5 With a computer interface, the exact pH value can be displayed on the computer screen when the pH meter is placed in the solution.

Measurement of pH Value of a Solution 1 The pH value of a solution can be measured by using (a) universal indicator or pH paper (b) a pH meter (with or without a computer interface) 2 Universal indicator is a mixture of indicators that gives different colours corresponding to different pH values as shown in Table 7.7. 3 Universal indicator is used in the form of (a) solution, or (b) paper strips (also known as pH paper).

Table 7.7 Colours of universal indicator

pH value

0, 1, 2

Colour

red

3

4

5

6

7

orange orange orange yellow green red yellow

8

9

10

11

12, 13, 14

greenishblue

blue

blue

bluishpurple

purple

Activity 7.3

To measure the pH values of some solutions used in daily life Apparatus Beakers, universal indicator solution, dropper, standard colour chart of universal indicator.

Acids and Bases

Materials Soap solution, carbonated drink, tap water, orange fruit juice, distilled water, milk, tea, dilute sodium hydroxide and hydrochloric acid. 192

Procedure 1 About 10 cm3 of soap solution is placed in a small beaker. 2 Two drops of universal indicator solution are added to the soap solution. The solution is then stirred. 3 The colour of the solution produced is matched against the standard colour chart of universal indicator. The corresponding pH value of the colour is noted and recorded. 4 The experiment is repeated using carbonated drink, tap water, orange fruit juice, distilled water, milk, tea, dilute sodium hydroxide and hydrochloric acid in place of the soap solution. Results

pH value

Soap Carbonated Tap Orange solution drink water juice 10

5

6

Distilled water

Milk

Tea

Dilute sodium hydroxide

Dilute hydrochloric acid

7

6

5

13

1

4

7

Solution

Conclusion 1 Different solutions have different pH values. 2 The pH value of a solution can be measured using the universal indicator solution.

Degree of Dissociation

HCl → H+ + Cl– HNO3 → H+ + NO3– H2SO4 → 2H+ + SO42–

1 The strength of an acid or an alkali depends on the degree of dissociation (also known as the degree of ionisation). 2 The degree of dissociation measures the percentage or fraction of molecules that dissociates into ions when dissolved in water. 3 For example, the degree of dissociation of hydrochloric acid is 100% or 1. This means that all the hydrogen chloride molecules in hydrochloric acid will ionise to form H+ ions and Cl– ions when dissolved in water. 4 In a 1.0 mol dm–3 aqueous ethanoic solution, only 4 out of 1000 molecules of ethanoic acid dissociate to form ions. Degree of dissociation of a 1.0 mol dm–3 aqueous ethanoic solution is 4 — — — — = 0.004 or 0.4%. 1000

(The one–way dissociation)

→

indicates

complete

3 Complete dissociation (100%) in water by a strong acid produces a high concentration of H+ ions and hence a low pH. 4 Weak acids are chemicals that dissociate partially (incomplete dissociation) into hydrogen ions H+ in water. 5 Most of the organic acids such as ethanoic acid, ethanedioic acid, methanoic acid, citric acid and tartaric acid are weak acids. CH3COOH H2C2O4

5 Acids can be divided into 2 categories: strong acids and weak acids, depending on their degree of dissociation. 6 Alkalis can be divided into 2 categories: strong alkalis and weak alkalis, depending on their degree of dissociation.

Strong and Weak Acids

arrow

(The two–way arrow reaction)

CH3COO– + H+ 2H+ + C2O42– indicates reversible

Examples of weak acids Ethanoic acid, CH3COOH Methanoic acid, HCOOH Ethanedioic acid, H2C2O4 Carbonic acid, H2CO3 Phosphoric acid, H3PO4 Chromic acid, H2CrO4 Nitrous acid, HNO2 Sulphurous acid, H2SO3

SPM

’08/P2, ’09/P1, ’10/P3

1 A strong acid is a chemical substance that dissociates completely (degree of dissociation is 100%) into hydrogen ions, H+ in water. 2 Mineral acids such as hydrochloric acid, nitric acid and sulphuric acid are strong acids. 193

Acids and Bases

6 In a weak acid solution, a big portion of the weak acid exists as molecules and only a small portion dissociates to ions.

Strong and Weak Alkalis

1 A strong alkali is a chemical substance that dissociates completely (100%) to hydroxide ions, OH– in water. 2 Examples of strong alkalis are sodium hydroxide and potassium hydroxide.

A concentrated acid does not mean that it is a strong acid. For example, concentrated ethanoic acid is still a weak acid. Consequently, a dilute acid does not mean that it is a weak acid. For example, dilute hydrochloric acid is still a strong acid.

NaOH → Na+ + OH– KOH → K+ + OH–

7

(The one–way dissociation)

Test

0.1 mol dm HCl

Ca(OH)2

Magnesium ribbon

Hydrogen gas evolves vigorously

Solution Test

CO2 gas evolves vigorously

CO2 gas evolves slowly

Electrical conductivity

Light bulb lights up brightly

Light bulb lights up dimly

Conclusion

• High • Low concentration concentration of H+ ions of H+ ions • HCl is a strong • CH3COOH is a acid weak acid

Acids and Bases

Ca2+ + 2OH–

Table 7.9 Comparison of properties of a strong alkali (NaOH) with a weak alkali (NH3)

Hydrogen gas evolves slowly

Calcium carbonate

NH4+ + OH–

5 Partial dissociation of a weak alkali results in a low concentration of OH– ions. Hence, the pH value of a weak alkali is lower than that of a strong alkali with the same concentration. 6 Table 7.9 shows the comparison of properties of a strong alkali (sodium hydroxide solution) with a weak alkali (aqueous ammonia solution).

0.1 mol dm CH3COOH

Red colour, pH~1 Orange red colour, pH~4

complete

(The two–way arrow indicates partial dissociation)

–3

Universal Indicator

→ indicates

NH3 + H2O

Table 7.8 Comparison between a strong acid (HCl) and a weak acid (CH3COOH) –3

arrow

3 A weak alkali is a chemical substance that dissociates partially (incomplete dissociation) to hydroxide ions, OH– in water. 4 Examples of weak alkalis are aqueous ammonia, calcium hydroxide and magnesium hydroxide.

7 For two different acids of the same concentration, the acid with the lower pH value is the stronger acid. 8 Partial dissociation of a weak acid results in a low H+ ion concentration. 9 Strong and weak acids have the same chemical properties, but the rate of reaction and electrical conductivity of a weak acid is lower as shown in Table 7.8.

Solution

SPM

’08/P1, ’09/P1

194

0.1 mol dm–3 NaOH solution

0.1 mol dm–3 aqueous NH3

Universal Indicator

Purple colour, pH~13

Blue colour, pH~10

Electrical conductivity

Light bulb lights up brightly

Light bulb lights up dimly

Conclusion

• High • Low concentration concentration of OH– ions of OH– ions • NaOH is a • Aqueous NH3 strong alkali is a weak alkali

Decreasing pH 1

Increasing pH 7

Strong acids: • Complete ionisation • High H+ ion concentration • pH value: 1–2

Weak acids: • Partial ionisation • Low H+ ion concentration • pH value: 3–6

14

Weak alkalis: • Partial ionisation • Low OH– ion concentration • pH value: 8–12

Strong alkalis: • Complete ionisation • High OH– ion concentration • pH value: 13–14

SPM

’10/P1

7

To measure the pH values of solutions with the same concentration

Materials 0.1 mol dm–3 hydrochloric acid, 0.1 mol dm–3 ethanoic acid, 0.1 mol dm–3 aqueous ammonia and 0.1 mol dm–3 sodium hydroxide solution. Procedure 1 About 15 cm3 of 0.1 mol dm–3 hydrochloric acid is placed in a small beaker. 2 The probe of a pH meter is rinsed with distilled water. 3 The pH meter probe is then immersed in the acid in the beaker. The reading registered on the pH meter after it has stabilised is recorded. 4 Steps 1 to 3 of the experiment are repeated using ethanoic acid, ammonia solution and sodium hydroxide solution to replace the hydrochloric acid. Results Solution of 0.1 mol dm–3

pH value

Hydrochloric acid, HCl

1

• High concentration of H+ ions • A strong acid

Ethanoic acid, CH3COOH

3

• Low concentration of H+ ions • A weak acid

Aqueous ammonia, NH3

10

• Low concentration of OH– ions • A weak alkali

Sodium hydroxide, NaOH

13

• High concentration of OH– ions • A strong alkali

Inference

Conclusion 1 Acids have pH values of less than 7. 2 Different acids with the same concentration have different pH values. The pH of hydrochloric acid is lower than ethanoic acid of the same concentration. 3 Alkalis have pH values of more than 7. 4 Different alkalis with the same concentration have different pH values. The pH of sodium hydroxide solution is higher than aqueous ammonia of the same concentration. Discussion 1 Hydrochloric acid, HCl, and ethanoic acid, CH3COOH, are both acidic as their pH values are less than 7. 2 However, the pH value of HCl is less than that of CH3COOH with the same concentration indicating that the concentration of hydrogen ions in HCl is higher than that in CH3COOH. 3 HCl is an example of a strong acid which undergoes complete ionisation. CH3COOH is an example of a weak acid which undergoes partial ionisation. 4 Ammonia, NH3, and sodium hydroxide solution are both alkaline as their pH values are more than 7. 5 However, the pH value of NaOH is more than that of NH3 with the same concentration indicating that the concentration of hydroxide ions in NaOH is higher than that in NH3. 6 NaOH is an example of a strong alkali which undergoes complete ionisation. NH3 is an example of a weak alkali which undergoes partial ionisation. 7 The strength of an acid (strong or weak) and the strength of an alkali (strong or weak) depends on the degree of dissociation. 195

Acids and Bases

Activity 7.4

Apparatus 50 cm3 beaker and pH meter.

3

7.2

’03

1 Six solutions A, B, C, D, E and F with concentration of 1.0 mol dm–3, have pH values as shown in the table below:

Information about two solutions is given below:

7

Concentration of nitric acid = 1.0 mol dm–3 Concentration of carbonic acid = 1.0 mol dm–3 Which of the following statements are true about the two solutions given above? I Nitric acid is a stronger acid than carbonic acid. II The pH value of nitric acid is higher than carbonic acid. III The degree of dissociation of nitric acid in water is higher than that of carbonic acid. IV The concentration of H+ ions in nitric acid is higher than that in carbonic acid. A I and II only B III and IV only C I, III and IV only D I, II, III and IV

B

C

D

E

F

pH values

13

7.0

10

4.0

1.0

3.5

2 Using suitable examples, explain the terms strong acid and weak acid. Predict the difference in pH values of the two acids with the same concentration. 3 The degree of dissociation of ethanoic acid is higher than that of propanoic acid but is lower than that of methanoic acid. (a) Arrange the above three acids in ascending order of the strength of acidity. (b) If the pH value of 1.0 mol dm–3 ethanoic acid is 3, predict the pH value of 1.0 mol dm–3 methanoic acid and propanoic acid.

’07

7.3

Which of the following statements describe a strong alkali? I Has a high pH value II Ionises completely in water III Has a high concentration of hydroxide ions IV Exists as molecules in water A I and II only B III and IV only C I, II and III only D I, II, III and IV

Concentration of Acids and Alkalis

The Meaning of Concentration and Molarity, and Their Relationship 1 A solution is formed when a solute dissolves in a solvent. solute + solvent → solution For example, when sodium hydroxide (solute) dissolves in water (solvent), sodium hydroxide solution is formed. 2 Concentration and molarity are measurements of the amount of solutes dissolved in a given volume of solvent when a solution is formed. 3 The amount of a solute can be measured in the unit of ‘gram’ or ‘mole’. The quantity of a solution produced is usually measured in the unit of volume, dm3.

Comments A strong alkali ionises completely in water to produce a high concentration of hydroxide ions in water and hence a high pH value. (I, II and III are correct) A strong alkali exists as ions in water. (IV is incorrect) Answer C

Acids and Bases

A

(a) Which of the above solutions has (i) the highest concentration of H+ ions? (ii) the highest concentration of OH– ions? (b) Which of the above solutions is (i) a strong acid? (iii) a weak acid? (ii) a strong alkali? (iv) a weak alkali? (c) Which of the above solutions may be (i) sodium chloride solution? (ii) hydrochloric acid? (iii) aqueous ammonia? (iv) sodium hydroxide solution?

Comments Nitric acid is a stronger acid (I correct), has a higher degree of dissociation (III correct) and hence a higher degree of H+ ions concentration (IV correct) but a lower pH value than carbonic acid, which is a weak acid (II incorrect). Answer C

4

Solution

196

Solution Mass of copper(II) sulphate = 5.00 g Convert volume from Volume of solution = 500 cm3 cm3 to dm3 500 =— — — — — dm3 = 0.5 dm3 1000 Hence, concentration of copper(II) sulphate solution 5.00 g =— — — — — — 0.5 dm3 Concentration (g dm–3) Mass of solute dissolved (g) = 10.0 g dm–3 =— — — — — — — — — — — — — — — — — — — — — — — —

4 The concentration of a solution is the mass (in grams) or the number of moles of solute dissolved in a solvent to form 1.0 dm3 (1000 cm3) of solution. Hence the concentration of a solution can be defined in two ways: Mass of solute dissolved (g) Concentration = — — — — — — — — — — — — — — — — — — — — — — — — — — Volume of solution (dm3) –3 (g dm )



Number of moles of solute (mol) Concentration = — — — — — — — — — — — — — — — — — — — — — — — — — Volume of solution (dm3) –3 (mol dm )



Volume of solution (dm3)

2

5 For example, • a 23.0 g dm–3 NaOH solution has 23.0 g of NaOH in 1.0 dm3 solution. • a 0.5 mol dm–3 NaOH solution has 0.5 mol of NaOH in 1.0 dm3 solution. 6 Concentration in terms of mol dm–3 is more commonly known as molarity. In chemistry, the measurement of concentration in mol dm–3 (molarity) is more useful because all changes in chemical reactions occur in terms of moles. 7 Concentration (in g dm–3) can be converted SPM to molarity by dividing concentration (in ’08/P1 g dm–3) by the molar mass. The molar mass is the mass of 1 mol of substance.

7



What is the mass of sodium carbonate required to dissolve in water to prepare a 200 cm3 solution that contains 50 g dm–3? Solution Volume of solution = 200 cm3 200 Convert volume =— — — — dm3 = 0.2 dm3 from cm3 to dm3 1000 Concentration mass of Na2CO3 dissolved (g) =— — — — — — — — — — — — — — — — — — — — — — — — (g dm–3) volume of solution (dm3) Mass of Na2CO3 required = 50 g dm–3  0.2 dm3 Mass = Concentration (g dm–3)  Volume of solution (dm3) = 10 g

Concentration (g dm–3) Molarity (mol dm–3) = — — — — — — — — — — — — — — — — — — — — — Molar mass (g mol–1)

3 Calculate the number of moles of ammonia in 150 cm3 of 2 mol dm–3 aqueous ammonia. Solution M = molarity, V = volume in cm3 MV Number of moles = — — — — 1000 150 Number of moles of ammonia = 2  — — — — = 0.3 1000

The relationship between the number of mols with molarity, M and volume, V can be represented by the formula below: MV Number of moles = — — — — — — — — — — 1000 where M = molarity of solution (mol dm–3) V = volume of solution (cm3).

4 A 250 cm3 solution contains 0.4 mol of nitric acid. Calculate the molarity of the nitric acid.

Calculations Involving Concentrations and Molarity

Solution MV 250 Number of moles = — — — — =M— — — — 1000 1000

1

0.4  1000 Molarity of nitric acid, M = — — — — — — — — — 250 = 1.6 mol dm–3

5.00 g of copper(II) sulphate is dissolved in water to form 500 cm3 solution. Calculate the concentration of copper(II) sulphate solution in g dm–3. 197

Acids and Bases

5

Concentration (g dm–3) Molarity = — — — — — — — — — — — — — — — — — — Molar mass (g mol–1) 16 Convert g dm–3 =— — — to mol dm–3 106 –­3 = 0.15 mol dm

Calculate the volume in dm of 0.8 mol dm sulphuric acid that contains 0.2 mol of H2SO4. 3

–3

Solution MV Number of moles Number of moles = — — — — 1000 0.2 mol Volume of sulphuric acid = — — — — — — — — — — 0.8 mol dm–3 = 0.25 dm3 Molarity, M

8

7

The concentration of a potassium hydroxide solution is 84.0 g dm–3. Calculate the number of moles of potassium hydroxide present in 300 cm3 of the solution. [Relative atomic mass: H, l; O, 16; K, 39]

6

Solution Molar mass of KOH = 39 + 16 + 1 = 56

Dilute hydrochloric acid used in school laboratories usually has a concentration of 2.0 mol dm–3. Calculate the mass of hydrogen chloride in 250 cm3 of the hydrochloric acid? [Relative atomic mass: H, l; Cl, 35.5]

Molarity of KOH Molarity (mol dm–3) 84.0 Concentration (g dm–3) =— — — =— — — — — — — — — — — — — — — — — — — — 56 Molar mass (g mol–1) = 1.5 mol dm–3 Number of moles of KOH 1.5  300 =— — — — — — — — MV 1000 Number of moles = — — — — 1000 = 0.45

Solution Number of moles of HCl 2.0  250 MV =— — — — — — — — Number of moles = — — — — 1000 1000 = 0.5 Molar mass of HCl = 1 + 35.5 = 36.5 g mol–1 Mass of HCl = 0.5  36.5 g = 18.25 g Mass = Number of moles  Molar mass

9 Calculate the number of moles of hydrogen ions present in 200 cm3 of 0.5 mol dm–3 sulphuric acid. Solution Number of moles of H2SO4 0.5  200 =— — — — — — — — MV Number of moles = — — — –— 1000 1000 = 0.1

7 4.0 g of sodium carbonate powder, Na2CO3, is dissolved in water and made up to 250 cm3. What is the molarity of the sodium carbonate solution? [Relative atomic mass: C, 12; O, 16; Na, 23]

H2SO4 → 2H+ + SO42– Sulphuric acid is a diprotic acid, which means 1 mol of sulphuric acid will produce 2 mol of H+ ions. Hence, 0.1 mol of sulphuric acid will produce 0.1  2 = 0.2 mol of H+ ions.

Solution Volume of sodium carbonate solution = 250 cm3 = 0.25 dm3 Convert volume from cm3 to dm3

Mass (g) Concentration = — — — — — — — — — — — Volume (dm3) 4.0 Convert mass to =— — — concentration 0.25 –3 =16 g dm

Preparation of Standard Solutions 1 A standard solution is a solution with a known concentration. 2 A volumetric flask (also known as standard flask) is an apparatus with a known volume. Examples are: 100 cm3, 200 cm3, 250 cm3, 500 cm3 and 1000 cm3.

Molar mass of Na2CO3 = (2  23) + 12 + (3  16) = 106 g mol–1 Acids and Bases

198

3 Volumetric flasks are used to prepare standard solutions. Beakers are not suitable for this purpose because volumes measured by beakers and measuring cylinders are not very accurate. 4 A volumetric flask can measure the volume of a liquid accurately, up to one decimal point.

1 Calculate the mass (m g) of the chemical required to prepare v cm3 of solution where v is the volume of the volumetric flask. 2 Weigh out the exact mass (m g) of the chemical accurately in a weighing bottle using an electronic balance. 3 Dissolve m g of the chemical in a small amount of distilled water. 4 Transfer the dissolved chemical into the volumetric flask. 5 Add enough water until the graduation mark.

Figure 7.10 A 100 cm3 volumetric flask

To prepare 100 cm3 of 2.0 mol dm–3 aqueous sodium hydroxide solution

Materials Sodium hydroxide solid and distilled water. Procedure 1 The mass of sodium hydroxide (NaOH) required to prepare 100 cm3 of 2.0 mol dm–3 aqueous sodium hydroxide is calculated as follows: Mass of NaOH required = Number of moles  molar mass of NaOH MV = (— — — —)  (23 + 16 + 1) 1000 2.0  100 =— — — — — — — — 40 1000 = 8.0 g 2 8.0 g of sodium hydro­ xide, NaOH solid is weighed accurately in a weighing bottle using an electronic balance. 3 Sodium hydroxide solid is transferred to a small beaker. Sufficient dis­ tilled water is added to dissolve all the solid sodium hydroxide.

SPM

’06/P2 Q4

4 Using a filter funnel and a glass rod, the dissolved sodium hydroxide is transferred to a 100 cm3 volumetric flask. 5 The small beaker, the weighing bottle and the filter funnel are all rinsed with distilled water and the contents are transferred into the volumetric flask. 6 Distilled water is then added slowly until the water level is near the level mark of the volumetric flask. A dropper is then used to add water drop by drop to finally bring the volume of solution to the 100 cm3 graduation mark. 7 The volumetric flask is closed with a stopper. The volumetric flask is then shaken several times to mix the solution completely. The solution prepared is 100 cm3 of 2.0 mol dm–3 aqueous sodium hydroxide. 199

Activity 7.5

Apparatus Electronic balance, 100 cm3 volumetric flask, filter funnel, dropper and washing bottle.

7

The steps involved in the preparation of a standard solution

Acids and Bases

7

The Correct Techniques Used in the Preparation of Standard Solution

2 When a solution is diluted, the volume of solvent increases but the number of moles of solute remains constant. Hence the concentration of the solution decreases. 3 If a solution with volume of V1 cm3 and molarity of M1 mol dm–3 is diluted to become V2 cm3, the new concentration of the diluted solution, M2 mol dm–3 can be determined as follows:

1 The chemical is weighed in a weighing bottle and not on a piece of filter paper. Some chemicals such as sodium hydroxide can absorb moisture from the air and become wet and may stick to paper. 2 After transferring the dissolved solute to the volumetric flask, the weighing bottle, the small beaker that contained the solution as well as the filter funnel are rinsed with distilled water. The content is then transferred to the volumetric flask to ensure that all the mass of the chemical that has been weighed is transferred to the volumetric flask. 3 The addition of distilled water to the volumetric flask must be carried out carefully so that the level of the solution does not exceed the graduation mark of the volumetric flask. The last few cm3 of water should be added drop by drop using a dropper. 4 A volumetric flask and not a beaker must be used to prepare a standard solution because a volumetric flask is calibrated to a high degree of accuracy. 5 The volumetric flask is stoppered after the standard solution is prepared to prevent the evaporation of water which can change the concentration of the solution prepared.

Number of moles of M1V1 =— — — — — solute before dilution 1000 Number of moles of M2V2 = ——— —— solute after dilution 1000 However, the number of moles of solute before dilution is the same as the number of moles of solute after dilution, M2V2 M1V1 ————— = ——— —— or 1000 1000

M1V1 = M2V2

The steps involved in the preparation of a standard solution by dilution 1 The volume of the stock solution, V1 required is calculated. 2 The required volume of stock solution is pipetted into a volumetric flask. 3 Enough distilled water is added to the volumetric flask to the required volume, V2.

Preparation of a Solution with a Specified Concentration Using the Dilution Method

The formula used in dilution is M1V1 = M2V2 where M1 = M2 = V1 = V2 =

1 Dilution is a process of diluting a concentrated solution by adding a solvent such as water to obtain a more diluted solution.

Initial molarity of solution Final molarity of solution Initial volume of solution Final volume of solution

To prepare 100 cm3 0.2 mol dm–3 sodium hydroxide from a 2.0 mol dm–3 sodium hydroxide solution by the dilution method M1V1 = M2V2

Activity 7.6

Apparatus 100 cm3 volumetric flask, 10 cm3 pipette, pipette filler, filter funnel, dropper and washing bottle.

2.0 3 V1 = 0.2 3 100 0.2 3 100 V1 = — — — — — — — — = 10 cm3 1.0

Materials 2.0 mol dm–3 sodium hydroxide solution and distilled water.

where M1 = Initial molarity of alkali M2 = Final molarity of alkali V1 = Initial volume of alkali V2 = Final volume of alkali

Procedure (A) To calculate the volume of sodium hydroxide solution that is required for dilution Acids and Bases

200

3 The flask is stoppered and is inverted several times to mix the solution. The solution prepared is 0.2 mol dm–3 sodium hydroxide solution. Conclusion A 0.2 mol dm–3 sodium hydroxide solution can be prepared by diluting 10 cm3 of 2.0 mol dm–3 of sodium hydroxide solution to 100 cm3.

(a) The higher the degree of dissociation of an acid, the lower the pH value of the acid. (b) The higher the degree of dissociation of an alkali, the higher the pH value of the alkali. 3 For an acid or alkali, its pH value depends on the molarity of the solution. (a) The higher the molarity of an acid, the lower the pH value. (b) The higher the molarity of an alkali, the higher the pH value.

Relationship between pH Values and Molarities of Acids or Alkalis 1 The pH value of an acid or an alkali depends on two factors, that is (a) degree of dissociation and (b) molarity or concentration. 2 At the same concentration, the pH value of an acid or an alkali depends on the degree of dissociation.

7.3

7

(B) To prepare 100 cm3 0.2 mol dm–3 sodium hydroxide by the dilution method 1 Using a pipette and a pipette filler, 10.0 cm3 of 2.0 mol dm–3 sodium hydroxide solution is transferred to a 100 cm3 volumetric flask. 2 Using a washing bottle, distilled water is added to the alkali in the volumetric flask until near the graduation mark. A dropper is then used to add water slowly in the volumetric flask up to the graduation mark.

SPM

’09/P3

To investigate the relationship between pH values and the molarity of an acid or an alkali Problem statement What is the relationship between pH values and the molarity of an acid or an alkali?

0.001 mol dm–3 hydrochloric acid as shown in Figure 7.11. 3 The pH value shown on the pH meter is recorded. 4 The pH values of hydrochloric acid solutions with Figure 7.11 Using a pH different molarities meter to measure the pH are measured one value by one in dry beakers as in Steps 1 to 3. 5 The experiment is repeated using sodium hydroxide solutions with different molarities to replace the hydrochloric acid.

Hypothesis (A) When the molarity of an acid increases, its pH value decreases. (B) When the molarity of an alkali increases, its pH value increases. Variables (a) Manipulated variable : Molarity of acid or alkali (b) Responding variable : pH values (c) Constant variable : Type of acid or alkali used Apparatus pH meter, 100 cm3 beakers and 100 cm3 measuring cylinders. Materials Hydrochloric acids and sodium hydroxide solutions with molarities of 0.001 mol dm–3, 0.01 mol dm–3, 0.05 mol dm–3, 0.08 mol dm–3 and 0.10 mol dm–3.

Molarities of 0.001 0.01 HCl (mol dm–3) pH values

Procedure 1 30 cm3 of 0.001 mol dm–3 hydrochloric acid is put in a dry beaker. 2 The probe of a pH meter that has been washed with distilled water is immersed in 30 cm3 of the

3.0

2.0

0.05

0.08

0.10

1.3

1.1

1.0

Molarities of 0.001 0.01 0.05 0.08 0.10 NaOH (mol dm–3) pH values 201

11.0 12.0 12.7 12.9 13.0 Acids and Bases

Experiment 7.3

Results

7

4 The graph of pH values versus molarity of an alkali is an increasing curve as shown in Figure 7.13. 5 When the molarity of an alkali increases, the concentration of OH– ions in the alkali increases and the solution becomes more alkaline. Hence the pH value increases.

Discussion 1 The graph of pH values versus molarity of an acid is a decreasing curve as shown in Figure 7.12. 2 When the molarity of an acid increases, the concentration of H+ ions in the acid increases and the solution becomes more acidic. Hence the pH value decreases. 3 From the graph, we can predict (a) the pH value, if the concentration of H+ ions of the solution is known. (b) the concentration of H+ ions in the solution, if the pH value is known.

Figure 7.13 Graph of pH versus molarity of NaOH

Conclusion 1 The higher the molarity of hydrochloric acid, the lower the pH value. The pH value of an acid decreases with the increase in molarity. 2 The higher the molarity of an alkali, the higher the pH value. The pH value of an alkali increases with the increase in molarity. The hypothesis is accepted.

Figure 7.12 Graph of pH versus molarity of HCl

0.8  250 M2 = — — — — — — — = 0.2 mol dm–3 1000 Molarity of potassium hydroxide solution produced is 0.2 mol dm–3.

Numerical Problems Involving Molarity of Acids and Alkalis 1 The molarity of an acid will change when (a) water is added to it, (b) an acid of a different concentration is added to it, (c) an alkali is added to it.

11 What is the volume of distilled water required to be added to 60 cm3 of 2.0 mol dm–3 sulphuric acid to produce a 0.3 mol dm–3 sulphuric acid?

10 What is the molarity of the potassium hydroxide solution produced when 750 cm3 of distilled water is added to 250 cm3 of potassium hydroxide solution of 0.8 mol dm–3?

Solution M1V1 = M2V2 2.0  60 = 0.3  V2 2.0  60 V2 = — — — — — — — — 0.3 3 = 400 cm

Solution Final volume of alkali, V2 = 250 cm3 + 750 cm3 = 1000 cm3 M1V1 = M2V2 0.8  250 = M2  1000

Acids and Bases

Calculate the total volume of acid produced

Volume of distilled water needed to be added to 60 cm3 H2SO4 = (400 – 60) cm3 = 340 cm3. 202

12 500 cm3 of a solution that contains 2.0 mol sodium hydroxide is added to 1500 cm3 of a solution that contains 4.0 mol sodium hydroxide. Calculate the molarity of the sodium hydroxide solution produced.

Two units for concentration

mol dm–3

 molar mass

determined by

determined by

Mass of solute (g) — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — Volume of solution (dm3)

Total volume of NaOH = (500 + 1500) cm3 = 2000 cm3 Calculate the total volume of alkali from the two solutions = 2 dm3

o. of moles of solute (mol) N — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — Volume of solution (dm3)

7.3

Molarity of Number of moles =— — — — — — — — — — — — — — NaOH produced Volume (dm3) 6.0 mol =— — — — — — 2 dm3 = 3.0 mol dm–3

1 Calculate the mass of potassium hydroxide required to produce (a) 2.0 dm3 of 46.4 g dm–3 solution (b) 100 cm3 of 2.0 mol dm–3 solution [Relative atomic mass: H, 1; O, 16; K, 39] 2 2.12 g of sodium carbonate is dissolved in 500 cm3 of distilled water. What is the concentration of the solution in (a) g dm–3? (b) mol dm–3? [Relative atomic mass: C, 12; O, 16; Na, 23]

13 200 cm3 of 2.0 mol dm–3 hydrochloric acid is added to 300 cm3 of 0.5 mol dm–3 hydrochloric acid. Calculate the molarity of the hydrochloric acid produced.

3 The concentration of sodium hydroxide solution is 8.0 g dm–3. (a) What is the molarity of the solution? (b) What is the molarity of the solution produced when 100 cm3 of distilled water is added to 100 cm3 of this solution? (c) What is the molarity of the solution produced when 20 cm3 of 2.0 mol dm–3 sodium hydroxide solution is added to 100 cm3 of this solution? [Relative atomic mass: H, 1; O, 16; Na, 23]

Solution Number of moles in 200 cm3 of 2.0 mol dm–3 HCl MV 2.0  200 =— — — —= — — — — — — — — 1000 1000 = 0.4 Calculate the total Number of moles in 300 cm3 of 0.5 mol dm–3 HCl MV 0.5  300 =— — — —= — — — — — — — — 1000 1000 = 0.15

 molar mass

g dm–3

number of moles of acid from the two solutions.

7.4

Total number of moles of HCl = 0.4 + 0.15 = 0.55

Neutralisation

The Meaning of Neutralisation and the Equation for Neutralisation

Total volume of solution = (200 + 300) cm3 = 500 cm3 Calculate the total volume of = 0.5 dm3 acid from the two solutions.

1 Neutralisation is the reaction between an acid and a base to produce salt and water only. 2 In a neutralisation reaction, the acidity of an acid is neutralised by an alkali. At the same time, the alkalinity of the alkali is neutralised by the acid. Salt and water are the only products of neutralisation.

Molarity of Number of moles of HCl =— — — — — — — — — — — — — — — — — — — — HCl produced Volume of HCl 0.55 mol =— — — — — — — — = 1.1 mol dm–3 0.5 dm3 203

Acids and Bases

7

Solution Total number of moles of NaOH = 2.0 + 4.0 Calculate the total number of moles = 6.0 of alkali from the two solutions.

3 Some examples of neutralisation reactions are as follows: acid

base

salt

The ionic equation for neutralisation between strong acids and strong alkalis is

water

H+(aq) + OH–(aq) → H2O(l)

HCl + NaOH → NaCl + H2O H2SO4 + CuO → CuSO4 + H2O 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O

7 In a neutralisation reaction, H+ ions from the acid react with OH– ions from the base to produce water. The pH value for water is 7, and a neutral condition is achieved.

4 In the reaction between ammonia and acid such as hydrochloric acid, ammonium salt is formed.

5

7

NH3 + HCl → NH4Cl

Which of the following pairs of compounds will react in a neutralisation reaction? I Hydrochloric acid and potassium hydroxide II Sulphuric acid and solid copper(II) oxide III Nitric acid and solid calcium carbonate IV Hydrochloric acid and zinc metal A I and II only B III and IV only C I, III and IV only D I, II, III and IV

Although there is no water formed in the above equation, in actual fact there is a little water formed together with NH4Cl. This is because a portion of ammonia, NH3 exists as NH4+ ions and OH– ions in aqueous solution. OH– ions and H+ ions (from acid) react to produce water, H2O. 5 Acids, bases and salts dissociate to form free ions. Only water exists as molecules. For example,



Comments Neutralisation is a reaction between an acid and a base to form salts and water only. Reactions in I and II are neutralisation because the two reactants react to form salts and water only. Reaction III is not neutrali­­ sa­ tion because carbon dioxide gas is formed in addition to salt and water. Reaction IV is not neutrali­sa­tion because hydrogen gas is formed in addition to salt. Answer A

H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → from HCl

from NaOH

Na+(aq) + Cl–(aq) + H2O(l)



from NaCl

’03

water molecule

6 Since Na+ ions and Cl– ions do not undergo any changes in the reaction, these ions can be omitted from the equation. Thus, the equation can be simplified.

Applications of Neutralisation in Daily Life Neutralisation is used in various fields such as agriculture, health and industries. In agriculture 1 Controlling the acidity of water is important in the rearing of freshwater fish and prawns. Lime (consisting of calcium oxide, CaO) which produces calcium hydroxide in water is used to control the acidity in aqua farming. 2 Plants do not grow well in acidic soil or basic soil. Lime or calcium carbonate is used to neutralise acidic soil. Basic soil is treated with compost which can release acidic gas to neutralise the alkali in basic soil.

Acids and Bases

SPM

’09/P1, ’10/P1

Calcium oxide and calcium hydroxide are used to neutralise acidic soil

204

SPM

In health

produced. This will prevent the corrosion of teeth enamel. 2 Antacids are medicine which contain bases such as magnesium hydroxide, aluminium hydroxide, calcium carbonate and calcium bicarbonate. Magnesia milk contains magnesium hydroxide. Antacids and milk of magnesia are used to neutralise the excess hydrochloric acid in the stomachs of gastric patients. 3 Alkaline creams or baking powder are applied to cure bee stings and ant bites which are acidic. Vinegar which contains ethanoic acid is used to cure alkaline wasp stings.

Toothpastes contain magnesium hydroxide to neutralise acids in teeth

1 Food trapped in gaps between teeth decompose into organic acids by bacteria. An alkaline compound such as magnesium hydroxide in toothpastes neutralises the organic acids

In industries 1 Bacteria in latex produces organic acids which coagulate latex. Ammonia is used to neutralise the organic acids produced by bacteria to prevent coagulation, so that latex can remain in the liquid state. 2 Calcium carbonate is used as a base to remove acidic gas such as sulphur dioxide emitted by power stations and industries. 3 Effluent from factories which is acidic is treated with lime, which will neutralise the acids in it before being discharged. 4 Neutralisation reaction is also used in the industry to produce manufactured products such as fertilisers, soaps and detergents.

Acidic gas emitted by factories must be removed before being discharged

(a) The use of acid–base indicators such as methyl orange, phenolphthalein and litmus paper which changes colour at the end point. (b) Measurement of the pH values during titration using a computer interface. 5 Titration technique can be used to determine the concentration of an acid (or an alkali).

Acid-base Titration 1 Titration is a quantitative analysis that involves the gradual addition of a chemical solution from a burette to another chemical solution of known quantity in a conical flask. 2 In acid-base titration, the volume of the alkali is measured using a pipette and transferred into a conical flask. The acid solution from a burette is then added slowly to the alkali in the conical flask until neutralisation occurs. 3 The end point of a titration is when neutralisation occurs, that is, when the acid has completely neutralised the alkali. 4 Since both the reactants (acid and alkali) and the products formed (salt and water) are all colourless, the end point of neutralisation is determined as follows:

The Use of Acid-base Indicators in Titrations 1 Acid–base indicators are chemicals that show different colours when the pH value of the solution changes. 2 Table 7.10 shows the colour of three common types of indicators at different pH values. 205

Acids and Bases

7

’08/P1

3 The pH meter records changes in pH values during titration. The information is linked to a computer by the interface. The change of pH values along the progress of titration is displayed on the computer screen.

Table 7.10 Colours of three common types of indicators in alkaline, neutral and acidic conditions

Indicator

Colour in alkali

Methyl orange

Yellow

Colour in Colour in neutral acid solution Orange

Red

Phenolphthalein Light pink Colourless Colourless

7

Litmus

Blue

Purple

Red

3 Methyl orange shows a yellow colour in an alkaline solution. At the point of neutralisation, the colour changes from yellow to orange. 4 Phenolphthalein shows a light pink in an alkaline solution. The first drop of acid that decolourises the light pink colour of phenolphthalein indicates the end point of titration.

In computer interface, the pH is displayed on a screen

4 A graph of pH value versus volume of alkali added is shown as in Figure 7.14. It is found that the pH value of the solution changes sharply at the end point of neutralisation. The neutralisation point can be determined from the midpoint of the sharp pH change.

The Use of Computer Interface in Titration 1 When an alkaline solution is added slowly from a burette to an acid in a conical flask, the pH value of the mixture solution increases slowly. 2 In titrations using a computer interface, the probe of a pH meter, immersed in the solution to be titrated, is connected to the pH module of the computer interface.

Figure 7.14 Graph of pH value versus volume of alkali added

To find the end point of an acid-base titration during neutralisation using an acid-base indicator Apparatus 25 cm3 pipette, pipette filler, 50 cm3 burette, retort stand, retort clamp, conical flask, filter funnel and white tile.

Activity 7.7

Materials Sulphuric acid of unknown concentration, 1.0 mol dm–3 potassium hydroxide and methyl orange. Procedure 1 A clean 25 cm3 pipette is rinsed with distilled water and then rinsed with a little of the potassium hydroxide solution. 2 25 cm3 of 1.0 mol dm–3 potassium hydroxide is transferred using the pipette to a clean conical flask. Three drops of methyl orange indicator are added to the alkali and the colour of the solution is noted. Acids and Bases

SPM

’10/P2

3 A 50 cm3 burette is rinsed with distilled water and then rinsed with a little of the sulphuric acid. 4 The burette is then filled with sulphuric acid and is clamped to a retort stand. The initial burette reading is recorded. 5 The conical flask containing 25 cm3 of potassium hydroxide is placed below the burette. A piece of white tile is placed below the conical flask for clearer observation of the colour change (Figure 7.15). 6 Sulphuric acid is added slowly from the burette to the potassium hydroxide solution in the conical flask while swirling the flask gently. 7 Titration is stopped when the methyl orange changes colour from yellow to orange. The final burette reading is recorded. 206

Figure 7.15 Titration of sulphuric acid with potassium hydroxide

Results Volume of sulphuric acid

Rough

Accurate

Final burette reading (cm3)

21.00

40.95 20.15

Initial burette reading (cm )

0.00

21.00 0.10

Volume of sulphuric acid used (cm3)

21.00

19.95 20.05

3

Conclusion 1 The volume of sulphuric acid used is calculated as follows: Volume of sulphuric acid used = Final burette reading – Initial burette reading 2 Average volume of sulphuric acid used

19.95 + 20.05 =— — — — — — — — — — — = 20.00 cm3 2 Hence, 20.00 cm3 of H2SO4 is required to completely neutralise 25.0 cm3 of 1.0 mol dm–3 KOH. Discussion 1 In this experiment, the pipette has to be rinsed with potassium hydroxide solution so that water droplets on the inner wall of the pipette do not dilute the concentration of the potassium hydroxide used. 2 The burette is rinsed with sulphuric acid so that water droplets at the inner wall of the burette do not dilute the concentration of the sulphuric acid used. 3 The conical flask does not need to be rinsed with potassium hydroxide so that the volume of the potassium hydroxide in the conical flasks will accurately be 25.0 cm3. Otherwise, droplets of potassium hydroxide in the conical flask may cause the volume of potassium hydroxide to exceed 25.0 cm3. 4 The end point of titration is when the colour of the indicator changes sharply. The colour of methyl orange is yellow in potassium hydroxide solution (because pH > 7). At the end point, the colour of methyl orange changes to orange (pH = 7). If methyl orange changes to a red colour, excess sulphuric acid has been added. 5 In acid-base titrations, only 2 or 3 drops of indicator should be used. This is because most of the indicators are weak acid or base that will affect the pH of the solution if used in excess.

7

8 Steps 1 to 7 are repeated until accurate titration values are obtained, that is, until the difference in the volumes of sulphuric acid used in two consecutive experiments is less than 0.10 cm3.

Apparatus 25 cm3 pipette, pipette filler, 50 cm3 burette, retort stand, retort clamp, 250 cm3 beaker, filter funnel, magnetic stirrer, magnetic stirrer bar, pH meter, computer interface and computer. Materials 1.0 mol dm–3 hydrochloric acid and 1.0 mol dm–3 sodium hydroxide. Procedure 1 A 25 cm3 pipette is rinsed with distilled water and then with a little of the hydrochloric acid. 2 25 cm3 of 1.0 mol dm–3 hydrochloric acid is transferred using the pipette to a clean beaker.

3 A 50 cm3 burette is rinsed with distilled water and then with a little of the sodium hydroxide solution. 4 The burette is then filled with sodium hydroxide solution and is clamped to a retort stand. 5 A magnetic stirrer bar is placed in the beaker containing 25 cm3 of hydrochloric acid. The beaker is then placed on a magnetic stirrer below the burette. 6 A pH meter is connected to a computer using a computer interface. The pH meter probe is then dipped into the acid. The magnetic stirrer is switched on and the computer is set to record and display the pH (Figure 7.16). 7 Sodium hydroxide solution is added drop by drop from the burette at a constant rate, into the acid in the beaker. 207

Acids and Bases

Activity 7.8

To find the end point of acid–base titration during neutralisation using a computer interface

7

8 A graph of pH change against the volume of sodium hydroxide in cm3 is printed using the computer printer when 50 cm3 of sodium hydroxide is added to the beaker.

Figure 7.16 Using a pH meter and a computer interface to measure pH changes during neutralisation

Results 1 A graph of pH against the volume of sodium hydroxide in cm3 as displayed by the computer is shown in Figure 7.17.

Conclusion 1 The pH value of hydrochloric acid is 1.0 at the beginning of titration. As sodium hydroxide is added to the acid, the pH value of the solution increases. 2 The pH value increases sharply at the end point of neutralisation. The midpoint of the sharp pH change is 7. At pH 7, the volume of sodium hydroxide used from the graph is 25.0 cm3. 3 When 1.0 mol dm–3 sodium hydroxide is titrated against 25.0 cm3 of 1.0 mol dm–3 hydrochloric acid, the end point of titration during neutralisation occurs at pH 7 when 25 cm3 of 1.0 mol dm–3 of sodium hydroxide solution has been added. 4 By using a computer interface to measure pH changes, the end point of titration during nuetralisation can be determined accurately.

Figure 7.17 Graph of pH against the volume of sodium hydroxide in cm3

6

’95

Describe an experiment to determine the concentration of a solution of sulphuric acid by titrating it with a 1.0 mol dm–3 potassium hydroxide solution. Solution A titration experiment similar to Activity 7.7 is carried out in which sulphuric acid of unknown concentration is titrated against 25.0 cm3 of 1.0 mol dm–3 potassium hydroxide using methyl orange (or phenolphthalein) as an indicator. Let's say the volume of sulphuric acid required to neutralise completely 25.0 cm3 of 1.0 mol dm–3 potassium hydroxide is V cm3. Number of moles of KOH in 25.0 cm3 of MV 1.0 3 25.0 1.0 mol dm–3 solution = ——— = ——————— = 0.025 1000 1000 Acids and Bases

The equation for the neutralisation reaction between potassium hydroxide and sulphuric acid is 2KOH + H2SO4 → K2SO4 + 2H2O According to the equation, 2 mol of KOH requires 1 mol of H2SO4 for complete neutralisation. 0.025 mol KOH will require 1 0.025 3 — = 0.0125 mol H2SO4 2 MV Number of moles of H2SO4 = ——— 1000 Molarity of sulphuric acid, M = 0.0125 3 1000/V 12.5 = ——— mol dm–3 V

208

Calculation Involving Neutralisation Using Balanced Equations

14

SPM

’09/P2

SPM

’11/P1

In an experiment, 25.0 cm3 of a sodium hydroxide solution of unknown concentration required 26.50 cm3 of 1.0 mol dm–3 sulphuric acid for complete reaction in titration. Calculate the molarity of sodium hydroxide.

1 Say, in a balanced equation, a mol of acid reacts with b mol of base as represented by the equation below: aA + bB → products

Solution b=2

MBVB 2 — — — —= — MAVA 1

MB  25.0 2 — — — — — — — — — = — 1.0  26.50 1

MAVA Number of moles of acid = — — — — 1000

where MB = molarity of NaOH MA = molarity of H2SO4 VB = volume of NaOH VA = volume of H2SO4

26.50 — — — — MB = 2  — 25.0 = 2.12 mol dm–3

MBVB Number of moles of base = — — — — 1000

Hence, the molarity of sodium hydroxide solution is 2.12 mol dm–3.

In the stoichiometry equation, a mol of acid HA reacts completely with b mol of base, M(OH)X. Hence the mole ratio of acid to base is MAVA — — — — 1000 a — — — —= — or MBVB b — — — — 1000

a=1

SPM

15

’10/P1, ’11/P2

What is the volume of 1.5 mol dm–3 aqueous ammonia required to completely neutralise 30.0 cm3 of 0.5 mol dm–3 sulphuric acid?

MAVA a — — — —— = — MBVB b

Solution

3 From the above relationship, the ratio of a and b can be obtained from the balanced equation. Any one of the four variables: MA, VA, MB, VB, can be determined if three of the other variables are known.

b=2

2NH3 + H2SO4 → (NH4)2SO4 MBVB 2 — — — —= — MAVA 1

a=1

1.5  VB 2 — — — — — — — —= — 0.5  30.0 1 1 The volume of a solution in a burette is read from the top to the bottom. 2 The accuracy of a burette reading is until 2 decimal places. 3 The accuracy of a pipette reading is until 1 decimal place.

where MB = molarity of NH3 MA = molarity of H2SO4 VB = volume of NH3 VA = volume of H2SO4

2  30.0  0.5 VB = — — — — — — — — — — — — = 20 cm3 1.5 Hence, the volume of aqueous ammonia required is 20 cm3.

209

Acids and Bases

7

2NaOH + H2SO4 → Na2SO4 + 2H2O

2 Say, the molarity of an acid is MA mol dm and the molarity of a base is MB mol dm–3. If in a titration, VA cm3 of acid neutralises VB cm3 of base –3

16 Calculate the volume (cm3) of 2.0 mol dm–3 hydrochloric acid that is required to react completely with 2.65 g of sodium carbonate. [Relative atomic mass: C, 12; O, 16; Na, 23]

• In an aqueous solution, the concentration of H+ ions in CH3COOH is lower than HCl of the same concentration. This is because CH3COOH dissociates partially in water. • However, both CH3COOH and HCl of the same concentration require the same amount of NaOH for complete neutralisation. This is because both are monoprotic (monobasic) acids. In the presence of NaOH, CH3COOH dissociates completely to react with NaOH. • In an acid-base titration, the preferred chemical to be put in the burette is the acid. This is because alkali can react with silica in glass to form silicate, thus dissolving the thin glass wall of the burette slowly.

7

Solution Molar mass of Na2CO3 = (23  2) + 12 + (16  3) = 106 g mass Number of moles of Na2CO3 = — — — — — — — — — — molar mass 2.65 =— — — — = 0.025 106 Na2CO3 + 2HCl → 2NaCl + CO2 + H2O From the equation, 1 mol of Na2CO3 reacts with 2 mol of HCl. Hence, 0.025 mol of Na2CO3 will react with 0.025  2 = 0.05 mol of HCl. where MV Number of moles =— — — — 1000 M = molarity of HCl V = volume of HCl 2.0 V 0.05 = — — — — — — — 1000 0.05  1000 V=— — — — — — — — — — = 25 cm3 2.0 Hence, the volume of hydrochloric acid required is 25 cm3.

7.4 1 Complete the blanks in the equations below: (a) 2HCl + Mg → (b) H2SO4 + Zn(OH)2 →

+

(c)

+ NaOH → CH3COONa + H2O

(d)

+

→ CaSO4 + 2H2O

2 Nitric acid reacts with magnesium hydroxide solution to produce magnesium nitrate, Mg(NO3)2 and water. (a) Write a balanced equation for the reaction between nitric acid and magnesium hydroxide. (b) 10.0 cm3 of nitric acid is required to neutralise 0.001 mol of magnesium hydroxide. Calculate the concentration of the nitric acid in mol dm–3.

17 15 cm3 of an acid with the formula HaX of 0.1 mol dm–3 required 30 cm3 of 0.15 mol dm–3 sodium hydroxide solution for complete neutralisation. Calculate the value of a and hence determine the basicity of the acid.

3 Sodium carbonate reacts with hydrochloric acid as represented by the equation below:

Solution HaX + aNaOH → aH2O + NaaX

Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

MAVA 1 — — — — —= — MBVB a

Calculate the volume (cm3) of 1.25 mol dm–3 hydrochloric acid that is required to react completely with 25.0 cm3 of 1.0 mol dm–3 sodium carbonate.

0.1 3 15 1 — — — — — — — —= — 0.15 3 30 a

4 15.0 cm3 of sulphuric acid neutralises 25.0 cm3 of 2.0 mol dm–3 aqueous ammonia. Calculate the molarity of the sulphuric acid used.

a=3 Hence HaX is a tribasic acid.

Acids and Bases

+

210

Acids

Alkalis

Acids are sour in taste

Alkalis are bitter in taste and feel soapy

pH values of less than 7

pH values of more than 7

Changes blue litmus paper to red

Changes red litmus paper to blue

7 Chemical properties of an acid: (a) Reacts with a base to produce a salt and water. (b) Reacts with a reactive metal to produce a salt and hydrogen gas is evolved.



MB = molarity of alkali VB = volume of alkali

7 Multiple-choice Questions 7.1

Characteristics and Properties of Acids and Bases

1 Chemical X is an acid. Which of the following may not be true about the property of X? A A pink colour is produced when phenolphthalein is added to a solution of X. B Hydrogen gas is evolved when zinc powder is added to a solution of X. C A solution of X reacts with an alkali to produce salt and water.

D X dissolves in water to produce H+ ions. 2 Which of the following statements is true of all bases? ’07 A Dissolve in water B Contain hydroxide ions C Have a pH value of between 12 and 13 D Produce ammonia gas when heated with ammonium salts 3 The diagram shows the set-up of apparatus for the reaction ’06 between solution Z and magnesium ribbon.

211

Which of the following is solution Z? A Glacial ethanoic acid B Ethanoic acid in methylbenzene C Concentrated ethanoic acid D Hydrogen chloride dissolved in propanone Acids and Bases

7

(c) Reacts with a metal carbonate to produce a salt, carbon dioxide gas and water. 8 Chemical properties of an alkali: (a) Reacts with an acid to produce a salt and water. (b) When heated with an ammonium salt, ammonia gas is produced. 9 The pH scale is a set of numbers range from (0 to 14) used to measure acidity or alkalinity of a substance. 10 The pH value of an acid or alkali depends on (a) the degree of dissociation (strength of acid or alkali), (b) the concentration of the acid or alkali. 11 A standard solution is a solution of known concentration. 12 The concentration of a solution is measured in g dm–3 or mol dm–3. 13 Neutralisation is the reaction between an acid and a base to produce a salt and water only. 14 When a mol of acid reacts completely with b mol of alkali in a reaction: MAVA a M = molarity of acid ———– = —— , where A b VA = volume of acid MBVB

1 An acid is a chemical compound that produces ­ hydrogen ions, H+ or hydroxonium ions, H3O+ when it dissolves in water. 2 Dry acids without water do not show any acidic property because they do not dissociate to H+ ions. 3 A base is defined as a chemical substance that can neutralise an acid to produce a salt and water. Most bases are not soluble in water. Bases that are soluble in water are known as alkalis. 4 An alkali is defined as a chemical compound that dissolves in water to produce free moving hydroxide ions, OH–. 5 Dry alkalis do not show alkaline property. 6 Physical properties of acids and alkalis:

4 The colours of indicator X in solutions of different pH values are shown below.

pH value 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ←⎯→ ←⎯→ ←⎯⎯→ ←⎯→ ←⎯⎯⎯⎯⎯→ ←⎯→ ←⎯→ Colour red orange yellow green bluish-green blue purple

7

Indicator X will show a green colour in a solution of 0.1 mol dm–3 of A sodium hydroxide solution B sodium chloride solution C aqueous ammonia D ethanoic acid 5 Which of the following substances can change red litmus paper to blue when dissolved in water? A Carbon dioxide gas B Glacial ethanoic acid C Solid sodium oxide D Solid sodium sulphate 6 The acidity of hydrogen chloride gas cannot be shown when it dissolves in the following solvents: I Water II Ethanol III Methylbenzene IV Propanone A I and II only B III and IV only C I and III only D II, III and IV only 7 Zinc carbonate powder is added to liquid X. A gas which turns limewater milky is evolved. X could be A glacial ethanoic acid. B aqueous citric acid. C hydrogen chloride gas dissolved in methylbenzene. D hydrogen chloride gas dissolved in propanone. 8 The table shows the pH for four aqueous solutions, W, X, Y and Z. Aqueous solutions

W

X

Y

Z

pH

2

7

12

13

When two of the solutions with the same volume are added together, which mixture will react Acids and Bases

with magnesium powder to liberate hydrogen gas? A W and X B X and Y C X and Z D Y and Z 9 Sulphuric acid is known as a diprotic acid because A there are two hydrogen atoms in one molecule of sulphuric acid. B one mole of sulphuric acid contains two moles of hydrogen atoms. C one mole of sulphuric acid dissociates into two moles of hydrogen ions in water. D sulphuric acid can be neutralised by two types of bases. 10 Which of the following particles in a solution of ammonia is responsible for its alkaline properties? A NH3 B OH– C H+ D NH4+

7.2

The Strength of Acids and Alkalis

11 A chemical that dissociates completely in water to produce hydroxide ions is a A strong acid B weak acid C strong alkali D weak alkali 12 The table shows the degree of dissociation of four solutions of acids which have the same concentration Solution

Degree of dissociation

W

High

X

Medium

Y

Very high

Z

Low

Which solution has the highest pH value? A W C Y B X D Z

212

13 Ethanoic acid is a weak acid because A it is an organic acid. B it dissolves slightly in water. C it is a weak conductor of electricity. D it ionises partially to form hydrogen ions in water. 14 Which of the following solutions has the highest pH value? A Ethanoic acid, 0.01 mol dm–3 B Hydrochloric acid, 0.01 mol dm–3 C Aqueous ammonia, 0.01 mol dm–3 D Sodium hydroxide solution, 0.01 mol dm–3 15

Which of the following statements is true of the two aqueous solutions shown above? A Both solutions are strong acids. B The pH of both solutions are equal. C Both solutions are strong electrolytes. D 25.0 cm3 of each solution requires 25.0 cm3 of 1.0 mol dm–3 sodium hydroxide to be neutralised. 16 Which of the following solutions contains the highest number of ’11 hydrogen ions? A 50 cm3 of 2 mol dm–3 ethanoic acid B 40 cm3 of 1 mol dm–3 sulphuric acid C 30 cm3 of 2 mol dm–3 nitric acid D 50 cm3 of 1 mol dm–3 hydrochloric acid 17 Which of the following is true when water is added to an aqueous sodium hydroxide solution?

18 An aqueous solution of Q has a pH value of 1. Q may be I 0.1 mol dm–3 hydrochloric acid II 0.001 mol dm–3 sulphuric acid III 0.1 mol dm–3 nitric acid IV 0.1 mol dm–3 ethanoic acid A I and III only B II and IV only C I, II and III only D I, III and IV only 19 Which of the following statements is true of both nitric acid and sulphuric acid? A Both are organic acids. B Both are diprotic acids. C Both undergo complete dissociation in water. D Both react with copper to produce hydrogen gas.

7.3

Concentration of Acids and Alkalis

20 Steps I to V below show the five steps that are involved in the preparation of a standard solution of sodium hydroxide, NaOH which may not be arranged in correct order. I Add distilled water until the graduation mark II Weigh the mass of sodium hydroxide III Transfer the solid sodium hydroxide into the volumetric flask IV Rinse the weighing bottle and pour the solution into the volumetric flask V Shake the volumetric flask Which of the following is the correct order of steps in the preparation? A I, II, III, IV, V B II, III, I, IV, V C II, III, IV, I, V D II, I, III, IV, V

21 Calculate the mass of potassium hydroxide that is required to prepare 250 cm3 of 2.0 mol dm–3 solution. [Relative atomic mass: H, 1; O, 16; K, 39] A 22.4 g C 56 g B 28 g D 112 g 22 0.2 mol dm–3 sulphuric acid is added slowly to a conical flask containing 20.0 cm3 of 0.2 mol dm–3 potassium hydroxide and 10 cm3 of water. What is the total volume (in cm3) of the solution in the flask when the solution is completely neutralised? A 30 C 50 B 40 D 60 23 The table shows four different test tubes P, Q, R and S containing different acids. Test tube

Content

P

15 cm of 0.5 mol dm–3 hydrochloric acid

Q

15 cm3 of 1.0 mol dm–3 ethanoic acid

R

10 cm3 of 1.0 mol dm–3 nitric acid

S

10 cm3 of 0.5 mol dm–3 sulphuric acid

3

Which test tube will produce the biggest volume of hydrogen gas with excess magnesium? A P C R B Q D S 24 Which of the following sodium hydroxide solutions have a concentration of 0.5 mol dm–3? [Relative atomic mass: H, 1; O, 16; Na, 23] I 5 g NaOH in 250 cm3 of water II 20 g NaOH in 1 dm3 of water III 250 cm3 of 2 mol dm–3 NaOH to which distilled water is added until it becomes 1 dm3 IV 1 mol dm–3 NaOH diluted to twice its volume A I and III only B II and III only C III and IV only D I, II, III and IV

213

25 When 2.8 g of potassium hydroxide is dissolved in 250 cm3 of distilled water, which of the following are true about the solution produced? [Relative atomic mass: H, 1; O, 16; K, 39] I It has a molarity of 0.05 mol dm–3. II It contains 11.2 g in 1 dm3. III The solution produces 0.2 mol of hydroxide ions. IV It contains 2 mol in 10 dm3 of solution. A I and II only B I and III only C III and IV only D II and IV only 26 What is the mass of sodium hydroxide contained in 50 cm3 of 0.4 mol dm–3 sodium hydroxide solution? [Relative atomic mass: H, 1; O, 16; Na, 23] A 0.4 g B 0.8 g C 1.6 g D 3.2 g 27 Calculate the number of moles of hydroxide ions in 2 dm3 of calcium hydroxide solution with a concentration of 14.8 g dm–3. [Relative atomic mass: H, 1; O, 16; Ca, 40] A 0.20 B 0.26 C 0.40 D 0.44 28 Which of the following substances will react with glacial ethanoic acid? A Zinc metal B Ammonia gas C Potassium hydroxide solid D Aqueous sodium carbonate solution

7.4

Neutralisation

29 The ionic equation for the reaction between nitric acid and sodium hydroxide is represented by A 2H2 + O2 → 2H2O B H+ + OH– → H2O C 2H+ + O2– → H2O D Na+ + NO3–→ NaNO3 Acids and Bases

7

A The pH value decreases. B The degree of ionisation decreases. C The hydroxide ion concentration increases. D The alkalinity increases.

7

30 Plants do not grow well in acidic soil. Which of the following are used to neutralise acidic soil? I Sodium hydroxide II Calcium hydroxide III Potassium hydroxide IV Calcium oxide A I and III only B II and IV only C I, II and III only D I, II, III and IV 31 Antacid is used to neutralise excess acid in the stomach. Which of the following chemicals is found in antacid? A Sodium hydroxide B Potassium hydroxide C Magnesium hydroxide D Ammonia 32 20 cm3 of 0.5 mol dm–3 sulphuric acid is added to 20 cm3 of 0.5 mol dm–3 sodium hydroxide. Which of the following statements are true? I Neutralisation reaction takes place. II The solution produced is acidic. III 0.02 mol of water is produced. IV The solution contains sodium sulphate and water only. A I and II only B III and IV only C II and III only D I, III and IV only 33 The equation shows the reaction between calcium carbonate and hydrochloric acid. CaCO3 + 2HCl → CaCl2 + CO2 + H2O What is the mass of calcium carbonate required to react completely with 10 cm3 of 2.0 mol dm–3 hydrochloric acid? [Relative atomic mass: C, 12; O, 16; Ca, 40] A 0.5 g B 1.0 g C 2.0 g D 4.0 g

Acids and Bases

34

Fe + 2HCl → FeCl2 + H2 Based on the equation above, calculate the mass of iron that will react with excess hydrochloric acid to produce 60 cm3 of hydrogen gas at room temperature. [Relative atomic mass: Fe, 56; 1 mol of gas occupies 24 dm3 at room temperature] A 0.14 g B 0.28 g C 3.36 g D 22.4 g

35

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) The equation above shows the neutralisation reaction between sodium hydroxide and sulphuric acid. Calculate the number of moles of sodium hydroxide that is required to neutralise 25 cm3 of 2.0 mol dm–3 sulphuric acid. A 0.05 mol B 0.10 mol C 0.50 mol D 1.00 mol

36 10.0 cm3 of a certain 0.5 mol dm–3 acid requires 50.0 cm3 of 0.3 mol dm–3 sodium hydroxide solution for complete neutralisation. Which of the following is the possible molecular formula for this acid? A HNO3 B H2SO4 C H3PO4 D CH3COOH 37 Which of the following reactions represent neutralisation? I CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) II H+(aq) + OH–(aq) → H2O(l) III Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)

214

IV CuO(s) + H2(g) → Cu(s) + H2O(l) A I and III only B II and III only C I, II and III only D II, III and IV only 38 The reaction between dilute hydrochloric acid and calcium carbonate is represented by the equation as follows. CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) What is the minimum volume of 2 mol dm–3 hydrochloric acid that is required to react completely with 5 g of calcium carbonate? [Relative atomic mass: C, 12; O, 16; Ca, 40] A 5 cm3 B 25 cm3 C 50 cm3 D 100 cm3 39 In a titration process, 0.1 mol dm–3 sulphuric acid from a ’03 burette is added slowly to 20 cm3 of 0.1 mol dm–3 aqueous ammonia solution in a conical flask with methyl orange indicator until neutralisation occurs. What is the total volume in the conical flask at the end point of titration? A 10 cm3 B 20 cm3 C 30 cm3 D 40 cm3 40 The equation below represents the neutralisation reaction of ’10 aqueous W hydroxide and hydrochloric acid. W(OH)2 + 2HCl → WCl2 + 2H2O What is the volume of 0.5 mol dm–3 hydrochloric acid needed to neutralise 25 cm3 of 0.2 mol dm–3 aqueous W hydroxide? A 10 cm3 B 20 cm3 C 30 cm3 D 40 cm3

Structured Questions [Relative molecular mass of NaOH = 40]

1 Diagram 1 shows the arrangement of apparatus used to prepare hydrogen chloride in methylbenzene and in water respectively.

[2 marks]

(c) (i) Explain if a measuring cylinder is suitable to be used to measure the volume of water in the preparation of the standard solution. [1 mark]



(ii) Name a suitable apparatus that is required to be used in the preparation of the standard solution. [1 mark]

(d) What are the two parameters that should be measured accurately to prepare the standard solution of sodium hydroxide? Parameter I: Parameter II: [2 marks] (e) State two steps that should be taken to ensure that the standard sodium hydroxide solution is exactly 100 cm3 with a molarity of 0.5 mol dm–3. [2 marks]

3 An experiment is carried out in the laboratory to determine the concentration of a strong diprotic acid, H2 A, by titration. A few drops of phenolphthalein indicator is added to 25.0 cm3 of 0.5 mol dm–3 potassium hydroxide solution and then titrated with the acid, H2 A, of unknown concentration. The results obtained are shown in Table 1.

Diagram 1

(a) What is the purpose of using the filter funnels in Diagram 1? [1 mark] (b) (i) What is observed when a piece of magnesium ribbon is placed in beakers A and B respectively? [2 marks] (ii) State the reason for your answer in (i).



(c) Name the particles present in (i) beaker A (ii) beaker B

Experiment

[2 marks]

I

II

III

Final burette reading (cm3)

26.55

36.15

27.20

Initial burette reading (cm3)

0.50

10.00

1.10

Volume of H2 A used (cm )

…

…

…

Volume of H2 A

[2 marks]

(d) The magnesium ribbon is removed. Water is added to the solution in beaker A and the mixture is then shaken. When sodium carbonate powder is added, effervescence occurs. (i) Name the gas and suggest a suitable test to identify the gas evolved. [1 mark] (ii) State the role of water in the reaction that caused the evolution of the gas. [1 mark] (iii) Write an ionic equation for the reaction involving the evolution of the gas. [1 mark] (iv) What is the conclusion that can be made from the observation? [1 mark]

3

Table 1

(a) What is meant by a diprotic acid?

[1 mark]

(b) Write an equation for the neutralisation reaction between the acid, H2 A and potassium hydroxide.

[1 mark]

(c) State the colour change of the phenolphthalein indicator at the end point of titration. [1 mark] (d) (i) Calculate the volume of the acid, H2 A used in the titration and complete Table 1.

2 A student is required to prepare a standard solution of 100 cm3 sodium hydroxide solution with a molarity ’06 of 0.5 mol dm–3 in the laboratory. He is given all the necessary apparatus required.

[1 mark]



(a) State the meaning of (i) a standard solution. [1 mark] (ii) molarity of the solution. [1 mark]

(ii) Calculate the average volume of the acid, H2 A used in the experiment. [1 mark]

(e) Calculate the concentration of the acid H2 A used in the experiment. [2 marks] (f) Draw a diagram to show the arrangement of the apparatus used in the above experiment.

(b) Calculate the mass of sodium hydroxide that the student needs to prepare a 100 cm3 solution with a molarity of 0.5 mol dm–3.

[2 marks]

215

Acids and Bases

7



4

Experiment I

Excess magnesium ribbon is put in 10.0 cm3 of 1.0 mol dm–3 sulphuric acid

Experiment II

Excess magnesium ribbon is put in 10.0 cm3 of 1.0 mol dm–3 ethanoic acid Table 2

7

Table 2 shows two experiments carried out. (a) The reactions in experiment I and II can be represented by the same ionic equation, involving a certain particle in both acids. (i) Write the formula of this particle. [1 mark] (ii) Write the ionic equation for the reactions that occur in both experiments. [1 mark]

Diagram 2

(b) The rates of reactions are different in experiments I and II. (i) Which reaction is more vigorous? [1 mark] (ii) Give a reason for your answer in (i).

(a) Mark the pH value on the graph in Diagram 2 when complete neutralisation occurs. What is the pH value? [2 marks] (b) Mark the volume of the nitric acid required for complete neutralisation on the graph in Diagram 2. What is this volume? [2 marks]

[3 marks]

(c) Compare and explain the difference between the pH values of sulphuric acid and ethanoic acid. [2 marks]

(c) Write a balanced equation for the reaction between nitric acid and sodium hydroxide.

(d) What will be the change in pH value if 10.0 cm of water is added to the sulphuric acid before the magnesium ribbon is added in experiment I? Explain your answer. [2 marks]

3

[1 mark]

(d) If methyl orange is added to the sodium hydroxide solution at the initial stage of the experiment, what is the change in colour that will take place at the end point of titration? [1 mark]

5 An experiment was carried out to determine the end point of titration during neutralisation using a computer interface. Nitric acid is added 0.5 cm3 by 0.5 cm3 from a burette to 25.0 cm3 of 0.5 mol dm–3 sodium hydroxide in a beaker. The graph in Diagram 2 shows the change in pH value of the solution in the beaker against the volume of nitric acid used in cm3 as measured by the computer interface.

(e) Calculate the concentration of nitric acid used in this experiment. [2 marks] (f) If the experiment is to be repeated by titrating sodium hydroxide against nitric acid, sketch a graph of the change in pH value against the volume of sodium hydroxide that will be obtained. [2 marks]

Essay Questions the chemical formula of magnesium hydroxide and explain its function in antacid. Name another chemical found in antacid. [4 marks]

1 (a) Using suitable examples, explain what is meant by neutralisation. [4 marks] (b) Explain why sodium hydroxide solution and aqueous ammonia of the same concentration have different pH values. [6 marks]

(b) Diagram 1 shows two beakers containing 0.1 mol dm–3 solution X and solution Y and their pH readings.

(c) Explain how you would prepare 250 cm3 of 1.0 mol dm–3 potassium hydroxide, starting from solid potassium hydroxide. Subsequently, explain how you would prepare 250 cm3 of 0.1 mol dm–3 potassium hydroxide from the above solution. [Relative atomic mass: H, 1; O, 16; K, 39] [10 marks]

2 (a) Magnesium hydroxide is one of the chemical compounds found in antacid medicine. Write Acids and Bases

Diagram 1

216





(i) Compare and contrast the two solutions X and Y in terms of their physical and chemical properties. Give a suitable example for each of the solutions X and Y. [10 marks] (ii) Predict and explain, with suitable ionic equations, what will happen when equal volumes of solution X and solution Y are mixed together. [4 marks]



(iii) Calculate the concentration of the solution that will be produced when 80 cm3 of water is added to 20 cm3 of solution X. [2 marks]

3 (a) Using a suitable example, explain the role of water in causing the acidic properties of an aqueous solution of an acid. [8 marks] (b) Briefly describe three tests (other than the use of an indicator) that can be used to confirm an acidic solution. Explain your tests with suitable equations. [12 marks]

7

Experiments 1 An experiment is carried out to determine the relationship between the concentrations of H+ ions and the pH values of nitric acid solutions. The pH values of six nitric acid solutions with concentrations of 0.100 mol dm–3, 0.060 mol dm–3, 0.040 mol dm–3, 0.025 mol dm–3, 0.015 mol dm–3 and 0.010 mol dm–3 are each measured using a pH meter. The corresponding pH values and the concentrations of the nitric acid solutions are shown in Diagram 1.

Diagram 1

(a) State the variables involved in this experiment. • Manipulated variable: • Responding variable: • Constant variable:

[3 marks]

(b) State the hypothesis for this experiment.

[3 marks]

(c) Construct a table to record the results of this experiment.

[3 marks]

(d) Based on the results of this experiment, draw a graph of pH value versus concentration of H ions on a graph paper. [3 marks] +

(e) Using the graph you have drawn in (d), predict the pH value of a 0.020 mol dm–3 nitric acid solution. [3 marks] 2 The volume of a sample of sulphuric acid required to neutralise a potassium hydroxide solution can be determined by titration. Design a laboratory experiment to determine the volume of the sulphuric acid required to neutralise 25.0 cm3 of 0.5 mol dm–3 potassium hydroxide solution. In designing your experiment, the following items must be included. (a) Problem statement (b) All the variables involved (c) Statement of the hypothesis (d) List of materials and apparatus (e) Procedure (f) Tabulation of data

217

[17 marks]

Acids and Bases

FORM 4 THEME: Interaction between Chemicals

CHAPTER

8

Salts

SPM Topical Analysis 2008

Year 1

Paper

2 A

Section Number of questions

2

2009

–

1

3

B

C

–

1 — 2

2 A

–

–

2010

–

3

B

C

–

1 — 3

–

1

3

2011

2

3

A

B

C

1

–

–

–

1

3

2

3

A

B

C

–

–

–

ONCEPT MAP

SALTS preparation

Soluble salts by

Reaction of acids with (a) alkalis (b) metal oxides (c) metal carbonates (d) metals

qualitative analysis

Insoluble salts by

Precipitation in double decomposition reactions

Anions test by

(a) Heating and identifying (i) the gases evolved (ii) the colour change of products formed (b) Reagents such as barium chloride, silver nitrate and brown ring test

Cations test by

(a) Sodium hydroxide (b) Aqueous ammonia (c) Specific reagents

1

Salts Generally, the formula of salt is

The Meaning of Salts

Cations Anions Salt = + (other than H+) (other than O2– or OH–)

1 Salt is an ionic compound that is formed when the hydrogen ion in an acid is replaced by a metal ion or ammonium ion (NH4+). 2 The chemical formula of a salt is comprised of a cation (other than hydrogen ion) and an anion (other than oxide ion and hydroxide ion). 3 The cations and the anions of a salt are bonded by strong ionic bonds.

3 Diprotic acids and triprotic acids contain more than one H+ ions that can be replaced. Hence, it is possible for these acids to form more than one type of salt as shown in Table 8.2. Table 8.2 Examples of diprotic and triprotic salts

Type of acid

Example of acid

Types of salts that can be formed

1 Examples of salts formed from their corresponding acids are shown in Table 8.1.

Diprotic acid

H2SO4

2

NaHSO4, Na2SO4

Table 8.1 Examples of salts formed from their corresponding acids

Triprotic acid

H3PO4

3

NaH2PO4, Na2HPO4, Na3PO4

Examples of Salts

Acid

General name of salt

Example of salt

Example of salt

NaCl, KCl, CuCl2, ZnCl2, NH4Cl

Hydrochloric acid, HCl

Chloride salts

Nitric acid, HNO3

Nitrate salts NaNO3, KNO3, Mg(NO3)2, Pb(NO3)2, NH4NO3

Sulphuric acid, Sulphate H2SO4 salts

Na2SO4, K2SO4, FeSO4, CaSO4, (NH4)2SO4

Carbonic acid, Carbonate H2CO3 salts

Na2CO3, CaCO3, MgCO3, ZnCO3, PbCO3

A salt consists of cations and anions, but the anions must not be oxide ion or hydroxide ion. This is because if the anions are O2– or OH– ions, the compound is a base, not a salt.

Generally, the type of salts can be classified according to the cation or anion. For example, sodium chloride is known as a type of sodium salt or alternatively it can also be classified as a chloride salt.

2 Chloride salts are formed when H+ ions in hydrochloric acid, HCl is replaced by a metal ion or ammonium ion (NH4+). KCl (potassium chloride) H+ replaced by K+ NaCl (sodium chloride)

H+ replaced by Na

+

HCl (hydrochloric acid)

H+ replaced by Mg2+

H+ replaced by NH

+ 4

NH4Cl (ammonium chloride)

H+ replaced by Zn2+

MgCl2 (magnesium chloride)

ZnCl2 (zinc chloride) 219

Salts

8

8.1

Uses of Salts in Daily Life In food preparation 1 Sodium chloride (NaCl), table salt, is used for seasoning food. 2 Monosodium glutamate (M.S.G.) is used to enhance the taste of food. 3 Self-raising flour contains sodium bicarbonate (NaHCO3) which helps breads and cakes to rise. 8

Sodium chloride is used to flavour food

In food preservation so that food can be kept longer without spoiling. 1 Sodium chloride is used as a food preservative in food such as salted fish and salted eggs. 2 Sodium benzoate (C6H5COONa) is used as a food preservative in food such as tomato sauce, oyster sauce and jam. 3 Sodium nitrite (NaNO2) is used to preserve processed meat such as burgers, sausages and ham.

Salts play an important role in our daily life. Here are few examples of salts and their uses.

Sodium benzoate is used as a food preservative in sauce

In agriculture: to increase the production of food. 1 Nitrate salts such as potassium nitrate (KNO3), sodium nitrate (NaNO3) and ammonium salts such as ammonium sulphate (NH4)2SO4, ammonium nitrate (NH4NO3), ammonium phosphate (NH4)3PO4 are nitrogenous fertilisers. 2 Salts such as copper(II) sulphate (CuSO4), iron(II) sulphate (FeSO4) and mercury(I) chloride (HgCl) are used as pesticides.

Salts

Copper(II) sulphate is a pesticide used to kill fungi

Fertilisers used in agriculture are ammonium salts

220

In medicine patients to be seen clearly in X-ray films. 7 Iron pills containing iron(II) sulphate are taken to increase the supply of iron for anaemic patients.

8

1 Antacid medicine contain calcium carbonate (CaCO3), and calcium hydrogen carbonate Ca(HCO3)2 that are used to reduce acidity in the stomachs of gastric patients. 2 Smelling salts contain ammonium chloride (NH4Cl). 3 Plaster of Paris, used to support fractured bones, contains calcium sulphate. 4 Epsom salts (magnesium sulphate heptahydrate) and Glauber salt (sodium sulphate decahydrate) are used as laxatives to clear the intestines. 5 Potassium permanganate (KMnO4) is used as an antiseptic to kill germs. 6 Barium sulphate, BaSO4, enables the intestines of suspected stomach cancer

Plaster of Paris consists of a calcium sulphate

Other uses

Salts play an important role in our daily life. Here are few examples of salts and their uses.

1 Fluoride toothpaste contains tin(II) fluoride, SnF2, that is used to prevent tooth decay. Silver bromide, AgBr, is used to produce black 2 and white photographic films. Sodium hypochlorite, NaOCl, is used as 3 a bleaching agent in soap powders and detergents.

Fluoride salt in toothpaste prevents tooth decay

Naturally occuring salts • Lead(II) sulphide (galena, PbS), calcium fluoride (fluorite, CaF2) and magnesium sulphate (Epsime, MgSO4) exist as minerals in the earth’s crust. • Corals, stalactites, stalagmites and limestone consist of calcium carbonate, (CaCO3).

Stalactites and stalagmites

221

Salts

Soluble Salts and Insoluble Salts 1 Solubility is the ability of a compound to dissolve in a solvent. Some salts are soluble in water while others are not. 2 The solubility of a salt in water depends on the types of cations and anions present as shown in Table 8.3. Table 8.3 Types of salt and their solubility in water

8

Type of salt

Figure 8.1 Separation of a soluble salt and an insoluble salt by filtration

Solubility in water

Sodium, potassium and ammonium salts

All are soluble

Nitrate salts

All are soluble

Chloride salts

All are soluble except PbCl2, AgCl and HgCl

Sulphate salts

All are soluble except PbSO4, BaSO4 and CaSO4

Carbonate salts

All are insoluble except Na2CO3, K2CO3 and (NH4)2CO3

5 The methods of preparing salts depend on the solubility of salts. 6 Soluble salts can be prepared in the laboratory by four methods as follows: (a) Reaction between an acid and an alkali (b) Reaction between an acid and a metal (c) Reaction between an acid and a metal carbonate (d) Reaction between an acid and a metal oxide or hydroxide 7 Insoluble salts can be prepared by precipitation in double decomposition reactions.

3 Information on the solubility of salts is useful in (a) the separation of salts in a salt mixture. (b) choosing the methods to prepare a salt. (c) identifying the types of ions in a salt in the qualitative analysis of salts. 4 Filtration can be used to separate an insoluble salt (as the residue) from a soluble salt (as the filtrate) as shown in Figure 8.1.

• If a salt is soluble in water, a solution is formed. The solution may be coloured if the cation is cop­per(II), iron(II) or iron(III). Other­­wise, a colourless solution is formed. • If a salt is insoluble in water, a cloudy mixture is formed when stirred. The undissolved salt will settle down as a precipitate.

8.1 To study the solubility of nitrate, sulphate, carbonate and chloride salts Problem statement Are nitrate, sulphate, carbonate and chloride salts soluble in water?

Materials Various types of salts and distilled water. Procedure 1 0.2 g of copper(II) nitrate is put in a test tube using a spatula. 2 5 cm3 of distilled water is added to the above test tube. The mixture is stirred and the solubility of the salt is noted. 3 Steps 1 and 2 are repeated using magnesium nitrate, zinc nitrate, lead(II) nitrate, calcium nitrate, copper(II) sulphate, magnesium sulphate, zinc sulphate, lead(II) sulphate, barium sulphate, calcium sulphate, copper(II) chloride, magnesium

Experiment 8.1

Hypothesis Some salts are soluble in water while some are not. Variables (a) Manipulated variable : Types of salts (b) Responding variable : Solubility in water (c) Constant variable : Quantity of salts, volume and temperature of water Apparatus Test tubes, glass rods, spatulas and test tube holder. Salts

222

Type of salt

Formula of salt

Carbonate Na2CO3, K2CO3, (NH4)2CO3

Results

Soluble

CuCO3, MgCO3, ZnCO3

Type of salt

Formula of salt

Solubility in water

Nitrate

Cu(NO3)2, Mg(NO3)2, Zn(NO3)2, Pb(NO3)2, Ca(NO3)2

Soluble

Sulphate

CuSO4, MgSO4 and ZnSO4

Soluble

PbSO4, BaSO4, CaSO4

Insoluble

CuCl2, MgCl2, ZnCl2

Soluble

PbCl2, AgCl, HgCl

Insoluble

Chloride

Insoluble

Conclusion 1 All nitrate salts are soluble in water. 2 All sulphate salts are soluble in water except lead(II) sulphate, PbSO4, barium sulphate, BaSO4 and calcium sulphate, CaSO4. 3 All chlorides salts are soluble in water except lead(II) chloride, PbCl2, silver chloride, AgCl, and mercury(I) chloride, HgCl. 4 All carbonates are insoluble in water except sodium carbonate, Na2CO3, potassium carbonate, K2CO3 and ammonium carbonate, (NH4)2CO3. 5 Some salts are soluble while some are not. The hypothesis is accepted.

Table 8.4 Some examples of acids and alkalis used in the preparation of soluble salts

Preparation of Soluble Salts The preparation of soluble salts can be divided into two categories. (a) Soluble salts of sodium, potassium and ammonium. (b) Soluble salts which are not salts of sodium, potassium and ammonium.

Type of soluble salt

Soluble Salts of K+, Na+ and NH4+

Type of alkali used

Example of salt

Type of acid used

Salts of sodium

NaCl HCl Sodium hydroxide, NaOH CH COONa CH COOH 3 3

Salts of potassium

Potassium hydroxide, KOH

Aqueous Salts of ammo­nium ammonia, NH3

1 The cation of a salt comes from the alkali while the anion comes from the acid. Hence the type of salt produced depends on the acid and alkali used. For example:

K2SO4

H2SO4

KNO3

HNO3

NH4NO3

HNO3

(NH4)2SO4

H2SO4

5 Impure soluble salts can be purified by recrystallisation. When an impure soluble salt is dissolved in enough distilled water, the insoluble impurities can be removed by filtration. The filtrate is evaporated to remove excess water to form a saturated solution. When the saturated solution is cooled to room temperature, the salt will recrystallise to form pure crystals. 6 Soluble salts in a mixture of a few salts can also be purified by recrystallisation. This is because salts have different solubility in water. A salt with a lower solubility will recrystallise earlier than a salt with a higher solubility. The process of recrystallisation may be repeated a few times to obtain a pure salt.

Na2SO4 from NaOH

Solubility in water

from H2SO4

2 Soluble salts of sodium, potassium and ammonium can be prepared from the reaction between an acid and an alkali, NaOH, KOH or NH3(aq) as in Table 8.4. 3 Titration method is used to ensure that all the acid is completely reacted with the alkali. 4 The flowchart in Figure 8.2 shows the steps involved in the preparation of soluble salts of sodium, potassium and ammonium. 223

Salts

8

chloride, zinc chloride, lead(II) chloride, silver chloride, mercury(I) chloride, sodium carbonate, potassium carbonate, ammonium carbonate, copper(II) carbonate, magnesium carbonate and zinc carbonate to replace the copper(II) nitrate.

1

Acid + alkali The alkali in the conical flask is titrated with acid in the burette 1 titration method

Dilute salt solution 8

After titration, the dilute salt solution is heated to hasten evaporation

2 evaporation 2 Saturated salt solution The saturated salt solution is then cooled to precipitate out the salt

3 cooling 3

Salt crystals in saturated salt solution The salt crystals are then filtered out from the solution using filter paper

4 filtration 4 Salt crystals The salt crystals are then dried with more filter paper

5 drying Dry salt crystals

5

The resulting salt crystals of sodium, potassium or ammonium are produced Figure 8.2 Preparation of soluble salts of sodium, potassium and ammonium

Salts

224

Preparation of Soluble Salts of Sodium, Potassium and Ammonium

Materials 2 mol dm–3 hydrochloric acid and 2 mol dm–3 potassium hydroxide and phenolphthalein indicator. Procedure 1 25 cm3 of potassium hydroxide is pipetted into a clean conical flask. 2 Three drops of phenolphthalein indicator are added to the alkali and the colour of the solution is noted. 3 A 50 cm3 burette is then filled with hydrochloric acid and is then clamped to a retort stand. The initial burette reading is recorded. 4 Hydrochloric acid is added gradually from the burette to the potassium hydroxide solution in the conical flask while swirling the flask gently. 5 Titration is stopped when phenolphthalein changes from a light pink colour to colourless. The final burette reading is recorded. 6 The volume of hydrochloric acid used is calculated as follows: V cm3 = Final burette – Initial burette reading reading

by dropping a drop of the solution on a piece of glass plate. If crystals are formed, then the solution is saturated. 9 The saturated solution is then cooled to allow crystallisation to occur. 10 The white crystals formed are then filtered, rinsed with a little distilled water and dried by pressing between filter paper.

Figure 8.3 Titration of potassium hydroxide with hydrochloric acid

Discussion 1 In the preparation of potassium chloride, the acid used is hydrochloric acid and the alkali used is potassium hydroxide. KOH + HCl → KCl + H2O

7 The experiment is repeated by adding V cm3 of hydrochloric acid to 25 cm3 of potassium hydroxide in a beaker without using phenolphthalein as an indicator. 8 The colourless solution in the beaker is evaporated to form a saturated solution (to about 1 — of the original volume). This can be tested 3

2 Phenolphthalein is used as an indicator at the beginning of the experiment to determine the volume of hydrochloric acid that is required to react with 25 cm3 of potassium hydroxide. However, the experiment is repeated without using phenolphthalein so that the salt prepared will not be contaminated by the indicator. 3 The salt solution is not heated until dry because the salt may decompose when heated strongly.

In the preparation of soluble salts of Na+/K+/NH4+, titration is carried out to determine the exact amount of acid required to neutralise all the alkali to form a

neutral salt using an indicator. Once the exact volumes of acid and alkali required are known, the indicator is not required to prepare the pure salt.

225

Salts

Activity 8.1

Apparatus 25 cm3 pipette, pipette filler, 50 cm3 burette, retort stand, retort clamp, conical flask, filter funnel, filter paper, beaker, tripod stand, wire gauze and Bunsen burner.

SPM

’04/P2

8

To prepare potassium chloride by the reaction between an acid and an alkali

To purify potassium chloride by recrystallisation Apparatus Beaker, glass rod, Bunsen burner, conical flask, spatula, filter funnel and filter paper.

8

Materials Impure potassium chloride and distilled water. Procedure 1 Impure potassium chloride is placed in a beaker. 2 A little distilled water, enough to cover the crystals is added. The mixture is heated while stirring, and more distilled water is added slowly until all the crystals are dissolved.

3 The hot solution is filtered into a clean conical flask to remove the impurities. 4 The filtrate is evaporated until a saturated solution is formed. 5 The saturated solution is then allowed to cool to room temperature for crystallisation. 6 The filtered crystals are then rinsed with distilled water and dried between two pieces of filter paper. Conclusion Impure potassium chloride can be purified by recrystallisation.

Soluble Salts which are not salts of Na+, K+, NH4+

ensure that all the acid is reacted completely, excess solids are used. The non-reacted excess solids can be removed by filtration.

1 Three methods are used to prepare soluble salts which are not salts of sodium, potassium and ammonium. This involves the reaction between (a) an acid and a metal (b) an acid and a metal carbonate (c) an acid and a metal oxide or hydroxide 2 Table 8.5 shows the chemicals suitable for the preparation of soluble salts which are not salts of sodium, potassium and ammonium. 3 In the reaction of an acid with a metal, metals that are less electropositive than hydrogen such as copper and silver do not react with dilute acids. 4 Metals, metal oxides and metal carbonates are solids that do not dissolve in water. Hence, to

Table 8.5 Examples of some salts and chemicals used in their preparation

Example of salt

Type of acid used

ZnCl2

Chemicals that react with the acid Metal oxide

Metal

Metal carbonate

HCl

ZnO

Zn

ZnCO3

Mg(NO3)2

HNO3

MgO

Mg

MgCO3

CuSO4

H2SO4

CuO

_

CuCO3

Preparation of Soluble Salts which are Not Salts of Sodium, Potassium and Ammonium

Activity 8.2 & 8.3

To prepare copper(II) nitrate by the reaction between an acid and a metal oxide

SPM

’04/05 P2

Apparatus Beaker, glass rod, 100 cm3 measuring cylinder, wire gauze, tripod stand, Bunsen burner, conical flask, spatula, filter funnel and filter paper. Materials 1 mol dm–3 nitric acid and copper(II) oxide powder.

Salts

226

Procedure 1 About 30 cm3 of 1 mol dm–3 nitric acid is put in a beaker and is heated. 2 Using a spatula, copper(II) oxide powder is added a little at a time, to the hot nitric acid while stirring continuously with a glass rod. The addition of copper(II) oxide is stopped when some black solids remain undissolved. 3 The mixture is filtered to remove the excess copper(II) oxide. 4 The filtrate is evaporated until a saturated solution is formed. 5 The saturated solution is then allowed to cool to room temperature. 6 The blue crystals formed are removed by filtration, rinsed with a little distilled water and dried between filter paper.

1

8

2

Discussion 1 Copper(II) oxide is a black powder. It dissolves in nitric acid to form a blue solution. The equation for the reaction is CuO(s) + 2HNO3(aq) → Cu(NO3)2 (aq) + H2O(l)

3

2 Neutralisation reaction takes place between copper(II) oxide and nitric acid. The ionic equation for the reaction is CuO(s) + 2H+(aq) → Cu2+(aq) + H2O(l) 3 The salt solution is not heated until dry because the salt may decompose when heated strongly. 4 The copper(II) nitrate crystals prepared may be purified by recrystallisation. 5 Copper(II) nitrate can also be prepared by the reaction between nitric acid and copper(II) carbonate. However, copper metal does not react with dilute nitric acid because copper is below hydrogen in the electrochemical series.

4

5

Conclusion Copper(II) nitrate can be prepared by the reaction between copper(II) oxide and nitric acid.

227

Salts

To prepare iron(II) sulphate by the reaction of an acid and a metal Apparatus Beaker, glass rod, 100 cm3 measuring cylinder, Bunsen burner, conical flask, spatula, filter funnel and filter paper.

8

Materials Iron powder and 2 mol dm–3 sulphuric acid. Procedure 1 30 cm3 of 2 mol dm–3 sulphuric acid is put in a beaker. 2 Iron powder is gradually added to the sulphuric acid while stirring continuously with a glass rod, until a slight excess of iron is present. 3 The mixture is filtered to remove the excess iron powder. 4 The filtrate is evaporated until a saturated solution is formed. The saturated solution is then allowed to cool to room temperature. 5 The green crystals formed are removed by filtration, rinsed with a little distilled water and dried between filter papers.

Discussion 1 Iron metal is grey in colour. It dissolves in sulphuric acid to form a green solution with effervescence. The gas evolved is hydrogen gas. The equation for the reaction is Fe(s)+ H2SO4(aq) → FeSO4(aq) + H2(g) 2 The ionic equation for the reaction between iron and H+ ion in acid is Fe(s)+ 2H+(aq) → Fe2+(aq) + H2(g) 3 Iron(II) sulphate can also be prepared by the reaction of sulphuric acid with iron(II) oxide or iron(II) carbonate. Conclusion Iron(II) sulphate can be prepared by the reaction between iron metal and sulphuric acid.

Activity 8.4 & 8.5

To prepare magnesium chloride by the reaction of an acid and a metal carbonate Apparatus Beaker, glass rod, 100 cm3 measuring cylinder, Bunsen burner, conical flask, spatula, filter funnel and filter paper. Materials Magnesium carbonate powder and 2 mol dm–3 hydrochloric acid. Procedure 1 Magnesium carbonate powder is added a little at a time, to 30 cm3 of 2 mol dm–3 hydrochloric acid in a beaker while stirring continuously. The addition is stopped when there is no more effervescence, and a little magnesium carbonate powder remains undissolved. 2 The mixture is filtered to remove the excess magnesium carbonate powder. 3 The filtrate is evaporated until a saturated solution is formed. 4 The saturated solution is then allowed to cool to room temperature. 5 The white crystals formed is removed by filtration, rinsed with a little distilled water and dried between filter papers. Salts

Discussion 1 Magnesium carbonate is white in colour. Effervescence occurs when it dissolves in hydrochloric acid to form a colourless solution. The gas evolved is carbon dioxide gas. The equation for the reaction is MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l) 2 The ionic equation for the reaction between magnesium carbonate and H+ ion in acid is MgCO3(s) + 2H+(aq) → Mg2+(aq) + CO2(g) + H2O(l) 3 Magnesium chloride can also be prepared by the reaction of hydrochloric acid with magnesium oxide or magnesium metal. Conclusion Magnesium chloride can be prepared from the reaction between magnesium carbonate and hydrochloric acid. 228

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’10/P1

1 Insoluble salts can be prepared by precipitation in double decomposition reactions. 2 In the precipitation method, an insoluble salt is precipitated when two aqueous solutions containing the cations and the anions are mixed together. The precipitate is then obtained by filtration. 3 In double decomposition, one of the aqueous solutions contains the cations of the insoluble salt, while the other aqueous solution contains the anions of the salt.

Can sodium nitrate be prepared by adding sodium chloride solution to nitric acid? Comments There is no reaction between sodium chloride solution and nitric acid. NaCl(aq) + HNO3(aq) NaNO3(aq) + HCl(aq) Solution consist of Solution consist of Na+, Cl–, H+ and Na+, Cl­–, H+ and – NO3 ions NO3– ions Sodium nitrate is usually prepared by the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH). H2O formed does not dissociate.

Cation M+ (from a + soluble salt solution)

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)

Insoluble salt, Anion X– MX (from a soluble salt → (formed as precipitate) solution)

4 In double decomposition, the ions of the two aqueous solutions interchange to produce a new compound which is insoluble. 5 The general equation can be represented as follows

Physical Characteristic of Salt Crystals 1 Salt crystals are formed when a saturated salt solution is cooled. 2 A salt is made up of positive ions and negative ions. When these ions are packed closely with a regular and repeated arrangement in fixed positions, a solid with definite geometry known as crystal lattice is formed. 3 The repeating basic unit in this orderly structure is called a unit cell. 4 All crystals have these physical characteristics: (a) Fixed geometrical shapes (e.g. cubic, hexagonal or rhombic). (b) Flat surface, straight edges and sharp angles. (c) Fixed angle between two adjacent surfaces. 5 All crystals of the same salt have the same shape although the sizes may be different. 6 The size of a crystal formed depends on the rate of crystallisation. Fast crystallisation (from fast cooling) will yield smaller crystals than slow crystallisation. 7 Crystals are hard and brittle, and can be cut into different shapes. This is because the particles of salt crystals are arranged in regular layers.

MY(aq) + NX(aq) → MX(s) + NY(aq) solution solution precipitate solution

6 The precipitate produced is obtained by filtration. The residue is the insoluble salt, which is then rinsed with distilled water to remove any other ions as impurities. Example Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq) precipitate

7 Precipitation of lead(II) chloride can be simplified as follows: Pb2+(aq) + 2Cl–(aq) → PbCl2(s) Sodium ions, Na+ and nitrate ions, NO3– do not undergo any change in the reaction. They are known as spectator ions and can be ignored in the ionic equation.

229

Salts

8

Preparation of Insoluble Salts

1

8 The following guidelines show the steps used in writing an ionic equation for the formation of an insoluble salt. Step 1

Step 2

Step 3

Identify the insoluble salt and write it as the product on the right-hand side of the equation.

Separate the cations and anions of the salt and write them as the reactants on the left-hand side of the equation.

Balance the charges of the cations and anions by adding the correct coefficient as the number of moles reacting.

Pb2+(aq) + Cl–(aq) → PbCl2(s)

Pb2+(aq) + 2Cl–(aq) → PbCl2(s)

8

→ PbCl2(s)

9 The following guidelines show the steps used in the selection of aqueous solutions in the preparation of an insoluble salt. Step 1

Step 2

Step 3

Identify the cation and anion of the insoluble salt. Example PbCl2: Cation = Pb2+ Anion = Cl–

Identify a soluble salt that can supply the cation, example, a nitrate salt (all nitrate salts are soluble). Example Pb(NO3)2 solution or Pb(CH3COO)2 solution

Identify a soluble salt that can supply the anion, example, a sodium or potassium salt (all sodium or potassium salts are soluble in water). Example NaCl or KCl solution

PbCl2 from Pb(NO3)2 or Pb(CH3COO)2

from NaCl or KCl or HCl

Table 8.6 Some examples of insoluble salts

Barium salt

Lead(II) salt

Silver salt

Name

Formula

Name

Formula

Name

Formula

Lead(II) chloride Lead(II) bromide Lead(II) iodide Lead(II) sulphate Lead(II) chromate(VI) Lead(II) carbonate

PbCl2 PbBr2 PbI2 PbSO4 PbCrO4 PbCO3

Barium sulphate Barium chromate(VI) Barium carbonate

BaSO4 BaCrO4 BaCO3

Silver chloride Silver bromide Silver iodide Silver carbonate

AgCl AgBr AgI Ag2CO3

2

’04

Can lead(II) sulphate be prepared by adding sulphuric acid to lead(II) oxide? Comments Both lead(II) sulphate and lead(II) oxide are insoluble in water. Hence, when lead(II) sulphate is formed, it cannot be separated from the mixture of lead(II) sulphate and lead(II) oxide. Lead(II) sulphate as an insoluble salt, is usually prepared from double decomposition reaction between Salts

two aqueous solutions, one containing the lead(II) ions (example, lead(II) nitrate) and the other containing the sulphate ions (example, sodium sulphate). Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq) Similarly, lead(II) sulphate cannot be prepared by adding sodium sulphate solution to lead(II) chloride. Both lead(II) sulphate and lead(II) chloride are insoluble in water.

230

To prepare insoluble salts: lead(II) iodide, lead(II) chromate(VI) and barium sulphate by precipitation reaction

Procedure (A) Preparation of lead(II) iodide 1 20 cm3 of 0.5 mol dm–3 potassium iodide solution is added to 20 cm3 of 0.5 mol dm–3 lead(II) nitrate solution in a beaker. 2 The mixture is stirred thoroughly with a glass rod. A yellow preci­pitate is formed immediately. 3 The mixture is filtered to obtain the yellow solids of lead(II) iodide as the residue. 4 The residue is rinsed with distilled water to remove any trace of other ions in it. 5 The yellow solid is dried by pressing between two pieces of filter papers. (B) Preparation of lead(II) chromate(VI) 1 20 cm3 of 0.5 mol dm–3 lead(II) nitrate solution is added to 20 cm3 of 0.5 mol dm–3 potassium chromate(VI) solution in a beaker. 2 The mixture is stirred thoroughly with a glass rod. A yellow precipitate is formed immediately. 3 The mixture is filtered to obtain the yellow solids of lead(II) chromate(VI) as the residue. 4 The residue is rinsed with distilled water and dried using filter papers. (C) Preparation of barium sulphate 1 20 cm3 of 0.5 mol dm–3 barium chloride solution is added to 20 cm3 of 0.5 mol dm–3 sodium sulphate solution in a beaker. 2 The mixture is stirred thoroughly with a glass rod. A white precipitate is formed immediately. 3 The mixture is filtered to obtain the white barium sulphate as the residue.

Discussion 1 The chemical equation for the reaction that occur in the preparation of lead(II) iodide is Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) 8

Materials 0.5 mol dm–3 solutions of lead(II) nitrate, potassium iodide, potassium chromate(VI), sodium sulphate and barium chloride.

4 The residue is rinsed with distilled water and dried using filter papers.

The ionic equation for the reaction is Pb2+(aq) + 2I–(aq) → PbI2(s) Lead(II) iodide is a yellow precipitate. 2 The chemical equation for the reaction that occurs in the preparation of lead(II) chromate(VI) is Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2KNO3(aq) The ionic equation for the reaction is Pb2+(aq) + CrO42–(aq) → PbCrO4(s) Lead(II) chromate(VI) is a yellow precipitate. 3 The chemical equation for the reaction that occurs in the preparation of barium sulphate is BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) The ionic equation for the reaction is Ba2+(aq) + SO42–(aq) → BaSO4(s) Barium sulphate is a white precipitate. Conclusion Insoluble salts of lead(II) iodide, lead(II) chromate(VI) and barium sulphate can be prepared by precipitation in double decomposition reactions.

Preparation of a Specified Salt 2 The following flowchart shows the procedure for the selection of the methods of preparing a specified salt (Figure 8.4).

1 The method of preparing a salt depends on (a) whether or not the salt is soluble in water, (b) whether the salt is a salt of sodium, potassium or ammonium for soluble salts. 231

Salts

Activity 8.6

Apparatus Beakers, glass rods, conical flasks, filter funnels and filter paper.

SPM

’08/P2

Method of preparing a salt Is the salt soluble in water?

Yes

No Is it the salt of sodium, potassium or ammonium?

No

Precipitation by double decomposition reaction

Neutralisation using titration

Reaction of an acid with • a metal • a metal oxide/metal hydroxide • a metal carbonate

8

Yes

Select two aqueous solutions that can supply the cations and the anions of the salt

Filtration to remove excess solids (keep the filtrate) Salt solution

Filtration (keep the residue)

1 evaporation 2 cooling (crystallisation) 3 filtration 4 recrystallisation (if necessary) Salt crystals 1 rinse with distilled water

2 dry with filter paper

Pure salt crystals Figure 8.4

3

’03

You are supplied with sodium carbonate solution, sulphuric acid and magnesium nitrate solution. Plan a scheme to prepare a sample of magnesium sulphate using the above chemicals. Write equations for the reactions involved. Comments Magnesium sulphate is a soluble salt which is not a salt of sodium, potassium and ammonium. Magnesium sulphate can be prepared by the reaction of sulphuric acid with magnesium metal/magnesium oxide/magnesium carbonate. Hence, a scheme of preparing magnesium sulphate is as follows:

Salts

Step 1 Magnesium nitrate is converted to magnesium carbonate by double decomposition between sodium carbonate solution and magnesium nitrate. Mg(NO3)2(aq) + Na2CO3(aq) → MgCO3(s) + 2NaNO3(aq) Step 2 Magnesium carbonate that is produced is then added to sulphuric acid until in excess.

232

MgCO3(s) + H2SO4(aq) → MgSO4(aq) + CO2(g) + H2O(l)

Ionic Equations of Insoluble Salts

Table 8.7 Formation of ionic equations from the mole ratio of ions

1 An ionic equation for the formation of a salt can be written if (a) the formula of the salt is known (from the charges of cation and anion), (b) the number of moles of ions required to form the salt is known. 2 The following guidelines show the construction of the ionic equation for the formation of an insoluble salt from the charges of cation and anion.

No. of moles No. of moles of cation of anion 2 mol Cl–

Pb2+(aq) + 2Cl–(aq) → PbCl2(s)

1 mol Pb2+

1 mol CrO42– Pb2+(aq) + CrO42–(aq) → PbCrO4(s)

2 mol Ag+

1 mol CrO42– 2Ag+(aq) + CrO42–(aq) → Ag2CrO4(s)

4 The number of moles of the cation and anion can be calculated if the volume and molarities MV are known using the formula — — — —. 1000

Step 1 If the charge of cation M is b and the charge of anion X is a, the formula of the salt is MaXb, MaXb

1

charge of X

charge of M

6.0 cm3 of 0.2 mol dm–3 Xn+ solution reacts completely with 4.0 cm3 of 0.1 mol dm–3 Ym– solution to form a salt XmYn. Write the ionic equation and hence determine the empirical formula of the salt in this reaction.

For example, the formula of iron(III) carbonate is Fe2(CO3)3.

Solution 0.2 6 Number of moles of Xn+ ions = ––––––– 1000 = 0.0012

Step 2

0.1  4 Number of moles of Ym– ions = ––––––– 1000 = 0.0004

This show that a mol of M ions has combined with b mol of Xa– ions. b+

Fe2(CO3)3 shows that 2 mol of Fe3+ ions combines with 3 mol of CO32– ions.

Mole ratio of



MV — — — — 1000

MV — — — — 1000

Xn+ ions : Ym– ions = 0.0012 : 0.0004 0.0012 : 0.0004 = –––––––– ––––––– 0.0004 0.0004 = 3 : 1

Hence 3 mol of Xn+ react with 1 mol of Ym–.

Step 3

Ionic equation is : 3Xn+ + 1Ym– → X3Y

Thus the ionic equation for the formation of MaXb is

Empirical formula of the salt is X3Y.

aMb+(aq) + bXa–(aq) → MaXb(s)

Constructing Ionic Equations Using the Continuous Variation Method

The ionic equation for the formation of Fe2(CO3)3 is

1 The mole ratio of ions that react to form a salt can be determined from an experiment through the continuous variation method. 2 In this method, fixed volumes of a reactant X are added to varying volumes of a second reactant Y in different test tubes. If the salt formed is an insoluble salt, the amount of

2Fe3+(aq) + 3CO32–(aq) → Fe2(CO3)3(s) 3 The examples in Table 8.7 show the method of writing ionic equations based on the simplest mole ratio of cations to anions combined to form the salts. 233

Salts

8

1 mol Pb2+

Ionic equation

precipitate produced will increase until all of the ions in solution X have reacted completely. The height of the precipitate will remain constant despite the increasing volumes of solution Y. 3 The flowchart of Figure 8.5 shows the steps involved in the continuous variation method. To determine the ionic equation of the reaction between X ions and Y ions Carry out an experiment to investigate the reaction between • fixed volumes of solution X and • different and varying volumes of solution Y

8

Determine the volume of Y ions that reacts with all of the X ions Calculate the number of moles of X ions and the number of moles of Y ions that have reacted using the formula: MV Number of moles = — — — — 1000 Determine the simplest mole ratio of X ions to the Y ions in the reaction to construct the ionic equation Figure 8.5 Flowchart for the steps in the continuous variation method

8.2

SPM

’11/P3

Experiment 8.2

To construct a balanced ionic equation for the precipitation of lead(II) chromate(VI) using the continuous variation method Problem statement How to determine the ionic equation for the precipitation of lead(II) chromate(VI)? Hypothesis The height of precipitate will increase with the increase in volume of lead(II) nitrate solution until all the potassium chromate(VI) has reacted. Variables (a) Manipulated variable : Volumes of lead(II) nitrate solution (b) Responding variable : Height of yellow precipitate (c) Constant variable : Volume of potassium chromate(VI) solution and the size of test tubes Apparatus Test tubes of the same size, test tube rack, 50 cm3 burette, retort stand with clamp and ruler. Materials 0.5 mol dm–3 lead(II) nitrate solution and 0.5 mol dm–3 potassium chromate(VI) solution. Salts

Procedure 1 A burette is filled with 0.5 mol dm–3 lead(II) nitrate solution and another burette is filled with 0.5 mol dm–3 potassium chromate(VI) solution. 2 Eight test tubes are labelled 1 to 8 and placed in a test tube rack. 3 5.00 cm3 of potassium chromate(VI) solution from the burette is placed in every test tube. Potassium chromate(VI) solution is yellow in colour. 4 Using another burette, 1 cm3 of 0.5 mol dm–3 of lead(II) nitrate solution is added to the first test tube. Progressively increase the volume of the lead(II) nitrate solution by 1 cm3 to the rest of the test tubes until 8 cm3 of lead(II) nitrate solution is added to the eighth test tube (Figure 8.6(a)). 5 Every test tube is well shaken in order to mix the solutions completely. The test tubes are then allowed to stand for 20 minutes for the yellow precipitate, lead(II) chromate(VI) to settle (Figure 8.6(b)).

234

6 The height of the precipitate formed in every test tube is measured accurately using a ruler. The colour of the solution above the precipitate is noted. 7 The result obtained is recorded in Table 8.8.

8

1.1

Figure 8.6 Continuous variation method

Results Table 8.8

Test tube number Volume of potassium chromate(VI) solution (cm3) Volume of lead(II) nitrate solution (cm3) Height of precipitate (cm) Colour of solution

1 5.0

2 5.0

3 5.0

4 5.0

5 5.0

6 5.0

7 5.0

8 5.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

0.6 yellow

0.9 yellow

1.8 yellow

2.2 yellow

2.8

2.8 2.8 colourless

2.8

MV 0.5  5.0 = ­­­­­ — — — — =— — — — — — — — = 2.5 3 10–3 1000 1000

Calculation 1 A graph showing the height of precipitate versus the volume of lead(II) nitrate solution is drawn (Figure 8.7).

Number of moles of CrO42– ions in 5.0 cm3 of 0.5 mol dm–3 potassium chromate(VI) solution MV 0.5  5.0 = ­­­­­ — — — — =— — — — — — — — = 2.5 3 10–3 1000 1000 4 Hence, 2.5  10–3 mol of Pb2+ ions react completely with 2.5  10–3 mol of CrO42– ions. ∴ 1.00 mol of Pb2+ ions will react completely with 1.00 mol of CrO42– ions. The ionic equation for the reaction is

Figure 8.7 Graph of height of precipitate versus the volume of lead(II) nitrate solution

Pb2+(aq) + CrO42–(aq) → PbCrO4(s)

2 From the graph, it is found that the height of precipitate increases as the volume of lead(II) nitrate increases. However, a constant height is reached when 5.0 cm3 of lead(II) nitrate solution is added. Thereafter, the height remains constant despite further increases in the volume of lead(II) nitrate solution. 3 This means that when 5.0 cm3 of 0.5 mol dm–3 lead(II) nitrate solution is used, all the chromate(VI) ions in 5 cm3 of 0.5 mol dm–3 potassium chromate(VI) solution has been precipitated. Number of moles of Pb2+ ions in 5.0 cm3 of 0.5 mol dm–3 lead(II) nitrate solution

5 Consequently, the balanced chemical equation for the reaction is

1 mol

1 mol

1 mol

Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2KNO3(aq) Conclusion 1 Since the diameter of the test tubes are the same, the height of the precipitate is directly proportional to the mass of precipitate formed. 2 The ionic equation for the precipitate of lead(II) chromate(VI) is Pb2+ + CrO42– → PbCrO4. The hypothesis is accepted. 235

Salts

Discussion 3 From test tubes 6 to 8, the heights of precipitate formed remains constant because all the chromate(VI) ions in the test tubes have been precipitated. There is an excess of Pb2+ ions in the test tubes. The clear solution above the precipitate which is colourless contains Pb2+ ions, K+ ions and NO3– ions.

8

1 From test tubes 1 to 4, the 2+ increase of Pb ions from the increase in volumes of lead(II) nitrate solution added, increases the mass of precipitate formed. There are excess (un­reacted) CrO42– ions in the test tubes which produces the yellow colour of the solutions above the precipitate. The yellow solution contains CrO42– ions, K+ ions and NO3– ions. The yellow colour became paler as more CrO42– ions have reacted.

2 In test tube 5, the reaction is completed when the precipitate formed reaches a maximum height. All the chromate(VI) ions have reacted with all the lead(II) ions. The clear solution contains K+ ions and NO3– ions.

4

’97

You are supplied with a lead(II) ions solution and a 0.1 mol dm–3 potassium chromate(VI) solution. Explain how you can determine the concentration of the lead(II) ions solution using the precipitation method. Comments An experiment using the continuous variation method of precipitating lead(II) chromate(VI), using a constant volume of potassium chromate(VI) solution and different volumes of lead(II) ions as in Experiment 8.2 is carried out. A graph of the height of precipitate against the volume of lead(II) ions solution will be obtained as follows:

Salts

From the graph, V cm3 of lead(II) ions solution is required to react completely with 5 cm3 of potassium chromate(VI) solution, when the height of the precipitate becomes constant. Calculation: 0.1 35 • Number of moles of CrO42– ions = ————— 1000 = 0.0005 M 3V • Number of moles of Pb2+ ions = —————, 1000 where M is the concentration. • Ionic equation of the reaction 2+ 2– Pb + CrO4 → PbCrO4 • From the equation, 1 mol of Pb2+ ions react with 1 mol 2– of CrO4 ions. 2+ 2– Mole ratio of Pb : CrO4 = 1:1, 0.0005 1 that is —————— = — 0.001MV 1 • Concentration of Pb2+ ions, M = 0.5/V mol dm–3

236

1 Hence, 0.2 mol of H2SO4 produce 0.2  — 3 = 0.067 mol of Al2(SO4)3.

SPM Numerical Problems Involving ’09/P1 Calculation of Quantities of Reactants or Products in Stoichiometric Reactions

1 A balanced equation gives information regarding the number of moles of reactants in a reaction and the number of moles of products formed. 2 For example, the equation for the reaction between magnesium and hydrochloric acid is

Type 2: Calculation involving quantities in mass If the quantities of reactants/products are given in terms of mass in gram, the quantity of solid can be converted to moles by the following relationship

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) 1 mol

2 mol

1 mol

1 mol

The coefficients before the reactants and products indicate the number of moles of chemicals involved in the reaction. The equation above shows that 1 mol of magnesium reacts with 2 mol of hydrochloric acid to produce 1 mol of magnesium chloride and 1 mol of hydrogen gas. 3 A balanced chemical equation (stoichiometric reaction) can be used to calculate the stoichiometric quantities of the reactants or products in terms of: • Mass • Volume and concentration (of aqueous solution) • Volume (of gas) 4 Since the quantities of chemicals involved in a reaction are in terms of moles, the quantities of reactants or products (the quantity of chemicals in terms of volume of gas, volume of solution, mass, numbers of molecules or atoms), must first be converted to moles in the initial step of the calculation regarding quantities of reactants and products.

SPM

’08/P1

2.0 g of sodium hydroxide reacts with excess sulphuric acid. What is the mass of sodium sulphate produced? [Relative atomic mass: H, l; O, 16; Na, 23; S, 32] Solution 2NaOH + H2SO4 → Na2SO4 + 2H2O Molar mass of NaOH = 23 + 16 + 1 = 40 g mol–1 2g 2 g of NaOH = — — — — — — — — ‑ 40 g mol–1 = 0.05 mol

Step 1: Write a balanced equation.

Step 2: Convert mass to mol.

From the equation, 2 mol of NaOH produces 1 mol of Na2SO4. Step 3: Get the mole ratio of NaOH and Na2SO4

Type 1: Calculation involving quantities in moles

1 Hence, 0.05 mol of NaOH produce — 0.05 2 = 0.025 mol of Na2SO4

2 Calculate the number of moles of aluminium sulphate produced by the reaction of 0.2 mol of sulphuric acid with excess aluminium oxide.

Molar mass of Na2SO4 = 2(23) + 32 + 4(16) = 142 g mol–1

Step 1: Write a balanced equation. Solution 3H2SO4 + Al2O3 → Al2(SO4)3 + 3H2O

1 mol

Mass (g) Number of moles = — — — — — — — — — — — — — — — — — — — Molar mass (g mol–1)

3

Examples of Calculation

3 mol



Step 2: Get the mole ratio of H2SO4 and Al2(SO4 )3 .

0.025 mol of Na2SO4 = 0.025 mol  142 g mol–1 = 3.55 g

From the equation, 3 mol of H2SO4 produce 1 mol of Al2(SO4)3 237

Step 4: Relate the number of moles of chemicals in the equation to that in the question.

Step 5: Convert mol to mass.

Salts

8

Step 3: Relate the number of moles of chemicals in the equation to that in the question.

Type 3: Calculation involving volumes of gas If the quantities of reactants or products are given in terms of volumes of gas, the volume of gas can be converted to moles by the following relationship:

Or: Number of moles = molarity of solution (mol dm–3)  volume of solution (dm3)

5

At s.t.p. (0°C and 1 atm):

At room conditions (25°C and 1 atm) :

What is the mass of magnesium required to react with 20 cm3 of 2 mol dm–3 hydrochloric acid to produce 120 cm3 of hydrogen at room temperature? [Relative atomic mass: Mg, 24; 1 mol of gas occupies 24 dm3 at room temperature]

Volume of gas (dm3) Number of moles = — — — — — — — — — — — — — — — — — 24.0 dm3 mol–1

Solution Mg + 2HCl → MgCl2 + H2

8

Volume of gas (dm3) Number of moles = — — — — — — — — — — — — — — — — — 22.4 dm3 mol–1

1 mol of gas occupies 24 dm3.

What is the volume of carbon dioxide gas evolved at s.t.p. when 2.1 g of magnesium carbonate reacts with excess nitric acid? [Relative atomic mass: C, 12; O, 16; Mg, 24; 1 mol of gas occupies 22.4 dm3 at s.t.p.]

Step 3: Get the mole ratio of Mg and H2.

Hence, 0.005 mol of H2 is produced by 0.005 mol of Mg

Step 1: Write a balanced equation.

Step 4: Relate the number of moles of chemicals in the equation to that in the question.

MgCO3 + 2HNO3 → Mg(NO3)2 + CO2 + H2O Molar mass of MgCO3 = 24 + 12 + 3(16) = 84 g mol–1 2.1 2.1 g MgCO3 = — — = 0.025 mol 84

0.005 mol of Mg = 0.005  24 g = 0.12 g

Step 2: Convert mass to mol.

What is the volume of 2 mol dm–3 hydrochloric acid required to dissolve 10 g of marble (calcium carbonate)? [Relative atomic mass: H, 1; O, 16; C, 12; Ca, 40]

Step 3: Get the mole ratio of MgCO3 and CO2.

Hence, 0.025 mol of MgCO3 produce 0.025 mol of CO2. Step 4: Relate the number of moles of chemicals in the equation to that in the question.

Solution Step 1: CaCO3 + 2HCl → CaCl2 + CO2 + H2O

Step 5: Convert mol to volume.

10 10 Step 2: 10 g of CaCO3 = — — — — — — — — — — — —= — — — 40 + 12 + 3(16) 100 = 0.1 mol

Hence, 0.025 mol of gas CO2 gas occupies 0.025  22.4 = 0.56 dm3 or 560 cm3.

Step 3: From the equation, 1 mol of CaCO3 requires 2 mol of HCl for a complete reaction.

Type 4: Calculation involving volumes and molarities of solutions If the quantities of reactants or products involves solutions, the quantity of chemicals in a solution can be converted to moles using the following relationship MV Number of moles = ———— 1000

Step 4: Hence, 0.1 mol of CaCO3 requires 0.1  2 = 0.2 mol of HCl for a complete reaction Step 5: Number of moles = molarity  volume (dm3) Volume of HCl Number of moles of HCl 0.2 mol =— — — — — — — — — — — — — — — — — — — —= — — — — — — — — — Molarity of HCl 2 mol dm–3 3 3 3 = 0.1 dm = 0.1  1000 cm = 100 cm

Where M = molarity of solution (mol dm–3) V = volume of solution (cm3) Salts

Step 5: Convert mol to mass.

6

From the equation, 1 mol of MgCO3 produces 1 mol of CO2.

1 mol of gas occupies 22.4 dm3.

Step 2: Convert volume to mol.

120 — — — — — — — — mol = 0.005 mol 120 cm3 gas = — 24  1000 From the equation, 1 mol of H2 is produced by 1 mol of Mg.

4

Solution

Step 1: Write a balanced equation.

238

7

SPM

5

’08/P1

4.0 g of magnesium oxide is added to 30.0 cm of 2.0 mol dm–3 hydrochloric acid. What is the mass of magnesium oxide that does not dissolve in this reaction? [Relative atomic mass: O, 16; Mg, 24]

’02

3

Pb2+(aq) + 2I–(aq) → PbI2(s) hat is the molarity of the potassium iodide W solution? MV

MV Number of moles = — — — — 1000

MgO + 2HCl → MgCl2 + H2O

Number of moles = — — — —

1000 Comments 20.0 Number of moles of Pb2+ ions = 0.25  — — — — 1000 = 0.005

2.0  30 Number of moles of HCl = — — — — — — — — = 0.06 1000 From the equation, 2 mol of HCl will dissolve 1 mol of MgO.

8

Solution

5.0 cm3 of a potassium iodide solution requires 20.0 cm3 of 0.25 mol dm–3 lead(II) nitrate solution to react completely according to the equation below.

Hence, 0.06 mol of HCl will dissolve

From the equation, 1 mol of Pb2+ ions reacts with 2 mol of I– ions.

1 0.06  — = 0.03 mol of MgO. 2

ence, 0.005 mol of Pb2+ ions react with (0.005  2) H = 0.01 mol of I– ions. 1000 M = Number of moles 3 — — — — V

0.03 mol of MgO = 0.03  (24 + 16) g = 1.2 g

0.01  1000 Molarity of KI solution = — — — — — — — — — — = 2 mol dm–3 5

∴ Mass of MgO that does not dissolve = Initial mass of MgO – mass of MgO dissolved = 4.0 g – 1.2 g = 2.8 g

8

Mass, m

What is the mass of copper(II) carbonate that is produced when 60 cm3 of 1 mol dm–3 sodium carbonate is added to 50 cm3 of 2 mol dm–3 copper(II) sulphate? [Relative atomic mass H, 1; O, 16; Cu, 64]

n  molar mass

Mole, n

Solution CuSO4 + Na2CO3 → CuCO3 + Na2SO4 Number of moles of CuSO4 2  50 =— — — — — = 0.1 1000

V  molar volume

Calculate the number of moles of both reactants to check which of the reactants is used up in the reaction. The number of moles of product formed depends on the number of moles of reactant that is used up (the limiting reactant).

V1  molarity

n  molarity

n  molar volume

Volume of gas, V

Number of moles of Na2CO3 1  60 =— — — — — = 0.06 1000 From the equation, 1 mol of CuSO4 reacts with 1 mol of Na2CO3 to produce 1 mol of CuCO3.

Volume of solution, V1

8.1 1 Suggest suitable methods and reactants for the preparation of the following salts. (a) Na2SO4 (b) (NH4)2SO4 (c) Al2(SO4)3 (d) Pb(NO3)2 (e) ZnCl2 (f) PbSO4 (g) AgCl

Hence, 0.06 mol of Na2CO3 will react completely, while 0.1 mol of CuSO4 is in excess. Thus, 0.06 mol of Na2CO3 will produce 0.06 mol of CuCO3.

m  molar mass

Na2CO3 is the limiting factor.

2 Suggest chemicals that can react with nitric acid to produce magnesium nitrate. Write equations for the reactions that take place.

0.06 mol of CuCO3 = 0.06  (64 + 12 + 3(16)) g = 0.06  124 g = 7.44 g

239

Salts

8

3 State the mole ratio of the ions in the following salts: (a) CaSO4 (b) Al(OH)3

Inference from the Colours of Salts or Salt Solutions

4 Write the ionic equations for the following reactions: (a) 2 mol of silver ions react with 1 mol of chromate(VI) ions (b) 0.3 mol of lead(II) ions react with 0.6 mol of bromide ions

1 Initial observation of the physical properties of a salt such as colour and solubility in water enables us to make inferences regarding the possible cations or anions present. However, the presence of the cations or anions needs to be confirmed by other tests. 2 Most salts are white in colour and when dissolved in water, will form colourless aqueous solutions. 3 Cations of transition elements have specific colours. 4 Table 8.9 below gives the colours of different cations in the solid form or in aqueous solutions.

5 5 cm3 of 0.2 mol dm–3 barium chloride solution reacts completely with 10 cm3 of 0.1 mol dm–3 sodium chromate(VI) solution. Calculate the mole ratio of the ions involved in the formation of barium chromate precipitate. Subsequently, write the balanced equation for the reaction that occurs.

chemical

6 In an experiment, 10 cm3 of 0.5 mol dm–3 silver nitrate reacts completely with 5 cm3 of 0.5 mol dm–3 potassium carbonate. Determine the ionic equation for the precipitation above.

Table 8.9 Colours of cations

7 Magnesium oxide reacts with excess phosphoric acid to produce 1.2 mol of magnesium phosphate. (a) Write a balanced equation for the reaction that occurs. (b) Calculate the number of moles of magnesium oxide that is used in the reaction above.

8.2

Qualitative Analysis of Salts

The Meaning of Qualitative Analysis 1 Qualitative analysis is a chemical technique used to determine the identities of chemical substances present in a mixture but not their quantity. 2 Qualitative analysis of salt is a scheme of tests carried out to identify the cation and anion present in the salt. 3 The technique of qualitative analysis includes: (a) Observing the colour of the salt or colour of the aqueous salt solution. (b) Observing the solubility of the salt in water. (c) Observing the effect of heat on the salt. (d) Identifying the gas evolved when a test is performed on the salt. (e) Identifying the precipitate formed when a specific chemical reagent is added to the aqueous salt solution. (f) Carrying out confirmatory tests, which are specific chemical tests to confirm the identity of a cation or an anion present in a salt. Salts

Solution

Colour

Solid

White or colourless

Salts of Na , K , NH4+, Mg2+, Ca2+, Ba2+, Al3+, Pb2+, Zn2+ (if the anions are colourless)

Na , K+, NH4+, Mg2+, Ca2+, Ba2+, Al3+, Pb2+, Zn2+

Yellow

PbO, PbI2, PbCrO4, BaCrO4

Fe3+, CrO42–

Blue

Hydrated Cu2+ salt

Cu2+

Green

Hydrated Fe2+ salt, CuCO3 and CuCl2

Fe2+

Black

Cu2+, Fe2+ oxide or sulphide

–

Brown/ orange

Hydrated Fe3+ salt

Fe3+, CrO72–

+

+

+

5 Table 8.10 shows the solubility of different types of salts in water. Table 8.10 Solubility of salts in water

Type of Salt

Solubility in water

Salts of Na+, All are soluble K+, NH4+

240

Nitrate

All are soluble

Sulphate

All common sulphates are soluble except BaSO4, PbSO4 and CaSO4

Chloride

All common chlorides are soluble except AgCl, HgCl and PbCl2(soluble in hot water)

Tests of Gases

Solubility in water

Carbonate

All common carbonates are insoluble except Na2CO3, K2CO3 and (NH4)2CO3

Oxide

All oxides are insoluble except Na2O, K2O and CaO (slightly soluble)

Hydroxide

All hydroxides are insoluble except KOH, NaOH, Ca(OH)2 and Ba(OH)2

1 Certain gases may be evolved when a chemical substance is (a) heated, (b) reacted with a dilute or concentrated acid, (c) heated with an alkali. 2 Based on the gas evolved, information about the types of ions present can be deduced. For instance, if carbon dioxide gas is evolved in a reaction, carbonate ions are present in the salt. 3 The physical properties and chemical tests for a few gases are summarised in Table 8.11.

Lead halides PbCl2, PbBr2 and PbI2 are insoluble in cold water but soluble in hot water

Table 8.11 Physical properties and tests on gases

Name of gas Colour of gas

Smell of gas

Oxygen, O2

Colourless

No smell No effect

When a glowing wooden splint is lowered into the test tube of oxygen, the glowing splint is lighted

Hydrogen, H2

Colourless

No smell No effect

When a lighted wooden splint is pla­ced near the mouth of the test tube of hydrogen, a ‘pop’ sound is produced

Carbon dioxide, CO2

Colourless

No smell Moist blue litmus turns to red

When carbon dioxide gas is bubbled into limewater using a delivery tube, the limewater becomes milky

Ammonia, NH3

Colourless

Pungent Moist red litmus turns to blue

When a glass rod dipped into concentrated hydrochloric acid is placed near the mouth of the test tube with ammonia, white fumes are formed

Chlorine, Cl2

Greenishyellow

Choking Decolourises moist red or blue litmus

Hydrogen chloride, HCl

Colourless

Pungent Moist blue litmus turns to red

When a glass rod dipped into concentrated ammonia is placed near the mouth of the test tube with hydrogen chloride, white fumes are formed

Sulphur dioxide, SO2

Colourless

Pungent Moist blue litmus turns to red

When sulphur dioxide gas is bubbled into acidified potassium manganate(VII) solution, the purple colour is decolourised (or when it is bubbled into acidified potassium dichromate(VI) solution, the colour changes from orange to green)

Nitrogen dioxide, NO2

Brown

Pungent Moist blue litmus turns to red

–

Effect on damp litmus

Confirmatory test on gas

–

(NH4)2CO3(s) → 2NH3(g) + H2O(l) + CO2(g)

Heating Test on Salts

(NH4)2SO4(s) → 2NH3(g) + H2SO4(l)

1 All ammonium, carbonate, nitrate and some sulphate salts will decompose when heated. 2 All ammonium salts liberate ammonia gas when heated. Examples NH4NO3(s) → NH3(g) + HNO3(g)

3 All carbonates except potassium carbonate and sodium carbonate produce carbon dioxide gas when heated. Table 8.12 shows the effect of heating on metal carbonates. 241

Salts

8

Type of Salt

Table 8.12 Effect of heat on carbonate salts

Carbonate salt Potassium carbonate Sodium carbonate

Effect of heat Will not decompose on heating

8

Decompose to metal oxide and carbon dioxide gas Calcium carbonate

CaCO3(s) → CaO(s) + CO2(g)

Magnesium carbonate

MgCO3(s) → MgO(s) + CO2(g)

Aluminium carbonate

Al2(CO3)3(s) → Al2O3(s) + 3CO2(g)

Zinc carbonate

ZnCO3(s) → ZnO(s) + CO2(g)

Iron(III) carbonate

Fe2(CO3)3(s) → Fe2O3(s) + 3CO2(g)

Lead(II) carbonate

PbCO3(s) → PbO(s) + CO2(g)

Copper(II) carbonate

CuCO3(s) → CuO(s) + CO2(g) Decompose to metal, carbon dioxide gas and oxygen gas

Mercury(II) carbonate

2HgCO3(s) → 2Hg(l) + 2CO2(g) + O2(g)

Silver carbonate

2Ag2CO3(s) → 4Ag(s) + 2CO2(g) + O2(g)

Gold(I) carbonate

2Au2CO3(s) → 4Au(s) + 2CO2(g) + O2(g) Decompose to carbon dioxide gas, ammonia and water vapour without any residue

Ammonium carbonate

(NH4)2CO3(s) → 2NH3(g) + H2O(g) + CO2(g)

4 All nitrates decompose when heated. Table 8.13 shows the effect of heating on metal nitrates. (a) Sodium nitrate and potassium nitrate produce oxygen gas and nitrites when heated. (b) Other metal nitrates produce oxygen gas, nitrogen dioxide gas and metal oxides when heated. Table 8.13 Effect of heat on nitrate salts

Nitrate Salt

Effect of heat Decompose to metal nitrite and oxygen gas

Potassium nitrate

2KNO3(s) → 2KNO2(s) + O2(g)

Sodium nitrate

2NaNO3(s) → 2NaNO2(s) + O2(g) Decompose to metal oxide, oxygen gas and nitrogen dioxide gas

Calcium nitrate

2Ca(NO3)2(s) → 2CaO(s) + 4NO2(g) + O2(g)

Magnesium nitrate

2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g)

Aluminium nitrate

4Al(NO3)3(s) → 2Al2O3(s) + 12NO2(g) + 3O2(g)

Zinc nitrate

2Zn(NO3)2(s) → 2ZnO(s) + 4NO2(g) + O2(g)

Iron(III) nitrate

4Fe(NO3)3(s) → 2Fe2O3(s) + 12NO2(g) + 3O2(g)

Lead(II) nitrate

2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g)

Copper(II) nitrate

2Cu(NO3)2(s) → 2CuO(s) + 4NO2(g) + O2(g) Decompose to metal, nitrogen dioxide gas and oxygen gas

Mercury(II) nitrate

Hg(NO3)2(s) → Hg(l) + 2NO2(g) + O2(g)

Silver nitrate

2AgNO3(s) → 2Ag(s) + 2NO2(g) + O2(g)

Gold(I) nitrate

2AuNO3(s) → 2Au(s) + 2NO2(g) + O2(g) Decompose to nitrous oxide gas, water vapour without any residue

Ammonium nitrate Salts

NH4NO3(s) → N2O(g) + 2H2O(g) 242

5 Most sulphate salts do not decompose when heated. Only a few sulphates such as iron(II) sulphate, zinc sulphate and copper(II) sulphate decompose to sulphur dioxide or sulphur trioxide gas when heated. Examples 2FeSO4(s) → Fe2O3(s) + SO2(g) + SO3(g)

Figure 8.8 Table 8.14 Deduction of types of ion present from gas produced

Type of gas produced

CuSO4(s) → CuO(s) + SO3(g) 6 All chloride salts are stable on heating except ammonium chloride. Ammonium chloride sublimes and decomposes to produce ammonia gas and hydrogen chloride gas. NH4Cl(s) → NH3(g) + HCl(g) 7 The deduction of the types of ions present based on the gas produced is shown in Table 8.14. 8 When a salt is heated, (a) the type of gas evolved has to be identified. This will give information to the type of anion (or cation, NH4+) present. (b) the colour change of the solid in the test tube must be recorded. This will give information regarding the type of cation present.

Type of ion

CO2

Carbonate ion, CO32– (except Na2CO3 and K2CO3)

O2

Nitrate ion, NO3–

NO2 and O2

Nitrate ion, NO3– (except NaNO3 and KNO3)

SO2

Sulphate ion, SO42–

NH3

Ammonium ion, NH4+

8

ZnSO4(s) → ZnO(s) + SO3(g)

9 Most salts that decompose produced metal oxides as residue. The change of colour during heating gives a good indication towards the type of metal oxide formed as shown in Table 8.15.

Table 8.15 Colour change of salts on heating

Colour of residue after heating

Metal oxide produced

Cation present in salt

White

Yellow when hot, white when cold

ZnO

Zn2+

White

Brown when hot, yellow when cold

PbO

Pb2+

Blue/green

Black

CuO

Cu2+

Green/yellow

Brown

Fe2O3

Fe2+/Fe3+

To study the effect of heat on carbonate salts Apparatus Boiling tubes, test tubes, test tube holder, delivery tube with rubber stopper, spatula and Bunsen burner. Materials Potassium carbonate, sodium carbonate, calcium carbonate, magnesium carbonate, zinc carbonate, lead(II) carbonate, copper(II) carbonate and limewater.

SPM

’10/P2

Procedure 1 One spatula of potassium carbonate powder is placed in a dry boiling tube and the colour of the solid is recorded. 2 The boiling tube is fitted with a stopper with a delivery tube. 3 The carbonate salt is heated slowly and then strongly. 243

Salts

Activity 8.7

Original colour of salt

8

4 Any gas evolved is passed through the delivery tube into the limewater. The effect on limewater is recorded (Figure 8.9). 5 When there is no further change, the colour of the residue when it is hot is recorded. The colour of the residue when it is cooled to room temperature is also recorded. 6 Steps 1 to 5 of the experiment is repeated using other carbonate salts as shown in Table 8.16.

stopped. Otherwise the limewater will be sucked back into the hot boiling tube.

Precaution Make sure that the end of the delivery tube is removed from the limewater before heating is

Figure 8.9 Heating test on carbonate salts

Results Table 8.16 Heating test on carbonate salts

Colour of residue

Colour of salt before heating

When hot

When cold

Potassium carbonate, K2CO3

White

White

White

No visible change

Sodium carbonate, Na2CO3

White

White

White

No visible change

Calcium carbonate, CaCO3

White

White

White

Limewater turns milky

Magnesium carbonate, MgCO3

White

White

White

Limewater turns milky

Zinc carbonate, ZnCO3

White

Yellow

White

Limewater turns milky

Lead(II) carbonate, PbCO3

White

Brown

Yellow

Limewater turns milky

Copper(II) carbonate, CuCO3

Green

Black

Black

Limewater turns milky

Carbonate salt

Discussion 1 In this experiment, limewater is used to test for the presence of carbon dioxide gas. Carbon dioxide gas turns limewater milky because calcium carbonate is formed as a white precipitate.

4 Note that if excess carbon dioxide gas is passed into the limewater, the white precipitate, CaCO3 formed will dissolve to form calcium hydrogen carbonate. The limewater will turn clear again. CaCO3(s) + H2O(l) + CO2(g) → Ca(HCO3)2(aq)

CO2(g) + Ca(OH)2(aq) → CaCO3(s) + H2O(l) 2 When zinc carbonate is heated, zinc oxide and carbon dioxide gas are produced. ZnCO3(s) → ZnO(s) + CO2(g) Zinc oxide is yellow when hot and white when cooled. 3 When lead(II) carbonate is heated, lead(II) oxide and carbon dioxide gas are produced.

white precipitate

colourless solution

Conclusion 1 Potassium carbonate and sodium carbonate will not decompose on heating. 2 Other metal carbonates decompose on heating to produce metal oxides and carbon dioxide gas. A general equation representing the decomposition of carbonate salt by heat is MCO3(s) → MO(s) + CO2(g)

PbCO3(s) → PbO(s) + CO2(g) Lead(II) oxide is brown when hot and yellow when cooled.

Salts

Effect on limewater

244

metal metal carbonate oxide

To study the effect of heat on nitrate salts Apparatus Boiling tubes, litmus paper, test tube holder, wooden splint, spatula and Bunsen burner. Materials Potassium nitrate, sodium nitrate, magnesium nitrate, zinc nitrate, aluminium nitrate, lead(II) nitrate and copper(II) nitrate.

4 When there is no further change, the colour of the residue is recorded when it is hot. The colour of the residue is recorded again when the residue is cooled to room temperature. 5 Steps 1 to 4 of the experiment are repeated using other nitrate salts as shown in Table 8.17.

8

Procedure 1 One spatula of potassium nitrate is placed in a dry boiling tube and the colour of the solid is noted. 2 The nitrate salt is heated slowly and then strongly. 3 Any gas evolved is tested by a glowing wooden splint (Figure 8.10(a)) and moist blue litmus paper (Figure 8.10(b)). The results are recorded.

Figure 8.10 Heating test on nitrate salt

Table 8.17 Heating test on nitrate salts

Test on gas

Colour of residue When hot

When cold

Effect on Effect on Colour of glowing wooden moist blue gas splint litmus paper

Potassium nitrate, KNO3

White

White

White

Colourless

Rekindles

No change

Sodium nitrate, NaNO3

White

White

White

Colourless

Rekindles

No change

Magnesium nitrate, Mg(NO3)2

White

White

White

Brown

Rekindles

Turns red

Aluminium nitrate, Al(NO3)3

White

White

White

Brown

Rekindles

Turns red

Zinc nitrate, Zn(NO3)2

White

Yellow

White

Brown

Rekindles

Turns red

Lead(II) nitrate, Pb(NO3)2

White

Brown

Yellow

Brown

Rekindles

Turns red

Copper(II) nitrate, Cu(NO3)2

Blue

Black

Black

Brown

Rekindles

Turns red

Discussion 1 The brown gas that changed moist blue litmus paper to red is nitrogen dioxide gas. 2 The gas that rekindles a glowing wooden splint is oxygen gas. Conclusion 1 All nitrate salts decompose on heating. Potassium nitrate and sodium nitrate decompose to oxygen gas when heated.

Generally, 2MNO3(s) → 2MNO2(s) + O2(g), metal nitrate

metal nitrite

where M = K or Na. 2 Other metal nitrates decompose to metal oxides, nitrogen dioxide gas and oxygen gas when heated. A general equation representing the decomposition of other nitrate salts by heat is

245

2M(NO3)2(s) → 2MO(s) + 4NO2(g) + O2(g) metal nitrate

metal oxide

Salts

Activity 8.8

Nitrate salt

Colour of salt before heating

Test for carbonate ions, CO32–

Test for nitrate ions, NO3–(brown ring test)

1 When dilute acid (hydrochloric acid, nitric acid or sulphuric acid) is added to an aqueous carbonate solution (or solid carbonate), effervescence occurs. 2 The gas evolved turns limewater milky. Carbon dioxide gas is produced, indicating the presence of carbonate ions.

1 When dilute sulphuric acid and iron(II) sulphate, FeSO4 solution are added to an aqueous nitrate solution, followed by concentrated sulp­huric acid added slowly along the side of the test tube, a brown ring is formed in the middle section of the solution mixture. 2 The formation of the brown ring (a complex) indicates the presence of nitrate ions.

8

CO32–(aq) + 2H+(aq) → CO2(g) + H2O(l)

Figure 8.11 Acid test for carbonate ions

Figure 8.12 Brown ring test for nitrate ions

Tests for the Presence of Anions in Aqueous Solutions

SPM

’05/P2 Q8

The presence of anions, CO32–, NO3–, SO42–, and Cl– can be identified by conducting specific tests on the aqueous salt solution.

Test for sulphate ions, SO42–

SPM

Test for chloride ions, Cl–

’10/P2

1 When dilute hydrochloric acid (or nitric acid) is added to an aqueous sulphate solution followed by barium chloride solution, BaCl2 (or barium nitrate solution, Ba(NO3)2), a white precipitate is formed. 2 The white precipitate is barium sulphate, BaSO4.

1 When dilute nitric acid is added to an aqueous chloride solution followed by silver nitrate solution, AgNO3, a white precipitate is formed. 2 The white precipitate is silver chloride, AgCl. Ag+(aq) + Cl–(aq) → AgCl(s)

Ba2+(aq) + SO42–(aq) → BaSO4(s) 3 BaNO3 or BaCl2 solution provide the Ba2+ ions to react with the SO42– ions to produce the insoluble BaSO4 salt.

3 AgNO3 solution provides the Ag+ ions to react with the Cl– ions to produce the insoluble AgCl salt.

H2SO4 followed by Ba(NO3)2 or BaCl2 solution is not a suitable test for the presence of SO42– ions. This is because the SO42– ions in H2SO4 will give a positive test with Ba(NO3)2 or BaCl2 solution.

HCl followed by AgNO3 solution is not a suitable test for the presence of Cl– ions. This is because the Cl– ions in HCl will give a positive test with AgNO3 solution.

Salts

246

To test for the presence of anions in aqueous salt solution Apparatus Test tubes, delivery tube with rubber stopper, spatula, test tube holder and Bunsen burner.

sulphate solution, 2 mol dm–3 hydrochloric acid and nitric acid, concentrated sulphuric acid and 0.1 mol dm–3 silver nitrate solution.

Materials 1 mol dm–3 sodium carbonate solution, sodium chloride solution, sodium sulphate solution, sodium nitrate solution, barium chloride solution and iron(II)

Procedure The steps to test the aqueous salt solutions as planned in Table 8.18 are carried out. The observations and inferences are tabulated. 8

Results Table 8.18 Tests for the presence of anions in aqueous salt solutions

Test

2 (a) About 2 cm3 of sodium chloride solution is placed in a test tube. (b) About 2 cm3 of dilute nitric acid is added to the chloride solution followed by about 2 cm3 of silver nitrate solution. 3 (a) About 2 cm3 of sodium sulphate solution is placed in a test tube. (b) About 2 cm3 of dilute nitric acid is added to the sulphate solution followed by about 2 cm3 of barium chloride solution. 4 (a) About 2 cm3 of sodium nitrate solution is placed in a test tube. (b) About 2 cm3 of dilute sulphuric acid is added to the nitrate solution followed by about 2 cm3 of iron(II) sulphate solution. The mixture is shaken to mix well. (c) Concentrated sulphuric acid is added care­ fully to the mixture along the wall of the tilted test tube without shaking the mixture. Conclusion 1 The presence of carbonate ions can be identified by the addition of a dilute acid where carbon dioxide gas evolved will turn the limewater milky. 2 The presence of chloride ions can be identified by the addition of silver nitrate solution to an acidified solution when a white precipitate is produced.

Inference

Effervescence occurs.

A gas is evolved.

Limewater becomes milky.

Carbon dioxide gas is evolved. Carbonate ion, CO32–, is present.

A white precipitate is produced.

Chloride ion, Cl–, is present.

A white precipitate is produced.

Sulphate ion, SO42–, is present.

A brown ring is formed Nitrate ion, NO3–, is in the middle of the present. solution.

3 The presence of sulphate ions can be identified by the addition of barium chloride solution to an acidified solution when a white precipitate is produced. 4 The presence of nitrate ions can be identified by the brown ring test. 247

Salts

Activity 8.9

1 (a) About 2 cm3 of sodium carbonate solution is placed in a test tube. (b) About 2 cm3 of dilute hydrochloric acid is added to the carbonate solution. (c) Any gas produced is bubbled into limewater.

Observation

8

Tests for Cations 1 The cations usually tested are: Al3+, Pb2+, Zn2+, Mg2+, Ca2+, Fe3+, Fe2+, Cu2+ and NH4+ ions. 2 An aqueous solution of the cation is prepared by (a) dissolving the salt in water (if the salt is soluble in water). (b) dissolving the salt in dilute acid and then filtering (if the salt is insoluble in water). The filtrate contains the cation. 3 The aqueous cation solution is then tested with (a) sodium hydroxide solution, NaOH, (b) aqueous ammonia solution, NH3(aq), (c) a specific reagent as a confirmatory test. 4 Sodium hydroxide and aqueous ammonia supply hydroxide ions, OH– to produce metal hydroxide as precipitate with cation solutions except Na+, K+ and NH4+ ions.

5 Transition element cations produce specific coloured metal hydroxide, whereas the other cations produce white precipitate as shown in Table 8.19.

Cu(OH)2

Cu2+

Dirty green precipitate

Fe(OH)2

Fe2+

Fe3+

–

Na+, K+, NH4+

9 All three Cu(OH)2, Fe(OH)2 and Fe(OH)3 do not dissolve in excess sodium hydroxide solution. However, Cu(OH)2 (a blue precipitate) dissolves in excess aqueous ammonia to form a dark blue solution. 10 A summary of the sodium hydroxide tests and aqueous ammonia tests for cations are shown in Table 8.20.

SPM

Blue precipitate

Fe(OH)3

NH4+(aq) + OH–(aq) → NH3(g) + H2O(l)

’11/P1

Cation present

Brown precipitate

Some metal hydroxides are soluble in excess sodium hydroxide or aqueous ammonia to form complexes. 6 Al3+, Pb2+ and Zn2+ ions form metal hydroxides which are white precipitates that are soluble in excess sodium hydroxide. Furthermore, zinc hydroxide is also soluble in excess aqueous ammonia. 7 Mg2+ and Ca2+ ions form metal hydroxides which are white precipitates with sodium hydroxide, but only Mg2+ ions will produce white precipitate with aqueous ammonia. 8 Ammonium ion, NH4+ does not produce any precipitate with sodium hydroxide or aqueous ammonia. However, if a mixture of ammonium ion and sodium hydroxide is heated, ammonia gas is produced.

Example: Pb2+(aq) + 2OH–(aq) → Pb(OH)2(s) Al3+(aq) + 3OH–(aq) → Al(OH)3(s)

Formula of metal hydroxide

Cation present

No precipitate

precipitate

Observation

Formula of metal hydroxide

White precipitate Al(OH)3, Pb(OH)2, Al3+, Pb2+, Zn(OH)2, Mg(OH)2, Zn2+, Mg2+, Ca(OH)2 Ca2+

Generally, Mn+(aq) + nOH–(aq) → M(OH)n(s)

Table 8.19 Colours of metal hydroxides

Observation

SPM

Table 8.20 Hydroxide tests for cations

Cation

A little sodium hydroxide, NaOH(aq)

Excess sodium hydroxide, A little aqueous NaOH(aq) ammonia, NH3(aq) No precipitate forms

’10/P1

Excess aqueous ammonia, NH3(aq)

NH4+

No precipitate forms. No precipitate forms. NH3 NH3 gas evolves when gas evolves when heated heated

Pb2+

White precipitate

White precipitate soluble in White precipitate excess NaOH

White precipitate insoluble in excess NH3(aq)

Zn2+

White precipitate

White precipitate soluble in White precipitate excess NaOH

White precipitate soluble in excess NH3(aq)

Al3+

White precipitate

White precipitate soluble in White precipitate excess NaOH

White precipitate insoluble in excess NH3(aq)

Salts

248

No precipitate forms

A little sodium hydroxide, NaOH(aq)

Excess sodium hydroxide, A little aqueous NaOH(aq) ammonia, NH3(aq)

Mg2+

White precipitate

White precipitate insoluble in excess NaOH

White precipitate

White precipitate insoluble in excess NH3(aq)

Ca2+

White precipitate

White precipitate insoluble in excess NaOH

No precipitate forms

No precipitate forms

Cu2+

Blue precipitate

Blue precipitate insoluble in Blue precipitate excess NaOH

Fe2+

Dirty green precipitate Dirty green precipitate insoluble in excess NaOH

Fe3+

Brown precipitate

Dirty green precipitate

Excess aqueous ammonia, NH3(aq)

Blue precipitate soluble in excess NH3(aq) to form a dark blue solution Dirty green precipitate insoluble in excess NH3(aq)

Brown precipitate insoluble Brown precipitate Brown precipitate insoluble in excess NH3(aq) in excess NaOH

Confirmatory tests for Fe2+, Fe3+,NH4+ and Pb2+ ions 1 Potassium hexacyanoferrate(II), K4Fe(CN)6 solution, potassium hexacyanoferrate(III), K3Fe(CN)6 solution and potassium thiocyanate, KSCN, solution can be used to confirm the presence of Fe2+ and Fe3+ ions. The observation is shown in Table 8.21. 2 Fe2+ ions can also be confirmed by acidified potassium manganate(VII), KMnO4 solution. If a few drops of KMnO4 solution acidified by dilute H2SO4 are added to a solution, and the purple colour of the manganate(VII) ion is decolouris­ed, then the solution contains Fe2+ ions. 3 Pb2+ ions can be confirmed by adding

(a) an iodide solution (e.g. KI), produces a yellow precipitate of PbI2. (b) a chloride solution (e.g. NaCl), produces a white precipitate of PbCl2. (c) a sulphate solution (e.g. H2SO4), produces a white precipitate of PbSO4. 4 Both lead(II) chloride, PbCl2 and lead(II) iodide, PbI2 are soluble in hot water and recrystallise when cooled. 5 Nessler reagent is a special reagent to test the presence of ammonium ion. This reagent forms a brown precipitate with ammonium ion. 6 A summary of the confirmatory tests for Pb2+, NH4+, Fe2+ and Fe3+ ions is shown in Table 8.21. SPM

’11/P1

Table 8.21 Confirmatory tests for Pb2+, NH4+, Fe2+ and Fe3+ ions

Cation

Specific reagent

Lead(II) ions, Pb2+

KI, NaI

Yellow precipitate, soluble in hot water and recrystallises when cooled

KCl, NaCl, HCl

White precipitate, soluble in hot water and recrystallises when cooled

K2SO4, Na2SO4, H2SO4

White precipitate, insoluble in hot water

Ammonium ions, NH4+ Nessler reagent Iron(II) ions, Fe

2+

Iron(III) ions, Fe3+

Observation

Brown precipitate

Potassium hexacyanoferrate(II), K4Fe(CN)6

Light blue precipitate

Potassium hexacyanoferrate(III), K3Fe(CN)6

Prussian blue (dark blue) precipitate

Acidified KMnO4

Purple colour decolourises

Potassium thio­cy­­anate, KSCN

Blood red colour

Potassium hexacyanoferrate(II), K4Fe(CN)6

Turnbull’s blue (dark blue) precipitate

Potassium hexacyanoferrate(III), K3Fe(CN)6

Greenish-brown solution

249

Salts

8

Cation

Flowchart for the analysis of cations in salts Unknown cation solution Test 1: Add NaOH(aq) solution

White precipitate

Coloured precipitate

Zn , Al3+, Pb2+, Ca2+, Mg2+ 2+

Add excess NaOH(aq) solution

8

No precipitate

Blue Green Brown Cu2+ Fe2+ Fe3+

NH4+ On heating

Cation solution

Precipitate dissolves

Precipitate does not dissolve

Zn2+, Al3+, Pb2+

Ca2+, Mg2+

Test 2:

Add aqueous NH3

NH4+

Coloured precipitate Cation solution Test 2:

White precipitate dissolves in excess NH3

White precipitate insoluble in excess NH3

Test 2:

White precipitate is formed

Mg2+ present Zn2+ present

Blue Green Brown Cu2+ Fe2+ Fe3+

Cation solution

Add excess aqueous NH3

Add excess aqueous NH3

Add excess aqueous NH3

No precipitate

Ca2+ present

Al3+ or Pb2+

Cation solution Cation solution Test 3:

Blood red colour

Cu present

Fe present 3+

Cation solution

Cation solution

Test 3: Add K3Fe(CN)6

Test 3: Add KI

Dark blue precipitate

Salts

Yellow precipitate

Al3+ present

Pb2+ present

Add KSCN

Dark blue solution

2+

No precipitate

Gas turns red litmus to blue. NH3 gas evolves.

Fe2+ present

250

Test 2

Add Nessler reagent

Brown precipitate

NH4+ present

Flowchart for the analysis of anions in salts Unknown solid anion Test 1: Heating the solid

Gas rekindles glowing wooden splint and a brown gas is evolved. O2 and NO2 gas evolve.

Gas evolves and decolourises acidified KMnO4 solution. SO2 or SO3 evolves.

No gas evolves

CO32– ion

NO3– ion

SO42– ion

SO42– or Cl–

Solid salt or salt solution

Anion solution

Anion solution

Anion solution

8

Gas evolves and turns limewater milky. CO2 gas is evolved.

Test 2 Test 2

Add dilute acid

Add FeSO4, dilute H2SO4 and then concentrated H2SO4 slowly

Test 2

Add HNO3 and Ba(NO3)2

Brown ring

White precipitate

NO3– ion present

SO42– ion present

Gas evolves and turns limewater milky. CO2 evolves.

Test 2

Add HNO3 and Ba(NO3)2

No precipitate. SO42– ion is not present.

Anion solution CO32– ion present

Test 3

Add HNO3 and AgNO3

White precipitate

Cl– ion present

Qualitative Analysis to Identify the Ions Present in a Salt 1 Two types of salt analysis: SPM (A) To confirm the cation and anion present ’09/P2 in a named salt Example Compound X is lead(II) carbonate. How do you carry out tests to confirm the cation and anion in compound X?

compounds. Carry out chemical tests to identify the anion and cation in solid Y and solution Z. 2 An example of problem A, to confirm the cation and anion in compound X, is shown in Activity 8.10 To confirm the cation, Pb2+ ions and the anion, CO32– ions in compound X. 3 An example of problem B, a planned analysis, is shown in Activity 8.11 To identify the cations and anions in unknown salt Y and salt Z.

(B) To identify the cation and anion present in an unknown salt Example You are supplied with solid Y and solution Z. Both Y and Z are ionic 251

Salts

To confirm the cation, Pb2+ ions and the anion, CO32– ions in compound X Apparatus Test tubes, boiling tube, test tube holder, delivery tube with stopper, spatula and Bunsen burner.

8

Materials Solid lead(II) carbonate, 2 mol dm–3 nitric acid, 2 mol dm–3 sodium hydroxide solution, limewater and 0.5 mol dm–3 potassium iodide solution.

Procedure 1 The steps in the experiment which are supplied in Table 8.22 are carried out. 2 The observations are recorded and inferences are made. 3 Care is taken to ensure that all test tubes and spatula are clean to prevent contamination.

Results Table 8.22

Experiment 1 (a) A spatula of compound X is heated in a boiling tube, gently at first and then strongly. (b) The gas produced is passed into lime­water in a test tube using a delivery tube. 2 5 cm3 of dilute nitric acid is added to a quarter spatula of compound X in a test tube. The gas evolved is tested with limewater. 3 The mixture in step 2 is filtered. The filtrate which contains the cation is divided into two portions in two test tubes. (a) For the first portion, sodium hydroxide solution is added gradually until in excess.





(b) For the second portion, a little potassium iodide solution is added. (i) A little distilled water is added to the mixture and then heated. (ii) The mixture is allowed to cool under running water.

Observation The residue is brown when hot and yellow when cold. The gas evolved turns limewater milky. The gas evolved turns limewater milky.

Inference Lead(II) oxide is formed. Pb2+ ions may be present. Carbon dioxide gas is evolved. CO32– ions may be present. Carbon dioxide gas is evolved. CO32– ions are confirmed to be present.

A white precipitate is produ­ ced which is soluble in excess sodium hydroxide solution. A yellow precipitate is produced. The yellow precipitate dissolves in hot water to form a colourless solution. Upon cooling, golden yellow crystals are reformed.

Pb2+, Al3+ or Zn2+ ions may be present.

Lead(II) iodide may be formed. The yellow precipitate is lead(II) iodide. Lead(II) ions, Pb2+ is confirmed to be present.

Activity 8.10 & 8.11

Conclusion It is confirmed that compound X contains lead(II) ions and carbonate ions.

To identify the cations and anions in unknown salt Y and salt Z Apparatus Test tubes, boiling tube, test tube holder, delivery tube with stopper, wooden splint and Bunsen burner.

Salts

Materials Unknown salt Y and salt Z, 2 mol dm–3 sodium hydroxide solution, aqueous ammonia, dilute sulphuric acid, dilute nitric acid, dilute hydrochloric acid, concentrated sulphuric acid, silver nitrate solution, barium chloride solution and iron(II) sulphate solution. 252

Procedure (A) Tests on salt Y Experiment

Observation

Inference

A brown gas is produced. A gas that rekindles a glowing wooden splint is also produced. A white residue is formed.

Nitrogen dioxide gas and oxygen gas are produced. Nitrate ion, NO3– is present.

2 The residue formed is cooled and then dissolved in dilute nitric acid. The resulting solution is divided into 3 portions.

The white residue dissolves in nitric acid.

The solution contains cations.

(a) Sodium hydroxide solution is added to the first portion of solution Y until in excess.

White precipitate that dissolves in excess sodium hydroxide is formed.

Pb2+, Al3+ or Zn2+ ions may be present.

(b) Aqueous ammonia is added to the second portion of solution Y until in excess.

White precipitate that does not dissolve in excess aqueous ammonia is formed.

Zn2+ ions are not present. Pb2+ ions or Al3+ ions may be present.

(c) Potassium iodide solution is added to the third portion of solution Y.

No noticeable change.

Pb2+ ions are not present. Al3+ ions are present.

A brown ring is formed.

NO3– ions are present.

3 Salt Y is dissolved in distilled water. A little iron(II) sulphate and dilute sulphuric acid is added to solution Y. Concentrated sulphuric acid is then added slowly along the side of the test tube to the mixture.

8

1 Salt Y is heated strongly.

(B) Tests on salt Z Experiment

Observation

Inference

1 Salt Z is dissolved in distilled water. The solution is divided into 4 portions.

Z dissolves in water.

Z is a soluble salt.

2 Sodium hydroxide solution is added to the first portion of solution Z until in excess.

A white precipitate that does not dissolve in excess sodium hydroxide is formed.

Mg2+ ions or Ca2+ ions may be present.

3 Aqueous ammonia is added to the second portion of solution Z until in excess.

A white precipitate that does not dissolve in excess aqueous ammonia is formed.

Mg2+ ions are present.

4 Dilute hydrochloric acid is added to the third portion of solution Z followed by barium chloride solution.

No noticeable change.

SO42– ions are not present.

5 Dilute nitric acid is added to the fourth portion of solution Z followed by silver nitrate solution.

A white precipitate is formed.

Cl– ions are present.

Conclusion Compound Y contains Al3+ ion and NO3– ion. Compound Z contains Mg2+ ion and Cl– ion.

253

Salts

8

In qualitative analysis of unknown ions, you should not test for the ions present in the reagents used in analysis. For example, if the salt is dissolved in dilute hydrochloric acid, do not test for the presence of chloride ion. In the same way, if aqueous ammonia is added to the salt solution, do not test for ammonium ion.

solution is added to Cl– ions solution followed by an acid, a white precipitate that is insoluble in acids will be formed. 3 Ba(NO3)2 or BaCl2 solutions produce insoluble salts with both CO32– ions and SO42– ions. However, BaCO3 is soluble in acids. Hence, the formation of white precipitate with Ba2+ ions solution in the presence of acids confirms the presence of SO42– ions. If Ba2+ ions solution is added to SO42– solution followed by an acid, a white precipitate that is insoluble in acids will be formed.

1 Pb(NO3)2 solution produces insoluble salts as precipitate with CO32– ions, SO42– ions and Cl– ions. Hence, Pb(NO3)2 solution is not a good test for the presence of the above three types of ions. However, a negative test may indicate the presence of NO3– ions. 2 AgNO3 solution produces insoluble salts with both CO32– ions and Cl– ions. However, Ag2CO3 is soluble in acids. Hence, the formation of white precipitate with AgNO3 solution in the presence of acids confirms the presence of Cl– ions. If AgNO3

6

’05

Describe chemical tests that can be used to verify the cations and anions in beaker 1 and beaker 2. Comments • Add a little sodium carbonate powder to 5 cm3 of solution from beaker 1. The evolution of a gas that turns limewater milky will verify the presence of hydrogen ions in the acid.

• Add 2 cm3 of barium nitrate solution to 5 cm3 of solution from beaker 1. The formation of a white precipitate will verify the presence of sulphate ions. • Add 1 cm3 of Nessler reagent to 5 cm3 of solution from beaker 2. The formation of a brown precipitate will verify the presence of ammonium ions. • Add a little of iron(II) sulphate and dilute sulphuric acid to 5 cm3 of solution from beaker 2, followed by concentrated sulphuric acid added slowly. The formation of a brown ring will verify the presence of nitrate ions.

8.2 2 Solid Y is not soluble in water but dissolves in dilute nitric acid and gives out a gas that turns limewater milky. The solution produced is yellow in colour and forms a brown precipitate when sodium hydroxide solution is added. (a) Give the name of salt Y. (b) Write an equation for the reaction between salt Y and dilute nitric acid. (c) Predict the reaction that would occur when salt Y is heated strongly. (d) Predict the observation that will occur when potassium thiocyanate solution is added to the yellow solution produced from the addition of nitric acid to salt Y.

1 The formulae of a few salts are given below: PbCl2, ZnSO4, Fe(NO3)3, ZnCO3, Na2CO3, Al2(SO4)3, CuCl2, CuCO3, Mg(NO3)2, Cu(NO3)2 Which of the above salts (a) is a solid that is insoluble in water? (b) is a white solid? (c) is soluble in water to produce a blue solution? (d) forms a white precipitate with barium chloride that is insoluble in acid? (e) forms a white precipitate that dissolves in sodium hydroxide solution? (f) forms a white precipitate with aqueous ammonia but is not soluble in excess aqueous ammonia?

Salts

254

(e) Write an equation for the formation of the brown precipitate when sodium hydroxide solution is added to an acidic solution of salt Y.

4 You are given three types of acids: sulphuric acid, nitric acid and hydrochloric acid. Using suitable chemical tests, describe briefly how you can identify the three types of acids.

3 Dilute nitric acid is added to aqueous solutions of unknown salts P, Q and R respectively. It is found that solutions P and Q do not show any noticeable change, whereas effervescence occurs in solution R. The gas that is evolved from solution R turns limewater milky. When silver nitrate solution is added to solution P, a white precipitate is formed. When barium nitrate is added to solution Q, a white precipitate is produced. Identify the anions that are present in solutions P, Q and R.

5 You are given three types of salts: zinc nitrate, lead(II) nitrate and calcium nitrate. Using suitable chemical tests, describe briefly how you can differentiate between the three types of salts.

8

6 An experiment was carried out to identify the cation and anion present in an unknown salt Z. The tests and observations are tabulated below. Fill in the correct inferences in the table and deduce the cation and anion present in salt Z.

Test

Observation

1 NaOH solution is added gradually to a little Z solution until in excess.

A white precipitate which is soluble in excess NaOH is produced.

2 Aqueous NH3 is added gradually to a little Z solution until in excess.

A white precipitate which is soluble in excess aqueous NH3 is produced.

3 Solid Z is heated slowly and then strongly.

A brown gas and a gas that rekindles a glowing wooden sp­lin­t are produced. The residue form­ed is yellow when hot and white when cooled.

4 Dilute nitric acid followed by BaCl2 solution is added to a little Z solution.

No noticeable change occurs.

8.3

Inference

Practising Systematic and Meticulous Methods when Carrying Out Activities

1 Correct methods in preparing salt crystals (a) A salt solution is evaporated until saturated so that solid salt may be crystallised. The salt is not heated until dry to prevent decomposition of the salt. (b) The salt crystals formed are removed by filtration and then rinsed with distilled water to remove any foreign ions present in the salt. 2 Correct methods in salt analysis (a) All observations must be recorded carefully and immediately after every test. The inferences are then recorded as soon as possible. (b) During heating, do not direct the mouth

of the test tube towards yourself or towards your fellow students. (c) Gases such as sulphur dioxide, nitrogen dioxide, chlorine, hydrogen chloride and ammonia are poisonous. Use suitable quan­tities as instructed. Do not use excess chem­­ icals. This will cause wastage of chemicals and require a longer time to carry out the experiment. Making accurate observations will be difficult. Handle all apparatus and chemicals carefully in the laboratory. (d) Addition of a reagent to a salt solution should be carried out drop by drop while shaking the test tube until no further change occurs.

255

Salts

1 A salt is an ionic compound that is formed when the hydrogen ion in an acid is replaced by a metal ion or ammonium ion (NH4 +). 2 The solubility of a salt in water depends on the types of cations and anions present.

8

Type of salt

3 Filtration can be used to separate an insoluble salt (as the residue) from a soluble salt (as the filtrate). 4 The methods of preparing salts depend on the solubility of salts. 5 Insoluble salts can be prepared by double decomposition reaction. Two aqueous solutions containing the cations and the anions are mixed together. The precipitate is then obtained by filtration. 6 The mole ratio of ions that react to form a salt can be determined from an experiment through the continuous variation method. 7 Qualitative analysis of salt is a scheme of tests carried out to identify the cation and anion present in the salt.

Solubility in water

Sodium, potassium and ammonium and nitrate salts

All are soluble

Chloride salts

All are soluble except PbCl2, AgCl and HgCl

Sulphate salts

All are soluble except PbSO4, BaSO4 and CaSO4

Carbonate salts

All are insoluble except NaCO3, K2CO3 and (NH4)2CO3

8 Multiple-choice Questions 8.1

Salts

1 Which of the following salts is insoluble in water? A Lead(II) nitrate B Calcium chloride C Magnesium carbonate D Iron(III) sulphate

4 Which is the best method to prepare ammonium nitrate? A Double decomposition B Neutralisation between an acid and an alkali C Reaction between an acid and a metal D Reaction between an acid and a metal carbonate

2 Which of the following chemicals is most suitable to ’06 react with HCl to prepare AgCl? A Silver carbonate B Silver nitrate C Silver metal D Silver oxide

5 A sample of zinc(II) sulphate bought contained some impurities. The impure salt can be purified by a process known as A distillation B evaporation C crystallisation D recrystallisation

3 Which following chemicals can be added to nitric acid to prepare copper(II) nitrate? A Copper metal B Copper(II) carbonate C Copper(II) chloride D Copper(II) sulphate

6 Which of the following salts can be prepared by double decomposition reaction? A Lead(II) nitrate B Aluminium sulphate C Silver chloride D Sodium sulphate

Salts

256

7 Which of the following pairs of solutions when added together will produce a precipitate? I Potassium carbonate and silver nitrate II Lead(II) nitrate and sodium chromate(VI) III Sodium carbonate and copper(II) sulphate IV Sodium sulphate and potassium carbonate A I and II only B II and III only C I, II and III only D I, III and IV only 8 Which of the following equations represent a double decomposition reaction? I Mg(NO3)2(aq) + Na2CO3(aq) → MgCO3(s) + 2NaNO3(aq) II CaCO3(s) + H2SO4(aq) → CaSO4(aq) + CO2(g) + H2O(l) III BaCl2(aq) + H2SO4(aq) → BaSO4(s) + 2HCl(aq)

9 If 20 cm3 of 0.5 mol dm–3 aqueous sodium chloride solution is added to 20 cm3 of 1.0 mol dm–3 silver nitrate solution, which of the following ions are present in the solution produced? I Na+ III NO3– II Ag+ IV Cl– A I and III only B II and III only C I, II and III only D I, II and IV only 10 4.2 g of magnesium carbonate reacts with excess hydrochloric acid to produce a salt. Which of the following are true about the reaction? [1 mol of gas occupies 24 dm3 at room temperature and pressure. Relative atomic mass: C, 12; O, 16; Mg, 24; Cl, 35.5] I Neutralisation reaction takes place. II 1.2 dm3 of gas is released. III Mass of salt formed is 4.75 g. IV 4.2 mol of water is formed. A I and II only B III and IV only C II and III only D I and IV only

B Nitric acid and barium chloride solution C Nitric acid and silver nitrate solution D Sodium hydroxide solution 14 Aluminium sulphate solution and zinc sulphate solution can be differentiated by the A addition of silver nitrate solution. B addition of aqueous ammonia. C addition of sodium hydroxide solution. D addition of barium chloride solution. 15 When lead(II) nitrate solution is added to solution X, a white precipitate is produced. Solution X may be I H2SO4 II HNO3

Qualitative Analysis of Salts

11 When solid X is heated strongly, a gas that turns limewater milky is produced, leaving a white residue. Which of the following may be solid X? A Lead(II) carbonate B Sodium carbonate C Zinc carbonate D Magnesium carbonate 12 Which of the following salts will produce a brown gas on heating? A Lead(II) bromide B Ammonium nitrate C Potassium nitrate D Zinc nitrate 13 Which of the following is used to test for sulphate ions? A Iron(II) sulphate solution and concentrated sulphuric acid

16 Solution M reacts with sodium hydroxide solution to form a white precipitate that is insoluble in excess sodium hydroxide solution. Solution M most probably contains I lead ion II calcium ion III aluminium ion IV magnesium ion A II and III only B I and III only C II and IV only D I, III and IV only

17 A student wants to identify cation X that is present in a salt solution. When ammonia solution is added into the salt solution, a green precipitate is ’11 formed. What is the next test that is needed and the expected observation to confirm cation X? Test

Observation

A

Add potassium iodide solution

Yellow precipitate is formed

B

Add potassium thiocyanate solution

Blood red colour is formed

C

Add potassium hexacyanoferrate(II) solution Add Nessler reagent

Light blue precipitate is formed

D

8.2

III HCl IV BaCl2 A I and III only B II and IV only C I, II and III only D I, III and IV only

Brown precipitate is formed

18 When a gas Z is passed into copper(II) sulphate solution, a blue precipitate is produced. Gas Z may be A ammonia C hydrogen chloride B chlorine D sulphur dioxide 19 When aqueous iron(III) chloride solution is added to reagent X, a blood red colour is produced. Reagent X may be A ammonium sulphite C potassium thiocyanate B potassium iodide D potassium hexacyanoferrate(II) 20 Hydrochloric acid can be differentiated from sulphuric acid by adding I barium nitrate III silver nitrate II barium hydroxide IV sodium carbonate A III only C II and IV only B I and II only D I, II and III only 21 When lead(II) nitrate is heated strongly in a test tube, the following can be observed. I A brown gas is evolved. II A gas that rekindles a glowing wooden splint is evolved. III A gas that changes moist blue litmus to red is evolved. IV A white residue is formed.

257

Salts

8

IV 2Zn(NO3)2(s) → 2ZnO(s) + 4NO2(g) + O2(g) A I and II only C II and III only B I and III only D II and IV only

8

A B C D

I and II only III and IV only I, II and III only I, II, III and IV

22 Which of the following ions will form a precipitate that dissolves in excess aqueous ammonia? I Copper(II) ions II Aluminium ions III Lead(II) ions IV Zinc ions A I and IV only B II and III only C II and IV only D II, III and IV only 23 Excess powdered carbonate of metal Z is added to sulphuric acid and stirred. After a few minutes, a light green solution is formed. Z could be a carbonate of A iron(II) C copper(II) B iron(III) D lead(II) 24 Which of the following reagents can be used to differentiate between sodium nitrate and potassium sulphate? I Lead(II) nitrate solution II Barium chloride solution III Silver nitrate solution IV Sodium hydroxide solution A I and II only B I and III only C II and IV only D I and IV only 25 When solution X is added to iron(III) sulphate solution, a brown precipitate is produced.

Solution X may be I sodium hydroxide II aqueous ammonia III Nessler reagent IV iron(III) nitrate A I and II only B III and IV only C I and IV only D I, II and III only

B A white precipitate, which dissolves in excess ammonia, is formed when aqueous ammonia is added. C A white precipitate is formed when lead(II) nitrate solution is added. D A white precipitate is formed when silver nitrate solution is added.

26 When solid X is heated strongly, a brown gas that turned moist blue litmus paper to red is evolved and a black residue is formed. Which of the following may be solid X? A Copper(II) oxide B Sodium nitrate C Copper(II) nitrate D Magnesium nitrate

29 When solution X is added to sodium chloride solution, a white precipitate is formed. The precipitate dissolves when it is heated with a little distilled water. Which of the following will be observed when solution X is added to a solution of sodium iodide solution? A A white precipitate is formed. B A yellow precipitate is formed. C A brown solution is formed. D A purple solution is formed.

27 Lead(II) nitrate solution and aluminium sulphate solution can be distinguished respectively by adding I sodium hydroxide solution II potassium sulphate solution III barium nitrate solution IV sodium iodide solution A I and III only B II and III only C II and IV only D III and IV only

30 Which of the following reagents can be used to differentiate Fe2+ ions from Fe3+ ions? I Potassium iodide solution II Potassium thiocyanate solution III Potassium hexacyanoferrate(III) solution IV Potassium manganate(VII) solution A I and II only B III and IV only C I, II and III only D II, III and IV only

28 Which of the following observations is true for both sodium chloride solution and zinc sulphate solution? A A white precipitate is formed when barium nitrate solution is added.

Structured Questions 1 Diagram 1 is a flowchart showing a series of reactions starting from lead(II) oxide. Lead(II) oxide

Precipitate F

sodium hydroxide

substance A Lead(II) nitrate

heat

Gas B + Gas C + Solid D

potassium iodide Precipitate E Diagram 1 (a) (i) Name substance A that is used to react with lead(II) oxide to produce lead(II) nitrate. (ii) Write a chemical equation for the reaction that takes place in (i). Salts

258

[1 mark] [1 mark]

(b) Gas B is a brown gas while gas C is colourless. (i) Identify gas B and solid D. (ii) Suggest a test that you can use to test the presence of gas C. (c) (i) Name precipitate E. What is the colour of precipitate E? (ii) Write the ionic equation for the formation of precipitate E. (d) (i) Name precipitate F. (ii) Predict the observation that will occur if excess sodium hydroxide is added to precipitate F. (e) Name another chemical that can replace lead(II) oxide to react with substance A to produce lead(II) nitrate.

[2 marks] [1 mark] [1 mark] [2 marks] [1 mark] [1 mark] [1 mark]

2 Diagram 2 is a flowchart showing a series of reactions starting from magnesium oxide. Excess magnesium oxide

8

+ Nitric acid stir and filter Residue A

Filtrate Process 1: evaporate, cool and filter

Process 2: heated strongly

Crystal B

Solid C + Gas D + Gas E Diagram 2

(a) Write a chemical equation showing the reaction between magnesium oxide and nitric acid. [1 mark] (b) If 50 cm3 of 2.0 mol dm–3 nitric acid is added to excess magnesium oxide, calculate the maximum mass of salt that can be produced. [Relative atomic mass: H, 1; N, 14; O, 16; Mg, 24] [3 marks] (c) The filtrate formed from the reaction above is colourless. (i) Predict what will be observed if aqueous sodium hydroxide is added to the filtrate gradually until in excess. [2 marks] (ii) Name crystal B. [1 mark] (d) In Process 2, the gas D produced is colourless while gas E is brown. (i) Identify gas E and name solid C. [2 marks] (ii) Suggest a test that you can use to test the presence of gas D. [1 mark] 3 The flowchart of Diagram 3 shows a series of reactions I to IV carried out to identify the ions present in compound A.

Diagram 3 (a) (i) Identify gas C. (ii) What is the anion present in compound A? (b) Based on the observation obtained in reaction III and reaction IV, predict the cation present in compound A. (c) Based on (a) and (b), write a balanced equation for the action of heat on compound A. (d) Predict the colour of residue B when it is hot and when it is cooled. (e) (i) Write a balanced equation for the formation of solution D in reaction II. (ii) State an observation that can be noted in reaction II.

259

[1 mark] [1 mark] [1 [1 [1 [1 [1

mark] mark] mark] mark] mark] Salts

(f) (i) Name the white precipitate F formed in reaction IV. (ii) Write an ionic equation for the formation of the white precipitate F. (g) Predict the observation that can be obtained when aqueous sodium hydroxide solution is added until in excess to solution E.

[1 mark] [1 mark] [1 mark]

4 Diagram 4 shows the steps involved in the preparation of zinc carbonate. Zinc oxide

Step 1 add hydrochloric acid

Step 2

Salt solution P

Zinc carbonate

add solution Q

8

Diagram 4 (a) Write a balanced equation for the formation of salt solution P. (b) Explain briefly how you can obtain a solution of salt P. (c) (i) Name solution Q that is required to be added to salt solution P in Step 2 to produce zinc carbonate. (ii) Name the type of reaction involved in Step 2. (iii) Write an ionic equation for the formation of zinc carbonate. (d) 30 cm3 of 2.0 mol dm–3 hydrochloric acid is reacted with excess zinc oxide. [Relative atomic mass: C, 12; O, 16; Zn, 65] (i) Calculate the number of moles of salt P that is formed. (ii) Calculate the maximum mass of zinc carbonate that is produced. (e) Suggest how you would convert zinc carbonate back to zinc oxide.

[1 mark] [2 marks] [1 mark] [1 mark] [1 mark]

[2 marks] [2 marks] [1 mark]

5 The flowchart in Diagram 5 shows the result of a qualitative analysis that is carried out on a mixture of metal Q and a water soluble salt P.

Salt P + metal Q

Gas R

add NaOH and heat

Salt P

add HNO3 + BaCl2

Process A Metal Q

add HCl

White precipitate S

Gas T + Solution U

Diagram 5 (a) Name the process A that is used to separate salt P and metal Q. (b) Gas R is a gas that can change red litmus paper to blue. Name gas R. Consequently what is the cation present in solution P? (c) Name the white precipitate S. What is the anion present in solution P? (d) When gas R is passed into solution U, a white precipitate is first formed but dissolves when excess gas R is passed through. Identify the cation present in solution U. (e) From your answer in (d), determine the identity of metal Q and gas T.

[1 mark] [2 marks] [2 marks] [1 mark] [2 marks]

Essay Questions 1 (a) The following are three examples of salts that can be prepared in the laboratory. ’07 • Sodium sulphate, Na2SO4 • Lead(II) chloride, PbCl2 • Magnesium nitrate, Mg(NO3)2 (i) From these examples, identify the soluble and insoluble salts. [3 marks] (ii) State the reactants for the preparation of the insoluble salt in (i). [3 marks] (b) With the aid of a labelled diagram, explain the crystallisation method for preparing a soluble salt from its unsaturated solution. [6 marks] (c) Table 1 shows the observations from tests carried out on salt X. Salts

Test

Procedure

Observation

I

Heating of salt X solid

A brown gas is given off and a residue that is brown when hot, yellow when cooled is formed.

II

Add excess aqueous sodium hydroxide solution to salt X solution until in excess

A white precipitate which is soluble in excess sodium hydroxide solution is formed.

Table 1

260



Describe a laboratory experiment to prepare the salt. In your description, include the chemical equations involved. [8 marks] (c) Three beakers with solutions labelled X, Y and Z may contain the following salt solutions:

Based on the information in Table 1: (i) Identify an anion that is present in Test I and describe a chemical test to verify the anion. [4 marks] (ii) Identify three cations that are present in Test II. Describe a chemical test to verify the cation present in salt X based on Test I and Test II. [5 marks]

• Zinc sulphate • Zinc nitrate • Magnesium sulphate

2 (a) What is meant by a salt? [2 marks] (b) You are required to prepare a sample of dry lead(II) carbonate salt. The following chemicals are supplied: • Sodium carbonate solution • Dilute nitric acid • Lead metal powder

You are provided only with ammonia and barium nitrate solutions. Describe how you could differentiate between the 3 salt solutions by using the two reagents provided. Include your observations and conclusions. [10 marks]

Experiment 1 Barium nitrate solution, Ba(NO3)2 and aqueous potassium chromate(VI), K2CrO4 solution react to produce a yellow preci­pi­tate with the formula BaCrO4. A student has carried out an experiment to measure the height of ’03 the precipitate produced by the reaction between 5.0 cm3 of 0.50 mol dm–3 aqueous barium nitrate solution and aqueous potassium chromate(VI) solution of unknown concentration at different volumes. The data of the experiment obtained is shown in Table 1. Test tubes

Volume of barium nitrate solution (cm3)

Volume of potassium chromate(VI) solution (cm3)

Height of precipitate (cm)

1

5.0

1.0

0.5

2

5.0

2.0

1.0

3

5.0

3.0

1.4

4

5.0

4.0

1.9

5

5.0

5.0

2.4

6

5.0

6.0

2.4

7

5.0

7.0

2.4

Table 1 (ii) Calculate the number of moles of barium ions that is used in this experiment. [3 marks]

(a) Suggest a suitable apparatus that can be used to measure the volume of barium nitrate solution and potassium chromate(VI) solution in this experiment. [3 marks]

(e) Based on the formula of the yellow precipitate given, calculate the number of moles of chromate(VI) ions used in the volume in (d) (i). Hence calculate the molarity of the aqueous potassium chromate(VI), K2CrO4, solution used in this experiment. [3 marks]

(b) State the manipulated variable, responding variable and constant variable in this experiment. [3 marks]

(c) Draw a graph of the height of the precipitate against the volume of potassium chromate(VI) solution. [3 marks]

(f) Explain why the height of the precipitate does not change when 5.0 cm3, 6.0 cm3 and 7.0 cm3 of potassium chromate(VI) solution is added to 5.0 cm3 of 0.5 mol dm–3 barium nitrate solution in Table 1. [3 marks]

(d) (i) What is the minimum volume of potassium chromate(VI) solution required to react completely with 5.0 cm3 of 0.5 mol dm–3 barium nitrate solution?

261

Salts

8



FORM 4 THEME: Production and Management of Manufactured Chemicals

CHAPTER

9

Manufactured Substances in Industry SPM Topical Analysis 2008

Year 1

Paper Section Number of questions

3

2009

2 A

B

C

1

–

–

3

1

–

2

2010

2 A

B

C

1

–

–

3

1

–

5

2011

2 A

B

C

–

–

–

3

1

–

3

3

2 A

B

C

–

1

–

ONCEPT MAP

MANUFACTURED SUBSTANCES IN INDUSTRY

Sulphuric acid Uses: To make fertilisers, polymers, detergents, pigments Manufactured by the Contact process: • Temperature: 450–550 °C • Pressure: 1 atmosphere • Catalyst: V2O5

Ammonia Uses: To make fertilisers, nitric acid and as a cooling agent (refrigerant) Manufactured by the Haber process: • Temperature: 450–550 °C • Pressure: 200–500 atmospheres • Catalyst: Iron

Synthetic polymers Examples and uses: • Polythene: To make plastic bags, plastic containers and toys. • Polypropene: To make plastic bottles, tables and chairs • PVC: To make water pipes, raincoats and wire casing

Glass and ceramics Uses: Construction materials, household items, laboratory apparatus Types of glass: • Fused glass • Soda glass • Borosilicate glass • Lead glass

Alloys Purposes: • To increase hardness and strength • To prevent corrosion • To improve the appearance Examples: Steel, bronze, brass, magnalium, pewter

Composite materials Examples: • Reinforced concrete • Superconductor • Fibre optic • Fibreglass • Photochromic glass

–

9.1

Sulphuric Acid

1 The manufacture of sulphuric acid is one of the most important chemical industries at the present time.

2 Sulphuric acid, H2SO4 is a non-volatile diprotic acid. 3 Concentrated sulphuric acid is a viscous colourless liquid.

Manufacture of fertilisers 1 Calcium dihydrogen phos­ phate (super­ phosphate) is prepared from the reaction between sulphuric acid and tricalcium phosphate:

ammonium sulphate

3 Potassium sulphate is prepared from the reaction between sulphuric acid and potassium hydroxide.

2H2SO4 + Ca3(PO4)2 → Ca(H2PO4)2 + 2CaSO4 calcium dihydrogen phosphate

H2SO4 + 2KOH → K2SO4 + 2H2O

2 Ammonium sulphate is prepared from the reaction between sulphuric acid and aqueous ammonia. Manufacture of detergents (synthetic cleaning agents) Sulphuric acid reacts with hydrocarbon to produce sulphonic acid. Sulphonic acid is then neutralised with sodium hydroxide to produce the detergent.

potassium sulphate

Manufacture of white pigment in paint barium sulphate, BaSO4

The Uses of Sulphuric Acid in Daily Life

Manufacture of synthetic fibres (polymers) Rayon is an example of a synthetic fibre that is produced from the action of sulphuric acid on cellulose.

The Industrial Process in the Manufacture of Sulphuric Acid

In school laboratories 1 As a strong acid 2 As a drying or dehydrating agent 3 As an oxidising agent 4 As a sulphonating agent 5 As a catalyst

The neutralisation between sulphuric acid and barium hydroxide produces barium sulphate.

sulphur or metal sulphide

SPM

’07/P2

burned in air sulphur dioxide, SO2

1 Sulphuric acid is manufactured by the Contact process in industry. 2 The raw materials used in the Contact process are sulphur (or sulphide minerals), air and water. 3 The flowchart of the Contact process is shown in Figure 9.1. It describes how sulphur or metal sulphide is converted to concentrated sulphuric acid, H2SO4 through regulated steps in the process.

(i) V2O5 as the catalyst (ii) temperature of 450 °C – 550 °C (iii) pressure of 1 atmosphere sulphur trioxide, SO3 dissolved in concentrated H2SO4 oleum, H2S2O7 diluted with equal volume of H2O concentrated sulphuric acid, H2SO4

Figure 9.1 Flowchart of the Contact process

263

Manufactured Substances in Industry

9

H2SO4 + 2NH3 → (NH4)2SO4

The Contact process involves three stages I

SPM

’04/P2

II

III

sulphur ⎯→ sulphur dioxide ⎯→ sulphur trioxide ⎯→ sulphuric acid

9

Figure 9.2 shows the three stages in the manufacture of sulphuric acid by the Contact process in industry.

Figure 9.2 The production of concentrated sulphuric acid in industry.

Stage I

Stage II

Conversion of sulphur dioxide to sulphur trioxide, SO3 1 The sulphur dioxide gas is dried and purified before being added to dry air to produce sulphur trioxide gas. This is (a) to remove water vapour in the air (the reaction of water with SO3 will produce heat that will vaporise the acid), (b) to remove contaminants such as arsenic compounds (found in the sulphur or sulphide minerals) that will poison the catalyst and make it ineffective. 2 Pure and dry sulphur dioxide with excess dry oxygen (from air) are passed through a converter. 3 A high percentage (98%) of sulphur dioxide is converted into suphur trioxide under the following conditions: (i) The presence of vanadium(V) oxide, V2O5, as a catalyst (ii) A temperature of between 450°C – 550°C (iii) A pressure of one atmosphere

Production of sulphur dioxide gas, SO2 This can be done by two methods: 1 Burning of sulphur in dry air in the furnace. S + O2 → SO2 2 Burning of metal sulphide such as zinc sulphide or iron(III) sulphide in dry air. 2ZnS + 3O2 → 2SO2 + 2ZnO

The production of sulphuric acid is known as the Contact process because the molecules of sulphur dioxide, SO2 and oxygen, O2 are in contact with the catalyst, V2O5. The use of V2O5 has replaced the use of platinum as a catalyst because (a) platinum is much more expensive, (b) platinum is easily poisoned (lose its catalyst effect) by arsenic compounds.

Manufactured Substances in Industry

SPM

’09/P1

2SO2 + O2

264

2SO3

Stage III Production of sulphuric acid 1 In the absorber, sulphur trioxide is dissolved in concentrated sulphuric acid to produce oleum, H2S2O7, a viscous liquid. SO3 + H2SO4 → H2S2O7

H2S2O7 + H2O → 2H2SO4 3 The two reactions in stage III are equivalent to adding sulphur trioxide to water. SO3 + H2O → H2SO4 However, sulphur trioxide is not dissolved directly in water to produce sulphuric acid. This is because (a) SO3 has a low solubility in water. (b) SO3 reacts violently in water, producing a large amount of heat which will vapourise sulphuric acid to form acid mist. The mist is corrosive, pollutes the air and is difficult to condense.

The burning of fossil fuels causes environmental pollution, producing air pollutants such as carbon monoxide, nitrogen monoxide, nitrogen dioxide and sulphur dioxide. Scientists are searching for alternative sources of energy to replace fossil fuels. Other than wind energy, solar energy, geothermal energy (energy from earth’s internal heat) and nuclear energy, scientists are also researching into the production of fuels from natural products such as palm oil.

Environmental Pollution by Sulphur Dioxide 1 Sulphur dioxide is the intermediate product of the Contact process. 2 Sulphur dioxide is also produced during volcanic eruptions. 3 However, the main source of sulphur dioxide is from the burning of fossil fuels such as petroleum. Most of the fossil fuels contain some sulphur. Hence sulphur dioxide is produced when fossil fuels are burned. 4 Waste gases from factories and extraction of metal from their sulphide ores also release sulphur dioxide into the atmosphere. 5 The burning of products manufactured from sulphuric acid such as rayon will also produce sulphur dioxide gas. 6 Sulphur is a poisonous and acidic gas that can cause environmental pollution. Inhaling sulphur dioxide affects the respiratory system. It can cause lung problems such as coughing, chest pains, shortness of breath and bronchitis. 7 Sulphur dioxide gas dissolves in atmospheric water to produce sulphurous acid, H2SO3 and sulphuric acid, H2SO4. The presence of these acids in rain water causes acid rain.

265

SO2 + H2O ⎯⎯→ H2SO3 2SO2 + O2 + 2H2O ⎯⎯→ 2H2SO4 8 The effects of acid rain are as follows: (a) Corrodes concrete buildings and metal structures (b) Destroys trees and plants in forest (c) Decreases the pH of the soil which becomes acidic, unsuitable for growth of plants and destroys the roots of plants (d) Reacts with minerals in the soil to produce salts which are leached out of the top soil; essential nutrients for plants growth are depleted (plants die of malnutrition and diseases) (e) Acid rain flows into lakes and rivers. This increases the acidity of water and may kill fish and other aquatic living things. 9 Two methods to reduce sulphur dioxide from the atmosphere: (a) Use low sulphur fuels to reduce the emission of sulphur dioxide in exhaust gases. (b) Remove sulphur dioxide from waste air by treating it with calcium carbonate before it is released.

Manufactured Substances in Industry

9

2 Oleum is then diluted with an equal volume of water to produce concentrated sulphuric acid (98%).

The industrial synthesis of sulphuric acid is represented by the flowchart. (a) Name this process. (b) Name gas X, gas Y and liquid Z. (c) Write balanced equations for Step 1 and Step 2. (d) State the optimum conditions involved in Step 2. (e) Why is gas Y not dissolved in water to produce sulphuric acid?

9

Manufacture by Contact process: • 2SO2 + O2 2SO3 • Temperature: 450 °C – 550 °C • Pressure : 1 atm • Catalyst : V2O5 Uses of Sulphuric Pollution by SO2: sulphuric acid: • Forms acid acid • Making paint rain pigments, detergents • Causes breathing and fertilisers problems and • As an electrolyte lung diseases in accumulators

9.2

1 Ammonia, NH3 is a very important compound in industry. 2 The main uses of ammonia: (a) To manufacture nitrogenous fertilisers such as ammonium sulphate, ammonium nitrate and urea (b) The liquid form is used as a cooling agent (refrigerant) in refrigerators (c) as a raw material for the manufacture of nitric acid in the Ostwald process (d) to be converted into nitric acid used for making explosives (e) as an alkali to prevent the coagulation of latex so that latex can remain in the liquid form (f) to produce ammonium chloride used as an electrolyte in dry cells (g) as a cleaning agent to remove grease (h) used in the manufacture of synthetic fibres such as nylon 3 The manufacture of nitrogenous fertilisers: SPM (a) Ammonium sulphate ’07/P2 Ammonia reacts with sulphuric acid by neutralisation to produce ammonium sulphate.

9.1 1 An important use of sulphuric acid is in the production of fertilisers. (a) Name the fertilisers produced and write the equations involved when sulphuric acid reacts with (i) aqueous ammonia (ii) potassium hydroxide (b) Barium sulphate, BaSO4, is a white pigment in paint. Write an equation for the formation of barium sulphate from the reaction between sulphuric acid and barium hydroxide. 2 Give three uses of sulphuric acid in daily life. 3 Concentrated sulphuric acid can be used as a dehydrating agent. (a) What is the function of a dehydrating agent? (b) When concentrated sulphuric acid is added to glucose, C6H12O6, a black residue is formed after a few minutes. (i) What is the black residue? (ii) Write a balanced equation for the reaction that has taken place. 4

2NH3 + H2SO4 ⎯⎯→ (NH4)2SO4 ammonium sulphate

Sulphur +

step 1 heating

Gas X

step 2

(b) Ammonium nitrate Ammonia reacts with nitric acid by neutralisation to produce ammonium nitrate.

Gas Y

Oxygen Sulphuric acid

Manufactured Substances in Industry

dilute with water

Ammonia and Its Salts

NH3 + HNO3 ⎯⎯→ NH4NO3

Liquid Z

ammonium nitrate

266

(c) Urea Ammonia reacts with carbon dioxide at a temperature of 200 °C and a pressure of 200 atmospheres to produce urea.

NH3 + H2O

NH4+ + OH–

4 As ammonia is very soluble in water, an inverted filter funnel is used to prevent the suction of water (Figure 9.3).

2NH3 + CO2 ⎯→ (NH2) 2CO + H2O urea

9

4 Liquid ammonia is suitable for use as a cooling agent (refrigerant) in refrigerators because it has a low boiling point and is very volatile. 5 In the Ostwald process, ammonia is converted into nitric acid by the following steps: (a) Ammonia is oxidised to nitrogen monoxide gas in the presence of platinum as the catalyst.

Figure 9.3 To dissolve ammonia gas in water

platinum

5 Ammonia gas reacts with hydrogen chloride gas to form white fumes of ammonium chloride. (This is used as a test for ammonia gas).

4NH3 + 5O2 ⎯⎯→ 4NO + 6H2O

(b) Nitrogen monoxide is further oxidised to nitrogen dioxide.

NH3 + HCl ⎯⎯→ NH4Cl

2NO + O2 ⎯⎯→ 2NO2

6 Ammonia is alkaline in property and reacts with dilute acids in neutralisation to produce salts. For example:

(c) Nitrogen dioxide is dissolved in water to produce nitric acid. 4NO2 + O2 + 2H2O ⎯⎯→ 4HNO3

2NH3 + H2SO4 → (NH4)­2SO4 NH3 + HNO3 → NH4NO3

6 Nitric acid is used to make explosives such as TNT when nitric acid reacts with organic substances such as methylbenzene (common name: toluene). 7 Ammonia can neutralise the organic acids that are produced by microorganisms in latex. Thus it is used to maintain latex in the liquid form. 8 Ammonia reacts with hydrogen chloride to produce ammonium chloride which is used as the electrolyte in dry cells.

7 Aqueous solutions of ammonia produces OH– ions to react with metal ions (except Na+ ion, K+ ion and Ca2+ ion) forming precipitates of metal hydroxides. Fe3+ + 3OH– → Fe(OH)3

brown precipitate

Mg2+ + 2OH– → Mg(OH)2

white precipitate

NH3 + HCl → NH4­Cl

8 Some metal hydroxides such as zinc hydroxide and copper(II) hydroxide dissolve in excess aqueous ammonia to form complexes. For example:

The Properties of Ammonia 1 Ammonia is a colourless and pungent gas. It is less dense than air. 2 Ammonia changes moist red litmus paper to blue. Thus ammonia is an alkaline gas. 3 Ammonia dissolves in water to produce a weak alkali. A 0.1 mol dm–3 ammonia solution has a pH value of about 10.

Zn(OH)2 + 4NH3 → [Zn(NH3)4]2+ + 2OH– Cu(OH)2 + 4NH3 → [Cu(NH3)4]2+ + 2OH–

267

Manufactured Substances in Industry

To investigate the properties of ammonia gas Apparatus pH paper, red litmus paper, beaker, glass rod, test tubes, U-tube with soda lime, beaker and delivery tube.

9

Materials Ammonium chloride, calcium hydroxide, concen­ trated hydrochloric acid, distilled water. Procedure 1 A spatula of ammonium chloride and a spatula of calcium hydroxide are put in a test tube. The test tube is then connected to a U-tube with soda lime as shown in Figure 9.4.

2 The mixture in the test tube is heated. The ammonia gas produced is tested in turn by using (a) a piece of moist red litmus paper, (b) a glass rod dipped in concentrated hydro­ chloric acid. 3 Ammonia gas is collected in inverted test tubes using downward displacement of air. The test tubes are then stoppered immediately. 4 A little distilled water is added to the test tube of ammonia gas which is then shaken. The solution formed is tested with a piece of pH paper. 5 A test tube of ammonia is inverted with its mouth below a beaker of water. The stopper of the test tube is then removed as shown in Figure 9.5. Observation made is recorded.

Figure 9.5 Testing the solubility of ammonia in water

Figure 9.4 Preparation of ammonia gas

Results

Activity 9.1

Test

Observation

Inference

1 With moist red litmus paper

The red litmus paper changed to a blue colour

Ammonia gas is alkaline

2 With hydrogen chloride vapour

White fumes are formed

Ammonia gas forms white fumes of ammonium chloride with hydrogen chloride gas

3 With pH paper

pH paper changed to a blue colour, corresponding to a pH value of about 10

Aqueous ammonia is a weak alkali

4 A test tube of ammonia is inverted in a beaker of water

Water rushes in and fills up the whole test tube

Ammonia gas is very soluble in water

Discussion 1 Ammonia gas is produced when an ammonium salt such as ammonium chloride is heated with an alkali such as calcium hydroxide. (a) The equation for the reaction between ammonium chloride and calcium hydroxide is 2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2O Manufactured Substances in Industry

(b) The ionic equation for the reaction between ammonium salt and alkali is NH4+ + OH– → NH3 + H2O 2 Soda lime in the U-tube is used as a drying agent to dry the ammonia gas produced. In place of soda lime, anhydrous calcium oxide can also be used as a drying agent. However, concentrated sulphuric acid cannot be used to dry ammonia gas as it will react with ammonia in a neutralisation reaction. 268

3 Ammonia gas is collected in inverted test tubes using downward displacement of air because ammonia is less dense than air. 4 Ammonia gas is very soluble in water. As such it cannot be collected by a downward displacement of water. 5 Ammonia gas reacts with hydrogen chloride to form white fumes of ammonium chloride.

Conclusion 1 Ammonia is an alkaline gas which turns red litmus paper to blue and dissolves in water to form a weak alkali. 2 Ammonia gas reacts with hydrogen chloride to form white fumes of ammonium chloride. 3 Ammonia gas is very soluble in water.

The Industrial Process in the Manufacture of Ammonia

4 In the Haber process: SPM (a) A mixture consisting of one volume of ’06/P2 nitrogen gas and three volumes of pure and dry hydrogen gas is compresssed to a pressure between 200 – 500 atmospheres. (b) The gas mixture is passed through a catalyst of powdered iron at a temperature of 450 – 550 °C. (c) At this optimum temperature and pressure, ammonia gas is produced.

1 The Haber process is the industrial method used to prepare ammonia gas on a large scale using nitrogen gas and hydrogen gas.

N2 + 3H2

2NH3

5 The gas mixture produced consists of about 17% ammonia gas. The ammonia gas is liquefied when the gas mixture is cooled. The unreacted nitrogen gas and hydrogen gas are pumped back to the catalytic column to be reacted again.

In 1918, Fritz Haber was awarded the Nobel prize for his discovery of the in­dustrial manufacture of ammonia gas from hydrogen gas and nitrogen gas

2 Nitrogen gas used in the Haber process is obtained from the fractional distillation of liquid air. 3 Hydrogen gas used in the Haber process can be obtained by two methods: (a) The reaction between steam and heated coke (carbon). H2O + C ⎯⎯→ CO + H2 This mixture is known as water gas (b) The reaction between steam and natural gas (consists mainly of methane, CH4).

Figure 9.6 The manufacture of ammonia gas by the Haber process

2H2O + CH4⎯⎯→ CO2 + 4H2

269

Manufactured Substances in Industry

9

NH3 + HCl → NH4Cl

(a) Based on the graph, what is the effect of temperature and pressure on the percentage of ammonia produced? (b) State one problem each that is accompanied with the effects mentioned in (a) when the factory wants to increase the percentage of ammonia produced. (c) Discuss how the problems in (b) may be overcome by using an optimum temperature and pressure.

SPM

’06/P2 Q5

1 The reaction between nitrogen gas and hydrogen gas to produce ammonia gas is a reversible reaction. This means that not all the nitrogen gas will react with hydro­gen gas completely. The optimum temperature and pressure are used to ensure a satisfactory yield of ammonia. 2 The percentage yield of ammonia gas depends on the temperature and pressure used, as shown in Figure 9.7 below.

9

Solution (a) Lower temperature will increase the percentage of ammonia produced. Higher pressure will increase the percentage of ammonia produced. (b) Lower temperature will lower the rate of reaction. Higher pressure will increase the cost of production. (c) Using an optimum temperature of 450-550 °C, the rate of reaction will not be too low and yet produce a high yield of ammonia. Using an optimum pressure of 200-500 atm, the cost of increasing the pressure and using stronger pipes will not be too high to produce a high yield of ammonia.

Figure 9.7 Effects of temperature and pressure on the yield of ammonia (a) The lower the temperature, the higher the yield of ammonia. (b) The higher the pressure, the higher the yield of ammonia. (c) However, low temperature will lower the rate of reaction. High pressure will increase the cost of production. (d) Hence, the most suitable temperature and pressure to produce a high yield of ammonia is a temperature of 450 °C–550 °C and a pressure of 200–500 atmospheres.

1

Ammonium Fertilisers 1 Plants require nitrogen to produce protein. Nitrogen is absorbed by plants in the form of nitrates, NO3– which are soluble in water. 2 Ammonium fertilisers are chemical fertilisers added to the soil to replace the elements in soil used up by plants. 3 Ammonium fertilisers contain ammonium ions, NH4+, that can be converted into nitrate ions by bacteria living in the soil. 4 The effectiveness of ammonium fertilisers is determined by the percentage of nitrogen by mass in them. The fertiliser with a higher percentage of nitrogen is more effective. 5 The percentage of nitrogen by mass can be calculated using the following formula:

’06

A factory that is manufacturing ammonia carried out a test to determine the percentage of ammonia that can be produced at two different conditions: I and II. The results are shown in the graph below.

Percentage of nitrogen by mass

Manufactured Substances in Industry

270

Mass of nitrogen = —————————————————————  100% Molar mass of fertiliser

1

SPM

’10/P1

Calculate the percentage by mass of nitrogen in ammonium sulphate, (NH4)2SO4. [Relative atomic mass: N,14; H.1; S,32; O,16]

= 2[14 + 4 (1)] + 32 + 4(16) = 132 1 mol (NH4)2SO4 consists of 2 mol atoms of nitrogen. Percentage of nitrogen in 1 mol of (NH4)2SO4 2(14) =— — — — 100% = 21.2% 132

Solution Relative molecular mass of (NH4)2SO4

(B) To prepare ammonium sulphate crystals 1 25 cm3 of 2.0 mol dm–3 aqueous ammonia solution is pipetted into a clean conical flask. 2 (V2 – V1) cm3 of sulphuric acid is added from the burette to the aqueous ammonia solution. 3 The mixture in the conical flask is transferred to a beaker and is slowly evaporated until a saturated solution is formed. 4 The saturated solution is left to cool. White crystals of ammonium sulphate are produced.

5 The ammonium sulphate crystals are then removed by filtration, washed with distilled water and dried between filter papers.

Figure 9.8 Titration of sulphuric acid with ammonia solution

Discussion 1 The equation for the neutralisation of aqueous ammonia and sulphuric acid is 2NH3 + H2SO4 → (NH4)2SO4 2 The first titration (Experiment A) is carried out to determine the volume of sulphuric acid required to completely neutralise 25 cm3 of aqueous ammonia. 3 In the second titration (Experiment B), the methyl orange indicator is not used so that the salt produced is not contaminated with the indicator. 4 The ammonium sulphate solution produced is not evaporated until dry because ammonium sulphate solid will decompose when heated. 5 Usually, the mass of ammonium sulphate crystals obtained from the experiment is less than the theoretical value because not all ammonium sulphate crystals can be crystallised from the solution. Some ammonium sulphate remains dissolved in the solution. Conclusion Ammonium sulphate, an example of ammonium fertiliser, can be prepared by the neutralisation reaction between aqueous ammonia and dilute sulphuric acid. 271

Manufactured Substances in Industry

Activity 9.2

Apparatus 25 cm3 pipette, 50 cm3 burette, dropper, retort stand with clamp and white tile. Materials 1 mol dm–3 sulphuric acid, 2 mol dm–3 aqueous ammonia solution and methyl orange indicator. Procedure (A) To determine the volume of sulphuric acid re­ quired to neutralise 25 cm3 of ammonia solution 1 25 cm3 of 2.0 mol dm–3 aqueous ammonia solution is transferred using a 25 cm3 pipette to a clean conical flask. Three drops of methyl orange indicator are added to the alkali and the colour of the solution is noted. 2 A 50 cm3 burette is filled with sulphuric acid and clamped to a retort stand. The initial burette reading (V1) is recorded. 3 The conical flask containing 25 cm3 of the aqueous ammonia is placed below the burette. A piece of white tile is placed below the conical flask for clearer observation of the change in colour (Figure 9.8). 4 Sulphuric acid is added slowly from the burette to the aqueous ammonia solution in the conical flask while the flask is gently swirled. 5 Titration is stopped when methyl orange changes colour from yellow to orange. The final burette reading (V2) is recorded. 6 The volume of sulphuric acid required to neutralise 25.0 cm3 of ammonia solution is (V2 – V1) cm3.

9

To prepare ammonium sulphate, (NH4)2SO4, an ammonium fertiliser

Ammonium sulphate is an acidic salt. Hence, long term use of ammonium sulphate as a fertiliser will increase the acidity in the soil. This can be overcome by adding quicklime (calcium oxide) to neutralise the acidic soil.

9

Manufacture of ammonia: Haber Process N2(g) + 3H2(g) 2NH3(g) • Temperature: 450 °C–550 °C • Catalyst: iron powder • Pressure: 200–500 atm. Uses • Manufacture of fertilisers, nitric acid

Ammonia, NH3

Tests • Turns moist red litmus paper blue • Forms white fumes with HCI gas Properties • Colourless • A weak alkali • Pungent smell • Very soluble in water

Reactions • Produce ammonium salts with acids • Produce metal hydroxide as precipitate

9.2 The given diagram shows the production of a fertiliser, ammonium nitrate. Step 2 is known as the Ostwald Process. (a) Name gas A, gas B and acid C in the diagram. (b) Name the industrial process in the production of gas B in step 1. (c) State the source from which gas A is obtained. (d) What will be observed if gas B comes in contact with hydrogen chloride gas? (e) Write a balanced equation for step 3. (f) Calculate the percentage by mass of nitrogen in ammonium nitrate. [Relative atomic mass: H,1; N,14; O,16]

1 Ammonia is commercially produced by the Haber process. (a) Name the raw materials used in the production of ammonia gas. (b) Name the catalyst used in the Haber process. (c) State the optimum temperature and pressure used for this process. (d) Write a balanced equation for this process. (e) State two uses of ammonia in daily life. 2

Gas A +

step 1

gas B

H2

step 2

acid C Ostwald process step 3 ammonium nitrate

9.3

2 Pure metals are weak and soft because the arrangement of atoms in pure metals makes them ductile and malleable. 3 Arrangement of pure metal atoms A pure metal contains atoms of the same size arranged in a regular and organised closedpacked structure (Figure 9.9).

Alloys

Meaning and Purpose of Making Alloys 1 An alloy is a mixture of two or more elements with a certain composition in which the major component is a metal. Manufactured Substances in Industry

272

(b) In an alloy, these atoms of different sizes disrupt the orderly arrangement of the metal atoms and also fill up any empty spaces in the metal crystal structure. (c) Hence, the layers of metal atoms are prevented from sliding over each other easily. This makes the alloy harder and stronger, less ductile and less malleable than its pure metals.

Figure 9.9 Arrangement of atoms in a pure metal

9

4 Weakness of pure metal (a) Ductility Pure metals are soft because the orderly arrangement of atoms of the same size enables the layers of atoms to slide over each other easily when an external force is applied on them. This makes the metals ductile and metals can be drawn to form long wires (Figure 9.10).

Figure 9.12 Arrangement of atoms in alloys

6 The properties of a pure metal are thus improved by making them into alloys. There are three aims of alloying a pure metal: (a) To increase the hardness and strength of a metal (b) To prevent corrosion or rusting (c) To improve the appearance of the metal surfaces, with a better finish and lustre

Figure 9.10 Ductility of pure metal

(b) Malleability There are imperfections in the natural arrangement of metal atoms. Empty space exists in the structures of pure metals. When ham­ mered or pressed, groups of metal atoms may slide into new positions in these empty spaces. This makes metals malleable, able to be made into different shapes or pressed into thin sheets (Figure 9.11).

Aluminium and copper wires can be made because of the ductility of metals Figure 9.11 Malleability of pure metal

Source: Wikimedia; Rosmaniakos

Golden drinking vessel from Iran was made in about 5 B.C., proving the use of metal alloys in early civilization

Pure gold is too soft to make jewellery. 916 gold consiststs of 91.6% gold SPM

’08/P1, ’09/P2,

5 The making of alloys ’11/P2 (a) In the process of alloying, one or more foreign elements are added to molten metal. When the alloy hardens, the positions of some of the metal atoms are replaced by atoms of foreign elements, with sizes bigger or smaller than the original metal atoms.

Pure metals have the following physical properties: (a) Ductile (can be drawn into a wire) (b) Malleable (can be shaped by hammering) (c) High melting and boiling points (d) High density (e) Good conductors of electricity

273

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Gold is one of the most ductile and malleable metal. Pure gold is too soft and is not suitable for making any type of jewellery. Gold is thus usually alloyed with copper, silver or palladium. The carat unit is used to measure the purity of gold. The carat value is the number of parts of gold in 24 parts of the alloy. Hence, 24-carat gold is pure gold. 18-carat gold consists of 75% of gold by weight.

9

Aims of alloying

To increase the hardness and strength Alloying improves the hardness and strength of a metal. 1 The addition of a little carbon to iron metal produces steel which is a very hard alloy of iron. 2 The addition of magnesium to aluminium metal produces an alloy called magnalium. Magnalium is harder than aluminium but still retains the low density of aluminium metal. 3 The addition of tin to copper metal produces bronze. Bronze is an alloy harder than both tin and copper.

The body of airplanes are made of magnalium which is harder than pure aluminium

Manufactured Substances in Industry

To prevent corrosion Pure metals such as tin and iron are easily corroded in damp, polluted or acidic air. 1 The addition of carbon, nickel and chromium to iron metal produces stainless steel. Stainless steel is an alloy which can resist rusting. The chromium and nickel form chromium(III) oxide and nickel(IV) oxide which prevents the iron from rusting. 2 The addition of tin to copper produces bronze which is able to resist corrosion and tarnish.

Stainless steel kitchenware resists rusting

274

To improve the appearance

SPM

’05/P1

Metals are easily tarnished because of the formation of metal oxides on the metal surfaces. The process of alloying can maintain the lustre on the surface of the metal. 1 Stainless steel is more shiny than pure iron. 2 Adding a little copper and antimony to tin produces the alloy pewter which is harder and shinier, and not so easily tarnished. 3 Alloy wheels made from aluminium and other elements improve the look of vehicles.

Malaysian pewter is suitable for making shiny and attractive ornamental objects

9.1 SPM

’05/P3 ’04/P2

To compare the hardness of a pure metal and its alloy

9

Problem statement Are alloys harder than pure metals? Hypothesis Bronze is harder than copper. When a weight is dropped onto a ball bearing placed on a metal block made of copper or bronze, a larger dent will be produced on the softer copper metal block than on the bronze block. Variables (a) Manipulated variable : Types of materials (copper or bronze) to make the metal block (b) Responding variable : Diameter of the dent made by a steel ball bearing (c) Constant variable : Size of steel ball bearing, mass of weight used, height from which it is dropped Materials Copper block, bronze block, ball bearing, 1 kg weight, metre ruler, retort stand with clamp, cellophane tape and thread. Procedure 1 A metre ruler is clamped to a retort stand, and a piece of copper block is placed on the base of the retort stand. 2 A steel ball bearing is placed on the copper block and a piece of cellophane tape is used to hold the ball bearing in place. 3 A 1 kg weight is hung at a height of 50 cm above the copper block (Figure 9.13). 4 The weight is dropped onto the ball bearing placed on the copper block. 5 The diameter of the dent made by the ball bearing is measured. 6 The experiment is repeated three times using different areas on the surface of the copper block. 7 The average diameter of the dent is calculated. 8 Steps 1 to 7 are repeated using a piece of bronze block.

Figure 9.13 To compare the hardness of an alloy with its pure metal

Results I

II

III

Average

Copper

3.2

3.3

3.2

3.23

Bronze

2.4

2.5

2.5

2.47

Discussion 1 The bigger the average diameter of the dents produced by the steel ball bearing on the metal means that it has been pressed deeper into the metal surface. 2 Thus copper is softer than bronze because the steel ball bearing has been pressed deeper into the surface of copper metal than that of bronze. 3 Bronze is a type of alloy formed from copper and tin. The tin atoms are larger than the copper atoms. They distort the orderly structure of the copper atoms so that the layers of copper atoms can no longer slide easily over one another. This makes bronze harder than copper. Conclusion 1 The average diameter of the dents made by the steel ball bearing on the copper block is bigger than that on the bronze block. 2 Hence, bronze, a type of alloy, is harder than pure copper metal. The hypothesis is accepted.

275

Manufactured Substances in Industry

Experiment 9.1

Diameter of the dent (mm)

Metal block

9.2 To compare the rates of rusting of iron, steel and stainless steel

Procedure 1 Three test tubes are half-filled with jelly solution and are labelled as A, B and C. 2 1 cm3 of potassium hexacyanoferrate(III) solution is added to every test tube. 3 An iron nail, a steel nail and a stainless steel nail are polished with sandpaper to remove any rust formed. The nails are then placed in the three test tubes labelled A, B and C respectively. 4 All three test tubes are allowed to stand for 5 days before they are examined.

Problem statement How do the rates of rusting of iron, steel and stainless steel differ? Hypothesis Pure iron rusts faster than steel while stainless steel does not rust easily. 9

SPM

’05/P3

Variables (a) Manipulated variable : Types of nails (iron, steel and stainless steel) (b) Responding variable : Rate of rusting (c) Constant variable : Size of nails, duration of rusting and conditions of experiment (temperature, supply of water and air) Materials Iron nail, steel nail, stainless steel nail, 5% jelly solution and potassium hexacyanoferrate(III) solution and sandpaper.

Figure 9.14 To compare the rates of rusting of iron, steel and stainless steel

Results Test tube

Type of nail

Observation

A

Iron nail

Blue colour is formed around the nail

Rusting occurs

B

Steel nail

A slight blue colour is formed

A little rusting occurs

C

Stainless steel nail

No blue colour is observed

No rusting occurs

Discussion 1 When iron rusts, iron(II) ion, Fe2+ is produced.

Conclusion 1 The formation of a blue colour shows that rusting of iron (corrosion) has occurred. 2 The presence of a blue colour shows that iron nail rusts easily (corroded), steel nail rusts slightly and stainless steel does not rust at all. The hypothesis is accepted.

Fe → Fe2+ + 2e–

Experiment 9.2

Inference

2 Potassium hexacyanoferrate(III) solution is used to test the presence of iron(II) ion. A dark blue colour will be formed. The intensity of the blue colour indicates the rate of rusting. 3 A stainless steel alloy which resists rusting, is pro­duced by adding nickel and chromium to iron metal.

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276

The Composition, Properties and Uses of Some Common Alloys Uses

Properties

Composition

Carbon steel

99% iron 1% carbon

Hard and strong

Stainless steel

74% iron 18% chromium 8% nickel 90% copper 10% tin

Shiny, strong and resists rusting

Brass

70% copper 30% zinc

Hard and shiny

Magnalium

70% aluminium 30% magnesium 95% aluminium 3% copper 1% magnesium 1% manganese 97% tin 3% copper and antimony 50% tin 50% lead copper, nickel (percentage according to colour)

Light, hard and strong

Bronze

Duralumin

Pewter

Solder Cupro-nickel

Hard, strong and shiny

Light, hard and strong

Lustrous and strong

Hard, shiny and with low melting point Hard, shiny and resists corrosion

• Frameworks of buildings and bridges • In the making of tools, framework of heavy machinery and body of vehicles • In the making of cutlery and kitchenware • In the making of machine parts and surgical instruments • In the making of kitchenware and ships’ propellers • In the making of decorative ornaments, statues and art crafts • In the making of electrical connectors and musical instruments • In the making of kitchenware and decorative ornaments • In the making of aircraft body frames • In the making of rims of racing car tyres • In the making of the bodies of aircrafts and bullet trains • In the making of racing bicycles, fan blades, light electrical cable • In the making of mugs, candlesticks, decorative ornaments and souvenirs • In the making of solder for electrical wires and metal pipes • To make coins of 10 sen, 20 sen, 50 sen

An alloy is a mixture of two or more elements in which the major component is a metal.

Malaysian coins are made from cupro-nickel alloy (75% Cu, 25% Ni)

9.3 (a) Name the elements used to make the alloy pewter. (b) What are the advantages of pewter compared to its pure metal? (c) State a use of pewter.

1 Steel and stainless steel are two examples of iron alloys. (a) What is an alloy? (b) What is the element that is added to iron to form (i) steel? (ii) stainless steel? (c) Draw the arrangement of particles in (i) pure iron (ii) steel (d) Explain why stainless steel and not iron is used to make cutlery.

3 Copper is one of the metals used since ancient times. (a) Explain why copper alloys are more commonly used than its pure form. (b) Name two examples of copper alloys. (c) Pure copper is ductile and malleable. Explain this property in terms of the arrangements of atoms.

2 Pewter is an important alloy made in Malaysia.

277

Manufactured Substances in Industry

9

Alloy

SPM

’10/P1, ’11/P1

9.4

The Meaning of Polymers

9

3 Protein is formed by the polymerisation of monomers known as amino acids.

Synthetic Polymers

polymerisation

SPM

amino acids ⎯⎯⎯⎯⎯⎯→ protein

’11/P1

(monomers) (polymer)

1 The word polymer originated from the Greek polumeros which means ‘having many parts’. 2 Polymers are large molecules made up of many smaller and identical repeating units joined together by covalent bonds. These small molecules that are joined into chains are called monomers. 3 Polymerisation is the chemical process by which the monomers are joined together to form the big molecule known as the polymer.

4 Carbohydrates such as starch and cellulose consist of monomers known as glucose joined together chemically. polymerisation

glucose ⎯⎯⎯⎯⎯⎯→ carbohydrate

(monomers)

5 Natural rubber found in latex consists of monomers known as isoprene (2-methylbuta1,3-diene) joined together chemically.

polymerisation

A A A A A A ⎯→ –A–A–A–A–A–A– monomers

H CH3 H H H CH3 H H ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ ⎮ nC=C–C=C → —C–C=C–C — ⎮ ⎮ ⎮ ⎮ H H H H n

polymer

Or n A → (–A–)n where A = monomer n = a big number

isoprene (monomer) natural rubber (polymer)

4 A polymer is a macromolecule (a very big molecule). Hence, the relative molecular mass of a polymer is large. 5 The properties of a polymer are different from its monomers. 6 Polymers can be divided into 2 types: (a) Naturally occurring polymers Polymers that exist in living things in nature (plants and animals) (b) Synthetic polymer Polymers that are man-made by chemical processes in the laboratories.

Synthetic Polymers 1 Synthetic polymers are polymers made in the industry from chemical substances. 2 Through scientific research, scientists are able to copy the structures of natural polymers to produce synthetic polymers. 3 Many of the raw materials for synthetic polymers are obtained from petroleum, after the refining and cracking processes. 4 The types of synthetic polymers include (a) plastics (b) fibres (c) elastomers 5 Plastics (a) Thermoplastic is a polymer which, when subjected to heat, becomes soft so they can be moulded into various shapes. (b) The properties of plastics are: light, strong, inert to chemicals such as acids and alkali and are insulators of electricity and heat. (c) Examples of plastics are polythene (polyethylene), polyvinylchloride (PVC), polypropene (polypropylene), polysty­rene, Perspex and Bakelite. 6 Synthetic fibres (a) Synthetic fibres are long chained polymers that withstand stretching.

Naturally Occurring Polymers 1 Naturally occurring polymers exist in plants or animals. 2 Examples of naturally occurring polymers are (a) protein : in muscles, skin, silk, hair, wool and fur. (b) carbohydrates : in starch and cellulose. (c) natural rubber : in latex.

Carbohydrates such as starch and cellulose are polymers Manufactured Substances in Industry

(polymer)

278

(b) Condensation polymerisation occurs when the monomers with two functional groups combine to form the polymer through a condensation reaction. Examples of condensation polymers are nylon and Terylene. In general:

nA

monomer monomer A B

H ⎮ n C = ⎮ CH3

H H H H ⎮ ⎮ n C = C ⎯→~C — C~ ⎮ ⎮ H Cl H Cl n Substance X

H ⎮ C = ⎮ Cl

Substance Y

Which of the properties is identical for substance X and Y? A Density C Relative molecular mass B Boiling point D Empirical formula Comments Substance X is a monomer while substance Y is a polymer. A polymer has a higher density, melting point and boiling point than its monomer. The relative molecular mass of a polymer is more than that of a monomer but the percentage composition and empirical formula are the same. Answer D

1 Nylon is a type of polyamide polymer, a polymer with the amide (–CONH–) group. 2 Terylene is a type of polyester polymer, a polymer with the ester (–COO–) group. 3 Neoprene is a type of synthetic rubber made from the monomer, chloroprene. 4 Styrene-butadiene rubber (SBR) is a type of synthetic rubber made from 2 types of monomers, styrene and butadiene.

n = a big number

H H H ⎮ ⎮ ⎮ C ⎯→ — C — C — ⎮ ⎮ ⎮ H CH3 H n

propene, C3H6 polypropene

n

’05

The diagram below shows a polymerisation process.

H H H H ⎮ ⎮ ⎮ ⎮ n C = C ⎯→ — C — C — ⎮ ⎮ ⎮ ⎮ H H H H n polythene

condensation polymer

2

There are two types of polymerisation. (a) Addition polymerisation occurs when the monomers with double bonds combine to form the polymer through an addition reaction. Examples of addition polymers are polythene, polypropene, PVC, polystyrene and Perspex. Examples:

ethene

→ —( A – B — ) n + nH2O

+ nB

H H H ⎮ ⎮ ⎮ C ⎯→ — C — C — ⎮ ⎮ ⎮ H Cl H n The making of Terylene fibre

chloroethene, polyvinylchloride C2H3Cl (PVC)

279

SBR is used in making tyres Manufactured Substances in Industry

9

(b) Examples of synthetic fibres are nylon and Terylene. (c) Nylon is used to make ropes, fishing lines, stocking, clothing and parachutes. (d) Terylene is used to make clothing, sleeping bags and fishing nets. Clothes made from Terylene do not crease easily. 7 Elastomer (a) An elastomer is a polymer that can regain its original shape after being stretched or pressed. (b) Both natural rubber and synthetic rubber are examples of elastomers. (c) Examples of synthetic rubbers are neoprene and styrene-butadiene rubber (SBR). (d) SBR is used to make car tyres. 8 There are two types of polymerisation processes: (a) Addition polymerisation (b) Condensation polymerisation 9 Plastics such as polythene and PVC are produced by addition polymerisation, whereas synthetic fibres such as nylon and Terylene are made by condensation polymerisation.

Table 9.1 Some examples of synthetic polymers, their monomers and uses

9

Synthetic Polymer

Uses

Monomer

1 Polyethylene (PE) IUPAC name: polythene

Ethene, C2H4

Plastic bags, shopping bags, plastic containers, plastic toys, plastic cups and plates

2 Polypropylene (PP) IUPAC name: polypropene

Propene, C3H6

Plastic bottles, bottle crates, plastic tables and chairs, car battery cases and ropes

3 Polyvinylchloride (PVC) IUPAC name: polychloroethene

Chloroethene, C2H3Cl

Water pipe, shoes, bags, rain clothes, artificial leather and wire casing

4 Polystyrene (PS) IUPAC name: polyphenylethene

Phenylethene, C6H5CH = CH2

Packaging materials, heat insulators, toys, disposable cups and plates

5 Perspex (PP) IUPAC name: poly(methyl-2methylpropenoate)

Methyl-2-methyl propenoate (methylmetacrylate) CH2 = C(CH3)CO2CH3

Safety glass, airplane windows, car lamps, traffic signs, lens, reflectors and toys

Tetrafluoroethene, C2F4 6 Teflon (PTFE) IUPAC name: polytetrafluoroethene

Coatings for non-stick frying pans and electrical insulators

7 Terylene

Hexane-1, 6-diol and benzene-1, 4-dicarboxylic acid

Clothing, sleeping bags, sails, ropes and fishing net. Clothes made from Terylene do not crease easily

8 Nylon

Hexane-1, 6-diamine and hexane-1, 6-dioic acid

Ropes, fishing lines, stocking, clothing, carpets and parachutes

Issues of the Use of Synthetic Polymers in Every Day Life

1 Synthetic polymers have been used widely to replace natural materials such as metals, wood, cotton, animal skin and natural rubber because of the following advantages: (a) Strong and light (b) Cheap (c) Able to resist corrosion (d) Inert to chemical reactions (e) Easily moulded or shaped and be coloured (f) Can be made to have special properties according to specific needs 2 The use of synthetic polymers, however, results in environmental pollution problems.







Pollution Problem Caused by Synthetic Polymers

1 Most polymers are non-biodegradable, that is, they cannot be decomposed by bacteria or

Manufactured Substances in Industry

280

other micro-organisms. This will cause disposal problems as the polymers will not decay like other organic garbage. 2 Discarded plastic items may cause blockage of drainage systems and rivers thus causing flash floods. 3 Plastic bottles and containers that are not buried in the ground will become breeding grounds for mosquitoes which will cause diseases such as dengue. 4 Small plastic items that are thrown into the rivers, lakes and seas are sometimes swallowed by aquatic animals. These animals may die from choking. 5 The open burning of polymers may release harmful and poisonous gases that will cause air pollution. For example, the burning of PVC will release hydrogen chloride gas which contributes to the acid rain problem. The burning of some polymers will release toxic gas such as hydrogen cyanide. 6 The main source of raw materials for the making of synthetic polymers is petroleum. Petroleum is a non-renewable resource.

Glass and Ceramics

1 The raw materials for making glass and ceramics are obtained from the Earth’s crust. 2 The main component of both glass and ceramics is silica or silicon dioxide, SiO2. 3 Both glass and ceramics are used widely in our daily life to replace metals because of the advantages above as well as their low cost of production. 4 Both glass and ceramics have the same properties as follows: (a) Hard but brittle (b) Inert to chemical reactions (c) Insulators of electricity (d) Poor conductors of heat and electricity (e) Withstand compression but not tension (stretching) (f) Can be easily cleaned 5 The use of glass and ceramics also depends on their differences. Table 9.2 shows the differences between glass and ceramic.

1 Reduce, reuse and recycle synthetic polymers (a) Reduce the use of non-biodegradable polymers. (b) Polymers are collected and reused or reprocessed to make new items. The biggest problem is the collection and separation. Not only must the plastics be separated from other types of solid waste but the different types of polymers must be separated from each other. 2 Develop biodegradable polymers These polymers can be decomposed by bacteria, other microorganisms or simply by sunlight (photodegradable). One type of biodegra­ dable polymer was developed by incorpora­ ting starch molecules into the plastic materials so that they can be decomposed by bacteria. However, biodegradable polymers are usually more expensive.

Table 9.2 The differences in properties between glass and ceramic

Glass

9.4 1 (a) What is a polymer? (b) Name two natural polymers that are used to make clothing. (c) Name two synthetic polymers that are used to make clothing.

Ceramic

Transparent

Opaque

Softens when heated

High melting point, hence retains shape on heating

Impermeable

Usually porous except when glazed

2 Fill in the blanks below. Monomer

Polymer In silicon dioxide, every silicon atom is bonded covalently to 4 oxygen atoms in a tetrahedral shape. Every oxygen atom is also bonded to two silicon atoms to form a giant covalent molecule (Figure 9.15).

Ethene Chloroethene Polypropene Polystyrene

3 State two properties of synthetic polymers that will cause environmental pollution in the disposal of syn­ the­ tic polymers. State two methods to overcome these problems.

Figure 9.15 Structure of silicon(IV) oxide

281

Manufactured Substances in Industry

9

9.5

Methods to overcome environmental problems of polymers

Table 9.3 The uses of glass

Property of glass

Uses

Examples

Inert

Household materials

Lamp, bottles, glasses, pla­tes, bowls and kitchen wares

Transparent

Building materials

Mirrors and window glass

Industrial materials

Bulbs, glass tubes for radios, radars and televisions

Scientific apparatus

Lens, burettes, beakers, test tubes, conical flasks, glass tubes and prisms

Inert and easily cleaned 9

Table 9.4 Uses of ceramics

Property of ceramics

Glass is transparent

SPM

’05/P1

Uses

Examples

Hard and strong

Building materials

Long lasting and noncorrosive

Materials for decorative Plates, bowls, cooking utensils, items porcelain and vases

Electrical insulators

To make electrical insulatin­g parts

Insulators in toasters and irons, spark plugs in car engines

Inert and hard

In surgical and dental apparatus

Artificial hands, legs and teeth

Semiconductor type of ceramics

As microchips

To make microchips in computers, radios and televisions

Bricks, tiles and cement

Ceramic is opaque and has higher melting point than glass

the addition of chromium ions will give the glass a green hue, cobalt ions will give a blue hue while manganate ions will give a purple hue to the glass.

Types, Properties, Composition and Uses of Glass 1 Fused glass is the simplest type of glass, which consists mainly of silica or silicon dioxide. Occasionally a little boron oxide is added. 2 Other types of glass are mainly metal silicates. 3 Various types of glass can be produced by changing the composition of glass. Different types of glass have different properties and they are used for various specific purposes. The chemical composition, specific properties and uses of four types of glass are summarised in Table 9.5. 4 Coloured glass is produced by adding traces of transition metal oxides to it. For example,

Composition of Ceramics 1 Ceramic is a manufactured substance made from clay that is dried and then baked in a kiln at high temperature. 2 The main constituent of clay is aluminate, silica and feldspar. 3 Kaolinite is an example of high quality white clay that consists of hydrated aluminosilicate crystals. 4 Red clay contains iron(III) oxide which gives its red colour.

Table 9.5 Properties, composition and uses of different types of glass

Name of glass

Properties

Chemical composition

Fused glass

• Very high softening point (1700 °C), hence highly heat-resistant • Transparent to ultraviolet and infrared light • Difficult to be made into different shapes • Does not crack when temperature changes (very low thermal expansion coefficient) • Very resistant to chemical reactions

SiO2 (99%) B2O3 (1%)

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282

Examples of uses Telescope mirrors, lenses, optical fibres and laboratory glass wares

Properties

Chemical composition

• Low softening point (700 °C), hence does not withstand heating • Breaks easily • Cracks easily with sudden temperature changes (high thermal coefficient of expansion) • Less resistant to chemical reactions • Easy to make into different shapes • Quite high softening point (800 °C), hence it is heat-resistant • Does not crack easily with sudden change in temperature • Transparent to ultraviolet light • More resistant to chemical reactions • Does not break easily

SiO2 (70%) Na2O (15%) CaO (10%) Others (5%)

Bottles, windowpanes, light bulbs, mir­rors, flat glass, glass-plates and bowls. (The most widely used type of glass)

SiO2 (80%) B2O3 (15%) Na2O (3%) Al2O3 (1%)

Laboratory apparatus, cooking utensils, electrical-tubes and glass pipelines

• Low softening point (600 °C) • High density • High refractive index • Reflects light rays and appears sparkling

SiO2 (55%) PbO (30%) K2O (10%) Na2O (3%) Al2O3 (2%)

Decorative items, crystal glasswares, lens, prisms and chandeliers

Soda lime glass

Borosilicate glass SPM

’09/P1, ’10/P1

Lead glass

Examples of uses

The Uses of Improved Glass and Ceramics for Specific Purposes Improved glass

Examples of Improved Glass and Ceramics

Photochromic glass • Photochromic glass is a type of glass that is sensitive to light intensity. The glass darkens when exposed to sunlight but be­­comes clear when light intensity decreases. • Photochromic glass is produced when a dispersion of silver chloride, AgCl or silver bromide, AgBr is added to normal glass. Conducting glass • Conducting glass is a type of glass that can conduct electricity. • Conducting glass is produced by embedding a thin layer of conducting material in glass. • A type of conducting glass is produced by adding a layer of indium tin(IV) oxide (ITO) that acts as an electrical conductor. This type of glass is used in the making of LCD (liquid crystal display) panel. • Another type of conducting glass is made by embedding thin gold threads in glass. Water condenses as ice on the window panes of aircraft at high altitudes and this obstructs the vision of the pilot. Hence, windows of aircraft are heated by passing electric current through the gold threads embedded in the glass. 283

Improved Ceramics

Superconductor • Superconductors are a class of ceramics that conducts electricity without resistance and without loss of electrical energy. • Superconductor ceramics are used to make light magnets, electrical generators and electric motors. Ceramic car engine block • Clay heated with magnesium oxide produces a type of ceramic that has a high thermal resistance. • This type of ceramic is used for making car engine blocks because it can resist high temperatures. • At a higher temperature, the combustion of fuel becomes more efficient and produces more energy with less pollution. • Ceramic engines offer great advantages in terms of fuel economy, efficiency, weight savings and performance.

Manufactured Substances in Industry

9

Name of glass

9

Main component is silica, SiO2

Glass

Ceramic

Types of glass • Fused glass (high heat resistance) • Soda lime glass (cannot withstand high temperatures) • Borosilicate glass (can withstand high temperatures) • Lead glass (high refractive index)

Examples of ceramics • Tiles • Cement • Bricks • Porcelain

Common properties • Hard but brittle • Inert to chemicals • Heat and electrical resistance

• Resist compression • Can be easily cleaned

3 In the making of composites, substances (known as components) are combined to form new types of materials that can overcome the limitations of the original materials. 4 Most of the composite materials are comprised of two phases: a continuous phase (also known as the base) and the dispersed phase (also known as the matrix). 5 Composite materials are harder, stronger, lighter (lower density), more resistant to heat and corrosion and also made for specific purposes. 6 A few types of composite materials and their components are shown in Table 9.6.

9.5 1 Glass is a manufactured substance in industry. (a) What is the major component of glass? (b) State a cheap source of this component. (c) State four types of glass. 2 Glass and ceramics are both manufactured from materials in the Earth's crust. (a) What is the common component found in both glass and ceramics? (b) State the three similarities and three differences between glass and ceramics. (c) State five uses of glass. (d) State two examples of the use of ceramics in the building industry.

Table 9.6

3 State one main difference between soda lime glass and borosilicate glass in terms of (a) composition and (b) property.

9.6

Differences • Glass is transparent, ceramic is opaque • Ceramic can withstand a higher temperature than normal glass

Composite materials

Reinforced concrete Concrete (cement, sand and small pebbles) and steel Superconductor

Yttrium oxide (Y2O3), barium carbonate (BaCO3) and copper(II) oxide (CuO)

Fibre optic

Glass {silica (SiO2), sodium carbonate (Na2CO3) and calcium oxide (CaO)} with different refractive indices

Fibreglass

Glass fibre and polyester (a type of plastic)

Photochromic glass

Glass and silver chloride or silver bromide

Composite Materials

The Meaning of Composite Materials 1 A composite material is a structural material formed by combining two or more materials with different physical properties, producing a complex mixture. 2 The composite material produced will have different properties far more superior to the original materials. Manufactured Substances in Industry

Components

284

(c) Glass and ceramics are brittle. (d) Plastic and glass cannot temperatures.

1 Most types of material used in our daily life have certain limitations. For example: (a) Metals can be easily corroded and are malleable and ductile. (b) Metals are good electrical conductors but the existence of resistance results in the loss of a big amount of electrical energy as heat.

resist

high

2 With knowledge of the compositions, structures and properties of these materials, chemists are able to develop new materials to suit specific purposes.

Reinforced concrete 1 Concrete is a composite material made from a mixture of sand and small stones bound by cement. Concrete is strong in compression but brittle and weak in tension. Concrete cannot withstand vibrations and will crack under the action of bending forces. 2 Steel is strong in tension (high tensile strength). But using thick steel columns to support a heavy load is expensive. Furthermore, steel corrodes easily. 3 Reinforced concrete is made by adding the concrete mixture of cement, water, sand, chips and small stones into a frame of steel bars or steel wire netting (Figure 9.16). When set, a composite material is formed. 4 Reinforced concrete is a stronger building material as it combines the compressive strength of concrete and tensile strength of steel. In addition, it does not corrode easily. Reinforced concrete is also relatively cheap and can be moulded into any shape. 5 Steel and concrete have about the same coefficient of expansion. Hence they are good composite components and do not crack when mixed. 6 Reinforced concrete can withstand very high applied forces (high pressure) and

SPM

’07/P1, ’08/P1

can support very heavy loads. It is used to construct framework for highways, brid­ ges, oil platforms and high-rise buildings.

Figure 9.16 The formation of reinforced concrete

Dams are constructed by reinforced concrete which is very strong

Superconductors 1 In normal electrical conductors such as copper metal, the existence of resistance causes the loss of electrical energy as heat. Furthermore, resistance increases as temperature increases. 2 Superconductors can conduct electricity with zero resistance when they are cooled to extremely low temperatures. Thus, superconductors conduct electricity without any loss of energy.

(known as the transition temperature). This low temperature can only be achieved using liquid helium which is expensive. 4 When a mixture of copper(II) oxide (CuO), barium oxide (BaO) and yttrium oxide (Y2O3) is heated up, a type of ceramic with the formula YBa2Cu3O7 is produced. This type of ceramic, known as perovskite or YBCO, can attain superconductivity at 90 K (–183°C). This temperature can easily be attained by using the cheaper liquid nitrogen.

3 Metals such as copper, can only achieve superconductivity at a very low temperature 285

Manufactured Substances in Industry

9

Comparison of the Properties of Composite Materials and Their Original Components

5 The metal oxides (CuO, Y2O3 and BaO) are all electrical insulators. However when they are combined to form a composite, the composite is a superconductor that can conduct very high current over long distance without any loss of energy. 6 Superconductors are used to make more efficient generators, magnetic energy-storage systems, transformers, electric cables,

amplifiers and computer parts. They are also used in magnetic resonance imaging (MRI) – a type of medical imaging device. Superconductors are also used to make stronger, lighter and more powerful electromagnets. High – speed levitated trains (trains that float on the railway track) involve the use of electromagnets and superconductors.

9

Fibre optics (also known as optical fibres) 1 Optical fibres are bundles of glass tubes with very small diameters. They are finer than human hair and are very flexible. 2 Fibre optics is a composite material that can transmit electronic data or signals, voice and images in a digital format, in the form of light along the fine glass tubes at great speeds. 3 Fibre optics consists of a core of glass of higher refractive index enclosed by a glass cladding of lower refractive index. A light wave entering the fibre will travel along the glass tube due to total internal reflection (Figure 9.17). 4 In the field of telecommunications, fibre optics is used to replace copper wire in long distance telephone lines, mobile phones, video cameras and to link computers within local area networks (LAN). Fibre optics uses light instead of electrons to carry data. Fibre optics carry more data (higher transmission capacity) with less interference, has a higher chemical stability and a lower material cost compared to metal communication cables such as copper. Fibre optics can also send

SPM

’11/P1

signals faster than metal cables and occupies less space. 5 In the field of medicine, a laser beam can be channelled through fibre optics in operations to remove unwanted Fibre optics tissues. Fibre optics is also used in endoscopes: instruments that are inserted into the body through the nose, mouth or ear, for doctors to examine the internal organs. 6 Fibre optics is also used in instruments to inspect the interior of manufactured products.

Figure 9.17 Cross section of a fibre optic

Fibreglass 1 Plastic is light (with a low density), elastic, flexible, but is brittle, not very strong and is inflammable (can catch fire). 2 Glass is hard and strong but is brittle, heavy (with a relatively high density) and has a low compressive strength. 3 When glass fibre filaments are embedded in polyester resin (a type of plastic), fibreglass which is light, strong, tough, resilient, inflammable, flexible with a high tensile strength is produced. It can also be easily coloured, moulded and shaped. A resilient material is one that returns to its original shape after bending, twisting, stretching and compression.

Manufactured Substances in Industry

Boats built from fibreglass is light and strong.

4 Fibreglass is an ideal material for making water storage tanks, boat hulls, swimming pool linings, food containers, fishing rods, car bodies, roofing, furniture and pipes.

286

Photochromic glass 7 Silver atoms and bromine gas recombine according to the following reaction

1 Glass is transparent and is not sensitive to light intensity. 2 Silver chloride or silver bromide is sensitive to light. When exposed to light, these compounds decompose to form dark silver particles. 3 In photochromic glass, silver chloride (AgCl) or silver bromide (AgBr) and a little copper(I) chloride is embedded into the structure of glass. 4 When exposed to ultraviolet light, the AgCl or AgBr decomposes to form silver and halogen atoms. The fine silver which is deposited in the glass is black and the glass is darkened. For example:

Br2 + 2Cu+ → 2Br– + 2Cu2+ … (1) Cu2+ + Ag → Cu+ + Ag+ … (2) 8 The overall reaction is

9 Photochromic glass is used to make camera lens, car windshields, information display panels, light intensity meters and optical switches.

uv light

2AgBr ⎯⎯⎯⎯→ 2Ag + Br2

5 Photochromic glass has the ability to change colour and become darker when exposed to ultraviolet light. 6 The photochromic glass will automatically become clear again when the light intensity is lowered, whereby silver is converted back to silver halides.

Photochromic glass is used to make lenses that change from light to dark, eliminating the neccessity for a separate pair of sunglasses.

Properties of composite materials compared to their components and the uses of composites

Composite material Reinforced concrete

Super­ conductor

Component

Properties of component

Properties of composite

Concrete

Hard but brittle, with low tensile strength

Steel

Hard with high tensile strength but expensive and can corrode

Copper(II) oxide, yttrium oxide and barium oxide

Insulators of electricity

287

Uses of composites

Stronger, higher tensile strength, not so brittle, does not corrode easily, can withstand higher applied forces and loads, relatively cheaper

Construction of framework for highways, bridges and high-rise buildings

Conducts electricity without resistance when cooled by liquid nitrogen

To make more efficient generators, transformers, electric cable, amplifiers, computer parts, stronger and lighter electromagnets

Manufactured Substances in Industry

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2Ag + Br2 → 2AgBr

Composite material Fibre optics

Properties of component

Properties of composite

Transparent, does not reflect light rays

Reflect light rays and allow light rays to travel along the fibre

Transmit data in the form of light in telecommunications

Glass

Heavy, strong but brittle and nonflexible

Polyester plastic

Light, flexible, elastic but weak and inflammable

Light, strong, tough, resilient and flexible, with high tensile strength, not inflammable

Water and food storage containers, boats, swimming pool linings, fishing rods, car bodies and roofing

Glass

Transparent and not sensitive to light

Silver chloride or silver bromide

Sensitive to light

Sensitive to light: darkens when light intensity is high, becomes clear when light intensity is low

Photochromic optical lens, camera lens, car windshields, optical switches, information display panels and light intensity meters

Component Glass of low refractive index Glass of higher refractive index

9

Fibreglass

Photo­ chromic glass

2 New materials are required to overcome new challenges and problems we face in the changing world. 3 Synthetic materials are developed cons­tantly due to the limitation and shortage of natural materials. 4 New needs and new problems will stimulate the development of new synthetic materials. For example, the use of new plastic compo­ site material will replace metal in the making of a stronger and lighter car body. This will save fuel and improve speed. Plastic compo­ site materials may one day be used to make organs for organ transplants in human bodies. New superconductors made from composite materials are developed. 5 The understanding of the interaction between different chemicals is important for both the development of new synthetic materials and the disposal of such synthetic materials as waste. 6 A responsible and systemic method of handling the waste of synthetic materials and their by– products is important to prevent environmental pollution. The recycling and development of environ­­ mentally friendly synthetic material should be enforced.

9.6 1 (a) State what is meant by the term composite materials. (b) Give five examples of composite materials and name one use for each example. 2 (a) What is fibreglass? Explain how the properties of fibreglass are superior to that of its original components. (b) What is reinforced concrete? Give two reasons why reinforced concrete is a strong construction material. 3 What is the advantage of using photochromic glass in the making of spectacles? Briefly explain how this glass works.

9.7

Appreciating Various Synthetic Industrial Materials

1 Continuous research and development (R & D) is required to produce better materials used to improve our standard of living.

Manufactured Substances in Industry

Uses of composites

288

9 Multiple-choice Questions 9.1

Sulphuric Acid

1 The uses of substance X is given below. • To make fertilisers • To manufacture detergents • To make paints What of the following substances could be X? A Nitric acid B Sulphuric acid

C Ammonia D Ammonium sulphate 2 Which of the following is a use of sulphuric acid? A As an electrolyte in dry cells B As a raw material in making explosives C As an electrolyte in the electroplating of metals D To clean the metal oxide layer of metals before electroplating

289

3 Which of the following is true about the manufacture of sulphuric acid by the Contact process? A Sulphur and vanadium(V) oxide are the raw materials. B Sulphur is converted into sulphur dioxide and then into sulphur trioxide. C Sulphur trioxide dissolves in water to form sulphuric acid. D A pressure of 200 atmosphere is used in the process. Manufactured Substances in Industry

9

9 The three aims of alloying are: (a) To increase the hardness and strength of a metal (b) To prevent corrosion or rusting (c) To improve the appearance of the metal surfaces 10 Polymers are large molecules made up of many smaller and identical repeating units (monomers) joined together by covalent bonds. 11 Some examples of synthetic polymers are polythene, polypropene, P.V.C., polystyrene, perspex, Terylene and nylon. 12 Synthetic polymers are strong and light, cheap, resist corrosion and inert to chemical attacks. However, they are nonbiodegradable and cause environmental pollution problems. 13 The main component of glass is silica or silicon dioxide, SiO2. The main constituents of ceramics are clay (aluminosilicate), sand (silica) and feldspar. 14 Both glass and ceramics have the following properties: (a) Hard but brittle (b) Inert toward chemicals (c) Insulators or bad conductors of heat and electricity 15 Some examples of glass are fused glass, soda lime glass, borosilicate glass and lead glass. 16 A composite material is a structural material formed by combining two or more materials with different physical properties to produce a complex mixture. 17 Some examples of composites are reinforced concrete, superconductors, fibre optic, fibreglass and photochromic glass.

1 Sulphuric acid is used to make other manufactured substances such as fertilisers, detergents, pesticides, polymers and paint pigments. 2 Sulphuric acid is manufactured by the Contact process using vanadium(V) oxide as a catalyst. The process involves three stages. I II Sulphur ⎯→ Sulphur dioxide ⎯→ Sulphur trioxide III ⎯→ Sulphuric acid 3 Sulphur dioxide gas can cause environmental pollution such as acid rain. 4 Ammonia is used to make nitrogenous fertilisers and nitric acid, and is used as a cooling agent in refrigerators. 5 Ammonia is produced in the industry by the Haber process with hydrogen gas and nitrogen gas and using iron powder as a catalyst. 6 An alloy is a mixture of two or more elements with a certain fixed composition in which the major component is a metal. 7 A pure metal is weak and soft because it contains atoms of the same size in an orderly arrangement. This enables the layers of atoms to slide over each other easily. 8 In an alloy, foreign atoms of different sizes disrupt the orderly arrangement of the metal atoms. This prevents the layers of metal atoms from sliding over each other easily.

9

4 The manufacturing of sulphuric acid in the Contact process ’06 involves several reactions. Which of the following is the reaction that requires a temperature of 450 – 550°C. A S + O2 → SO2 B 2SO2 + O2 2SO3 C SO3 + H2SO4 → H2S2O7 D H2S2O7 + H2O → 2H2SO4 5 In the Contact process, oleum is produced when A sulphur dioxide reacts with oxygen. B sulphur dioxide dissolves in water. C sulphur trioxide dissolves in concentrated sulphuric acid. D sulphur trioxide dissolves in water. 6 Which of the following gas ’08 dissolves in rainwater and consequently kills trees and corrodes concrete buildings? A Ammonia B Sulphur dioxide C Carbon monoxide D Carbon dioxide 7 Which of the following are caused by the presence of sulphur dioxide in the atmosphere? I Certain lung diseases and bronchitis II An increase of acidity in blood III An increase of acidity in rain water IV Corrodes concrete buildings A I and II only B III and IV only C I, II and III only D I, III and IV only

9.2

Ammonia and its Salts

8 Which of the following chemicals is manufactured using ammonia in the industry? A Sodium hydroxide B Hydrochloric acid C Sulphuric acid D Nitric acid Manufactured Substances in Industry

9 Which of the following are true of ammonia gas? I It is a colourless and odourless gas. II It is lighter than air. III It is sparingly soluble in water. IV It produces white fumes with hydrogen chloride gas. A I and IV only B II and III only C II and IV only D III and IV only 10 Which of the following is a source of hydrogen for the Haber process? A The decomposition of water B Fractional distillation of liquid air C Reaction of coke or natural gas with steam D Reaction of zinc metal with sulphuric acid 11 Which of the following is the effect of using iron powder as a catalyst in the production of ammonia in the Haber process? A The quantity of ammonia produced is increased. B The temperature required for Haber process is reduced. C The pressure required for Haber process is reduced. D The rate of reaction between hydrogen gas and nitrogen gas is increased. 12 Which of the following compounds reacts with ammonia to produce urea, a type of fertiliser? A Carbon dioxide B Sulphuric acid C Phosphoric acid D Ethanoic acid 13 Which of the following is the function of iron powder in the Haber process? A To speed up the rate of production of ammonia B To increase the percentage yield of ammonia C To lower the cost of production of ammonia

290

D To lower the pressure required in the production of ammonia 14 Which of the following are the conditions in the production of ammonia by the Haber process? I A temperature of about 450 °C II A pressure of one atmosphere III Equal volumes of nitrogen gas and hydrogen gas IV The use of iron powder as a catalyst A I and II only B I and IV only C II and III only D I, II and IV only 15 Ammonium nitrate is used as a fertiliser. What is the ’10 percentage by mass of nitrogen in ammonium nitrate? [Relative atomic mass: H, 1; N, 14; O, 16] A 17.5% B 17.7% C 28.6% D 35.0%

9.3

Alloys

16 The alloying process increases the hardness of a metal. This is because the foreign atoms added to a metal in the alloying process A increases the orderliness of the arrangement of the metal atoms. B strengthens the bond between the metal atoms. C forms strong bonds between the metal atoms and the foreign atoms. D makes it difficult for the layers of metal atoms to slide over each other. 17 Which of the following is not the aim of alloying iron to form steel? A To make it harder B To make it stronger C To make it more resistant to rusting D To increase the melting point

’11

Alloy

Uses

A

Duralumin

Building of monuments

B

Brass

Bodies of aeroplanes

C

Bronze

Frameworks of buildings

D

Stainless steel

Making of surgical instruments

19 The alloy produced by the addition of tin to copper metal is known as A bronze B brass C pewter D duralumin 20 Which of the following alloys is suitable for the making of an aircraft body? A Duralumin B Bronze C Brass D Cupro-nickel 21 Iron is alloyed to produce steel. Which of the following property is not true of steel compared to iron? A Harder B More resistant to rusting C More presentable D A better electrical conductor

9.4

Synthetic Polymers

22 The polymer formed from the polymerisation of phenylethene is known as A polythene B polypropene C polystyrene D polyvinyl chloride 23 Which of the following are the correct pairs of polymer and monomer?

Polymer

Monomer

I

Natural rubber Isoprene

II

Carbohydrate Sucrose

III

Polypropene

Methylmethacrylate

IV

PVC

Chloroethene

A B C D

II and III only I and IV only I, II and III only I, II, and IV only

24 The diagram shows the repeating unit of a synthetic polymer.

H H ⎮ ⎮ — C – C — ⎮ ⎮ CH3 CH3 Which of the following are true of this synthetic polymer? I It is a type of addition polymer. II It dissolves easily in hot water. III It burns in air to produce carbon dioxide and water. IV It has a high relative molecular mass. A I and II only B III and IV only C II and III only D I, III and IV only 25 Which of the following are true about Terylene, a type of synthetic polymer? I It is easily biodegradable. II It burns easily. III It is a type of fibre. IV Its monomer is ethene. A I and IV only B II and III only C II, III and IV only D I, II and III only 26 Uncontrolled disposal of synthetic polymers will cause environmental pollution. Which of the following are the characteristics of synthetic polymers that causes this environmental pollution? I Polymers are nonbiodegradable. II Polymers increase the pH of the water when dissolved in water.

291

III Polymers promote excessive growth of algae in water. IV Polymers release toxic gases when burned. A I and IV only B II and III only C I, III and IV only D II, III and IV only

9.5

Glass and Ceramics

27 A transparent solid is formed when molten sand at high temperature is cooled quickly. What is this solid? A Ceramic B Fused glass C Soda lime glass D Borosilicate glass 28 Which of the following are true for both glass and ceramic? I They contain a common component, silica. II They are electrical insulators. III They are resistant to chemicals. IV They can resist compression. A I and II only B III and IV only C I, II and III only D I, II, III and IV 29 Material Y has the following properties: ’09

• Resistance towards chemical substances • Low coefficient of thermal expansion What is material Y ? A Bronze B Polystyrene C Borosilicate glass D Conducting glass 30 Which of the following glass has a low softening point and can be easily moulded into different shapes? A Fused quartz glass B Soda lime glass C Borosilicate glass D Photochromic glass 31 Lead glass is very suitable for making fine glassware and art objects because this type of glass Manufactured Substances in Industry

9

18 Which alloy is correctly matched to its uses?

9

A has a high refractive index. B has a high softening point. C is most transparent to ultraviolet and infrared rays. D is a good heat insulator.

B Photochromic glass C Lead glass D Borosilicate glass

9.6

Composite Materials

32 What is the purpose of adding feldspar to kaolin in the making ’05 of porcelain? A To make kaolin softer B To make kaolin smoother C To make kaolin harder D To make porcelain that is inert towards chemicals

35 Which composite material is made of glass of different ’08 refractive index? A Fibreglass B Fibre optics C Photochromic glass D Borosilicate glass

33 The raw materials for making soda lime glass are I sodium carbonate II calcium carbonate III silicon dioxide IV boron oxide A I and II only B II and III only C I, II and III only D I, II, III and IV

36 The supporting pillars of flyovers of highways are made of ’06 substance X. Substance X has the following properties : strong, not brittle, can withstand erosion and can withstand weathering. Which of the following is substance X? A Concrete B Steel C Marble D Reinforced concrete

34 A decorative glassware manufacturer wants to make a display panel that is sensitive to the intensity of light. What is the most suitable material? A Conducting glass

37 The diagram shows the formation of a composite ’07 substance from its original components.

Based on the diagram, why is reinforced concrete often used more to build buildings compared to concrete? A The steel bars cannot stretch and make it tough. B The concrete and the steel bars can slide over each other and make it flexible. C The steel bars fix the position of the concrete particles and make it hard. D The concrete particles are evenly dispersed among the steel bars and make it able to withstand vibrations.

Structured Questions (c) (i) Name liquid Z. [1 mark] (ii) How is liquid Z formed from gas Y?

1 Diagram 1 shows a series of steps involved in the production of sulphuric acid in industry starting from sulphur. Sulphur

Sulphuric acid

I Oxygen IV

Gas X

Liquid Z

[1 mark]

(d) Sulphuric acid is also formed when gas Y dissolves in water. Why is this not done in the manufacturing of sulphuric acid in industry? [1 mark]

II Oxygen III

(e) State two uses of sulphuric acid. Gas Y

[2 marks]

2 Diagram 2 shows how a type of fertiliser is produced.

Diagram 1

’07

(a) (i) Name gas X. [1 mark] (ii) Write a balanced equation for the reaction in step I. [1 mark]

Process P Process Q

Sulphuric acid Ammonia

Fertilliser X

Diagram 2

(b) (i) Name gas Y. [1 mark] (ii) State the conditions used in step II in order to produce a high percentage yield of gas Y.

(a) Process P and process Q are industrial processes. What are the names of each of these processes?

[2 marks]

[2 marks]

Manufactured Substances in Industry

292



[1 mark] (ii) State a natural source of silica. (iii) W is a an important component of [1 mark] borosilicate glass. What is W? (b) (i) Name compound X. [1 mark] (ii) Compound X can be used to make plastic chairs and tables. State one advantage of this type of material as compared to metals.

[3 marks]

(c) (i) Name fertiliser X. (ii) Write a balanced equation for the formation of fertiliser X. [2 marks] (d) Calculate the mass of ammonia that is required to react with 0.2 mol of sulphuric acid. [Relative atom mass: H, 1; N, 14] [2 marks]

[1 mark]

(e) Name another type of fertiliser that is produced when ammonia reacts with carbon dioxide.

(c) (i) Identify component Y. [1 mark] (ii) Explain why magnalium is harder than pure [2 marks] aluminium.

[1 mark]

3 The flowchart in Diagram 3 shows the conversion of nitrogen into various substances in steps I, II, III and IV. Nitrogen

I

II

Ammonia

+ Substance X

III

(d) Z can withstand high pressures and can very heavy loads. What is Z? (e) (i) Identify the type of compound T. (ii) State a property of compound T.

Nitric acid IV

5 Bronze, brass and duralumin are examples of alloys.

+ Substance Y

(a) What is meant by alloy?

Diagram 3

(c) What type of particles are present in pure [1 mark] copper?

(a) Step I is an industrial process. State the conditions used for a good yield of ammonia in this process.

(d) Draw a diagram to show the arrangement of particles in (i) pure copper (ii) bronze [2 marks]

[3 marks]

(b) Step II is also an industrial process. (i) Name this process. [1 mark] (ii) State the catalyst used in this process.

(e) (i) Name the elements that are used to make [1 mark] the alloy duralumin. (ii) What is the advantage of duralumin compared [1 mark] to its main component? (iii) State one use of the alloy duralumin. [1 mark]

[1 mark]

(c) State one physical property and one chemical property of ammonia. Physical property: Chemical property: [2 marks] (d) Name a possible compound for substance X in step III and hence write a balanced equation that occurs in step III. [2 marks] (e) (i) Besides ammonia, name a compound that may be substance Y in step IV. [1 mark] (ii) Write a balanced equation for the reaction in step IV. [1 mark]

6 Polyethene and polyvinyl chloride are examples of synthetic polymers that are widely used in daily life. (a) Name the process in which monomers are [1 mark] joined together to form polymers. (b) (i) Name the monomer that is used for the [1 mark] making of polyethene. (ii) Give a use of polyethene. [1 mark]

4 Table 1 shows the examples and components of five different types of manufactured substances in industry.

Glass Polymer Alloy

Example

Components

Borosilicate glass

Silica, sodium oxide and W

X

(c) The following diagram shows a part of the molecular structure of polyvinyl chloride.

Propene

Magnalium

Composite material

Z

T

Photochromic glass

[1 mark]

(b) Name the main element added to copper to form [1 mark] (i) brass: (ii) bronze: [1 mark]

Nitrate salt

Type

support [1 mark] [1 mark] [1 mark]



Aluminium and Y Concrete (cement, sand and small pebbles) and steel



Glass and silver chloride

H Cl H Cl H Cl ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⎯ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ H H H H H H (i) Draw the structure of its monomer. [1 mark] (ii) Polyvinyl chloride is widely used to make water pipes. State two advantages of PVC pipes compared with iron pipes. [2 marks] (iii) State two ways in which polyvinyl chloride can cause environmental pollution. [2 marks]

(d) What is the source of the raw material used in producing polyethene and polyvinyl chloride?

Table 1

(a) (i) Give the chemical name for silica. [1 mark]

[1 mark]

293

Manufactured Substances in Industry

9

(b) State the conditions for process Q.

Essay Questions 1 (a) Using polyethene as an example, explain the terms polymer and monomer. [4 marks]

What are composite materials? Use two suitable examples to explain the above statement. [10 marks]

(b) Name three examples of (i) natural polymers and (ii) synthetic polymers. State a use for each type of polymer. [6 marks]

(b) Mr Vellu found that an art sculpture made of pure metal is easily dented in his workshop but if he were to use an alloy, the sculpture will not be dented. Using one suitable example, describe an experiment to show how you can compare the hardness of an alloy with that of a pure metal.

(c) Explain briefly how sulphuric acid is manufactured in the industry. [10 marks]

9

2 (a)

[10 marks]

Composite materials are produced for the purpose of improving the original materials and to fulfill specific needs.

Experiments (a) Measure the diameters of the two depressions accurately and record in the spaces provided in Diagram 2. [3 marks]

1 Brass is a copper alloy that is used to make souvenirs and decorative items. Diagram 1 shows the experimental set-up used to compare the hardness of pure copper and brass.

(b) Construct a table to show all the data in the experiment. [3 marks] (c) State the operational definition for alloy. [3 marks] (d) What is the relationship between the diameter of the depression and the hardness of the materials? [3 marks] (e) Referring to results obtained from the experiment, state the conclusion that can be drawn from the experiment. [3 marks] (f) State three variables that must be kept constant in this experiment. [3 marks]

Diagram 1

The 1 kg weight is dropped onto the steel ball bearing and the diameter of the depression formed on the block is measured. The experiment is repeated using a copper block to replace the brass block.

2 Iron nails that are used in the construction of buildings rust more than stainless steel nails when exposed to rain.

The cross section of the diameter of depression of the two materials is shown in Diagram 2.

Referring to the situation above, plan an experiment to compare the rate of rusting of an iron nail and a stainless steel nail. Your explanation should have the following items: (a) Statement of problem (b) All the variables (c) Statements of hypothesis (d) List of materials and apparatus (e) Procedure (f) Tabulation of data [17 marks]

Diagram 2

Manufactured Substances in Industry

294

FORM 5 THEME: Interaction between Chemicals

CHAPTER

1

Rate of Reaction

SPM Topical Analysis 2008

Year Paper

1

Section Number of questions

2

2009

2 A

B

C

1

–

–

3

1

–

6

2010

2 A

B

C

1

–

–

3

1

1

5

2011

2 A

B

C

–

1

–

3

1

1

4

3

2 A

B

C

1

–

–

1

ONCEPT MAP Rate of reaction • Average speed is the amount of reactant used up or the product formed per unit time. 1 • Rate is proportional to ————————— . time taken

Measuring the speed of reaction • from changes in the mass of reactant or product against time. • from changes in the volume of gas produced against time.

Applications in daily activities • Combustion of charcoal • Keeping food in a refrigerator • Cooking food in a pressure cooker

Concentration An increase in concentration increases the speed of reaction.

Particle size A decrease in the particle size (larger total surface area) increases the speed of reaction.

Concentration-time graph • The gradient of the graph indicates the rate of reaction. • The rate of reaction decreases as the reaction proceeds.

RATE OF REACTION

Temperature An increase in temperature increases the speed of reaction.

Collision theory • Explains rate of reaction in terms of effective collisions between reactant particles. • For effective collisions, the particles must have energy equal to or greater than the activation energy. • Any factor that increases the rate of effective (successful) collisions will increase the speed of reaction.

Pressure An increase in pressure increases the speed of reaction (applies only to gases). Catalyst Catalyst increases the rate of reaction.

Uses of catalysts in industry • Iron in the Haber process N2 + 3H2 2NH3 • V2O5 in the Contact process 2SO2 + O2 2SO3 • Pt in the Ostwald process 4NH3 + 5O2 4NO + 6H2O

1.1

Reactants: CaCO3(s) + 2HCl(aq) → Products: CaCl2(aq) + CO2(g) + H2O(l)

Rate of Reaction

The Meaning of Rate of Reaction

1 During a chemical reaction, the reactants are used up as the products are formed.



Example: CaCO3 + 2HCl → CaCl2 + H2O + CO2



Thus, the amounts of reactants decrease (Figure 1.1(a)) while the amounts of products increase as the reaction proceeds (Figure 1.1(b)).



1

Figure 1.1 The graph of amount of substance (mol) against time (minutes)



(b) During the reaction, the following observable changes take place. (i) The mass of calcium carbonate (the reactant) decreases. (ii) The concentration of hydrochloric acid (the reactant) decreases. (iii) The volume of carbon dioxide (the product) produced increases. (c) Thus, the rate of reaction between calcium carbonate and hydrochloric acid can be determined by measuring (i) the decrease in mass of calcium carbonate per unit time, or (ii) the increase in volume of carbon dioxide per unit time. That is, Mass of CaCO3 reacted Reaction rate = — — — — — — — — — –— — — — — — — — — — — , or Time taken Volume of CO2 produced Reaction rate = — — — — — — — — — –— — — — — — — — — — — — — Time taken

2 Definition The rate of reaction is defined as the amount of a reactant used up or the amount of a product obtained per unit time. The gas produced during a reaction can be collected by using a burette or a gas syringe.

Amount of reactant used up Rate of reaction = — — — — — — — — — — — — — — — — — — — — — — — — Time taken

5 The rate of reaction is inversely proportional to the time taken for the reaction to be completed.

or

1 Reaction rate ∝ — — — — — — — — — Time taken

Amount of product obtained Rate of reaction = — — — — — — — — — — — — — — — — — — — — — — — — — Time taken

The reaction is fast if it takes a short time to complete. Conversely, the reaction is slow if it takes a long time for the reaction to complete. 6 Example of a reaction involving a change in colour (a) The reaction between potassium manganate(VII), KMnO4, and ethanedioic acid, H2C2O4, can be represented by the ionic equation below.

3 Methods of measuring reaction rates SPM The amount of a reactant used up or a product ’07/P1, ’09/P1, obtained can be measured in terms of ’11/P1 (a) changes in the mass or concentration of the reactant or product (b) volume of gas produced (c) changes in colour (d) formation of precipitate (e) changes in mass of the reaction mixture 4 Reaction between calcium carbonate and dilute hydrochloric acid (a) The reaction between calcium carbonate (marble chips) and dilute hydrochloric acid can be represented by the equation: Rate of Reaction

5C2O42–(aq) + 16H+(aq) + 2MnO4–(aq) ethanedioate ion manganate(VII) ion (purple) → 10CO2(g) + 8H2O(l) + 2Mn2+(aq) colourless 296

Table 1.1 Example of some fast reactions

(b) Observable changes: When excess ethanedioic acid solution is added to an aqueous solution of potassium manganate(VII), KMnO4, the purple colour of KMnO4 decolourises slowly at room temperature.

Type of reaction

SPM

’09/P1

Example

Neutralisation Reaction between an acid and an alkali. HCl + NaOH → NaCl + H2O

1 Reaction rate ∝ — — — — — — — — — — — — — — — — — — — — — — Time taken for the purple colour to disappear

Reaction between silver nitrate Double decomposition solution and sodium chloride solution to form silver chloride precipitate. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) Combustion

A concentrated solution of manganese(II) ions, Mn , is pink in colour. However, a very dilute solution of Mn2+ ions appears colourless. 2+

7 Example of a reaction involving the formation of a precipitate (a) The reaction between sodium thiosulphate and dilute hydrochloric acid is a slow reaction.

Burning fuel to form carbon dioxide and water. CH4 + 2O2 → CO2 + 2H2O

Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g) + S(s)

Table 1.2 Example of slow reactions

yellow precipitate

Iron rusting

SPM

’09/P1

Example

Type of reaction

(b) Observable changes: When dilute hydro­­­ chloric acid is added to sodium thiosul­ phate solution, the solution becomes cloudy because sulphur is precipitated. Sulphur is a yellow solid, but in small quantities, it appears yellowish-white.

1

Other fast reactions include • burning of magnesium 2Mg + O2 → 2MgO • reaction of sodium or potassium with water 2Na + 2H2O → 2NaOH + H2

Rusting takes place slowly in the presence of oxygen and water. 4Fe + 3O2 + 2H2O → 2Fe2O3•H2O rust

Fermentation of In the presence of yeast, glucose solution fermentation of glucose solution produces alcohol and carbon dioxide. C6H12O6 → 2C2H5OH + 2CO2

1 Rate of reaction ∝ — — — — — — — — — — — — — — — — — — — Time taken for a given amount of sulphur precipitate to form

glucose

Photosynthesis

8 The units used for the rate of reaction will depend on the changes measured. For example, (a) cm3 per unit time (second or minute) for a gas evolved (b) g per unit time or mol per unit time for a solid reactant (c) mol dm–3 per unit time for a reactant in aqueous solution 9 Different chemical reactions take place at different rates. Some reactions occur rapidly and some slowly. Table 1.1 shows some examples of fast reactions. Table 1.2 shows some examples of slow reactions.

alcohol (ethanol)

During photosynthesis, carbon dioxide reacts with water to form glucose and oxygen gas. 6CO2 + 6H2O → C6H12O6 + 6O2

glucose

• The reactions of Groups 1 and 2 metals with oxygen is a fast reaction. However, the reactions of other metals (such as copper) with oxygen are slow reactions. • The rate of decay of the radioactive carbon-14 is very low. For example, 1.0 g of carbon-14 takes 5730 years to disintegrate (decay) to 0.50 g. The rate of decay of carbon-14 is used in archaeology to estimate the age of ancient artifacts. This method is called carbon dating.

297

Rate of Reaction

2 A piece of magnesium ribbon weighing 0.1 g is added to dilute hydrochloric acid. After 5 seconds, all the magnesium had dissolved. What is the average rate of reaction?

The rate of a reaction depends on various factors (Section 1.2). For example, the rate of rusting of iron is increased if the iron is exposed to acid (such as polluted air in industrial areas) or to the salt, sodium chloride in sea air.

Solution Average rate of Mass of magnesium reacted = reaction Time taken 0.1 g = = 0.02 g s–1 5s

1

Measuring Reaction Rates 1 The rate of reaction can be expressed in two SPM ways: ’04/P1, ’10/P1, (a) the average rate of reaction over a period ’11/P1 of time, or (b) the rate of reaction at any given time. 2 The average rate of reaction is the average of the reaction rates over a given period of time. We can measure the average rate of reaction by measuring the change in amount (or concentration) of a reactant or a product over a period of time. 3 For example, the average rate of reaction between magnesium and hydrochloric acid can be determined by measuring the time taken for a piece of magnesium to dissolve completely in the acid.

1

The value obtained is the average rate of reaction over a period of 5 seconds.

4 The average rate of reaction can also be determined from the graph. Based on Figure 1.2, the average rate of the first t1 second V1 =— — cm3 s–1 t1

The average rate of reaction from t1 to t2 (V2 – V1) =— — — — — — — cm3 s–1 (t2 – t1)

SPM

’04/P1

Calcium carbonate reacts with dilute hydrochloric acid to form carbon dioxide. After 1.2 minutes, the volume of gas produced is 100 cm3. Calculate the average rate of reaction in units of (a) cm3 min–1, (b) cm3 s–1.

5 The rate of reaction at any given time is the actual rate of reaction at a given time. The reaction rate at any given time is also known as the instantaneous rate of reaction. 6 The rate of reaction at a given time can be determined by the following methods. (a) By measuring the gradient of the graph of mass of reactant against time (Figure 1.3).

Solution Volume of CO2 produced (a) Average reaction rate = — — — — — — — — — — — — — — Time taken 100 cm3 = — — — — — — — 1.2 min = 83.3 cm3 min–1 Volume of CO2 produced (b) Average reaction rate = — — — — — — — — — — — — — Time taken 83.3 cm3 min–1 = — — — — — — — — — — — — 60 s min–1 = 1.39 cm3 s–1

Rate of Reaction

Figure 1.2

Figure 1.3 Measuring the rate of reaction involving a change in mass at a given time

298

Determining the gradient of the tangent at time, t: The following steps are used to determine the gradient of the tangent at time, t (Figure 1.3).

Analysing a reaction rate curve (i) The steeper the gradient, the higher the rate of reaction.

Step 1 Draw the tangent XY at point P.

Figure 1.5 Comparing the rates of reaction for a given reaction at different times

Step 2 Complete the right-angled triangle XYZ.

(ii) Figure 1.6 shows the graph of volume of hydrogen against time for the reaction between excess zinc powder and dilute hydrochloric acid. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Step 3

1

Measure the lengths of XZ and ZY.

Step 4 Find the gradient of the line XY.

Figure 1.6

a Gradient of the line XY = — b = rate of reaction at time, t (g s–1) (b) By measuring the gradient of the graph of volume of gas produced (product) versus time (Figure 1.4).

(a) Initially (t1), • the graph is steep, • the rate of reaction is high.



(b) As the reaction proceeds (t2), • the graph is less steep, • the rate of reaction decreases because the concentration of hydrochloric acid decreases.

(c) Finally (t3), • the graph becomes horizontal, • the gradient of the graph becomes zero, • the reaction stops because all the hydrochloric acid has reacted.

Figure 1.4 Measuring the rate of reaction at a given time involving a change in the volume of a gas

a Gradient = — b = rate of reaction at time, t (cm3 min–1) 299

Rate of Reaction

3 Consider the reaction between excess magnesium and dilute sulphuric acid: Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g) The reaction stops at 20 seconds. Plot the graph of (a) mass of magnesium against time, (b) concentration of sulphuric acid against time, (c) concentration of magnesium sulphate against time, (d) volume of hydrogen gas against time. Solution (a)

(c)

(d)

1

(b)

There are some parameters which cannot be measured accurately to determine the instantaneous rate of reaction, for example the change in colour or the formation of precipitate.

To find the reaction rates at (a) 90 s, (b) 180 s and (c) the average rate of the reaction between zinc and dilute sulphuric acid Procedure 1 The burette is filled with water and inverted over a basin of water. 2 Using a measuring cylinder, 20.0 cm3 of 0.3 mol dm–3 sulphuric acid is measured out and poured into a conical flask. 3 5.0 g of granulated zinc is then added to the sulphuric acid in the conical flask. 4 The conical flask is then closed and the hydrogen gas produced is collected in the burette by the displacement of water as shown in Figure 1.7. 5 The stopwatch is started immediately. 6 The volume of hydrogen gas collected in the burette is recorded at 30-second intervals.

Figure 1.7

Apparatus

Conical flask, measuring cylinder, delivery tube, burette, basin, retort stand, retort clamp and stopwatch.

Materials

Granulated zinc and 0.3 mol dm–3 sulphuric acid.

Results

Activity 1.1

Time (s)

0

30

60

90

120

150

180

210

240

270

300

330

360

Burette reading (cm3)

50.00 33.00 24.50 18.00 13.00 10.00 6.50 5.00 4.00 3.50 3.00 3.00 3.00

Volume of H2 released (cm3)

0.00 17.00 25.50 32.00 37.00 40.00 43.50 45.00 46.00 46.50 47.00 47.00 47.00

Based on the experimental results, a graph of the volume of hydrogen released against time is plotted. Rate of Reaction

300

Calculation (a) The rate of reaction at 90 s = gradient of the curve at 90 s YZ (52 – 20) cm3 = — — — =— — —— — –— — — — — — XY (180 – 30) s 32 cm3 = — — — — — = 0.213 cm3 s–1 150 s (b) The rate of reaction at 180 s = gradient of the curve at 180 s QR (48 – 30) cm3 = — — — =— — — — — — — — — — — PQ (240 – 18) s 18 cm3 = — — — — — = 0.081 cm3 s–1 222 s (c) The average rate of reaction Total volume of H2 produced = — — — — — — — — — — — — — — — — — — — — — — — — Total time taken 47 cm3 = — — — — — = 0.157 cm3 s–1 300 s

1

Conclusion The rate of reaction decreases as the reaction proceeds.

To measure the rate of reaction between calcium carbonate (CaCO3) and excess hydrochloric acid Apparatus

Conical flask, electronic balance, measuring cylinder and stopwatch.

Materials

Calcium carbonate (CaCO3) pieces, 2.0 mol dm–3 hydrochloric acid and cotton wool.

Figure 1.8

Procedure 1 Using a measuring cylinder, 50 cm3 of 2 mol dm–3 hydrochloric acid is measured out and poured into a dry conical flask. The mouth of the conical flask is covered with some cotton wool. The cotton wool is inserted into the mouth of the conical flask to prevent liquid from splashing out during the reaction. 2 The conical flask is placed on the electronic balance as shown in Figure 1.8. 3 The mass of conical flask, calcium carbonate, hydrochloric acid and cotton wool is recorded. 4 The calcium carbonate is then transferred to the hydrochloric acid in the conical flask and the stopwatch is started immediately. 5 The mass of the conical flask (and its contents) is recorded at one-minute intervals.

Results

Mass of conical flask + contents (g)

0

1

2

3

4

5

6

7

8

60.0 59.1 58.3 57.9 57.4 57.0 56.8 56.5 56.3

Based on the experimental results, a graph of the mass of conical flask and its contents against time is plotted (Figure 1.9). 301

Rate of Reaction

Activity 1.2

Time (min)



0.9 g =— — — — — –— 1.0 min = 0.9 g min–1

(b) The average rate of reaction between 1.4 minutes and 2.2 minutes. Rate of decrease in mass (58.8 – 58.3) g =— — — — — — — — — — — — From the graph (Figure 1.9) (2.2 – 1.4) min = 0.625 g min–1

1

Figure 1.9

Calculation (a) The average rate of reaction for the first minute. Decrease in mass = mass of carbon dioxide produced = (60.0 – 59.1) g See table of results. = 0.9 g Average rate of reaction for the first minute Mass of CO2 produced = — — — — — — — — — — — — — — — — — — – Time taken

(c) The reaction rate at the 5th minute a = gradient of the graph at the 5th minute = — b a = 57.5 – 56.4 = 1.1 g b = 7.0 – 3.4 = 3.6 minutes 1.1 g Gradient = ————— 3.6 min = 0.306 g min–1 Conclusion The rate of reaction decreases as the reaction proceeds. Finally, the reaction will stop when all the calcium carbonate added have reacted.

Solving Numerical Problems Involving Rate of Reaction

SPM

’08/P2

The graph below shows the total volume of oxygen gas produced against time for the decomposition of hydrogen peroxide.

1 The rate of reaction can be stated in terms of (a) the average rate of reaction for a given period of time, or (b) the rate of reaction at any given time (instantaneous rate). 2 The average rate of reaction can be calculated (a) directly from the data given (Example 2) or (b) from the graph drawn (Example 4). 3 The reaction rate at a given time can only be obtained from the gradient of the graph at the given time (Example 5).

At time, t, the maximum volume of oxygen is collected. The gradient of the curve at time, t is zero. Hence, the rate of reaction is zero, that is, the reaction has stopped at time, t.

Rate of Reaction

The rate of reaction is useful to a chemist because he is not satisfied with merely converting one substance to another. In most cases, he wants to obtain the products in the fastest and most economical way.

302

4

5

3.0 g of excess marble (CaCO3) are added to 100 cm3 of dilute hydrochloric acid. Figure 1.10 shows the graph of volume of carbon dioxide produced against time.

Hydrogen peroxide decomposes as represented by the equation: 2H2O2(aq) → 2H2O(l) + O2(g) The results of an experiment on the decomposition of hydrogen peroxide are given below. Time (s)

0

15 30 45 60 90

Volume of O2 (cm3)

0

16 30 40 48 56

Calculate the rate of reaction at 40 seconds in units of (a) cm3 s–1, (b) cm3 min–1.

1

Solution (a)

Figure 1.10

Calculate (a) the average rate of reaction, (b) the concentration of hydrochloric acid in mol dm–3. [1 mol of any gas occupies 24 dm3 at room conditions] Solution (a) Total volume of carbon dioxide evolved = 360 cm3 Time taken = 8.0 minutes 360 cm3 Average rate of reaction = — — — — — — — 8 min = 45 cm3 min–1 (b) Number of moles of CO2 evolved 360 cm3 =— — — — — — — — — — — — — — = 0.015 (24  1000) cm3 The rate of reaction at 40 s

CaCO3 + 2HCl → CaCl2 + H2O + CO2 Mole ratio of HCl : CO2 =2:1 ? : 0.015

= gradient at 40 s a (49 – 21) cm3 =—=— — — — — — — — — — — b (58 – 18) s

From equation

According to the equation, number of moles of hydrochloric acid used = 2  0.015 = 0.03 mol. Concentration of hydrochloric acid Number of moles 0.03 mol =— — — — — — — — — — — — — — =— — — — — — — Volume (in dm3) 0.1 dm3 = 0.3 mol dm–3

obtained from the graph

= 0.70 cm3 s–1 (b) 1 minute = 60 seconds ∴Rate of reaction in cm3 min–1 = 0.70 cm3 s–1  60 s min–1 = 42 cm3 min –1

From the question: 100 cm3 = 0.1 dm3

303

Rate of Reaction

1

’09

Which of the following is correctly matched with its rate of reaction? High reaction rate

Low reaction rate

A

Combustion of fuels

Respiration

B

Combustion of fuels

Double decomposition between silver nitrate and sodium chloride solution

C

Rusting of iron

Fermentation of glucose solution

D

Respiration

Neutralisation reaction between an acid and an alkali

1

Comments Combustion, double decomposition and neutralisation are fast reactions. Rusting and respiration are slow reactions. Answer A

1.1 1 Which of the following reactions occur at (a) a high rate, (b) a low rate? I Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s) II 2Cu(s) + O2(g) → 2CuO(s) III S2O32–(aq) + 2H+(aq) → S(s) + H2O(l) + SO2(g) IV 4K(s) + O2(g) → 2K2O(s) 2 You are given the chemicals and apparatus as listed below. • A piece of zinc of mass 2.0 g • A beaker containing sulphuric acid • A stopwatch (a) Using the chemicals and apparatus given, describe an experiment to measure the rate of reaction between zinc and sulphuric acid. (b) State the units for the rate of reaction. (c) State two assumptions for this experiment.

Figure 1.11

3 A student intends to study the rate of reaction between iron and dilute sulphuric acid. The equation for the reaction is as follows.

(a) What is the total time required for the magnesium ribbon to react completely with hydrochloric acid? (b) Based on the graph, is the reaction rate at the first minute higher or lower than the reaction at the second minute? Explain your answer. (c) Is this a normal behaviour? Suggest one reason for this behaviour. (d) The reaction between hydrochloric acid and another metal produces 12 cm3 of hydrogen after 1.0 minute. Is this reaction rate higher or lower than the reaction between magnesium and hydrochloric acid? Explain your answer.

Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g) Suggest two methods that he can use to measure the rate of reaction. 4 The graph in Figure 1.11 shows the results of an experiment to measure the rate of reaction of magnesium ribbon with an excess of dilute hydrochloric acid.

Rate of Reaction

304

What is the average rate of reaction? (b) 4.0 g of magnesium is added to excess dilute sulphuric acid. If the average rate of reaction is 0.0030 mol s–1, what is the mass of magnesium unreacted after 0.5 minute? [Relative atomic mass of Mg = 24]

5 The table below shows the results for two experiments carried out under room conditions. Reaction

Result

I

1 g of nickel powder + 50 cm3 of 1 mol dm–3 hydrochloric acid

Time taken to collect 60 cm3 of hydrogen gas = 120 s

1 g of zinc powder + 50 cm3 of 1 mol dm–3 hydrochloric acid

Time taken to collect 45 cm3 of hydrogen gas = 56 s

II

7 In the presence of manganese(IV) oxide, hydrogen peroxide decomposes according to the equation: MnO2 2H2O2(aq) ⎯⎯→ 2H2O(l) + O2(g) A sample of hydrogen peroxide decomposed in the presence of a catalyst and the volume of oxygen gas produced was collected at regular time intervals. The results of the experiment were recorded in the following table.

Based on the information given in the table above, predict which metal is more reactive, nickel or zinc? 6 (a) The volume of hydrogen gas collected at regular intervals for the reaction between granulated zinc and dilute hydrochloric acid is shown below.

1.2

Time (s)

Volume of H2 (cm3)

0

0

20

16

40

26

60

32

70

36

80

36

1 1— 2

1 2— 2

Time (min)

0

1

Volume of O2 (cm3)

0

32 46 56 64 69 74 74

2

3

4

5

1

Experiment

Calculate (a) the average rate of reaction for the first 144 seconds. (b) the average rate of reaction for the overall reaction in cm3 s–1. (c) the average rate of reaction between the first minute and the 3rd minute. (d) the rate of reaction at the 150th second.

Factors that Affect the SPM Rate of Reaction

’08/P1

Uranium is the radioactive isotope used for making nuclear bombs. Uranium decays slowly to form lead. The decay of uranium and other radioactive elements is unique. These nuclear reactions are not influenced by factors such as surface area, temperature and catalyst.

1 The rate of reaction is affected by the following factors: (a) Total surface area (or particle size) of the solid reactant (b) Concentration of reactant (c) Temperature of reaction (d) Use of catalyst (e) Pressure (for reactions involving gases) 2 When the condition of reaction changes, the rate of reaction also changes. 3 Table 1.3 explains briefly how these conditions of reaction affect the rate of reaction between zinc metal and dilute sulphuric acid.

4 Reaction involving gases (a) Changes in pressure will not affect reactions in aqueous solutions. (b) Changes in pressure will only affect reactions involving gases. (c) Increasing the pressure will compress more gas molecules into a given space. Hence the gaseous particles will collide more frequently and the rate of reaction increases. 305

Rate of Reaction

Table 1.3

Surface area (particle size)

• When zinc foil is broken into smaller pieces, the total surface area increases. • The smaller the size of zinc foil, the greater the total surface area exposed to the hydrogen ions. Hence the rate of reaction increases.

Concentration of reactant

1

• In dilute acid, there are not so many hydrogen ions present. • In more concentrated acid, there are more hydrogen ions in the solution. Hence the rate of reaction increases.

Temperature of reaction

• When the temperature of a reaction increases, the particles move faster because they have higher kinetic energy. Hence the rate of reaction increases.

Catalyst Some catalysts used in industry. The catalysts are in pellet form for larger surface area.

SPM

’10/P2

• Manipulated variable: Size (total surface area) of magnesium • Responding variable: Time taken to collect 60 cm3 of hydrogen gas • Constant variables: Temperature, concentration and volume of sulphuric acid as well as mass of magnesium 2 The results of the experiments are shown below.

Factors that Influence the Rate of Reaction Effect of Surface Area on the Rate of Reaction 1 Two experiments are carried out to study the rate of reaction between magnesium and dilute sulphuric acid under different conditions. Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g) Experiment

Conditions of experiment

I

1.0 g of magnesium ribbon and 50 cm3 of 1.0 mol dm–3 sulphuric acid 1.0 g of magnesium powder and 50 cm3 of 1.0 mol dm–3 sulphuric acid

II

The size (surface area) of magnesium is manipulated

Rate of Reaction

A catalyst will increase the rate of reaction. This will be explained in Section 1.3.

306

Time taken to collect 60 cm3 of H2 (s) 30

Average rate of reaction (cm3 s–1) 2

10

6

3 The results show that the time taken to collect 60 cm3 of gas using magnesium powder is shorter than using magnesium ribbon. This is due to the smaller size of particles (total surface area is greater) in magnesium powder than in magnesium ribbon.

1 mol dm–3 hydrochloric acid. This means that the higher the concentration of hydrochloric acid, the higher the rate of reaction. 4 (a) It is important to know • how to plot graphs (on the same axis), or • how to interpret graphs on rate of reaction if information on the conditions of reaction are given. (b) The two features on the graph to be considered are • the gradient of the graph which shows the rate of reaction, • the height of the graph which shows the total amount of product formed.

Effect of Concentration of Reactant on the Rate of Reaction 1 Two experiments are carried out to study the SPM rate of reaction between magnesium and ’04/P1 ’05/P3 hydrochloric acid. ’06/P1

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) • Manipulated variable: Concentration of hydrochloric acid • Responding variable: Time taken for magnesium to dissolve completely • Constant variables: Size of magnesium ribbon, volume of hydrochloric acid and temperature of experiment 2 The results of the experiments are shown below. Conditions of experiment

Experiment

1 Two experiments are carried out to study the rate of reaction between zinc and sulphuric acid at different temperatures. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) • Manipulated variable: Temperature of sulphuric acid • Responding variable: Volume of hydrogen gas evolved • Constant variables: Mass of zinc, concentration and volume of sulphuric acid 2 The results are shown in Figure 1.12.

Time taken for magnesium to dissolve completely (s)

I

5 cm magnesium ribbon and 50 cm3 of 1 mol dm–3 hydrochloric acid

78

II

5 cm magnesium ribbon and 50 cm3 of 2 mol dm–3 hydrochloric acid

39

Concentration of acid is manipulated

Figure 1.12 Comparing the rates of reaction at different temperatures

3 The shorter the time taken for a reaction to complete, the higher the rate of reaction. The results show that the time taken for magnesium to react completely in 2 mol dm–3 hydrochloric acid is shorter than that in

3 The results show that the higher the temperature of sulphuric acid, the steeper the graph and hence, the higher the rate of reaction.

Experiment

Mass of zinc

Volume of sulphuric acid

Concentration of sulphuric acid

Temperature of sulphuric acid

I

1.0 g of zinc powder

20 cm3

0.1 mol dm–3

28 °C

II

1.0 g of zinc powder

20 cm

0.1 mol dm

35 °C

3

Conditions remain unchanged

307

–3

Temperature of acid is manipulated

Rate of Reaction

1

Effect of Temperature on the Rate of Reaction

4 In general, the rate of reaction increases if the temperature of the reactants is increased.

occurs. In this reaction, manganese(IV) oxide acts as a catalyst and catalyses the decomposition of hydrogen peroxide to give oxygen gas and water.

Effect of Catalysts on the Rate of Reaction

MnO2

1 Definition A catalyst is a substance that increases the rate of a reaction but is itself chemically unchanged at the end of the reaction. 2 In contrast, a substance that decreases the rate of a chemical reaction is called an inhibitor. 3 At room temperature, hydrogen peroxide decomposes very slowly. But when a very small amount of manganese(IV) oxide is added to hydrogen peroxide, a vigorous effervescence

1

The Characteristics of Catalysts Only a small amount of catalyst is needed to increase the rate of reaction.

The physical appearance of a catalyst may change at the end of the reaction. For example, small pieces of catalyst may become fine powder after the reaction.

2H2O2(aq) ⎯⎯⎯→ 2H2O(l) + O2(g)

4 The reaction between zinc and dilute acid is a slow reaction. Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) When it is catalysed by copper(II) sulphate solution, the reaction speeds up.

SPM

’06/P1, ’08/P1, ’10/P1, ’11/P1

• The catalyst remains chemically unchanged after the reaction. • Thus the chemical properties, mass and chemical composition of the catalyst remain unchanged at the end of the reaction.

Characteristics of a catalyst

A catalyst increases the rate of a chemical reaction but it does not change (increase or decrease) the yield of a chemical reaction. CuSO4

Zn(s) + H SO (aq) ⎯⎯⎯→ ZnSO4(aq) + H2(g)

2 4 catalyst

A catalyst lowers the activation energy of a reaction (see Section 1.3 – The Collision Theory)

(a) In general, catalysts are highly specific. For example, iron catalyses the reaction: N2 + 3H2 2NH3 but not the reaction: 2SO2 + O2 2SO3 (b) However, some catalysts can catalyse several different reactions. For example, MnO2 can catalyse the following reactions: MnO2

2H2O2(aq) ⎯⎯⎯→ 2H2O(l) + O2(g)

MnO2

2KClO3(s) ⎯⎯⎯→ 2KCl(s) + 3O2(g)

Catalysts can be poisoned by impurities. When a catalyst is poisoned, its effectiveness as a catalyst is decreased.

Rate of Reaction

308

Examples of Catalysts 1 Transition metals and compounds of transition metals are often used as catalysts for industrial processes. 2 Table 1.4 shows some examples of catalysts and the reactions catalysed by them. Table 1.4 Some common catalysts

Type of reaction

Catalyst used

(a) Haber process for the manufacture of ammonia. N2(g) + 3H2(g)

Iron, Fe

2NH3(g)

(b) Contact process for producing sulphur trioxide. 2SO2(g) + O2(g)

Vanadium(V) oxide, V2O5

2SO3(g)

Sulphur trioxide is used for the manufacture of sulphuric acid.

(c) Ostwald process for producing nitrogen monoxide. 4NH3(g) + 5O2(g)

Platinum, Pt

4NO(g) + 6H2O(g)

Nitrogen monoxide is used for the manufacture of nitric acid. Nickel, Ni

(e) Cracking process When big alkane molecules are passed over a catalyst at 600 °C, a mixture of small alkane and alkene molecules is produced. This process is called catalytic cracking. (Refer Sections 2.2 and 2.3 on alkanes and alkenes)

Aluminium oxide, Al2O3 or Silicon(IV) oxide, SiO2

2

1

(d) Manufacture of margarine In the presence of a catalyst at 200 °C, vegetable oils react with hydrogen to produce margarine. This process is called hydrogenation.

Effect of Pressure on the Rate of Reaction

’06

1 The changes in pressure will only affect reactions involving gases. An increase in pressure increases the rate of reaction. In contrast, a decrease in pressure decreases the rate of reaction. In the following reactions involving gases, the rate of reaction increases if the pressure is increased.

Which of the following statements about catalysts are true? I A catalyst is specific in its reaction. II A catalyst changes the quantity of product formed. III Only a small amount of a catalyst is needed to change the rate of reaction. IV The chemical properties of the catalyst remain unchanged at the end of a reaction. A I and II only C I, II and III only B II and IV only D I, III and IV only Answer D

N2(g) + 3H2(g)

2NH3(g) (Haber process)

2SO2(g) + O2(g)

2SO3(g) (Contact process)

2 Pressure has no effect on reactions involving only solids or liquids. For example, the following reactions are not affected by changes in pressure. CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

A catalyst takes part in a chemical reaction. In actual fact, a catalyst combines with the reactants to form an unstable intermediate species. This species then decomposes to re-form the catalyst and to produce the products.

2H2O2(aq) → 2H2O(l) + O2(g)

309

Rate of Reaction

1.1 To investigate the effect of the surface area of a reactant on the rate of reaction

SPM

’06/P2

Procedure

Problem statement How does the surface area of a solid reactant affect the rate of reaction?

1

Hypothesis The smaller the size of the marble chips, that is, the larger the total surface area of the marble chips, the higher the rate of reaction. Variables (a) Manipulated variable : Size of the marble chips used (b) Responding variable : Volume of gas given off at 30-second intervals (c) Constant variables : Temperature of the experiment, mass of marble chips, concentration and volume of hydrochloric acid Apparatus

Conical flask, delivery tube, retort stand and clamp, burette, measuring cylinder and stopwatch.

Materials

Marble chips, powdered marble and 0.08 mol dm–3 hydrochloric acid.

Figure 1.13

1 A burette is filled with water and inverted over a basin containing water. The burette is clamped to the retort stand. The water level in the burette is adjusted and the initial burette reading is recorded. 2 5.0 g of marble chips are placed in a small conical flask. 3 50 cm3 of 0.08 mol dm–3 hydrochloric acid is added to the marble chips. 4 The conical flask is then stoppered and the stopwatch is started immediately (Figure 1.13). 5 The burette readings are recorded at 30-second intervals.

Experiment I The rate of reaction using large marble chips Results Time (s)

0

30

60

90

120

150

180

210

240

Burette reading (cm3)

50.00 45.50 41.50 38.00 35.00 33.00 31.00 29.00 28.00

Volume of gas (cm3)

0.00

Experiment II

4.50

8.50 12.00 15.00 17.00 19.00 21.00 22.00

The rate of reaction using powdered marble

Constant variable is also known as fixed variable or controlled variable.

Experiment 1.1

Procedure 1 Steps 1 to 4 in Experiment I are repeated using 5.0 g of powdered marble. All other conditions such as temperature, volume and concentration of hydrochloric acid are kept constant. 2 The results of the experiment are recorded in the following table. Results Time (s)

0

30

60

90

120

150

180

210

240

Burette reading (cm3)

50.00 42.00 35.00 29.50 25.50 22.00 19.50 17.50 16.00

Volume of gas (cm3)

0.00

Rate of Reaction

8.00 15.00 20.50 24.50 28.00 30.50 32.50 34.00 310

volume of the hydrochloric acid used in both the experiments are the same. 3 The gradients of the graphs for Experiments I and II become less steep as the reactions proceed. This shows that the rates of reaction (a) are very high at the beginning of the reactions, (b) decrease as the reactions proceed, (c) become zero when the reactions are completed. At this time, the graphs become horizontal. 4 The rate of reaction between the marble and hydrochloric acid decreases because (a) the mass of the remaining unreacted marble decreases, (b) the concentration of hydrochloric acid decreases. 5 The reaction in Experiment I stops after t2 minutes while the reaction in Experiment II stops after t1 minutes, where t1 < t2. This shows that the rate of reaction for Experiment II (powdered marble) is higher than the rate of reaction for Experiment I (marble chips). 6 The total volume of carbon dioxide collected in the burette is usually slightly less than the theoretical value (48 cm3 for the experiment above). This is because carbon dioxide is slightly soluble in water. To overcome this problem, a gas syringe is used to collect carbon dioxide released during the experiment (Figure 1.16).

Figure 1.14

Discussion 1 Figure 1.15 shows the graphs that will be obtained if the reactions in Experiments I and II are completed.

Figure 1.16 Measuring the volume of gas using a gas syringe

Same maximum volume of CO2 collected because mass of CaCO3, concentration and volume of HCl are kept constant.

Conclusion Graph II is steeper than graph I. This shows that the rate of reaction in Experiment II is higher than the rate of reaction in Experiment I as powdered marble is used in Experiment II. Thus, the rate is higher with powdered marble than with marble chips. Hence, we can conclude that the smaller the particle size, the larger the total surface area exposed for reaction and the higher the rate of reaction. The hypothesis is accepted.

Figure 1.15

2 Figure 1.15 shows that both graphs level off at the same value. This indicates that the maximum volume of carbon dioxide collected at the end of reaction for both Experiments I and II are the same (that is, 44 cm3). This happens because the mass of the marble, concentration and

311

Rate of Reaction

1

Based on the results obtained, a graph of the total volume of carbon dioxide produced against time for each experiment is plotted on the same axes (Figure 1.14).

1.2 To study the effect of concentration on the rate of reaction between sodium thiosulphate solution and dilute sulphuric acid

SPM

’07/P1, ’11/P2

Apparatus 10 cm3 and 100 cm3 measuring cylinders, 100 cm3 conical flask, white paper marked with a cross ‘X’ and stopwatch.

Problem statement How does the concentration of a reactant affect the rate of reaction between sodium thiosulphate and dilute sulphuric acid?

Materials 0.2 mol dm–3 sodium thiosulphate solution, 1.0 mol dm–3 sulphuric acid and distilled water. Procedure 1 50 cm3 of 0.2 mol dm–3 sodium thiosulphate solution is measured out using a 100 cm3 measuring cylinder. The solution is then poured into a clean, dry conical flask. 2 The conical flask is placed on a piece of paper with a cross ‘X’ marked on it (Figure 1.17). 3 5 cm3 of dilute sulphuric acid is measured out by using a 10 cm3 measuring cylinder. The acid is then quickly poured into sodium thiosulphate solution. The stopwatch is started immediately. 4 The reaction mixture is swirled once and the cross ‘X’ is viewed from above. A yellow precipitate will appear slowly in the conical flask. 5 The stopwatch is stopped as soon as the cross disappears from view and the time taken is recorded. 6 Steps 1 to 5 are repeated with different mixtures of sodium thiosulphate solution and distilled water as shown in the following table.

1

Figure 1.17

Hypothesis The more concentrated the sodium thiosulphate solution, the higher the rate of reaction. Variables (a) Manipulated variable: Concentration of sodium thiosulphate solution (b) Responding variable: Time taken for the cross ‘X’ to disappear (c) Constant variables: Concentration and volume of dilute sulphuric acid as well as the temperatures of the solutions Results Experiment

1

2

3

4

5

Volume of Na2S2O3 (cm )

50

40

30

20

10

Volume of water (cm )

0

10

20

30

40

Volume of H2SO4 (cm )

5

5

5

5

5

0.20

0.16

0.12

0.08

0.04

24

30

42

62

111

3

3

3

Concentration of Na2S2O3 (mol dm–3) Time taken (s)

Experiment 1.2

1 — — — — (s–1) Time

0.042 0.033 0.024 0.016 0.009 The ionic equation is as follows:

Discussion 1 The following equation shows the reaction between sodium thiosulphate, Na2S2O3 and dilute sulphuric acid: Na2S2O3(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) + SO2(g) + S(s) Rate of Reaction

Different volumes (V1) of Na2S2O3 solution are diluted with water to make up to 50 cm3 solution (V2).

S2O32–(aq) + 2H+(aq) → S(s) + SO2(g) + H2O(l) The sulphur is precipitated as fine yellow particles that cause the solution to turn cloudy. 312

2 As the amount of sulphur increases, the cross ‘X’ becomes more and more difficult to see. Finally, the cross ‘X’ disappears from view when a certain mass of sulphur is precipitated. Hence, the time recorded for the disappearance of the cross ‘X’ is the time taken for the formation of a fixed mass of sulphur. Mass of sulphur produced 3 Rate of reaction = — — — — — — — — — — — — — — — — — — — — — Time taken 1 Hence, rate of reaction ∝ — — — — — — — — — — — — — — — — — — — Time taken for the cross ‘X’ to disappear 4 The concentration of sodium thiosulphate solution after mixing with water can be obtained by using the following formula: M1V1 Concentration of Na2S2O3 = — — — — — V2

7 The conical flask used for each experiment must have the same size (for example, 100 cm3 volume). If the conical flask of a larger size is used, the time, t, taken for the cross ‘X’ to disappear will increase. Conversely, if a smaller conical flask is used, the time taken for the cross to disappear will be shorter. 8 If the experiment is repeated with dilute sulphuric acid of different concentrations, but the concentration of sodium thiosulphate is kept constant, the rate of reaction will also be directly proportional to the concentration of the acid used.

0.2  Volume of Na2S2O3 used =— — — — — — — — — — — — — — — — — — — — — — — — — mol dm–3 50 5 Based on the results obtained, two graphs can be plotted. (a) The graph of concentration of sodium thiosulphate against time (Graph I, Figure 1.18).

Conclusion 1 (a) From graph I, we can conclude that the higher the concentration of sodium thiosulphate, the shorter the time taken for a certain mass of sulphur to be precipitated, that is, for the cross ‘X’ to disappear from view. (b) This means that the higher the concentration of sodium thiosulphate, the higher the rate of reaction. 2 From graph II, it can be concluded that the concentration of sodium thiosulphate is directly 1 proportional to — — —. time 1 Concentration of Na2S2O3 ∝ — — — …(1) time 1 3 But the rate of reaction is ∝ — — — … (2) time Hence, combining equations (1) and (2), we have, 1 — — ∝ reaction rate concentration of Na2S2O3 ∝ — time That is, rate of reaction ∝ concentration of Na2S2O3 solution. The hypothesis is accepted.

Figure 1.18

(b) The graph of concentration of sodium 1 thiosulphate against — — — (Graph II, Figure 1.19). time 6 Different volumes of distilled water are added to sodium thiosulphate solution so that the final volume of the diluted sodium thiosulphate solution is 50 cm3 in each experiment. Hence, the concentration of sodium thiosulphate solution is directly proportional to its volume before dilution. 313

Rate of Reaction

1

Figure 1.19

3

’06

Which of the following reactants produces the highest rate of reaction with magnesium powder? A 50 cm3 of 0.5 mol dm–­3 nitric acid B 50 cm3 of 0.5 mol dm–­3 ethanoic acid C 50 cm3 of 0.5 mol dm–­3 sulphuric acid D 50 cm3 of 0.5 mol dm–3 hydrochloric acid Comments Magnesium reacts with the hydrogen ions of the acids.

Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) Ethanoic acid is a weak acid and produces very few H+ ions. HNO3, H2SO4 and HCl are strong acids. But each mole of H2SO4 produces two moles of H+ ions. Hence, 50 cm3 of 0.5 mol dm–3 sulphuric acid contains the highest concentration of H+ ions. Answer C

Experiment 1.3

1

1.3 SPM To study the effect of temperature on the rate of reaction between sodium ’05/P1 thiosulphate solution and dilute sulphuric acid 5 The stopwatch is started immediately and the Problem statement conical flask is swirled gently. How does temperature affect the rate of reaction between 6 The cross ‘X’ is viewed from above. The stopwatch sodium thiosulphate solution and sulphuric acid? is stopped as soon as the cross disappears from view Hypothesis The higher the temperature of the and the time taken is recorded. reactant, the higher the rate of 7 The solution in the conical flask is poured out. reaction. The conical flask is washed thoroughly and dried. 50 cm3 of 0.1 mol dm–3 sodium thiosulphate Variables solution is poured into the conical flask. (a) Manipulated variable: The temperature of 8 The solution is heated over a wire gauze until the sodium thiosulphate solution temperature reaches about 45 °C (Figure 1.21). (b) Responding variable: The time taken for the cross ‘X’ to disappear (c) Constant variables: The concentrations and volumes of both sodium thiosulphate solution and dilute sulphuric acid

Apparatus

Conical flask, 10 cm3 measuring cylinder, thermometer, stopwatch, white paper marked with a cross ‘X’, wire gauze, tripod stand and Bunsen burner.

Materials

0.1 mol dm–3 sodium thiosulphate solution and 1.0 mol dm–3 sulphuric acid.

Figure 1.20

Procedure 1 50 cm3 of 0.1 mol dm–3 sodium thiosulphate solution is poured into a clean, dry conical flask. 2 The temperature of the sodium thiosulphate solution is measured with a thermometer. 3 The conical flask is placed on a white paper marked with a cross ‘X’ (Figure 1.20). 4 5 cm3 of 1 mol dm–3 sulphuric acid is quickly poured into the sodium thiosulphate solution. Rate of Reaction

Figure 1.21

314

9 The hot conical flask is placed over a white paper marked with a cross ‘X’. 10 5 cm3 of 1 mol dm–3 sulphuric acid is measured out using a 10 cm3 measuring cylinder. 11 When the temperature of sodium thiosulphate solution falls to 40°C, the sulphuric acid is quickly poured into the thiosulphate solution. 12 The stopwatch is started immediately and the conical flask is swirled gently. 13 The cross ‘X’ is viewed from the top and the time taken for the cross to disappear from view is recorded. 14 Steps 7 to13 are repeated at higher temperatures as shown in the following table.

Experiment

1

2

3

4

5

Temperature (°C)

30

40

50

55

60

Time (s)

52

27

16

13

10

1 — — — — (s–1) Time

Discussion 1 The graph shows that the temperature of sodium thiosulphate solution is proportional (but not 1 linearly) to — — —. time 1 2 Temperature ∝ — — — … (1) time 1 But rate of reaction ∝ — — — … (2) time Combining equations (1) and (2), we have, Rate of reaction ∝ temperature

0.019 0.037 0.063 0.077 0.100

Based on the results of the experiment, a graph of temperature of sodium thiosulphate solution against 1 — — — is plotted (Figure 1.22). time

Conclusion The higher the temperature of the experiment, the higher the rate of reaction.

4

’05

The rate of reaction between sodium thiosulphate solution and dilute sulphuric acid can be determined by using the arrangement of apparatus as shown below. Which of the following conditions will cause the mark ‘X’ to take the shortest time to disappear from sight? Sulphuric acid

A B C D

Sodium thiosulphate solution

Volume (cm3)

Concentration (mol dm–3)

Volume (cm3)

Concentration (mol dm–3)

Temperature (°C)

5 5 5 10

1.0 1.0 0.5 0.5

45 45 45 40

0.5 0.5 0.5 0.5

30 35 30 35

315

Rate of Reaction

1

1 Figure 1.22 Graph of temperature against — — — time

Results

Comments The shorter the time taken for the mark ‘X’ to disappear, the faster the reaction. The rate of reaction is affected by temperature and concentration. The higher the temperature, the faster the reaction. (Answer B or D is correct). The higher the concentration of sulphuric acid or sodium thiosulphate in the reaction mixture, the faster the reaction (Answer D is incorrect). Answer B

1.4 To study the effect of a catalyst on the rate of decomposition of hydrogen peroxide

1

Problem statement How do catalysts affect the rate of decomposition of hydrogen peroxide? Hypothesis Manganese(IV) oxide increases the decomposition of hydrogen peroxide.

rate

4 0.5 g of manganese(IV) oxide, MnO2, is added to hydrogen peroxide and shaken. The changes that take place in the test tube and on the glowing splint are recorded.

of

Results

Variables (a) Manipulated variable: The presence of manganese(IV) oxide (b) Responding variable: The release of oxygen gas (c) Constant variables: Volume and concen­tration of hydrogen peroxide

Observation Experiment

Apparatus Test tube and wooden splint Materials Hydrogen peroxide and manganese(IV) oxide Procedure 1 A test tube is half-filled with hydrogen peroxide. 2 A glowing splint is placed at the mouth of the test tube to test for the gas evolved (Figure 1.23).

Inside the test tube

On the glowing splint

H2O2 without No effervescence MnO2

The glowing splint does not light up.

H2O2 with MnO2

The glowing splint is rekindled and burns brightly.

Bubbles of oxygen gas are produced

Discussion 1 The following equation shows the decomposition of hydrogen peroxide: 2H2O2(aq) → 2H2O(l) + O2(g)

Experiment 1.4

2 The glowing splint is rekindled in the presence of oxygen gas. Conclusion The rate of evolution of oxygen gas increases when manganese(IV) oxide is added to hydrogen peroxide. This proves that manganese(IV) oxide acts as a catalyst and speeds up the decomposition of hydrogen peroxide to water and oxygen. The hypothesis is accepted.

Figure 1.23 The effect of a catalyst on the decomposition of hydrogen peroxide

3 The changes that take place inside the test tube and on the glowing splint are recorded.

Rate of Reaction

316

reaction in Experiment II. We can therefore conclude that the higher the concentration of hydrogen peroxide, the higher the rate of reaction. (c) The maximum volume of oxygen gas produced in Experiment I is twice that produced in Experiment II. This is because the number of moles of hydrogen peroxide used in Experiment I is twice that used in Experiment II.

The reaction mixture remaining after Experiment 1.4 can be filtered to obtain the manganese(IV) oxide. It is found that (a) the mass of manganese(IV) oxide before and after the experiment is the same (0.5 g), (b) the chemical properties of manganese(IV) oxide remain unchanged.

SPM

’05/P1

Explaining the Effectiveness of Different Catalysts on the Rate of Decomposition of Hydrogen Peroxide

1 The graphs in Figure 1.24 show the effect of concentration of hydrogen peroxide on the rate of decomposition of hydrogen peroxide.

1 Figure 1.25 shows the results of an experiment carried out to study the effect of different catalysts (of the same mass) on the rate of decomposition of hydrogen peroxide.

Graph I: more O2 produced and higher rate of reaction because larger volume and higher concentration of H2O2 is used.

Maximum volumes of O2 collected are the same for Experiments I and II because the concentration and volume of H2O2 used are the same.

Figure 1.24 The effect of concentration of hydrogen peroxide on the rate of decomposition of hydrogen peroxide

Figure 1.25 The effect of different catalysts on the rate of reaction

In Experiment I, 50 cm of 0.14 mol dm of hydrogen peroxide and 0.2 g of manganese(IV) oxide are used. In Experiment II, a solution containing 25 cm3 of the same hydrogen peroxide mixed with 25 cm3 of water and 0.2 g of manganese(IV) oxide are used. For both the experiments, the temperature is kept constant. 2 (a) For Experiment I Concentration of H2O2 = 0.14 mol dm–3 higher concentration 3

–3

In Experiment I, 50 cm3 of hydrogen peroxide and 0.5 g of manganese(IV) oxide are used. In Experiment II, 50 cm3 of hydrogen peroxide and 0.5 g of iron(III) oxide are used. For both the experiments, the concentration and volume of hydrogen peroxide as well as the temperature are kept constant. 2 Analysis of the reaction rate curve in Figure 1.25. (a) At any particular instant, the gradient of graph I is greater than the gradient of graph II. This means that the rate of reaction in Experiment I is higher than the rate of reaction in Experiment II. Thus, the experiment proves that manganese(IV) oxide is a more effective catalyst than iron(III) oxide in the decomposition of hydrogen peroxide. (b) The maximum volumes of oxygen gas collected in both the experiments are the same because the volume and concentration of hydrogen peroxide used are the same. This experiment shows that a catalyst does not change the yield of the products.

For experiment 2, hydrogen peroxide is diluted. (M1V1)before dilution = (M2V2)after dilution

Concentration of H2O2 after dilution 0.14 mol dm–3  25 cm3 =— — — — — — — — — — — — — — — — — — — — 50 cm3 lower concentration = 0.07 mol dm–3 (b) At any particular instant, the gradient of graph I is greater than the gradient of graph II. This means that the rate of reaction in Experiment I is higher than the rate of 317

Rate of Reaction

1

The Effect of Concentration of Hydrogen Peroxide on the Rate of Reaction

Effect of surface area

Effect of temperature

Reaction of HCl(aq) with marble chips

Reaction of HCl(aq) with zinc powder

Graph I : Large marble chips mass of Graph II : Small marble chips marble is Graph III : Powdered marble the same

Graph I : at 30°C mass of zinc (in excess), volume Graph II : at 40°C and concentration of HCl are kept constant

Comment: The smaller the size of marble chips, the higher the reaction rate.

Comment: When the temperature is raised, the rate of reaction also increases.

Factors affecting the rate of reaction Effect of catalyst

1

Effect of concentration of reactant Reaction of HCl(aq) with magnesium

Decomposition of hydrogen peroxide

Graph I : 0.5 mol dm–3 HCl mass of Mg and Graph II : 1.0 mol dm–3 HCl volume of HCl (in excess) Graph III : 2.0 mol dm–3 HCl are kept constant

Graph I : Fe2O3 used as catalyst Graph II : MnO2 used as catalyst Graph III : No catalyst

Comment: When the concentration of hydrochloric acid increases, the rate of reaction also increases.

Comment: A catalyst increases the reaction rate. MnO2 is a more effective catalyst than Fe2O3.

Effect of concentration and volume of acid used Reaction of HCl(aq) with magnesium

Reaction of HCI(aq) with magnesium

Graph I : Mg in excess 20 cm3 of 0.2 mol dm–3 HCI Graph II : Mg in excess 20 cm3 of 0.1 mol dm–3 HCl

Graph I : Mg in excess 10 cm3 of 0.2 mol dm–3 HCI Graph II : Mg in excess 30 cm3 of 0.1 mol dm–3 HCl

Comment: Graph Reaction rate Amount of H2 released

Rate of Reaction

I

II

Higher

Lower

0.002 mol (48 cm3)

0.001 mol (24 cm3)

Comment: Graph Reaction rate Amount of H2 released

318

I

II

Higher

Lower

0.001 mol (24 cm3)

0.0015 mol (36 cm3)

is kept in a refrigerator will last longer because the decaying reaction that destroys the food can be slowed down. 4 In the supermarkets, fish, meat and other types of fresh foods are kept in deep-freeze compartments where the temperature is about –20 °C. This keeps the food fresh for a few months because the very low temperature slows down the chemical reactions that cause the food to decay.

Applications of Factors that Affect Rates of Reaction in Daily Life and in Industrial Processes Combustion of Charcoal 1 Combustion of charcoal in excess oxygen produces carbon dioxide and water. Heat energy is released during combustion. 2 Large pieces of charcoal will not burn easily because the total surface area exposed to oxygen is small. 3 If small pieces of charcoal are used, they can burn easily. This is because the total surface area exposed to the air increases. Thus, the rate of reaction with oxygen (combustion) increases.

1 Pressure cookers are used to speed-up cooking. 2 In the pressure cooker, the higher pressure enables SPM water or oil to boil at a temperature higher than ’06/P1 their normal boiling points. Furthermore, an increase in pressure causes an increase in the number of water molecules or cooking oil molecules coming into contact and colliding with the food particles. 3 At a higher temperature and pressure, the rate of reaction becomes higher. Thus, food cooks faster in pressure cookers.

Coal is mainly carbon. Coal mining is dangerous because coal dust present in the coal mine catches fire very easily. Because of this, serious accidents in coal mines can happen due to the explosion of coal dust. Human lives are often lost in such explosions.

Uses of Catalysts in Industry

Storing Food in Refrigerators

1 Catalysts do not increase the yields of reactions. However, catalysts are used widely in industrial processes to increase the rates of reactions so that the same amount of products can be obtained in a shorter time. As a result, the use of catalysts brings down the cost of production. 2 In the chemical industry, small pellets of solid catalysts are used instead of big lumps. This is to give a larger surface for catalytic reaction to occur and hence a faster reaction will result. 3 The table below summarises the raw materials and the conditions needed for the Haber, Contact and Ostwald processes.

1 The decomposition and decay of food is a SPM chemical reaction caused by the action of ’05/P2 /SB microorganisms such as bacteria and fungi. These microorganisms multiply very rapidly at the temperature range of 10–60 °C. 2 Room temperature is the optimum temperature for the breeding of microorganisms in food. As a result, food turns bad quickly at room temperature. 3 At low temperatures, for example, 5 °C (the normal temperature of a refrigerator), the activities of bacteria are slowed down. Hence, food that Industrial process

Substances

Manufacture of ammonia (Haber process)

Nitrogen and hydrogen

Optimum conditions/equation reaction Temperature: 450–500 °C Pressure: 250 atmospheres Catalyst: Finely divided iron (Fe)

Manufacture of sulphuric acid (Contact process)

Sulphur (to make SO2), air and water

N2(g) + 3H2(g)

450 °C, 250 atm Fe (catalyst)

2NH3(g)

Temperature: 400–450 °C Pressure: 1–2 atmospheres Catalyst: Vanadium(V) oxide, V2O5 319

Rate of Reaction

1

Cooking Food in Pressure Cookers

Industrial process

Substances

Optimum conditions/equation reaction • The following reaction scheme shows the steps involved in the manufacture of sulphuric acid: oxidation

oxidation

S ⎯⎯→ SO2 ⎯⎯→ SO3 ⎯→ H2S2O7 ⎯→ H2SO4 step 1

step 2

step 3

step 4

• In step 2, sulphur dioxide is oxidised to sulphur trioxide.

450 °C, 1 atm

2SO2(g) + O2(g)

2SO3(g)

V2O5 (catalyst)

Manufacture of nitric acid (Ostwald process)

Ammonia, air and water

Temperature: 900 °C Pressure: 5 atmospheres Catalyst: platinum • The following reaction scheme shows the steps involved in the manufacture of nitric acid. oxidation oxidation oxidation

1



NH3 ⎯⎯⎯→ NO ⎯⎯⎯→ NO2 ⎯⎯⎯→ HNO3 step 1

step 2

step 3

In step 1, ammonia is oxidised to nitric oxide.

4NH3(g) + 5O2(g)

900 °C, 5 atm Pt (catalyst)

4NO(g) + 6H2O(g)

Solving Problems on Rate of Reaction

6 Curve I in Figure 1.26 is obtained by treating 5.0 g of granulated zinc with 2.0 mol dm–3 sulphuric acid (in excess) at 30 °C.

Mass and nature of Zn C D

Figure 1.26

A B

Rate of Reaction

Temperature

2 mol dm–3

40 °C

1 mol dm–3

30 °C

Comments Curves I and II show that: (a) The total volume of hydrogen produced in Experiment II is the same as that produced in Experiment I. This means that the amount of zinc used is 5 g and not 2.5 g. Answer A is incorrect. (b) Reaction II is slower than reaction I. This means that zinc powder or a higher temperature of 40 °C is not used in Experiment II. Answers B and C are incorrect. The low rate is achieved by using sulphuric acid more dilute than 2 mol dm–3 (1 mol dm–3). Answer D

Which of the following conditions will produce graph II? Concentration Mass and of H2SO4 nature of Zn 2.5 g of 2 mol dm–3 granulated Zn 5.0 g of Zn 2 mol dm–3 powder

5.0 g of granulated Zn 5.0 g of granulated Zn

Concentration of H2SO4

Temperature 30 °C 30 °C

320

7 Two experiments were carried out to determine the rate of oxygen gas production during the decomposition of hydrogen peroxide. In Experiment I, 20 cm3 of 2 mol dm–3 hydrogen peroxide were used and the results of the experiment are shown on graph I in Figure 1.27.

(b) Differences in terms of rate of reaction Graph II is steeper than graph I because the rate of reaction in Experiment II is expected to be higher than Experiment I. When the concentration of hydrogen peroxide is increased from 2 mol dm–3 to 4 mol dm–3, the rate of reaction also increases. Difference in terms of volume of oxygen released Step 1 To calculate the volume of oxygen produced in Experiment I 2H2O2(aq) → 2H2O(l) + O2(g)

Figure 1.27

(a) Sketch a graph on the same axes to show the results of the experiment that will be obtained if 5 cm3 of 4 mol dm–3 hydrogen peroxide were used for the reaction. (b) Explain your answer in (a). (c) State the constant variables for both the experiments.

Step 2 To calculate the volume of oxygen produced in Experiment II Number of moles of H2O2 used in Experiment II 4  5 =— — — — — = 0.02 1000 ∴ Volume of oxygen collected at room temperature in Experiment II 1 1 = —  0.02  24 000 = 240 cm3 (— V cm3) 2 2

Solution (a)

(c) Constant variables: In both the experiments, the same mass of the catalyst and the same temperature of reaction are used.

1.2 1 State three ways that can be used to increase the rate of reaction of zinc powder with dilute sulphuric acid. 2 Excess calcium carbonate is added to hydrochloric acid at room temperature. The volume of carbon dioxide collected is recorded at regular time intervals. The results of the experiment are shown in Figure 1.28. (a) At what time does the reaction stop? (b) Why does the reaction stop at this particular time? (c) The experiment is repeated by using the same hydrochloric acid but at a lower temperature than room temperature. On the same axes, plot a graph to show the results of the second experiment. (d) State the constant variables for both the experiments.

Figure 1.28

321

Rate of Reaction

1

Number of moles of H2O2 used in Experiment I 2  20 =— — — — — — = 0.04 1000 ∴ Volume of oxygen collected at room temperature in Experiment I 1 = —  0.04  24 000 = 480 cm3 (V cm3) 2

3 Hydrogen peroxide decomposes as represented by the equation:

(c) Give one inference that can be made from the results in Experiment I. (d) Explain why the initial readings on the electronic balance are different for the three experiments.

2H2O2(aq) → 2H2O(l) + O2(g) (a) On the same axes, sketch two graphs of total amount (in mol) of oxygen gas given off against time to show the results of Experiments I and II under the conditions stated below. Experiment I: 100 cm3 of 1.0 mol dm–3 H2O2 Experiment II: 300 cm3 of 0.2 mol dm–3 H2O2

5 Four experiments are carried out to study the rate of reaction between zinc (in excess) and sulphuric acid at different conditions. In each experiment 80 cm3 of 0.1 mol dm–3 sulphuric acid is used. The time taken to collect 192 cm3 of hydrogen gas produced are shown below. Experiment

1

(b) Explain your answer based on the graphs that you have sketched. 4 In an experiment carried out at room temperature (28 °C), 8.0 g of marble chips are added to 100 cm3 of dilute hydrochloric acid in a conical flask. The mass of the conical flask and its contents is determined using an electronic balance at the beginning of the experiment (that is, as soon as the marble chips are added) and then after 1 minute.

Electronic Electronic balance balance Experiment Temperature reading at reading (°C) the after 1 beginning (g) minute (g) I

28

270.35

270.04

35

271.42

270.01

III

40

268.20

266.00

(a) State a hypothesis for the experiment. (b) State the constant variables for the experiment.

1.3

I

Zinc + H2SO4

35

40

II

Zinc + H2SO4

38

18

III

Zinc + H2SO4 + 1 cm3 of CuSO4

35

12

IV

Zinc + H2SO4 + 1 cm3 of Na2SO4

35

40

particles (atoms, molecules or ions) collide with each other. However, not all collisions will result in a chemical reaction to form the products of the reaction. It is likely that the particles collide and bounce back without producing any changes. The collisions that are successful in producing a chemical reaction are called effective collisions. 4 Collisions of particles that are unsuccessful in producing a chemical reaction are called ineffective collisions. 5 The collision theory states that for a chemical reaction to occur, the reacting particles must (a) collide with each other so that the breaking and formation of chemical bonds can occur. SPM

The Collision Theory

’05/P1

1 According to the kinetic theory of matter, all matter is made up of tiny, discrete particles. These particles are continually moving and so have kinetic energy. 2 Based on the assumption that the particles in matter are moving all the time and collide with each other, the collision theory was introduced to explain (a) how chemical reactions occur, and (b) the factors (such as particle size, concentration, temperature, catalyst and pressure) affecting the rates of reactions. 3 When the reactants are mixed, the reactant Rate of Reaction

Temperature Time (°C) (s)

(a) Sketch the graphs of total volume of hydrogen released against time for Experiments I, II and III on the same axes. (b) Explain why the reaction rate for (i) Experiment I is different from that of Experiment II, (ii) Experiment II is different from that of Experiment III. (c) What conclusion can you make by comparing Experiments III and IV? (d) The reaction mixture in Experiment III is filtered. Excess sodium hydroxide is added to the filtrate. (i) Predict what you would observe. (ii) Write an ionic equation for the reaction. (e) Based on your answer in (d), what inference can you make with regards to the property of a catalyst?

The experiment is repeated at 35 °C and 40 °C. The experimental results are shown below.

II

Substances

322

5 (a) If two molecules with sufficient energy (that is, energy equal to or more than the activation energy) collide in the correct orientation, the chemical bonds in the reactant molecules will break and reaction will occur to form new bonds in the product molecules (Figure 1.31). For example,

Activation Energy 1 Activation energy is the minimum energy that the reactant particles must possess at the time of collision in order for a chemical reaction to take place. 2 The activation energy can also be considered as an energy barrier that must be overcome by the colliding particles in order that collision will result in the formation of product molecules. 3 Figure 1.29 shows the energy profile diagram for the exothermic reaction: A + B → C + D. An exothermic reaction is the reaction that releases heat energy (Refer Section 4.1). In the energy profile diagram, the y-axis represents the energy content of the reactants and products, while the x-axis represents the progress of the reaction (reaction coordinate).

H2(g) + I2(g)

2HI(g)

Figure 1.31 Effective collision (sufficient energy and correct orientation)

(b) However, if two molecules, with energy equal to or more than the activation energy, but collide with each other in an incorrect orientation, then reaction will not occur. (c) If two reactant molecules, with energy less than the activation energy, collide in the correct orientation, then reaction will also not occur. The colliding molecules will simply rebound and move away from each other.

SPM

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The energy of the products is lower than the energy of the reactants. Therefore heat is released during the reaction.

Relating the Frequency of Effective Collisions with Factors Influencing the Rate of Reaction

Figure 1.29 Energy profile diagram for an exothermic reaction

4 Figure 1.30 shows the energy profile diagram for the endothermic reaction: E + F → G + H. An endothermic reaction is a reaction that absorbs heat energy.

1 Based on the collision theory, two important factors that determine the rate of a chemical reaction are (a) the frequency of effective collisions and (b) the magnitude of the activation energy. 2 Frequency of effective collisions For a given reaction, if the frequency of collisions between the reactant molecules is high, it follows that the frequency of effective collisions that causes a reaction to occur will also be high. As a result, the rate of reaction increases. 3 Magnitude of activation energy Reactions that have high activation energy will occur at a slow rate. This is because only a small fraction of the molecules possess sufficient energy to overcome the activation energy for the reaction to occur. In contrast, reactions that

The energy of the products is higher than the energy of the reactants. Therefore heat is absorbed during the reaction.

Figure 1.30 The energy profile diagram for an endothermic reaction

323

Rate of Reaction

1

(b) possess energy that is equal to, or more than the minimum energy called the activation energy. (c) collide in the correct orientation.

Effect of Concentration on the Rate of Reaction

possess low activation energy will occur at a high rate. This is because most of the molecules have sufficient energy to overcome the activation energy and this enables the reaction to occur. 4 In general, any factor that increases the rate of effective collisions will also increase the rate of reaction.

1 Magnesium reacts with dilute hydrochloric acid as represented by the equation Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) When the concentration of hydrochloric acid increases, the rate of reaction also increases. 2 Figure 1.33 shows the arrangement of particles in 1 mol dm–3 hydrochloric acid and 2 mol dm–3 (more concentrated) hydrochloric acid.

Effect of Size of Reactant (Surface Area) on the Reaction Rate

1

1 The sodium chloride crystal as shown in Figure 1.32(a) has a total surface area of 16 cm2. When this crystal is divided into smaller crystals as shown in Figure 1.32(b), the total surface area is increased to 24 cm2. Total surface area of the NaCl crystal in Figure 1.32(a) = (1  2)  4 + (2  2)  2 = 16 cm2

Figure 1.33 Arrangement of particles in dilute and concentrated solutions

When the concentration of hydrochloric acid increases, the number of particles per unit volume also increases and the particles are closer together. 3 When the number of particles increases, the frequency of collisions also increases. As a result, the frequency of effective collisions increases. This causes the rate of reaction to increase.

Total surface area in Figure 1.32(b) = (1  1)  6  4 = 24 cm2

5

’05

The decomposition of hydrogen peroxide produces oxygen gas. Curve P is obtained when 25 cm3 of 0.1 mol dm–3 hydrogen peroxide undergoes decomposition.

Figure 1.32

The smaller the particle size, the greater the total surface area exposed for reaction to occur. 2 Dilute sulphuric acid reacts with magnesium as represented by the equation Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g) If small pieces of magnesium or magnesium powder are used, the rate of reaction between magnesium and sulphuric acid will increase. 3 The smaller the size of the solid, the larger the total surface area exposed for collisions. This means that the frequency of effective collisions (that is, collisions with the correct orientation and with energy equal to or greater than the activation energy) between reacting particles will increase. As a result, the rate of reaction also increases. Rate of Reaction

If the experiment is repeated using another hydrogen peroxide solution, which solution will produce curve Q? A 10 cm3 of 0.25 mol dm–3 hydrogen peroxide B 15 cm3 of 0.15 mol dm–3 hydrogen peroxide C 20 cm3 of 0.20 mol dm–3 hydrogen peroxide D 25 cm3 of 0.15 mol dm–3 hydrogen peroxide Comments • Steepness of curves P and Q Curve Q is more steep than curve P. This means 324

that the rate of reaction is higher. That is, the hydrogen peroxide used for curve Q is more concentrated. • Maximum volume of oxygen gas produced in curve Q is less than that in curve P. This means that the number of moles of hydrogen peroxide used in curve Q is less than that in curve P. Number of moles of hydrogen peroxide concentration (mol dm–3)  volume (cm3) =— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — 1000 Relative concentration

Number of moles of H2O2

For curve P A B C D

0.1 mol dm–3 More concentrated More concentrated More concentrated More concentrated

2.5  10–3 2.5  10–3 2.25  10–3 4.0  10–3 3.75  10–3

An increase in pressure will not increase the speed of the reacting particles. In actual fact, the increase in rate at high pressures is caused by the particles being squeezed closer together. This increases the frequency of effective collisions and hence the rate increases.

Effect of Temperature on the Rate of Reaction 1 Calcium carbonate reacts with hydrochloric acid to form carbon dioxide as represented by the following equation CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

Answer B

Figure 1.35 shows the graphs of total volume of carbon dioxide given off against the time taken for the reaction between calcium carbonate and dilute hydrochloric acid at 25 °C and 30 °C.

Effect of Pressure on the Rate of Reaction 1 In chemical reactions involving gases, increasing the pressure increases the rate of reaction. Conversely, decreasing the pressure decreases the rate of reaction. For example, the rate of reaction between nitrogen and oxygen to produce nitrogen monoxide can be increased by increasing the pressure.

SPM

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N2(g) + O2(g) → 2NO(g) Figure 1.35 Effect of temperature on the rate of reaction

2 At low pressures, the gaseous molecules are spread far apart (Figure 1.34(a)). When the pressure is increased, the volume of the gas is reduced (Figure 1.34(b)).

2 At low temperatures, particles of reactants move at a slower speed. However, when the temperature is increased, the particles absorb the heat energy. As a result, the kinetic energy of the particles increases. Hence, (a) the reacting particles move faster, and (b) the number of reacting particles with the activation energy required for the reaction increases. 3 Consequently, the frequency of effective collisions increases and hence, the rate of reaction also increases. 4 Temperature has a great effect on the rate of reaction. For most reactions, the rate of reaction approximately doubles when the temperature of reaction increases by 10 °C.

Figure 1.34 Effect of pressure on gaseous molecules

This means that at high pressures, (a) the number of gaseous molecules per unit volume is increased, and (b) the gaseous molecules are packed closer together. 325

Rate of Reaction

1

Solution

3 As a result, increasing the pressure causes the gaseous molecules to collide more frequently. Consequently, the frequency of effective collisions increases and the rate of reaction also increases.

Effect of Catalysts on Reaction Rates

4 A catalyst provides an alternative reaction route (or pathway) for the reaction to occur. In the presence of a positive catalyst, this alternative route has a lower activation energy. In other words, a positive catalyst lowers the activation energy required for the reaction (Figure 1.37). As a result, more reacting particles possess sufficient energy to overcome the lower activation energy required for effective collisions. Hence, the frequency of effective collisions increases and the rate of reaction increases.

1 The decomposition of hydrogen peroxide to water and oxygen occurs very slowly at room temperature. 2H2O2(aq) → 2H2O(l) + O2(g) In the presence of a catalyst, the decomposition of hydrogen peroxide occurs rapidly. 2 Figure 1.36 shows the rate of evolution of oxygen for the decomposition of hydrogen peroxide without a catalyst and in the presence of a catalyst such as manganese(IV) oxide or iron(III) oxide.

Ea = Activation energy without catalyst Ea’ = Activation energy with catalyst Ea’ < Ea

1

Figure 1.37 Effect of catalyst on the activation energy of a reaction Figure 1.36 The effect of a catalyst on the decomposition of hydrogen peroxide

3 A chemical reaction occurs when reactant particles collide with one another. In the presence of a catalyst, the reactant particles can collide with the catalyst and also with each other. This causes the reactants to react in a different way. Thus, the activation energy of the reaction can be increased or decreased depending on the type of catalyst used.

Effective collisions: Collisions that produce a reaction. Reaction rate increases when the effective collisions increase.

Enzymes are biological catalysts. Enzymes are protein molecules produced in living cells. The enzyme, catalase, is found in the liver. This enzyme can catalyse the decomposition of hydrogen peroxide (a toxic substance produced by metabolism in human bodies) to harmless substances, that is, water and oxygen. Enzymes are also used in detergents to remove protein stains (for example, food or bloodstains) on clothings.

Collision theory: a reaction only occurs if the particles (a) have sufficient energy to overcome the activation energy and (b) collide in the correct orientation.

Activation energy: the minimum energy the reac­ting substances must possess before reaction can occur.

is used to explain the following factors particle size When the particle size is decreased, the total surface area exposed for reaction increases.

concentration/pressure When the concentration/ pressure is increased, the number of particles per unit volume increases.

temperature

When the temperature is increased, the number of particles with the activation energy required increases.

Frequency of effective collisions increases

Rate of Reaction

326

catalyst A catalyst will lower the activation energy required for the reaction by providing an alternative route with a lower activation energy.

Rate of reaction increases

6

’03

Experiment Experimental set-up

Time taken for all the zinc to dissolve (s) Temperature

I

II

30

12

35 °C

35 °C

By using collision theory, explain why the time taken for Experiment II is different from that of Experiment I.

Comments In these two experiments, the constant variables are • volume, concentration and temperature of sulphuric acid used, • mass and surface area of zinc used. Solution The time taken for Experiment II is shorter. This implies that the reaction for Experiment II is faster. Thus, copper(II) sulphate acts as the catalyst. • If a catalyst is added, the rate of reaction increases because the catalyst provides an alternative route with a lower activation energy for the reaction to occur. • Hence, the minimum energy required for the reaction is less. As a result, more reacting particles possess sufficient energy to overcome the lower activation energy required for effective collisions. Hence, the frequency of effective collisions increases and the rate of reaction increases.

1.3 1 Which of the following changes will increase the rate of reaction between sodium thiosulphate solution and sulphuric acid? I By using sulphuric acid of a higher concentration II By increasing the temperature of sodium thiosulphate solution used III By increasing the volume of sodium thiosulphate used IV By adding a small amount of sodium hydroxide solution to the reaction mixture Explain your answer in terms of the collision theory. 2 An experiment is carried out to study the decomposition of hydrogen peroxide. In this experiment, 2.0 g of manganese(IV) oxide is added to 30 cm3 of 0.2 mol dm–3 hydrogen peroxide. The volume of oxygen produced is recorded at 30-second intervals. (a) Calculate the maximum volume of oxygen gas produced in the experiment at room conditions. [1 mol of gas occupies 24 dm3 at room conditions] (b) (i) Sketch a graph of volume of oxygen produced against time. (ii) Explain the shape of the graph. (c) (i) What is the function of manganese(IV) oxide in this experiment? (ii) Based on collision theory, explain the effect of manganese(IV) oxide on the decomposition of hydrogen peroxide.

3 Four experiments were carried out to study the rate of reaction between nitric acid and calcium carbonate of different sizes. In Experiment I, V cm3 of 1.0 mol dm–3 nitric acid is added to big lumps of excess calcium carbonate in a conical flask. The total volume of carbon dioxide produced is plotted against time taken (Figure 1.38).

Figure 1.38

(a) (i) What is the difference between the rate of reaction at the first minute and the rate of reaction at the second minute? (ii) Explain this difference in terms of collision theory. (b) The experiment was repeated three times by changing the reaction conditions for each experiment as shown below.

327

Rate of Reaction

1

Two experiments were carried out to study the rate of reaction between zinc and excess sulphuric acid at room temperature. The table below shows the results of the experiments.

Experiment II

Nitric acid at a lower temperature

III

Small lumps of calcium carbonate but of the same mass

IV

2.0 mol dm–3 nitric acid but of the same volume (V cm3)

1

1.4

(i) Sketch the graphs for Experiments II, III and IV on Figure 1.38 and label each of these graphs. (ii) Explain the difference between the reaction rates for Experiments I and II in terms of collision theory.

Change in conditions of reaction

4 With the aid of an energy profile diagram, explain how a negative catalyst (inhibitor) affects the rate of reaction.

4 Nowadays, enzymes are used extensively in industry to enable reactions to proceed rapidly at room temperature and pressure. 5 In our daily living, we face many social and environmental issues that threaten the quality of living. For example, air and water pollution, food shortages, diseases and so on. 6 We must use science and technology to overcome these problems in a rational and systematic way so that the quality of life can be improved. 7 We should be thankful for the contribution of scientists in enhancing the quality of life in modern living.

Practising Scientific Knowledge to Enhance Quality of Life

1 In our homes, we require machines to increase the rates of reactions and machines to reduce the rates of reactions. Examples of such machines are microwave ovens and refrigerators respectively. 2 In the hospitals, oxygen tents are used to save lives. The high concentration of oxygen helps patients with difficulty in breathing to breathe normally. 3 In human bodies, enzymes (biological catalysts) are needed to catalyse complex biochemical reactions.

• Particle size (surface area) of solid reactant • Concentration of reactants • Temperature of reactants • Presence of catalyst • Pressure (for reactions involving gases) 7 Transition metals (Fe, Ni and Pt) and compounds of transition metals (MnO2 and V2O5) are often used as catalysts. 8 According to the collision theory, a reaction will occur if the reacting particles • collide with each other • possess activation energy • collide in the correct orientation 9 The collisions that are successful in producing a chemical reaction are called effective collisions. 10 Any factor that increases the rate of effective collisions will also increase the rate of reaction.

1 The rate of reaction is defined as the amount of a reactant used up or the amount of a product obtained per unit time. 2 The rate of a reaction is inversely proportional to the time taken for the reaction. 3 Different chemical reactions take place at different rates. A fast reaction takes a shorter time to complete than a slow reaction. 4 The rate of reaction can be determined in the school laboratory by measuring the • changes in the mass of the reactant, • changes in volume of gas produced, • time taken for formation of precipitate. 5 The rate of reaction can be expressed in terms of (a) the average rate of reaction over a period of time, or (b) the instantaneous rate of reaction at a given time. 6 The rate of a reaction is affected by the following factors:

Rate of Reaction

328

1 Multiple-choice Questions 1.1

Rate of Reaction

1 Calcium carbonate reacts with dilute hydrochloric acid to produce carbon dioxide. The total volume of carbon dioxide collected during the reaction is ’11 shown below. Time (s)

0

5

10

15

20

25

30

35

4 The diagram below shows the apparatus set-up used to determine the rate of reaction.

40

The apparatus set-up is not suitable to be used for determining the rate of reaction for A Na2SO3(s) + H2SO4(aq) → Na2SO4(aq) + H2O(l) + SO2(g)

What is the overall average rate of reaction? A 0.825 cm3 s–1 C 1.100 cm3 s–1 3 –1 B 0.943 cm s D 1.300 cm3 s–1 2 The average rate of reaction of calcium carbonate with hydrochloric acid is 0.0080 mol s–1. What is the time taken for 9.60 g of calcium carbonate to completely react with excess hydrochloric acid? [Relative atomic mass: C, 12; O, 16; Ca, 40] A 12 s B 24 s C 120 s D 240 s 3 Calcium carbonate reacts with dilute hydrochloric acid to produce carbon dioxide gas. The plot of volume of CO2 produced against time is shown as follows.

1

Volume of 0.00 12.00 20.00 26.50 31.00 32.50 33.00 33.00 33.00 CO2 (cm3)

B Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) C NaHCO3(s) + HNO3(aq) → NaNO3(aq) + H2O(l) + CO2(g) MnO2(s)

D 2H2O2(aq) ⎯⎯⎯→ 2H2O(l) + O2(g)

5 The graph shows the total volume of carbon dioxide evolved when 10.0 g of calcium carbonate (in excess) reacts with 20.0 cm3 of 1.0 mol dm–3 dilute hydrochloric acid. volume of gas(cm3) Z Y X

0

Which of the following can be deduced from the graph? I The average rate of reaction is 1.5 cm3 s–1. II The rate of reaction decreases with time. III The rate of reaction at 35 seconds is zero. IV The gradient of the curve decreases because the concentration of acid decreases. A I and II only C I, II and III only B III and IV only D II, III and IV only

329

t1

TC 55

t2

t3 time(s)

Which of the following statements is correct? A The reaction is faster at point Y than at point X. B The reaction is fastest at point Z. C The reaction reaches completion at time t3. D The total volume of carbon dioxide evolved is the same if 12.0 g of calcium carbonate is used.

Rate of Reaction

6 Two experiments on the decomposition of hydrogen peroxide were carried out. The graphs in the following diagram show the total volume of oxygen collected against time for each of the experiments. The rate of reaction is best determined by A measuring the volume of SO2 produced at regular time intervals. B measuring the concentration of hydrochloric acid at regular time intervals. C recording the time as soon as the ‘cross’ mark disappears. D recording the time as soon as precipitate appears.

1.2

Factors that Affect the Rate of Reaction

8 Ammonia is produced using the Haber process. The table shows the mass of ammonia produced at four conditions of temperature and pressure.

1

’04

Which of the following graphs shows how the rates of reaction vary with time for the experiments? A

Condition Mass of ammonia produced (kg) Time taken

I

II

III

IV

300

250

150

200

4 hours

2 ½ hours

8 minutes

12 minutes

At which condition is the rate of production of ammonia the highest? A Condition I C Condition III B Condition II D Condition IV 9 Zinc powder reacts with hydrochloric acid according to the equation:

B

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) In order to have the highest initial rate, which of the following solutions should be used for the reaction with zinc powder? A 30 g of HCl in 1000 cm3 of water C 15 g of HCl in 100 cm3 of water B 20 g of HCl in 1000 cm3 of water D 4.0 g of HCl in 50 cm3 of water 10 Two experiments were carried out at 25 °C to study the rate of reaction between magnesium carbonate powder (in excess) and an acid. The volume of carbon dioxide liberated was measured at regular intervals.

C

D

Experiment

Acid used

I

100 cm3 of 0.5 mol dm–3 hydrochloric acid

II

100 cm3 of 0.5 mol dm–3 sulphuric acid

Which of the following graphs represents the results obtained in Experiments I and II? A C

7 The diagram shows the apparatus set-up for an experiment to ’07 determine the rate of reaction between sodium thiosulphate and hydrochloric acid

D

B

S2O32–(aq) + 2H+(aq) → H2O(l) + SO2(g) + S(s) Rate of Reaction



330

Temperature (°C)

Concentration (mol dm–3)

Form of zinc

A

30

0.5

Small pieces

B

25

1.0

Powder

C

35

1.0

Small pieces

D

35

1.0

Powder

12 For the following reaction: Zn + H2SO4 → ZnSO4 + H2 , which factor does not affect the rate of reaction? ’08 A Surface area of zinc C Volume of sulphuric acid B Concentration of sulphuric acid D Temperature of sulphuric acid 13 Graph X in the diagram below shows the result of the decomposition of 10 cm3 of 0.4 mol dm–3 hydrogen peroxide. The experiment was carried out at 30 °C.

Which of the following conditions produces graph Y if 0.1 g of manganese(IV) oxide was used as the catalyst for both experiments? Volume of H2O2 (cm3)

Concentration of H2O2 (mol dm–3)

Temperature (°C)

A

10

0.25

30

B

12.5

0.40

30

C

20

0.25

30

D

20

0.40

40

14 The graph in the diagram below shows the changes in the concentration of hydrogen peroxide, H2O2, when powdered manganese(IV) oxide is ’04 added to it.

The gradients of the graph at times t1 and t2 are different because the I concentration of hydrogen peroxide decreases. II volume of hydrogen peroxide decreases.

331

III temperature of hydrogen peroxide decreases with time. IV mass of manganese(IV) oxide decreases. A I only B I and II only C III and IV only D I, III and IV only 15 The following equation shows the reaction between powdered ’03 calcium carbonate and dilute hydrochloric acid CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) The production of carbon dioxide can be slowed down by I reducing the temperature of hydrochloric acid used. II adding distilled water to hydrochloric acid before the reaction. III using larger pieces of calcium carbonate. IV reducing the pressure on the reaction mixture. A I, II and III only B I, III and IV only C I, II and IV only D I, II, III and IV 16 A piece of magnesium ribbon is allowed to react with 100 cm3 of 1.0 mol dm–3 hydrochloric acid. Which of the following changes will increase the rate of reaction? I Increasing the temperature of hydrochloric acid. II Replacing the magnesium ribbon with magnesium powder. III Replacing the acid with 50 cm3 of 2.0 mol dm–3 hydrochloric acid. IV Adding 50 cm3 of 1.0 mol dm–3 hydrochloric acid. A I and II only B III and IV only C II and IV only D I, II and III only 17 What is the most suitable method for cooking 100 g of ’06 potatoes within a short time? A Steam the potatoes in a steamer B Fry the potatoes in a copper pot Rate of Reaction

1

11 2.5 g of zinc were allowed to react with 100 cm3 of hydrochloric acid under different conditions as shown below. Under which conditions will hydrogen gas be given off at the highest rate?

C Boil the potatoes in a saucepan D Boil the potatoes in a pressure cooker

1

18 Iron(III) oxide is a brown solid and iron(III) salts are brown in colour. When iron(III) oxide is added to hydrogen peroxide solution in a test tube, a fast reaction occurs and oxygen gas is liberated. What is left in the test tube at the end of the reaction? A A brown solution only. B A brown solid and a brown solution. C A brown solid and a colourless solution. D A white solid and a colourless solution . 19 The diagram below shows the total volume of carbon dioxide given off when dilute hydrochloric acid reacts with calcium carbonate powder.

As the reaction proceeds, the gradient of the graph becomes less steep because I the mass of calcium carbonate decreases. II the total surface area of calcium carbonate decreases. III the volume of hydrochloric acid decreases. IV the temperature of the mixture increases. A I and II only B I, II and III only C II, III and IV only D I, III and IV only 20 Which of the following reactions takes place in the Ostwald process? A N2(g) + 3H2(g) 2NH3(g) B 2SO2(g) + O2(g) 2SO3(g) C 2NH3(g) + O2(g) 2NO(g) + 3H2(g) D 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l) Rate of Reaction

21 Which of the following pairs of catalyst and processes are correctly matched? Catalyst

Process

I

Iron

Manufacture of ammonia in the Haber process

II

Nickel

Manufacture of nitric acid in the Ostwald process

III

Vanadium(V) oxide

Manufacture of sulphuric acid in the Contact process

IV

Lead(IV) oxide

Production of oxygen by the decomposition of hydrogen peroxide

A I and II only B II and III only 22 Which of the following reactions require a catalyst to speed up the reaction? I 2H2O2 → 2H2O + O2 II 2SO2 + O2 → 2SO3 III SO3 + H2SO4 → H2S2O7 IV Na2S2O3 + H2SO4 → Na2SO4 + H2O + SO2 + S A I and II only B I and III only C I, III, and IV only D II, III and IV only 23 Which of the following reactions require a catalyst to speed up the reaction? I N2 + 3H2 → 2NH3 II SO3 + H2SO4 → H2S2O7 III 2H2O2 → 2H2O + O2 IV Na2S2O3 + 2HCl → 2NaCl + H2O + SO2 + S A I and II only B I and III only C I, III and IV only D II, III and IV only 24 The reaction between sulphuric acid and magnesium carbonate is carried out at different conditions. Which reaction is fastest? A

B

332

C I, II and IV only D I, III and IV only C

D

25 Magnesium ribbons of the same length are added separately to each of the following solutions of hydrochloric acid. In which solution will the magnesium ribbon disappear first? [The hydrochloric acid used is in excess] Volume Concentration Temperature of HCI of HCI of HCI (cm3) (mol dm–3) (°C) A

300

1.0

30

B

200

1.0

25

C

100

2.0

30

D

200

2.0

25

26 The energy profile diagram for an uncatalysed reaction is shown below.

∆H

Ea

A

No change

Decrease

B

Decrease

No change

C

Decrease

Decrease

D

Increase

Increase

27 Carbon dioxide is produced when magnesium carbonate reacts with dilute hydrochloric acid. MgCO3(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) + CO2(g) Which of the following changes will increase the initial rate of carbon dioxide production? A Heat the reaction mixture B Increase the size of solid magnesium carbonate C Increase the volume of hydrochloric acid D Increase the pressure on the reaction mixture 28 Calcium carbonate is added to excess hydrochloric acid at 30 °C. The experiment is repeated at 40 °C. The volume of carbon dioxide released for each experiment is measured at room temperature and pressure. Which of the following graphs represents the results of these two experiments? A

B

C

1.3 The Collision Theory

D

29 Three experiments are carried out to study the rate of decomposition of hydrogen peroxide by the catalyst, manganese(IV) oxide. In all these three experiments, the mass of the catalyst used is the same. The experimental results are shown in the following diagram. Solutions of hydrogen peroxide used Solution P: 50 cm3 of 2.0 mol dm–3 H2O2 Solution Q: 100 cm3 of 1.0 mol dm–3 H2O2 Solution R: 100 cm3 of 3.0 mol dm–3 H2O2

30 Based on the collision theory, what are the effects of a rise ’06 in temperature on the reactant particles? I The kinetic energy of the reactant particles increases. II The number of reactant particles per unit volume increases. III The frequency of collisions between reactant particles increases. IV The activation energy of the reactant particles increases. A I and II only B I and III only C II and IV only D III and IV only 31 When zinc powder is added to dilute sulphuric acid, gas bubbles ’11 are produced slowly. When a few drops of copper(II) sulphate are added to the reaction mixture, gas bubbles are produced vigorously. Which statement best explains the effect of copper(II) sulphate on the reaction? A It lowers the activation energy. B It increases the collision frequency between the reacting particles. C It increases the concentration of sulphate ions and hence increases the rate of reaction. D It causes the reacting particles to collide in the correct orientation. 32 The diagram below shows the energy profile for the following reaction

Which of the following statements are correct? I The curve X is obtained using solution R. II The curve Y is obtained using solution P. III The curve Z is obtained using solution Q. IV The curve Z is obtained using solution R. A I and II only B II and IV only C I, II and III only D I, II and IV only

333

X(g) + Y(g) → Z(g) by by by by

The curves, P and Q, represent two different paths for this Rate of Reaction

1

The reaction was repeated using a catalyst. What is the effect of the catalyst on the heat of reaction (∆H) and activation energy (Ea) for the reaction?

reaction. What conclusion can be drawn based on the diagram? A The reaction by path P occurs at a higher temperature. B The reaction by path P occurs at a higher rate than by path Q. C The activation energy for path P is (x + y) kJ. D The activation energy for path Q is (x – y) kJ. 33 Consider the reaction between magnesium and dilute hydrochloric acid.

1

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Which of the following will increase the frequency of collisions between the reactants? I Increase the concentration of hydrochloric acid. II Increase the temperature of reaction. III Use magnesium ribbon instead of magnesium powder. IV Remove the hydrogen gas produced.

A I and II only B III and IV only C I, II and III only D II, III and IV only 34 The reaction between Fe3+ and SO32– is represented by the ’06 equation: 2Fe3++ SO32–+ H2O → 2Fe2+ + H2SO4 brown green It is found that the change of colour from brown to green occurs at a higher rate when the reaction mixture is heated. This is due to the I decrease in activation energy. II increase in the frequency of collisions between Fe3+ and SO32– ions. III increase in the kinetic energy of Fe3+ and SO32– ions. IV increase in the frequency of effective collisions. A I and II only B II and III only C I, III and IV only D II, III and IV only 35 Iron is used in the Haber process to manufacture ammonia from nitrogen and hydrogen.

Why is iron used in this process? A To increase the rate of reaction between nitrogen and hydrogen B To absorb the smell of ammonia C To oxidise nitrogen to form ammonia D To increase the yield of ammonia 36 The rate of reaction between 1 mol dm–3 hydrochloric acid and 3 g of magnesium powder is higher than the rate of reaction between 1 mol dm–3 ethanoic acid and 3 g of magnesium powder. What is the explanation for this observation? A Hydrochloric acid is more soluble in water than ethanoic acid. B The kinetic energy of hydrochloric acid is higher than ethanoic acid. C Hydrochloric acid forms a soluble salt whereas ethanoic acid forms an insoluble salt. D The concentration of H+ ions in hydrochloric acid is higher than ethanoic acid.

Structured Questions 1 Diagram 1 shows two experiments to investigate one factor that affects the rate of reaction between zinc ’06 and dilute sulphuric acid.

Experiment II

Experiment I

Diagram 1

(a) What is the factor that affects the rate of reaction in Experiments I and II? [1 mark] (b) State two constant variables in both experiments.

[2 marks]

(c) The following equation represents the reaction that occurs in both the experiments. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) Rate of Reaction

334

recorded. The experiment is repeated using different acids as shown below. All experiments are carried out at the same temperature. Experiment

Reactants

I

50 cm of 1.0 mol dm HCl + 5.0 cm of magnesium ribbon

t1

II

50 cm3 of 1.0 mol dm–3 CH3COOH + 5.0 cm of magnesium ribbon

t2

(d) Graph 1 shows the results for both experiments.



Graph 1

Based on Graph 1: (i) Which experiment has a higher rate of reaction? [1 mark] (ii) How do you come to this conclusion? [1 mark] (iii) Explain what happens after time t. [1 mark] (iv) Why are both curves at the same level after time t? [1 mark]



Time taken (s) –3

(i) Write the ionic equation for the reaction between magnesium and an acid. [1 mark] (ii) Which is shorter t1 or t2? Why? [3 marks]

(c) Three experiments are carried out to investigate the factors that affect the rate of the following reaction: ’08 Mg + 2HCl → MgCl2 + H2 The conditions of this experiment are shown below.

(e) State the conclusion for the experiments. [1 mark] (f) Another experiment was carried out using excess zinc powder and dilute sulphuric acid with different concentrations. Sketch the curve of concentration of dilute sulphuric acid against the time taken to collect a fixed quantiy of the product.



3

[2 marks]

2 (a) Propane, C3H8, burns in excess oxygen to form carbon dioxide and water as represented by the equation

Temperature (°C)

Experiment

Reactants

I

Excess magnesium powder + 25 cm3 1.0 mol dm–3 HCl

35

II

Excess magnesium ribbon + 25 cm3 0.5 mol dm–3 HCl

25

III

Excess magnesium ribbon + 25 cm3 1.0 mol dm–3 HCl

35

The results of this experiment are shown in Diagram 2.

C3H8 + 5O2 → 3CO2 + 4H2O At time t, the rate of reaction of propane is 0.20 mol s–1.

Calculate (i) the rate of consumption (using up) of oxygen, and (ii) the rate of production of carbon dioxide at time t. [2 marks]

(b) 50 cm3 of 1.0 mol dm–3 of hydrochloric acid is poured into a conical flask. A piece of 5.0 cm magnesium ribbon is added to the acid. The time taken to dissolve the magnesium completely is

Diagram 2



335

(i) Which curves (X, Y or Z) represent the results [3 marks] of Experiments I, II and III? Rate of Reaction

1

(i) Choose one of the products shown in the equation that is most suitably used to measure the rate of reaction. [1 mark] (ii) Give one reason for your answer in (i). [1 mark]

(a) What is the average rate of reaction for Experiment II? [2 marks]

(ii) Give one reason why the final volume of gas obtained in curve X is half the final volume of gas in curve Z. [1 mark]



(b) Use the collision theory to explain why the time taken for Experiment II is shorter than for Experiment I. [3 marks]

3 Three experiments were carried out to investigate the factors influencing the rate of reaction. Three pieces of 0.12 g of magnesium ribbon are added separately to excess hydrochloric acid. The time taken for all the magnesium to dissolve is taken. Table 1 shows the results of the experiments. Experiment

I

II

III

Reactants

0.12 g Mg + excess HCl(aq)

0.12 g Mg + excess HCl(aq)

0.12 g Mg + excess HCl(aq) + CuSO4(aq)

Temperature (oC)

30

35

35

Time (s)

60

32

10

(c) Explain why the time taken for Experiment III is longer than for Experiment II. [3 marks] (d) Suggest another method that can be used to increase the rate of decomposition of hydrogen in Experiment II. [1 mark] (e) Write a chemical equation for the catalytic decomposition of hydrogen peroxide. [1 mark] (f) Experiments II and III were allowed to continue until the decomposition of hydrogen peroxide was completed. Sketch a graph of total volume of gas against time for Experiments II and III on the same axes. [2 marks] 5 Dilute sulphuric acid reacts with sodium thiosulphate solution to produce sulphur. The presence of sulphur causes the solution to become cloudy. Five experiments were carried out to study the rate of reaction between dilute sulphuric acid and sodium thiosulphate solution. The reaction takes place in a conical flask placed over a white paper marked with a cross, ‘X’. The time taken for the cross to disappear from view was recorded. In each experiment, the volume and concentration of sodium thiosulphate solution were kept constant. The experimental results are shown in Table 3.

1

Table 1

(a) Write the chemical equation for the reaction between magnesium and hydrochloric acid. [1 mark] (b) Calculate the maximum volume of hydrogen gas produced in Experiment I. [2 marks] (c) Predict the maximum volume of hydrogen gas produced in Experiment III. Explain your answer. [2 marks] (d) Calculate the average rate of reaction in Experiment II. [1 mark] (e) (i) Which experiment has the highest rate? Justify your answer. (ii) Sketch the graphs for the volume of hydrogen gas against time for Experiments I, II and III on the same axes. [3 marks] [Relative atomic mass: Mg = 24; Molar gas volume, 24 dm3 at r.t.p.]

Experiment

4 Three experiments were carried out to study the effect of iron(III) oxide, Fe2O3, on the rate of decomposition of 0.5 mol dm–3 hydrogen peroxide. Table 2 shows the mixtures of substances used and the time taken to collect 30 cm3 of the colourless gas given off in each experiment. Experiment

Mixture of substances

Time taken

I

20.0 cm3 of 0.5 mol dm–3 H2O2

A few weeks

II

20.0 cm3 of 0.5 mol dm–3 H2O2 + 0.2 g of Fe2O3

35 s

III

20.0 cm3 of 0.5 mol dm–3 H2O2 + 25.0 cm3 of water + 0.2 g of Fe2O3

45 s

Time taken (s)

A

0.15

20

65

B

0.10

30

45

C

0.10

20

85

D

0.05

30

55

E

0.05

20

105

Table 3

(a) The reaction between dilute sulphuric acid and sodium thiosulphate (Na2S2O3) produces sulphur, sodium sulphate, sulphur dioxide and water.





Table 2 Rate of Reaction

Concentration Temperature of acid (°C) –3 (mol dm )

336

Complete the following equations: (i) Na2S2O3 + H2SO4 → ____ + ____ + ____ (ii) S2O32– + 2H+ → ____ + ____ + ____

[2 marks]

(b) (i) Which of the experiments shown above should be chosen to compare the effect of concentration of the acid on the rate of reaction? [1 mark] (ii) Give one reason why you chose these experiments. [1 mark] (iii) What conclusion can be made from the experiments in (i)? [1 mark]

(c) (i) Which of the experiments should be chosen to compare the effect of temperature on the rate of reaction? [1 mark] (ii) Give one reason for your choice. [1 mark] (iii) What conclusion can be made from the experiments you have chosen in (i)? [1 mark]



(a) Table 5 shows the conditions used for carrying out Experiment 1. Complete Table 5 to predict the conditions used for obtaining the results of Experiments 2, 3 and 4. In each case, state the constant variables and manipulated variables used and briefly explain your answer. Experi­ ment 1

Experi­ ment 2

Experi­ ment 3

Experi­ ment 4

Volume of hydrogen peroxide (cm3)

40

40

…

…

Volume of water (cm3)

40

40

40

0

Temperature (°C)

30

30

32

30

Mass of MnO2 used (g)

1.0

…

1.0

1.0

(d) Explain your answer in (c)(iii) based on collision theory. [2 marks] 6 Four experiments were carried out to investigate the decomposition of hydrogen peroxide to form water and oxygen gas. MnO2

2H2O2(aq) ⎯⎯⎯→ 2H2O(l) + O2(g)

The total volume of oxygen evolved at one-minute intervals were recorded in Table 4. Volume of oxygen gas released (cm3) Time (min) Experiment Experiment Experiment Experiment 1 2 3 4 0

0

0

0

0

1

18

0

25

34

2

33

0

35

60

3

36

0

37

69

4

37

0

38

75

5

38

0

38

76

6

38

0

38

76



Table 5

[5 marks]

(b) Copper(II) oxide is a less effective catalyst than manganese(IV) oxide for the decomposition of hydrogen peroxide. If copper(II) oxide is used to replace manganese(IV) oxide in Experiment 1, what is the effect of this change on (i) the volume of oxygen collected at 1.0 minute? (ii) the volume of oxygen collected after the reaction has completed? Explain your answers. [2 marks] (c) ‘A catalyst remains chemically unchanged at the end of the reaction’. (i) What is meant by chemically unchanged? (ii) How would you prove that your answer in (i) is correct? [3 marks]

Table 4

Essay Questions 1 (a) Two experiments are carried out to study the rate of reaction between iron and dilute acids. Experiment

Reactants

I

1.12 g of iron and 50 cm3 of 2.0 mol dm–3 sulphuric acid

II

1.12 g of iron and 50 cm3 of 2.0 mol dm–3 hydrochloric acid

The following graphs show the results of the experiments.

337

Rate of Reaction

1



Based on the graph: (i) Calculate the average rate of reaction for Experiment I. [2 marks] (ii) Explain the difference in the rate of reaction between Experiment I and Experiment II before 100 s. [6 marks]

(c) Ethanedioic acid, H2C2O4, decolourises potassium manganate(VII) slowly at room temperature. The reaction is catalysed by manganese(II) sulphate, MnSO4. Describe how you would prove that the sulphate ions, SO42–, do not act as a catalyst in this reaction. [3 marks]

(b) Describe an experiment to show that lead(IV) oxide is a more effective catalyst than copper(II) oxide for the decomposition of hydrogen peroxide. Your answer should include a labelled diagram on the apparatus set-up for the experiment. [12 marks]

4 (a) Two experiments are carried out to study the rate of reaction between zinc and two acids, R and T. ’07 The data for the experiments are shown below.

’07



Experiment

Reactants

Observation

Products

I

Excess zinc and 25 cm3 of 1.0 mol dm–3 acid R

The temperature of the reaction mixture increases

Zinc sulphate and hydrogen

II

Excess zinc and 25 cm3 of 1.0 mol dm–3 acid T

The temperature of the reaction mixture increases

Zinc chloride and hydrogen

2 (a) Iron powder will dissolve in cold dilute hydrochloric acid while coarse iron filings do not dissolve until the acid is heated. Explain these observations. [7 marks]

1

(b) There is a high risk of explosions occurring in coal mines. Explain why this is so. [6 marks] (c) ‘Temperature is important in preserving food’. Give one example from your daily life to justify this statement. [4 marks] (d) When a drop of blood is added to hydrogen peroxide solution, a vigorous effervescence occurs. Explain this observation. [3 marks]



3 (a) (i) Define the term rate of reaction? [2 marks] (ii) At high temperatures and pressures, nitrogen reacts with hydrogen to form ammonia. N2(g) + 3H2(g)



2NH3(g)

Use the collosion theory to explain how high pressure increases the rate of reaction between nitrogen and hydrogen. [5 marks] (b) Describe an experiment to demonstrate the effect of the concentration of sodium thiosulphate on the rate of reaction between hydrochloric acid and sodium thiosulphate. Draw the apparatus used for this experiment and state the hypothesis and variables in this experiment. [10 marks]

(i) State the names of the acids used in experiments I and II. [2 marks] (ii) Write the chemical equation for the reaction that occurs in Experiment I. [2 marks] (iii) Draw the energy profile diagram for the reaction in Experiment II. On the energy profile diagram, show the • activation energy without catalyst, Ea, • activation energy with a catalyst, Ea’, • heat of reaction, ΔH. Explain the energy profile diagram. [10 marks]

(b) Explain the factors that affect the rate of reaction in the following daily activities: (i) Combustion of charcoal (ii) Cooking food in a pressure cooker [6 marks]

Experiments The experiment was repeated using sodium thiosulphate solutions at 30 °C, 35 °C, 40 °C and 45 °C. Diagram 1 shows the stopwatch readings for each of the experiments.

1 Sodium thiosulphate solution reacts with dilute sulphuric acid to produce a yellow precipitate of sulphur. 50 cm3 of 0.10 mol dm–3 sodium thiosulphate solution at 25 °C was measured into a 250 cm3 conical flask. The conical flask was placed on a white paper marked with the ‘X’ sign. 5 cm3 of 0.50 mol dm–3 sulphuric acid was added to the sodium thiosulphate solution and the mixture shaken. At the same time, the stopwatch was started. The time was taken as soon as the ‘X’ sign was no longer visible.

Rate of Reaction

338

at 25 °C

at 30 °C

at 35 °C

Time, t1 ______ s Time, t2 ______ s

at 45 °C

at 40 °C

Time, t3 ______ s

Time, t4 ______ s





Time, t5 ______ s

Diagram 1

(a) Record the readings of the stopwatch in the spaces provided in Diagram 1.

[3 marks]

(b) State the variables in this experiment. Manipulated variable: Responding variable: Constant variable:

[3 marks]

1 (d) (i) Draw a graph of temperature against ————— on a graph paper. time

[4 marks]



[3 marks]

(ii) Based on the graph in (i), what conclusion can be drawn from this experiment?

(e) Predict the time taken for the ‘X’ sign to disappear if the experiment is repeated at 50 °C.

[3 marks]

(f) State the hypothesis for this experiment.

[3 marks]

(g) Based on your hypothesis, explain why meat and fish are always kept in refrigerators.

[3 marks]

2 An experiment was carried out to investigate the rate of reaction between granulated zinc and dilute hydrochloric acid. The results of the experiment are shown below. ’09

Time (s)



0

30

40

Burette reading 50.00 40.00 33.50 (cm3)

……

……

21.50 20.00

Volume of gas evolved (cm3)

……

……

28.50 30.00 32.50 32.50

0.00

10

20

10.00 16.50

50

60

70

80

17.50

17.50

Diagram 2 shows the burette readings at 30 seconds and 40 seconds respectively. 29

25

TC 56

28

24

At 30 s

At 40 s

Diagram 2

(a) Based on this experiment, what is meant by the rate of reaction? (b) Based on Diagram 2, what are the volumes of gas evolved at 30 seconds and 40 seconds? (c) State one conclusion, based on the experimental results.

339

[3 marks] [3 marks] [3 marks] Rate of Reaction

1

1 (c) Construct a table containing the information on temperature, time and ————— for the experiments. time [2 marks]

FORM 5 THEME: Interaction between Chemicals

CHAPTER

2

Carbon Compounds

SPM Topical Analysis 2008

Year 1

Paper

2 A

Section Number of questions

4

2009

1

B –

3

1

C –

2 A

1

2

2010

1

B 1

3

1

C –

–

7

2011

2

3

A

A

C

–

1 — 2

–

1

–

4

2

3

A

B

C

–

–

1

–

ONCEPT MAP CARBON COMPOUNDS

Organic carbon compounds: Produce CO2 and H2O on complete combustion

Hydrocarbons (elements C and H only) Physical properties: Insoluble in water, low melting and boiling points, non-conductors of electricity

Alkanes (saturated) Reactions: • Combustion • Substitution

hydrogenation

Alkenes (non-saturated) Reactions: • Combustion • Addition • Polymerisation

Isomerism (same molecular formula, different structural formulae)

Natural rubber (natural polymers) • Coagulation • Vulcanisation

Non-hydrocarbons (Elements: C, H, O)

hydration dehydration

Alcohols Reactions: • Combustion • Oxidation • Dehydration • Esterification

oxidation

Carboxylic acids Reactions: • With metals/metal carbonates/alkalis to form salts • Esterification

esterification

Esters Physical properties: • Sweet fruity smell • Insoluble in water Fats and Oils • Fats: saturated, higher melting point • Oils: unsaturated, lower melting point

8 Almost all organic compounds contain the elements carbon and hydrogen. Hence the complete combustion of carbon compounds produces carbon dioxide and water.

Carbon Compounds

1 Carbon compounds are compounds that contain the element carbon. 2 Carbon compounds can be classified into two groups: inorganic compounds and organic compounds. 3 Organic compounds are carbon compounds in which carbon is bonded to other elements by covalent bonds. Examples of organic compounds are hydrocarbons, alcohols, carboxylic acids and esters. 4 Most carbon compounds are derived from living organisms. Nowadays many organic compounds can be synthesised in laboratories. 5 Most inorganic compounds do not contain carbon. Examples of inorganic compounds containing the element carbon are carbonates, hydrogen carbonates, oxides of carbon and cyanides. 6 Carbon atom (proton number 6) has the electron arrangement: 2.4. Hence, each carbon atom can form four covalent bonds in organic compounds. 7 The covalent bonds in carbon compounds may be (a) single bond, (b) double bond or (c) triple bond.

Carbon compounds

Organic compounds

Example: • Hydrocarbons • Alcohols • Carboxylic acids • Esters • Carbohydrates

Inorganic compounds

Example: • Hydrogen carbonates • Carbonates • Carbides • Oxides of carbon • Cyanides

2

2.1

Organic compounds are the largest group of chemicals we know today, numbering in thousands. All the food and medicines we consume are organic compounds as well as most of the synthetic products such as clothing and household materials.

To investigate the products formed by complete combustion of organic compounds Apparatus

Ethanol and palm oil, limewater, ice and water, anhydrous cobalt(II) chloride paper.

Procedure

Filter funnel, test tubes, delivery tubes, spirit lamp, suction pump and beaker.

1 A filter funnel is connected to the suction pump via test tubes A and B where test tube A is dipped in ice water and test tube B contains limewater. 2 A spirit lamp filled with ethanol is lit and placed under the filter funnel and the suction pump turned on. 3 The changes in test tubes A and B are noted. 4 The liquid collected in test tube A is tested with anhydrous cobalt(II) chloride paper. 5 Steps 1 to 4 are repeated using a spirit lamp with palm oil.

Figure 2.1 Combustion of organic compounds

341

Carbon Compounds

Activity 2.1

Materials

Results Test tube

Observation

Inference

A

A colourless liquid is formed and it changes anhydrous cobalt(II) chloride paper from blue to pink

The colourless liquid formed in test tube A is water

B

Limewater turns milky

Carbon dioxide gas is produced

Conclusion 1 The combustion of organic compounds such as ethanol and palm oil produces water and carbon dioxide. 2 During the combustion of an organic compound, (a) the carbon combines with oxygen to form carbon dioxide, (a) the hydrogen combines with oxygen to form water.

excess oxygen (complete combustion), carbon dioxide and water are produced. 8 Incomplete combustion of hydrocarbons will produce water, carbon dioxide, carbon monoxide and carbon (as soot).

2

Hydrocarbons 1 Hydrocarbons are organic compounds that contain the elements carbon and hydrogen only. 2 Hydrocarbons that have only single covalent bonds between all the carbon atoms in the molecule are called saturated hydrocarbons. Example Propane

Hydrocarbon



H H H carbon-carbon | | | single bond H—C—C—C—H | | | H H H 3 Hydrocarbons that have at least one carboncarbon double bond (C = C) or triple bond (C ≡ C) in the molecule are called unsaturated hydrocarbons. Example Propene

Saturated hydrocarbons have only single bonds between the carbon atoms. Examples: Ethane, propane

1









H H H carbon-carbon | | | double bond H—C=C—C—H | H 4 The main sources of hydrocarbons are: (a) Petroleum (crude oil) (b) Natural gas (c) Coal 5 Petroleum is a complex mixture of hydrocarbons. 6 Fractions of hydrocarbons are separated by a process called fractional distillation. The fractions are separated based on the difference in boiling points. The fractions with lower boiling points will be distilled off earlier. 7 Hydrocarbons contain carbon and hydrogen only. Thus when hydrocarbons are burnt in

Carbon Compounds

Unsaturated hydrocarbons have double or triple bonds between the carbon atoms. Examples: Ethene, propene

SPM

’10/P1

The complete combustion of 0.1 mol of a hydrocarbon Z in excess oxygen produces 0.3 mol of carbon dioxide and 0.4 mol of water. Determine the molecular formula of hydrocarbon Z. Solution Since 0.1 mol of Z produces 0.3 mol of carbon dioxide and 0.4 mol of water, 1 mol of Z will produce 3 mol of carbon dioxide and 4 mol of water. C + O2 → CO2 The number of moles of carbon in 1 mol of Z = the number of moles of carbon dioxide = 3. 1 2H + —O → H2O 2 2 The number of moles of hydrogen in 1 mol of Z = 2 3 the number of moles of water = 2 3 4 = 8. Hence the molecular formula of Z is C3H8. 342

7 Table 2.1 shows the prefixes used to indicate the number of carbon atoms per molecule of an organic compound, the names and molecular formulae of the first ten alkanes.

2.1 1 Classify the following substances into organic compounds and inorganic compounds.

Table 2.1 Prefixes used to indicate the number of carbon atoms, names and molecular formulae of alkanes

2 (a) What is meant by hydrocarbon? (b) State three sources of hydrocarbon.

Number of carbon atoms per molecule

Prefix

Name of alkane

1

Meth

Methane

CH4

2

Eth

Ethane

C2H6

3

Prop

Propane

C3H8

4

But

Butane

C4H10

5

Pent

Pentane

C5H12

6

Hex

Hexane

C6H14

7

Hept

Heptane

C7H16

8

Oct

Octane

C8H18

9

Non

Nonane

C9H20

10

Dec

Decane

C10H22

3 State two main products that are formed when rubber is burnt in excess air. Explain your answer.

2.2

Alkanes

1 Alkanes are saturated hydrocarbons with the general formula Cn H2n+2, where n = 1, 2, 3… 2 Alkanes are called saturated hydrocarbons SPM because the molecules contain only single ’11/P1 covalent bonds between carbon atoms. 3 The molecular formula is a chemical formula that shows the actual number of atoms of each element present in one molecule of the substance. 4 The molecular formula of an alkane can be obtained by substituting n in the general formula Cn H2n+2 with the number of carbon atoms. 5 In the naming of alkanes according to the IUPAC system, all members of the alkane series have their names ending with -ane (IUPAC is the abbreviation for International Union of Pure and Applied Chemistry). 6 The first part (prefix) of the name of an alkane depends on the number of carbon atoms in the molecule.

Molecular formula

8 The structural formula of an organic compound is the chemical formula that shows the arrangement of atoms and covalent bonds between atoms in a molecule of the compound. 9 Note that when writing the structural formula of alkanes, (a) each carbon atom should have four single covalent bonds. (b) each hydrogen atom should have one single covalent bond. (c) the carbon atoms are connected by single bonds. 10 The structural formulae of the first ten straight chain alkanes are shown as follows:

Table 2.2 The structural formulae of the first ten straight chain alkanes

H | H–C–H | H methane (CH4)









H H H H | | | | H–C–C–C–C–H | | | | H H H H butane (C4H10)



H H | | H–C – C–H | | H H ethane (C2H6)

H H H H H | | | | | H–C–C–C–C–C–H | | | | | H H H H H pentane (C5H12) 343



H H H | | | H–C–C–C–H | | | H H H propane (C3H8)



H H H H H H | | | | | | H–C–C–C–C–C–C–H | | | | | | H H H H H H hexane (C6H14) Carbon Compounds

2

Rubber, sugar, limestone, carbon dioxide, calcium carbonate, sand, sodium chloride, vinegar, polyvinylchloride, urea, ammonium sulphate.



H H H H H H H | | | | | | | H–C–C–C–C–C–C–C–H | | | | | | | H H H H H H H heptane (C7H16) H H H H H H H H H | | | | | | | | | H–C–C–C–C–C–C–C–C–C–H | | | | | | | | | H H H H H H H H H nonane (C9H20)



H H H H H H H H | | | | | | | | H–C–C–C–C–C–C–C–C–H | | | | | | | | H H H H H H H H octane (C8H18)



H H H H H H H H H H | | | | | | | | | | H–C–C–C–C–C–C–C–C–C–C–H | | | | | | | | | | H H H H H H H H H H decane (C10H22)

Physical Properties of Alkanes 1 On going down the alkane series, the physical properties change gradually as shown in Table 2.3.

2

Table 2.3 Physical properties of alkanes

Name of alkane

Molecular formula

Relative molecular Melting point Boiling point mass (°C) (°C)

Physical state

Density (g cm–3)

Methane

CH4

16

–182

–162

Gas

—

Ethane

C2H6

30

–183

–89

Gas

—

Propane

C3H8

44

–188

–42

Gas

—

Butane

C4H10

58

–138

–0.5

Gas

—

Pentane

C5H12

72

–130

36

Liquid

0.626

Hexane

C6H14

86

–95

69

Liquid

0.659

Heptane

C7H16

100

–90

98

Liquid

0.684

Octane

C8H18

114

–57

126

Liquid

0.703

Nonane

C9H20

128

–54

151

Liquid

0.718

Decane

C10H22

142

–30

174

Liquid

0.730

2 Melting and boiling points of alkanes (a) Alkanes exist as simple covalent molecules. Alkanes have low melting and boiling points because of the weak van der Waals forces between molecules. Little energy is required to overcome the weak forces of attraction. (b) When the number of carbon atoms per molecule of alkane increases, the relative molecular mass increases and the melting point and boiling point increase. This is because the larger the molecular size, the stronger the van der Waals forces of attraction between the molecules. 3 The physical states of alkanes The first four members are gases as their boiling points are below room temperature (25 °C). The alkanes from C5 Carbon Compounds

Figure 2.2 The boiling points of alkanes increase with relative molecular mass

344

to C18 are liquids with the rest of the alkanes being solids. 4 Densitites of alkanes Alkanes are less dense than water. The density of alkanes increases gradually down the series as the molecular mass increases. 5 Solubility of alkanes (a) All alkanes are insoluble in water. When liquid alkane is shaken with water, two separate layers of liquids are formed. (b) Alkanes are soluble in organic solvents such as propanone. 6 Electrical conductivity of alkanes All alkanes do not conduct electricity because they are covalent compounds, consisting of molecules.

SPM

’05/P1

CH4 + Cl2 → CH3Cl + HCl chloromethane

CH3Cl + Cl2 → CH2Cl2 + HCl dichloromethane

CH2Cl2 + Cl2 → CHCl3 + HCl trichloromethane

CHCl3 + Cl2 → CCl4 + HCl tetrachloromethane

Chemical Properties of Alkanes

H Cl Cl | Cl2 | Cl2 | H — C — H ⎯→ H — C — H ⎯→ H — C — Cl | | | H H H CH3Cl CH2Cl2 1 H substituted 2 H substituted

1 Reactivity of alkanes Alkanes are saturated hydrocarbons and are less reactive compared to unsaturated hydrocarbons. 2 Combustion of alkanes (a) Alkanes undergo complete combustion in the presence of excess air or oxygen to produce carbon dioxide and water. SPM For example ’05/P1, CH4 + 2O2 → CO2 + 2H2O ’11/P2

Cl Cl Cl2 | Cl2 | ⎯→ H — C — Cl ⎯→ Cl — C — Cl | | Cl Cl CHCl3 CCl4 3 H substituted 4 H substituted

methane

2C2H6 + 7O2 → 4CO2 + 6H2O ethane

(b) The combustion of alkanes is highly exothermic, that is, produces a lot of heat energy. Hence alkanes are used as fuels. (c) Incomplete combustion of alkanes will produce carbon (black smoke), carbon monoxide and water. (d) The larger the molecular size of the alkane molecule, (i) the smokier or sootier the flame, (ii) the more heat produced on complete combustion. 3 Substitution reactions SPM (a) When a mixture of alkane and chlorine ’10/P1 is exposed to sunlight or ultraviolet light, substitution reaction occurs slowly and a mixture of organic compounds and hydrogen chloride is produced. (b) In a substitution reaction, the hydrogen atoms in an alkane are replaced gradually by chlorine atoms. For example The reaction between methane and chlorine

The Effects of Methane on Everyday Life 1 Methane, commonly known as natural gas, is used as a fuel. 2 It is produced by the anaerobic decay of plants and organic matter by bacteria. Hence methane is found in landfills and peat swamps. 3 Methane is a greenhouse gas. It can trap radiation energy from the sun and also contribute to global warming.

Methane is also known as marsh gas because it is found in marshes and stagnant ponds. It can cause fires in garbage landfills and peat swamps.

345

Carbon Compounds

2

In the substitution reaction of CH4 with Cl2, the carbon atom still has four covalent bonds, either bonded to H atom or Cl atom.

9 The names and molecular formulae of the first nine members of the alkene series are shown in Table 2.4.

2.2 1 Name and give the molecular formula of an alkane with (a) three carbon atoms (b) five carbon atoms (c) six carbon atoms

Table 2.4 The names and molecular formulae of the first nine members of the alkene series

Number of carbon atoms per molecule

Prefix

Name of alkene

2

Eth

Ethene

C2H4

3 Ethane reacts with chlorine under certain conditions to form chloroethane. (a) State the condition for reaction to take place. (b) Name the reaction that takes place. (c) Write an equation for the reaction that occurs.

3

Prop

Propene

C3H6

4

But

Butene

C4H8

5

Pent

Pentene

C5H10

4 A saturated hydrocarbon X has a relative molecular mass of 58. Identify X. [Relative atomic mass: H, 1; C, 12]

6

Hex

Hexene

C6H12

7

Hept

Heptene

C7H14

8

Oct

Octene

C8H16

9

Non

Nonene

C9H18

10

Dec

Decene

C10H20

2

2 Petrol used in cars consists mainly of octane. (a) Give the molecular formula of octane. (b) Write an equation for the complete combustion of octane.

2.3

Alkenes

1 Alkenes are hydrocarbons with the general formula CnH2n, where n = 2, 3, 4…. 2 Alkenes are unsaturated hydrocarbons and contain at least one double bond between carbon atoms. 3 The functional group in alkenes is the carboncarbon double bond (C=C). 4 The molecular formula of an alkene can be obtained by substituting n in the general formula CnH2n with the number of carbon atoms. 5 In the naming of alkenes according to the IUPAC system, all members of the alkene series have their names ending with -ene. 6 The first part (prefix) of the name of an alkene depends on the number of carbon atoms in the molecule. 7 The first member of the alkene series is ethene, C2H4 because there must be a minimum of two carbon atoms to form a carbon-carbon double bond (C=C). Hence methene does not exist. 8 Note that when writing the structural formula of alkenes, (a) there is a carbon-carbon double bond (C=C) in the chain. (b) each carbon atom forms four bonds (four single bonds or one double bond + two single bonds). (c) each hydrogen atom should have one single covalent bond. Carbon Compounds

Molecular formula

10 The structural formulae of the first nine straight chain alkenes with one terminal double bond (double bonds at the end of a chain) are shown as follows:

H H | | H–C=C–H ethene



H H H | | | H–C–C=C–H | H propene



346

H H H H | | | | H–C–C–C=C–H | | H H butene H H H H H | | | | | H–C–C–C–C=C–H | | | H H H pentene

2 Similarly, the carbon chain of an alkane drawn as a straight chain is the same as that drawn in a bent chain.

H H H H H H | | | | | | H–C–C–C–C–C=C–H | | | | H H H H hexene





H—

H H H H H H H | | | | | | | H–C–C–C–C–C–C=C–H | | | | | H H H H H heptene





H H H H H H H H | | | | | | | | H–C– C–C–C–C–C–C=C–H | | | | | | H H H H H H octene





Physical Properties of Alkenes 1 The physical properties of alkenes are the same as that of alkanes. Low melting and boiling points due to weak van der Waals forces between molecules

H H H H H H H H H | | | | | | | | | H–C–C–C–C–C–C–C–C=C–H | | | | | | | H H H H H H H nonene

H H H H H H H H H H | | | | | | | | | | H–C–C–C–C–C–C–C–C–C=C–H | | | | | | | | H H H H H H H H decene

Do not conduct electricity because they consist of covalent molecules

Less dense than water hence will float on the surface of water

2 The physical properties change gradually on going down the alkene series as shown in Table 2.5. 3 Similar to that of alkanes, the melting points and boiling points of alkenes increase down the homologous series. When the number of carbon atoms per molecule increases, the molecular size increases. Thus the van der Waals forces of attraction between molecules increases, increasing the amount of heat needed to overcome the forces of attraction during melting or boiling. 4 Similar to that of alkanes, the density of alkenes increases gradually down the series as the molecular mass increases.

1 In the drawing of the structural formula of an alkene, the hydrogen atom bonded to the carbon atom can be written as (a) bonded to the side or (b) to the top or (c) bottom of the carbon atom. This is because the single covalent bond can be rotated freely.



Insoluble in water but soluble in organic solvents such as propanone

Physical properties of alkenes





H H H H | | | | C — C — C — C — H is the same as | | | | H H H H H | H—C—H H H | | H—C—C—C—H | | | H H H



H H H H | | | | H — C = C — H or C = C or H — C = C — H | | | | H H H H

347

Carbon Compounds

2



Table 2.5 Physical properties of alkanes

Name of alkene

Molecular formula

Relative molecular mass

Melting point (°C)

Boiling point (°C)

Density (g cm–3)

Ethene

C2H4

28

–169

–104

Gas

–

Propene

C3H6

42

–185

–48

Gas

–

Butene

C4H8

56

–130

–6

Gas

–

Pentene

C5H10

70

–138

30

Liquid

0.64

Hexene

C6H12

84

–140

64

Liquid

0.67

Heptene

C7H14

98

–119

93

Liquid

0.70

Octene

C8H16

112

–104

122

Liquid

0.72

Nonene

C9H18

126

–94

146

Liquid

0.73

Decene

C10H20

140

–66

171

Liquid

0.74

(c) Alkenes undergo addition reactions. During addition reactions, the carbon-carbon double bond breaks open to form two new single bonds. In this process, the unsaturated hydrocarbon is converted to a saturated compound.

Chemical Properties of Alkenes 2

Physical state at room temperature

1 Combustion of alkenes SPM (a) Alkenes burns in excess air or oxygen to ’07/P1 form carbon dioxide and water. Heat energy is released during combustion. C2H4 + 3O2 → 2CO2 + 2H2O 2C3H6 + 9O2 → 6CO2 + 6H2O

| | | | —C=C—+X—Y→—C—C— alkene | | (unsaturated) X Y (saturated)



(b) The combustion of an alkene is more luminous and smokier than an alkane with the same number of carbon atoms. This is because the percentage by mass of carbon in an alkene is higher than that of an alkane. Example Percentage by mass of carbon in hexene (Mr of C6H12 = 84) 6 3 12 =— — — — — — 3 100 = 85.7% 84 Percentage by mass of carbon in hexane (Mr of C6H14 = 86) 6 3 12 =— — — — — — 3 100 = 83.7% 86 (c) Incomplete combustion of alkenes produces carbon (black smoke) and carbon monoxide. 2 Addition reactions of alkenes (a) Addition reactions are reactions in which an unsaturated organic compound combines with another compound to form a single new saturated compound. (b) Alkenes contain a double bond between carbon atoms (C = C bond) which is very reactive. Hence alkenes are more reactive than alkanes. Carbon Compounds



one double bond



two new single bonds

3 The following are examples of addition reactions between alkenes and other elements/ compounds: (a) Hydrogen (hydrogenation) (b) Halogens (halogenation) (c) Water (hydration) (d) Acidified potassium manganate(VII) solution (e) Hydrogen halides 4 Hydrogenation (a) Hydrogenation is the addition of a hydrogen molecule across a carbon-carbon double bond in the presence of nickel or platinum as a catalyst. An alkene is converted to an alkane. For example SPM

’07/P1





Ni or Pt

C3H6 + H2 ⎯⎯⎯→ C3H8 propene propane

(b) Hydrogenation is used to make margarine (in solid form) from vegetable oils (in liquid form). 348

5 Halogenation SPM (a) The addition reactions between alkenes ’10/P1, ’11/P2 and halogens (chlorine and bromine) are called halogenation. (b) Bromination is used as a chemical test to distinguish alkanes from alkenes. Alkenes decolourise the brown colour of liquid bromine whereas alkanes do not decolourise the brown colour of liquid bromine. For example: When ethene is passed into liquid bromine, the brown colour of bromine is decolourised immediately and a colourless organic liquid is formed.

(d) Like liquid bromine, acidified potassium manganate(VII) solution is used to distinguish between alkane and alkene compounds. An alkene decolourises the purple colour of acidified potassium manganate(VII) solution whereas an alkane does not. 8 Polymerisation Alkenes undergo polymerisation to form polymers. SPM (a) Ethene undergoes additional polymerisation ’07/P1, ’06/P1 to form polyethene.

room

(saturated)

6 Hydration (a) Hydration occurs when a water molecule is added across the double bond between the atoms in the presence of phosphoric(V) acid with H3PO4 as a catalyst at 300 °C. SPM (b) An alkene is converted to an alcohol in ’07/P2 hydration. For example

H H H H | | | | polymerisation nH — C = C — H ⎯⎯⎯⎯⎯→ — C — C — | | ethene H H n polyethene

H3PO4



2

(c) In the formation of a diol, two hydroxyl (–OH) groups are added across the double bond in the alkene molecule. For example: C2H4 + H2O + [O] → C2H4(OH)2 ethene ethane-1, 2-diol

⎯⎯⎯⎯→ C H Br2 2 4 temperature bromine 1,2-dibromoethane

C2H4 + Br2 ethene



| | —C ⎯ C— | | OH OH

C2H4 + H2O ⎯⎯⎯⎯⎯→ C2H5OH 300 °C, 60 atm ethanol

ethene

7 Reaction with acidified potassium manganate(VII) solution (a) When an alkene reacts with acidified potassium manganate(VII) solution, the purple colour of potassium manganate(VII) is decolourised immediately and an organic compound called diol is formed. (b) A diol is a saturated alcohol with two hydroxyl (–OH) groups on adjacent carbon atoms.

(b) Propene undergoes polymerisation to form polypropene. H H H | | | polymerisation nH — C = C — C — H ⎯⎯⎯⎯→ | H propene

H | C — | H

CH3 | C | H n

polypropene

To compare the chemical properties of alkanes and alkenes having the same number of carbon atoms Apparatus Porcelain dish, wooden splint, dropper and Bunsen burner.

SPM

’11/P2

Procedure A Combustion of alkanes and alkenes in air EXUQLQJVSOLQW

Materials Hexane, hexene, liquid bromine and acidified potassium manganate(VII) solution.

ILOWHUSDSHU

SRUFHODLQ GLVK

KH[DQH RUKH[HQH

KH[DQH RUKH[HQH

Figure 2.3 Combustion of alkane and alkene 7&

349

Carbon Compounds

Activity 2.2





3 The mixture is shaken gently. 4 The colour change that takes place in the test tube is recorded. 5 Steps 1 to 4 are repeated using hexene.

1 About 1 cm3 of hexane and hexene are placed separately into two separate porcelain dishes. 2 The organic liquids are ignited with a burning splint. 3 A piece of filter paper is held above the flames in each of the dishes to detect the amount of soot formed. 4 The sootiness of the flame and the amount of soot collected on the two pieces of filter papers are recorded.

C Reactions with acidified potassium manganate(VII) solution 1 A few drops of potassium manganate(VII) solution are placed in a test tube, followed by about 1 cm3 of dilute sulphuric acid. 2 About 2 cm3 of hexane is then added to the acidified potassium manganate(VII) solution. 3 The mixture is shaken gently. 4 The colour change that occurs in the test tube is recorded. 5 Steps 1 to 4 are repeated using hexene.

B Reactions with bromine 1 About 1 cm3 of liquid bromine is placed in a test tube. 2 About 2 cm3 of hexane is then added to the liquid bromine. Results

Observations

Test 2

Hexane

Hexene

Combustion

Burns in air with a sooty flame

Burns in air with a more sooty yellow flame

Reaction with liquid bromine

The brown colour of liquid bromine remains unchanged

The brown colour of liquid bromine is decolourised

Reaction with acidified potassium manganate(VII) solution

The purple colour of potassium manganate(VII) solution remains unchanged

The purple colour of potassium manganate(VII) solution is decolourised

Conclusion The chemical properties of hexene are different from those of hexane: (a) Both hexane and hexene undergo combustion but the flame of hexene is sootier than that of hexane. (b) Hexene decolourises the brown colour of liquid bromine whereas hexane does not. (c) Hexene decolourises the purple colour of acidified potassium manganate(VII) solution whereas hexane does not. Discussion 1 The combustion of hexene produces more soot than the combustion of hexane. This is because the percentage by mass of carbon in hexene is higher than that of hexane. 2 Hexene undergoes addition reaction with bromine: C6H12 + Br2

→

C6H12Br2

(brown) 1, 2-dibromohexane (colourless)

3 Hexene undergoes addition reaction with acidified potassium manganate(VII) C6H12 + H2O + [O]

Carbon Compounds

→ C6H12(OH)2

from KMnO4 hexane-1,2-diol (purple) (colourless)

350

Br H Br | | | C H — C — C — C — H | | | H H H H H H | | | D H — C — C — C — Br | | | H H H

In the addition reaction of an alkene for example with chlorine, the product formed has the two Cl atoms bonded to the two C atoms adjacent to each other. Products with two Cl atoms bonded to the same C atom or across another C atom are not formed. Example: Addition of chlorine to propene cannot form the following products: H H H | | | H—C—C—C—H | | | Cl H Cl

and

1

Solution Propene undergoes addition reaction with Br2 at the carbon-carbon double bond, hence the product formed has two Br atoms attached to two adjacent C atoms. Answer: B

Homologous Series

’05

1 A homologous series is a family of organic compounds with the same functional group and with similar chemical properties. 2 A functional group is an atom or a group of atoms that determines the chemical properties of an organic compound. 3 All members in the same homologous series have the same functional group and the same chemical properties. 4 All members in the same homologous series (a) have the same general formula (b) can be prepared using similar methods (c) show a gradual change in their physical properties (d) have similar chemical properties (e) differ from each other by a –CH2 group 5 General formulae of some homologous series are shown in Table 2.6.

The following is the equation that represents the reaction between propene and bromine. Propene + Br2 → P Which of the following is the structural formula of P? Br H H | | | A H — C — C — C — H | | | Br H H H H Br | | | B H — C — C — C — H | | | H Br H

Table 2.6 General formulae of some homologous series

Homologous series

General formula

SPM

’07/P2

Functional group

Alkanes

CnH2n+2

Alkenes

CnH2n

– C = C – (double bond)

Alcohols

CnH2n+1OH

– O – H (hydroxyl group)

Carboxylic acids

CnH2n+1COOH

O i – C – O – H (carboxyl group)

Esters

CnH2n+1COOCmH2m+1

O i – C – O – (carboxylate group)

–

351

Carbon Compounds

2

H H Cl | | | H—C—C—C—H | | | H H Cl

CnH2n+2 alkane Hydrogenation H2/Ni, 180 °C

– (CnH2n)n – polymer Addition polymerisation

CnH2n+1OH alcohol

Addition of halogen, X2

CnH2n alkene

Addition of hydrogen halide, HX

Addition of acidified KMnO4

Addition of water, H3PO4, 180 °C, 60 atm

CnH2nX2

CnH2n+1X

2

CnH2n(OH)2 diol

4 Isomers have different physical properties because they have different structural formulae. 5 For methane, ethane and propane, there is only one structure. Therefore methane, ethane and propane do not have isomers. All the other alkanes, have isomers. 6 There are two different ways of arranging the four carbon atoms and ten hydrogen atoms for butane, C4H10. Thus butane has two isomers as follows:

2.3 1 (a) What is the general formula of alkene? (b) Give the molecular formula of an alkene with (i) three carbon atoms (ii) five carbon atoms (iii) seven carbon atoms 2 Propane and propene are both hydrocarbons. (a) State two common physical properties between propane and propene. (b) State one common chemical property between propane and propene. 3 Write chemical equations for the reactions between propene with (a) chlorine (b) water (c) hydrogen (d) excess oxygen (e) acidified potassium manganate(VII)

2.4

Isomerism

(a) Straight chain (all 4 C atoms form a straight chain) H H H H | | | | H—C—C—C—C—H | | | | H H H H (b) Branched chain (3 C atoms form a straight chain with 1 C atom forming a branch) H | H—C—H H H | | H—C—C—C—H | | | H H H

SPM

’10/P1, ’10/P2, ’11/P2

1 Isomers are compounds which have the same molecular formula but with different structural formulae. 2 Isomerism is the existence of two or more compounds that have the same molecular formula but with different structural formulae. 3 Isomers will have the same chemical properties when they have the same functional group. Carbon Compounds

352

7 There are three different ways of arranging the five carbon atoms and twelve hydrogen atoms in pentane, C5H12. Thus C5H12 has three isomers.

H H H H H | | | | | H—C—C—C—C—C—H | | | | | H H H H H

(b) One branched chain (4 C atoms form a straight chain with 1 C atom forming a branch) H | H—C—H H H H | | | H—C—C—C—C—H | | | | H H H H

(c) Two branched chains (3 C atoms form a straight chain with 2 C atoms forming two branches) H | H—C—H H H | | H—C—C—C—H | | H H H—C—H | H

8 All the alkenes above propene have isomers. Butene, C4H8 has three isomers as follows: (a) Straight chain with a double bond at the end of the chain (4 C atoms form a straight chain with a double bond at the first C atom) H H H H | | | | H—C=C—C—C—H | | H H

(b) Straight chain with a double bond in the middle of the chain (4 C atoms form a straight chain with a double bond at the second C atom) H H H H | | | | H—C—C=C—C—H | | H H

(c) Branched chain (3 C atoms form a straight chain with a double bond and 1 C atom forming a branch) H | H—C—H H H | | H—C=C—C—H | H

9 Naming of branched isomers of alkanes according to the IUPAC system Step 1 Find the longest continuous chain of carbon atoms in the molecule and name the longest chain as the parent chain.

For example: H | H—C—H H H H | | | H—C—C—C—C—H | | | | H H H H The longest chain has four carbon atoms. The name of the parent chain is butane. 353

Carbon Compounds

2

(a) Straight chain (5 C atoms form a straight chain)

Step 2 1 Name the branched chain attached to the parent chain as alkyl group. 2 The alkyl groups are named according to the number of carbon atoms present.

Formula and name of alkyl groups Number of carbon atoms

Formula

Name

1

–CH3

Methyl

2

–C2H5

Ethyl

3

–C3H7

Propyl

The alkyl group has the general formula CnH2n+1 where n = 1, 2, 3…

H –CH3 (methyl group) | is attached to the H—C—H parent chain of four H H H carbon atoms | | | H—C—C—C—C—H | | | | H H H H 2

Step 3 Identify the position of the alkyl group that is attached to the parent chain by number. (a) This is done by numbering the carbon atom in the parent chain using the lowest number. (b) Use hyphens to separate words from numbers in the name, for example: 2 – methyl.

For example: H | H—C—H –CH3 (methyl group) is attached to carbon H H H number 2 | | | H — C1 — C2 — C3 — C4 — H | | | | H H H H The name of the alkane is 2-methyl butane carbon no. 2

side chain is –CH3

parent chain has four carbon atoms

Step 4 1 If there are more than one similar branch, use the following prefixes (a) di for two similar branched chains (b) tri for three similar branched chains (c) tetra for four similar branched chains 2 Name the positions of carbon atoms in the parent chain containing the branches. For example, the positions of two branches may be 2, 2 or 2, 3.

For example: H | H—C—H H H H | | | H—C—C—C—C—H | | | H H H H—C—H | H 2,2–dimethyl butane both –CH3 branches attached to carbon no. 2

Carbon Compounds

354

two –CH3 branches

parent chain has four carbon atoms

Step 5 If there are more than one alkyl group, list the names of the alkyl groups in alphabetical order.

For example: H —

H H C2H5 CH3 H H | | | | | | C — C — C — C — C — C — H | | | | | | H H H H H H 3-ethyl,4-methylhexane SPM

10 Naming of alkenes according to the IUPAC system

’11/P1

1 Select the longest carbon chain with the double bond (C=C) as the parent alkene. 2 Name the parent alkene according to the number of carbon atoms.

H H H H | | | | H — C1 — C2 == C3 — C4 — H | | H H Correct name but-2-ene

Step 2 1 Select the position of the double bond by choosing the smallest number for the carbon atom with the C=C bond. 2 Name the position of the double bond with a number followed by a hyphen, for example: 2-ene.

parents chain has 4 carbon atoms

2

For example H | H—C—H H H | | H—C=C—C—H | longest chain with C=C H has four carbon atoms

Step 1

double bond present

double bond at carbon no. 2

Wrong name but-3-ene 3, the larger number is not used

Example methyl group at carbon H no. 2 | H—C—H H H H | | | H—C—C=C—C—H | | H H

Step 3 Identify the alkyl group and its position in the parent chain as fixed in Step 2.

2-methylbut-2-ene –CH3 group at carbon no. 2

355

C=C at carbon no. 2

Carbon Compounds

1 To choose the longest carbon chain as the parent chain, count the number of carbon atoms which could be in a straight chain or a bent chain. For example: –C– This is the longest | – C – chain with 6 C atoms | –C–C=C–C–C–

2

2 The alkyl group bonded by a single covalent bond to a carbon atom can rotate in a molecule. It may be drawn as up or down or at the side in a 2-D structural formula. For example: These two structures below are not isomers, they are the same compound. H | H—C—H H H H H H H | | | | | | H — C — C = C — C — H H—C—C=C—C—H | | | | H H H H H—C—H | H 3 The following alkene does not exist because the central carbon atom has five covalent bonds. C | C=C—C | C 4 The following alkene structures are the same, they are not isomers. (a) C = C — C — C — C is the same as C — C — C — C = C (b) C = C — C — C; C — C — C = C; C — C — C — C are the same isomer. | | i C C C

2.4 1 Give the IUPAC name of the following compounds: (a) H | H—C—H H H H | | | H—C—C=C—C—H | | H H (b) H | H H — C — H H H H | | | | | H ⎯ C ⎯ C ⎯ C ⎯ C = C ⎯ H | | | H H H — C — H | H

Carbon Compounds

(c) H ⎯

H H H H | | | | C ⎯ C ⎯ C = C ⎯ C ⎯ H | | | | H H — C — H H — C — H H | | H H

2 Draw and name all the isomers for C5H10. 3 Ethane reacts with chlorine in the presence of sunlight to form a substituted product with the molecular formula of C2H4Cl2. Draw and name all the possible isomers of this product. Which of the isomer is also formed when ethene reacts with chlorine in the addition reaction?

356

Alcohols

7 All alcohols above ethanol have isomers. The position of the hydroxyl (–OH) group and the alkyl group are shown by numbering the carbon atoms from the end of the carbon chain which gives the smallest number to the –OH group. 8 Propanol, C3H7OH has two isomers:

1 Alcohols have the general formula CnH2n+1OH where n = 1, 2, 3… The CnH2n+1– group represents the alkyl group. 2 The functional group of alcohol is the hydroxyl (–OH) group. The hydroxyl (–OH) group is joined to the carbon atom in the alcohol molecule by a single covalent bond. 3 The molecular formula of an alcohol can be obtained by substituting n in the general formula CnH2n+1OH with the number of carbon atoms. 4 Based on the IUPAC system of naming straight chain alcohols, the letter e at the end name of alkane is replaced by the suffix ol. For example CH4 (methane) → CH3OH (methanol) C2H6 (ethane) → C2H5OH (ethanol)

Alkane

Alkane formula (CnH2n+2)

Alcohol

Alcohol formula

Methane

CH4

Methanol

CH3OH

Ethane

C2H6

Ethanol

C2H5OH

Propane

C3H8

Propanol

C3H7OH

Butane

C4H10

Butanol

C4H9OH

Structural formula (a)

H H H parent chain has three carbon atoms | | | H–C–C–C–H Correct name propan-1-ol | | | O H H position of the –OH group is at the first carbon atom | H Wrong name propan-3-ol



The –OH group is at the first carbon atom

(b) H H H | | | H–C–C–C–H | | | H O H | H



The parent chain has three carbon atoms

propan-2-ol position of the –OH group is at the second carbon atom

9 Butanol, C4H9OH has four isomers as follows: Structural formula (a) H H H H | | | | H–C–C–C–C—H | | | | O H H H | H (b) H H H H ⎮ ⎮ ⎮ ⎮ H–C–C–C–C—H ⎮ ⎮ ⎮ ⎮ H O H H ⎮ H

H H H ⎮ ⎮ ⎮ H — C — H H—C—C—H ⎮ ⎮ ⎮ O O H ⎮ ⎮ H H methanol

3, the bigger number is not used

The –OH group is at the second carbon atom

5 Note that when writing the structural formula of alcohols, (a) each carbon atom should have four single covalent bonds. (b) each hydrogen atom should have one single covalent bond. (c) each oxygen atom has two single covalent bonds. (d) the carbon atoms are connected by single bonds. 6 Methanol and ethanol has one structural formula each. Hence they have no isomers.



IUPAC name

ethanol

357

IUPAC name parent group has four carbon atoms

butan-1-ol position of the –OH group is at the first carbon atom

parent group has four carbon atoms

butan-2-ol position of the –OH group is at the second carbon atom

Carbon Compounds

2

2.5

Structural formula (c)

methyl group at second carbon atom

H | H—C—H H H | | H–C–C–C–H | | | O H H | H

Example

IUPAC name



methyl group at second carbon atom

H H H |

|

|

|

|



O O O



H H H



|

2-methylpropan-1-ol parent group has three carbon atoms

|

H—C—C—C—H |

|

three-OH groups

propan-1,2,3-triol

position of the –OH group is at the first carbon atom

parent group has three carbon atoms

three-OH groups at 1st, 2nd and 3rd carbon atoms

Industrial Production of Alcohols

2

(d) H | H — C — H H H | | H–C–C–C–H | | | H O H | H

methyl group at second carbon atom

1 Ethanol (C2H5OH) can be produced industrially by two processes: (a) The hydration of ethene (b) The fermentation of sugar or starch 2 Hydration of ethene When a mixture of ethene and steam is passed over the catalyst, phosphoric(V) acid (H3PO4) at 300 °C and high pressure (65 atm), ethanol is produced. 3 Fermentation When yeast is added to sugar or starch, ethanol and carbon dioxide are produced. The enzyme called zymase, breaks down the glucose molecules to form ethanol and carbon dioxide.

2-methylpropan-2-ol parent group has three carbon atoms

position of the –OH group at second carbon atom

10 In the naming of an alcohol with more than one –OH group: (a) di is used for two –OH groups (b) tri is used for three –OH groups

SPM

’06/P1

yeast

C6H12O6 ⎯⎯→ 2C2H5OH + 2CO2

glucose ethanol carbon dioxide

To prepare ethanol in the laboratory by fermentation and distillation 3 The yeast paste is added to the glucose solution and the mixture is stirred well. 4 The conical flask is closed with a rubber stopper fitted with a delivery tube. The other end of the delivery tube is dipped in limewater. 5 The apparatus is left in a warm place (about 25 – 35 °C) for about one week. The changes that occur in the conical flask and the test tube are recorded from time to time. 6 After about one week, the products in the conical flask are filtered. The filtrate obtained is transferred into a distillation flask. 7 The filtrate is distilled in a flask fitted with a fractionating column and Liebig condenser.

Apparatus Conical flask, boiling tube, thermometer, fractionating column, wire gauze, retort stand, distillation flask, rubber stopper, delivery tube, tripod stand and Bunsen burner.

Activity 2.3

Materials Glucose, yeast, distilled water and limewater. Procedure 1 About 20 g of glucose is dissolved in 200 cm3 of distilled water in a conical flask. 2 A little warm water (35 °C) is added to about 5 g of yeast in a small beaker. The mixture is stirred well to form a paste. Carbon Compounds

358

Figure 2.4 Fermentation process

Figure 2.5 Fractional distillation

Conclusion Ethanol can be prepared in the laboratory by the fermentation of glucose or any other carbohydrates.

2

Result 1 During fermentation, ethanol and carbon dioxide are produced. 2 The carbon dioxide produced causes limewater to turn cloudy. 3 The concentration of ethanol produced in fermentation can be increased by fractional distillation.

2 Ethanol is a volatile liquid because it has a low boiling point at 78 °C. 3 Ethanol is very soluble in water because of the presence of the hydroxyl group. (a) The hydrocarbon part of alcohol is insoluble in water. (b) Hence alcohol with a large hydrocarbon chain is insoluble in water. (c) The solubility of alcohols in water decreases as the molecular size increases. 4 Alcohols are neutral and have a pH of 7. 5 Alcohols are covalent compounds. They do not conduct electricity.

Yeast is a living organism. Ethanol is actually a by-product formed from the living process of yeast. The highest concentration of ethanol prepared by fermentation is only 14%. This is because yeast is killed when the ethanol formed exceeds 14%. Hence higher concentration of ethanol has to be obtained from fractional distillation.

Physical Properties of Ethanol and Other Alcohols 1 Ethanol is a colourless liquid and has a characteristic odour.

To investigate the chemical properties of ethanol Apparatus Evaporating dish, wooden splint, test tube, boiling tube, delivery tube with stopper, glass wool, porcelain chips, beaker, retort stand with clamp and test tube holder. Ethanol, concentrated sulphuric acid, potassium dichromate(VI), blue litmus paper and liquid bromine. (A) Combustion of ethanol in air Procedure 1 About 2 cm3 of ethanol is poured into an evaporating dish. 2 The ethanol is ignited using a lighted wooden splint. 3 The flammability of ethanol and the sootiness of the flame are recorded. 359

Carbon Compounds

Activity 2.4

Materials

3 Acidified potassium dichromate(VI) solution is an oxidising agent.

Result

2

Nature of combustion

Observation

Flammability

Easily burned

Colour of flame

Blue

Sootiness of flame

Non-sooty

Conclusion Ethanol burns readily in air. The combustion of ethanol produces a non-sooty, pale blue flame. The products of combustion are carbon dioxide and water. (B) Oxidisation of ethanol Procedure 1 About 1 cm3 of concentrated sulphuric acid and 5 cm3 of potassium dichromate(VI) solution are poured into a boiling tube. 2 About 5 cm3 of ethanol is added to the acidified potassium dichromate(VI) solution. 3 A rubber stopper fitted with a delivery tube is inserted into the boiling tube. The delivery tube is inserted into a test tube placed in a beaker half-filled with ice-cold water. 4 The mixture of ethanol and acidified potassium dichromate(VI) is heated gently. Any colour change in the mixture is noted. 5 The distillate is collected in the test tube and is tested with litmus paper.

(C) Dehydration of ethanol Procedure 1 Some glass wool is placed in a boiling tube. 2 About 2 cm3 of ethanol is poured into the boiling tube to soak the glass wool. 3 Some porcelain chips are placed in the midsection of the boiling tube. 4 The boiling tube is closed with a rubber stopper fitted with a delivery tube. The other end of the delivery tube is placed under an inverted test tube filled with water in a beaker. 5 The porcelain chips are heated strongly until red hot. The Bunsen burner flame is then shifted to the glass wool to vaporise the ethanol absorbed in it. 6 The gas is collected in two test tubes. The gas produced is collected by displacement of water and tested with (a) a few drops of bromine water and shaken (b) a few drops of acidified potassium manganate(VII) solution and shaken SPM

’05/P1

Figure 2.7 Dehydration of ethanol

Result

Figure 2.6 Oxidation of ethanol

Test on gas collected

Result Compound Reactant mixture

Distillate

With bromine water

Observation The acidified potassium dichromate(VI) solution changes from orange to green

Brown colour of bromine is decolourised

With acidified potassium Purple colour of manganate(VII) solution potassium manganate(VII) is decolourised

A colourless liquid (a vinegary smell) which changes blue litmus paper to red

Conclusion 1 When ethanol vapour is passed over heated porcelain chips (aluminium oxide), dehydration occurs and ethene is produced. 2 Ethene is confirmed to be present by the decolourisation of the brown colour of bromine water and the purple colour of acidified potassium manganate(VII) solution.

Conclusion 1 When ethanol is boiled with acidified potassium dichromate(VI) solution, it is oxidised to ethanoic acid. 2 Ethanoic acid is a colourless liquid with a vinegary smell and turns blue litmus paper red. Carbon Compounds

Observation

360

(b) Dehydration is carried out by (i) passing alcohol vapour over heated porcelain chips or (ii) refluxing alcohol with concentrated sulphuric acid (acts as a dehydrating agent). (c) Methanol does not undergo dehydration since there is no alkene with one carbon atom.

Chemical Properties of Ethanol and Other Alcohols 1 (a) Alcohol undergoes complete combustion SPM when it burns in excess air to produce carbon ’08/P2 dioxide and water. For example C2H5OH + 3O2 → 2CO2 + 3H2O ethanol 9 C3H7OH + — O2 → 3CO2 + 4H2O 2 propanol

The alcometer (breath tester) used for testing the breath of suspected drink-driver contains potassium dichromate(VI). The chemical will oxidise any ethanol present in the breath and produces an electric current.

(b) Combustion of alcohol such as ethanol gives out a lot of heat energy. Hence ethanol is a good fuel. (c) If the supply of oxygen is insufficient, incomplete combustion of ethanol occurs. Carbon monoxide gas, carbon (black soot) and water are produced. 2 (a) Oxidation of alcohols produces the corresponding carboxylic acids. For example

1 Alcohols as fuels When alcohol is burn in air, carbon dioxide and water are produced, and a large quantity of heat energy is released. Ethanol is a clean fuel because it does not release toxic gases in combustion. 2 Alcohols as solvents Alcohols are good solvents for organic compounds such as shellac, varnishes, paints, perfumes and dyes. 3 Uses of alcohols in medicines (a) Ethanol is used as a mild antiseptic for skin infection and disinfection. (b) Propan-2-ol is used as rubbing alcohol to reduce fever. 4 Uses of alcohols in cosmetics (a) Ethanol is used to make aftershave lotion and nail polish. (b) Propan-1,2,3-triol (common name is glycerol) is used in moisturiser. (c) Alcohols are used as solvents for perfumes. 5 Alcohols as a source of chemicals (a) Ethanol is oxidised to make vinegar. (b) Methanol is used to make formalin and polymers. 6 The misuse and abuse of alcohols (a) Ethanol is a component of alcoholic drink. Excess drinking of alcohol increases the risk of heart disease, kidney disease and liver disease. (b) Alcoholism is an addiction caused by excessive drinking of alcohol for a prolonged period of time.

K2Cr2O7/H2SO4

CH3CH2OH + 2[O] ⎯⎯⎯⎯⎯→ ethanol from K2Cr2O7 CH3COOH + H2O ethanoic acid CH3CH2CH2OH + 2[O] ⎯→ propanol CH3CH2COOH + H2O propanoic acid (b) Oxidation is carried out by heating the alcohols with oxidising agents such as acidified potassium manganate(VII) solution or acidified potassium dichromate(VI) solution. (c) When oxidation reaction occurs: (i) The colour of potassium dichromate(VI) changes from orange to green. (ii) The colour of potassium manganate(VII) changes from purple to colourless (decolourisation). 3 (a) Dehydration of alcohols (except methanol) produces the corresponding alkenes. For example SPM C2H5OH ⎯→ C2H4 + H2O ’07/P2, ’06/P1 ethanol ethene C3H7OH ⎯→ C3H6 + H2O propanol

propene

361

Carbon Compounds

2

Uses of Alcohol in Everyday Life

(c) Driving after drinking too much alcohol can cause road accidents.

Number of carbon atoms

Alkane

Carboxylic acid

1

Methane

Methanoic acid

2

Ethane

Ethanoic acid

3

Propane

Propanoic acid

2.5 1 Compound X has a molecular formula of C4H8O. When compound X is refluxed with acidified potassium dichromate(VI), the mixture changes colour from orange to green. (a) Name the funtional group of compound X. (b) What is the general formula of the homologous series in which compound X is a member. (c) Draw and name all the isomers of compound X. (d) Write the chemical equation for the reaction between compound X and acidified potassium dichromate(VI).

4 The name of the carboxylic acid will depend on the number of carbon atoms in its molecule. The carbon atom of the functional group is counted as part of the carbon chain. 5 The molecular formula of a carboxylic acid can be obtained by substituting n in the general formula CnH2n+1COOH with the number of carbon atoms. 6 In the writing of the structural formula of a carboxylic acid, (a) the –COOH group is always at the terminal carbon atom (at the end of the chain). (b) the carboxyl group consists of a carbon atom which forms a double bond with oxygen atom and a single covalent bond with the hydroxyl (–OH) group.

2

2 Identify the compounds P, Q, R, S and T from the reaction scheme given below:

Glucose

KMnO4/

yeast

Compound H2SO4 P

oxygen, heat

Compound Q

porcelain chips, heat

Compound R + compound S



Compound T



3 State the type of reactions that occur and give the molecular formulae of compounds W, X, Y and Z in the following equations.

concentrated H SO

Table 2.7 The names and formulae of the first four members of carboxylic acids

phosphoric acid

(b) X + H2O ⎯⎯⎯⎯⎯⎯→ C3H7OH

K Cr O /H SO

2 2 7 2 4 (c) C4H9OH + 2[O] ⎯⎯⎯⎯⎯→ Y + H2O

Name

(d) Z + 6O2 ⎯→ 4CO2 + 5H2O

2.6

Molecular formula

Methanoic HCOOH acid

Carboxylic Acids

Ethanoic acid

1 Carboxylic acids are organic acids that have the general formula CnH2n+1COOH, where n is 0, 1, 2, 3 …. 2 The functional group of carboxylic acids is the carboxyl group, –COOH. 3 Based on the IUPAC system of naming, a carboxylic acid is named by replacing the final letter e in the name of the corresponding alkane with oic acid. Carbon Compounds

single bond

7 The molecular and structural formulae of the first four members of carboxylic acids are shown in Table 2.7.

2 4 (a) W ⎯⎯⎯⎯⎯⎯⎯⎯→ C3H6 + H2O



O double bond i –C–O–H

Structural formula O i H — C — OH

CH3COOH O i CH3 — C — OH

Propanoic C2H5COOH O i acid CH3 — CH2 — C — OH Butanoic acid

362

C3H7COOH O i CH3 — CH2 — CH2 — C — OH

2 Ethanoic acid has a vinegary smell. 3 Ethanoic acid is soluble in water. 4 On going down the homologous series of the carboxylic acids, (a) the solubility in water decreases, (b) the boiling point increases.

Preparation of Carboxylic Acid 1 Carboxylic acid is prepared in the laboratory by the oxidation of the corresponding alcohol. Example oxidation Ethanol ⎯⎯⎯→ ethanoic acid oxidation Propanol ⎯⎯⎯→ propanoic acid oxidation Butanol ⎯⎯⎯→ butanoic acid

Chemical Properties of Ethanoic Acid 1 Ethanoic acid is a weak acid because it undergoes partial ionisation in water. Only a small percentage of the ethanoic acid molecules ionise to form hydrogen ions, H+. Most of the ethanoic acid remains as molecules.



CH3COOH

CH3COO– + H+

2 Aqueous ethanoic acid turns blue litmus paper red and is an electrolyte. 3 Ethanoic acid reacts with bases to form salts and water in neutralisation. The salts produced in the reaction are known as ethanoate. For example CH3COOH + NaOH → CH3COONa + H2O ethanoic acid sodium ethanoate

2CH3COOH + CuO → (CH3COO)2Cu + H2O

ethanoic copper copper(II) acid oxide ethanoate (black) (blue)

4 Ethanoic acid reacts with metal carbonates SPM to form salts, carbon dioxide and water. For ’05/P1 example 2CH3COOH + Na2CO3 ⎯→ 2CH3COONa + CO2 + H2O

K2Cr2O7/H2SO4

CH3CH2OH + 2[O] ⎯⎯⎯⎯⎯⎯→ CH3COOH + H2O



sodium ethanoate

2CH3COOH + CaCO3 ⎯→ (CH3COO)2Ca + CO2 + H2O calcium ethanoate

5 Ethanoic acid reacts with reactive metals such as magnesium and zinc to form salts and hydrogen gas. For example 2CH3COOH + Mg ⎯→ (CH3COO)2Mg + H2

Figure 2.8 Preparation of ethanoic acid by reflux



magnesium ethanoate

2CH3COOH + Zn ⎯→ (CH3COO)2Zn + H2

Physical Properties of Ethanoic Acid and Other Carboxylic Acids



zinc ethanoate

6 Ethanoic acid reacts with an alcohol to form ester and water. For example Ethanoic acid + ethanol → ethyl ethanoate + water

1 Ethanoic acid is a colourless liquid at room temperature. Pure ethanoic acid is known as glacial ethanoic acid because it freezes to form colourless crystals which look like ice.

(ester)

363

Carbon Compounds

2

2 In the laboratory, ethanoic acid is prepared by the oxidation of ethanol using an oxidising agent such as (a) acidified potassium dichromate(VI) or (b) acidified potassium manganate(VII) 3 The method of heating a mixture of ethanol and the oxidising agent in a flask fitted with an upright Liebig condenser is known as reflux. 4 Heating under reflux is used (a) to prevent volatile substances (ethanol and ethanoic acid) from escaping into the atmosphere, (b) to ensure that the reactants go to complete reaction. 5 In the oxidation of ethanol by acidified potassium dichromate(VI), (a) the colour of potassium dichromate(VI) solution changes from orange to green. (b) the ethanoic acid produced has a vinegary smell.

To study the chemical properties of ethanoic acid Results

Apparatus Test tubes, beakers, evaporating dish, delivery tube and wooden splints.

Test

Materials Ethanoic acid, sodium hydroxide solution, sodium carbonate, limewater, magnesium ribbon, glacial ethanoic acid and concentrated sulphuric acid.

2

Procedure 1 About 3 cm3 of dilute ethanoic acid is placed in a test tube. 2 About 3 cm3 of sodium hydroxide is added to the ethanoic acid and the mixture is shaken. 3 The reaction mixture is poured into an evaporating dish and is heated until it becomes dry. 4 The residue left in the evaporating dish is observed.

Effervescence occurs and sodium carbonate dissolves

(b) Gas released + limewater

Gas produced turns limewater cloudy

(C) Reaction between ethanoic acid and a metal SPM Procedure ’11/P2 1 About 5 cm3 of ethanoic acid is placed in a test tube. 2 A piece of magnesium ribbon is added to the ethanoic acid. 3 A lighted splint is placed near the mouth of the test tube to test the gas liberated.

Results

Observation A white powder was left in the evaporating dish.

Test

Conclusion Ethanoic acid reacts with a base, sodium hydroxide to produce a salt and water. (B) Reactions between ethanoic acid and a metal carbonate

Observation

(a) Ethanoic acid + magnesium

Effervescence occurs and the magnesium ribbon dissolves

(b) Gas released + lighted splint

Gas burns with a ‘pop’ sound when the lighted splint is placed at the mouth of the test tube

Conclusion Ethanoic acid reacts with magnesium, a metal to form a salt and hydrogen gas.

Procedure 1 About 5 cm3 of ethanoic acid is placed in a test tube. 2 A spatula of sodium carbonate is added to the ethanoic acid. 3 The gas released is passed into limewater.

Activity 2.5

(a) Ethanoic acid + sodium carbonate

Conclusion Ethanoic acid reacts with sodium carbonate, a metal carbonate to form a salt, carbon dioxide and water.

(A) Reaction of ethanoic acid with a base

(D) Reactions between ethanoic acid and alcohol Procedure 1 2 cm3 of glacial ethanoic acid is placed into a test tube. 2 4 cm3 of pure ethanol is added to the ethanoic acid. 3 About 1 cm3 of concentrated sulphuric acid is added slowly and carefully to the mixture using a dropper. 4 The reaction mixture is shaken and heated slowly for about 3 minutes. 5 The content of the test tube is then poured into a beaker half-filled with water.

Figure 2.9 Reaction with ethanoic acid and a metal carbonate

Carbon Compounds

Observation

364

2 An ester is an organic compound formed when a carboxylic acid reacts with an alcohol. 3 Esterification is the reaction between a carboxylic acid and an alcohol to produce ester and water. 4 Concentrated sulphuric acid acts as a catalyst to speed up esterification. 5 When ethanoic acid is warmed with ethanol in the presence of a few drops of concentrated sulphuric acid, esterification occurs. The ester known as ethyl ethanoate and water are produced.

Observation 1 An oily product forms a layer on top of the water’s surface. 2 The product is colourless and has a sweet fruity smell. Conclusion 1 Ethanoic acid reacts with ethanol, an alcohol to form an ester and water. 2 Ester has a sweet fruity smell and is insoluble in water.



conc. H2SO4

CH3COOH + C2H5OH ⎯⎯⎯⎯→ CH3COOC2H5 + H2O ethyl ethanoate

Discussion 1 The sweet smelling oily product is an ester. Chemical properties of other carboxylic acids

General 2RCOOH + CaCO3 → (RCOO)2Ca + H2O + CO2 Example 2HCOOH + CaCO3 → (HCOO)2Ca + H2O + CO2 2C2H5COOH + CaCO3 → (C2H5COO)2Ca + H2O + CO2 (d) react with alcohols to form esters and water. SPM General ’06/P1 RCOOH + R′OH → RCOOR′ + H2O Example HCOOH + C2H5OH → HCOOC2H5 + H2O C2H5COOH + C2H5OH → C2H5COOC2H5 + H2O

1 All members of the carboxylic acids have similar chemical properties because they have the same functional group, –COOH. 2 The general formula of a carboxylic acid can also be written as RCOOH where R is H or an alkyl group. 3 All members of this carboxylic group will (a) react with alkalis to form salts and water General RCOOH + NaOH → RCOONa + H2O Example HCOOH + NaOH → HCOONa + H2O C2H5COOH + NaOH → C2H5COONa + H2O (b) react with active metals to form salts and hydrogen gas General 2RCOOH + Mg → (RCOO)2Mg + H2 Example 2HCOOH + Mg → (HCOO)2Mg + H2 2C2H5COOH + Mg → (C2H5COO)2Mg + H2 (c) react with metallic carbonates to form salts, carbon dioxide and water

Uses of Carboxylic Acids in Daily Life Carboxylic acid

365

Use

Methanoic acid

To coagulate latex

Ethanoic acid

To make vinegar

Benzoic acid

Used as a food preservative

Carbon Compounds

2

Figure 2.10 Reaction of ethanoic acid and ethanol

cracking

Alkane CnH2n+2

hydrogenation

Alkene CnH2n

hydration dehydration

combustion

combustion

Alcohol CnH2n+2OH

combustion

2

3 The functional group of esters is the carboxylate group, –COO–. 4 The name of an ester is derived from the alcohol and the carboxylic acid used to prepare it. (a) The first part of the name of the ester is taken from the alkyl group of the alcohol. Example

’11/P1

1 Compound X with an empirical formula of CH2O has a relative molecular mass of 60. (a) Calculate the molecular formula of X and draw its structural formula. (b) Identify the functional group of X and give its general formula. (c) Write a balanced equation between X and calcium carbonate. Predict the observation that will take place. [Relative atomic mass: of H,1; C,12; O,16] 2 (a) Give the name and molecular formula for a carboxylic acid that has five carbon atoms. (b) Give the molecular formula of the organic compound formed when the carboxylic acid named in (a) reacts with (i) sodium hydroxide (ii) magnesium metal

Step I: K2Cr2O7/H2SO4

Alcohol

Alkyl group

Methanol

Methyl

Ethanol

Ethyl

Propanol

Propyl

Butanol

Butyl

(b) The second part of the name comes from the carboxylic acid. The ending –oic of carboxylic acid is replaced by –oate. Example

3 Compound P, with a molecular formula of C3H8O is converted to compounds Q and R in the following reaction scheme. Q

Step II

R (a) Identify compound Q and compound R. (b) State the condition of reaction in Step II. (c) Name the types of reactions that have occurred in Steps I and II.

Carboxylic acid

Carboxylate group

Methanoic acid

Methanoate

Ethanoic acid

Ethanoate

Propanoic acid

Propanoate

(c) In general, the names of esters are of the form ‘alkyl carboxylate’ where alkyl comes from the alcohol used to prepare the ester. Example

Esters

1 The general formula for an ester is CnH2n+1COOCmH2m+1, where n = 0, 1, 2, 3, 4… and m = 1, 2, 3, 4…. 2 The general formula for an ester can also be written as R–COO–R′ where R and R′ are alkyl groups. Carbon Compounds

esterification

SPM

2.6

2.7

Carboxylic Acid CnH2n+1COOH

Ester RCOOR′

Carbon dioxide and water

P

oxidation

366

Alcohol

Carboxylic acid

Names of ester

Methanol

Methanoic acid

Methyl methanoate

Methanol

Ethanoic acid

Methyl ethanoate

Ethanol

Ethanoic acid

Ethyl ethanoate

Ethanol

Propanoic acid

Ethyl propanoate

5 In the writing of the structural formula of an ester using the general formula R–COO–R′, the part R– CO comes from the carboxylic acid and the part of O– R′ comes from the alcohol. 6 Generally, O i R – C – OR′

Hence, the name of the ester is ethyl propanoate. 7 The reaction between an alcohol and a carboxylic acid to produce an ester and water is known as esterification. General Alcohol + carboxylic acid → ester + water Example C2H5OH + CH3CH2COOH →

from carboxylic acid from alcohol



Example O i CH3CH2C O – CH2CH3

propanoic acid



from propanoic acid from ethanol

ethyl propanoate

SPM

’07/P1, ’06/P1

Example:

Step 1

the alcohol part (bonded to –O) is –CH3, hence the prefix is methyl

O i CH3 — CH2 — C — O — CH3

Identify the alcohol part of the ester. The alcohol part is the alkyl part bonded to oxygen atom by single bond.

Step 2

O i CH3 — CH2 — C — O — CH3

Identify the carboxylic acid part of the ester. The acid part is the alkyl part bonded to the carbon atom with a double bond with oxygen.

the carboxylic part (with – C=O) has altogether 3 C atoms, hence it is from propanoic acid

Step 3 Name of ester is methyl propanoate

Combine the two parts to name the ester. The alcohol part is named first.

2

’07

The molecular formula below represents an organic compound.

Solution

O i CH3 — C — O — CH2 — CH2 — CH3

O i CH3 — C — O — CH2 — CH2 — CH3

What is the name of the compound?



this is the acid part as it is bonded to C=O

from ethanoic acid (ester is ethanoate)

this is the alcohol part as it is bonded to O atom

from propanol (prefix is propyl)

The name of the compound is propyl ethanoate.

367

Carbon Compounds

2

Steps in naming an ester

ethanol

O i CH3CH2C O–CH2CH3 + H2O

Writing the structural formula of an ester

SPM

’07/P1

To predict the formula of an ester prepared from a named alcohol and a named carboxylic acid.

Example Write the structural formula of the ester produced from butanol and propanoic acid. General formula is

Step 1 Write the general formula of ester in the form R – COO – R′

O i R – C – O – R′

Alcohol is butanol, with four C atoms. Hence R′ is –CH2CH2CH2CH3.

Step 2 Write down the structural formula of the alcohol part to replace –R′(R bonded to O by single bond) 2

O i R – C – O – CH2CH2CH2CH3

Carboxylic acid is propanoic acid, with three C atoms. Hence RCO– is CH3CH2CO–.

Step 3 Write down the structural formula of the acid part to replace R– (R bonded to C=O)

O i CH3CH2C–

Structural formula of ester is: O i CH3 — CH2 — C — O — CH2 — CH2 — CH2 — CH3

Step 4 Replace –OR′ with the alcohol part and the R–CO– with the carboxylic part in R – COO – R′ for the full structure of the ester.

3

Preparation of Ethyl Ethanoate

’07

1 Small quantities of ethyl ethanoate can be prepared by heating a mixture of glacial ethanoic acid with pure ethanol in the presence of a small quantity of concentrated sulphuric acid in a boiling tube. 2 To prepare large quantity of esters, the alcohol and carboxylic acid need to be heated under reflux. 3 Heating under reflux is necessary as ethanol, C2H5OH is very volatile. If the mixture is not heated under reflux, the ethanol C2H5OH will vaporise and escape into the atmosphere before it can react with ethanoic acid, CH3COOH.

Write the chemical equation for the reaction between methanol and ethanoic acid. Solution Methanol, an alcohol reacts with ethanoic acid, a carboxylic acid to produce an ester and water. The ester will be methyl ethanoate. CH3OH + CH3COOH → CH3COOCH3 + H2O

Carbon Compounds

368

SPM

To prepare ethyl ethanoate

’04/P3

Apparatus Round bottomed flask, Liebig condenser, oil bath, porcelain chips Materials Pure ethanol, glacial ethanoic acid and concentrated sulphuric acid Procedure 1 About 30 cm3 of pure ethanol and 25 cm3 of glacial ethanoic acid are placed in a round bottom flask. Small pieces of porcelain chips are added to prevent bumping and to ensure smooth boiling. 2 About 5 cm3 of concentrated sulphuric acid is added to the reaction mixture. The mixture is shaken gently to ensure complete mixing. 3 The Liebig condenser is fitted vertically to the round bottom flask. The mixture is boiled under reflux for about 30 minutes. 4 After boiling, pure ethanol is obtained by distillation.

Figure 2.11 Preparation of ethyl ethanoate by reflux

Conclusion Ethyl ethanoate is produced when ethanoic acid and ethanol are heated in the presence of concentrated sulphuric acid as a catalyst. CH3COOH + C2H5OH → CH3COOC2H5 + H2O

Natural Sources of Esters

2.7 1 Name the following esters and identify the alcohols and carboxylic acids required to prepare these esters. (a) HCOOCH3 (b) CH3COOC3H7 (c) C3H7COOCH3

1 Most of the simple esters exist naturally in flowers and fruits. For example, pentyl ethanoate is found in bananas, octyl ethanoate in lime and methyl butanoate in apples. 2 These volatile esters are responsible for the fragrant smell of flowers and fruits. 3 Vegetable oils and animal fats are esters with large molecules. For example, coconut oil and palm oil. 4 Waxes such as beeswax, wax found on leaves and candle wax are solid esters.

3 Methanol reacts with butanoic acid under certain conditions to produce an ester. (a) Write a balanced equation for the reaction that occurs. (b) State the conditions for the reaction that took place. (c) State two physical properties of the ester produced.

Used as solvents for many organic compounds

Uses of esters To make synthetic polymers

Esters in oils are used to make soap

Used as medicine, for example, aspirin 369

Carbon Compounds

Activity 2.6

2 Name and draw the structural formulae of the organic compounds produced from the reactions between the following pairs of organic compounds. (a) Methanol + propanoic acid (b) Propan-1-ol + ethanoic acid (c) Propan-1-ol + methanoic acid

Uses of Esters in Daily Life To make perfumes, cosmetics and artificial food flavourings

2

Observation A colourless liquid with a fragrant odour is obtained.

2.8

stearic acid is C17H35COOH or CH3(CH2)16COOH

Oils and Fats

2 A fatty acid with a carbon-carbon double bond is SPM an unsaturated acid. ’08/P1 For example: Oleic acid is C17H33COOH or CH3(CH2)7CH = CH(CH2)7COOH

1 Oils and fats are naturally occurring esters and are found in animals and plants. 2 Fats (for example, butter) are found in animals. Oils are usually found in plants and fish. 3 Fats are solids at room temperature. In contrast oils are liquids at room temperature. Hence fats have higher melting points than oils. 4 When fats and oils are hydrolysed, glycerol SPM and long chain carboxylic acids are formed. ’08/P1 Hydrolysis means the decomposition (breaking up) of a chemical substance by water.

The importance of oils and fats for body processes Sources of energy: Fats are high energy food, they provide energy for our bodies.

2

Ester + water → carboxylic acid + glycerol 5 The carboxylic acids produced from fats are also known as fatty acids. Fatty acids usually contain 16 or 18 carbon atoms per molecule. 6 Glycerol is an alcohol that contains three hydroxyl (–OH) groups per molecule.

Importance of oils and fats in our bodies

H H H | | | H—C—C—C—H | | | OH OH OH



Source of nutrients: Fats are required to enable the human body to absorb vitamins A, D, E and K

Thermal insulation: The layer of fat beneath the skin regulates body temperature

propane-1,2,3-triol (glycerol)

7 There are two types of carboxylic acids, namely, saturated carboxylic acids and unsaturated carboxylic acids. (a) Saturated carboxylic acids do not contain double bonds. (b) Unsaturated carboxylic acids contain double bonds. 8 Animal fats usually contain a high percentage of saturated carboxylic acids whereas vegetable oils and fish oil contain a high percentage of unsaturated carboxylic acids upon hydrolysis. 9 Saturated fats are fats that contain saturated carboxylic acids. Unsaturated fats are fats that contain unsaturated carboxylic acids. 10 The presence of double bonds in unsaturated fats causes them to have lower melting points than saturated fats. Hence unsaturated fats (commonly known as oils) exist as liquids at room temperature.

Protection: The layer of fats around the vital organs acts as a protective cushion

Conversion of Unsaturated Fats to Saturated Fats 1 Vegetable oils can be converted to saturated fats by hydrogenation. 2 Hydrogenation is the chemical process in which hydrogen is added to the double bond between carbon atoms (C = C bond). 3 The hydrogenation process will change liquid vegetable oils to solid fats. 4 In the manufacture of margarine from vegetable oil, hydrogenation is carried out by passing hydrogen gas into palm oil at 200 °C and 4 atmospheres in the presence of nickel powder as catalyst. Ni

Palm oil + hydrogen ⎯⎯⎯⎯⎯→ 200 °C, 4 atm margarine (fat)

1 A fatty acid with the general formula of CnH2n+1COOH is a saturated acid. For example: Palmitic acid is C15H31COOH or CH3(CH2)14COOH and Carbon Compounds

5 The hardness of margarine formed depends on the degree of hydrogenation. Partial hydrogenation will produce soft margarine. 370

Effects of Eating Food High in Fats on Health

2.8 1 State one similarity and one difference between fats and oils (a) in terms of their molecular structures. (b) in terms of their physical properties.

1 Food high in fats is high in calories. Hence high comsumption can lead to obesity. 2 Saturated fats contain a high percentage of cholesterol. Excess intake of saturated fats increase the risk of (a) hypertension (high blood pressure) (b) cardiovascular disease (heart disease) (c) stroke

2 Stearic acid is a saturated fatty acid whereas oleic acid is an unsaturated fatty acid. (a) Explain what is meant by a saturated fatty acid and an unsaturated fatty acid. (b) Give a test to differentiate stearic acid from oleic acid.

1 There are two types of oil extracted from fresh oil palm fruits. (a) Palm oil from the flesh of the fruit (b) Palm kernel oil from the kernel or seed 2 The flowchart shows the steps involved for the extraction of palm oil in the palm oil mill. Sterilisation

Fruit branches are sterilised to kill fungus and bacteria

Stripping

Fruits are separated from the branches

Digestion

Fruits are heated to break down the oil-bearing cells

Pressing

Purification

2.9

Natural Polymers 1 Polymers can be classified into two broad categories: natural polymers and synthetic polymers. 2 Examples of naturally-occurring polymers are natural rubber, carbohydrates and proteins. 3 The monomer of natural rubber is isoprene, C5H8. Hence natural rubber is polyisoprene. Natural rubber is produced by the addition polymerisation of isoprene.

Oil is pressed out from fruits

H CH3 H H | | | | addition polymerisation n(H — C = C — C = C — H) ⎯⎯⎯⎯⎯⎯⎯⎯→ isoprene (monomer) H CH3 H H | | | | — ( C ⎯ C = C ⎯ C— )n | | H H

Mixture is filtered to separate the oil. Oil is then dried.

Advantage of palm oil as a vegetable oil Rich in Vitamins A and E

Lower the LDL or bad cholesterol and raise the HDL or good cholesterol in the body

poly(isopropene) – natural rubber

4 The monomers for carbohydrates such as sugar, starch and cellulose is glucose. When starch is heated with dilute acid, glucose is produced. This reaction is called hydrolysis.

Advantage of palm oil as a vegetable oil Withstand heat and resistant to oxidation, hence suitable to be used for deep frying

Natural Rubber



Highly competive in price. Cheaper than other types of vegetable oils

– (C6H10O5)n– + nH2O → nC6H12O6 starch glucose

5 The monomers of proteins are amino acids. Amino acids are joined together by peptide linkages. Hence proteins are polypeptides. 371

Carbon Compounds

2

3 (a) Name the process in which palm oil is converted to margarine. (b) State the conditions required for this conversion. (c) State one difference in physical property between palm oil and margarine.

Industrial Extraction of Palm Oil

(a) the hydrogen ions from the acid neutralise the negative charges on the membranes’ surfaces of the colloidal particles. (b) When the neutral rubber particles collide, the membranes will break, releasing the rubber polymers to form lumps. Hence the latex solidifies. 8 The coagulation of latex can occur without the addition of acid if the latex is exposed to air for a few days. This is because (a) the bacteria present in the latex produces organic acids. (b) the hydrogen ions from the acids produced neutralises the negative charges on the rubber particles. 9 Coagulation of latex can be prevented by the addition of aqueous ammonia because (a) the hydroxide ions from aqueous ammonia neutralise the acids produced by bacteria. (b) the negative charges at the membranes of rubber particles are maintained.

2

Coagulation of Latex 1 The milky fluid from tapped rubber trees is called latex. 2 The conversion of latex to the solid form is known as coagulation. 3 Latex is a colloidal solution containing an aqueous suspension of rubber particles. 4 Each rubber particle contains rubber polymers enclosed with a protein membrane with negative charge. 5 The negative charge on the membrane’s surface repel colloidal particles from one another, preventing the rubber polymers from combining together. Hence the latex remains in liquid form. 6 Coagulation of rubber can be speeded up by the addition of acids. 7 The addition of a weak acid on latex causes coagulation because

add a weak

coagulation

acid

of latex

• Latex is coagulated by adding a weak acid such as methanoic acid or ethanoic acid. When an acid is added to latex, the hydrogen ions from the acid will neutralise the negative charges on the surfaces of the colloidal particles. As a result, the particles become neutral and can come closer and collide with one another.

• The collisions between latex particles break open the protein membrane and releases the rubber polymers.

• The rubber polymers can coalesce (combine) and form lumps of rubber. This process of forming lumps of rubber is called coagulation of latex. • The lumps of rubber are white solids and are quite elastic.

Figure 2.12 The coagulation of latex

Activity 2.7

To investigate the coagulation of latex and methods to prevent coagulation Apparatus Beaker, glass rod and dropper

Procedure 1 About 50 cm3 of latex are placed in three beakers labelled A, B and C respectively. 2 Using a dropper, dilute ethanoic acid is added to the latex in beaker B. The mixture is stirred with a glass rod until the latex becomes acidic (blue litmus paper turns red).

Materials Fresh latex, dilute ethanoic acid, dilute aqueous ammonia and blue litmus paper.

Carbon Compounds

SPM

’08/P3

372

3 Aqueous ammonia is slowly added to beaker C until the latex becomes alkaline (red litmus paper turns blue). 4 The three beakers are left overnight. The changes that occurred are recorded.

Beaker

Results

Latex only

B

Latex + acid

Coagulation of latex occurs rapidly

C

Latex + aqueous ammonia

Latex does not coagulate (no visible changes)

Observation Coagulation occurs slowly

Properties of Natural Unvulcanised Rubber and Vulcanised Rubber

2

A

Chemicals added to latex

6 Vulcanisation is the process of hardening rubber by heating it with sulphur or sulphur compounds. 7 Vulcanisation of rubber is carried out by (a) heating natural rubber with sulphur at about 140 °C, using zinc oxide as catalyst, (b) immersing rubber in a solution of disulphur dichloride (S2Cl2) in methylbenzene. 8 In vulcanised rubber, the sulphur atoms form cross-links between long chains of rubber polymers.

SPM

’10/P1

1 Natural rubber has the following properties: (a) Quite elastic (b) Water repellent (c) Does not conduct electricity 2 Elasticity is the ability of an object to be stretched and then returned to its original shape when the stress is removed. 3 Natural rubber before treatment with sulphur is unvulcanised rubber. 4 Unvulcanised natural rubber has few practical uses because (a) it is not elastic enough (b) it becomes soft and sticky when heated (c) it becomes brittle and crack easily when oxidised by oxygen. 5 When unvulcanised rubber is stretched, the coiled rubber molecule is lengthened and straightened.

c s s c

rubber polymer

c s s c

c s s c

sulphur cross-links

TC 2/15

c c Figure 2.15 Vulcanised rubber has sulphur cross-links

9 Vulcanised rubber has many uses because of improved properties. Vulcanisation makes the rubber more elastic, stronger and more resistant to heat and oxidation. Improved properties of vulcanised rubber Stronger and harder

polymer chain of rubber under ordinary conditions

Observation

Conclusion 1 Acids such as ethanoic acid speed up the coagulation of latex. 2 Alkalis such as aqueous ammonia slow down the coagulation of latex. 3 Coagulation of latex can occur by itself slowly.

Figure 2.13 To investigate the coagulation of latex

Beaker

Chemicals added to latex

polymer chain of rubber when straightened

Figure 2.14 Unvulcanised rubber is not very elastic

373

Explanation The sulphur cross-links prevent the polymer chains from slipping past one another when stretched Carbon Compounds

Improved properties of vulcanised rubber

Improved properties of vulcanised rubber

Explanation

More elastic

The sulphur cross-links pull the chains back to their original arrangement when released

More resistant to heat

The presence of sulphur increases the melting point of rubber

More resistant to oxygen

Explanation Sulphur cross-links reduce the number of double bonds in the molecules of vulcanised rubber

2

To produce vulcanised rubber and to compare the properties of vulcanised rubber and unvulcanised natural rubber (A) Preparation of natural rubber and vulcanised rubber

Materials Vulcanised rubber and unvulcanised rubber.

Apparatus Glass plate, beaker, a pair of tongs and razor blade.

Procedure 1 A strip of vulcanised rubber is hung using a clip. 2 The original length of the vulcanised rubber strip is measured. 3 A 20 g weight is hung on the strip of vulcanised rubber. 4 The increase in length of the vulcanised rubber strip is measured. 5 The weight is removed and the final length of the vulcanised rubber strip is measured again. 6 Steps 1 to 5 are repeated using the unvulcanised rubber strip of the same length to replace the vulcanised rubber strip.

Materials Disulphur dichloride in methylbenzene and rubber latex. Procedure 1 A small quantity of latex is poured on a glass plate. 2 A glass rod is used to level the latex to produce a flat, thin layer of latex about 1 mm thick. 3 The glass plate is put aside for one day for the latex to coagulate. 4 The rubber produced is cut into two strips of rubber of equal size using a razor blade. 5 One of the strips of rubber is dipped into a solution of disulphur dichloride in methylbenzene for 2-3 minutes to produce a strip of vulcanised rubber. 6 The strip of vulcanised rubber is then removed from the solution and dried with filter paper. (B) To compare the elasticity of vulcanised rubber and unvulcanised rubber Apparatus Clip, retort stand with clamp, metre rule and weight.

FOLS UHWRUWVWDQG

UXOHU

WKUHDG JZHLJKW

SPM

’06/P3

Figure 2.16 To compare the elasticity of rubber

Results

Activity 2.8

UXEEHUVWULS

Original length (cm)

Stretched length with weight (cm)

Increase in length (cm)

Final length after weight is removed (cm)

Unvulcanised rubber

X

X1

X1 – X = Y1

Y3

Vulcanised rubber

X

X2

X2 – X = Y2

Y4

Type of rubber

Carbon Compounds

374

Discussion 1 The increase in length of vulcanised rubber (stretched length) is less than the increase in length of unvulcanised rubber (that is, Y2 < Y1). This shows that vulcanised rubber is harder and stronger than unvulcanised rubber. 2 The vulcanised rubber has a higher ability to return to its original length after the weight

is taken off (that is Y4 < Y3). This shows that the vulcanised rubber is more elastic than unvulcanised rubber. Conclusion Vulcanised rubber is harder, stronger and more elastic than unvulcanised rubber.

Uses of Natural Rubber 2 Unvulcanised natural rubber is used for making adhesive (glue) and as crepe rubber in insulating blankets. 3 Vulcanised rubber has many uses in industries and home.

Vehicle tyres

Surgical gloves and protective gloves

Shock absorbers Insulating layer for electric cables and equipment

Shoe soles 2

1 Unvulcanised natural rubber has limited uses as it is soft, has poor heat resistance and does not wear well. It becomes soft and sticky when heated. It also becomes hard and brittle due to oxidation.

Rubber hoses

Things made from vulcanised natural rubber

Conveyor belts Balloons

Rubber mattresses

Rubber bands

Research on Natural Rubber in Malaysia 1 In Malaysia, the research and development on rubber is conducted by the (a) Rubber Research Institute of Malaysia (RRIM) (b) Malaysian Rubber Producer Research Association (MRPRA) (c) Rubber Board of Malaysia

2 The scope of research and development activities include (a) finding new uses of rubber and rubber products (b) improving the quality of natural rubber (c) automating the tapping system so as to overcome labour shortage.

375

Carbon Compounds

2.9

2.11

1 Complete the following table. Natural polymer

Monomer

Uses of Various Organic Materials in Everyday Life

Protein Isoprene

1 Organic compounds were originally extracted from living things, products or remains of living things. The term organic actually means ‘derived from living organisms’. Living things are made of complex organic compounds that have structural, chemical or genetic functions. 2 However since the nineteenth century, organic compounds have been synthesised in the laboratory from inorganic materials. In 1828, the German chemist Friedrich Wohler was able to synthesise urea (an organic compound) from ammonium cyanate (an inorganic compound). 3 In present times, thousands of organic compounds are synthesised every year to fulfill our needs in the modern society. These synthetic organic compounds include synthetic polymers, vitamins, medicines, cosmetics, pesticides, paints, varnishes, glues, adhesives, synthetic fibers for clothing materials and others. 4 Organic synthesis is the preparation of specific and desired organic compounds from readily available resources. 5 Research and development towards natural organic compounds enable us to (a) simulate the structure of natural organic compounds and make useful synthetic organic compounds which are imitations of natural compounds. Examples: Dyes, food flavours, fragrances and medicines. (b) extract the active ingredients from traditional medicines. Cheaper and more effective medicines without side effects can then be made commercially. Generic medicines are made to lower the cost of medicines for consumers. (c) produce seeds of higher quality and more resistant towards pests, so as to increase the yield of food production. (d) find new uses for agriculture products. For example, oil palm waste is used to produce biomass fuel and make composite construction materials. 6 The economical development of our country depends heavily on products of organic

Carbohydrate Starch 2 State the chemicals used in the following processes in rubber industries and their effects on the properties of rubber. (a) Coagulation (b) Vulcanisation

2

3 What causes latex to coagulate under natural conditions? Suggest a method to prevent this phenomenon.

2.10

Order in Homologous Series

1 Organic compounds are grouped in families called the homologous series to make the study of organic chemistry more systematic and orderly. 2 Alkanes, alkenes, alcohols, carboxylic acids and esters are examples of homologous series. 3 The chemical properties of members of a homologous series are the same as they have the same functional group. 4 The physical properties of members of a homologous series show a regular pattern and change gradually as the number of carbon atoms increases.

Descending the homologous series as the number of carbon atoms increases

• Relative molecular mass increases • Melting point increases • Boiling point increases • Volatility decreases • Density increases • Solubility decreases

5 The order in the physical properties of members in a homologous series enables us to predict the properties of an unknown member in the series. Carbon Compounds

The Variety of Organic Materials in Nature

376

8 In 2012, universities and research institutes in collaboration with Agensi Innovasi Malaysia have successfully developed products such as disease-resistant chilli, lumber from oil palm, coconut body armour, biopesticide to control mosquito larvae and mosquito repellant gel.

1 Organic carbon compounds include hydrocarbons such as alkanes and alkenes, alcohols, carboxylic acids, esters, fats and oils and natural rubber. 2 Physical properties of alkanes and alkenes: (a) Low melting and boiling points (b) Insoluble in water, soluble in organic solvents (c) Non-conductors of electricity 3 Chemical properties of alkanes (general formula: CnH2n+2): (a) Combustion in excess oxygen to form CO2 and H2O (b) Substitution reactions with halogen under ultraviolet light 4 Chemical properties of alkenes (general formula: CnH2n): (a) Combustion in excess oxygen to form CO2 and H2O (b) Addition reaction (i) with H2 (hydrogenation) to form alkanes (ii) with H2O (hydrolysis) to form alcohols (iii) with KMnO4 to form diols (iv) with Cl2 or Br2 to form dichloroalkanes or dibromoalkanes (c) Polymerisation to form polymers 5 Chemical properties of alcohols (general formula: CnH2n+1OH): (a) Combustion in excess oxygen to form CO2 and H2O (b) Oxidation by acidified KMnO4 or K2Cr2O7 to form carboxylic acids (c) Dehydration by heated porcelain to form alkenes (d) Esterification with carboxylic acids (with conc. H2SO4 as a catalyst) to form esters 6 Preparation of ethanol, C2H5OH: (a) Fermentation of glucose by yeast

(b) Hydrolysis of ethene (phosphoric acid as a catalyst) 7 Chemical properties of carboxylic acids (general formula: CnH2n+1COOH): (a) Reacts with reactive metals to form salts and H2 gas (b) Reacts with metal carbonates to form salts and CO2 gas (c) Reacts with alkalis to form salts and H2O (d) Esterification with alcohols to form esters and H2O 8 Physical properties of esters (general formula: RCOOR′): (a) Have sweet/ fragrant/ fruity smells (b) Insoluble in water 9 Fats and oils are esters. (a) Fats are solids (with lower melting points) and are saturated (without C = C bonds). (b) Oils are liquids (with higher melting points) and are unsaturated (with C = C bonds). (c) Oils can be converted to margarine by hydrogenation. 10 Hydrolysis of fats (or oils) will produced fatty acids (long chained carboxylic acids) and glycerol. 11 Coagulation is the conversion of liquid latex to rubber solid. It (a) can be speeded by the addition of acids. (b) can occur slowly when the bacteria present produce acids. (c) can be prevented by the addition of aqueous ammonia. 12 Vulcanisation of rubber is the conversion of natural rubber to vulcanised rubber by forming sulphur cross-links between rubber polymers. Vulcanised rubber is stronger, harder, more elastic and more resistant to heat and oxidation.

377

Carbon Compounds

2

compounds such as petroleum, palm oil and natural rubber. 7 Presently, research and development of natural resources in our country is encouraged to produce new and useful organic compounds. New synthetic medicines are developed to combat disease and new polymers are synthesised to replace the use of metals and even replace organs in our bodies.

2 Multiple-choice Questions 2.1

Carbon Compounds

2

1 Which of the following is an organic compound? A Calcium carbonate B Glucose C Carbon monoxide D Copper(II) oxide 2 Which of the following products are formed from the complete combustion of all organic compounds? A Water and carbon dioxide B Carbon dioxide and carbon monoxide C Water, carbon dioxide and nitrogen dioxide D Carbon and water

2.2

Alkanes

3 Which of the following is a saturated hydrocarbon? ’06 A Alkane B Alkene C Alcohol D Carboxylic acid 4 Which of the following substances can undergo substitution reaction ’06 with chlorine in sunlight? A Ethane B Ethene C Ethanol D Ethanoic acid 5 What are the products formed when ethane is burnt with ’07 excess oxygen? A Carbon and hydrogen B Carbon dioxide and water C Carbon monoxide and water D Carbon monoxide and hydrogen Carbon Compounds

6 Which of the following produces the most soot when burned in air? A Methane B Ethane C Butane D Hexane

2.3

Alkenes

7 Which of the following substances can be used to ’04 differentiate propene from propane? A Limewater B Bromine water C Sodium hydroxide solution D Potassium dichromate(VI) solution 8 Which of the following is the structural formula of an ’05 unsaturated hydrocarbon? A H —

H H H H | | | | C—C=C—C—H | | H H

B H —

H H H H | | | | C—C=C—C—H | | H OH

C H —

H O | i C—C—O—H | H

D H —

H H H H | | | | C—C—C—C—H | | | | H H H H

378

9 Propene can be transformed to propane by the process of ’07 A hydration B oxidation C dehydration D hydrogenation 10 A hydrocarbon compound is burnt completely in air to form ’05 7.2 cm3 of carbon dioxide gas and 7.2 cm3 of water vapour. What is the molecular formula of the hydrocarbon compound? [Given that 1 mol of gas occupies 24 dm3 at room temperature] A C2H6 B C3H6 C C3H8 D C6H6 11 What is the product formed when propene is shaken with ’10 chlorine water? A 1, 1- dichloropropane B 2, 2 -dichloropropane C 1, 2- dichloropropane D 1, 3- dichloropropane 12 The diagram below shows the structural formula of a ’11 compound. H | H—C—H H H H | | | H—C=C—C—C—H | | H H What is the name of this compound? A Pentene B Methylbutene C 2-methylbut-1-ene D 2-methylbut-2-ene

H H H H | | | | A — C — C — C — C — | | | | H H H H H CH3 H CH3 | | | | B — C — C — C — C — | | | | H H H H H CH3 H H | | | | C — C — C — C — C — | | | | H CH3 H H

A Pentane B Methylbutane C 2-methylbutane D 3-methylbutane 16 Which of the following pairs of compounds are isomers? A Butane and butene B Butane and 2-methylbutane C But-1-ene and 2-methylpropene D 2,2-dimethyl propane and 2-methylpropane

2.5

Alcohols

17 The diagram shows the set-up of apparatus for the reaction of ’06 yeast with glucose solution.

CH3 H H H | | | | D — C — C — C — C — | | | | H H H H

2.4

Isomerism

14 Which of the following is true of all the isomers of a hydrocarbon? A They have the same structural formula. B They have the same functional group. C They have the same chemical properties. D They produce the same number of moles of carbon dioxide and water on complete combustion. 15 The structural formula of a compound is given below. ’05

H | H—C—H H H H | | | H—C—C—C—C—H | | | | H H H H Which of the following is the IUPAC name for the compound?

What is the name of the reaction that has taken place in the conical flask after a few days? A Oxidation B Hydration C Hydrolysis D Fermentation 18 The following chemical equation shows the conversion of ethanol ’04 to ethanoic acid. C2H5OH + 2[O] → CH3COOH + H2O What is the name of the process shown by the above equation? A Dehydration B Reduction C Oxidation D Fermentation 19 The diagram shows the structural formula of compound X.

379

H | H—C—H H H | | H—C—C=C—H | H What is the name of the compound formed when compound X reacts with potassium permanganate? A Butanol B 2-methylpropanol C Butan-1,2-diol D 2-methylpropan-1,2-diol 20 The diagram below shows the structural formula of ’04 pent-1-ene. H H H H H | | | | | H—C=C—C—C—C—H | | | H H H Which of the following are the possible alcohols that can produce pent-1-ene on dehydration? I Pentan-1-ol II Pentan-2-ol III Pentan-3-ol IV 2-methylbutan-1-ol A I and II only B II and IV only C I and IV only D I, II and III only 21 The diagram below shows the set-up of the apparatus for a ’05 reaction. glass wool soaked with ethanol

porcelain chips

compound X

heat heat

water

Which of the following is compound X? A Ethane TC 2/18 B Ethene C Ethanoic acid D Carbon dioxide Carbon Compounds

2

13 Polypropene is a polymer that is formed from the combination ’07 of propene molecules. Which of the following represents part of the structure of polypropene?

22

CH3 | H3C – C – OH | H

2

Which of the following may be true of the compound represented in the figure above? A Decolourise the brown colour of bromine water. B Decolourise the purple colour of cold aqueous potassium permanganate. C Changes the orange colour of potassium dichromate to green when heated. D Produces a sweet-smelling liquid when heated with ethanol and concentrated sulphuric acid.

2.6

fermentation

X oxidation

Y Which of the following pairs of compounds may be compound X and compound Y? X

23 Compound X has the following properties: • Reacts with sodium carbonate to produce a gas that turns limewater milky. • Reacts with ethanol to produce a sweet-smelling compound. Which of the following compounds may be compound X? A Carbonic acid B Ethanoic acid C Propanol D Ethyl ethanoate 24 A liquid produced effervescence when reacted with zinc metal. ’05 Which of the following may be the molecular formula of the liquid? A HCOOH B HCOOCH3 C CH3OH D CH3COONa 25 The diagram shows the conversion of glucose to compound X and subsequently to compound Y. Carbon Compounds

Y

A

Ethanol

Ethanoic acid

B

Ethanol

Ethene

C

Propanol

Propanoic acid

D

Yeast

Carbon dioxide

2.7 Carboxylic Acids

Which of the following is the structural formula for compound Z? O i A C2H5 — C — O — CH3

Glucose

Esters

26 Scented flowers contain naturally occurring esters. Which of the ’04 following is a property of an ester? A Soluble in water B Low boiling point C Higher density than water D Change blue litmus to red 27 The diagram below represents the structural formula of a ’04 carbon compound. O i CH3 — CH2 — C — O — CH2 — CH2 — CH2 — CH3 The compound is produced by the reaction between A ethanol and butanoic acid B butanol and ethanoic acid C butanol and propanoic acid D propanol and butanoic acid 28 The diagram below shows the process of producing compound ’05 Z. C3H6

O i B CH3 — C — O — C3H7 O i C C2H5 — C — O — C2H5 O i D C2H5 — C — O — C3H7

29 Which of the following equations can produce a product with a ’06 sweet fruity scent? I CH3COOH + NaOH → CH3COONa + H2O II CH3OH + C2H5COOH → C2H5COOCH3 + H2O III C2H5OH + 2[O] → CH3COOH + H2O IV C5H11OH + CH3COOH → CH3COOC5H11 + H2O A I and III only B II and IV only C III and IV only D I, II and IV only

30 The diagram shows the conversion of ethanol to compound X and subsequently to compound Y. acidified KMnO4 C2H5OH ⎯⎯⎯⎯⎯⎯→ X



CH3OH, concentrated

⎯⎯⎯⎯⎯⎯⎯⎯⎯→ Y H2SO4, reflux

+ steam

Ethanol + C2H5OH

Y

oxidation

Compound Z

380

X

Which of the following may be compound Y? A Ethanoic acid B Ethyl ethanoate C Ethyl methanoate D Methyl ethanoate

A I only B III only

Oils and Fats

31 Which of the following belong to the homologous series of esters? I Palm oil II Margarine III Sodium butanoate IV Glycerol A I and II only B I and III only C III and IV only D I, II and III only

C I, II and III only D I, II, III and IV

34 The diagram shows the change in structure of natural rubber after process P. c c

c c

c c

c c

c c

c c

c s s c

c s s c

c c

rubber X

32 Which of the following molecular formulae of fatty acids is formed from the hydrolysis of a nonsaturated fat? A C15H31COOH B C17H35COOH C C17H33COOH D C23H47COOH

c s s c c c

rubber Y

Which of the following statements is true about the change? A Rubber X is more elastic than rubber Y. TCY.2/19 B Rubber X is stronger than rubber C Rubber X is more resistant to heat than rubber Y. D Rubber X is more easily oxidised than rubber Y. 35 A rubber tapper faces a problem of transporting latex to a glove-making factory in liquid form. To solve the problem he has to add a substance ’06 into the latex to prevent the coagulation of the latex. Choose the correct substance and the explanation to solve the problem. Substance

2.9

process P

Natural Rubber

33 Which of the following chemicals can cause latex to coagulate? I Methanoic acid II Sulphuric acid III Ethanoic acid IV Nitric acid

Explanation

A

Water

To make the latex more dilute

B

Ethanoic acid

Contains H+ ions that neutralise the negative charge on the membrane of the rubber particle

C

Ammonia solution

Contains OH– ions that neutralise the H+ ions from the acids produced by bacteria

D

Sodium chloride solution

As a preservative to maintain the original state of the latex

Structured Questions (b) Complete Table 1 by filling in the molecular formula of propane and the name of C4H10.

1 Table 1 shows the names, molecular formulae, melting points and boiling points of a few straight chain members of a homologous series. Name Ethane

Molecular formula

Melting point (°C)

Boiling point (°C)

C2H6

–183

–89

–188

–42

C4H10

–138

–0.5

C5H12

–130

36

Propane

Pentane



[2 marks]

(c) Draw the structural formula of the first member of this homologous series. [1 mark] (d) Predict the physical states of (i) propane (ii) pentane

[1 mark] [1 mark]

(e) Give the name and the molecular formula of the member of the same homologous series after pentane. [2 marks] (f) Write a balanced equation for the complete combustion of ethane. [1 mark]

Table 1

(a) Give the name and the general formula of this homologous series. [2 marks]

381

Carbon Compounds

2

2.8

(b) State the function of concentrated sulphuric acid [1 mark] in this reaction. (c) (i) Name the reaction for the preparation of [1 mark] ethyl ethanoate. (ii) Write the chemical reaction for the reaction [1 mark] in (i).

2 Propene is an important hydrocarbon in the petrochemical industries. Diagram 1 shows the conversion of propene into other organic compounds. Polypropene process II

(d) The experiment is repeated by replacing ethanol with methanol. [1 mark] (i) Name the ester formed. (ii) Draw the structural formula of the ester [1 mark] formed. (iii) State one physical property of the ester.

process process Propene Propan-1-ol I III

Propane

process IV

Compound X

[1 mark]



Diagram 1

(e) The flowchart shows the conversion of ethene to ethanol and then to ethanoic acid.

Based on Diagram 1, answer the following questions.

2

(a) State the homologous series of propene.

[1 mark]

(b) Name process I.

[1 mark]

Ethene

(c) Under certain conditions, propene reacts to form polypropene. Write an equation for the formation of polypropene in process II. [1 mark] (d) (i) Explain briefly how process III is carried out in industries. [2 marks] (ii) Draw the structural formula of propan-1-ol.



[1 mark]

(e) Acidified potassium manganate(VII) is added to propene in process IV. (i) Predict the observation that will take place.

process I

Ethanol

process II

Ethanoic acid

Based on the flowchart, write the chemical equation for [1 mark] (i) process I (ii) process II [1 mark] (iii) State a suitable chemical that can be used [1 mark] to carry out process II.

4 Diagram 3 shows conversions I, II and III starting with glucose.

[1 mark]



(ii) Write a balanced equation for the reaction that has occurred. [1 mark]

Glucose

(f) Both propene and propane are combustible in air. Compare and explain the difference in the quantity of soot produced by the two compounds during combustion. [2 marks]

II

Liquid B

reagent

concentrated H2SO4 asid

Liquid C Diagram 3

(a) (i) Name conversion I. [1 mark] (ii) Draw the structural formula of liquid A.

water out

[1 mark]

(b) Liquid B has a vinegary smell. (i) Name the type of reaction that takes place [1 mark] in conversion II. (ii) Write a chemical equation for conversion II. [1 mark] TC 2/20

water in

(c) Give one chemical test that can be used to [2 marks] distinguish liquid B from liquid A.

ethanol, ethanoic acid and concentrated sulphuric acid



(d) (i) Name the homologous series of which [1 mark] liquid C is a member. (ii) Name liquid C. [1 mark] (iii) State one use of liquid C. [1 mark]

heat

Diagram 2

(e) Suggest another method to produce liquid A other than from glucose in conversion I.

(a) Why is ethanol and ethanoic acid heated under reflux? [1 mark] Carbon Compounds

yeast

Liquid A

III

3 Diagram 2 shows the set-up of apparatus for the preparation of ethyl ethanoate by heating ethanol and ethanoic acid under reflux.

xxxxxxxxxxxxxx

I



382

[1 mark]

(c) Name compound X.

5 Diagram 4 shows the conversion of latex to compound X.

Latex

process I

Natural process II rubber

c s s c

c s s c

c s s c

(d) State two differences in physical properties between natural rubber and compound X. TC 2/21

[2 marks]



(e) How is process II carried out in the industry? [1 mark] (f) (i) Name a chemical that can be used to retain [1 mark] latex in the liquid form. (ii) Explain the function of the chemical named [1 mark] in (i).

c c

compound X Diagram 4

(a) Name process I and process II.

[1 mark]

[2 marks]

(b) Name a chemical that can be used to carry out process I. [1 mark]

(g) Give one use of compound X.

[1 mark]

Essay Questions 2 (a) Diagram 1 shows the formation of carboxylic acids from alcohols.

(b) The information below is referring to carbon compound X.

Alcohols Diagram 1

• Carbon 40.0% • Hydrogen 6.7% • Oxygen 53.3% • Relative molecular mass = 60 • Relative atomic mass of H = 1, C = 12 and O = 16



Carboxylic acids

2

1 (a) Draw the structural formulae of two isomers of but-1-ene and give their IUPAC names. [4 marks]

Using suitable reagents and with the help of a labelled diagram, describe how you can prepare a named carboxylic acid in the laboratory. Include the observation and a test to show that the product formed is an acid. Write a chemical equation for the reaction involved. [10 marks]

Based on the information of the carbon compound X, (i) determine the molecular formula of X. (ii) draw the structural formula of X. (iii) name the carbon compound X. (iv) write the general formula for its homologous series. [8 marks]

(b) Many artificial flavours used in the food industry are esters. Various types of esters can be formed from the esterification between an alcohol and a carboxylic acid. Name one possible ester that can be formed and describe how you can prepare the named ester in the laboratory. Name the alcohol and carboxylic acid that is used and the chemical equations involved.

(c) Margarine can be made from palm oil. Compare and contrast margarine and palm oil in terms of their structures and physical properties. Briefly describe how palm oil can be converted to margarine. [8 marks]



[10 marks]

Experiment 1

Your (i) (ii) (iii) (iv) (v) (vi)

Hexane is a saturated hydrocarbon whereas hexene is an unsaturated hydrocarbon. Both are colourless liquids. However they undergo different reactions toward addition reaction. You are required to plan an experiment to differentiate the two compounds.

383

planning should include the following: Statement of the problem All the variables Hypothesis List of materials and apparatus Procedure Tabulation of data [17 marks]

Carbon Compounds

FORM 5 THEME: Interaction between Chemicals

CHAPTER

3

Oxidation and Reduction SPM Topical Analysis 2008

Year 1

Paper

2009 3

2

Section

A

B

C

Number of questions

1 — 6

–

1

5

–

1

2010

2

4

3

A

B

C

–

–

1

1

–

4

2011

2

3

A

B

C

–

–

–

–

1

3

2

3

A

B

C

1

–

–

–

ONCEPT MAP Oxidising agent A substance that accepts (gains) electrons and is itself reduced in the process

Reducing agent A substance that donates (loses) electrons and is itself oxidised in the process

Oxidation • Gain of oxygen/increase in oxidation number • Loss of hydrogen/loss of electrons

Reduction • Loss of oxygen/decrease in oxidation number • Gain of hydrogen/gain of electrons

Rusting of iron Fe → Fe2+ + 2e– → Fe2O3.xH2O OXIDATIONREDUCTION (REDOX) REACTIONS

Oxidation of Fe2+ ions: Fe2+ → Fe3+ + e– Reduction of Fe3+ ions: Fe3+ + e– → Fe2+ Combustion of metals

Reactivity series of metals K > Na > Ca > Mg > Al > Zn > Fe > Sn > Pb > Cu > Ag > Au

Prevention of rusting • Galvanising • Sacrificial metal • Alloying

Displacement of halogens from the halides e.g. Cl2 + 2KBr → 2KCl + Br2

Displacement of metals from their salts e.g. Zn + CuSO4 → ZnSO4 + Cu

Heating metal oxides with carbon and hydrogen

Electrochemical series K > Na > Ca > Mg > Al > Zn > Fe > Sn > Pb > H > Cu > Ag

Position of C and H in the reactivity series of metals K > Na > Ca > Mg > Al > C > Zn > H > Fe > Sn > Pb > Cu > Ag

Electrolytic cell • Electrons flow from anode (+ve) to cathode (–ve) in the external circuit

• Extraction of iron 2Fe2O3 + 3C → 4Fe + 3CO2 • Extraction of tin SnO2 + C → Sn + CO2

Chemical cell • Electrons flow from anode (–ve) to cathode (+ve) in the external circuit

Redox Reactions

Oxidation and Reduction Reactions

Oxidation in Terms of Loss of Hydrogen 1 Oxidation can also be defined as the loss of hydrogen from a substance. If a substance loses hydrogen during a reaction, it is said to be oxidised. 2 When hydrogen sulphide gas is mixed with chlorine gas at room temperature, a yellow precipitate of sulphur is formed and hydrogen chloride gas is released.

SPM

’08/P1, ’09/P1

1 Oxidation can be defined as (a) acceptance (gain) of oxygen, (b) donation (loss) of hydrogen, (c) loss of electrons and, (d) increase in the oxidation number of the element. 2 In contrast, reduction can be defined as (a) loss of oxygen, (b) gain of hydrogen, (c) gain of electrons and, (d) decrease in the oxidation number of the element.

H2S(g) + Cl2(g) → 2HCl(g) + S(s) loss of hydrogen (oxidation)

In this reaction, hydrogen sulphide loses hydrogen and is oxidised to sulphur. 3 When ammonia gas is passed over hot copper(II) oxide, the following reaction occurs.

A newly cut apple turns yellow on exposure to air. This is due to the oxidation of apples by oxygen. The reaction is catalysed by the enzymes present in apples.

2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(l) oxidation

Oxidation in Terms of Gain of Oxygen

Ammonia undergoes oxidation because it loses hydrogen. In other words, ammonia is oxidised to nitrogen.

1 Oxidation is a chemical reaction in which oxygen is added to a substance. If a substance (element or compound) gains oxygen during a reaction, it is said to be oxidised. 2 When calcium burns in oxygen, the following reaction occurs:

2Ca(s) + O2(g) → 2CaO(s)

Reduction in Terms of Loss of Oxygen 1 A reduction reaction is the reverse process of an oxidation reaction. Reduction is defined as the loss of oxygen from a substance. If a substance loses oxygen during a reaction, it is said to be reduced. 2 When a mixture of zinc powder and copper(II) oxide is heated, the following reaction occurs.

addition of oxygen (oxidation)

(a) This process is known as oxidation. (b) Calcium is oxidised to calcium oxide because it gains oxygen in this reaction. 3 Methane burns in air as represented by the equation: oxidation



CH4(g) + 2O2(g) ⎯⎯⎯→ CO2(g) + 2H2O(g) oxidation



In this reaction, (a) the carbon atom in methane is oxidised to carbon dioxide, (b) the hydrogen atoms in methane is oxidised to water. Therefore, combustion is an oxidation reaction.

loss of oxygen (reduction)

Zn(s) + CuO(s) → ZnO(s) + Cu(s)

3 In this reaction, copper(II) oxide has lost its oxygen. It is said to be reduced to metallic copper.

385

Oxidation and Reduction

3

3.1

oxidation

Reduction in Terms of Gain of Hydrogen 1 Reduction can also be defined as the addition of hydrogen to a substance. If a substance gains hydrogen during a reaction, it is said to be reduced. 2 When a mixture of hydrogen and chlorine is exposed to sunlight, a vigorous reaction occurs and white fumes of hydrogen chloride are produced.

Mg(s) + H2O(g) → MgO(s) + H2(g)

reduction

4 In this reaction, magnesium has gained oxygen and is oxidised. In contrast, water has lost its oxygen and is reduced.



addition of hydrogen (reduction) Respiration is a redox process. When respiration occurs, the food is oxidised and oxygen molecules accept electrons and are reduced to water. Photosynthesis is also a redox reaction.

H2(g) + Cl2(g) → 2HCl(g) 3 In this reaction, chlorine has gained hydrogen. This means that chlorine has been reduced. 3

6CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g) sugar Electrons are removed from water molecules and are used to reduce carbon dioxide to sugar.

Oxidising and Reducing Agents

1 An oxidising agent is a substance that brings about (causes) oxidation in another substance. In bringing about oxidation, the oxidising agent is itself reduced. 2 A reducing agent is a substance that brings about reduction in another substance and is itself oxidised. 3 The following are examples of oxidising and reducing agents.

Antoine Lavoisier (1743–1794) Antoine Lavoisier is known as the Father of Modern Chemistry. He was the first chemist who explained the oxidation and reduction reactions that occur during combustion.

Redox Reactions 1 Oxidation and reduction always take place together. A redox reaction is defined as a reaction in which both oxidation and reduction take place simultaneously. 2 In a redox reaction, when one substance in a reaction is oxidised, the other substance is reduced. 3 When steam is passed over heated magnesium, magnesium oxide and hydrogen are produced. Oxidation and Reduction

SPM

’08/P1, ’09/P1

Oxidising agents

Reducing agents

• Chlorine and bromine • Acidified potassium manganate(VII) • Acidified potassium dichromate(VI) • Concentrated nitric acid

• Metals such as sodium, magnesium, zinc and aluminium • Sulphur dioxide gas and hydrogen sulphide gas • Sodium sulphite and sodium thiosulphate

4 In a redox reaction involving A and B: if A is an oxidising agent, then B is the reducing agent and vice versa. 386

Reaction The reaction between copper(II) oxide and carbon: oxidised (gain oxygen)



2CuO(s) + C(s) → 2Cu(s) + CO2(g)

Oxidising agent

Reducing agent

Copper(II) oxide • Copper(II) oxide oxidises carbon to carbon dioxide. • It is reduced to copper.

Carbon • Carbon reduces copper(II) oxide to copper. • It is oxidised to carbon dioxide.

Chlorine • Chlorine oxidises hydrogen sulphide to sulphur. • It is reduced to hydrogen chloride.

Hydrogen sulphide • Hydrogen sulphide reduces chlorine to hydrogen chloride. • It is oxidised to sulphur.

reduced (loss of oxygen)

The reaction between chlorine and hydrogen sulphide: oxidised (loss of hydrogen)

Cl2(g) + H2S(g) → 2HCl(g) + S(s)

3

reduced (gain hydrogen)

1 The equation for the reaction between iron(III) oxide and carbon monoxide is shown below.

In many cases, oxidation or reduction reactions are accompanied by colour changes. Br2(brown) → Br–(colourless) … reduction MnO4–(purple) → Mn2+ (colourless) … reduction Cr2O72–(orange) → Cr3+ (green) … reduction Fe2+ (green) → Fe3+ (brown/yellow) … oxidation 2I– (colourless) → I2(brown) … oxidation

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Identify the oxidising and reducing agents in this reaction. Solution Fe2O3 is an oxidising agent because it oxidises CO to CO2 and is itself reduced to Fe. oxidation



Methylhydrazine, CH3NHNH2 is a powerful reducing agent.

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

4CH3NHNH2 + 5N2O4 → 4CO2 + 9N2 + 12H2O It is used as rocket fuel for the Apollo 11 project which landed the first man on the moon on 21 July 1969. The compound is toxic and carcinogenic (cancer causing). It is also found in trace amounts in raw common mushrooms.

reduction

CO is a reducing agent because it reduces Fe2O3 to Fe and is itself oxidised to CO2.

387

Oxidation and Reduction

3.1 To identify oxidation and reduction processes in the reaction between metal oxides and carbon Problem statement In the reaction between metal oxide and carbon, which reagent undergoes oxidation and which reagent undergoes reduction?

Materials Powdered carbon, powdered copper(II) oxide, iron(III) oxide and lead(II) oxide.

3

Hypothesis (a) Carbon undergoes oxidation to form carbon dioxide gas. (b) Copper(II) oxide, iron(III) oxide and lead(II) oxide undergo reduction to form copper, iron and lead respectively.

Figure 3.1 Heating copper(II) oxide with carbon

Procedure 1 A spatula of copper(II) oxide is placed in a crucible. 2 Another spatula of powdered carbon is added to the copper(II) oxide. 3 The two substances are mixed thoroughly and the mixture is then heated strongly. 4 The observations are recorded in the table given below. 5 Steps 1 to 4 are repeated using iron(III) oxide and lead(II) oxide in place of copper(II) oxide.

Variables (a) Manipulated variable : Type of metal oxide (b) Responding variable : Reaction products (c) Constant variables : Carbon and the conditions of reaction Apparatus Crucible, clay-pipe triangle, tripod stand, spatula and Bunsen burner. Results Metal oxide

Colour of metal oxide before heating

Observation

(a) Copper(II) oxide

Black

Brown spots (copper) are formed in the black mixture.

(b) Iron(III) oxide

Brown

Grey solid (iron) is formed.

(c) Lead(II) oxide

Yellow

Greyish-black solid (lead) is formed.

Experiment 3.1

Discussion In all the reactions above, metal oxides have lost oxygen to form the metals. This shows that carbon has reduced metal oxides to the corresponding metals. Conclusion Carbon undergoes oxidation and the metal oxides undergo reduction. The hypothesis is accepted. The chemical equations for the redox reactions are (a) 2CuO(s) + C(s) → 2Cu(s) + CO2(g) (b) 2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g) (c) 2PbO(s) + C(s) → 2Pb(s) + CO2(g)

Oxidation and Reduction

388

Table 3.1 Oxidation numbers of elements in the free state

SPM

’08/P1

he oxidation number of an element is an T arbitrary charge assigned to the element according to a set of rules. Oxidation number is also known as the oxidation state. Oxidation numbers of elements in ionic compounds 1 An ionic compound can contain monatomic ions (for example, Na+ and Cl– ions) or polyatomic ions (for example, NH4+ or SO42– ions). 2 For a monatomic ion in an ionic compound, the oxidation number is the charge on the ion. Magnesium oxide, MgO is an ionic compound. In magnesium oxide, magnesium exists as magnesium ions, Mg2+ and oxygen exists as oxide ions, O2–. Thus, magnesium is said to have the oxidation number of +2 and oxygen has the oxidation number of –2.

Element

Formula

Oxidation number

Hydrogen

H2

0

Oxygen

O2

0

Chlorine

Cl2

0

Sulphur

S

0

Iron

Fe

0

Copper

Cu

0

2 For monatomic ions, the oxidation number equals to the charge on the ion (Table 3.2). Table 3.2 Oxidation numbers of monatomic ions

Simple ion

Formula of ion Oxidation number

Hydrogen ion

H+

Sodium ion

Na+

+1

Magnesium ion

Mg

+2

Aluminium ion

Al3+

+3

Oxidation numbers of elements in covalent compounds

Chloride ion

Cl–

–1

Oxide ion

O2–

–2

Carbon dioxide, CO2 is a covalent compound. However, when determining the oxidation numbers of carbon and oxygen, you will have to consider that this molecule exists as ions. • Each oxygen atom is considered as an oxide ion (O2–) and carries a charge of –2. So two oxide ions carry a total charge of –4. • As a result, each carbon ion carries a charge of +4 so that CO2 exists as a neutral molecule.

Nitride ion

N3–

–3

+1

2+

3 The sum of the oxidation states of all the atoms present in the formula of a compound is zero. The compound can be an ionic compound or a covalent compound. For example, (a) in calcium carbonate, CaCO3 CaCO3

CO2

(+2) + (+4) + 3(–2) = 0 (b) in aluminium nitrate, Al(NO3)3

(+4) + 2  (–2) = 0

Al(NO3)3

Therefore, the oxidation number of carbon is +4 and the oxidation number of oxygen is –2.

(+3) + 3(+5) + 9(–2) = 0 4 For a polyatomic ion (that is, an ion that contains a few atoms), the sum of the oxidation numbers of all the atoms equals the charge on the ion. For example, (a) in a sulphate ion, SO42–

Rules for Assigning Oxidation Numbers To work out the oxidation number, the following rules must be applied. 1 An atom or a molecule of an element in the SPM free state (that is, not combined with other ’06/P1 ’07/P1 elements) has an oxidation number of zero (Table 3.1).

SO42– (+6) + 4(–2) = –2 389

sum of the oxidation numbers is the same as the charge on the polyatomic ion

Oxidation and Reduction

3

Oxidation Number

3

8 The oxidation number of hydrogen in all its compounds is +1 except in metal hydrides. A metal hydride is a compound consisting of hydrogen and a metal only. The oxidation number of hydrogen in metal hydrides is –1. (a) Non-metallic hydride

(b) in an ammonium ion, NH4+ NH4+ sum of the oxidation numbers is the same as the charge on the polyatomic ion (–3) + 4(+1) = +1 (c) in a nitrate ion, NO3– NO3– sum of the oxidation numbers is the same as the charge on the polyatomic ion (+5) + 3(–2) = –1 (d) in a dichromate(VI) ion, Cr2O72– Cr2O72– sum of the oxidation numbers is the same as the charge on 2(+6) + 7(–2) = –2 the polyatomic ion 5 In a given compound, the more electronegative atom is given a negative oxidation number and the less electronegative (or more electropositive) atom has a positive oxidation number.

HCl (+1)



NaH (+1) (–1)

(+1) (+5) 3(–2)

Oxidation and Reduction

2(–1) (+3)

3(–1)

(–2) (+2)

(–2)

H2O2 BaO2

6 The oxidation number of fluorine remains unchanged in all its compounds and is always –1. This is because fluorine is the most electronegative element. F2O BrF3 Na3AlF6 2(–1) (+2) (+3) 3(–1) 3(+1) (+3) 6(–1) Notice that (a) the oxidation number of oxygen in F2O is +2 and not –2 (b) the oxidation number of bromine in BrF3 is +3 and not –1. 7 Chlorine, bromine and iodine usually have the oxidation number of –1 except when combined with a more electronegative element. For example, HClO Cl2O (+1) (+1) (–2) 2(+1) (–2)

(+1) (–1)

(+2)

AlH3

(b) Peroxides

electronegativity increases



CaH2

H2O BaO 2(+1)

KIO3

(–2)

9 The oxidation number of oxygen in all its compounds is –2 except in fluorine compounds (see point 6) and in peroxides. The oxidation number of oxygen in peroxides is –1. (a) Oxides



KI

2(+1)

(b) Metal hydrides

I, Br, Cl, N, O, F



(–1)

H2S



2(+1) 2(–1) (+2)

2(–1)

10 Metals usually have positive oxidation numbers. For example, the oxidation number of a Group 1 element in a compound is always +1. The oxidation number of a Group 2 element in a compound is always +2. 11 Some metals show different oxidation numbers in their compounds. For example, manganese shows oxidation numbers of +2, +4, +6 and +7 (Table 3.3). Table 3.3 Oxidation numbers of manganese

Compound

MnSO4

Oxidation number of manganese

+2

MnO2 K2MnO4 KMnO4 +4

+6

+7

12 Non-metals usually have negative oxidation numbers. However, chlorine, bromine, iodine and nitrogen can have positive or negative oxidation number (Table 3.4) depending on the elements which combine with them. 390

Table 3.4 Oxidation numbers of chlorine and nitrogen

Chlorine compound Oxidation number of chlorine Nitrogen compound Oxidation number of nitrogen

HCl

HClO

HClO2

ClO2

HClO3

HClO4

–1

+1

+3

+4

+5

+7

NH3

N2O

NO

NO2–

NO2

NO3–

–3

+1

+2

+3

+4

+5

Calculating the Oxidation Numbers of Elements in Compounds and Ions­

2

4 What is the oxidation number of sulphur in the thiosulphate ion, S2O32–?

Solution Let the oxidation number of manganese = x Oxidation number of K = +1 Oxidation number of O = –2 KMnO4

Solution Let the oxidation number of sulphur = x Oxidation number of O = –2 S2O32– 3

What is the oxidation number of manganese in the compound, KMnO4?

2(x) 3(–2) Sum of the oxidation numbers of all atoms in the polyatomic ion is equal to the charge on the ion. 2x + 3(–2) = –2 x = +2 The oxidation number of sulphur in S2O32– ion is +2.

+1 x 4(–2) Sum of the oxidation numbers of all the elements in the neutral compound, KMnO4, is zero. (+1) + x + 4(–2) = 0 x = +7 The oxidation number of manganese in KMnO4 is +7.

3

1,1,1-trichloroethane (C2H3Cl3) is used as a solvent for halogens. The oxidation numbers of carbon, hydrogen and chlorine are shown below.

Calculate the oxidation number of nitrogen in nitric acid, HNO3.

C2H3Cl3

Solution Let the oxidation number of nitrogen = x Oxidation number of H = +1 Oxidation number of O = –2 HNO3

2(0) + 3(+1) + 3(–1) = 0 The oxidation number of an element in the free state is always zero. However, in some cases (for example, carbon in the C2H3Cl3), the oxidation number of the element in the compound can also be zero.

+1 x 3(–2) Sum of the oxidation numbers of all the elements in the compound is zero. (+1) + x + 3(–2) = 0 x = +5 The oxidation number of nitrogen in HNO3 is +5.

IUPAC Nomenclature of Inorganic Compounds 1 The IUPAC system is used to name inorganic compounds in order to avoid confusion that may arise due to elements having different oxidation numbers. 2 For example, there are two oxides of copper: copper(I) oxide, Cu2O, and copper(II) oxide, CuO. Copper(I) oxide is a brown powder whereas copper(II) oxide is a black powder. The Roman numerical figures (I) and (II) refer to the oxidation numbers of copper in the compound.

The oxidation number of nitrogen in HNO3 is +5 and not 5. The oxidation number of the element must be accompanied by a positive or negative sign on the left of the number.

391

Oxidation and Reduction

Oxidation Number and IUPAC Nomenclature

Formula Oxidation of number of compound metal

3

1 (a) For an ionic compound or a covalent compound that contains a metal with more than one oxidation number, the Roman numerical figure is stated in brackets after the name of the metal to show the oxidation number of the metal. (b) For example, tin forms two types of chlorides, SnCl2 (ionic) and SnCl4 (covalent). SnCl2 is called tin(II) chloride and SnCl4 is called tin(IV) chloride. (c) Similarly, lead(II) oxide refers to the compound with the formula PbO whereas lead(IV) oxide refers to the compound PbO2. 2 Table 3.5 shows the formulae and the IUPAC names of some compounds containing metals.

IUPAC name

FeCl2

+2

Iron(II) chloride

FeCl3

+3

Iron(III) chloride

CuCl

+1

Copper(I) chloride

CuSO4

+2

Copper(II) sulphate

Mn(NO3)2

+2

Manganese(II) nitrate

MnO2

+4

Manganese(IV) oxide

Oxidation and Reduction

+1

Magnesium nitrate (NOT magnesium(II) nitrate)

AlCl3

+3

Aluminium chloride (NOT aluminium(III) chloride)

Formula Oxidation of number of metal compound in negative ion

Table 3.6 Naming inorganic compounds containing elements of Groups 1, 2 and 3

K2SO4

+2

Table 3.7 Name of compounds containing metals in negative ions

3 The metallic elements in Groups 1, 2 and 3 of the Periodic Table always have the oxidation numbers +1, +2 and +3 respectively. According to the IUPAC nomenclature, the Roman numerical figure is not used in naming a compound if the metal shows only one oxidation state in its compounds. 4 Table 3.6 shows some examples of naming compounds containing elements of Groups 1, 2 and 3.

Formula Oxidation of number of compound metal

Mg(NO3)2

5 For a negative ion that contains a metal with more than one oxidation state, the Roman number is stated in brackets after the name of the metal, and the name of the metal ends with -ate. For example, (a) manganate(VII) refers to the negative ion containing the manganese metal with oxidation number +7, that is, MnO4–. (b) chromate(VI) refers to the negative ion containing chromium metal with oxidation number +6, that is, CrO42–. (c) dichromate(VI) refers to the negative ion containing two chromium atoms with oxidation number +6, that is, Cr2O72–. (d) hexacyanoferrate(III) refers to the negative ion containing six cyano (CN–) groups and iron metal with oxidation number +3, that is, [Fe(CN)6]3–. Table 3.7 shows the names of some compounds containing metals in negative ions.

Table 3.5 Naming inorganic compounds containing metals

Formula Oxidation of number of compound metal

Name of compound

K2MnO4

+6

Potassium manganate(VI)

KMnO4

+7

Potassium manganate(VII)

K2CrO4

+6

Potassium chromate(VI)

K2Cr2O7

+6

Potassium dichromate(VI)

K4Fe(CN)6

+2

Potassium hexacyanoferrate(II)

K3Fe(CN)6

+3

Potassium hexacyanoferrate(III)

Name of compound Potassium sulphate (NOT potassium(I) sulphate) 392

IUPAC name

Notice that in K4Fe(CN)6, the negative ion is [Fe(CN)6]4– while in K3Fe(CN)6, the negative ion is [Fe(CN)6]3–. 6 Oxoanions are anions (negative ions) that consist of an oxygen atom and another nonmetallic atom. Examples of oxoanions are nitrate ion, NO3– and sulphate ion, SO42–. For a

non-metal that shows more than one oxidation number in its oxoanion, the Roman number stated in brackets refers to the oxidation number of the non-metal. 7 Table 3.8 shows the common names and the IUPAC names for some compounds containing oxoanions.

Table 3.8 Common names and IUPAC names for some compounds

Oxidation number

IUPAC name

Na2SO3

+4 (for S)

Sodium sulphate(IV)

Sodium sulphite

Na2SO4

+6 (for S)

Sodium sulphate(VI)

Sodium sulphate

NaNO2

+3 (for N)

Sodium nitrate(III)

Sodium nitrite

NaNO3

+5 (for N)

Sodium nitrate(V)

Sodium nitrate

NaClO

+1 (for Cl)

Sodium chlorate(I)

Sodium hypochlorite

NaClO3

+5 (for Cl)

Sodium chlorate(V)

Sodium chlorate

HNO2

+3 (for N)

Nitric(III) acid

Nitrous acid

HNO3

+5 (for N)

Nitric(V) acid

Nitric acid

H2SO4

+6 (for S)

Sulphuric(VI) acid

Sulphuric acid

1

Common name of compound

3

Molecular formula of compound

(a) the changes in oxidation numbers or (b) the transfer of electrons; that is, acceptance (gain) or donation (loss) of electrons. 3 Using oxidation numbers to identify redox reactions (a) Oxidation is a process in which the oxidation number of the element is increased. (b) Reduction is a process in which the oxidation number of the element is decreased. 4 The following equation shows the reaction between iron and chlorine.

’06

What is the oxidation number for oxygen in the thiosulphate ion, S2O32–? A –2 B –3 C +2 D +3 Solution The oxidation number for oxygen in a polyatomic ion, such as NO3–, S2O32– or Cr2O72– is always –2. Answer A Comment The oxidation number of oxygen in covalent compounds, such as CO2 or SO3 is also –2.



Oxidation and Reduction in Terms of Changes in Oxidation Numbers

1 Most redox reactions occur without involving hydrogen or oxygen. For example, the reaction between chlorine and iron(II) chloride is a redox reaction:

(oxidation number increases) oxidation 0

0

+3 –1

2Fe(s) + 3Cl2 → 2FeCl3 reduction (oxidation number decreases)

In this redox reaction, (a) the iron metal is oxidised to chloride because its oxidation increases from 0 to +3. (b) chlorine (Cl2) is reduced to ion (Cl–) because its oxidation decreases from 0 to –1.

2FeCl2(aq) + Cl2(g) → 2FeCl3(aq) 2 For reactions that do not involve hydrogen or oxygen, an oxidation or reduction reaction is discussed in terms of 393

iron(III) number chloride number

Oxidation and Reduction

(c) chlorine acts as the oxidising agent because it oxidises iron and is itself reduced. (d) iron acts as the reducing agent because it reduces chlorine and is itself oxidised. 5 When chlorine gas is passed into potassium bromide solution, the following reaction occurs:

Sulphur dioxide usually acts as a reducing agent. However, in the following reactions, sulphur dioxide acts as an oxidising agent because hydrogen sulphide and carbon are stronger reducing agents than sulphur dioxide.



(a) oxidation number of S increases (oxidation)

(oxidation number increases) oxidation

0

–1

–1



0

Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(aq)



–2

0

SO2(g) + 2H2S(g) ⎯⎯⎯→ 2H2O(l) + 3S(s)

+4

0





oxidation number of S decreases (reduction)

reduction (oxidation number decreases)

(b) oxidation



3

The oxidation number of potassium in the above reaction does not change because potassium does not take part in the reaction.

7 A reaction is not a redox reaction if the substances involved in the reaction do not undergo any changes in oxidation numbers. For example, +1 –2 +1



+1 –2

0

Oxidation and reduction always take place together. A redox reaction must have • a substance that undergoes oxidation and acts as the reducing agent, and • another substance that undergoes reduction and acts as the oxidising agent.

reduction (oxidation number decreases)

In this reaction, (a) ammonia is oxidised to nitrogen because the oxidation number of nitrogen increases from –3 to 0. (b) copper(II) oxide is reduced to copper because the oxidation number of copper decreases from +2 to 0. (c) copper(II) oxide acts as an oxidising agent and ammonia acts as a reducing agent. (d) the oxidation numbers of hydrogen and oxygen remain unchanged. Oxidation and Reduction

+1 +6 –2

The reaction between sodium hydroxide (NaOH) and sulphuric acid (H2SO4) is a neutralisation reaction and not a redox reaction. As a result, the oxidation numbers of all the elements (sodium, oxygen, hydrogen and sulphur) are the same before and after the reaction.

2NH3(g) + 3CuO(s) → N2(g) + 3H2O(l) + 3Cu(s)



+1 +6 –2

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)



0

0

reduction

(oxidation number increases) oxidation +2

+4

+4

In this reaction, (a) bromide ion (Br–) is oxidised to bromine because the oxidation number of bromine increases from –1 to 0. (b) chlorine (Cl2) is reduced to chloride ion (Cl–) because the oxidation number of chlorine decreases from 0 to –1. (c) chlorine acts as an oxidising agent and potassium bromide acts as a reducing agent. 6 Ammonia reacts with copper(II) oxide as represented by the equation:

–3

0

SO2(g) + C(s) ⎯⎯⎯→ CO2(g) + S(s)

Oxidation and Reduction in Terms of Electron Transfer 1 In terms of electron transfer, (a) oxidation is defined as the loss of electrons SPM from a substance. If a substance loses ’05/P1 electrons during a reaction, it has been oxidised. 394

5

(b) reduction is defined as the gain of electrons by a substance. If a substance gains electrons, it has been reduced. 2 During a redox reaction, transfer of electrons occurs between the reactants.

Write the half-equation for the reduction of acidified manganate(VII) ion (MnO4–) to manganese(II) ion (Mn2+) in the presence of acid. Solution Step 1: Write the reactants and products involved in the reaction.

An oil rig is used for getting oil and gas out of the ground in the petroleum industry. Use ‘OIL RIG’ to help you remember oxidation and reduction in terms of electron transfer. OIL : OXIDATION IS LOSS OF ELECTRONS RIG : REDUCTION IS GAIN OF ELECTRONS



Step 2: Balance the number of atoms on both sides of the equation:

MnO4– + 8H+ → Mn2+ + 4H2O

Step 3: Balance the number of charges on both sides of the equation:

3 The reactant that loses electrons undergoes oxidation and acts as a reducing agent. For example,



Na(s) ⎯⎯⎯→ Na+(aq) + e– … (1)



oxidation

MnO4– + 8H+ → Mn2+ + 4H2O total charge total charge = (–1) + (+8) = +7 = +2 3

In this reaction, (a) sodium atoms undergo oxidation by losing electrons to form sodium ions (Na+). (b) sodium acts as the reducing agent. 4 A substance that accepts electrons undergoes reduction and acts as an oxidising agent. For example,

MnO4– + H+ → Mn2+ + H2O

to balance the charges, add 5e– to the left of the equation

MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)

Cl2(g) + 2e– ⎯⎯⎯→ 2Cl–(aq) … (2) reduction

6 (a) Zinc reacts with hydrochloric acid as represented by the equation

In this reaction, (a) each chlorine molecule (Cl2) accepts two electrons to form two chloride ions (Cl–). (b) chlorine acts as the oxidising agent and is itself reduced. 5 Balancing half-equations for oxidation and SPM reduction ’06/P1 Equations (1) and (2) as shown above are known as half-equations. Half-equations must be balanced in terms of (a) the number of atoms, and (b) the number of charges.

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) The ionic equation for the reaction is oxidation (loss of electrons)



Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)





reduction (gain of electrons)

(b) The transfer of electrons can be represented by the following half-equations: Photographic films are coated with silver bromide, AgBr. When the film is exposed to light, the following redox reaction occurs:

Zn(s) → Zn2+(aq) + 2e– ... oxidation

2Ag+ + 2Br–­­­ → 2Ag + Br2

2H+(aq) + 2e– → H2(g) ... reduction

reducing agent

The amount of silver produced depends on how much light gets through the camera lens. In this reaction, silver ions are reduced to silver by the gain of electrons and bromide ions are oxidised to bromine by the loss of electrons.

oxidising agent

(c) In the reaction between hydrochloric acid and zinc, zinc is oxidised to zinc chloride whereas hydrochloric acid is reduced to hydrogen. 395

Oxidation and Reduction

9 Combustion of metals in chlorine Figure 3.3 shows the combustion of copper in chlorine. When the hot copper foil is placed in a gas jar of chlorine, a vigorous reaction occurs and a green precipitate of copper(II) chloride, CuCl2 is formed.

(d) Hydrochloric acid acts as an oxidising agent by accepting electrons and is itself reduced. Conversely, zinc acts as the reducing agent by donating electrons and is itself oxidised. 7 In terms of transfer of electrons, oxidising agents are electron acceptors while reducing agents are electron donors. 8 If a coil of copper is placed in a solution of silver nitrate, the copper slowly dissolves and the solution turns blue. At the same time, the copper coil becomes coated with a layer of silver metal (Figure 3.2).

Figure 3.3 Combustion of copper in chlorine

3

(a) In the reaction between copper and chlorine to form copper(II) chloride, a transfer of electron occurs between copper metal and chlorine gas.

Figure 3.2 Oxidation of copper

(a) The overall equation for the reaction is

Cu(s) + Cl2(g) → CuCl2(s)

Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)



Cu(s) → Cu2+(s) + 2e– … oxidation Cl2(g) + 2e– → 2Cl– (s) ... reduction (c) In the reaction between copper and chlorine, the copper atom (Cu) (i) loses electrons (ii) undergoes oxidation (iii) is oxidised to copper(II) ion, Cu2+ (iv) acts as a reducing agent (d) Conversely, the chlorine molecule (Cl2) (i) gains electrons (ii) undergoes reduction (iii) is reduced to chloride ions, Cl– (iv) acts as an oxidising agent 10 Combustion of metals in oxygen When metals burn in oxygen, (a) the metals undergo oxidation by losing electrons to form metal ions, (b) the oxygen undergoes reduction by gaining electrons to form oxide ions (O2–). The combustion of lead in oxygen is studied in Activity 3.1.

Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

reduction (gain of electrons)

(b) In this reaction, each silver ion (Ag+) accepts one electron to form a silver atom (Ag). Ag+(aq) + e– → Ag(s) ... reduction

An oxidising agent is an electron acceptor. Hence, silver ion acts as an oxidising agent in this reaction. (c) Conversely, each copper atom donates two electrons and are converted to copper(II) ion (Cu2+) in the aqueous solution. Cu(s) → Cu2+(aq) + 2e– ....oxidation

Reducing agents are electron donors. Hence, copper acts as a reducing agent.

Oxidation and Reduction

reduction (gain of electrons)

(b) The transfer of electrons can be represented by the half-equations as shown below:

The reaction can be represented by the ionic equation:

oxidation (loss of electrons)

oxidation (loss of electrons)

396

To investigate the combustion of metals in oxygen and chlorine Apparatus

Gas jar, tongs, combustion spoon, gas jar and Bunsen burner.

Materials

Magnesium ribbon, sodium and chlorine.

Procedure

2 The magnesium ribbon is held with a pair of tongs and lit in the Bunsen burner. 3 It is quickly placed into a gas jar filled with oxygen. 4 Any changes that occur are recorded. (B) Reaction of sodium with chlorine

Figure 3.4 The combustion of magnesium in oxygen

(A) Combustion of magnesium in oxygen 1 A piece of 5 cm magnesium ribbon is cleaned with sandpaper.

1 A small piece of sodium metal is placed in a combustion spoon and heated. 2 When the sodium metal starts to burn, it is quickly placed in a gas jar filled with chlorine gas. 3 Any changes that occur are recorded.

3

Figure 3.5 The combustion of sodium in chlorine

Observation Experiment

Observation

Combustion of magnesium in oxygen

• The magnesium ribbon burns with a bright white flame. • White fumes are produced. • A white powder is formed.

Reaction of sodium with chlorine

• The sodium metal burns with a yellow flame. • White fumes are produced. • A white powder is formed.

oxidation (loss of electrons)

2Mg(s) + O2(g) → 2MgO(s)

5 In this reaction, • magnesium acts as the reducing agent because it reduces oxygen to oxide ion. • oxygen acts as the oxidising agent because it oxidises magnesium to magnesium ion. (B) Combustion of sodium in chlorine 1 The combustion of sodium in chlorine produces sodium chloride (white powder). 2 Sodium is oxidised by losing electrons to form sodium ions, Na+. Half-equation: Na(s) → Na+(s) + e– 3 Chlorine is reduced by gaining electrons to form chloride ion, Cl–. Half-equation: Cl2(g) + 2e– → 2Cl–(s)



reduction (gain of electrons)

397

Oxidation and Reduction

Activity 3.1

Discussion (A) Combustion of magnesium in oxygen 1 The combustion of magnesium in oxygen produces magnesium oxide (white powder). 2 Magnesium is oxidised by losing electrons to form magnesium ions, Mg2+. Half-equation: Mg(s) → Mg2+(s) + 2e– 3 Oxygen is reduced by gaining electrons to form oxide ion, O2–. Half-equation: O2(g) + 4e– → 2O2–(s) 4 The overall equation for the reaction is

4 The overall equation for the reaction is oxidation (loss of electrons)

2Na(s) + Cl2(g) → 2NaCl(s)



reduction (gain of electrons)

5 In this reaction, • sodium acts as the reducing agent because it reduces chlorine to chlorine ion. • chlorine acts as the oxidising agent because it oxidises sodium to sodium ion.

Conclusion 1 In the combustion of magnesium in oxygen, (a) magnesium undergoes oxidation to form Mg2+ ions, (b) oxygen undergoes reduction to form O2– ions. 2 In the combustion of sodium in chlorine, • sodium act as reducing agent by losing electrons, • chlorine act as oxidising agent by gaining electrons.

3

Conversion of Fe2+ Ions to Fe3+ Ions and Vice Versa

SPM

’08/P2, ’09/P1

There are some chemicals such as hydrogen peroxide (H2O2) and nitric(III) acid (nitrous acid, HNO2) which can act as an oxidising agent or a reducing agent depending on the conditions of the reaction. For example, in reaction (1), hydrogen peroxide acts as an oxidising agent, but in reaction (2), it acts as a reducing agent.

1 Iron metal (Fe) exhibits two oxidation states, +2 and +3. 2 Fe2+ ions can be converted to Fe3+ ions. Similarly, Fe3+ ions can be converted to Fe2+ ions.

H2O2 + 2I– + 2H+ → I2 + 2H2O … (1)

Oxidation of Fe2+ to Fe3+

oxidising agent

reducing agent

1 Iron(II) ion, Fe2+, can be converted to iron(III) ions, Fe3+, by oxidation reaction.

5H2O2 + 2MnO4– + 6H+ → 2Mn2+ + 8H2O + 5O2 … (2) reducing agent

oxidation (loss of electrons)



oxidising agent

Fe2+(aq) → Fe3+(aq) + e–

2 Potassium manganate(VII) is an oxidising agent that can oxidise Fe2+ ions to Fe3+ ions. 3 (a) When acidified potassium manganate(VII) solution is added to a solution of iron(II) salt, decolourisation occurs. MnO4– ions are reduced to Mn2+ ions while Fe2+ ions (pale green) are oxidised to Fe3+ ions (brown). oxidation (loss of electrons)

MnO (aq) + 8H (aq) + 5Fe (aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) +

– 4

2+

green reduction (gain of electrons)

brown

A catalytic converter Catalytic converters are fitted to the exhaust pipes of cars to reduce air pollution. In the catalytic converter, the following redox reactions take place to convert poisonous gases (NO, CO and petrol vapour) to nonpoisonous gases. For example,

Dilute sulphuric acid is always used to acidify KMnO4 solution.

(b) The half-equations for the reactions are:

2NO(g) + 2CO(g) → N2(g) + 2CO2(g)

Fe2+(aq) → Fe3+(aq) + e–

oxidising reducing agent agent

Oxidation and Reduction



398

(oxidation – loss of electrons)

MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)



(reduction – gain of electrons)







Fe3+(aq) + e– → Fe2+(aq) (reduction – gain of electrons)



SO32–(aq) + H2O(l) → SO42–(aq) + 2H+(aq) + 2e– (oxidation – loss of electrons)



3 The formation of Fe2+ ions can be confirmed by using sodium hydroxide solution. When sodium hydroxide solution is added to the reaction product, a dirty green precipitate of iron(II) hydroxide, Fe(OH)2, insoluble in excess NaOH(aq), is obtained.

Br2(l) + 2Fe2+(aq) → 2Fe3+(aq) + 2Br–(aq) (c) Acidified potassium dichromate(VI) solution (acidified with dilute sulphuric acid) Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

Fe2+(aq) + 2NaOH(aq) → Fe(OH)2(s) + 2Na+(aq)

(d) Concentrated nitric acid

4 Other reducing agents that can be used to reduce Fe3+ ions to Fe2+ ions include the following. (a) Metals more reactive (electropositive) than iron. For example, zinc.

HNO3(aq) + 3Fe (aq) + 3H (aq) → 3Fe3+(aq) + 2H2O(l) + NO(g) +

(e) Acidified hydrogen peroxide

Zn(s) + 2Fe3+(aq) → 2Fe2+(aq) + Zn2+(aq)

H2O2(aq) + 2H (aq) + 2Fe (aq) → 2Fe3+(aq) + 2H2O(l) 2+

(b) Sulphur dioxide SO2(g) + 2H2O(l) + 2Fe3+(aq) → 2Fe2+(aq) + 2H+(aq) + H2SO4(aq)

Reduction of Fe to Fe

2+

(c) Potassium iodide

1 Iron(III) ions, Fe3+, can be converted to iron(II) ions, Fe2+, by reduction.



–

(b) Liquid bromine

3+

oxidation (lose electron)

2KI(aq) + 2Fe3+(aq) → 2Fe2+(aq) + 2K+(aq) + I2(s)

reduction (gain electron)

(d) Hydrogen sulphide Fe3+(aq) + e– → Fe2+(aq)

H2S(aq) + 2Fe3+(aq) → 2Fe2+(aq) + 2H+(aq) + S(s)

2 (a) When sodium sulphite (Na2SO3) solution is added to iron(III) chloride, and the mixture is acidified with dilute sulphuric acid, the colour of the solution changes from brown to light green.

(e) Tin(II) chloride solution Sn2+(aq) + 2Fe3+(aq) → 2Fe2+(aq) + Sn4+(aq)

399

Oxidation and Reduction

3

Cl2(aq) + 2Fe (aq) → 2Fe (aq) + 2Cl (aq)

+

green

Sodium sulphite acts as the reducing agent and reduces iron(III) ions to iron(II) ions and is itself oxidised to sulphate ions (SO42–). (b) The half-equations for the reaction are:

5 Other oxidising agents that can be used to oxidise Fe2+ to Fe3+ are as follows. (a) Chlorine gas or chlorine water

2+

brown



Fe3+(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3Na+(aq)

3+

SO32– + H2O + 2Fe3+ → 2Fe2+ + H2SO4

4 The formation of Fe3+ ions can be confirmed by using sodium hydroxide solution. When sodium hydroxide solution is added to the reaction product, a brown precipitate of iron(III) hydroxide, Fe(OH)3, insoluble in excess NaOH(aq) is obtained.

2+

reduction (gain electron)

3

To study the oxidation of Fe2+ ions to Fe3+ ions and the reduction of Fe3+ ions to Fe2+ ions

SPM

’09/P2

Apparatus

Test tubes and droppers.

Materials

FeSO4, KMnO4, FeCl3, Na2SO3, dilute H2SO4 and dilute NaOH solution.

4 Sodium hydroxide solution is then added to the reaction mixture slowly until in excess. 5 The observations are recorded in the table below.

Procedure (A) Conversion of Fe2+ ions to Fe3+ ions 1 About 2 cm3 of iron(II) sulphate solution is poured into a test tube. 2 About 2 cm3 of potassium manganate(VII) solution is poured into another test tube, followed by about 2 cm3 of dilute sulphuric acid. 3 Using a dropper, about 2 cm3 of the acidified potassium manganate(VII) solution is added slowly to the iron(II) sulphate solution. The mixture is shaken gently.

(B) Conversion of Fe3+ ions to Fe2+ ions 1 About 2 cm3 of iron(III) chloride solution is added to a test tube. 2 Sodium sulphite (Na2SO3) solution is added to iron(III) chloride solution, followed by dilute sulphuric acid. The mixture is shaken gently. 3 Sodium hydroxide solution is then added slowly to the reaction mixture until in excess. 4 The observations are recorded in the table below.

Observations Solution

Test

Observation

FeSO4(aq)

(a) Fe2+ ion + acidified KMnO4

• The light green iron(II) sulphate solution changes to yellow. • The purple colour of acidified KMnO4 solution turns colourless (decolourised).

(b) Add excess NaOH(aq) to (a)

• A brown precipitate, insoluble in excess NaOH(aq) is formed.

(c) Fe3+ ion + Na2SO3(aq)

• The colour of the solution changes from yellow to light green.

(d) Add excess NaOH(aq) to (c)

• A dirty green precipitate, insoluble in excess NaOH(aq) is obtained.

FeCl3(aq)

Conclusion 1 Fe2+ ions are oxidised to Fe3+ ions by the acidified KMnO4 solution. 2 Fe3+ ions are reduced to Fe2+ ions by the sodium sulphite (Na2SO3) solution. 3 Electropositive metals are strong reducing agents. In contrast, the metallic ions of electropositive metals are weak oxidising agents. Figure 3.6 shows that in the electrochemical series, (a) the strength of a metal as a reducing agent increases on going up the electrochemical series, (b) the strength of the metallic ion as an oxidising agent increases on going down the series.

Activity 3.2

Displacement of Metals from Their Salt Solutions 1 The arrangement of metals according to their SPM tendency to lose electrons to form positive ’04/P1 ’07/P1 ions is called the electrochemical series. 2 The higher the position of the metal in the electrochemical series, (a) the more electropositive the metal, (b) the more readily the metal donates electrons to form positive ions, (c) the more easily the metal will undergo oxidation. Oxidation and Reduction

400

• Tendency of a metal to ionise (by donating electrons) increases.

• Tendency of an ion to accept electrons increases.

• Strength of a metal as a reducing agent increases.

• Strength of an ion as an oxidising agent increases.

4 Consider the formation of sodium ions (Na+) from sodium metal (Na).

position in the electrochemical series) will displace a less electropositive metal (lower position in the electrochemical series) from the salt solutions of the less electropositive metal. 6 Transfer of electrons occurs during displacement reactions. (a) The more electropositive metal donates electrons and acts as a reducing agent. The metal undergoes oxidation and is oxidised to its metal ions. (b) The metal ion (from the less electropositive metal) in aqueous solution acts as an oxidising agent. The metal ions undergo reduction and is reduced to its metal.

Na metal has a strong tendency to lose an electron to form sodium ion

⎯⎯⎯⎯⎯⎯⎯⎯→ Na(s) → Na+(aq) + e– ←⎯⎯⎯⎯⎯⎯⎯⎯ +

Na ion has a weak tendency to accept an electron to form Na metal

(a) Sodium metal is placed at a high position in the electrochemical series. (b) This means that sodium metal donates electrons very easily. As a result, sodium is a strong reducing agent. (c) Conversely, sodium ions (Na+) have a weak tendency to accept electrons. Since oxidising agents are electron acceptors, sodium ions are weak oxidising agents. 5 A displacement reaction is a reaction in which one element (metal or non-metal) displaces another element (metal or non-metal) from its salt solution. In the displacement reactions of metals, the more electropositive metal (higher

3

Figure 3.6 Electrochemical series

A more electropositive metal is also a more reactive metal. We can therefore state that a more reactive metal will displace a less reactive metal from the solution of its salts.

3.2 To study the redox reaction in terms of displacement reaction of a metal from its salt solution Problem statement How does redox reaction occur in a displacement reaction in which a metal is displaced from its salt solution?

Variables (a) Manipulated variable : A pair of metals and salt solutions (b) Responding variable : Precipitation of metal and colour changes in the solutions (c) Constant variables : Volumes and concentrations of solutions containing the metal ions Apparatus Beakers 401

Oxidation and Reduction

Experiment 3.2

Hypothesis (a) The metal that acts as a reducing agent will form metal ion. (b) The metal ion that acts as an oxidising agent will be precipitated as metal.

Materials Copper(II) sulphate solution and silver nitrate solution, zinc plate and copper plate.

Procedure 1 A strip of zinc plate and a strip of copper plate are cleaned with sandpaper. 2 The zinc plate is then immersed in copper(II) sulphate solution (beaker P) and the copper plate is immersed in silver nitrate solution (beaker Q). 3 The mixture is left aside for half an hour. 4 The changes that take place on the zinc plate, the copper plate, and in the copper(II) sulphate solution and the silver nitrate solution are recorded.

Figure 3.7 The displacement of a metal from its salt solution

Observation Reactants

Observation

Explanation

• A section of the zinc plate dissolves. • Brown precipitate is deposited on the zinc plate. • The blue solution fades until it becomes colourless.

• Zinc displaces copper metal (brown precipitate) from copper(II) sulphate solution. • Copper metal is deposited on the zinc plate. • The blue colour fades as the concentration of Cu2+ ions decreases.

(b) Copper plate in silver nitrate solution

• The copper plate dissolves. • A greyish-black precipitate is deposited on the copper plate. • The colourless solution turns blue.

• Copper displaces silver metal (greyishblack) from the silver nitrate solution. • Silver metal is precipitated on the copper plate. • The colourless solution turns blue because of the formation of Cu2+ ions.

3

(a) Zinc plate in copper(II) sulphate solution

Conclusion During the displacement reaction, the more electropositive metal will act as a reducing agent. It reduces the metal ion (oxidising agent) which is less electropositive to form metal. The hypothesis is accepted.

Displacement of Copper by Zinc from Copper(II) Sulphate Solution

The reaction can be represented by the following half-equations: (a) Zn(s) → Zn2+(aq) + 2e–

SPM

’08/P2

1 The following equation shows the reaction between copper(II) sulphate solution and zinc.

(oxidation – loss of electrons) Zinc acts as a reducing agent (electron donor)

(b) Cu2+(aq) + 2e– → Cu(s)

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

(reduction – gain of electrons) Copper ion acts as an oxidising agent (electron acceptor)

Zinc is more electropositive than copper. It displaces copper from its salt (that is, copper(II) sulphate). 2 A displacement reaction is a redox reaction.

3 When copper(II) ion is displaced, the concentration of Cu2+ ions in the solution decreases. This causes the blue colour to fade.

oxidation (loss of electrons)

Displacement of Silver by Copper from Silver Nitrate Solution



Zn(s) + CuSO4(aq) ⎯⎯→ ZnSO4(aq) + Cu(s)





1 The displacement reaction between copper and silver nitrate solution is shown as follows.

reduction (gain of electrons)

Oxidation and Reduction

402



(a) Cu(s) → Cu2+(aq) + 2e–

oxidation (loss of electrons)



(oxidation – loss of electrons) Copper acts as a reducing agent (electron donor).

Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)

(b) Ag+(aq) + e– → Ag(s)



(reduction – gain of electrons) Silver ion acts as an oxidising agent (electron acceptor).

reduction (gain of electrons)

Copper is more electropositive than silver. It displaces silver from its salt (that is, silver nitrate). 2 The reaction can be represented by the following half-equations:

3 When copper dissolves in silver nitrate solution, the formation of copper(II) ion causes the solution to turn blue. The intensity of blue colour increases as more copper is dissolved.

2

’07

Solution Metal

AgNO3

Q

Pb(NO3)2

FeSO4

MgSO4

No change

No change

No change

No change

No change

X

Silver is displaced

Y

Silver is displaced

Lead is displaced

Z

Silver is displaced

Lead is displaced

What is the correct position of the metals, in ascending order, of the tendency of the metals to be oxidised? A Q, X, Y, Z C X, Y, Z, Q B Q, X, Z, Y D Y, Z, X, Q Comments The most reactive metal is the strongest reducing agent, that is, it has the highest tendency to form metal ions by losing electrons, that is to be oxidised.

Iron is displaced No change

Q is the least reactive. It has no reactions with Pb2+, Fe2+ and Mg2+. Y is the most reactive. It can displace three metals from their solutions. Z is more reactive than X. It can displace two metals from their salt solutions. Answer B

Displacement of Halogens from Halide Solutions

halogen will displace a less reactive halogen from the solution of its halide ions. 3 Hence, chlorine displaces bromine from an aqueous solution of bromide ions. It also displaces iodine from an aqueous solution of iodide ions. Similarly, bromine displaces iodine from an aqueous solution of iodide ions.

1 In general, the stronger the oxidising strength SPM of the halogen, the weaker the reducing ’04/P1 ’05/P1 strength of the corresponding halide ion is. Thus, chlorine is a stronger oxidising agent than iodine, but the iodide ion is a stronger reducing agent than the chloride ion. Figure 3.8 shows the trend in the reactivity and oxidising strength of the halogens and the reducing strength of the halide ions. 2 The reactivity of the halogens can be used to predict whether the displacement reactions of halogens can occur or not. A more reactive

Cl2(aq) + 2KBr(aq) → 2KCl(aq) + Br2(aq) Cl2(aq) + 2KI(aq) → 2KCl(aq) + I2(aq) Br2(aq) + 2KI(aq) → 2KBr(aq) + I2(aq) 4 Conversely, bromine cannot displace chlorine from an aqueous solution of chloride ions and iodine cannot displace bromine or chlorine 403

Oxidation and Reduction

3

An experiment is carried out to determine the positions of the metals, Q, X, Y, Z in the electrochemical series. The results of the experiment on displacement reactions are shown in the table below.

from an aqueous solution of bromide ions or chloride ions respectively.

1,1,1-trichloroethane. The colours of halogens in 1,1,1-trichloroethane are shown in Table 3.10. Table 3.10 The colours of halogens in 1,1,1trichloroethane

Br2(aq) + KCl(aq) → No reaction I2(aq) + KCl(aq) → No reaction I2(aq) + KBr(aq) → No reaction 5 The colours of halogens in water are shown in Table 3.9.

Halogen

Colour

Chlorine

Colourless

Bromine

Brown

Iodine

Purple

3

Table 3.9 The colour of halogens in water

Halogen

Concentrated aqueous solution

Dilute aqueous solution

Chlorine

Greenish-yellow

Colourless

Bromine

Brown

Yellow

Iodine

Brown

Yellow

The structural formula of 1,1,1-trichloroethane is: H Cl ⎮ ⎮ H ⎯ C ⎯ C ⎯ Cl ⎮ ⎮ H Cl

6 Halogens can be identified by adding 1,1,1trichloroethane (CH3CCl3) to its aqueous solution. Water and 1,1,1-trichloroethane are immiscible and two layers are formed. The upper layer is water and the lower layer is

It is very volatile and is used as a solvent in paper correction fluid. It is produced as one of the organic products when ethane reacts in chlorine in the presence of sunlight.

A The tendency of electrons being removed from halide ions to form halogens increases

A Reactivity of halogens increases A Oxidising strength of halogens increases

A The strength of halide ion as a reducing agent increases

Figure 3.8

3.3 To study the displacement reactions between halogens and halide ions Problem statement How do redox reactions occur in displacement reactions between halogens and aqueous solutions of halide ions?

Experiment 3.3

Hypothesis A more reactive halogen will displace a less reactive halogen from an aqueous solution of its halide ions. Variables (a) Manipulated variable : A pair of halogens and their halide ions (b) Responding variable : Changes in colour in 1,1,1-trichloroethane, CH3CCl3 (c) Constant variable : Volume of reaction mixture Apparatus Test tubes Oxidation and Reduction

404

2 2 cm3 of chlorine water is added to the potassium bromide solution. The mixture is shaken gently. 3 2 cm3 of 1,1,1-trichloroethane (CH3CCl3) is then added to the mixture obtained in step 2. The mixture is then shaken vigorously. 4 The test tube is allowed to stand for a few minutes and the colour of the 1,1,1-trichloroethane layer is recorded. 5 Steps 1 to 4 are repeated using the following mixtures. (a) Chlorine water and potassium iodide, KI solution (b) Liquid bromine and potassium chloride, KCl solution (c) Liquid bromine and potassium iodide solution (d) Iodine solution and potassium bromide solution (e) Iodine solution and potassium chloride solution

Materials 1,1,1-trichloroethane, potassium bromide, KBr(aq), potassium chloride, KCl(aq), potassium iodide, KI(aq), chlorine water, liquid bromine and iodine solution.

Figure 3.9 Displacement of bromine from potassium bromide solution

Procedure 1 A test tube is filled with 2 cm3 of potassium bromide, KBr solution.

3

Observation Mixture

Colour of CH3CCl3 layer

Halogen in CH3CCl3 layer

Has displacement reaction occurred?

Cl2(aq) + KBr(aq)

Brown

Bromine

Yes

Cl2(aq) + KI(aq)

Purple

Iodine

Yes

Br2(l) + KCl(aq)

Brown

Bromine

No

Br2(l) + KI(aq)

Purple

Iodine

Yes

I2(aq) + KBr(aq)

Purple

*Iodine

No

I2(aq) + KCl(aq)

Purple

*Iodine

No

ψ

The bromine present in the CH3CCl3 layer is due to the bromine added. *The iodine present in the CH3CCl3 layer is due to the iodine added.

ψ



Discussion 1 When chlorine water is added to potassium bromide solution, the colour of the solution changes from colourless to brown because chlorine displaces bromine from an aqueous solution of bromide ions.

oxidation

Cl2(aq) + 2KBr(aq) → 2KCl(aq) + Br2(l)

reduction

4 The half-equations for the reactions are as follows: (a) 2Br–(aq) → Br2(l) + 2e–

Cl2(aq) + 2KBr(aq) → 2KCl(aq) + Br2(l) 2 If 1,1,1-trichloroethane is added to the reaction mixture and shaken, two liquid layers are formed. The lower organic layer (1,1,1-trichloroethane) has a brown colour and shows the presence of bromine. This means that the bromide ions have been oxidised to bromine. 3 Displacement reactions can also be considered in terms of oxidation and reduction. Chlorine oxidises bromide ion (Br–) to bromine and chlorine is itself reduced to chloride ion (Cl–).

(oxidation – loss of electrons) (b) Cl2 (aq) + 2e– → 2Cl–(aq) (reduction – gain of electrons) Chlorine accepts electrons and acts as an oxidising agent. Bromide ion (Br–) loses electrons and acts as a reducing agent. 5 Other displacement reactions that occur in this experiment are 405

Oxidation and Reduction



(b) I2(aq) + KBr(aq) → No reaction I2(aq) + KCl(aq) → No reaction This is because iodine is less reactive than bromine and chlorine.

oxidation Cl2(aq) + 2KI(aq) → 2KCl(aq) + I2(aq)



Conclusion 1 Chlorine displaces bromine from potassium bromide solution and iodine from potassium iodide solution. Bromine displaces iodine from iodide solution but does not displace chlorine from chloride solution. Iodine does not displace chlorine from chloride solution or bromine from bromide solution. 2 The experimental results prove that a more reactive halogen can displace a less reactive halogen from its halide solution. The hypothesis is accepted.

reduction oxidation Br2(l) + 2KI(aq) → 2KBr(aq) + I2(aq)



reduction

6 The following displacement reactions do not occur. (a) Br2(l) + KCl(aq) → No reaction This is because bromine is less reactive than chlorine.

3

Redox Reactions by the Transfer of Electrons at a Distance In Figure 3.9 (Experiment 3.3), the aqueous layer contains KCl but the organic layer (CH3CCl3) does not contain KCl. This is because KCl is an ionic compound which is soluble in water but not in organic solvent.

3

1 If a solution containing an oxidising agent is SPM separated from a solution containing a reducing ’06/P2, ’11/P2 agent by an electrolyte, the redox reaction can still occur by transfer of electrons at a distance. The apparatus set-up is shown in Figure 3.10.

’04

Which of the following equations represent redox reactions? I Cu(s) + Cl2(g) → CuCl2(s) II Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s) III Cu(OH)2(aq) + 2HCl(aq) → CuCl2(aq) + 2H2O(l) IV Cu(NO3)2(aq) + K2CO3(aq) → CuCO3(s) + 2KNO3(aq) A I and II only B III and IV only C I, II, III only­ D I, II, III and IV

Figure 3.10 The apparatus set-up in which the transfer of electrons occurs at a distance

Comments Combustion of metals in chlorine or oxygen are redox reactions (I is correct). Displacement reactions are redox reactions (II is correct). Neutralisation reactions are not redox reactions (III is incorrect). Double decomposition reactions are not redox reactions (IV is incorrect). Answer A

Oxidation and Reduction

2 In Figure 3.10, the electrons are transferred by the connecting wire in the external circuit, from a reducing agent to an oxidising agent. 3 The electrode placed in a solution containing a reducing agent acts as the negative terminal. The reducing agent undergoes oxidation with the loss of electrons. 4 The electrons produced will flow through the connecting wire to the electrode placed in a solution of oxidising agent. This electrode acts as the positive terminal. 406

2 The deflection of the galvanometer needle shows that electrons flow in the external circuit from the carbon electrode immersed in the potassium iodide solution (negative electrode) to the carbon electrode immersed in the acidified potassium manganate(VII) solution (positive electrode). 3 Changes at the negative electrode (a) The colourless layer of potassium iodide solution slowly changes to yellow. (b) The oxidation reaction occurs at the negative electrode.

Redox reactions can occur under two conditions as shown below. Conditions

Experiment

(a) When the oxidising agent and the reducing agent are in contact.

Mixing the oxidising and reducing agents in a test tube.

(b) When the oxidising agent and the reducing agent are not in contact.

Transfer of electrons at a distance in a redox cell.

2I–(aq) → I2(aq) + 2e–

(c) The electrons released during oxidation then flow through the connecting wire (external circuit) to the positive electrode and are accepted by the MnO4– ions. Hence, MnO4– ion acts as the oxidising agent. 4 Changes at the positive electrode (a) The purple layer of potassium man­ ganate(VII) slowly becomes colour­­less (decolourises). (b) The reduction reaction occurs at the positive electrode. MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l) (c) Overall reaction The ionic equation for the redox reaction is shown below.

Reaction between Potassium Iodide and Acidified Potassium Manganate(VII) by Transfer of Electrons at a Distance



1 Figure 3.11 shows the apparatus set-up using acidified potassium manganate(VII) as an oxidising agent and potassium iodide as a reducing agent.

oxidation

2MnO4– + 10I– + 16H+ → 2Mn2+ + 5I2 + 8H2O



reduction

reducing agent oxidising agent

Figure 3.11 Apparatus set-up consisting of KI and KMnO4

407

Oxidation and Reduction

3

5 Sulphuric acid acts as the salt bridge. The functions of a salt bridge are (a) to separate the oxidising agent from the reducing agent, (b) to complete the electric circuit so that ions can move through it. 6 Besides dilute sulphuric acid, the following strong electrolytes can also be used as salt bridges: (a) Sodium or potassium chloride solution (b) Sodium or potassium nitrate solution

3

To study the redox reactions by the transfer of electrons at a distance Apparatus

U-tube, dropper, carbon electrodes, galvanometer, retort stand with clamp and connecting wire with crocodile clips.

A few drops of starch solution are added to the potassium iodide solution. 9 Steps 1 to 6 are again repeated using chlorine and potassium bromide solution.

Materials

Liquid bromine, potassium dichromate(VI) solution, chlorine, iron(II) sulphate solution, potassium iodide solution, potassium bromide solution, sulphuric acid, sodium hydroxide solution and starch solution.

Results 1 Reaction of iron(II) sulphate solution with liquid bromine

Procedure 1 The U-tube is half-filled with dilute sulphuric acid. 2 Using a dropper, iron(II) sulphate solution is added to the left arm of the U-tube until a 2 cm high layer is formed on top of the dilute sulphuric acid. 3 Using the same method, liquid bromine is added to the right arm of the U-tube. 4 Two carbon electrodes are then immersed into the upper layer of the U-tube. The electrodes are connected to the galvanometer by electric wire (Figure 3.12).

Experiment

Observation

(a) Colour change in iron(II) sulphate solution

The colour of the solution changes from light green to yellow.

(b) Colour change in liquid bromine

The colour of solution changes from brown to colourless.

(c) Reaction with sodium hydroxide solution

A brown precipitate insoluble in excess NaOH(aq) is produced.

2 Reaction of potassium iodide solution with acidified potassium dichromate(VI) solution Experiment

reducing agent

oxidising agent

Activity 3.3

Figure 3.12 Oxidation of iron(II) sulphate by liquid bromine

5 The direction of the deflection of the galvanometer needle is observed to determine the direction of the flow of electrons. 6 The apparatus is left aside for a few minutes and the changes that occur at the carbon electrodes are recorded. 7 Using a clean dropper, a small quantity of the iron(II) sulphate solution is removed and placed in a test tube. Sodium hydroxide solution is then added to the iron(II) sulphate solution. 8 Steps 1 to 6 are repeated using acidified potassium dichromate(VI) solution and potassium iodide solution. A small quantity of the potassium iodide solution is removed and placed in a test tube. Oxidation and Reduction

Observation

(a) Colour change in potassium iodide solution

Colour changes from colourless to brown

(b) Colour change in potassium dichromate(VI) solution

Colour changes from orange to green

(c) Reaction with starch solution

Blue-black colour is produced.

3 Reaction of potassium bromide solution with chlorine

408

Experiment

Observation

(a) Colour change in potassium bromide solution

Colour changes from colourless to brown

(b) Colour change in chlorine

Colour changes from greenish-yellow to colourless

Conclusion (a) Bromine oxidises iron(II) sulphate (b) Acidified potassium dichromate(VI) oxidises potassium iodide (c) Chlorine oxidises potassium bromide by transfer of electrons at a distance

oxidation

Discussion 1 Reaction of iron(II) sulphate solution and bromine (a) At the negative terminal/pole/electrode (i) Oxidation of Fe2+ ions to Fe3+ ions occurs and electrons are released.





+2 +3

Fe → Fe + e green brown 2+











3+

Oxidation number of iron increases from +2 to +3. Fe2+ ion is a reducing agent.

(ii) The carbon electrode dipped in iron(II) sulphate solution is negatively-charged because electrons released during the reaction gather at the electrode. (iii) The electrons flow through the electric wire (external circuit) to the positive terminal and are accepted by the bromine molecules. (iv) The colour of the solution changes from pale green to yellow or brown when Fe2+ ions are oxidised to Fe3+ ions. (b) At the positive terminal/pole/electrode (i) Reduction of bromine molecules to bromide ions occurs. Electrons are accepted by the bromine molecules.

0

–1

Br2(aq) + 2e– → 2Br–(aq) brown colourless



–

The oxidation number of bromine decreases from 0 to –1. Bromine is an oxidising agent.

(ii) The colour of the solution changes from brown to colourless when bromine molecules are reduced to bromide ions. (c) Overall reaction The ionic equation for the reaction between bromine and iron(II) sulphate is Br2(aq) + 2Fe2+(aq) → 2Fe3+(aq) + 2Br–(aq)

2 Reaction of potassium iodide and acidified potassium dichromate(VI) (a) At the negative terminal/pole/electrode Oxidation of iodide ions (I–) to iodine occurs.

–1 0

2I–(aq) → I2(aq) + 2e– colourless brown



The oxidation number of iodine increases from –1 to 0. Iodide ion (I–) is a reducing agent.

(b) At the positive terminal/pole/electrode (i) Reduction of dichromate(VI) ions to chromium(III) ions occurs.

reduction +6

+3

Cr2O72–(aq) + 14H+(aq) + 6e– → 2Cr3+(aq) + 7H2O(l) orange green

(ii) The oxidation number of chromium decreases from +6 to +3. (c) Overall reaction Cr2O72–(aq) + 14H+(aq) + 6I–(aq) → 2Cr3+(aq) + 7H2O(l) + 3I2(aq)

3 Reaction of potassium bromide solution and chlorine (a) At the negative terminal/pole/electrode At the negative electrode, bromide ions (Br–) donate electrons and are themselves oxidised to liquid bromine (Br2). The oxidation number of bromine increases from –1 to 0. oxidation –1 0



2Br–(aq) → Br2(l) + 2e– colourless brown

(b) At the positive terminal/pole/electrode At the positive electrode, chlorine molecules, Cl2(aq) accept electrons and are themselves reduced to chloride ions, Cl–(aq).

Cl2(g) + 2e– → 2Cl–(aq) greenish-yellow colourless (c) Overall reaction The overall reaction can be represented by the following ionic equation:

oxidation

Cl2(aq) + 2Br–(aq) → 2Cl–(aq) + Br2(l)

reduction 409

Oxidation and Reduction

3





Oxidation reaction • Addition of oxygen • Removal of hydrogen • Increase in oxidation number • Loss of electrons

Reduction reaction • Removal of oxygen • Addition of hydrogen • Decrease in oxidation number • Gain of electrons

occur simultaneously

3

Redox reactions (a) Combustion of metals in oxygen or chlorine (b) Heating of metallic oxide with carbon (c) Changes of Fe2+ ions to Fe3+ ions and Fe3+ ions to Fe2+ ions (d) Displacement (i) of metal from its salt, (ii) halogen from its halide solution (e) Transfer of electrons at a distance

Redox cell (transfer of electrons at a distance) • The electrode immersed in a reducing agent is the negative electrode • The electrode immersed in an oxidising agent is the positive electrode • Electrons flow from negative electrode to positive electrode

Reducing agent • Brings about reduction in another substance • It is an electron donor • Examples: FeSO4, KBr, KI, C, CO, H2, metals

Oxidising agent • Brings about oxidation in another substance • It is an electron acceptor • Examples: Cl2, Br2, KMnO4, K2Cr2O7

Rules for determining oxidation numbers • The oxidation number of an uncombined element is zero. • The sum of the oxidation numbers of all the atoms in a polyatomic ion is the charge on the ion. • The sum of the oxidation numbers of all the atoms in a compound is zero.

4

’03

Which of the following reagents can be used to convert Fe3+ ions in solution to Fe2+ ions? I Magnesium powder II Potassium iodide solution III Potassium hexacyanoferrate(II) solution IV Acidified potassium manganate(VII) solution A I and II only B II and III only C I, II and III only D I, II, III and IV Comments The conversion of Fe3+ ions to Fe2+ ions is a reduction reaction. Hence, reducing agents are required for the conversion. Potassium hexacyanoferrate(II) is not a reducing agent. (III is incorrect) Acidified potassium manganate(VII) is an oxidising agent. (IV is incorrect) Mg is a more electropositive metal than iron. It

Oxidation and Reduction

reduces Fe3+ to Fe2+ by displacement reaction. (I is correct) reduction 0



+3

+2

+2

Mg(s) + 2Fe3+(aq) → 2Fe2+(aq) + Mg2+(aq)

oxidation

Potassium iodide is a reducing agent. (II is correct) reduction



+3

+2

2KI(aq) + 2Fe3+(aq) → 2Fe2+(aq) + I2(s) + 2K+(aq) Answer A

410

oxidation



3.1 (b) Explain your answer in terms of (i) transfer of electrons, (ii) change in oxidation number. 6 Zinc reacts with iron(III) chloride as represented by the following ionic equation Zn(s) + 2Fe3+(aq) → 2Fe2+(aq) + Zn2+(aq)

2 Calcium reacts with chlorine to form calcium chloride.

(a) Describe what you would see when zinc powder is added to iron(III) chloride solution. (b) Write the half-equations for the chemical changes that take place. (c) Is zinc being oxidised or reduced? Explain your answer in terms of transfer of electrons.

Ca(s) + Cl2(g) → CaCl2(s) (a) Identify the element that has been (i) oxidised and (ii) reduced. (b) Identify the oxidising and reducing agents in this reaction. 3 (a) State the oxidation numbers of the underlined elements in each of the following compounds: (i) CaCrO4 (ii) HNO2 (iii) NaClO3 (iv) Cr2(SO4)3 (v) FeCl3.6H2O (vi) K2MnO4 (b) Give the oxidation numbers of the underlined elements in the following ions: (i) SO42– (iii) ClO – (ii) CuCl43– (iv) NO3– 4 (a) Write the molecular formulae of the following compounds. (i) Copper(I) sulphate (ii) Manganese(II) chloride (iii) Nickel(II) sulphate (b) Write the names of the compounds with the following molecular formulae: (i) CrCl3 (ii) CoO (iii) Fe2(SO4)3 (c) The formulae of two compounds are shown below.

7 (a) Identify the redox reactions from the equations given below. (i) Cl2 + 2KBr → 2KCl + Br2 (ii) 4Fe + 3O2 → 2Fe2O3 (iii) 2NaCl + Pb(NO3)2 → PbCl2 + 2NaNO3 (iv) Mg(OH)2 + H2SO4 → MgSO4 + 2H2O (v) 2FeCl3 + SnCl2 → 2FeCl2 + SnCl4 (b) Write the half-equations for these redox reactions. (c) Identify the oxidising agents and reducing agents in the following reactions. Explain your answer in terms of oxidation number. (i) SnCl2 + 2FeCl3 → 2FeCl2 + SnCl4 (ii) Zn + Pb(NO3)2 → Zn(NO3)2 + Pb (iii) Cl2 + 2KBr → 2KCl + Br2 (iv) 2KI + H2O2 + H2SO4 → I2 + K2SO4 + 2H2O 8 The figure shows an experiment on the transfer of electrons at a distance. ’06

’06

Fe2O3 and Al2O3

(i) State the oxidation numbers for iron and aluminium in the compounds above. (ii) Name both the compounds using IUPAC nomenclature system. (iii) Explain the difference between the names of the two compounds.

(a) Name the oxidising agent and reducing agent in the experiment. (b) Write the half-equations for the reactions that occur at the negative and positive terminals. (c) Based on your answer in (b), describe the oxidation and reduction processes in terms of electron transfer at the negative and positive terminals. (d) State the changes that occur at the negative and positive terminals after 10 minutes.

5 The chemical equation for the combustion of magnesium in air is 2Mg(s) + O2(g) → 2MgO(s) (a) Is magnesium oxidised or reduced?

411

Oxidation and Reduction

3

1 Consider the following reactions and state whether the chemicals underlined have been oxidised or reduced? (a) 2Fe + 3Cl2 → 2FeCl3 (b) Zn + CuSO4 → ZnSO4 + Cu (c) PbO + CO → Pb + CO2 (d) 4NH 3 + 5O2 → 4NO + 6H2O

Rusting as a Redox Reaction

The golden mask of King Tutankhamun was buried in his tomb for more than 3000 years. It is still uncorroded. This is because gold is an unreactive metal and never corrodes.

Conditions for the Rusting of Iron 1 Rusting is a redox reaction between iron, oxygen and water to form a brown substance called rust. Rust is hydrated iron(III) oxide, Fe2O3.xH2O. The composition of water in rust is not constant.

2 Most metals corrode readily in air. When corrosion occurs, the metal surface loses its luster (shine) and becomes dull. This is because metals react slowly with oxygen in the air to from metal oxides on the metal surfaces. 3 Metals that are more electropositive (reactive) will corrode more readily.

4Fe(s) + 3O2(g) + 2xH2O(l) → 2Fe2O3.xH2O(s)

electropositivity of metal decreases

2 Two conditions required for rusting are (a) the presence of air (oxygen) and (b) the presence of water.

3

Corrosion of Metals 1 The corrosion of metals is a redox reaction in which a metal is oxidised spontaneously at room temperature with the release of electrons to form the metal ions. M(s) → Mn+(aq) + ne– metal metal ion

K Na Ca Mg Al Zn Fe Sn Pb Cu Hg Ag Au

tendency for corrosion decreases

3.2

very easy to corrode

very difficult to corrode

reactivity series

4 The higher the position of the metal in the reactivity series, the easier it is for the metal to donate its electrons and be corroded.

Thus, metal M is said to have corroded and the process is known as corrosion of metal. Group 1 metals in the Periodic Table (for example, sodium and potassium) are very reactive and must be kept in paraffin oil to protect them from oxidation by air and water. When sodium or potassium is exposed to the atmosphere, the metals corrode rapidly.

Aluminium corrodes rapidly in air to form a thin layer of aluminium oxide on its surface. This oxide layer is hard, impermeable (does not allow liquid or gas to pass through) and difficult to crack. Thus, the thin layer of aluminium oxide protects the aluminium below it from further corrosion.

Corrosion of metals and positions of metals in the reactivity series

Less reactive metals such as chromium, zinc and nickel also form hard metal oxides that are impermeable to water and air and resistant to cracks. These metal oxide layers can prevent and hence protect the metals from further corrosion. Oxidation and Reduction

412

Unreactive metals (such as gold and platinum) do not corrode because they are resistant to oxidation.

5 The layer of aluminium oxide can be made thicker by electrolysis. This process is called anodising. In industry, anodising is used to protect aluminium from rusting. Anodising is an electrolytic process using aluminium as the anode (positive electrode). During electrolysis (anodising), a layer of aluminium oxide is deposited on the surface of aluminium.

(b) The side of the drop of water This is the area in the metal surface where it is rich in oxygen. This area will act as the positive terminal (cathode). 3 The following stages are involved during the rusting of iron. (a) In the centre of the water droplet (anode) Iron rusts via the oxidation process to form iron(II) ions. Fe(s) → Fe2+(aq) + 2e– … oxidation

• Window and door frames made from anodised aluminium do not require painting because the aluminium oxide layer prevents them from attack by air and water. • Iron is less reactive than aluminium and window frames made from iron must be painted frequently. This is because the hydrated iron(III) oxide (rust) formed during corrosion is permeable to air and water, and can crack easily. Thus, rusting will continue below the rusted surface, until all the metal is ‘eaten up’ by rust.

(b) At the edge of water droplet (cathode) Oxygen accepts electrons from the oxidation of iron and is reduced to hydroxide ions. O2(g) + 2H2O(l) + 4e– → 4OH–(aq) … reduction

Rusting in Terms of Oxidation and Reduction (c) Formation of Fe(OH)2 The Fe2+ and OH– ions in the water droplet combine to form iron(II) hydroxide.

1 Rusting of iron is an electrochemical process SPM that occurs spontaneously. When iron is in ’05/P1 ’06/P1 contact with water, a simple chemical cell ’07/P2 is formed. Figure 3.13 shows the reactions involved in the formation of rust.

Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)

(d) Formation of rust The iron(II) hydroxide produced is oxidised by oxygen to form iron(III) hydroxide, which then decomposes to hydrated iron(III) oxide. oxidation

4Fe(OH)2 + 2H2O + O2 ⎯⎯⎯⎯→ 4Fe(OH)3

decomposition

Fe(OH)3 ⎯⎯⎯⎯⎯⎯→ Fe2O3.xH2O

rust Figure 3.13 Rusting of iron

4 (a) The equations for the redox reactions are shown below.

2 Consider a drop of water on the metal (iron) surface. (a) The centre of a drop of water This is the area in the metal surface where there is a lack of oxygen. This area will act as the negative terminal (anode).

Anode : 2Fe(s) → 2Fe2+(aq) + 4e– … oxidation Cathode : O2(g) + 2H2O(l) + 4e– → 4OH–(aq) … reduction 413

2Fe(s) + O2(g) + 2H2O(l) → 2Fe(OH)2(s) Oxidation and Reduction

3

The electrons flow to the edge of the water droplet through the iron surface.

(b) The overall equation for the rusting of iron is

Prevention of Rusting of Iron 1 The rusting of iron can be prevented if iron is in contact with a more electropositive metal. Conversely, the rate of rusting of iron is increased if the iron is in contact with a less electropositive metal.

3

4Fe(s) + 3O2(g) + 2xH2O(l) → 2Fe2O3.xH2O(s) 5 The rate of rusting of iron is increased if a strong electrolyte (such as salt and acid) is present. Thus, the rusting of iron occurs more rapidly in areas near the sea or in industrial areas. This is because sea air contains salts such as sodium chloride and magnesium chloride. In the industrial area, the air is polluted by acidic gases such as sulphur dioxide and nitrogen dioxide. These substances increase the electrical conductivity of water, thus, making water a better electrolyte. 6 The rate of rusting is also increased if iron is in contact with a metal less electropositive than iron. For example, if iron is in contact with copper (a metal less electropositive than iron), the rate of rusting is increased. This process is known as electrochemical corrosion. 7 Besides the corrosion of iron and steel, corrosion of other metals can also occur. The main causes of corrosion of metals are attack by chemicals, such as acids, damp air or electrochemical corrosion.

Iron does not corrode if it is in contact with Zn, Al or Mg (Refer to Experiment 3.4)

K

Na

Mg

Al

Zn

Fe

Iron corrodes rapidly if it is in contact with Sn, Pb or Cu

Sn

Pb

Cu

Electropositivity decreases

2 When two metals are in contact, the greater the difference in electropositivity between these two metals, the faster the more electropositive metal will rust. For example, a piece of iron joined to copper will corrode more rapidly than a piece of iron joined to tin. 3 Experiment 3.4 shows the experimental set-up for the study of the effect of other metals on the rate of rusting of iron.

3.4

Experiment 3.4

SPM To investigate the effect of other metals with different electropositivity on the ’04/P2 rusting of iron (a) Manipulated variable : Different metals used to Problem statement wrap around iron nails What is the effect of other metals with different (b) Responding variable : Colour change in the electropositivity on the rusting of iron? gelatin solution Hypothesis (c) Constant variable : Iron nails (a) A metal more electropositive than iron will Apparatus Test tubes protect iron from rusting. (b) A metal less electropositive than iron will Materials Iron nails, magnesium, zinc, tin increase the rate of rusting. and copper foils, gelatin, potassium hexacyanoferrate(III), phenolphthalein Variables indicator and sandpaper.

Procedure 1 Five pieces of iron nails are cleaned using sandpaper. 2 The first clean iron nail is placed in test tube A. 3 Strips of magnesium (Mg), zinc (Zn), tin (Sn) and copper (Cu) foils are cleaned with sandpaper. 4 Each iron nail is wrapped with a different metal foil and placed in test tubes B, C, D and E respectively.

Figure 3.14 Effect of contact with other metals on the rusting of iron Oxidation and Reduction

414

5 A solution of gelatin in hot water is prepared. A few drops of potassium hexacyanoferrate(III) solution, K3Fe(CN)6 and phenolphthalein indicator are added to the hot gelatin solution. 6 The mixture is stirred and then poured into each of the test tubes (Figure 3.14). 7 The test tubes are set aside for three days and then examined. The observations are recorded in the table below.

Test tube

A

B

C

D

E

Fe only

Fe + Mg

Fe + Zn

Fe + Sn

Fe + Cu

Intensity of blue colour

Low

None

None

High

High

Intensity of pink colour

None

High

High

Low

Low

Gas bubbles

None

Plenty

Plenty

Few

Few

Observation Metal

(c) At the anode: Magnesium foil and zinc foil The oxidation of magnesium to Mg2+ ions and the oxidation of zinc to Zn2+ ions occur. Electrons are released.

Discussion 1 (a) Potassium hexacyanoferrate(III) is used to detect the Fe2+ ions. Potassium hexacyanoferrate(III) produces a dark blue colour in the presence of Fe2+ ions. (b) Phenolphthalein is used to detect the OH– ions. Phenolphthalein produces a pink colour in the presence of OH– ions. (c) The bubbles of gas produced are hydrogen gas. 2 Reactions in test tube A (Fe only) (a) Test tube A is used as a control to study the effect of other metals on the rusting of iron. (b) In the presence of water and oxygen, rusting of iron occurs to produce iron(II) ions (Fe2+) and hydroxide ions (OH–).

oxidation Mg(s) ⎯⎯⎯⎯→ Mg2+(aq) + 2e– oxidation Zn(s) ⎯⎯⎯⎯→ Zn2+(aq) + 2e– Thus, magnesium and zinc are corroded instead of iron. The electrons released then flow to the iron nail and prevent it from forming Fe2+ ions, that is, prevent the rusting of iron. (d) At the cathode: Iron nails Water molecules dissociate to form hydrogen ions (H+) and hydroxide ions (OH–). H2O(l)

oxidation Fe(s) ⎯⎯⎯⎯→ Fe2+(aq) + 2e–

H+(aq) + OH–(aq)

Hydrogen ions accept electrons and are reduced to hydrogen gas.

reduction O2(g) + 2H2O(l) + 4e– ⎯⎯⎯⎯→ 4OH–(aq)

reduction 2H+(aq) + 2e– ⎯⎯⎯⎯→ H2(g)

(c) Fe ions react with potassium hexacyano­ ferrate(III) to produce a deep blue precipitate. (d) Phenolphthalein does not produce a pink colour because the OH– ions produced react with Fe2+ ions to form Fe(OH)2. 3 Reactions in test tubes B (Fe + Mg) and C (Fe + Zn) (a) Deep blue colour does not appear in test tubes B and C. This implies that Fe2+ ions are not produced, that is, the iron does not rust. (b) Magnesium and zinc are more electropositive than iron. Thus, magnesium and zinc act as the negative terminal (electrode) and iron acts as the positive terminal (electrode). 2+

(e) When hydrogen ions are discharged to form hydrogen gas, the concentration of hydroxide ions in water increases. Consequently, the area around the iron nail becomes alkaline and causes the colour of phenolphthalein to change from colourless to pink. 4 Reactions in test tubes D (Fe + Sn) and E (Fe + Cu) (a) Deep blue colour appears. This implies that Fe2+ ions are produced, that is, rusting of iron nails has occurred. The high intensity of the blue colour shows that the rusting of iron nail is speeded up. 415

Oxidation and Reduction

3

Results

(b) Iron is more electropositive than tin and copper. Hence, iron has a greater tendency to lose electrons and acts as the negative terminal (anode). Tin and copper, on the other hand, act as the positive terminal (cathode). (c) At the anode: Iron nail Oxidation of iron to Fe2+ ions occurs and electrons are released. oxidation Fe(s) ⎯⎯⎯⎯→ Fe2+(aq) + 2e–

Consequently, tin foil and copper foil do not corrode. (e) When hydrogen ions are discharged, the concentration of hydroxide ions (OH–) in water increases. However, most of the OH– ions produced will combine with Fe2+ ions to form rust. Consequently, the area around tin foil or copper foil is slightly alkaline and a slight pink colour is observed.

(d) At the cathode: Tin foil or copper foil Electrons released by the iron nail will flow to the tin or copper foil which acts as the positive terminal. Hydrogen ions from water will then accept these electrons and are themselves reduced to hydrogen gas. reduction 2H+(aq) + 2e– ⎯⎯⎯⎯→ H2(g)

Conclusion 1 The rusting of iron can be prevented if iron is in contact with more electropositive metals such as magnesium or zinc. 2 The rusting of iron is speeded up if iron is in contact with less electropositive metals such as tin or copper. The hypothesis is accepted. (ii) Another reason is that tin plating makes the articles shiny and more attractive in appearance. (iii) However, there is one disadvantage in tin plating because tin is less electro­positive than iron. If the tin coating is broken, the iron beneath it will rust even more rapidly because iron is more electropositive than tin. (b) Plating iron with chromium (i) Chromium is a metal that is resistant to rusting. When chromium is exposed to water and air, an impermeable, nonbrittle oxide layer is formed. (ii) The oxide layer acts as a protective layer to prevent iron beneath it from coming into contact with water and air in the atmosphere. (iii) Car bumpers, bicycle handles and pipes are chromium-plated to prevent corrosion. 3 Using more electropositive (reactive) metals SPM (a) Galvanising is the coating of iron or steel ’10/P1 with zinc for protection from corrosion. Galvanising is carried out by dipping the iron object into molten zinc or by electro­ plating. Zinc-coated iron is known as galvanised iron. (i) Even if the layer of zinc is scratched or broken, the iron beneath it does not rust. This is because zinc is more electro­ positive than iron and will corrode first.

3



Zinc is not used to coat food cans although zinc can prevent the rusting of iron. This is because zinc is a poisonous substance. Zinc is more reactive than tin and is susceptible to attack by acids such as fruit juices. Food cans are usually electroplated with tin because tin is non-poisonous. Furthermore, tin is resistant to oxidation by oxygen and water. Tin can prevent iron from rusting as long as the tin surface is not cracked.

Methods Used for the Prevention of Rusting

SPM

’11/P1

1 Using a protective layer (a) Rusting of iron and steel can be prevented by keeping them away from air (oxygen) and water. (b) A layer of paint, oil, grease or plastic coating protects the iron surface from coming into contact with air and water. Without the presence of both air and water, rusting of iron cannot occur. (c) Oil and grease are usually used for movable machine parts. Plastics are usually used for household articles such as flower pot stands, coat hangers and so on. Paints are used for bigger objects such as car bodies, fences, gates and bridges. 2 Using less electropositive metals (a) Plating iron with tin (i) Tin is a not an electropositive metal and is resistant to oxidation by water and air. Hence, iron plated with tin is used in making food cans. Oxidation and Reduction

Zn(s) → Zn2+(aq) + 2e– ... oxidation 416





(ii) This method of rust prevention is called cathodic protection. The metal zinc is known as sacrificial metal because zinc is sacrificed in the protection of iron from rusting. (iii) Galvanised iron are used for making zinc roof and gutter. (b) Rusting in ships is prevented by fixing bars of zinc to the part of the ship submerged in water (Figure 3.15). Zinc is oxidised (corroded) in preference to iron.

Magnesium is more electropositive than iron and will be corroded in preference to iron.

4 Using alloys (a) The best known rust-resistant alloy of iron is stainless steel. Stainless steel contains 10–20% nickel and 10–25% chromium. When exposed to the air, a hard layer of chromium(III) oxide is formed on the surface of iron and prevents the iron from rusting. (b) Stainless steel is used to make surgical instruments and kitchen wares such as knives, forks and spoons.

Figure 3.15 Using a sacrificial metal, zinc, to prevent the rusting of ships

(c) Rusting in underground iron pipes is prevented by having blocks of magnesium attached to the iron pipes (Figure 3.16).

Conditions for rusting of iron • Presence of air (oxygen) • Presence of water

Corrosion of metals and rusting of iron • Corrosion of metals is a redox reaction in which a metal is oxidised spontaneously by losing electrons to form metal ions. • When iron corrodes, the process is called rusting.

Methods for rust prevention • Using a protective layer such as paint, oil, grease, or plastic covering • Coating/plating iron with tin • Coating/plating iron with chromium • Using sacrificial metals • Using alloys

Explaining the rusting of iron • Corrosion of metals and rusting of iron are redox reactions. • At the anode (negative electrode), iron is corroded. Fe(s) → Fe2+(aq) + 2e– … oxidation • At the cathode (positive electrode), OH– ions are produced. O2 + 2H2O + 4e– → 4OH– … reduction • Fe2+ combines with OH– to form Fe(OH)2. • Fe(OH)2 is oxidised to Fe2O3.xH2O (rust).

417

Oxidation and Reduction

3

Figure 3.16 Using the sacrificial metal (magnesium blocks) to protect the underground pipes

4 Oxygen used for burning the metals is supplied by heating potassium manganate(VII), potassium nitrate or a mixture of potassium chlorate(V) and manganese(IV) oxide (catalyst). (a) 2KMnO4(s) ⎯⎯→ K2MnO4(s) + MnO2(s) heat + O2(g) (b) 2KNO3(s) ⎯⎯→ 2KNO2(s) + O2(g)

3.2 1 State the (a) physical changes, (b) chemical changes that occur when iron corrodes. 2 (a) Write the chemical equation for the rusting of iron. Assume the formula of rust as Fe2O3.H2O. (b) Explain why a layer of grease applied to an iron object will prevent iron from rusting.

heat MnO2

3 (a) What is meant by galvanised iron? (b) Explain why galvanised iron does not rust even though its surface is scratched.

(c) 2KClO (s) ⎯⎯→ 2KCl(s) + 3O2(g)

3 heat

Reactivity Series of Metals

4 Figure 3.17 shows the arrangement of apparatus to study the effect of magnesium on the corrosion of zinc.

1 The reactivity series is a list of metals arranged according to their chemical reactivity with oxygen. 2 Metals high in the reactivity series are very reactive. These metals react vigorously with oxygen. Potassium is the most reactive of these metals. It is placed at the top of the series. 3 Metals that are less reactive are placed at the lower part of the series. These metals react slowly with oxygen. Gold is the most unreactive of these metals. It is placed at the bottom of the series. 4 Figure 3.18 shows the reactivity series of metals that do not include hydrogen and carbon.

3

Figure 3.17 (a) Write the half-equations for the reactions that occur at (i) the zinc rod, (ii) the magnesium foil. (b) Explain the effect of magnesium on the corrosion of zinc.

3.3

The Reactivity Series of Metals and Its Applications

Reactivity of Metals with Oxygen 1 Most metals form metal oxides when heated SPM or burnt in air. For example,

’05/P2 /SA

oxidation





0

0

Figure 3.18 The reactivity series of metals +1 –2

The reactivity series of metals that includes hydrogen and carbon is shown in Figure 3.24. 5 The position of a metal in the reactivity series SPM can also be determined by the reaction between ’05/P1 the metal and the oxide of another metal (Figure 3.19).

4Na(s) + O2(g) ⎯⎯⎯→ 2Na2O(s)

reducing oxidising agent agent reduction

2 Different metals have different reactivity with oxygen. Reactive metals have strong affinity for oxygen. Hence, reactive metals will burn rapidly and vigorously in oxygen. Conversely, less reactive metals will react slowly with oxygen. 3 The reactivity of metals with oxygen can be compared by observing the flame or the glow produced when a metal is heated in oxygen. The more reactive the metal is with oxygen, the more brightly and rapidly the metal burns. Oxidation and Reduction

Figure 3.19 Heating a mixture of a metal and the oxide of another metal

418

8 Magnesium is more reactive than copper. Thus, magnesium can displace copper from copper(II) oxide. Conversely, copper is less reactive than magnesium. Thus, copper cannot displace magnesium from magnesium oxide. 9 The displacement reaction can be considered in terms of oxidation and reduction. For example,

more reactive

less reactive

→



→

X + oxide of Y ⎯⎯→ oxide of X + Y



7 For example, when a mixture of copper(II) oxide and magnesium is heated, the following reaction occurs because magnesium is more reactive than copper.



Zn(s) + PbO(s) ⎯⎯→ ZnO(s) + Pb(s) heat



more reactive less reactive

reduction

→

10 In the reaction between zinc and lead(II) oxide, zinc acts as a reducing agent and reduces lead(II) oxide to lead. Conversely, lead(II) oxide acts as an oxidising agent and oxidises zinc to zinc oxide. 11 The reactivity series can be used to predict the reaction between a metal and the oxide of another metal. A more reactive metal will displace a less reactive metal from its oxide.

Mg(s) + CuO(s) ⎯⎯→ MgO(s) + Cu(s)

→





heat

Conversely, when a mixture of magnesium oxide and copper is heated, reaction does not occur because copper is less reactive than magnesium. MgO(s) + Cu(s) → No reaction

6 Test 2 Metal R has no reaction with the oxide of P. Thus, R is less reactive than P.

When a mixture of the oxide of a metal P and the powdered metal Q is heated, there is a glow in the mixture. When the experiment is repeated by using metal R as a substitute for metal Q, no change occur. From these observations, arrange the reactivity of P, Q and R in ascending order.

Conclusion The reactivity of P, Q and R in ascending order is R Fe > Pb > Cu (Very reactive) unreactive) Reactivity decreases The hypothesis is accepted.

1 Burning of magnesium 2Mg(s) + O2(g) → 2MgO(s) Magnesium oxide is white in colour. 2 Burning of zinc 2Zn(s) + O2(g) → 2ZnO(s) Zinc oxide is yellow when hot and white when cold. 3 Burning of iron 4Fe(s) + 3O2(g) → 2Fe2O3(s) Iron(III) oxide is reddish-brown. 4 Burning of lead 2Pb(s) + O2 → 2PbO(s) Lead(II) oxide is brown when hot and yellow when cold. 5 Burning of copper 2Cu(s) + O2(g) → 2CuO(s) Copper(II) oxide is black.

Discussion (A) The intensity of flame/glow or the vigour of reaction 1 Magnesium burns very rapidly in oxygen and produces a very bright white flame. This shows that magnesium is very reactive. In contrast, copper only glows weakly when it reacts with oxygen. This shows that copper is very unreactive. 2 Comparing zinc with iron and lead, zinc burns rapidly and produces a bright glow. This shows that zinc is more reactive, compared with iron and lead. 3 Iron burns less rapidly and lead burns slowly. Thus, iron is more reactive than lead.

Oxidation and Reduction

420

agent and the oxide of metal X is reduced to metal X.

The Position of Carbon in the Reactivity Series of Metals

reduction

Heating Carbon with Metal Oxides

Carbon + oxide of metal X ⎯⎯→ metal X reducing + agent carbon dioxide

1 The position of carbon in the reactivity series can be determined by heating carbon with metal oxides. 2 When a mixture of carbon and the oxide of metal X is heated strongly, a reaction will occur if carbon is more reactive than the metal X. In this reaction, carbon acts as the reducing



oxidation

3 Conversely, if carbon does not remove oxygen from a metal oxide, this means that carbon is less reactive than the metal in the oxide.

3.6

Problem statement How is the position of carbon in the reactivity series of metals determined? Hypothesis (a) A reaction will occur if carbon is more reactive than the metal. (b) A reaction will not occur if carbon is less reactive than the metal. (c) Carbon is placed between aluminium and zinc in the reactivity series of metals. Apparatus Spatula, asbestos paper, wire gauze, tripod stand and Bunsen burner. Materials Powdered carbon, powdered zinc oxide, copper(II) oxide and aluminium oxide.

Variables (a) Manipulated variable : Type of metal oxide (b) Responding variable : Intensity of flame (c) Constant variable : Carbon powder

Procedure 1 Two spatulas of carbon powder are placed on a piece of asbestos paper. 2 One spatula of zinc oxide is added to the carbon powder. The zinc oxide and carbon powder are mixed uniformly. 3 The asbestos paper with its contents is placed on a wire gauze over a tripod stand. The mixture of

zinc oxide and carbon is heated strongly for a few seconds (Figure 3.21). 4 After this, the Bunsen flame is removed and the mixture examined to determine whether it will continue to glow. 5 Steps 1 to 4 are repeated using a mixture of carbon and (a) copper(II) oxide, (b) aluminium oxide.

3

To determine the position of carbon in the reactivity series of metals

Figure 3.21 The reaction between carbon and the metal oxide

Mixture

Observation

Reactivity of carbon

(a) C + ZnO

• The reaction mixture glows brightly. • A grey solid is formed.

Carbon is more reactive than zinc.

(b) C + CuO

• The reaction mixture burns with a bright flame. • A brown solid is obtained.

Carbon is more reactive than copper.

(c) C + Al2O3

• No visible change

Carbon is less reactive than aluminium. 421

Oxidation and Reduction

Experiment 3.6

Results

2 Carbon does not react with aluminium oxide. Thus, carbon is less reactive than aluminium.

Discussion 1 Carbon has reduced zinc oxide and copper(II) oxide to their respective metals.

C(s) + 2ZnO(s) ⎯⎯→ 2Zn(s) + CO2(g) Δ white grey



C(s) + 2CuO(s) ⎯⎯→ 2Cu(s) + CO2(g) Δ black brown



Conclusion The position of carbon is between aluminium and zinc. Thus, the position of carbon in the reactivity series is K > Ca > Mg > Al > C > Zn > Cu > Hg > Ag > Au

(Note: the symbol ‘Δ’ denotes heating.) This means that carbon is more reactive than copper and zinc.

Reactivity decreases

The hypothesis is accepted.

Heating Carbon Dioxide with Metals

3

C(s) + Al2O3(s) ⎯⎯→ No reaction Δ

4 When a piece of burning magnesium ribbon is put into carbon dioxide in a gas jar, the magnesium will continue to burn for a short time. Black specks of carbon can be seen on the sides of gas jar and magnesium burns to form a white powder (magnesium oxide).

1 The ability of a metal to remove oxygen from carbon dioxide can be used to determine the position of carbon in the reactivity series. 2 Sodium, potassium, calcium, magnesium and aluminium are more reactive than carbon. These metals will therefore react with carbon dioxide and remove oxygen from carbon dioxide.

2Mg(s) + CO2(g) → 2MgO(s) + C(s) 5 This reaction shows that magnesium (a) is more reactive than carbon with oxygen, (b) acts as the reducing agent, (c) reduces carbon dioxide to carbon, (d) is itself oxidised to magnesium oxide.

Metal + carbon dioxide → metal oxide + carbon 3 Figure 3.22 shows the arrangement of apparatus used to investigate the reaction between magnesium and carbon dioxide.





oxidation

2Mg(s) + CO2(g) → 2MgO(s) + C(s)

reducing agent reduction

6 The higher the element is in the reactivity series, the stronger it acts as a reducing agent in the redox reaction. 7 Conversely, if a metal does not remove oxygen from carbon dioxide, it implies that the metal is less reactive than carbon.

Figure 3.22 The burning of magnesium in carbon dioxide gas

5

’05

When powdered metal X is heated with the black oxide of metal Y, • a glow is seen, • the residue produced is yellow when it is hot and white when it is cold. Based on the above observations, it can be deduced that

Oxidation and Reduction

422

A metal Y can displace metal X from its salt solution. B metal X can react with magnesium oxide when heated. C the oxide of metal X can react with iron powder when heated. D the oxide of metal Y can react with carbon powder when heated.

Comments The residue is zinc oxide, which is yellow when hot and white when cold. Metal Y is less reactive than metal X. It cannot displace X from its salt solution. Metal X (zinc) is less reactive than magnesium. It

cannot react with heated magnesium oxide. Iron is less reactive than zinc. It cannot react with oxide of zinc when heated. The oxide of metal Y is copper(II) oxide (black). Carbon can reduce copper(II) oxide to copper. Answer D spread throughout the metal oxide and the metal is produced. 4 If hydrogen gas does not remove oxygen from the metal oxide, hydrogen is less reactive with oxygen than the metal. 5 Hydrogen used for reducing metal oxides to metals can be produced from the reaction between dilute sulphuric acid or dilute hydrochloric acid and zinc.

The Position of Hydrogen in the Reactivity Series of Metals 1 The position of hydrogen in the reactivity series SPM can be determined by passing dry hydrogen gas ’05/P1 over hot metal oxides. 2 If hydrogen is more reactive than metal X, hydrogen will reduce the oxide of metal X to metal X. reduction

Hydrogen + oxide of metal X → water + metal X

Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

reducing agent

3





oxidation

6 Hydrogen gas is dried by passing it through a drying agent such as concentrated sulphuric acid or anhydrous calcium chloride.

3 If the reaction between hydrogen gas and a metal oxide occurs, a flame or a glow will

3.7

Problem statement How is the position of hydrogen in the reactivity series of metals determined?

Safety precautions A mixture of hydrogen and air will explode when ignited. The following steps must be taken before the hydrogen is ignited. 1 Make sure that all the stoppers are fitted tightly so that there is no leakage. 2 Make sure that hydrogen gas flows through the apparatus continuously. 3 Make sure that all the air in the apparatus is removed before igniting hydrogen at the small hole in the combustion tube. This can be carried out as follows: (a) A sample of gas is collected from the small hole in the combustion tube. (b) The gas is then tested with a lighted wooden splint. If a ‘pop’ sound is heard, the test is repeated until no ‘pop’ sound is heard on ignition of the gas. 4 The combustion tube must be tilted downwards to prevent the flow of water formed back to the hot part of the tube, thus causing the combustion tube to crack.

Hypothesis Hydrogen is placed between zinc and iron in the reactivity series of metals. Variables (a) Manipulated variable : Different types of metal oxides (b) Responding variable : Intensity of flame (c) Constant variable : Hydrogen gas Apparatus A combustion tube with a small hole, roundbottomed flask, U-tube, delivery tube, retort stand with a clamp and Bunsen burner. Materials Zinc, dilute sulphuric acid, copper(II) oxide, lead(II) oxide, iron(III) oxide and zinc oxide. 423

Oxidation and Reduction

Experiment 3.7

To determine the position of hydrogen in the reactivity series of metals

round- bottomed flask. It is dried by passing through anhydrous calcium chloride in a U-tube. 3 The dry hydrogen gas is passed through the combustion tube to displace all the air in the apparatus. 4 When all the air in the combustion tube has been removed, the hydrogen gas coming out from the small hole of the test tube is ignited. 5 Copper(II) oxide is heated strongly while hydrogen gas is passed over it until the reaction is complete, that is, until no further changes in colour is observed. 6 Hydrogen gas is allowed to continue to flow through, in order to cool the apparatus and to prevent air from entering the combustion tube. 7 Steps 1 to 6 are repeated using (a) lead(II) oxide, (b) iron(III) oxide, (c) zinc oxide.

Procedure

Figure 3.23 Reaction between hydrogen and metal oxides

3

1 A small quantity of copper(II) oxide is put on an asbestos paper which is then placed in a combustion tube with a small hole at one end of the tube. 2 The hydrogen gas is produced by the reaction between zinc and dilute sulphuric acid in a Results Mixture

Observation

Reactivity of hydrogen

(a) H2 + CuO

A bright flame is produced. The black powder changes to brown.

Hydrogen is more reactive than copper.

(b) H2 + PbO

A bright flame is produced. The yellow powder changes to grey.

Hydrogen is more reactive than lead.

(c) H2 + Fe2O3

A bright glow spreads over iron(III) oxide. The brown powder changes to grey.

Hydrogen is more reactive than iron.

(d) H2 + ZnO

No glow is observed. The white powder becomes yellow when heated and white when cold.

Hydrogen is less reactive than zinc.

2 Hydrogen does not react with zinc oxide. Hence, hydrogen is less reactive than zinc.

Discussion 1 Hydrogen has reduced copper(II) oxide, lead(II) oxide and iron(III) oxide to their respective metals and is itself oxidised to water.

ZnO(s) + H2(g) ⎯⎯→ No reaction

yellow when hot

CuO(s) + H2(g) → Cu(s) + H2O(l) black brown

PbO(s) + H2(g) → Pb(s) + H2O(l) yellow grey

Conclusion Hydrogen is placed above iron but below zinc in the reactivity series of metals. That is,

Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(l)

Zn > H > Fe > Pb > Cu

brown grey

Reactivity decreases

This means that hydrogen is more reactive than copper, lead and iron.

Oxidation and Reduction

The hypothesis is accepted.

424

Positions of Carbon and Hydrogen in the Reactivity Series 1 The reactivity series that includes both carbon and hydrogen is shown in Figure 3.24. SPM

’07/P1

Figure 3.24 The reactivity series that includes both carbon and hydrogen

3

2 Table 3.11 summarises the reactions between (a) metal oxides and carbon, (b) metal oxides and hydrogen. Table 3.11 Reduction of metal oxides with carbon and hydrogen

Metal oxide

Reaction with carbon

Reaction with hydrogen

Potassium oxide (K2O) Sodium oxide (Na2O) Calcium oxide (CaO) Magnesium oxide (MgO) Aluminium oxide (Al2O3)

No reaction

No reaction

Zinc oxide, ZnO

2ZnO + C → 2Zn + CO2

Iron(III) oxide (Fe2O3) Tin(IV) oxide (SnO2) Lead(II) oxide (PbO) Copper(II) oxide (CuO) Silver(I) oxide (Ag2O)

2Fe2O3 + 3C → 4Fe + 3CO2 SnO2 + C → Sn + CO2 2PbO + C → 2Pb + CO2 2CuO + C → 2Cu + CO2 2Ag2O + C → 4Ag + CO2

No reaction Fe2O3 + 3H2 → 2Fe + 3H2O SnO2 + 2H2 → Sn + 2H2O PbO + H2 → Pb + H2O CuO + H2 → Cu + H2O Ag2O + H2 → 2Ag + H2O

In general, the electrochemical series is the same as the reactivity series. However, the position of hydrogen in these two series are different. In the electrochemical series, hydrogen is placed between lead and copper. In the reactivity series, the position of hydrogen is between zinc and iron.

1 The position of a metal in the reactivity series can be used to predict the ability of a metal to react with water. 2 Metals that are more reactive than hydrogen will reduce water to hydrogen. The higher the position of the metal in the reactivity series, the faster and more vigorous the metal will react with water. Conversely, metals that are less reactive than hydrogen do not react with water. 3 Hence, potassium and sodium react vigorously with cold water and magnesium reacts with steam but not cold water. Lead and copper do not react with water.

Extraction of Metals from Their Ores 1 Most metals in metal ores exist in the forms of oxides, carbonates and sulphides in the Earth’s crust. 425

Oxidation and Reduction

2 The extraction of metals involves the reduction of metal ores to metals. 3 Two main methods are used to extract metals from their ores. (a) Electrolysis of metal compounds in the molten state. (b) Reduction of metal oxides by carbon. 4 The important factor for determining the most suitable method in the extraction of metals is the position of the metal in the reactivity series. 5 For metals lower than carbon in the reactivity series of metals, the method used is the reduction of metal oxides by carbon. 6 For metals higher than carbon in the reactivity series, the extraction of metals must be carried out by the electrolysis of molten metal compounds. 7 Table 3.12 shows the methods used to extract some metals.

• Modern blast furnaces are very tall (50 – 70 m high). They are made of steel and lined with fireproof bricks. • Coke is a form of carbon made by heating coal in the absence of air. The process is called destructive distillation.

3

1 The important iron ores are haematite and magnetite. Haematite contains iron(III) oxide, Fe2O3 whereas magnetite contains triiron tetroxide, Fe3O4. 2 The extraction of iron from haematite or magnetite is carried out in the blast furnace (Figure 3.25) by reduction using carbon.

Table 3.12 Methods of metal extraction according to the position of the metal in the reactivity series

Metal

Method of extraction

Potassium, K Sodium, Na Calcium, Ca Magnesium, Mg

Electrolysis of metal chlorides in the molten state

Aluminium, Al

Electrolysis of Al2O3 in the molten state

Zinc, Zn Iron, Fe Tin, Sn Lead, Pb

Heating metal oxides with carbon

Copper, Cu Mercury, Hg

Heating metal sulphides in air

Silver, Ag Gold, Au

Exist as free elements in Earth’s crust

Figure 3.25 The blast furnace

(a) Raw materials required A mixture of iron are, coke (carbon) and limestone (calcium carbonate) is put in a blast furnace. Hot air is blown into the furnace from the bottom. (b) Production of carbon dioxide In the lower section of the blast furnace, the oxygen in hot air reacts with coke to form carbon dioxide.

Extraction of Iron from Its Ore

C(s) + O2(g) → CO2(g) At high temperatures, limestone decomposes into quicklime (calcium oxide, CaO) and carbon dioxide. CaCO3(s) → CaO(s) + CO2(g)

Molten iron flowing out from the blast furnace Oxidation and Reduction

426

(c) Production of carbon monoxide In the upper section of the blast furnace, carbon dioxide reacts with coke to produce carbon monoxide.

Waste gases • Limestone • Iron ore • Coke

C(s) + CO2(g) → 2CO(g)

Slag Blast furnace Molten iron Hot air

(d) Reduction of iron ore to iron • In the upper section of the blast furnace, where the temperature is about 400 – 800 °C, the iron ore is reduced by carbon monoxide to iron.

(b) The iron produced in this process is not pure iron and contains about 5% carbon. The iron is called cast iron.

Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

• Cast iron is hard and brittle. Cast iron is used for making the base of Bunsen burner, lamp posts or pipes. • Cast iron is usually converted into other more useful forms such as steel in an oxygen furnace.

haematite

Fe3O4(s) + 4CO(g) → 3Fe(l) + 4CO2(g) magnetite

3

Thermite Process

• In the lower section of the blast furnace, the iron ore is reduced by coke (carbon) to iron.

1 The thermite process is a displacement reaction between aluminium and iron(III) oxide to produce iron.

Fe2O3(s) + 3C(s) → 2Fe(l) + 3CO(g)

2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l)

Fe3O4(s) + 2C(s) → 3Fe(l) + 2CO2(g)

Thus, the thermite process can be used for the small-scale extraction of iron. 2 The thermite process can be carried out in the school laboratory by using the apparatus as shown in Figure 3.26. Magnesium acts as the fuse to ignite the mixture for this experiment.

• In these reactions, carbon and carbon monoxide act as the reducing agents. • The molten iron produced flows to the bottom of the blast furnace and is collected. The molten iron is poured into moulds and set aside to solidify. 3 Removal of impurities (a) In the blast furnace, calcium oxide is produced from the decomposition of limestone. It then reacts with silica (sand) to form slag (calcium silicate). CaO(s) + SiO2(s) → CaSiO3(s)

Figure 3.26 Thermite process

3 When the mixture of magnesium powder and barium peroxide (BaO2) burns, a large amount of heat is produced to initiate the thermite process to produce molten iron. 4 We can also consider the thermite process as a redox reaction:

(b) Molten slag floats on top of iron. The slag and iron are separated through a tap at the bottom of the furnace. (c) In this reaction, calcium oxide acts as a basic oxide, whereas silica acts as an acidic oxide. (d) The slag produced during the extraction of iron is used mainly for road surfacing. 4 (a) The extraction of iron can be summarised in the flowchart as follows.



oxidation

2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l) reduction

427

Oxidation and Reduction

5 The thermite process is highly exothermic, that is, it gives out a lot of heat during the reaction. It is used for welding steel objects, for example, railway lines.

(c) The carbon monoxide produced can also reduce tin(IV) oxide to tin. SnO2(s) + 2CO(g) → Sn(l) + 2CO2(g) (d) The molten tin is then tapped off and poured into a mould and solidified into ingots. 5 The extraction of tin can be summarised in the flowchart below.

Coke is a form of carbon. It is the solid substance left after destructive distillation of coal. The term ‘destructive distillation’ means heating a substance in the absence of air. Coke contains more than 80% of carbon. Coke is used as a reducing agent in the extraction of metals and as a non-smoky fuel.

tin ore water + oil

3

Extraction of Tin from Its Ore

soil

1 The most important tin ore is cassiterite. Cassiterite contains tin(IV) oxide, SnO2 and unwanted materials such as sand, soil, oil, sulphur and carbon. 2 The two main steps involved in the extraction of tin are (a) concentration process, (b) reduction process. 3 Concentration process (a) At the first stage of tin extraction, the tin ore is concentrated by froth floatation method. In this process, the tin ore is crushed to a fine powder and mixed with water and special oils (known as frothing agents) in a large tank. (b) The mixture is agitated by blowing air to form a froth. The unwanted materials sink to the bottom of the tank. (c) The froth contains particles of concentrated tin ore and floats to the top of the tank where it is removed. The concentrated tin ore is then dried and roasted to remove impurities such as carbon, sulphur and special oils. 4 Reduction process (a) The concentrated tin ore is mixed with coke (a form of carbon). The mixture is heated to a high temperature (about 1360 °C) in a furnace. (b) During heating, tin(IV) oxide is reduced by carbon to molten tin and carbon is oxidised to carbon dioxide and carbon monoxide.

sand

concentrated tin ore

carbon roasting sulphur furnace + coke

carbon dioxide carbon monoxide

tin ingots The Use of Carbon as the Main Reducing Agent in Metal Extraction The main reasons for using carbon as the main reducing agent in metal extraction are: 1 Chemical reason Carbon is more reactive than zinc, iron, tin and lead. Therefore, carbon can easily reduce the oxides of these metals. 2 Economic reason Carbon is cheap and can be obtained easily. Reduction of metal ores using coke (carbon) is cheaper than using electricity for the electrolysis of molten ores. 3 Environmental reason The carbon dioxide gas produced during metal extraction is non-poisonous and does not pollute the atmosphere.

SnO2(s) + C(s) → Sn(l) + CO2(g) SnO2(s) + 2C(s) → Sn(l) + 2CO(g) Oxidation and Reduction

froth floatation

428

Reactions of metals with oxygen The more rapidly the metal burns in oxygen and the brighter the flame produced, the more reactive the metal is with oxygen. is used to build up reactivity series of metals Position of carbon in the reactivity series • Carbon will reduce the oxide of metal X if carbon is more reactive than X. • The position of carbon is between aluminium and zinc.

The reactivity series of metals The reactivity series of metals is a series of metals arranged in the order of how vigorously the metals react with oxygen. K > Na > Ca > Mg > Al > C > Zn > H > Fe > Sn > Pb > Cu reactivity decreases →

Position of hydrogen in the reactivity series • The position of hydrogen can be determined by passing the hydrogen gas over hot metal oxides. • The position of hydrogen is between zinc and iron.

3

uses of reactivity series in the extraction of metals Extraction of iron • Raw materials: Iron ore (haematite or magnetite), limestone and air. • Iron is produced by the reduction of iron ore by coke (carbon) or carbon monoxide:

Extraction of tin • Raw materials: Tin ore (cassiterite) and coke. • The froth floatation method is used to produce concentrated tin ore. • Tin is produced by the reduction of cassiterite by coke or carbon monoxide:

Fe2O3(s) + 3C(s) → 2Fe(l) + 3CO(g)

SnO2(s) + C(s) → Sn(l) + CO2(g)

Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

SnO2(s) + 2CO(g) → Sn(l) + 2CO2(g)

• Limestone is used to form slag. CaCO3(s) → CaO(s) + CO2(g) CaO(s) + SiO2 → CaSiO3(s)

3.3 (a) Based on the experimental results, arrange the metals P, Q, R and T in descending order of the reactivity of the metals with oxygen.

1 The table below shows the experimental results when a mixture of a metal and the oxide of another metal is heated strongly. Mixture

(b) Predict whether metal Q will react with the oxide of metal P. Explain your answer.

Observation

Metal P and oxide of metal Q

The mixture glows

Metal Q and oxide of metal R

The mixture glows

Metal Q and oxide of metal T

The mixture does not glow

Metal P and oxide of metal T

The mixture does not glow

2 The oxide of metal X can be reduced by carbon but not hydrogen. (a) Identify the metal X.

(b) (i) Write the equation for the reaction between the oxide of metal X and aluminium powder. (ii) Identify the oxidising and reducing agents in this reaction.

3 State four different methods that can be used to obtain lead from lead(II) oxide. Write the equations for all the reactions that take place.

429

Oxidation and Reduction

3.4

(a) the external circuit, (b) the reactions occurring in the electrolyte, and (c) the reactions at the electrodes. 3 When electrolysis occurs, electrical energy is converted into chemical energy. The electrical energy is used to decompose the electrolyte in the electrolytic cell. 4 Inert electrodes such as carbon or platinum are usually used as electrodes in electrolytic cells. However, in some electrolytic cells, one of the electrodes is an inert electrode and the other electrode is a metal electrode. There are also electrolytic cells in which both the electrodes are of the same metal.

Redox Reactions in Electrolytic Cell and Chemical Cell

1 In Chapter 6 of Form 4, you have already studied the concepts of electrolytic cell and chemical cells. 2 Chemical cells are also known as voltaic cells. 3 The reactions that occur in electrolytic cells and chemical cells are redox reactions involving the transfer of electrons. 4 Whether it is an electrolytic cell or a chemical cell, oxidation occurs at the anode, and reduction occurs at the cathode.

Chemical Cells

3

Differences between an Electrolytic Cell and a Chemical Cell

SPM

’04/P1, ’05/P1 ’05/P2, ’07/P1

1 The chemical cell is set up by dipping two different metals in an electrolyte. The metals act as the electrodes in the chemical cell. 2 Figure 3.28 shows the apparatus in a chemical cell, using zinc and copper electrodes and dilute sulphuric acid as the electrolyte. 3 The basic structure of a chemical cell is as follows: (a) A connecting wire for electrons to flow through in the external circuit, (b) An electrolyte for electric current to flow through, (c) Two electrodes for the transfer of electrons from a reducing agent to an oxidising agent. 4 In a chemical cell, chemical energy is converted into electrical energy.

Electrolytic Cells 1 The basic structure of an electrolytic cell is as follows: (a) A battery to supply electrical energy, (b) An electrolyte to supply free (mobile) ions for conducting electric current, (c) Two electrodes for the transfer of electrons to occur. Electrons are transferred from the anions (negative ions) to the anode and from the cathode to the cations (positive ions). 2 Figure 3.27 shows the apparatus set-up in an electrolytic cell. The electrolysis process involves three main aspects, that is,

External circuit During electrolysis, electrons flow from the anode (positive terminal) to the cathode (negative terminal)

Anode (positive terminal)

Cathode (negative terminal)

During electrolysis, • anions (negative ions) move towards the anode • anions release electrons at the anode • oxidation occurs at the anode

During electrolysis, • cations (positive ions) move towards the cathode • cations accept electrons from the cathode • reduction occurs at the cathode Electrolyte

During electrolysis • cations move towards the cathode • anions move towards the anode • the flow of ions to the electrodes constitute the flow of electric current in the electrolyte

Figure 3.27 The movement of ions and electrons during electrolysis Oxidation and Reduction

430

External circuit Electric current is produced because electrons flow from the more electropositive electrode (negative terminal) to the less electropositive electrode (positive terminal).

Anode (negative terminal)

Cathode (positive terminal)

• Zinc dissolves to form zinc ions with the release of electrons.

• H+ ions accept electrons from zinc to form hydrogen gas. 2H+(aq) + 2e– → H2(g)

Zn(s) → Zn2+(aq) + 2e–

• Effervescence occurs around the copper electrode. • Reduction occurs at the cathode.

• The electrons flow through the external circuit to the copper electrode. • Oxidation occurs at the anode. Electrolyte

• The concentration of Zn ions increases, while the concentration of H+ ions decreases. • The overall reaction that occurs in the chemical cell is 2+

Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

3

• Zinc acts as the reducing agent and hydrogen ions act as the oxidising agent.

Figure 3.28 A simple chemical cell

Redox Reactions in Electrolytic Cells

2 The redox reactions that occur at the electrodes are shown in Table 3.13.

SPM

’09/P1

1 Ionic compounds, in the molten state, or dissolved in water, are electrolytes. 2 During electrolysis, oxidation occurs at the anode and reduction occurs at the cathode. (a) At the anode: An– → A + ne– … oxidation (b) At the cathode: Bn+ + ne– → B ... reduction 3 In an electrolytic cell, electrons flow from the anode (positive electrode) to the cathode (negative electrode) through the connecting wire (Figure 3.29).

Table 3.13 Electrolysis of molten lead(II) bromide

At the cathode (a) Pb ions gain electrons to form lead metal. 2+

Pb2+(l) + 2e– → Pb(l) (b) This is a reduction process. The oxidation number of lead decreases from +2 to 0.

At the anode (a) Br ions lose electrons to form bromine molecules. –

2Br–(l) → Br2(l) + 2e– (b) This is an oxidation process. The oxidation number of bromine increases from –1 to 0.

Overall reaction The overall reaction is the breakdown of lead(II) bromide to give lead and bromine: PbBr2(l) → Pb(l) + Br2(l) Figure 3.29 Electrolytic cell

Electrolysis of Molten Lead(II) Bromide

Electrolysis of Copper(II) Sulphate Solution using Inert Electrodes

SPM

’09/P1

1 When molten lead(II) bromide is electrolysed, the cations (lead(II) ions, Pb2+) are attracted to the cathode and the anions (bromide ions, Br–) are attracted to the anode.

1 Electrolysis of copper(II) sulphate solution can be carried out using the apparatus as shown in Figure 3.30. Platinum (Pt) electrodes are used as inert electrodes in this experiment. 431

Oxidation and Reduction

SPM

’10/P1

1 If the electrolysis of copper(II) sulphate solution is carried out using reactive electrodes such as copper electrodes, a different reaction occurs at the anode, that is, both the OH– ions and the SO42– ions are not discharged. Instead, the copper anode dissolves (corrodes) to form copper(II) ions.

Figure 3.30 Electrolysis of CuSO4 using inert electrodes

Cu(s) → Cu2+(aq) + 2e–

2 An aqueous solution of copper(II) sulphate contains four types of ions.

2 At the copper anode, three possible reactions that can occur are: (a) SO42– ions are discharged (b) OH– ions are discharged by donating (c) Copper metal is oxidised electrons to Cu2+ ions 3 Because copper is a reactive electrode, the reaction that can occur most easily is the conversion of copper to Cu2+ ions. Consequently, SO42– ions and OH– ions remain in the solution and are not discharged. Instead, the copper anode dissolves to form Cu2+(aq) ions and the electrode becomes thinner.

From CuSO4 : Cu2+(aq) and SO42–(aq)

3

From H2O

: H+(aq) and OH–(aq)

3 The cations, Cu2+ and H+ ions are attracted to the cathode and the anions, SO42– and OH– ions, are attracted to the anode. 4 Table 3.14 shows the redox reactions that occur at the electrodes. Table 3.14 Redox reactions at platinum electrodes during the electrolysis of CuSO4(aq)

At the cathode (negative electrode)

At the anode (positive electrode)

Cu(s) → Cu2+(aq) + 2e–

(a) OH– ion is below (a) Copper is below SO42– ion in the hydrogen in the electrochemical electrochemical series. Hence, OH– series. Hence, Cu2+ ions are preferentially ions are preferentially discharged at the discharged at the cathode. anode.

4 The redox reactions that occur at the electrodes during the electrolysis of copper(II) sulphate using copper electrodes are shown in Table 3.15. Table 3.15 Redox reactions at copper electrodes during electrolysis of CuSO4(aq)

reduction oxidation Cu2+(aq) + 2e– ⎯⎯⎯→ 4OH–(aq) ⎯⎯⎯→ Cu(s) O2(g) + 2H2O + 4e– (b) At the cathode, Cu2+ ions are reduced to copper metal. (c) H+ ions remain in the solution.



At the cathode

At the anode

Cu (aq) + 2e → Cu(s) … reduction

Cu(s) → Cu2+(aq) + 2e– … oxidation

2+

(b) At the anode, OH– ions are oxidised to oxygen gas. (c) SO42– ions remain in the solution.

–

Overall reaction • The overall reaction is the transfer of copper from the anode to the cathode. • The concentration of copper(II) sulphate does not change and the blue colour of the electrolyte does not fade.

Overall reaction Copper metal is deposited at the cathode, oxygen gas is given off at the anode and the solution becomes more and more acidic.

Electrolysis of Copper(II) Sulphate Solution using Copper Electrodes

Electrolysis of Concentrated Sodium Chloride Solution

reduction electrolysis

2CuSO4 + 2H2O ⎯⎯⎯⎯→ 2Cu + O2 + 2H2SO4

1 Figure 3.31 shows the apparatus set-up for the electrolysis of concentrated sodium chloride solution.

oxidation Oxidation and Reduction

432

A simple way to remember redox reactions at the electrodes is to remember ‘RED CAT’. REDuction occurs at CAThode Conversely, oxidation occurs at the anode.

Electrolysis of Dilute Sodium Chloride Solution 1 A dilute sodium chloride solution contains: (a) Na+(aq) and Cl–(aq) from NaCl(aq) (b) H+(aq) and OH–(aq) from H2O(l) 2 Table 3.17 shows the redox reactions that occur and the products obtained at the carbon electrodes.

Figure 3.31 Electrolysis of concentrated sodium chloride solution

2 An aqueous solution of sodium chloride contains four types of ions: (a) From NaCl: Na+(aq) and Cl–(aq) (b) From H2O: H+(aq) and OH–(aq) 3 Na+ ions and H+ ions are attracted to the cathode. Cl– ions and OH– ions are attracted to the anode. 4 Table 3.16 shows the redox reactions that occur at the anode and cathode when concentrated sodium chloride solution is electrolysed using inert electrodes.

At the cathode (Na+ and H+ ions are present)

OH– ions donate electrons H+ ions gain electrons from the cathode to form to the anode to form hydrogen gas. oxygen gas and water.

Table 3.16 Redox reactions during electrolysis of concentrated sodium chloride solution

At the cathode (negative electrode), H+ and Na+ ions are present

At the anode (positive electrode), Cl– and OH– ions are present

(a) Hydrogen ions (H ) are preferentially discharged at the cathode.

(a) Chloride ions (Cl ) are preferentially discharged at the anode.

+

2H+(g) + 2e– → H2(g)

2Cl–(aq) → Cl2(g) + 2e–

(b) At the cathode, H+ ions are reduced to hydrogen gas. (c) Na+ ions remain in the solution.

(b) At the anode, Cl– ions are oxidised to chlorine gas. (c) OH– ions remain in the solution.

4OH–(aq) → O2(g) + 2H2O(l) + 4e–

Overall reaction

–

2H+(aq) + 2e– → H2(g)

At the anode (Cl– and OH– ions are present)

oxidation electrolysis 2H2O(l) ⎯⎯⎯⎯⎯⎯⎯→ 2H2(g) + O2(g) reduction • Electrolysis of dilute sodium chloride solution produces two volumes of hydrogen at the cathode and one volume of oxygen at the anode. • Since water is being removed (by decomposition to form H2 and O2), the concentration of sodium chloride increases gradually.

Overall reaction oxidation electrolysis 2NaCl(aq) + 2H2O(l) ⎯⎯⎯⎯→ 2NaOH(aq) + H2(g) + Cl2(g)

Electrolysis of • molten NaCl produces Na(l) and Cl2(g) • dilute NaCl produces H2(g) and O2(g) • moderately concentrated NaCl produces H2(g) at the cathode, a mixture of O2(g) and Cl2(g) at the anode • concentrated NaCl produces H2(g) and Cl2(g).

reduction

Electrolysis of concentrated sodium chloride solution produces one volume of hydrogen at the cathode, one volume of chlorine at the anode and sodium hydroxide solution. 433

Oxidation and Reduction

3

Table 3.17 Redox reactions during electrolysis of dilute sodium chloride solution

5 At the negative electrode (zinc plate) (a) Zinc is more electropositive than copper. Hence, zinc has a greater tendency to donate its electrons compared to copper and is oxidised to zinc ions.

Redox Reactions in Chemical Cells 1 In chemical cells, the electric current is produced from chemical reactions that occur in the cell. Examples of chemical cells are: (a) Daniell cell (b) Dry cell/alkaline cell (c) Lead-acid accumulator 2 In chemical cells, oxidation occurs at the negative terminal (anode) while reduction occurs at the positive terminal (cathode).

Zn(s) → Zn2+(aq) + 2e– … oxidation (b) At the negative electrode, oxidation occurs and the zinc metal acts as the reducing agent. 6 At the positive electrode (copper plate) (a) At the positive electrode, Cu2+ ions gain electrons from zinc and is reduced to copper metal.

Daniell Cell

Cu2+(aq) + 2e– → Cu(s) … reduction

1 The Daniell cell is made up of a zinc plate dipped into zinc sulphate solution and a copper plate dipped into copper(II) sulphate solution (Figure 3.32). 3

(b) At the positive electrode, reduction occurs and Cu2+ ions act as the oxidising agent. 7 Overall reaction (a) The overall reaction that occurs in the Daniell cell is a redox reaction. oxidised

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) reduced

(b) The redox reactions that occur in the Daniell cell and many other chemical cells are displacement reactions. A metal higher up in the electrochemical series displaces another metal lower in the electrochemical series from an aqueous solution of its salt. (c) When the Daniell cell is in use, (i) the concentration of Zn2+ ions in the solution increases. (ii) the blue colour of copper(II) sulphate solution fades gradually as more copper is deposited and the concentration of Cu2+ decreases. (iii) the mass of zinc electrode decreases gradually. (iv) the mass of copper electrode increases gradually. 8 Cell symbols (a) The cell symbols are used to represent the chemical cells. For example, the cell symbol for Daniell cell is

Figure 3.32 Daniell cell

2 The function of the porous pot is to SPM (a) separate the zinc sulphate solution from ’11/P1 the copper(II) sulphate solution so that the solutions do not mix. (b) complete the electric circuit by allowing the ions to pass through it. 3 The Daniell cell can also be set up by using a salt bridge to replace the porous pot as shown in Figure 3.33.

Figure 3.33 Daniell cell

Zn(s)/Zn2+(aq) // Cu2+(aq)/ Cu(s)

4 Zinc is more electropositive than copper and is placed higher than copper in the electrochemical series. Hence, zinc plate acts as the negative electrode and copper plate acts as the positive electrode. Oxidation and Reduction

(b) According to the IUPAC rules, the negative electrode is written on the left of the cell symbol and the positive electrode is written on the right. The symbol ‘//’ represents the porous pot or the salt bridge. 434

9 Voltage of Daniell cell (a) If the concentrations of both ZnSO4 and CuSO4 solutions are 1.0 mol dm–3, the maximum voltage of the Daniell cell is 1.10 V. (b) The voltage of the cell will decrease with time when the cell is being used because the concentration of Cu2+ ions decreases.

anode cathode (negative electrode) (positive electrode)



Zn(s)/Zn2+(aq) // Cu2+(aq)/ Cu(s)



electrolyte at anode

electrolyte at cathode

salt bridge (or porous pot)

Apparatus Beaker, porous pot, sandpaper, voltmeter, connecting wires with crocodile clips and electronic balance. Materials 1.0 mol dm–3 copper(II) sulphate solution, 1.0 mol dm–3 zinc sulphate solution, zinc plate and copper plate. Procedure

Figure 3.34 The Daniell cell

1 A piece of zinc plate and a piece of copper plate are cleaned using sandpaper.

2 The electrodes are then washed with distilled water, dried and weighed. 3 A beaker is filled with 1.0 mol dm–3 copper(II) sulphate solution. 4 A porous pot is filled with 1.0 mol dm–3 zinc sulphate solution. 5 The zinc plate is dipped into the solution of zinc sulphate while the copper plate is dipped into the solution of copper(II) sulphate. 6 The zinc and copper electrodes are then connected to the voltmeter as shown in Figure 3.34. 7 The Daniell cell is allowed to operate for 30 minutes. 8 After 30 minutes, the zinc and copper electrodes are removed from the electrolytes. 9 The electrodes are rinsed with distilled water, dried and weighed again. 10 The changes that occur in the electrodes, the electrolytes and the voltmeter are recorded.

3

To study the reactions that occur in the Daniell cell

Results 1 The blue colour of copper(II) sulphate fades gradually until it becomes colourless. 2 The zinc plate becomes thinner and the copper plate becomes thicker. 3 The voltmeter needle is deflected. The deflection of the voltmeter needle shows that electrons flow from the zinc electrode to the copper electrode. 4 The changes in mass of the zinc and copper electrodes are shown below. Mass of electrode before experiment (g)

Mass of electrode after experiment (g)

Zinc plate

20.50

18.29

• The mass of zinc plate decreases. • Oxidation occurs at the anode.

Copper plate

25.74

27.89

• The mass of copper plate increases. • Reduction occurs at the cathode.

Electrode

435

Inference

Oxidation and Reduction

Activity 3.4



Discussion 1 Zinc metal is more electropositive than copper metal. Hence, zinc metal has a greater tendency to give up electrons to form zinc ions. Zn(s) → Zn (aq) + 2e … oxidation 2+

–

2 The electrons released are accepted by copper(II) ions to form copper metal.

3

Cu2+(aq) + 2e– → Cu(s) … reduction

3 The overall reactions is Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) … redox reaction 4 In this reaction, Cu2+ ions act as the oxidising agent and zinc acts as the reducing agent. Conclusion 1 Oxidation occurs at the anode (zinc plate). 2 Reduction occurs at the cathode (copper plate). 3 The reaction that occurs in the Daniell cell is a redox reaction.

Different types of chemical cells 1 There are two main classes of chemical cells: Primary cells and secondary cells. 2 Primary cells are not rechargeable. Examples of primary cells are dry cell, alkaline cell and mercury cell. 3 Secondary cells are rechargeable. Examples of secondary cells are lead-acid accumulator and nickel-cadmium (Ni-Cd battery).

Dry Cell

Figure 3.35 The structure of a dry cell

1 A dry cell is made up of a zinc container as the anode (negative terminal) and a carbon rod as the cathode (positive terminal). The electrolyte in the dry cell is a paste consisting of ammonium chloride, zinc chloride and a little water (Figure 3.35).

2 When the dry cell is used to generate electrical energy, oxidation occurs at the negative terminal (zinc container) and reduction occurs at the positive terminal (carbon rod). 3 The reactions at the electrodes are shown below.

Reactions at anode (negative terminal)

Reactions at cathode (positive terminal)

• Zinc is oxidised to Zn2+.

• NH4+ is reduced to NH3 and H2.

oxidation

reduction







0 +2

+1

Zn(s) → Zn2+(aq) + 2e–

• Electrons flow from the zinc container to the carbon rod.

• The H2 gas produced is removed by the reaction with MnO2. 2MnO2 + H2 → Mn2O3 + H2O

Summary • The overall reaction is a redox reaction. Zn + 2NH4+ + 2MnO2 → Zn2+ + 2NH3 + Mn2O3 + H2O • Oxidising agent: Ammonium ion, NH4+ Reducing agent: Zinc

Oxidation and Reduction

0

2NH4+(aq) + 2e– → 2NH3(g) + H2(g)

436

Alkaline Cell 1 Figure 3.36 shows the structure of an alkaline cell. An alkaline cell is also known as an alkaline battery. Negative terminal: Zinc container Positive terminal: Manganese(IV) oxide powder Electrolyte: Lithium hydroxide (LiOH) or potassium hydroxide (KOH) 2 The reactions at the electrodes during discharge are shown below.

Figure 3.36 The structure of an alkaline cell

Reactions at anode (negative terminal)

Reactions at cathode (positive terminal) • MnO2 is reduced to manganese(III) oxide, Mn2O3.

• Zinc is oxidised to Zn . 2+

oxidation

reduction







0 +2

Zn(s) → Zn2+(aq) + 2e–

+4

+3

2MnO2(s) + H2O(l) + 2e– → Mn2O3(s) + 2OH–(aq)

Summary • The overall reaction is a redox reaction. Zn + 2MnO2 + H2O → Zn2+ + Mn2O3 + 2OH–

3

• Electrons flow from the zinc container to MnO2. • Oxidising agent: Manganese(IV) oxide, MnO2 Reducing agent: Zinc

Mercury Cell 1 Figure 3.37 shows the structure of mercury cell. Positive terminal: Mercury(II) oxide, HgO Negative terminal: Zinc metal Electrolyte: Potassium hydroxide 2 The reactions at the anode and cathode of mercury cell are summarised below. Reactions at anode (negative terminal) • Zinc is oxidised to Zn2+. Zn(s) + 2OH–(aq) → Zn(OH)2(aq) + 2e–

Figure 3.37 The structure of mercury cell

Lead-acid Accumulator

Reactions at cathode (positive terminal)

SPM

’07/P1

1 A lead-acid accumulator is often known as a car battery. The accumulator is a battery (chemical cell) that can be recharged by passing a current through it from an external direct current (d.c.) supply. 2 Figure 3.38 shows the structure of a lead-acid accumulator.

• HgO is reduced to mercury. HgO(s) + H2O(l) + 2e– → Hg(l) + 2OH–(aq)

• Electrons flow from the zinc electrode to HgO. Summary • The overall reaction is a redox reaction. Zn(s) + HgO(s) + H2O(l) → Zn(OH)2(s) + Hg(l) • Oxidising agent: Mercury(II) oxide Reducing agent: Zinc

Figure 3.38 The structure of a lead-acid accumulator

437

Oxidation and Reduction

3 Table 3.18 shows the reactions at the electrodes during discharge, that is, when the battery supplies electricity.

Positive terminal: Lead plate coated with PbO2 Negative terminal: Lead plate Electrolyte: Sulphuric acid

Table 3.18 Redox reactions that occur during discharge in lead-acid accumulator

At the anode (negative terminal)

At the cathode (positive terminal) (a) Lead(IV) oxide is reduced to Pb2+ ions by accepting electrons.

(a) Lead is oxidised to lead(II) ions with the release of electrons. oxidation





reduction



0 +2

+4 +2

Pb(s) → Pb (aq) + 2e 2+

–

3

(b) The electrons given out at the cathode flow through the external circuit to the positive terminal. (c) A white precipitate is produced when Pb2+ ions react with SO42– ions in the sulphuric acid to form lead(II) sulphate. Pb2+(aq) + SO42–(aq) → PbSO4(s) (d) The negative electrode becomes white because white solid lead(II) sulphate is deposited on its surface. (e) The overall reaction at the negative electrode during discharge is Pb(s) → Pb2+(aq) + 2e–

PbO2(s) + 4H+(aq) + 2e– → Pb2+(aq) + 2H2O(l) (b) A white solid is produced when Pb2+ ions react with SO42– ions in sulphuric acid to form lead(II) sulphate. Pb2+(aq) + SO42–(aq) → PbSO4(s) (c) The white solid, lead(II) sulphate, then deposits on the surface of the positive electrode to form a white coating. (d) The overall reaction at the positive electrode during discharge is PbO2(s) + 4H+(aq) + 2e– → Pb2+(aq) + 2H2O(l) Pb2+(aq) + SO42–(aq) → PbSO4(s) PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– → brown PbSO4(s) + 2H2O(l) white

Pb2+(aq) + SO42–(aq) → PbSO4(s) Pb(s) + SO42–(aq) → PbSO4(s) + 2e– grey white Summary • The overall cell reaction is a redox reaction.

Pb(s) + PbO2(s) + 4H+(aq) + 2SO42–(aq) → 2PbSO4(s) + 2H2O(l) 2H2SO4(aq) • Oxidising agent: Lead(IV) oxide, PbO2 Reducing agent: Lead • During discharge, sulphuric acid is used up. (a) At the negative terminal Lead(II) sulphate is reduced to lead.

4 Recharging of lead acid accumulator When the battery is fully charged, the concentration of sulphuric acid is 4.1 mol dm–3 and the density is 1.3 g cm–3. When the density of sulphuric acid drops from 1.3 to 1.1 g cm–3, the accumulator must be recharged. 5 When an accumulator is recharged, the direct current is passed through it, in the direction which is opposite to the discharge. 6 The reactions that occur when the accumulator is recharged are as follows. Oxidation and Reduction

PbSO4(s) + 2e– → Pb(s) + SO42–(aq) ... reduction white grey

(b) At the positive terminal Lead(II) sulphate is oxidised to lead(IV) oxide. PbSO4(s) + 2H2O(l) → PbO2(s) + 4H+ + SO42–(aq) white brown + 2e– ...oxidation 438

(c) The overall reaction during recharging is 2PbSO4(s) + 2H2O(l)

Pb(s) + PbO2(s) + 4H+(aq) + 2SO42–(aq)

2H2SO4(aq)

6

Figure 3.39 Electrolysis of molten sodium chloride ’05

3 In the electrolytic cell, the electrode connected to the positive terminal of a chemical cell is called the anode. Conversely, the electrode connected to the negative terminal of the chemical cell is called the cathode. 4 (a) For both the chemical and electrolytic cells, (i) oxidation occurs at the anode, (ii) reduction occurs at the cathode. (b) In the electrolytic cell, the anions from the electrolyte donate the electrons and is oxidised at the anode. (c) In the electrochemical cell, the more electropositive metal is oxidised at the anode and donate the electrons to the cathode. (d) Table 3.19 compares the chemical cell and the electrolytic cell in terms of oxidation and reduction.

A chemical cell is shown below.

Which of the following occur in the chemical cell? I The iron rod becomes thinner. II The copper rod becomes thicker. III The intensity of the blue colour in beaker 1 decreases. IV The concentration of iron(II) ions (Fe2+) in beaker 2 decreases. A I and III only C I, II and III only B II and IV only D I, II, III and IV

Table 3.19 Comparison between chemical cell and electrolytic cell in terms of redox reactions

Comments At the iron rod: Fe(s) → Fe2+(aq) + 2e–

Chemical cell (for example, Daniell cell)

Therefore, the iron rod becomes thinner and the concentration of Fe2+ in FeSO4 solution increases. At the copper rod: Cu2+(aq) + 2e– → Cu(s) Therefore, the copper rod becomes thicker and the intensity of blue colour decreases as the concentration of Cu2+ decreases. Answer C

Electrolytic cell (for example, electrolysis of molten sodium chloride)

At the anode: oxidation occurs

At the anode: oxidation occurs

Zn(s) → Zn2+(aq) + 2e–

2Cl–(l) → Cl2(g) + 2e–

At the cathode: reduction At the cathode: reduction occurs occurs Cu2+(aq) + 2e– → Cu(s)

Compare and Contrast Electrolytic Cells and Chemical Cells in Terms of Redox Reactions

Na+(l) + e– → Na(s)

5 However, electrodes in chemical and electrolytic cells have different signs (positive or negative) as shown in Table 3.20. For example, the anode in a chemical cell is the negative electrode whereas the anode in an electrolytic cell is a positive electrode.

1 Redox reactions occur in both the chemical and electrolytic cells. 2 Figure 3.39 shows the arrangement of apparatus for the electrolysis of sodium chloride in molten condition. 439

Oxidation and Reduction

3



recharging

6 For both chemical and electrolytic cells, anions move to the anode while cations move to the cathode. Figure 3.40 shows the movement of ions in a chemical cell. Cations (Zn2+ and Cu2+) move to the cathode and the anions (SO42–) move to the anode.

Table 3.20 Positive and negative terminals for chemical and electrolytic cells

Chemical cell Anode: negative terminal In the chemical cell, electrons are released at the anode. Thus, the anode is the negative terminal. Cathode: positive terminal In the chemical cell, electrons are removed from the cathode by positive ions present in the electrolyte. Hence, the cathode is the positive terminal.

Electrolytic cell Anode: positive terminal In the electrolytic cell, electrons flow out from the anode to the battery. Thus, the anode is the positive terminal. Cathode: negative terminal In the electrolytic cell, electrons flow from the battery and enter the cathode. Hence, the cathode is the negative terminal.

Figure 3.40 The movement of cations and anions in a chemical cell.

7 3

’07

A chemical cell is shown below.

B The electrolyte is hydrochloric acid. C The positive terminal is lead plate. D The cell is non-rechargeable. Comments The electrolyte is sulphuric acid. The positive terminal (cathode) is the plate coated with PbO2. At the anode, reduction occurs:

Which of the following is true about the chemical cell? A At the positive terminal, lead(IV) oxide is reduced to Pb2+ ions.

PbO2 + 4H+ + 2e– → Pb2+ + 2H2O Answer A

3.4 1 Explain the redox reactions at the anode and cathode when electric current is passed into the following solutions. (a) Concentrated potassium iodide solution with carbon electrodes. (b) Copper(II) sulphate solution with copper electrodes. 2 The following figure shows the arrangement of apparatus for the electrolysis of iron(II) sulphate.

(b) What is the experiment?

energy

conversion

in

this

3 Nickel-cadmium cell (Ni-Cd battery) is a rechargeable battery. It consists of Anode(–): Cadmium Cathode (+): Nickel(IV) oxide Electrolyte: Potassium hydroxide The overall equation for the reaction that occurs when Ni-Cd battery is supplying current is Cd + NiO2 + 2H2O → Cd(OH)2 + Ni(OH)2 (a) Identify the oxidising and reducing agents in this reaction. (b) Write the half-equation for the reaction that occurs at (i) the anode, (ii) the cathode. (c) State the direction of flow of electrons in the external circuit.

(a) Describe (i) the oxidation and reduction processes, (ii) the transfer of electrons that occur at the carbon electrodes.

Oxidation and Reduction

440

Comparison between electrolytic cell and chemical cell

Electrolytic cell

Chemical cell

• Cathode (–) • Anode (+) – Reduction occurs – Oxidation occurs – Cations accept electrons – Anions release electrons from the cathode at the anode

• Cathode (+) • Anode (–) – Reduction occurs – Oxidation occurs – Accepts electrons – Releases electrons

Redox reactions in electrolytic and chemical cells

• Anode: Carbon Cathode: Carbon • At the anode: Oxidation occurs 2Cl– → Cl2 + 2e– • At the cathode: Reduction occurs 2H++ 2e– → H2

3.5

Chemical cell

Negative terminal (oxidation)

Positive terminal (reduction)

• Daniell cell

Zn(s) → Zn2+(aq) + 2e–

Cu2+(aq) + 2e– → Cu(s)

• Dry cell

Zn(s) → Zn2+(aq) + 2e–

2NH4+(aq) + 2e– → 2NH3(g) + H2(g)

• Lead-acid accumulator

Pb(s) → Pb (aq) + 2e

2MnO2(s) + H2(g) → Mn2O3(s) + H2O(l) 2+

–

Pb2+(aq) + SO42–(aq) → PbSO4(s)

PbO2(s) + 4H+(aq) + 2e– → Pb2+ + 2H2O(l) Pb2+(aq) + SO42–(aq) → PbSO4(s)



Appreciating the Ability of the Elements to Change their Oxidation Numbers

oxidation





0 +2

Fe2O3(s) + 3C(s) → 2Fe(s) + 3CO(g)

+3 0

reduction

Various Applications of the Changes of Oxidation Numbers in Substances

4 In the corrosion of iron, the following changes in oxidation numbers occur.

1 Groups 1 and 2 elements in the Periodic Table have fixed oxidation numbers of +1 and +2 respectively. However, most elements (metals and non-metals) have variable oxidation numbers. 2 The changes in the oxidation number of a substance can be applied in the following processes: (a) Extracting metal from its ore (b) Corrosion of metal (c) Preventing corrosion of metal (d) Generation of electricity by cells (e) Recycling of metals 3 In the extraction of iron from its ores, the changes in the oxidation numbers of both iron and carbon are shown as follows.

oxidation

0

+3

4Fe + 3O2 + 2xH2O → 2Fe2O3.xH2O 0

–2



reduction

5 The following chemical changes occur when zinc is used in the prevention of rusting. Zn → Zn2+ + e– ... (1) O2 + 2H2O + 4e– → 4OH– ... (2) The oxidation number of zinc changes from 0 to +2 while the oxidation number of oxygen changes from 0 to –2. 441

Oxidation and Reduction

3

Electrolysis of concentrated NaCl(aq)

6 When a Daniell cell is used to generate electricity, the overall cell reaction is

Methods for Rust Prevention Table 3.21 shows a summary of the common methods used for the prevention of metal corrosion, especially the rusting of iron.

Zn + Cu2+ → Zn2+ + Cu

Table 3.21 Methods for rust prevention

Method

The oxidation number of zinc changes from 0 to +2 while the oxidation number of copper changes from +2 to 0.

3

The Occurrence of Various Ores in Our Country 1 Gold Gold mines are found in Pahang (Lipis, Raub, Jerantut), Terengganu (Mandi), Sabah (Merungin and Paginatan) and Sarawak (Bau, Serian and Lund). 2 Iron, bauxite (Al2O3) and ilmenite (FeTiO3) Iron mines are found in Johor (Kota Tinggi), Kedah (Semeling), Pahang and Perak. Bauxite is aluminium ore and is found in Johor (Teluk Ramunia). Ilmenite is a titanium ore and is found in Terengganu and Pulau Pinang. 3 Tin Tin ores are found in Perak (Lembah Kinta) and Selangor (Lembah Langat). 4 Coal Coal mines are found in Sarawak (in the areas of Kapit, Lucky Selantek and Sri Aman). 5 Kaolin and barite (barytes) Kaolin is a type of clay used for making ceramics. Kaolin is found in Johor (Mersing) and Perak (Bidor-Tapah). Barite (BaSO4) is the chief ore of barium. It is found in Kelantan (Gua Musang) and Terengganu.

1

Paint

Big objects like motor vehicles, ships and steel bridges

2

Oil and grease

Tools and machine parts

3

Phosphoric acid (H3PO4)

The bottom (chassis) of cars

4

Galvanising (zincplating)

Buckets, ‘zinc’ roof

5

Tin-plating

Food cans

6

Chrome-plating

Taps, bicycle handle bars, car bumpers

7

Block of magnesium or zinc (sacrificial metals)

Underground pipes, ships

8

Stainless steel

Cutlery, surgical instruments

Chemical Cell as a Source of Electrical Energy 1 Chemical cells are alternative sources of renewable energy. Chemical cells are also known as batteries. Nowadays, there are many types of chemical cells. Examples are: zinc-carbon battery, mercury battery, alkaline battery, lithium battery, nickel-cadmium (Ni-Cd) battery, leadacid accumulator, fuel cell and photo/solar battery. 2 A fuel cell is a device in which the fuel is oxidised in a chemical cell so as to produce electricity directly. In the hydrogen-oxygen fuel cell, chemical energy from the redox reaction between hydrogen and oxygen to form water is used to generate electric current. Fuel cells are used widely in spacecraft. The water produced can be used for drinking. Fuel cells are also used to power electric cars. Fuel cells differ from the usual chemical (voltaic) cells in two ways:

The Contribution of the Metal Extraction Industry in Enhancing the National Economy 1 The metal extraction industry provides job opportunities and lowers the unemployment rate. The export revenue from tin and other metals enhances the national economy. 2 Our country exports tin ingots to countries like Japan, America and Britain. This will earn foreign exchange which can be used for national development. 3 The metal extraction industry produces metals as raw materials for many other industries, such as the motor and construction industries. Oxidation and Reduction

Objects

442

strikes the cell, the electrons flow from the donor to the acceptor crystal. Solar cells have been used for years to power spaceships. 4 Research is still being carried out on chemical cells to develop their potential as an alternative energy source, which is now produced chiefly by the burning of fossil fuels. 5 One of the industries that is actively engaged in developing a light, powerful, efficient and long-lasting battery is the automobile industry. In the near future, electric cars that use battery to power, instead of using petrol as the fuel will be on the road. If that happens, air pollution would be greatly reduced.

1 Oxidation is • gain of oxygen, or • loss of hydrogen, or • loss of electrons or • increase in the oxidation number of the element. 2 Reduction is • loss of oxygen, or • gain of hydrogen, or • gain of electrons or • decrease in the oxidation number of the element. 3 An oxidising agent is a substance that causes oxidation in another substance. 4 A reducing agent is a substance that causes reduction in another substance. 5 Oxidation and reduction take place simultaneously in a redox reaction. 6 A displacement reaction is a redox reaction. A more electropositive metal will displace a less electropositive metal from its salt solution. 7 In a redox reaction, electrons are transferred from the reducing agent to the oxidising agent. 8 Rusting is a redox reaction. For iron to rust, oxygen (air) and water must be present. 9 During rusting, iron reacts with oxygen and water to form a brown substance, called rust (Fe2O3. xH2O).

11 A more reactive metal will displace a less reactive metal from its oxide. For example, if P has a reaction with the oxide of Q, P is more reactive than Q.

3

(a) The fuel and oxygen are fed into the cell continuously, (b) The electrodes are made from inert material such as platinum that does not react during the process. 3 Solar cells (also known as photovoltaic cells) convert sunlight directly into electricity. Solar cells are made from extremely pure silicon crystals. One type of crystal has about 1 ppm (parts per million) of arsenic added to it. This crystal is called the donor crystal. Another crystal is made by adding about 1 ppm of boron. This crystal is called the acceptor crystal. The two crystals are connected by an external circuit. When sunlight

P + oxide of Q → Q + oxide of P 12 The reactivity series that includes both carbon and hydrogen is K, Na, Ca, Mg, Al, C, Zn, H, Fe, Sn, Pb, Cu, Ag ⎯⎯⎯⎯⎯ reactivity decreases ⎯⎯⎯⎯→ 13 Reactive metals such as K, Na, Ca, Mg and Al are extracted from their ores by electrolysis. 14 Metals such as Zn, Fe, Sn and Pb are extracted from their ores by heating the metal oxides with carbon. 15 The reactions that occur in electrolytic cells or chemical cells (voltaic cells) are redox reactions involving the transfer of electrons. 16 In an electrolytic cell: • At the anode (positive electrode), oxidation occurs and anions are discharged by losing electrons to form molecules. • At the cathode (negative electrode), reduction occurs and cations are discharged by gaining electrons to form metal or hydrogen gas. 17 In a chemical cell: • The more electropositive metal is the negative electrode and the less electronegative metal is the positive electrode. • Oxidation occurs at the anode (negative electrode). • Reduction occurs at the cathode (positive electrode).

4Fe + 3O2 + 2xH2O → 2Fe2O3.xH2O 10 Rusting can be prevented by • a protective layer (oil, grease or plastic layer), • coating/plating iron with tin/chromium • using sacrificial metals • using alloys

443

Oxidation and Reduction

3 Multiple-choice Questions 3.1

Redox Reactions

1 Which of the following substances are correctly ’09 matched with their roles as oxidising or reducing agent?

3

Oxidising agent

Reducing agent

A

Sulphur dioxide Chlorine

B

Chlorine

C

Potassium Bromine manganate(VII)

D Potassium bromide

Sulphur dioxide

Potassium manganate(VII)

2 The manufacture of nitric acid in the Ostwald process involves ’08 the following steps.

5 The following equation shows the reaction between metal X ’11 and the salt solution of metal Y. X(s) + Y(NO3)2(aq) → X(NO3)2(aq) + Y(s)

Which metals could be X and Y? A

X

Y

Copper

Tin

B

Copper

Iron

C

Iron

Tin

D

Iron

Zinc

7 Consider the following reaction:

Which of the following shows the correct sequence of changes in the oxidation number of nitrogen? A +3 → +2 → +4 → +3 B +3 → +2 → +4 → +5 C –3 → +2 → +4 → +3 D –3 → +2 → +4 → +5

2I–(aq) + 2Fe3+(aq) → ’03 2Fe2+(aq) + I2(aq)

4 Which of the underlined substances in the following equations acts as the oxidising agent? A Fe2O3 + 3CO → 2Fe + 3CO2 B H2S + Br2 → S + 2HBr C MgBr2 + Na2CO3 → MgCO3 + 2NaBr D Mg + H2SO4 → MgSO4 + H2 Oxidation and Reduction

10 The diagram below shows the set-up of apparatus to investigate the transfer of electrons at a distance.

6 Which of the following cannot occur during reduction? ’09 A Loss of oxygen B Gain of hydrogen C Donation of electrons D Decrease in oxidation number

NH3 → NO → NO2 → HNO3

3 Which of the following organic compounds is reduced in the following reactions? A CH3CH2OH → CH3COOH B CH3CH2OH → C2H4 + H2O C CH3CHO → CH3CH2OH D CH3CHO → CH3COOH

What would be observed if iodine solution is used instead of chlorine water? A Bromide ion is oxidised to bromine. B Iodine is reduced to iodide ion. C No change is observed. D A brown solution is formed.

Which of the following is true about the ionic equation? A Fe3+ is oxidised. B Fe3+ is a reducing agent. C I– is an oxidising agent. D I– donates electrons to Fe3+. 8 In which of the following reactions is copper oxidised? ’04 A The reaction of copper(II) oxide with magnesium. B The reaction of copper with silver nitrate solution. C Electrolysis of copper(II) sulphate solution using carbon electrodes. D Chemical cell with copper and magnesium foils in dilute sulphuric acid. 9 The following equation represents the reaction between ’05 chlorine and potassium iodide: Cl2(aq) + 2KBr(aq) → 2KCl(aq) + Br2(aq)

444

Which of the following statements about this experiment is correct? A The oxidation number of manganese decreases from +5 to +2. B The iodide ions act as a reducing agent. C The electrons are transferred from electrode X to electrode Y through dilute sulphuric acid. D At electrode Y, H+ ions from sulphuric acid are reduced to hydrogen gas. 11 The following reaction occurs when hydrogen peroxide is added to an aqueous solution of iron(II) salt. H2O2(aq) + 2Fe2+(aq) + 2H+(aq) → 2Fe3+(aq) + 2H2O(l) Which of the following statements are true?

12 The electronic configuration of four elements are shown below. Q: 2.8.2 X: 2.8.6

Y: 2.8.7 Z: 2.8.8

Which of these elements will act as a reducing agent? A Element Q C Element Y B Element X D Element Z 13 When an aqueous solution of compound X is added to acidified potassium dichromate(VI) solution, the solution changes from orange to green. What could be compound X? I Sulphur dioxide II Sodium sulphite III Chlorine IV Iron(III) chloride A I and II only B II and IV only C III and IV only D I, III and IV only 14 The diagram below shows the arrangement of the apparatus to study the redox reaction between chlorine and iron(II) sulphate solution.

Which of the following statements are true regarding the experiment shown above? I Electrons flow from X to Y through the external circuit. II Electrode X is the positive electrode. III The green colour of FeSO4 is changed to brownish-yellow.

IV Chlorine is oxidised. A I and II only B II and III only C III and IV only D I, II and III only 15 FeSO4 solution can be prepared from Fe2(SO4)3 solution by I adding zinc powder II adding potassium manganate(VII) solution III passing chlorine gas IV passing sulphur dioxide gas A I and III only B II and IV only C I and IV only D I, III and IV only

3.2

A magnesium B iron C lead D copper 19 Which of the following does not prevent the corrosion of iron? A Coating iron with grease B Coating iron with zinc C Fixing a copper bar to iron D Fixing an aluminium bar to iron 20 Consider the experiments shown in the diagram below.

Rusting as a Redox Reaction

16 The diagram below shows the rusting of iron.

3

I The solution becomes colourless. II Hydrogen peroxide molecule accepts electrons. III H+ ion acts as the oxidising agent. IV Iron(II) salt is oxidised. A I and III only B II and IV only C I, II and III only D II, III and IV only

’06

Which beaker contains the most rust? A Beaker I C Beaker III B Beaker II D Beaker IV 21 The diagram shows the set-up of apparatus to study the rusting of iron. Which of the following equations occurs at the cathode? A Fe2+ + 2e– → Fe B Fe → Fe2+ + 2e– C 4OH– → O2 + 2H2O + 4e– D O2 + 2H2O + 4e– → 4OH– 17 Which of the following statements about the rusting of iron are true? I Rusting requires both oxygen and water. II Rusting is accelerated by the presence of magnesium chloride. III An iron atom releases two electrons to form an iron(II) ion. IV The chemical formula of rust is Fe3O4. xH2O. A I and III only B II and IV only C I, II and III only D I, II, III and IV 18 A metal X is placed in zinc sulphate solution and a reaction occurs. The metal X is

445

The observations are recorded below. Test tube

Colour of solution

1

No change

2

Turns pale blue

3

Turns dark blue

Based on the observation, the reactivity of the metal increases in the order: A P < Q < R B R < Q < P C Q < P < R D Q < R < P Oxidation and Reduction

3

22 When metal X is twisted around an iron nail, the iron nail rusts rapidly. However, if metal Y is twisted around the iron nail, the iron nail does not rust. These observations show that I metal X is more reactive than metal Y. II metal X is more reactive than iron. III metal Y is more reactive than iron. IV metal Y can protect metal X from corrosion. A I and II only B III and IV only C I, II and III only D II, III and IV only 23 An iron nail, with a piece of zinc foil wrapped around it, is placed in a beaker of water. Which of the following ionic equations represent the reactions that occur after a few days? I Zn(s) → Zn2+(aq) + 2e– II Fe(s) → Fe2+(aq) + 2e– III 2H+(aq) + 2e– → H2(g) IV Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) A I and III only B II and IV only C I, II and III only D II, III and IV only

3.3

Reactivity Series of Metals and Its Applications

24 What is the position of carbon in the reactivity series of metals? ’07 A Between zinc and iron B Between aluminium and zinc C Between magnesium and aluminium D Between iron and lead 25 The following pairs of substances are heated strongly. Which pair will produce a glow? A Aluminium oxide and iron B Iron(III) oxide and lead C Lead(II) oxide and copper D Zinc oxide and aluminium 26 Which of the following reactions does not occur during the extraction of tin from tin ore which contains tin(IV) oxide? Oxidation and Reduction

A SnO2 B SnO2 C SnO2 D SnO2

+ + + +

C → Sn + CO2 2C → Sn + 2CO 2CO → Sn + 2CO2 CaO → CaSnO3

27 The diagram below shows the extraction of iron in a blast furnace.

30 Which of the following reactions occurs at the cathode and anode when concentrated copper(ll) chloride is electrolysed using carbon electrodes? Cathode

Anode

A

H ions reduced to hydrogen

Cl ions oxidised to chlorine

B

Cl– ions reduced to chlorine

Cu2+ ions oxidised to copper

C

Cu2+ ions reduced to copper

OH– ions oxidised to oxygen

D

Cu2+ ions reduced to copper

Cl– ions oxidised to chlorine

’06

What is substance X? A Coke B Platinum C Aluminium D Vanadium(V) oxide 28 Which of the following reactions occur in the blast furnace for the extraction of iron from its ore? I Coke (carbon) reduces iron ore to iron. II Iron ore acts as an oxidising agent. III Carbon dioxide is reduced by coke to carbon monoxide. IV Limestone (calcium carbonate) acts as a reducing agent. A I and III only B II and IV only C I, II and III only D I, II, III and IV

3.4

Redox Reactions in Electrolytic Cell and Chemical Cell

29 What are the reactions that occur at the anode and the cathode during the electrolysis of an aqueous solution of sodium sulphate? Cathode

Anode

A

Oxidation

Reduction

B

H2 gas is produced

O2 gas is produced

C

OH– ions are reduced

H+ ions are oxidised

D SO42– ions are Na+ ions are reduced oxidised

446

+

–

31 The diagram shows the set-up of apparatus for an experiment ’06 to study the electrolysis of copper(II) sulphate solution.

The intensity of the blue colour of copper(II) sulphate solution decreases as electrolysis proceeds. Which of the following statements best explains the observation? A Cu2+ ion is oxidised to form copper metal at the cathode. B Cu2+ ion is reduced to form copper metal at the cathode. C OH– ion is oxidised to form oxygen gas at the anode. D OH– ion is reduced to form oxygen gas at the anode. 32 The diagram below shows a simple chemical cell.

33 Which of the following statements is true? A In chemical cells, anodes are positively-charged. B In chemical cells, electrical energy is converted to chemical energy. C In electrolytic cells, anodes are negatively-charged. D In electrolytic cells, electrons move from the anode to the catho­­­­­­de through the external circuit. 34 A piece of tin and a piece of copper are used as electrodes in a chemical cell as shown in the diagram below.

Which of the following statements is true about this experiment? A The size of the tin plate becomes smaller. B Bubbles of gas are formed around the tin plate. C Electrons flow from the copper plate to the tin plate. D The copper plate is covered with a thin layer of a reddishbrown substance. 35 In a dry cell, A zinc ions are reduced to zinc metal. B carbon rod acts as the negative terminal. C manganese(III) oxide is oxi­dis­­ ed to manganese(IV) oxide D ammonium chloride is reduced to hydrogen and ammonia.

36 When a dry cell supplies electric current, A zinc container is oxidised. B the carbon rod acts as the anode. C the ammonium ion acts as the reducing agent. D ammonium ion is reduced to nitrogen. 37 The electroplating of an iron spoon with copper is carried out using the apparatus shown in the diagram below.

Which of the following redox reactions occurs at the cathode during electrolysis? A Cu → Cu2+ + 2e– … oxidation B Cu2+ + 2e– → Cu … reduction C Fe → Fe2+ + 2e– … oxidation D 4OH– →2H2O + O2 + 4e– … oxidation 38 The apparatus set-up for a simple cell is shown below.

What could be metals P and Q? Metal P A B C D

Metal Q

Copper

Lead

Lead

Copper

Iron

Tin

Aluminium

Tin

447

39 The structure of a lead-acid battery is described below. Cell terminal P: Lead plate Cell terminal Q: Lead plate coated with lead(IV) oxide. Electrolyte: Solution X Which of the following are true about lead-acid battery? I Cell terminal P is the positive terminal. II Lead is oxidised to Pb2+ ions at cell terminal P. III PbO2 is reduced to Pb2+ ions at cell terminal Q. IV Solution X is dilute hydrochloric acid. A I and II only B II and III only C III and IV only D I, II and III only 40 The diagram below shows a magnesium-silver cell. ’11

V Mg plate

MgSO4(aq)

Ag plate

AgNO3(aq)

Which of the statements about this voltaic cell is true? I Magnesium dissolves to form magnesium ions. II Magnesium acts as the reducing agent. III The concentration of silver nitrate decreases. IV The ions in the solution flow through the porous pot. A I and III only B II and IV only C I, II and III only D I, II, III and IV

Oxidation and Reduction

3

Which of the following undergoes oxidation when the cell is supplying electricity? A Zinc plate B Copper plate C Copper(II) ions D Hydrogen ions

Structured Questions 1 Excess iron powder is added to an aqueous solution containing magnesium chloride and copper(ll) nitrate. The mixture is shaken and filtered.

3 (a) Diagram 2 shows an experiment for investigating the positions of metal X, metal Y and hydrogen in the reactivity series.

(a) Write the ionic equation for the reaction that occurs. [1 mark] (b) Explain your answer in (a).

[2 marks]

(c) (i) Identify the substance in the residue after filtration. [1 mark] (ii) Describe what you see in this experiment. [2 marks]

Diagram 2

(d) Identify (i) the oxidising agent, (ii) the reducing agent in this reaction. [2 marks] (e) Explain your answer in (d).

The experimental results are shown in the following table.

[2 marks]

Metal oxide

2 (a) The following reaction occurs when ammonia is passed over heated oxide of copper. ’08

3

2NH3(g) + 3CuO(s) → N2(g) + 3H2O(l) + 3Cu(s)



(i) State the changes in oxidation number of copper in this reaction. [1 mark] (ii) Is ammonia oxidised or reduced in this reaction? Explain your answer. [1 mark]

Observation During heating

After heating

Oxide of metal X

Glows brightly The colour of the oxide changes from brown to grey

Oxide of metal Y

Does not glow The oxide is yellow when hot and white when cold



(i) With the help of a diagram, explain how dry hydrogen is prepared in the laboratory. [3 marks] (ii) Explain one safety precaution that must be taken in this experiment. [1 mark] (iii) Predict the melting point and the boiling point of liquid Z produced. [1 mark] (iv) Suggest the identity of the oxide of X. [1 mark] (v) Suggest the identity of the oxide of Y. [1 mark] (vi) Arrange X, Y and hydrogen in order of the reactivity series. [1 mark] (vii) Explain your answer in (vi). [2 marks]

(b) Diagram 1 shows the set-up of apparatus to investigate electron transfer through a solution.

(b) Diagram 3 shows the arrangement of apparatus in an experiment for investigating the redox reactions of three elements: magnesium, zinc and carbon.

Diagram 1

(i) What is meant by oxidising agent. [1 mark] (ii) Name the oxidising agent and the substance that is reduced in the experiment. [2 marks] (iii) Which is the positive electrode, X or Y ?





[1 mark]

(iv) Write the half-equations that take place at the negative electrode. [1 mark] (v) Name one substance that can be used to replace dilute sulphuric acid. [1 mark]

Diagram 3



(i) What is the function of potassium manganate(VII) in this experiment? [1 mark] (ii) Record the experimental results expected to be obtained in the following table.

(c) State the changes that can be observed at electrodes X and Y after 5 minutes. [2 marks] Oxidation and Reduction

448

6 The apparatus set-up of two cells, P and Q, are shown in diagram 4.

Observation Element

Nature of flame/ glow

Nature of residue (if any)

Magnesium Zinc Carbon [3 marks]



(iii) Describe the steps to be taken so that the [2 marks] redox reactions can occur.

4 (a) State two conditions for the rusting of iron. [1 mark] (b) Based on your answer in (a), draw a labelled diagram to show what happens when iron rusts. Your diagram should also include the ionisation [3 marks] of iron and the flow of electrons.

3

’07

(c) Describe the reactions that take place at the edge [3 marks] of a water droplet.

Diagram 4

(a) (i) Name the types of reactions that occur in [1 mark] cells P and Q. (ii) What is the difference between cell P and cell Q in terms of energy change? [1 mark]

5 When metal X is added to iron(III) chloride solution, a redox reaction occurs and a green solution is ’09 formed. (a) Which metal given in the list below is most likely to be metal X? Explain your answer.

Answer the following questions by referring to cell P. (b) Write the half-equations for the reaction at (i) the [2 marks] anode, (ii) the cathode. (c) (i) Write the chemical equation for the overall [1 mark] reaction. (ii) Explain the above reaction in terms of [2 marks] oxidation and reduction.

copper, lead, magnesium, potassium [3 marks] (b) (i) What is the change in oxidation number for [2 marks] both the reactants? (ii) Write the half-equations involved in the [2 marks] redox reaction.

Answer the following questions by referring to cell Q. (d) State the observations that occurred at (i) the copper plate, (ii) the zinc plate. (e) (i) Write the ionic equation for the overall [1 mark] reaction. (ii) Explain the above reaction in terms of [2 marks] oxidation and reduction.

(c) Explain the role that iron(III) chloride plays in this [2 marks] reaction?

Essay Questions 1 (a) Describe an industrial method for the extraction of iron from its ore.

[6 marks]

(b) In terms of electron arrangement, explain why metals (such as sodium, magnesium and aluminium) act as reducing agents and non-metals (such as oxygen and chlorine) act as oxidising agents. [10 marks] (c) Explain why galvanising can prevent iron from rusting.

[4 marks]

2 (a) Magnesium is more reactive than copper. Design two experiments to prove this statement. (b) Some substances act as an oxidising agent in one reaction but a reducing agent in another reaction. Using iron(II) sulphate as an example, describe how you would prove this statement.

449

[10 marks] [10 marks]

Oxidation and Reduction

Experiments 1 (a) Table 1 shows the set-up of apparatus and the result of the experiment to study the reaction of copper(II) oxide with carbon. Table 1

Set-up of apparatus

Observation on the mixture

Bright glow

What inference can you make based on the above observation.

[3 marks]

(b) The experiment is repeated using magnesium oxide, lead(II) oxide and sodium oxide mixed with carbon. The results of the experiment are shown in Table 2.

3

Table 2



Mixture of metal oxide with carbon

Observation on the mixture

Magnesium oxide + carbon

No change

Lead(II) oxide + carbon

Faint glow

Sodium oxide + carbon

No change

(i) State one hypothesis for the experiment. (ii) Based on Tables 1 and 2, construct a table to classify the metals into two groups.

[3 marks] [2 marks]

(c) Diagram 1 shows the set-up of apparatus to determine the reactivity of metals with oxygen.



Diagram 1

(i) What is the purpose of heating potassium manganate(VII)? (ii) The experiment is carried out using powders of copper, iron, lead and magnesium. The results of the experiments are shown in Table 3.

Oxidation and Reduction

450

[1 mark]

Table 3

Set-up of apparatus

Observation of the metal

Set-up of apparatus

Observation of the metal

Look at the flame or glow in each of the experiments. Then complete Table 3 by stating the observations of the reaction of metals with oxygen.

[3 marks]

(d) State one hypothesis for the experiment.

[3 marks]

(e) Based on the observations in Table 3, arrange the metals, copper, iron, lead and magnesium in descending order of reactivity with oxygen.

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ Reactivity of metal with oxygen decreases

[3 marks]

(f) Complete Table 4 by stating the action to be taken for each variable.

Table 4

Variables



Action to be taken

(i) Manipulated variable

(i) How to manipulate the variable

(ii) Responding variable

(ii) What to observe in the responding variable

(iii) Constant variable

(iii) How to maintain the constant variable



2 Steel but not iron is used to build bridges. This is because iron rusts easily when exposed to air and water but steel is more resistant to rusting. You are asked to plan an experiment to show that steel is more difficult to rust than iron. Your experiment should include the following aspects: (a) Statement of the problem (b) Statement of the hypothesis (c) All the variables (d) Lists of materials and apparatus (e) Procedure of the experiment (f) Tabulation of data

451

[3 marks]

[17 marks]

Oxidation and Reduction

3

Faint glow

FORM 5 THEME: Interaction between Chemicals

CHAPTER

4

Thermochemistry

SPM Topical Analysis 2008

Year 1

Paper Section Number of questions

3

2009

2 A

B

C

1

–

–

3

1

–

2

2010

2 A

B

C

1

–

–

3

1

–

3

2011

2 A

B

C

–

–

1

3

1

–

4

3

2 A

B

C

–

–

–

–

ONCEPT MAP Energy level diagram Example: Exothermic reaction N2 + 3H2 2NH3; ∆H = –92 kJ LULYN` 5/

THERMOCHEMISTRY • The branch of chemistry that studies the changes in heat energy in chemical reactions

¬/$¶ R1 5/

Calculations (a) Number of moles of reactants or products = x (b) Heat released or absorbed (m 3 c 3 θ ) = –––––––––– kJ = y kJ 1000 (c) From (a) and (b), y heat of reaction = – — kJ mol–1 x

Heat of precipitation Example: Heat of precipitation of silver chloride Ag+(aq) + Cl–(aq) → AgCl(s); ∆H = –63 kJ mol–1

Heat of displacement Example: Heat of displacement of copper by zinc Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq); ∆H = –210 kJ mol–1

Heat of neutralisation Example: Heat of neutralisation of hydrochloric acid with sodium hydroxide HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l); ∆H = –57 kJ mol–1

Heat of combustion Example: Heat of combustion of ethanol C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l); ∆H = –1371 kJ mol–1

Heat of neutralisation of a strong acid with a strong base is always –57 kJ mol–1 because strong acids and strong bases dissociate completely.

Heat of neutralisation of a strong acid with a weak base (or a weak acid with a strong base) is always less than –57 kJ mol–1 because weak acids and bases absorb heat energy on dissociation.

Choice of suitable fuels Important factors: • High fuel value (kJ/g) • Burns easily • Do not produce air pollutants • Cheap • Can be obtained easily

4.1

Exothermic Reactions

Energy Changes in Chemical Reactions

SPM

’09/P1, ’11/P1

1 Exothermic reactions are reactions that release heat energy to the surroundings. 2 Figure 4.1 shows a simple experiment for measuring the temperature change when anhydrous copper(II) sulphate solid is dissolved in water.

1 Energy exists in various forms. In chemistry, the important forms of energy are (a) heat energy, (b) light energy, (c) chemical energy, (d) electrical energy, (e) kinetic energy. 2 According to the law of conservation of energy, energy cannot be created or destroyed. However, energy can be converted from one form to another as shown in Table 4.1. Table 4.1 Examples of conversion of energy

Example

Figure 4.1 An exothermic reaction

(a) Electrical energy Electrolysis → chemical energy

3 In this reaction, heat energy is released to the surroundings (Figure 4.2).

(b) Chemical energy Chemical cells (batteries) → electrical energy (c) Light energy → chemical energy

The term ‘surroundings’ means the water in which the chemical is dissolved, the container in which the chemical reaction occurs, the air and the thermometer.

• Photosynthesis • Photochemical reactions CH4 + Cl2 → CH3Cl + HCl or action of sunlight on photochromic glass

SPM

’07/P1

3 Chemical energy is the energy stored in all chemical substances. 4 During a chemical reaction, the chemical energy stored in the reactants are converted mainly into heat energy and other forms of energy (such as light energy). 5 Thermochemistry is the branch of chemistry that studies the changes in heat energy in chemical reactions. 6 Chemical reactions can be divided into two classes based on the energy changes that occur during the reaction. (a) Exothermic reactions (b) Endothermic reactions

CuSO4 + H2O

Figure 4.2 Heat energy is released to the surroundings in an exothermic reaction

4 When an exothermic reaction occurs, (a) heat is released and is transferred from the reactants to the surroundings, (b) the reaction mixture and the container become hot, (c) the temperatures of the reaction mixture and the container rise, (d) the chemical energy is converted into heat energy. 5 Figure 4.3 shows the changes in temperature when an exothermic reaction occurs. Initially, the temperature of the reaction mixture rises until the highest temperature is reached. When

Heat energy can be transferred from the reactants or the products to the surroundings and vice versa. However, the total energy remains unchanged at the end of the reaction.

453

Thermochemistry

4

Energy change

(a) Combustion of fuels

the reaction is completed, the temperature of the reaction mixture falls until it reaches room temperature.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) (b) Oxidation of food in the respiration process C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O(g) glucose

(c) Rusting of iron 4Fe(s) + 3O2(g) + 2xH2O(l) → 2Fe2O3.xH2O rust

Figure 4.3 Variation of temperature with time for an exothermic reaction

(d) Dissolving soluble bases (metal oxides) in water

4

6 Examples of exothermic reactions involving physical changes (a) Condensation process (gas to liquid)

CaO(s) + H2O(l) → Ca(OH)2(aq)

H2O(g) → H2O(l)

(e) Neutralisation reactions between acids and bases

HCl(g) → HCl(l)

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) H2SO4(aq) + CuO(s) → CuSO4(aq) + H2O(l)

steam water

Condensation process (gas to solid)

(f) Reaction between acids and metals or metal carbonates

CO2(g) → CO2(s) dry ice

Zn(s) + H2SO4(aq) → ZnSO4(aq)+ H2(g) Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

I2(g) → I2(s) (b) Freezing (solidification) process (liquid to solid)

(g) Displacement reaction of a metal from its salt solution by a more reactive metal

H2O(l) → H2O(s) water ice

Fe(l) → Fe(s)

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

(c) Dissolving alkalis and acids (especially concentrated acids) in water.

(h) Haber process for the manufacture of ammonia

NaOH(s) + water → Na+(aq) + OH–(aq)

N2(g) + 3H2(g)

sodium hydroxide

(i) Contact process for the production of sulphur trioxide

H2SO4(l) + water → 2H+(aq) + SO42–(aq) sulphuric acid

2SO2(g) + O2(g)

(d) Dissolving anhydrous salts, such as anhydrous copper(II) sulphate (CuSO4) and anhydrous sodium carbonate (Na2CO3) in water.

Endothermic Reactions

2SO3(g) SPM

’09/P1

1 Endothermic reactions are reactions that absorb heat energy from the surroundings. 2 Figure 4.4 shows a simple experiment to measure the temperature change when copper(II) sulphate crystals are dissolved in water.

CuSO4(s) + water → Cu2+(aq) + SO42–(aq) Na2CO3(s) + water → 2Na+(aq) + CO32–(aq) 7 Examples of exothermic reactions involving chemical changes Thermochemistry

2NH3(g)

454

6 Examples of endothermic reactions involving physical changes (a) Dissolving ammonium salts such as ammonium chloride (NH4Cl), ammonium nitrate (NH4NO3) and ammonium sulphate, (NH4)2SO4 in water. NH4Cl(s) + water → NH4+(aq) + Cl–(aq) NH4NO3(s) + water → NH4+(aq) + NO3–(aq) (NH4)2SO4(s) + water → 2NH4+(aq) + SO42–(aq)

Figure 4.4 An endothermic reaction

(b) Dissolving crystalline salts such as hydrated copper(II) sulphate (CuSO4.5H2O), hydrated magnesium sulphate (MgSO4.7H2O) and hydrated sodium carbonate (Na2CO3.10H2O).

3 In this reaction, heat energy is absorbed from the surroundings and is transferred from water to the copper(II) sulphate crystals (Figure 4.5).

Na2CO3.10H2O + water → 2Na+(aq) + CO32–(aq) + 10H2O(l) (c) Melting process (solid to liquid)

Figure 4.5 Heat energy is absorbed from the surroundings in endothermic reactions

H2O(s) → H2O(l) ice water

Sn(s) → Sn(l)

4 When an endothermic reaction occurs, (a) heat energy is absorbed and is transferred to the reactants, (b) the reaction mixture and the container becomes cold, (c) the temperatures of the reaction mixture and the container fall, (d) the heat energy is converted into chemical energy. 5 Figure 4.6 shows the variation of temperature with time when an endothermic reaction occurs. Initially, the temperature of the reaction mixture falls until it reaches the minimum temperature. When the reaction is completed, the temperature of the reaction mixture rises until it reaches room temperature.

(d) Evaporation and boiling processes (liquid to gas) H2O(l) → H2O(g) water

water vapour

H2O(l) → H2O(g) water steam

7 Examples of endothermic reactions involving chemical changes (a) The reaction between acids and sodium or potassium hydrogen carbonate. HCl(aq) + NaHCO3(s) → NaCl(aq) + H2O(l) + CO2(g) (b) Thermal decomposition of metal carbonates, metal nitrates and ammonium chloride. ZnCO3(s) → ZnO(s) + CO2(g) 2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g) NH4Cl(s)

NH3(g) + HCl(g)

(c) Photosynthesis 6CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g) glucose

Figure 4.6 Variation of temperature with time for an endothermic reaction

455

Thermochemistry

4

CuSO4.5H2O + water → Cu2+(aq) + SO42–(aq) + 5H2O(l)

The reaction between an acid and sodium carbonate (Na2CO3) is an exothermic reaction, but the reaction between an acid and sodium hydrogen carbonate (NaHCO3) is an endothermic reaction.

When an endothermic reaction occurs, the heat energy is absorbed from the surroundings and converted into the chemical energy of the reactants. The loss of heat energy from the surroundings causes the temperature of the solution to fall.

4

To study exothermic and endothermic reactions Materials

Plastic cup, thermometer and spatula.

Apparatus

Sodium hydroxide solid, ammonium chloride solid, 1 mol dm–3 hydrochloric acid, 1 mol dm–3 sodium hydroxide solution and 2 mol dm–3 sodium hydrogen carbonate solution.

Figure 4.7 To investigate exothermic and endothermic reactions

(A) Dissolution of sodium hydroxide and ammonium chloride in water 1 Using a measuring cylinder, 50 cm3 of water is measured out and poured into a plastic cup. 2 A thermometer is placed in the water as shown in Figure 4.7 and the temperature of water is measured. 3 A spatula of sodium hydroxide solid is added to the water in the plastic cup. 4 The water is slowly stirred with the thermometer to dissolve the sodium hydroxide. The highest temperature obtained is recorded. 5 Steps 1 to 4 are repeated using ammonium chloride solid instead of sodium hydroxide solid. The lowest temperature obtained is recorded.

Activity 4.1

Reaction

Initial temperature (°C)

Lowest temperature (°C)

NH4Cl(s) + water

28

24

Discussion 1 When sodium hydroxide is dissolved in water, the temperature of water rises. This means that heat energy is given off to water. 2 When ammonium chloride is dissolved in water, the temperature of water falls. This means that heat energy is absorbed from water.

Procedure

Results (a) Dissolving sodium hydroxide solid in water Reaction

Initial temperature (°C)

Highest temperature (°C)

NaOH(s) + water

28

36

Thermochemistry

(b) Dissolving ammonium chloride solid in water

Conclusion 1 Dissolving sodium hydroxide solid in water is an exothermic process. 2 Dissolving ammonium chloride solid in water is an endothermic reaction. (B) The reactions between hydrochloric acid and (a) sodium hydroxide, (b) sodium hydrogen carbonate 1 Using a measuring cylinder, 40.0 cm3 of 1.0 mol dm–3 hydrochloric acid is measured out and poured into a plastic cup. 2 A thermometer is placed in the hydrochloric acid and the temperature is recorded. 3 Using another measuring cylinder, 30.0 cm3 of 1.0 mol dm–3 sodium hydroxide solution is measured out and poured into the hydrochloric acid. 4 The reaction mixture is stirred slowly with the thermometer and the highest temperature achieved is recorded. 5 Steps 1 to 4 are repeated using 50.0 cm3 of 1.0 mol dm–3 hydrochloric acid and 25.0 cm3 of 2.0 mol dm–3 sodium hydrogen carbonate solution. The lowest temperature obtained is recorded. 456

Reaction

Initial temperature (°C)

Highest temperature (°C)

HCl(aq) + NaOH(aq)

30

35

(b) Reaction between hydrochloric acid and sodium hydrogen carbonate Reaction

Initial Highest temperature temperature (°C) (°C)

HCl(aq) + NaHCO3(aq)

30

27

Discussion 1 When sodium hydroxide solution is added to hydrochloric acid, the temperature of the solution rises. This means that heat energy is released to the solution. 2 When sodium hydrogen carbonate solution is added to hydrochloric acid, the temperature of the solution falls. This means that heat energy is absorbed from the solution. Conclusion 1 The reaction between hydrochloric acid and sodium hydroxide solution is an exothermic reaction. 2 The reaction between hydrochloric acid and sodium hydrogen carbonate solution is an endothermic reaction.

7 For the general reaction, A+B→C+D Heat of reaction (ΔH) = Total energy content of the products (H1) – total energy content of the reactants (H2) = (HC + HD) – (HA + HB)

• Heat is the transfer of energy caused by the temperature difference between the reacting particles and the surroundings. • Temperature is the measure of the average kinetic energy of all the particles involved in the reaction. The unit of temperature is kelvin (K). • Heat will always flow from the region of high temperature to the region of low temperature.

Energy Level Diagram

4

Results (a) Reaction between hydrochloric acid and sodium hydroxide solution

That is, ΔH = Hproducts –Hreactants 8 (a) For an exothermic reaction, the value of ΔH is negative. That is, 1 H2(g) + — O2(g) → H2O(l); ΔH = –286 kJ mol–1 2

SPM

’08/P1

1 Energy is defined as the ability to do work. The unit of energy is joule (symbol J).

(b) This equation shows that when 1.0 mol of water is produced from hydrogen and oxygen, 286 kJ of heat energy are released. (c) When 2.0 mol of hydrogen react with 1.0 mol of oxygen to form 2.0 mol of water, (2  286 kJ) or 572 kJ of heat are released. That is,

1 kJ (kilojoule) = 1000 J 2 All substances contain energy. But different substances have different energy content. 3 The absolute energy content of a substance is given the symbol H (the unit of H is kJ or J). 4 The absolute energy content of a given substance cannot be determined but the changes in energy content that occur when the reactants are converted to the products can be determined and is given the symbol, ∆H. 5 When substances react to produce new substances, changes in energy content occur. This causes heat to be released or absorbed during the reaction. 6 The changes in the energy content in a chemical reaction that produces 1 mol of a product, ∆H, is called the heat of that chemical reaction. The unit for ∆H is kJ mol–1.

2H2(g) + O2(g) → 2H2O(l); ΔH = –572 kJ 9 (a) In endothermic reactions, the values of ΔH are positive. For example, 1 1 — N (g) + — O2(g) → NO(g); ΔH = +90 kJ mol–1 2 2 2

(b) The equation shows that when 1.0 mol 1 of nitrogen oxide is produced from — 2 1 mol of nitrogen and — mol of oxygen, 2 90 kJ of heat are absorbed. 457

Thermochemistry

(c) When 1.0 mol of nitrogen reacts with 1.0 mol of oxygen to form 2.0 mol of nitrogen oxide, 180 kJ of heat are absorbed. That is, N2(g) + O2(g) → 2NO(g); ΔH = +180 kJ 10 The changes in the energy content of exothermic and endothermic reactions can be represented by energy level diagrams. 11 The energy level diagram is the diagram that shows the total energy content of the reactants compared to the total energy content of the products.

Figure 4.9 Energy level diagram for the reaction between magnesium and oxygen

4 In the reaction between magnesium and oxygen, the total energy content of magnesium oxide is less than the total energy content of magnesium and oxygen. The excess energy is released as heat. Thus, the reaction is an exothermic reaction.

The Energy Level Diagrams for Exothermic

4

Reactions

1

1 Figure 4.8 shows the energy level diagram for an exothermic reaction. The difference between the total energy content of the reactants (H1) and the total energy content of the products (H2) gives the heat evolved (ΔH) during a reaction, that is, ΔH = H2 – H1. 2 For an exothermic reaction, the total energy SPM content of the products is lower than the total ’10/P1 energy content of the reactants. Hence, the ΔH value of an exothermic reaction is negative.

When 1.0 mol of zinc powder reacts with 2.0 mol of hydrochloric acid, 120 kJ of heat are released. Sketch the energy level diagram to represent this reaction. Solution

The Energy Level Diagrams for Endothermic Reactions

Figure 4.8 The energy level diagram for an exothermic reaction

SPM

’09/P1

1 Figure 4.10 shows the energy level diagram for an endothermic reaction.

ΔH = Hproducts – Hreactants = negative (if Hproducts < H reactants) 3 When magnesium reacts with oxygen, heat energy is released. 2Mg(s) + O2(g) → 2MgO(s); ΔH = –1204 kJ The energy level diagram for this exothermic reaction is shown in Figure 4.9.

Thermochemistry

Figure 4.10 The energy level diagram for an endothermic reaction

458

ΔH = Hproducts – Hreactants = positive (if Hproducts > H reactants)

2

Comments The energy content of the products is lower than that of the reactants. The reaction is exothermic. (I is correct) The activation energy is x kJ (II is correct). So the value of x decreases in the presence of a catalyst. But indirectly, the value of y also decreases. (IV is incorrect) The heat of reaction (ΔH) is –(y – x) kJ, the negative sign before (y – x) indicates an exothermic reaction. (III is incorrect) Answer A

State (a) two facts that are given, and (b) two facts that are not given by the energy level diagram in Figure 4.11.

Figure 4.11

Solution (a) The energy level diagram in Figure 4.11 shows that (i) the total energy content of the reactants is higher than the total energy content of the products. This implies that the reaction is exothermic, with negative ΔH value. (ii) when 1.0 mol of nitrogen reacts with 3.0 mol of hydrogen to form 2.0 mol of ammonia, 92 kJ of heat energy are released. N2(g) + 3H2(g) → 2NH3(g);

Relationship between Energy Change and the Formation and Breaking of Bonds 1 The energy change in a reaction is caused by (a) the formation of chemical bonds and (b) the breaking of chemical bonds. 2 When a reaction occurs, energy is absorbed to break the bonds that exist between atoms in the molecules of the reactants. Heat energy is then released when new bonds are formed to produce the products. This means that (a) the breaking of chemical bonds is an endothermic process, and (b) the formation of chemical bonds is an exothermic process. 3 Bond energies (a) Table 4.2 shows the bond energies for some chemical bonds.

ΔH = –92 kJ

(b) The energy level diagram in Figure 4.11 does not give information concerning (i) the rate of reaction, (ii) the conditions such as temperature, pressure or the catalyst required to carry out the reaction.

1

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Table 4.2 Bond energy

The energy profile of a reaction is shown below. SPM

Covalent bond

Bond energy (kJ mol–1)

C–C C=C H–H N≡N Cl – Cl

346 612 436 946 242

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459

Thermochemistry

4

Which of the following are true of this reaction? I The reaction is exothermic. II The activation energy is x kJ. III The heat of reaction is y kJ. IV The value of y increases in the presence of a catalyst. A I and II only B II and III only C I and IV only D I, II and IV only

2 For an endothermic reaction, the total energy content of the products is higher than the total energy content of the reactants. This means that the value of ΔH for an endothermic reaction is positive.

ΔH1 (bond breaking) = +678 kJ ΔH2 (bond forming) = –862 kJ ΔH (heat of reaction) = (+678) + (–862) = –184 kJ

(b) The bond energy is the energy required to break one mole of covalent bonds. The same amount of heat energy is released when 1.0 mol of the same covalent bonds is formed. (c) Table 4.2 shows that the bond energy of chlorine is +242 kJ mol–1. This implies that 242 kJ of heat energy are required to break the covalent bonds in one mole of chlorine molecules.

(c) The energy level diagram for the reaction between hydrogen and chlorine to form hydrogen chloride is shown in Figure 4.13.

Cl2(g) → 2Cl(g); ΔH = +242 kJ mol–1 Conversely, 242 kJ of heat energy are released when 1.0 mol of chlorine gas is formed from chlorine atoms.

4

2Cl(g) → Cl2(g); ΔH = –242 kJ mol–1

Figure 4.13

4 The heat of reaction is the sum of the energy absorbed in breaking the bonds and the energy released in forming the bonds.

6 (a) If the energy absorbed to break the bonds is more than the energy released to form new bonds, then the reaction is endothermic. That is,

ΔH (reaction) = ΔH1 (bond breaking) + ΔH2 (bond forming)

ΔH (reaction) is positive if ΔH1 (bond breaking) > ΔH2 (bond forming)

5 (a) If the energy absorbed to break the covalent bonds is less than the energy released to form the new covalent bonds, the reaction is exothermic. That is,

For example, the reaction between carbon and steam to form carbon monoxide and hydrogen is an endothermic reaction.

ΔH (reaction) is negative if ΔH1 (bond breaking) < ΔH2 (bond forming)

C(s) + H2O(g) → CO(g) + H2(g); ΔH = positive

For example, the reaction between hydrogen and chlorine to form hydrogen chloride is an exothermic reaction. H2(g) + Cl2(g) → 2HCl(g);

(b) Figure 4.14 shows the breaking and the formation of bonds when carbon reacts with steam. ΔH1 (bond breaking) = +1645 kJ ΔH2 (bond forming) = –1513 kJ ΔH (reaction) = ΔH1 + ΔH2 = (+1645) + (–1513) = +132 kJ

ΔH = negative

(b) Figure 4.12 shows the breaking and formation of bonds when hydrogen reacts with chlorine.

Figure 4.12 The breaking and formation of bonds in the reaction between hydrogen and chlorine Thermochemistry

Figure 4.14 The breaking and formation of bonds in the reaction between carbon and steam

460

(b) The reaction is very endothermic and the water temperature can drop as much as 18 oC, depending on the amount of ammonium nitrate used.

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The equation for the reaction between hydrogen and iodine is shown below. H2(g) + I2(s) → 2HI(g);

NH4NO3(s) + water → NH4NO3(aq); ΔH = +26 kJ mol–1

ΔH = +11.3 kJ

Which of the following shows that the reaction is endothermic? A The reaction releases 11.3 kJ of heat energy when 2 mol of hydrogen iodide are formed. B The total heat absorbed to break the bonds is more than the total energy released during the formation of hydrogen iodide. C The energy contained in hydrogen and iodine is higher than the energy contained in hydrogen iodide. D The reaction is a slow reaction.

4 (a) There are several varieties of hot packs. One variety of hot pack contains a small bag of water and a dry chemical such as anhydrous calcium chloride or anhydrous magnesium sulphate. When the pack is squeezed, the small bag breaks and the anhydrous salt dissolves in water. The dissolving process is very exothermic. CaCl2(s) + water → Ca2+(aq) + 2Cl–(aq); ΔH = –81 kJ mol–1

Comments This is an endothermic reaction. The reaction absorbs heat energy. In an endothermic reaction, the heat content of the products is higher than the heat content of the reactants.

MgSO4(s) + water → Mg2+(aq) + SO42–(aq); ΔH = –91 kJ mol–1 (b) Another variety of hot pack uses the oxidation of iron to produce heat. The hot pack contains wet iron powder and sodium chloride put in a perforated bag. When the perforated bag is taken out and exposed to the air, a very exothermic reaction occurs as iron reacts with oxygen.

Answer B

Applications of Exothermic and Endothermic Reactions in Everyday Life

4Fe(s) + 3O2(g) → 2Fe2O3(s); ΔH = –1648 kJ

1 Ice packs and hot packs are used to reduce swelling and pain due to muscle or joint sprain. In the hospitals, cold packs are put on the foreheads of patients to reduce fever. 2 In actual fact, an ice pack does not contain SPM ice but contains chemicals that can react ’11/P1 endothermally with water. Thus, ice packs are often called cold packs. 3 (a) Cold packs contain ammonium nitrate in a strong bag and water in a thin inner bag. When the cold pack is squeezed, the inner bag containing water will break. The water then reacts with ammonium nitrate (Figure 4.15).

The presence of water on the iron surface and sodium chloride increases the rate of reaction.

3

’06

The following chemical equation shows the redox reaction between zinc and copper(II) oxide: Zn(s) + CuO(s) → ZnO(s) + Cu(s); ΔH = –193 kJ mol–1 Which of the following is true about this reaction?

Type of reaction A Endothermic B Endothermic C Exothermic D Exothermic

Heat change Heat is released Heat is absorbed Heat is absorbed Heat is released

Comments ΔH has a negative value. So the reaction is exothermic. Answer D

Figure 4.15 The structure of a cold pack

461

Thermochemistry

4

2

Exothermic reactions • Exothermic reactions are reactions that release heat energy to the surroundings. • Energy level diagram • Energy profile diagram

• ΔH is negative because ΔH absorbed for bond breaking < ΔH released for bond forming. • Hot packs contain chemicals that react with water to give out heat.

Application • Cold pack and hot pack

4

Energy changes in chemical reactions

Endothermic reactions • Endothermic reactions are reactions that absorb heat energy from the surroundings. • Energy level diagram

• Energy profile diagram

• ΔH is positive because ΔH for bond breaking > ΔH released for bond forming. • Hot packs contain chemicals that react with water to absorb heat.

4.1 1 Classify the following as exothermic or endothermic reactions: (a) Dissolving ammonium sulphate solid in water (b) Dissolving sodium hydroxide in water (c) Adding sodium hydrogen carbonate to dilute hydrochloric acid (d) Adding sodium hydroxide solution to dilute hydrochloric acid 2 When sodium carbonate solid dissolves in water, the following reaction occurs:

Thermochemistry

water Na2CO3(s) ⎯⎯⎯⎯→ Na2CO3(aq); ΔH = –23 kJ mol–1 (a) Describe the changes in temperature of the solution. (b) Construct an energy level diagram for the reaction. 3 When a hot pack is squeezed, a chemical reaction occurs. (a) Is the reaction exothermic or endothermic? (b) Complete the table shown below describing the characteristic of this reaction.

462

Characteristic

(a) Identify the step in the reaction that is exothermic. (b) Identify the step in the reaction that is endothermic. (c) State two facts that are given in the above energy level diagram.

Reaction in the hot pack

(i) Temperature change (ii) Energy content of reactants and products

5 (a) Using the following data, calculate the mass of anhydrous copper(II) sulphate needed to dissolve in water so that 16.5 kJ of heat are released.

(iii) Energy involved in bond breaking and bond forming

CuSO4(s) + water → CuSO4(aq); ΔH = –66 kJ mol–1

4 The energy changes for the reaction between nitrogen and oxygen are represented in the energy level diagram as shown below.

[Relative atomic mass: Cu, 64; S, 32; O, 16] (b) The following equation shows the reaction between nitrogen and oxygen. N2(g) + 2O2(g) → 2NO2(g); ΔH = +66 kJ mol–1

4

Calculate the energy change when 12.0 dm3 of nitrogen reacts with 24.0 dm3 of oxygen.

4.2

N2(g) + 3H2(g)

Heat of Precipitation

ΔH = –92 kJ …(2)

The Concept of Heat of Reaction

Equation (1) shows that 46 kJ of heat energy are released when 1.0 mol of 1 ammonia is formed from — mol of nitrogen 2 3 and — mol of hydrogen. 2 Hence, the heat of reaction for the formation of ammonia is –46 kJ mol–1. (b) Equation (2) shows that when 2.0 mol of ammonia are produced from 1.0 mol of nitrogen and 3.0 mol of hydrogen, 92 kJ of heat energy are released. 3 (a) The heat of reaction also depends on the direction of the reaction. For example,

1 (a) The heat of reaction is the heat energy SPM absorbed or released when the number of ’06/P1 moles of reactants, as shown in the chemical equation, react to form the products. For example, Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g); ΔH = –27 kJ Hence, the heat of reaction (ΔH) is –27 kJ. (b) The chemical equation that contains the value of ΔH on the right of the equation is called the thermochemical equation. (c) The thermochemical equation in (a) shows that 27 kJ of heat are released, when 1.0 mol of iron(III) oxide reacts with 3.0 mol of carbon monoxide to form 2.0 mol of iron and 3.0 mol of carbon dioxide. 2 (a) The heat of reaction depends on the number of moles of reactants that are involved as shown by the equation. For example, 1 3 — N2(g) + — H2(g) 2 2

2NH3(g);

1 3 — N2(g) + — H2(g) NH3(g); 2 2 ΔH = –46 kJ mol–1 … (3) NH (g) 3

1 3 — N2(g) + — H2(g); 2 2 ΔH = +46 kJ … (4)

(b) If the formation of 1.0 mol of ammonia from nitrogen and hydrogen (equation (3)) releases 46 kJ of heat energy, then the decomposition of 1.0 mol of ammonia into nitrogen and hydrogen (equation (4)) will absorb 46 kJ of heat energy.

NH3(g);

ΔH = –46 kJ mol–1…(1) But, 463

Thermochemistry

4

For an exothermic reaction, θ = rise in temperature θ = T2 – T1

’06

What is meant by heat of reaction? A The energy needed to start a reaction. B The energy released on forming chemical bonds. C The change in the energy contained in the reactants and in the products. D The energy absorbed or released when matter changes its physical state.

For an endothermic reaction, θ = fall in temperature θ = T1 – T2 (T1 = initial temperature of solution, T2 = final temperature of solution) 4 In calculating heat of reaction, 1 cm3 of any aqueous solution is assumed to have a density of 1.0 g cm–3. That is, 1 cm3 of any aqueous solution has a mass of 1.0 g. 5 The other assumptions for calculating heat of reaction are as follows: (a) There is no loss of heat energy to the surroundings or gain of heat energy from the surroundings. (b) The container, the thermometer and all other apparatus used in the experiment absorb a negligible amount of heat.

Solution Heat of reaction (ΔH) = Hproducts – Hreactants where H is total heat content. Answer C

4

4 The heat of reaction can be further classified as shown below, depending on the type of reaction that occurs. Heat of reaction

Type of reaction

(a) Heat of precipitation (b) Heat of displacement

Precipitation of an ionic compound Displacement of a metal from its salt solution by another metal Neutralisation reaction between an acid and a base The reaction between an element or a compound with oxygen on burning

(c) Heat of neutralisation (d) Heat of combustion

3 When a spatula of ammonium chloride is dissolved in 300 cm3 of water at an initial temperature of 28 °C, the minimum temperature obtained is 23 °C. Calculate the heat change for the reaction. (The specific heat capacity of solution = 4.2 J g–1 °C–1). Solution Fall in temperature = 28 – 23 = 5 °C Heat energy absorbed = mcθ = 300  4.2  5 = 6300 J = 6.3 kJ Heat change = +6.3 kJ

SPM

Calculations Involving Heat Changes

’04/P1 ’05/P1 ’06/P1

1 The unit for measuring heat energy is joule (J). 2 The specific heat capacity of a solution is the heat energy required to raise the temperature of 1.0 g of the solution by 1.0 °C.

5

When 25.0 cm3 of dilute sulphuric acid is poured into 25.0 cm3 of sodium hydroxide solution, 4200 J of heat are released. What is the temperature change of the reaction mixture? [Specific heat capacity of the solution = 4.2 J g–1 °C–1; assume that the density of the solution = 1 g cm–3] A 2 °C B 5 °C C 10 °C D 20 °C

The specific heat capacity of water is 4.18 J g–1 °C–1. This means that 4.18 J of heat energy are required to raise the temperature of 1.0 g of water by 1.0 °C. 3 The amount of heat (in J) released or absorbed in a reaction can be determined by using the following formula: ΔH = mcθ

Solution ΔH = mcθ 4200 = (25 + 25) 3 4.2 3 θ θ = 20 °C Answer D

where m = mass of solution, c = specific heat capacity of solution, θ = change in temperature. Thermochemistry

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464

Solution Step 1: Calculate the heat energy released in the reaction (a) The average temperature of the solution before reaction

When 10 g of ammonium chloride is dissolved in 250 cm3 of distilled water, the temperature of the solution falls by 3 °C. Heat change = 250  4.2  3 J The heat absorbed on dissolving ammonium chloride lowers the temperature of water. The mass of ammonium chloride used is not included in the calculation.

= 30.0 °C Temperature change

SPM

’04/P1 ’06/P1 ’07/P1

= 35.6 – 30.0

1 The heat of precipitation is the heat change when 1.0 mol of a precipitate is formed from its ions. 2 The heat of precipitation of lead(II) sulphate is –51 kJ mol–1. Based on this information, we can write the thermochemical equation for the reaction as follows.

= 5.6 °C (b) Volume of reaction mixture = 25 + 25 = 50 cm3 Assuming that the density of solution = 1.0 g cm–3

Pb2+(aq) + SO42–(aq) → PbSO4(s); ΔH = –51 kJ mol–1

Heat energy released = mcθ = 50  4.2  5.6 = 1176 J = 1.176 kJ

3 The energy level diagram for the precipitation reaction is shown in Figure 4.16.

Step 2: Calculate the number of moles of AgCl produced Ag+(aq) + Cl–(aq) → AgCl(s) 0.023 mol of Ag+ ions react with 0.023 mol of Cl– ions to produce 0.023 mol of AgCl. Step 3: Calculate the heat of precipitation of AgCl The precipitation of 0.023 mol of AgCl releases 1.176 kJ of heat energy. Therefore, the precipitation of 1.0 mol of AgCl releases 1 1.176  — — — — 0.023 mol → 1.176 kJ 0.023 1.0 mol → ? kJ

Figure 4.16 The energy level diagram for the precipitation of lead(II) sulphate

4

= 51.1 kJ of heat energy.

25 cm of silver nitrate solution is added to 25 cm of potassium chloride solution. The results of the experiment are shown below. Initial temperature of potassium chloride solution = 30.5 °C Initial temperature of silver nitrate solution = 29.5 °C Highest temperature of reaction mixture = 35.6 °C Calculate the heat of precipitation of silver chloride. (Specific heat capacity of solution = 4.2 J g–1 °C–1; the solutions contain 0.023 mol of Ag+ ions and 0.023 mol of Cl– ions respectively) 3

Hence, heat of precipitation = –51.1 kJ mol–1

3

Determining the Heat of Precipitation

SPM

’04/P2 ’07/P1

1 The heat of precipitation for the ionic solid Mn+Xn– is determined by measuring the temperature change during precipitation when a given volume of a solution of Mn+ ions is added to a given volume of a solution of Xn– ions. The concentrations of both the solutions are known. 465

Thermochemistry

4

The Concept of Heat of Precipitation

30.5 + 29.5 =— — — — — — — — — 2

2 The following steps are involved when calculating the heat of precipitation. Step 1 Calculate the number of moles of solid precipitated (x mol)

Step 2

Step 3

Calculate the heat energy released mcθ ΔH = — — — — = y kJ 1000

Heat of precipitation –y =— — kJ mol–1 x

3 Activity 4.2 shows the experimental procedure for measuring the heat of precipitation of silver chloride.

Crystallisation and precipitation are exothermic reactions. In one type of hot pack, the crystallisation of sodium ethanoate is used to produce heat.

Precipitation, like crystallisation, is the reverse of dissolving. If a solid comes out of a solution slowly, crystals are formed and the process is called crystallisation. But if the solid is formed very quickly, many tiny particles are produced in the liquid. This process is called precipitation.

4

CH3COONa + 3H2O → CH3COONa.3H2O; ΔH = –37.9 kJ mol–1

To determine the heat of precipitation of silver chloride Apparatus

Measuring cylinders, thermometer and plastic cup.

Materials

0.5 mol dm–3 silver nitrate solution and 0.5 mol dm–3 sodium chloride solution.

5 The mixture is stirred with a thermometer throughout the experiment and the highest temperature obtained is recorded.

Activity 4.2

Procedure

Temperature

NaCl(aq)

Highest temperature obtained (°C)

32.0

Initial temperature of sodium chloride solution (°C)

29.0

Initial temperature of silver nitrate solution (°C)

28.0

Figure 4.17 To determine the heat of precipitation of silver chloride

Assumptions used for calculations The specific heat capacity of solution = 4.2 J g–1 °C–1 Density of solution = 1.0 g cm–3

1 25 cm3 of 0.5 mol dm–3 sodium chloride solution is measured and poured into a clean and dry plastic cup using a measuring cylinder. 2 The initial temperature of sodium chloride solution is measured and recorded. 3 Using another measuring cylinder, 25 cm3 of 0.5 mol dm–3 silver nitrate solution is measured. The initial temperature of the silver nitrate solution is measured and recorded. 4 The silver nitrate solution is poured quickly and carefully into the sodium chloride solution (Figure 4.17).

Calculations Step 1: Calculate the heat energy released in the experiment: (a) The average initial temperature of the solutions before mixing 29 + 28 =— — — — — — — = 28.5 °C 2 The highest temperature of the reaction mixture = 32.0 °C Rise in temperature = (32.0 – 28.5) °C = 3.5 °C

Thermochemistry

466

Discussion 1 The reaction between silver nitrate and sodium chloride solutions produces a white precipitate of silver chloride. 2 The reaction between silver nitrate and sodium chloride can be represented by a chemical equation or an ionic equation. (a) Chemical equation: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

Ag+(aq) + Cl–(aq) → AgCl(s); ΔH = –58.8 kJ mol–1 4 In actual fact, the theoretical heat of precipitation of silver chloride is –66 kJ mol–1. Ag+(aq) + Cl–(aq) → AgCl(s); ΔH = –66 kJ mol–1 5 The heat of precipitation obtained experimentally is usually less than the theoretical value. This is because the calculation of the theoretical value is based on two assumptions, namely, • there is no loss of heat to the surroundings, • the plastic cup and the thermometer used in the experiment do not absorb heat. 6 In order to prevent the loss of heat to the surroundings, the following precautionary steps must be taken when carrying out the experiment. (a) Plastic cups (for example, polystyrene or polythene cup) are used because plastics are poor conductors of heat. (b) The thermometer must be placed in the solution for a few minutes before taking the reading. This is to ensure that the solution has reached a uniform temperature. (c) The solutions containing the reactants are mixed quickly, so that the reaction can be completed within a short time. (d) The reaction mixture in the plastic cup must be stirred slowly and continuously so that the temperature of the solution is uniform. (e) The thermometer should not be taken out of the reaction mixture when taking the reading. (f) The thermometer reading should be observed throughout the reaction so that the highest temperature of the reaction mixture can be recorded accurately. 7 The energy level diagram for the precipitation of silver chloride is shown in Figure 4.18.

(b) Ionic equation: Ag+(aq) + Cl–(aq) → AgCl(s) The ionic equation does not contain nitrate ions (NO3–) and sodium ions (Na+). This is because NO3– and Na+ ions are spectator ions and do not participate in the precipitation reaction. 3 From the experimental results, the heat of precipitation of AgCl is –58.8 kJ mol–1. The thermochemical equation for the precipitation of silver chloride is AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq); ΔH = –58.8 kJ mol–1 or

Figure 4.18 Energy level diagram for the precipitation of silver chloride

Conclusion The heat of precipitation of silver chloride is –58.8 kJ mol–1. 467

Thermochemistry

4

(b) Heat energy released = mass of solution  specific heat capacity  rise in temperature = (25 + 25)  4.2  3.5 = 735 J = 0.735 kJ Step 2: Calculate the number of moles of AgCl formed (a) Number of moles of Ag+ ions used 25 = 0.5  — — — — = 0.0125 1000 (b) Number of moles of Cl– ions used 25 = 0.5  — — — — = 0.0125 1000 Ag+(aq) + Cl–(aq) → AgCl(s) From the equation, 1.0 mol of Ag+ ions react with 1.0 mol of Cl– ions to form 1.0 mol of AgCl. Therefore, 0.0125 mol of Ag+ ions react with 0.0125 mol of Cl– ions to form 0.0125 mol of AgCl. Step 3: Calculate the heat of precipitation of silver chloride From steps 1 and 2: Precipitation of 0.0125 mol of AgCl releases 0.735 kJ of heat energy. Precipitation of 1.0 mol of AgCl will release 1 0.735  — — — — — = 58.8 kJ of heat energy 0.0125

If potassium chloride solution is used to replace the sodium chloride in Activity 4.2, the heat of precipitation of silver chloride will still be the same. AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq);

ΔH = –58.8 kJ mol–1

This is because K+ ions, like Na+ ions, do not take part in the reaction.

Calculations Involving the Heat of Precipitation

5 Step 2: Calculate the heat energy released when 0.06 mol of PbSO4 is precipitated

Consider the following equation:

4

Pb2+(aq) + SO42–(aq) → PbSO4(s); ΔH = –50 kJ mol–1

The thermochemical equation shows that the heat of precipitation of PbSO4 is –50 kJ mol–1. This means that the precipitation of 1.0 mol of PbSO4 releases 50 kJ of heat energy. Therefore, heat energy released by the formation of 0.06 mol of PbSO4 = 0.06  50 = 3.0 kJ

Calculate the highest temperature reached when 60 cm3 of 1.0 mol dm–3 lead(II) nitrate solution reacts with 60 cm3 of 2.0 mol dm–3 sodium sulphate solution. The initial temperatures of both the solutions are 29 °C. [Specific heat capacity of solution = 4.2 J g–1 °C–1; density of solution = 1 g cm–3]

Step 3: Calculate the rise in temperature

Solution Step 1: Calculate the number of moles of lead(II) sulphate precipitated in this experiment

Let the rise in temperature = θ Heat energy released (in J) = mcθ = (60 + 60)  4.2  θ = 504  θ

Number of moles of Pb2+ ions used 1.0  60 =— — — — — — — = 0.06 1000

From step 2, heat energy released = 3.0 kJ = 3  1000 J

Number of moles of SO42– ions used 2.0  60 =— — — — — — — 1000 = 0.12

3  1000 = 504  θ 3  1000 θ= — — — — — — — 504 = 6 °C

Pb2+(aq) + SO42–(aq) → PbSO4(s) From the equation, 1.0 mol of Pb2+ ions reacts with 1.0 mol of SO42– ions to form 1.0 mol of PbSO4. 0.06 mol of Pb2+ ions react with 0.06 mol of SO42– ions to form 0.06 mol of PbSO4.

Thermochemistry

θ = final temperature – initial temperature 6 = final temperature – 29 Final temperature = 29 + 6 = 35 °C Highest temperature reached is 35 °C.

468

Ag+(aq) + Cl–(aq) → AgCl(s); ΔH = –66 kJ mol–1 That is, heat of precipitation of silver chloride is –66 kJ mol–1.

Constructing the energy level diagram

Calculations involving heat of precipitation (a) Calculate the number of moles of salt precipitated = x mol mcθ (b) Calculate the heat energy released (ΔH) = — — — — — — — — — — — kJ = y kJ 1000 (c) From (a) and (b): Precipitation of x mole of salt releases y kJ of heat energy. y (d) From (c): Heat of precipitation = – — — — kJ mol–1 x

4.2 1 The heat of precipitation of silver chloride is –66 kJ mol–1. In an experiment, 200 cm3 of 0.1 mol dm–3 AgNO3(aq) is added to 100 cm3 of 0.1 mol dm–3 CaCl2(aq). (a) Write (i) the chemical equation, (ii) the ionic equation, for the reaction between silver nitrate and calcium chloride. (b) Calculate the heat change in the experiment. 2 When 100 cm3 of 0.5 mol dm–3 of a nitrate salt, M(NO3)2 at 29.5 °C is added to 100 cm3 of 1.0 mol dm–3 sodium carbonate solution, the maximum temperature reached is 30.5 °C. [Specific heat capacity of solution = 4.2 J g–1 °C–1; density of solution = 1.0 g cm–3] (a) Write the ionic equation for the reaction. (b) Calculate the heat energy released in the experiment.

4.3

Heat of Displacement

(c) Calculate the heat of precipitation of the metal carbonate. (d) Sketch the energy level diagram for the reaction. 3 When 100 cm3 of 0.5 mol dm–3 metal nitrate solution, M(NO3)2, reacts with 100 cm3 of 1.0 mol dm–3 potassium chloride solution, the temperature of the reaction mixture increases by 10 °C. M(NO3)2(aq) + 2KCl(aq) → MCl2(s) + 2KNO3(aq) What is the effect (if any) on the temperature change if the experiment is repeated using (a) 200 cm3 of 0.5 mol dm–3 M(NO3)2 solution and 200 cm3 of 1.0 mol dm–3 KCl solution? (b) 100 cm3 of 1.0 mol dm–3 M(NO3)2 solution and 100 cm3 of 2.0 mol dm–3 KCl solution?

SPM

SPM

’11/P1

’08/P1, ’10/P1

1 The heat of displacement is the heat released when 1.0 mol of a metal is displaced from its salt solution by a more electropositive metal. 2 (a) The following thermochemical equation shows that the heat of displacement of iron by magnesium is –202 kJ mol–1.

Mg(s) + FeCl2(aq) → MgCl2(aq) + Fe(s); ΔH = –202 kJ mol–1

(b) When excess magnesium powder is added to iron(II) chloride solution, 202 kJ of heat are released when 1.0 mol of iron is displaced from iron(II) chloride solution. (c) In this reaction, the colour of the solution turns from green to colourless and iron powder (grey in colour) is produced. 469

Thermochemistry

4

The meaning of heat of precipitation • Heat of precipitation is the heat change when 1.0 mol of the precipitate is formed from its ions. Example:

Determining the heat of precipitation • Mix a known volume of solution of M n+ with a known volume of solution of X n–. The concentrations of the two solutions are known. • Measure the highest temperature reached.

3 Figure 4.19 shows the energy level diagram for the displacement reaction between magnesium and iron(II) chloride.

Determining the Heat of Displacement 1 The heat of displacement is determined by adding excess metal powder to a given volume of the salt of another metal and measuring the highest temperature reached. The metal chosen must be more electropositive than the metal to be displaced and the concentration of the salt solution is known. 2 Activity 4.3 shows the method used to determine the heat of displacement of copper by zinc and iron from copper(II) sulphate solution.

Figure 4.19 The energy level diagram for the displacement reaction of iron by magnesium

4

To determine the heat of displacement of copper by (a) zinc, (b) iron Apparatus

Materials

Thermometer, plastic cup, measuring cylinder, weighing bottle and electronic balance.

Results Type of metal

0.2 mol dm–3 copper(II) sulphate solution, zinc powder and iron powder.

Procedure

Activity 4.3

Figure 4.20 To determine the heat of displacement

1 Using a measuring cylinder, 25.0 cm3 of 0.2 mol dm–3 copper(II) sulphate solution is measured into a plastic cup. The temperature of the solution is measured with a thermometer. 2 About 0.5 g of zinc powder (in excess) is weighed out. 3 The zinc powder is poured into copper(II) sulphate solution in the plastic cup. The contents of the plastic cup is then stirred with a thermometer (Figure 4.20) and the highest temperature reached is recorded. 4 Steps 1 to 3 are repeated using iron powder to replace zinc powder. Thermochemistry

SPM

’05/P2 /SA

Zinc powder Iron powder

Highest temperature obtained (°C)

41

37.5

Initial temperature of copper(II) sulphate solution (°C)

31

30.5

Increase in temperature (°C)

10

7.0

Calculations Assumptions Specific heat capacity of solution = 4.2 J g–1 oC–1 Density of solution = 1.0 g cm–3 (A) Heat of displacement of copper by zinc Step 1: To determine the heat energy released in this experiment Heat energy released = mass of solution  specific heat capacity of solution  increase in temperature = 25  4.2  10 J = 1050 J = 1.05 kJ Step 2: To calculate the number of moles of copper displaced by zinc Number of moles of Cu2+ ions used 0.2  25 =— — — — — — — 1000 = 0.005 Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) 470

Step 3: To determine the heat energy released when 1 mol of copper is displaced From steps 1 and 2: Displacement of 0.005 mol of copper metal releases 1.05 kJ of heat. Therefore, heat energy released when 1.0 mol of copper is displaced 1 = 1.05  — — — — — 0.005 = 210 kJ (B) Heat of displacement of copper by iron Step 1: To determine the heat energy released in this experiment Heat energy released = mass of solution  specific heat capacity of solution  increase in temperature = 25  4.2  7 J = 735 J = 0.735 kJ Step 2: To calculate the number of moles of copper displaced by iron Number of moles of Cu2+ ions used 0.2  25 =— — — — — — — 1000 = 0.005 Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) From the equation, 1.0 mol of Cu2+ ions produces 1.0 mol of copper metal. Therefore, 0.005 mol of Cu2+ ions produce 0.005 mol of copper metal. Step 3: To determine the heat energy released when 1 mol of copper is displaced From steps 1 and 2: Displacement of 0.005 mol of copper metal releases 0.735 kJ of heat. Therefore, heat energy released when 1.0 mol of copper is displaced 1 = 0.735  — — — — — 0.005 = 147 kJ Discussion 1 Zinc and iron are more electropositive than copper. This means that zinc and iron will displace copper from an aqueous solution of copper salt. The displacement reaction is an exothermic reaction.

2 The copper metal displaced from the salt will be precipitated as a brown solid. Chemical equation: Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) Ionic equation: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) 3 Chemical equation for the displacement of copper by iron: Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s) Ionic equation: Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) 4 Excess zinc and iron are used to make sure that all the copper from the copper(II) salt are displaced. 5 When calculating the heat of displacement of copper, the heat absorbed by the solids (copper produced, excess zinc and iron remaining in the solution) is very small and can be ignored. 6 The following precautionary steps must be taken when carrying out the experiment: (a) The initial temperature of copper(II) sulphate solution is taken only after the thermometer has been placed in the solution for a few minutes. This is to make sure that the solution has reached a uniform temperature. (b) The reactants must be mixed quickly so that the reaction can be completed in the shortest time. In this way, the loss of heat energy to the surroundings can be minimised. (c) The reaction mixture in the plastic cup must be stirred slowly and continuously so that the temperature of the solution is uniform. (d) The thermometer reading must be observed continuously so that the highest temperature reached can be recorded. This is to ensure that the reaction has completed and the heat of reaction has all been released. (e) The zinc and iron metals used are in powdered form to ensure that the reaction proceeds rapidly and completely. In this way, the loss of heat to the surroundings can be reduced. Metal powder has a larger surface area compared with pieces of metals with the same mass. 7 Figure 4.21 shows the energy level diagram for the displacement reaction between zinc and copper(II) ions. 471

Thermochemistry

4

From the equation, 1.0 mol of Cu2+ ions produces 1.0 mol of copper metal. Therefore, 0.005 mol of Cu2+ ions produce 0.005 mol of copper metal.

8 Figure 4.22 shows the energy level diagram for the displacement of copper by iron.

Figure 4.21 Figure 4.22

Conclusion (a) The heat of displacement of copper by zinc from aqueous solution of Cu2+ ions is –210 kJ mol–1. (b) The heat of displacement of copper by iron from aqueous solution of Cu2+ ions is –147 kJ mol–1. Calculations Involving Heat of Displacement

6

4

Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s) … (1) Zn(s) + Fe2+(aq) → Zn2+(aq) + Fe(s) … (2)

SPM

’07/P1

When excess aluminium powder is added to 100 cm3 of iron(II) sulphate solution, 0.48 g of iron is produced and the temperature of the solution changes from 28 °C to 33.5 °C. Calculate the heat of displacement of iron by aluminium.

In reaction 1, 1 mol of iron is displaced by 1 mol of magnesium, but in reaction 2, 1 mol of iron is displaced by 1 mol of zinc. Because different reactants are used, the heat of displacement will not be the same for these two reactions.

3Fe2+(aq) + 2Al(s) → 3Fe(s) + 2Al3+(aq)

6

’05

[Specific heat capacity of solution = 4.2 J g–1 °C–1; density of solution = 1.0 g cm–3; relative atomic mass of iron = 56].

The following equations show the values of heat of displacement of copper by zinc, iron and metal M.

Solution Step 1: Calculate the heat energy released in this experiment (a) Rise in temperature = 33.5 – 28 = 5.5 °C (b) Heat energy released = mass of solution  specific heat capacity  rise in temperature = 100  4.2  5.5 J = 2310 J = 2.31 kJ

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s); ∆H = –210 kJ mol–1 Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s); ∆H = –150 kJ mol–1 M(s) + CuSO4(aq) → MSO4(aq) + Cu(s); ∆H = –100 kJ mol–1 Predict metal M by choosing from the list: aluminium, magnesium and tin. Comments Iron is less electropositive than zinc. The heat of displacement of copper by iron is less than the heat of displacement of copper by zinc. The heat of displacement of copper by metal M is less than the heat of displacement of copper by iron. Hence, metal M is less electropositive than iron.

Step 2: Calculate the number of moles of iron displaced Relative atomic mass of iron = 56 Number of moles of Fe produced mass =— — — — — — — — — — — — — — — — — relative atomic mass 0.48 =— — — — 56 = 0.00857

Solution Both Al and Mg are more electropositive than iron. Hence, metal M is tin, which is less electropositive than iron.

Thermochemistry

472

Step 3: Calculate the heat of displacement of iron by aluminium Displacement of 0.00857 mol of Fe releases 2.31 kJ of heat energy Heat released from the displacement of 1.0 mol of Fe 1 = 2.31  — — — — — — — = 269.5 kJ 0.00857 Heat of displacement of iron by aluminium is –269.5 kJ mol–1.

Comments Zn is oxidised. Oxidation number of zinc increases from 0 in Zn to +2 in ZnSO4. The negative value of ΔH shows that the reaction is exothermic. This means that the temperature increases during the reaction. The thermochemical equation shows that 210 kJ of heat are released when 1 mol (64 g) of copper is displaced.

’03

The equation below shows the displacement reaction of copper metal and its heat of reaction.

2.56 Number of moles of Cu displaced = — — — 64 = 0.04

Zn + CuSO4 → ZnSO4 + Cu; ΔH = –210 kJ mol–1

Heat released = 210  0.04 = 8.4 kJ That is, heat change (ΔH) is –8.4 kJ (not + 8.4 kJ). Answer C

[Relative atomic mass of copper is 64] Which of the following statements are true about the reaction represented by the above equation? I Zinc is oxidised. II The reaction is exothermic.

Heat of displacement • Heat of displacement is the heat released when 1.0 mol of a metal is displaced from its salt solution by another more electropositive metal. For example, Mg(s) + Fe2+(aq) → Fe(s) + Mg2+(aq);

∆H = –202 kJ mol–1

ΔH is the heat of displacement of iron by magnesium.

Method for finding heat of displacement • Add excess of a more electro­ positive metal (in pow­dered form) to a known volume of a salt solution of another metal of known concentration. • Measure the highest tempe­ rature reached.

Energy level diagram

Calculations (a) Calculate the number of moles of metal displaced = x mol (b) Calculate the heat energy evolved mcθ ΔH = ———— = y kJ 1000 (c) From (a) and (b): Displacement of x mol of metal produces y kJ of heat. (d) Heat of displacement y = – —— kJ mol–1 x

473

Thermochemistry

4

7

III The heat change if 2.56 g of copper metal is displaced is +8.4 kJ. IV The temperature of the reaction mixture increases during the reaction. A I and III only C I, II and IV only B II and IV only D I, III and IV only

4.3 1 You are asked to carry out an experiment to determine the heat of displacement of lead by magnesium. (a) (i) Name the chemicals needed for the experiment. (ii) Which of the chemicals must be used in excess? Why? (b) (i) Draw a labelled diagram to show the apparatus set-up for the experiment. (ii) What other apparatus are needed besides the ones shown in your diagram? (c) Based on your answer in (b), select the data that is (i) required, (ii) not required for calculating the heat of displacement.

If the initial temperature of the solution is 30 °C, what is the maximum temperature reached in this experiment? [Specific heat capacity of solution = 4.2 J g–1 °C–1; relative atomic mass of Fe = 56] 4 Five experiments were carried out to determine the heat of displacement of iron by magnesium. In each experiment, an increasing amount of magnesium was added to 50 cm3 of 0.25 mol dm–3 iron(II) chloride. The maximum increase in temperature of the solution were recorded.

2 The equation below shows a displacement reaction and its heat of reaction. ’08

4

Zn + CuSO4 → ZnSO4 + Cu; ΔH = –210 kJ mol–1 What is the change in heat energy if 2.56 g of copper is displaced? [Relative atomic mass of copper: 64] 3 An experiment is carried out by adding 0.7 g of iron powder to 50 cm3 of 0.50 mol dm–3 copper(II) chloride ’03 solution. The energy level diagram for the reaction between iron and copper(II) chloride solution is shown in Figure 4.23.

Heat of Neutralisation

Mass of magnesium (g)

Maximum increase in temperature (°C)

A

0.10

4

B

0.15

6

C

0.2

8

D

0.3

12

E

0.35

12

[Specific heat capacity of solution = 4.2 J g –1 °C–1; density of solution = 1.0 g cm–3] (a) Why was the rise in temperature increasing in the first four experiments? (b) Why was the rise in temperature for experiments D and E constant? (c) Calculate the heat of displacement of iron by magnesium. (d) Another experiment is carried out by adding magnesium powder to a solution of iron(II) chloride in excess. What is the mass of magnesium required to release 252 kJ of heat? [Relative atomic mass of magnesium = 24.3]

Figure 4.23

4.4

Experiment

SPM

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l); ΔH = –57.3 kJ mol–1

’05/P1, ’10/P1

The Meaning of Heat of Neutralisation

4 Figure 4.24 shows the energy level diagram for SPM the neutralisation reaction between nitric acid ’06/P1 and sodium hydroxide solution.

1 All neutralisation reactions are exothermic reactions, that is, ΔHneut has a negative value. 2 The heat of neutralisation is the heat released when 1.0 mol of H+ ions react with 1.0 mol of OH– ions to produce 1.0 mol of water molecules. H+(aq) + OH—(aq) → H2O(l); ΔH = negative value 3 For example, the heat of neutralisation of nitric acid with sodium hydroxide solution is –57.3 kJ mol–1. Thermochemistry

Figure 4.24 Energy level diagram for the reaction between HNO3(aq) and NaOH(aq)

474

5 For the neutralisation reaction between ethanoic acid and sodium hydroxide solution, the heat of neutralisation is –55.2 kJ mol–1 (Figure 4.25).

8

’07 ’08

2.4 g of magnesium is added to 100 cm3 of 2.0 mol dm–3 copper(II) chloride solution at 30 °C. The heat of reaction is –4.20 kJ mol–1. What is the highest temperature for this experiment? [Specific heat capacity of solution = 4.21 g–1 °C–1; relative atomic mass of magnesium = 24].

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l); ΔH = –55.2 kJ mol–1

Solution Number of moles of Mg 2.4 =— — 24 = 0.1 Number of moles of Cu2+ MV =— — — — 1000

4

2.0 3 100 =— — — — — — — — — 1000

Figure 4.25 Energy level diagram for the reaction between CH3COOH(aq) and NaOH(aq)

= 0.2 Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s) Number of moles of Cu displaced = 0.1

6 Hydrochloric acid, nitric acid and ethanoic acid are monobasic acids and sulphuric acid is a dibasic acid. The heat of neutralisation between sulphuric acid and sodium hydroxide solution is –57.3 kJ mol–1 and the thermochemical equation can be written as follows:

Heat of reaction = –4.2 kJ mol–1 0.1 ΔH for the reaction = — — — 3 4.2 kJ 1.0 = 0.42 kJ = 420 J Let temperature rise = θ °C ΔH = mcθ 420 = 100 3 4.2 3 θ θ = 1 °C

1 1 — H SO (aq) + NaOH(aq) → — Na2SO4(aq) 2 2 4 2 + H2O(l); ΔΔH = –57.3 kJ mol–1

Highest temperature = (30 + 1) °C = 31 °C (ΔH represents the heat released on the formation of 1 mol of water).

Determination of the Heat of Neutralisation The reaction between hydrogen chloride gas and sodium hydroxide solution is a neutralisation reaction. However, the heat energy released when 1 mol of HCl(g) reacts with 1 mol of NaOH(aq) is not –57.3 kJ mol–1. This is because the solubility of hydrogen chloride in water to form hydrochloric acid releases heat energy.

1 The heat of neutralisation can be determined by adding a known volume of acid (with a known concentration) to a known volume of an alkali (with a known concentration) and measuring the highest temperature obtained. 2 Experiment 4.1 shows the method used to determine the heat of neutralisation between sodium hydroxide with (a) hydrochloric acid (strong acid) and (b) ethanoic acid (weak acid).

HCl(g) + water → HCl(aq); ΔH = negative … (1) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l); ΔH = –57.3 kJ mol–1 … (2)

475

Thermochemistry

4.1 To determine the heat of neutralisation between an acid and an alkali Problem statement How are heats of neutralisation determined and compared?

Results (A) Reaction between hydrochloric acid and sodium hydroxide solution

Hypothesis The heat of neutralisation between hydro­ chloric acid and sodium hydroxide is higher than the heat of neutralisation between ethanoic acid and sodium hydroxide.

30 °C

Initial temperature of hydrochloric acid

31 °C

Highest temperature obtained

43.5 °C

Average initial temperature of solutions before neutralisation 30 + 31 =— — — — — — = 30.5 °C 2

Variables Manipulated variable : Different types of acids Responding variable : Heat of neutrali­sation Constant variable : Concentra­ tions and volumes of acid and alkali used 4

Initial temperature of sodium hydroxide solution

Rise in temperature during neutralisation = 43.5 – 30.5 = 13 °C

Apparatus Thermometer, plastic cup and measuring cylinder.

(B) Reaction between ethanoic acid and sodium hydroxide solution

Materials 2.0 mol dm–3 hydrochloric acid, 2.0 mol dm–3 ethanoic acid and 2.0 mol dm–3 sodium hydroxide solution. Procedure

Initial temperature of sodium hydroxide solution

30 °C

Initial temperature of ethanoic acid

30 °C

Highest temperature reached

42 °C

Average initial temperature of solutions before neutralisation = 30 °C Rise in temperature = 42 – 30 = 12 °C Assumptions Specific heat capacity of solution = 4.2 J g–1 °C–1 Density of solution = 1.0 g cm–3 Calculations Section (A) (a) Heat energy released (mcθ) = mass of solution  specific heat capacity  rise in temperature = (100 + 100)  4.2  13 = 10 920 J = 10.92 kJ

Experiment 4.1

Figure 4.26 To determine the heat of neutralisation

1 100 cm3 of 2 mol dm–3 sodium hydroxide solution is poured into a plastic cup by using a measuring cylinder. The initial temperature of the alkali is recorded (Figure 4.26). 2 Using another measuring cylinder, 100 cm3 of 2 mol dm–3 hydrochloric acid is measured. The initial temperature of the acid is recorded. 3 The hydrochloric acid is then poured quickly and carefully into the sodium hydroxide solution. The mixture is stirred with a thermometer and the highest temperature obtained is recorded. 4 Steps 1 to 3 are repeated using 100 cm3 of 2 mol dm–3 ethanoic acid instead of hydrochloric acid. Thermochemistry

(b) Number of moles of HCl used 2  100 =— — — — — — — = 0.2 1000 Number of moles of NaOH used 2  100 =— — — — — — — = 0.2 1000 H+(aq) + OH–(aq) → H2O(l) Number of moles of water molecules produced = 0.2 476

(c) From (a) and (b) Formation of 0.2 mol of H2O molecules releases 10.92 kJ of heat energy. (d) Amount of heat energy released when 1 mol of water is formed 1 = 10.92  — — — = 54.6 kJ 0.2

acid (–54.6 kJ mol–1) is higher than the heat of neutralisation between sodium hydroxide and ethanoic acid (–50.4 kJ mol–1). 2 The theoretical value for the heat of neutralisation of a strong acid and a strong alkali is –57.3 kJ mol–1 and the theoretical value for the heat of neutralisation of a weak acid (CH3COOH) and a strong alkali is –55.2 kJ mol–1.

Heat of neutralisation for strong acids and strong alkalis is –54.6 kJ mol–1.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l); ΔH = –57.3 kJ mol–1

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l); ΔH = –54.6 kJ mol–1 Section (B) (a) Heat energy released (mcθ) = (100 + 100)  4.2  12 = 10 080 J = 10.08 kJ

3 The heats of neutralisation determined by experiments are usually less than the theoretical values because some of the heat released are lost to the surroundings or absorbed by the apparatus (thermometer and plastic cup). Both these losses of heat are ignored in the calculation. 4 The following precautionary steps must be taken when carrying out the experiment. (a) The initial temperatures of the solutions of sodium hydroxide, hydrochloric acid and ethanoic acid are only taken after the thermo­meter is left in the solution for a few minutes. This is to ensure that the solutions have reached a constant temperature. (b) The reactants (hydrochloric acid and ethanoic acid) must be added quickly and stirred so that the reaction can be completed in a very short time. (c) The thermometer reading must be observed throughout the experiment so that the highest temperature reached can be recorded.

(b) Number of moles of CH3COOH used 2  100 =— — — — — — — = 0.2 1000 Number of moles of NaOH used 2  100 =— — — — — — — = 0.2 1000 H+(aq) + OH–(aq) → H2O(l) Number of moles of water molecules produced = 0.2 (c) From (a) and (b): Formation of 0.2 mol of H2O molecules releases 10.08 kJ of heat energy. (d) Amount of heat energy released when 1 mol of water is formed 1 = 10.08  — — — = 50.4 kJ 0.2 Heat of neutralisation for weak acids and strong alkalis is –50.4 kJ mol–1. CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l); ΔH = –50.4 kJ mol–1

Conclusion The heat of neutralisation for strong acids and strong alkalis is higher than the heat of neutralisation for weak acids and strong alkalis. The hypothesis is accepted.

Discussion 1 The experiment shows that the heat of neutralisation between sodium hydroxide and hydrochloric

The heat of neutralisation is NOT the heat released when 1.0 mol of an acid neutralises 1.0 mol of an alkali. The reaction between 1.0 mol of an acid with 1.0 mol of a base can produce 2.0 mol of water. For example, H2SO4 + Mg(OH)2 → MgSO4 + 2H2O;

ΔH = 2  heat of neutralisation

In actual fact, the heat of neutralisation is defined as the heat evolved when 1.0 mol of H+ ions react with 1.0 mol of OH– ions to form 1.0 mol of water.

477

Thermochemistry

4

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l); ΔH = –55.2 kJ mol–1

Comparing the Heats of Neutralisation of a Reaction Involving a Strong Acid and Strong Alkali with a reaction involving a weak acid and a weak alkali

SPM

CH3COOH(aq) 100 mol

’04/P1, ’05/P1, ’10/P2

This means that an aqueous solution of ethanoic acid contains mainly CH3COOH molecules and very few H+ ions. 7 (a) When neutralisation occurs, some of the heat released are absorbed by ethanoic acid molecules to break the O–H bonds in the molecules so that the acid will eventually dissociate completely.

4

1 The energy level diagram for the heat of neutralisation of a strong acid (HNO3) with a strong alkali (NaOH) is shown in Figure 4.24 and the heat of neutralisation of a weak acid (CH3COOH) with a strong alkali (NaOH) is shown in Figure 4.25. 2 For the neutralisation reaction between a strong acid and a strong alkali, the heat of neutralisation is –57.3 kJ mol–1. For example,



HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l); ΔH = –57.3 kJ mol–1 HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l); ΔH = –57.3 kJ mol–1

O i CH3 – C – O – H + Na+ + OH– → breaking O – H bond in CH3COOH molecule absorbs heat energy



3 This is because strong acids and strong alkalis dissociate completely in aqueous solutions to form hydrogen ions and hydroxide ions respectively.

O i CH3 – C – O–Na+ + H – O – H forming O – H bond in H2O molecule releases heat energy

(b) As a result, the heat of neutralisation of a weak acid with a strong alkali is less that –57.3 kJ mol–1. 8 (a) Ammonia solution is a weak alkali. This is because ammonia dissociates only partially in aqueous solution.

HCl(aq) → H+(aq) + Cl–(aq) 1 mol 1 mol NaOH(aq) → Na+(aq) + OH–(aq) 1 mol 1 mol 4 When a strong acid is neutralised by a strong alkali, the reaction that occurs is the bonding of H+ ions with OH– ions to form water molecules. Hence, the heat of neutralisation of strong acids and strong alkalis have the same value, that is, –57.3 kJ mol–1.

NH3(aq) + H2O(l)

NH4+(aq) + OH–(aq)

This means that an aqueous solution of ammonia contains mainly ammonia molecules and very few OH– ions. We can represent the reaction between aqueous ammonia and hydrochloric acid by the chemical equation

H+(aq) + OH–(aq) → H2O(l); ΔH = –57.3 kJ mol–1 strong strong acid alkali

NH4 +(aq) + OH–(aq) + HCl(aq) → (ammonia solution) NH4Cl(aq) + H2O(l)

5 The heat of neutralisation of a weak acid with a strong alkali is less than –57 kJ mol–1. For example, CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l); ΔH = –55.2 kJ mol–1

(b) The heat of neutralisation of hydrochloric acid and ammonia solution is –51.5 kJ mol–1. The energy level diagram for the heat of neutralisation of a strong acid (hydrochloric acid) with a weak alkali (ammonia solution) is shown in Figure 4.27.

6 Weak acids dissociate only partially in water to produce very few hydrogen ions in aqueous solutions. Thermochemistry

CH3COO–(aq) + H+(aq) 4 mol

478

9

’05

The energy level diagram of a reaction is shown below.

(c) The heat of neutralisation for a strong acid with a weak alkali is less than –57.3 kJ mol–1. This is because the weak alkali does not dissociate completely in aqueous solution. Therefore, when the neutralisation reaction occurs, some of the heat energy released is absorbed in the dissociation of the weak alkali. 9 (a) The heat of neutralisation for ethanoic acid and ammonia solution is –50.4 kJ mol–1. The energy level diagram for the heat of neutralisation of a weak acid (ethanoic acid) with a weak alkali (ammonia solution) is shown in Figure 4.28.

Which of the following pairs of acid and alkali can be used to replace hydrochloric acid and sodium hydroxide to obtain the same ΔH value? A Ethanoic acid (CH3COOH) and NaOH B Nitric acid (HNO3) and KOH C Carbonic acid (H2CO3) and NaOH D Hydrochloric acid (HCl) and ammonia solution Comments Sulphuric acid is a strong acid and sodium hydroxide is a strong alkali. The heat of neutralisation of a strong acid and a strong alkali is –57 kJ mol–1. Ethanoic acid and carbonic acid are weak acids and ammonia solution is a weak alkali. Answer B

Calculations Involving the Heat of Neutralisation

Figure 4.28 The energy level diagram for the heat of neutralisation of ethanoic acid and ammonia solution

7 When 100 cm3 of 1.2 mol dm–3 hydrochloric acid is added to 100 cm3 of 1.2 mol dm–3 ammonia solution, the temperature of the solution changes from 30 °C to 37.5 °C. Calculate the heat of neutralisation of hydrochloric acid with ammonia. [Specific heat capacity of solution = 4.2 J g–1 °C–1; density of solution = 1.0 g cm–3]

(b) The heat of neutralisation (ΔHneut) of a strong/weak acid with a strong/weak alkali decreases in the order: ΔHneut strong acid/strong alkali > ΔHneut strong acid/weak alkali or ΔHneut weak acid/strong alkali > ΔHneut weak acid/weak alkali

Solution (a) Calculate the heat energy released in this experiment Heat energy released (mcθ) = (100 + 100)  4.2  (37.5 – 30) = 6300 J = 6.30 kJ

(c) This is because when neutralisation between a weak acid and a weak alkali occurs, some of heat energy released is absorbed not only in the dissociation of the weak acid but also in the dissociation of the weak alkali. 479

Thermochemistry

4

Figure 4.27 The energy level diagram for the heat of neutralisation of hydrochloric acid and ammonia solution

Experiment 2 Since the same heat energy is released in Experiments 1 and 2, the number of moles of water molecules produced are the same. That is, 0.2 mol of water molecules are produced in Experiment 2.

(b) Calculate the number of moles of water molecules produced Number of moles of HCl used 1.2  100 =— — — — — — — — 1000 = 0.12 Number of moles of NH3 used 1.2  100 =— — — — — — — — 1000 = 0.12

2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l) The equation shows that 2 mol of HCl reacts with 1 mol of Ca(OH)2 to form 2 mol of H2O. Therefore, 0.2 mol of HCl will react with 0.1 mol of Ca(OH)2 to form 0.2 mol of H2O. That is, number of moles of Ca(OH)2 used = 0.1 … (1) Let concentration of calcium hydroxide, Ca(OH)2 = M mol dm–3 Number of moles of Ca(OH)2 used

4

HCl(aq) + NH4+(aq) + OH–(aq) → NH4Cl(aq) + H2O(l) 0.12 mol of H+ ions react with 0.12 mol of OH– ions to produce 0.12 mol of water molecules. (c) From (a) and (b) Formation of 0.12 mol of water molecules release 6.30 kJ of heat. (d) Heat released when 1.0 mol of water molecules are formed

M  100 =— — — — — — — — 1000 = 0.1 M … (2) From (1) and (2), we have, 0.1  M = 0.1 M = 1.0 mol dm–3

1 = 6.30  — — — — 0.12 = 52.5 kJ That is, heat of neutralisation of hydrochloric acid with ammonia is –52.5 kJ mol–1.

10 8

Which of the following neutralisation reactions releases the least heat? A 1 mol of nitric acid and 1 mol of potassium hydroxide B 1 mol of ethanoic acid and 1 mol of potassium hydroxide 1 C — mol of sulphuric acid and 1 mol of sodium 2 hydroxide D 1 mol of hydrochloric acid and 1 mol of sodium hydroxide

When 200 cm3 of 1.0 mol dm–3 NaOH is added to 100 cm3 of hydrochloric acid (in excess), x kJ of heat energy is released. The experiment is repeated using 100 cm3 of Ca(OH)2 solution and 100 cm3 of hydrochloric acid (in excess). The heat energy released is also x kJ. What is the concentration of calcium hydroxide? Solution Experiment 1 Number of moles of NaOH used 1.0  200 =— — — — — — — — 1000 = 0.2

Comments In each reaction, 1 mol of H2O is formed. Ethanoic acid is a weak acid, which dissociates partially in aqueous solution. Part of the heat of neutralisation released is used to dissociate ethanoic acid. Hence, the heat of neutralisation for the reaction between a weak acid and a strong alkali is less than that between a strong acid and a strong alkali. Answer B

H+(aq) + OH–(aq) → H2O(l) HCl (in excess) reacts with 0.2 mol NaOH to produce 0.2 mol of water molecules and x kJ of heat energy is released. Thermochemistry

’04

480

Method for determining heat of neutralisation • Add a known volume of an acid (of known concentration) to a known volume of an alkali (of known concentration) and measure the maximum temperature reached.

Heat of neutralisation

Energy level diagram

(a) Strong acid/strong alkali neutralisation: ΔH = –57.3 kJ mol–1 (b) Strong acid/weak alkali, weak acid/strong alkali and weak acid/weak alkali neutralisation: ΔH less than –57.3 kJ mol–1 (c) This is because weak acids/weak alkalis undergo partial dissociation in water. Part of the heat released during neutralisation is absorbed for the dissociation of the weak acid/weak alkali molecule.

Method of calculation (a) Number of moles of H2O produced = x mol (b) Heat energy released mcθ = ———— 1000 = y kJ (c) From (a) and (b): Formation of x mol of H2O releases y kJ of heat (d) From (c), –y Heat of neutralisation = —— kJ mol–1 x

4.4 (a) What is meant by ΔH = –114 kJ? (b) What is the heat of neutralisation between sulphuric acid and sodium hydroxide solution? (c) When 50 cm3 of 1.0 mol dm–3 nitric acid is added to 50 cm3 of a sodium hydroxide solution, the temperature rise is 6.5 °C. What is the concentration of sodium hydroxide solution?

Use the following data for all the calculations. Specific heat capacity of solution = 4.2 J g–1 °C–1; density of solution = 1.0 g cm–3 1 Figure 4.29 shows the energy level diagram for the reaction between sulphuric acid and sodium hydroxide solution.

Figure 4.29 Energy level diagram for the reaction between H2SO4 and NaOH

2 Consider five neutralisation reactions. I 50 cm3 of 1.0 mol dm–3 HNO3 + 25 cm3 of 1.0 mol dm–3 NaOH II 50 cm3 of 1.0 mol dm–3 HCl + 50 cm3 of 1.0 mol dm–3 NaOH III 50 cm3 of 1.0 mol dm–3 CH3COOH + 25 cm3 of 1.0 mol dm–3 NH3 IV 25 cm3 of 1.0 mol dm–3 CH3COOH + 25 cm3 of 2.0 mol dm–3 NaOH V 50 cm3 of 1.0 mol dm–3 H2SO4 + 25 cm3 of 1.0 mol dm–3 KOH

481

Thermochemistry

4

• Heat of neutralisation is the heat energy released when 1.0 mol of H+ ions react with 1.0 mol of OH– ions to form 1.0 mol of water molecules.

(a) Which two reactions will release the same amount of heat? (b) Which reaction will release the highest amount of heat? (c) Which reaction will release the lowest amount of heat? Briefly explain your answers. 3 When 100 cm3 of 1 mol dm–3 ethanoic acid is mixed with 100 cm3 of 1 mol dm–3 sodium hydroxide solution, the temperature of the reaction mixture increases by 5.5 °C. Calculate the rise in temperature

4.5

4 When 50.0 cm3 of 1 mol dm–3 sodium hydroxide solution is added to an excess of acid, HA, 2.65 kJ of heat is released. (a) Calculate the heat of neutralisation. (b) Is the acid, HA, a strong acid or a weak acid? (c) State the assumptions that you made in answer (b).

Heat of Combustion

The Meaning of Heat of Combustion

4

if 100 cm3 of 0.8 mol dm–3 ethanoic acid is mixed with 100 cm3 of 0.8 mol dm–3 sodium hydroxide solution.

6 The heat of combustion of the compounds, methanol and methane, are –715 kJ mol–1 and –890 kJ mol–1 respectively.

SPM

’04/P1

1 Combustion is a redox reaction in which a substance reacts rapidly with oxygen with the production of heat energy. Combustion is always accompanied by a flame and light. 2 During combustion, the elements are converted to their oxides. For example, carbon (C) burns to form carbon dioxide gas (CO2) while hydrogen burns to form water (H2O). 3 All combustion reactions are exothermic reactions. 4 The heat of combustion is the heat released when 1 mol of a substance is burnt completely in excess oxygen. 5 (a) The heat of combustion of the elements, carbon and hydrogen, are –392 kJ mol–1 and –286 kJ mol–1 respectively.



3 CH3OH(l) + — O2(g) → CO2(g) + 2H2O(l); 2 ΔH = –715 kJ mol–1 CH4 (g) + 2O2(g) → CO2(g) + 2H2O(l); ΔH = –890 kJ mol–1

Determining the Heat of Combustion 1 The heat of combustion can be determined SPM by burning a known mass of a fuel (such as ’11/P1 ethanol) and the heat evolved is used to heat water. From the increase in the temperature of water, the heat of combustion of the fuel can be determined. 2 The following relationship is used for calculation:

C(s) + O2(g) → CO2(g); ΔH = –392 kJ mol–1 1 H2(g) + — O2(g) → H2O(l); 2 ΔH = –286 kJ mol–1

Heat evolved during combustion of fuel

(b) Figure 4.30 shows the energy level diagram for the combustion of the element carbon.

Heat absorbed by water

3 Experiment 4.2 shows the simple apparatus used for determining the heat of combustion of alcohol. The copper container is called the calorimeter. Other metal calorimeters can also be used in the experiment. The calorimeter is made from substances that are good conductors of heat to minimise the loss of heat. Thus, the rise in temperature during combustion can be determined accurately.

SPM

’07/P1

Figure 4.30 Energy level diagram for the combustion of carbon Thermochemistry

=

482

4.2 To determine the heat of combustion of various alcohols Problem statement Hypothesis

How does the number of carbon atoms per molecule of an alcohol affect the heat of combustion?

As the number of carbon atoms per molecule in an alcohol increases, so does the heat of combustion.

Variables (a) Manipulated variable : Type of alcohol (b) Responding variable : Heat of combustion (c) Constant variable : Volumes of water and alcohol, metal container (calorimeter) and spirit lamp Apparatus

Copper container, spirit lamp, measuring cylinder, thermometer, electronic balance, tripod stand, asbestos screen, wooden block and Bunsen burner.

Materials

Methanol, ethanol, propan-1-ol and butan-1-ol.

4

Procedure

Figure 4.31 Determining the heat of combustion of ethanol

1 Using a measuring cylinder, 250 cm3 of water is measured into a copper container. 2 The copper calorimeter is placed on the tripod stand. The initial temperature of water is measured and recorded. 3 The spirit lamp is half-filled with ethanol. The spirit lamp and ethanol are weighed and the mass is recorded. 4 The spirit lamp is placed below the copper calorimeter and the wick is lighted (Figure 4.31). The flame of the spirit lamp is shielded from the draught (blow of wind) by using an asbestos screen. 5 The water in the calorimeter is stirred throughout the experiment. 6 When the temperature of water increases by about 30 °C, the spirit lamp is extinguished. 7 The spirit lamp and ethanol are weighed again and the mass is recorded. 8 The experiment is repeated using other alcohols as shown below to replace ethanol: (a) Methanol (b) Propan-1-ol (c) Butan-1-ol

Alcohol

Volume of water (cm3)

Initial temperature of water (°C)

Highest temperature of water (°C)

Rise in temperature of water (°C)

Initial mass of lamp + alcohol (g)

Final mass of lamp + alcohol (g)

Ethanol

250

30.5 (t1)

59.5(t2)

t2 – t1

218 (m1)

216.6 (m2)

Methanol

250

t3

t4

t4 – t3

m3

m4

Propan-1-ol

250

t5

t6

t6 – t5

m5

m6

Butan-1-ol

250

t7

t8

t8 – t7

m7

m8

Manipulated variable

Constant variable

Responding variable

483

Thermochemistry

Experiment 4.2

Results

4

(a) of the loss of heat to the surroundings, (b) the combustion of ethanol is incomplete, (c) some of the ethanol escapes due to evaporation. 3 (a) When ethanol burns, some of the heat released is absorbed by the copper calorimeter and some is lost to the surroundings. (b) In complete combustion, all carbon atoms in ethanol are converted into carbon dioxide.

Calculations Assumption Specific heat capacity of water = 4.2 J g–1 °C–1 Density of water = 1.0 g cm–3 (a) Calculate the heat energy released during the combustion of ethanol in the experiment Rise in the temperature of water = t2 – t1 = 59.5 – 30.5 = 29 °C Heat energy evolved = mass of water in the copper container  specific heat capacity  rise in temperature = 250  4.2  29 = 30 450 J = 30.45 kJ (b) Calculate the number of moles of ethanol burnt in the experiment Mass of ethanol burnt = (m1 – m2) = 218 – 216.6 = 1.4 g Relative molecular mass of ethanol (C2H5OH) = (2  12) + (1  6) + 16 = 46 Therefore, the number of moles of ethanol burnt mass of ethanol burnt 1.4 =— — — — — — — — — — — — — — — — — — — — — — — — — — — —= — — — relative molecular mass of ethanol 46 = 0.03 (c) Calculate the heat energy released when 1 mol of ethanol is burnt From (a) and (b), combustion of 0.03 mol of ethanol produces 30.45 kJ of heat energy. Therefore, heat evolved when 1 mol of ethanol burns 1 = 30.45  — — — — = 1015 kJ 0.03 That is, heat of combustion of ethanol = –1015 kJ mol–1. (d) The energy level diagram for the combustion of ethanol is shown in Figure 4.32.

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l); ΔH = –1371 kJ mol–1 In partial combustion, carbon atoms in ethanol are converted to carbon dioxide, carbon monoxide and also soot. As a result of incomplete combustion, less heat energy is released. (c) Ethanol is a volatile liquid. When combustion occurs, the temperature around the spirit lamp becomes higher. As a result, ethanol tends to evaporate off. Hence, the spirit lamp and the remaining alcohol after combustion must be weighed as soon as possible.

(e) Heat of combustion of the other alcohols can be obtained by the same method used for determining the heat of combustion of ethanol.

Precautionary steps 1 The following actions must be taken to avoid loss of heat to the surroundings. (a) An asbestos screen is placed around the copper calorimeter and the tripod stand. The asbestos screen protects the flame of the spirit lamp from the disturbance of air currents. (b) The spirit lamp is placed on a wooden block so that the spirit lamp is at a minimum distance from the copper calorimeter. In this way, a bigger area of the flame can be in contact with the calorimeter. 2 The copper calorimeter and the water in it are heated directly by spirit lamp. A wire gauze is not used because it is a good conductor of heat and will absorb heat energy given off by the fuel. 3 The water in the copper calorimeter must always be stirred so that its temperature is uniform. 4 Precautionary steps must also be taken when carrying out the experiment because the liquid fuel (ethanol) is very volatile and catches fire easily.

Sources of error 1 In fact, the actual value of the heat of combustion of ethanol is –1371 kJ mol–1. 2 The heat of combustion of ethanol obtained in this experiment is very much lower than the actual value (–1371 kJ mol–1) because

Conclusion 1 The heat of combustion of ethanol = –1371 kJ mol–1 2 The heat of combustion increases as the number of carbon atoms per molecule in the alcohol increases.

Figure 4.32

Thermochemistry

484

Heats of Combustion of Various Alcohols

3 The heat of combustion of ethanol is higher than the heat of combustion of methanol. This is because one molecule of ethanol contains one carbon atom and two hydrogen atoms more than one molecule of methanol. When carbon and hydrogen burns, heat energy is released. Hence, the greater the number of carbon atoms and hydrogen atoms per molecule, the higher the heat of combustion of the alcohol.

1 The heat of combustion of methanol, ethanol, propan-1-ol and butan-1-ol are shown in Table 4.3. When the relative molecular mass of the alcohol increases, the heat of combustion also increases. Table 4.3 Heat of combustion of some alcohols ’04/P1 ’05/P2 /SA ’07/P1

Alcohol

Heat of Equation for reaction combustion (kJ mol–1)

Methanol 3 CH3OH(l) + — O2(g) → 2 CO2(g) + 2H2O(l)

–715

Ethanol C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

–1371

Propanol 9 C3H7OH(l) + — O2(g) → 2 3CO2(g) + 4H2O(l)

–2010

Butanol C4H9OH(l) + 6O2(g) → 4CO2(g) + 5H2O(l)

–2673

The Fuel Values of Various Fuels 1 Fuels are substances that can burn easily in air to produce heat energy. 2 Fuels can be classified into three groups depending on their physical states. (a) Solid fuels: coal, charcoal, coke and peat. (b) Liquid fuels: petrol, diesel and kerosene. (c) Gaseous fuel: hydrogen, natural gas and coal gas. 3 Different fuels have different fuel values. Fuel value is the heat energy released when 1.0 g of fuel is burnt in excess oxygen. Fuel value is also known as heat value. The units for fuel value is kJ g–1. 4 Table 4.4 shows the fuel values of some fuels. For example, 1.0 g of hydrogen releases 143 kJ of heat energy and 1.0 g of coal releases only 34 kJ of heat energy.

2 If the heat of combustion of alcohols are plotted against the number of carbon atoms in the alcohol molecules, a straight line is obtained (Figure 4.33).

Table 4.4 Fuel values

Fuel

Fuel value (kJ g–1)

Hydrogen

143

Natural gas (methane)

52

Propane

51

Butane

50

Petrol

42

Coal

34

Wood

21

5 In terms of heat energy released per g of fuel, hydrogen is the best fuel and other fuels such as natural gas and petrol are better fuels than coal. 6 The fuel values are used in industry to compare the energy cost of various fuels.

Figure 4.33 The heat of combustion of alcohols against the number of carbon atoms per molecule

485

Thermochemistry

4

SPM

9

They are good fuels because they do not produce soot or poisonous gases that pollute the air. Liquid fuels and gaseous fuels do not leave ashes after combustion. They are therefore better fuels than solid fuels. (c) Cost per gram of fuel Coal is a cheap fuel and fuels such as butane and petrol are quite expensive. 2 In terms of the economy and environment, good fuels have the following characteristics: (a) Produce a large amount of heat energy when burnt (b) Do not cause pollution (c) Can be obtained cheaply (d) Can be obtained easily (e) Can be burnt easily (f) Can be kept and transported easily and safely 3 We use a particular fuel for a specific purpose. For example, charcoal is used to roast meat, coal is used to generate electricity in power stations and petrol is for car engines. 4 Besides traditional fuels such as coal and petrol, scientists continue to look for new fuels for various purposes. For example, hydrazine (N2H4) is produced by scientists as fuel for rockets.

The table below shows the fuel value (in kJ g–1) and the cost (in ringgit per kg) for two fuels, fuel X and fuel Y. Fuel

Fuel value (kJ g–1)

Cost of fuel

X

52

RM 0.35 per kg

Y

42

RM 1.00 per kg

4

Which is the more expensive fuel, fuel X or fuel Y, in terms of (a) fuel cost and (b) energy cost? Solution (a) Fuel cost Fuel Y is more expensive than fuel X in terms of cost per gram of fuel. (b) Energy cost of fuel X 1 kg of fuel X costs RM 0.35. 1 kg of fuel X produces 52  1000 J of heat energy. From fuel value Therefore, 52  1000 J heat energy costs RM 0.35. 0.35 That is, 1000 J of heat energy costs RM — — — 52 = RM 0.0067 … (1) Energy cost of fuel Y 1.0 kg of fuel Y costs RM1.00. 1.0 kg of fuel Y produces 42  1000 J of heat energy. From fuel value Therefore, 42  1000 J of heat energy costs RM1.00. 1.00 That is, 1000 J of heat energy costs RM — — — 42 = RM 0.024 … (2) Based on answers (1) and (2), fuel Y is more expensive than fuel X in terms of energy cost.

N2H4(l) + O2(g) → N2(g) + 2H2O(g); ΔH = –622 kJ mol–1 The combustion of acetylene in air is highly exothermic and gives a very hot flame. The oxyacetylene flame is used to weld and cut metals. Ethyne (C2H2) is commonly known as acetylene.

The Selection of Suitable Fuels for Specific Purposes 1 When selecting a suitable fuel for a specific purpose, three main factors must be considered. (a) Fuel value The higher the fuel value, the more energy is released per gram of the fuel. (b) Effect on the surroundings Most fuels produce a lot of soot when burnt and this causes air pollution. Fuels such as hydrogen are known as clean fuels. Thermochemistry

486

Solution (a) Calculate the heat energy released in this experiment Heat energy released (mcθ) = 250  4.2  10 J = 10 500 J = 10.5 kJ

Calculations Involving Heat of Combustion

10 When 2.7 g of glucose (C6H12O6) is burnt completely in excess oxygen, the heat released increases the temperature of 600 g of water by 12.5 °C. Calculate the heat of combustion of glucose. [Specific heat capacity of water = 4.2 J g–1 °C–1; density of water = 1.0 g cm–3; relative atomic mass: H, 1; C, 12; O, 16]

(b) Calculate the heat released by the combustion of 1.0 mol of methane 0.0125 mol of methane gives off 10.5 kJ of heat energy. Heat energy given off by 1.0 mol of methane 1 = 10.5  — — — — — 0.0125 = 840 kJ

Solution (a) Calculate heat energy released in this experiment Heat energy released = mass of water (m)  specific heat capacity (c)  rise in temperature (θ) = 600  4.2  12.5 J = 31 500 J = 31.5 kJ (b) Calculate the number of moles of glucose burnt in this experiment

(c) Calculate the heat released by the combustion of 1.0 g of methane

Relative molecular mass of glucose (C6H12O6) = (12  6) + 12 + (6  16) = 180

4

Relative molecular mass of methane (CH4) = 12 + (4  1) = 16 840 Fuel value = — — — — 16 = 52.5 kJ g–1

Number of moles of glucose burnt 2.7 =— — — 180 = 0.015 (c) Calculate the heat energy released when 1 mol of glucose is burnt From (a) and (b), 0.015 mol of glucose produces 31.5 kJ of heat energy. Therefore, heat energy released by 1 mol of glucose 1 = 31.5  — — — — 0.015 = 2100 kJ

11

’07

In which of the following is the heats of combustion for methanol, ethanol and propanol correctly matched?

That is, heat of combustion of glucose is –2100 kJ mol–1.

11 When 0.0125 mol of methane (CH4) is completely burnt, the heat energy released raises the temperature of 250 cm3 of water by 10 °C. What is the fuel value of methane? [Specific heat capacity of water = 4.2 J g–1 °C–1; density of water = 1.0 g cm–3; relative atomic mass: H, 1; C, 12]

Methanol (kJ mol–1)

Ethanol (kJ mol–1)

Propanol (kJ mol–1)

A

–715

–2010

–1321

B

–715

–1321

–2010

C

–2010

–1321

–715

D

–2673

–2010

–715

Comments The heat of combustion increases (more exothermic) from methanol to propanol. Answer B

487

Thermochemistry

Heat of combustion • Heat of combustion is the heat released when 1.0 mol of the substance is burnt completely in excess oxygen.

4

Method of determining heat of combustion • Find the mass of the fuel used. • Find the volume of water to be heated. • Find the initial and final temperatures of the water.

Differences in heats of combustion For various alcohols • CH3OH; ΔH = –715 kJ mol–1 • C2H5OH; ΔH = –1371 kJ mol–1 • Caused by an additional –CH2 group

Does it burn easily?

Calculation (a) Calculate number of moles of fuel used (x mol) (b) Calculate heat evolved mcθ ΔH = ———————— = y kJ 1000 (c) Heat of combustion –y = —— kJ mol–1 x

Selection of suitable fuel • For specific purposes • Examples: Coal, natural gas, petrol, hydrogen

Differences in fuel values (kJ g–1) • Coal: 34 kJ g–1 • Petrol : 42 kJ g–1 • Natural gas: 52 kJ g–1

Factors to consider

Is it easy to store?

Does it produce soot or leave residue after combustion?

Problems of safety

Problems of pollution

Advantages

Can it be obtained easily?

Is it cheap or expensive?

Disadvantages/limitations

Type of fuel recommended for the specific purpose

Thermochemistry

Energy level diagram

488

12

(b) If 1.14 g of propan-2-ol is used for complete combustion, what is the rise in temperature of 500 cm3 water?

’04

The heat of combustion of three alcohols are shown below. Alcohol

Heat of combustion (kJ mol–1)

CH3OH

–715

C2H5OH

–1371

C3H7OH

–2010

(c) What is the rise in temperature, if 500 cm3 of water is heated by the heat energy produced by the complete combustion of 2.28 g of propan-1-ol? (d) What is the rise in temperature if 250 cm3 of water is heated by the heat energy produced by the complete combustion of 0.57 g of propan-1-ol?

Which of the following factors cause the increase in the heat of combustion of alcohols? I The increase in the molecular size of alcohols. II The increase in the number of carbon atoms per molecule. III The increase in the number of hydrogen atoms per molecule. IV The increase in the forces of attraction between the molecules of alcohol. A I and IV only C I, II and III only B II and III only D I, II, III and IV

(e) 2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(l); ΔH = –x kJ Determine the value of x in the thermochemical equation given above.

’04

C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(l); ΔH = –5520 kJ mol–1 The combustion of heptane in excess oxygen releases 1104 kJ of heat energy. What is the mass of heptane used? [Relative atomic mass: C, 12; H, 1]

Comments The increase in molecular size and forces of attraction increase the boiling points of alcohols but not the heat of combustion. (I and IV are incorrect). The increase in the heat of combustion of alcohols is due to the combustion of carbon and hydrogen:

4 What are the advantages and disadvantages of using (a) coal, (b) natural gas as fuel?

4.6

C(s) + O2(g) → CO2(g); ΔH = –394 kJ mol 2H2(g) + O2(g) → 2H2O(l); ΔH = –572 kJ

–1

Answer B

Appreciating the Existence of Various Energy Sources

Various Sources of Energy 1 Fuels can be classified as renewable source of energy or non-renewable source of energy. 2 The non-renewable sources of energy are fossil fuels such as coal, petroleum and natural gases. The renewable sources of energy are: (a) Nuclear power (e) Solar energy (b) Hydroelectric power (f) Biomass and biogas (c) Geothermal energy (g) Energy from tides (d) Wind power

4.5 1 (a) A student wants to carry out an experiment to determine the heat of combustion of pentane (C5H12). State three measurements that he must take in order that the heat of combustion can be determined. (b) The heats of combustion of propane, carbon and hydrogen are –2220 kJ mol–1, –394 kJ mol–1 and –286 kJ mol–1 respectively. Predict the heat of combustion of butane.

Choosing an Energy Source

2 In an experiment, the complete combustion of 1.14 g of propan-1-ol (C3H7OH) increases the temperature of 500 cm3 of water by 18.2 °C. [Specific heat capacity of water = 4.2 J g–1 °C–1; density of water = 1.0 g cm–3; relative atomic mass: H, 1; C, 12; O, 16]

In choosing an energy source, we need to know (a) the advantages and disadvantages of using that particular energy source, (b) the costs involved, (c) the environmental impact and (d) whether there are suitable sites for harnessing this source of energy. For example, although

(a) Give the structural formula of propan-1-ol and propan-2-ol.

489

Thermochemistry

4

3 The thermochemical equation for the combustion of heptane is shown below.

5 Solar energy There are three methods to harness solar energy: (a) The solar energy can be absorbed on a solar collector placed on a rooftop. This method is called solar heating and is used for providing hot water or for heating the building. (b) The solar energy can be converted into electricity in photovoltaic cell or more simply, solar cells. Solar cells are now widely used to power small electronic devices such as electronic calculators. (c) The solar energy is concentrated to heat water and produce steam, which is used to produce electricity. This is called solar thermal electric power. 6 Biomass, biogas and biodiesel (a) Biomass is plant material and animal manure used as sources of energy. The energy produced by biomass is called biomass energy. (b) Most biomass is • burnt directly for heating, cooking or other industrial uses, or • converted to gaseous and liquid biofuels. (c) The synthetic natural gas (gaseous biofuel) produced from biomass is called biogas. Methane in biogas is produced from machines known as digesters. The digesters can convert animal manure, such as cow dung and chicken dung into methane gas. Methane gas produced in this way is called biogas. (d) Biodiesel is a biofuel made from vegetable oils such as palm oil. Biodiesel can be used to power motor vehicles. 7 Energy from the tides The power of tides is another potential source of energy. Twice a day during high and low tides, water that flows into and out of coastal bays can spin turbines to produce electricity.

tide-powered plants produce great amounts of energy, there are few suitable coastal sites that can be used to build such plants.

4

Technology for Harnessing Various Energy Sources 1 Nuclear energy Nuclear power stations use nuclear reactors to generate electricity. In the nuclear reactors, uranium-235 nuclei split to form lighter nuclei (fission reaction) and energy is released. 2 Hydroelectric power Hydroelectric power is electricity generated by the energy of water power. The most common form of hydroelectric power uses dams on rivers to create large reservoirs of water. Water released from the reservoirs flows through turbines, causing them to spin and to generate electricity when connected to generators. 3 Geothermal energy The layers of rock deep below the earth’s surface is very hot. Heat contained in underground rocks is an important source of energy. When water is pumped into the hot rocks, it is turned into steam which can be used to operate turbines and generate electricity. 4 Wind power Wind power can be used to generate electricity.

Wind turbines used to produce electricity individually or in groups called wind farms.

The Advantages and Disadvantages in using Various Energy Sources

The kinetic energy of the moving air is converted into mechanical energy for the turbines. The turbine then drives a generator that converts mechanical energy into electrical energy. Wind turbines usually have two or three blades. The longer the blades and the faster the wind speed, the more electricity the turbine generates. To produce the most electricity, wind turbines must be located in areas where the wind blows at a constant speed. Thermochemistry

1 The main advantages of these energy sources is that, they are renewable and they do not produce carbon dioxide gas (except biomass) that causes greenhouse effect or acidic gases that cause acidic rain or poisonous gases that cause air pollution. 2 Table 4.5 shows the disadvantages of these renewable energy resources. 490

Table 4.5

Disadvantages

Nuclear energy

(a) Nuclear reactors produce radioactive wastes, which are difficult to dispose off. (b) Accidents in nuclear reactors can cause thousands of deaths.

Hydroelectric power

(a) It provides low-cost electricity but the construction cost is high. (b) Danger of collapse which can cause catastrophic floods.

Geothermal energy

(a) The geothermal power is low in cost (second only to hydropower) but can only be obtained in areas where steam or hot water is at or near the surface. (b) The wastewater is quite salty and its disposal is a problem.

Wind power

(a) Steady wind needed. (b) Contributes to noise pollution when located near populated areas.

Solar energy

(a) The installation for solar heating is expensive. Solar collectors need to be exposed to the sun 60% of the time. (b) Solar cells are not very efficient. Generation of enough energy to meet demand would require large areas of land with solar cells.

Biomass, biogas and biodiesel

Carbon dioxide emissions.

Energy from the tides

Electricity can be generated only when the tide is coming in or going out.

θ (in °C) = change in temperature. 11 The heat of precipitation is the heat released when 1.0 mol of a precipitate is formed from its ions. 12 The heat of displacement is the heat released when 1.0 mol of a metal is displaced from its salt solution by another metal. 13 The heat of neutralisation is the heat released when 1.0 mol of H+ ions react with 1.0 mol of OH– ions to form 1.0 mol of water molecules. 14 For the neutralisation reaction between a strong acid and a strong alkali, the heat of neutralisation is –57 kJ mol–1. 15 For the neutralisation of weak acid and strong alkali or the neutralisation of strong acid and weak alkali, the heat of neutralisation is less than –57 kJ mol–1. This is because weak acid and weak alkali do not dissociate completely in aqueous solution. Some of the heat released is absorbed to dissociate the weak acid or weak alkali. 16 The heat of combustion is the heat released when 1.0 mol of a substance is burnt completely in excess oxygen. 17 The heat of combustion of alcohols increases as the relative molecular mass increases. 18 Fuel value is the heat energy released when 1.0 g of fuel is burnt in excess oxygen. 19 Hydrogen is the best fuel because it has the highest fuel value and is a clean fuel (that is, does not produce air pollutants). However, it is an expensive fuel.

1 An exothermic reaction is a reaction that releases heat energy to the surroundings. The temperature of the reaction mixture increases. ΔH has a negative sign. 2 An endothermic reaction is a reaction that absorbs heat energy from the surroundings. The temperature of the reaction mixture decreases. ΔH has a positive sign. 3 The energy level diagram shows the total energy content of the reactants and that of the products. 4 Exothermic reaction: Total energy content of products is lower than that of the reactants. 5 Endothermic reaction: Total energy content of products is higher than that of the reactants. 6 Bond breaking is an endothermic process while bond making is an exothermic process. 7 ΔH of reaction = ΔH1 (bond breaking) + ΔH2 (bond making) 8 Cold packs contain chemicals such as NH4NO3(s) and water. The reaction between NH4NO3(s) and water is endothermic. 9 Hot packs contain chemicals such as anhydrous CaCl2 and water. The reaction between CaCl2 and water is exothermic. 10 The heat change in a reaction can be calculated using the formula: ΔH = mcθ where ΔH (in J) = heat released or absorbed in aqueous solution, m (in g) = mass of solution, c (in J g–1 °C–1) = specific heat capacity of solution

491

Thermochemistry

4

Energy resource

4 Multiple-choice Questions

4

4.1

Energy Changes in Chemical Reactions

1 Which of the following operations absorbs heat energy ’08 from the surroundings? A Dissolving sodium hydroxide in water B Evaporating sodium chloride solution C Adding sodium chloride solution to silver nitrate solution D Adding ethanoic acid to ammonia solution

’11

fever. Which pair of substance is used in cold packs? A Ammonium nitrate and water B Anhydrous sodium carbonate and water C Fused calcium chloride and water D Zinc powder and copper(II) sulphate solution

5 A simple experiment as shown is carried out.

2 The energy profile diagram for the reaction, P → X + Y, is ’05 shown below.

Which of the following will occur during the experiment? A The pH of the solution increases. B Heat energy is absorbed from the surroundings. C The temperature of the solution decreases. D The total energy content of the products is lower than the total energy content of the reactants.

What is the heat of reaction? A –40 kJ C +40 kJ B +30 kJ D +70 kJ 3 Water can be converted to ice or steam as shown in the ’09 diagram below. steam

6 The energy level diagram of a reaction is shown below. ’09

energy water

In which conversion is heat energy absorbed from the surroundings? A Ice to steam B Steam to ice TC 58 C Water to ice D Steam to water 4 Cold pack is used to put on the forehead of a patient to reduce Thermochemistry

= + 579 kJ mol

–1

CaCO3

Based on the diagram, it can be deduced that TC 59 A it is an endothermic reaction. B the total energy of the

492

7 The reaction between methane and chlorine to form ’11 chloromethane and hydrogen chloride is exothermic. CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ΔH = –102 kJ Which statement is correct about this reaction? A Heat is absorbed during the reaction. B The heat absorbed on bond breaking is the same as the heat released on bond making. C The heat absorbed on bond breaking is more than the heat released on bond forming. D The heat absorbed on bond breaking is less than the heat released on bond forming. 8 When 25.0 cm3 of lime juice is added to 185.0 cm3 of milk to ’04 prepare yogurt, the temperature of yogurt increases by 2.4 °C. What is the total amount of heat released? [The specific heat capacity of yogurt is X J g–1 °C–1; the density of the solution is 1 g per cm3]. A 444X J C 504X J B 444X kJ D 504X kJ 9 The diagram shows an energy profile diagram for a chemical reaction.

CaO + CO2

ice

products is 570 kJ. C the reactant has more energy than the products. D more bonds are formed than are broken.

For the uncatalysed reaction Activation energy

Heat of reaction

A

ΔH1

ΔH3

B

ΔH1

ΔH3 – ΔH1

C

ΔH2

ΔH4

D

ΔH2

ΔH4 – ΔH2

4.2

What is the rise in temperature? [Specific heat capacity of solution = 4.2 J g–1 °C–1] A 7 °C C 12 °C B 10 °C D 14 °C 13 The following table shows the volumes of 1.0 mol dm–3 silver nitrate (AgNO3) solution and 1 mol dm–3 potassium chloride solution used for determining the heat of precipitation of silver chloride in four experiments. Which of the experiments gives the highest rise in temperature?

Heat of Precipitation

10 Ag+ (aq) + Cl–(aq) → AgCl(s) What is the heat change when 5.74 g of AgCl is precipitated when mixing solutions of silver nitrate and potassium chloride? [Relative atomic mass: Ag, 108; Cl, 35.5; ∆H = –58.8 kJ mol–1] A 2.35 kJ C 23.6 kJ B 3.38 kJ D 33.8 kJ 11 The following equation represents the reaction between Pb2+ ions and CrO42– ions. Pb2+(aq) + CrO42–(aq) → PbCrO4(s) When 50 cm3 of 0.5 mol dm–3 lead(II) nitrate solution is added to 50 cm3 of 0.5 mol dm–3 potassium chromate(VI) solution, the temperature rises by t °C. If the experiment is repeated using 50 cm3 of 1.0 mol dm–3 lead(II) nitrate solution and 50 cm3 of 1.0 mol dm–3 potassium chromate(VI) solution, what is the rise in temperature? 1 o A — t C C 2t °C 2 B t °C D 4t °C 12 The heat of precipitation of copper(II) hydroxide in the reaction between copper(II) sulphate solution and sodium hydroxide solution is –59 kJ mol–1. In one experiment, 50 cm3 of 2.0 mol dm–3 sodium hydroxide solution is poured into 50 cm3 of 1.0 mol dm–3 copper(II) sulphate solution.

Experi­­ment

Volume of Volume AgNO3 of KCl (cm3) (cm3)

A

1

B

2

50

10

C

3

40

20

D

4

35

25

55

5

14 The energy level diagram for the reaction between lead nitrate solution and sodium bromide solution is shown below. energy

Pb2+(aq) + 2Br–(aq) H = –x kJ mol–1 PbBr2(s)

What is the energy change when 1.0 mol of Pb2+ ions is added to 0.5 mol of Br – ions? 1 A –— x kJ C – x kJ 4 1 B –— x kJ D – 2x kJ 2 15 An experiment is carried out to determine the heat of precipitation of silver chloride for the reaction between silver nitrate solution and sodium chloride solution. Which of the following steps must be taken to obtain accurate results? I Stir the solution continuously. II Use a container that is a good conductor of heat. III Add the silver nitrate solution quickly to the sodium chloride solution.

493

IV Use a screen to block the wind. A I and III only B II and IV only C I, II and III only D I, II, III and IV 16 The equation shows the reaction between calcium chloride and sodium sulphate. CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + 2NaCl(aq); ΔH = –x kJ mol–1 Which of the following pairs of solutions will produce the same value for ∆H? (The concentration and volume of the solutions are kept constant). A Ca(NO3)2(aq) + H2SO4(aq) B Ca(NO3)2(aq) + NaCl(aq) C CaCl2(aq) + HNO3(aq) D CaCl2(aq) + HCl(aq)

4.3

4

Which energy changes are the activation energy and heat of reaction for the uncatalysed reaction?

Heat of Displacement

17 When 6 g of magnesium powder (in excess) is added to 50 cm3 of copper(II) sulphate solution containing 0.2 mol of Cu2+ ions, the rise in temperature is t °C. The heat of displacement of copper from copper(II) sulphate solution by magnesium is 50  4.2  t A – ( — — — — — — — — — — — — — — — — — — — — — — — — — — — — — ) kJ mol–1 1000 50  4.2  t B – ( — — — — — — — — — — — — — — — — — — — — — — — — — — — — ) kJ mol–1 0.2  1000 56  4.2  t C – ( — — — — — — — — — — — — — — — — — — — — — — — — — — — — ) kJ mol–1 1000 56  4.2  t D – ( — — — — — — — — — — — — — — — — — — — — — — — — — — — — ) kJ mol–1 0.2  1000 18 The following shows the energy level diagram for the displacement ’03 of iron by the metal, M.

What is the temperature reached if excess M is added to 50 cm3 of 0.2 mol dm–3 iron(II) sulphate solution at 30 °C? [Specific heat capacity of solution = 4.2 J g–1 °C–1] A 14.4 °C C 37.8 °C B 22.2 °C D 44.9 °C Thermochemistry

19 Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq); ΔH = –218 kJ mol–1

4

What is the minimum mass of zinc required in order that its reaction with excess copper(II) sulphate solution produces 10.9 kJ of heat? [Relative atomic mass: Zn, 65.4] A 1.7 g B 3.3 g C 5.0 g D 6.6 g 20 When 4.8 g of magnesium is added to 50 cm3 of 2.0 mol ’07 dm–3 copper chloride solution, the temperature of the mixture increases by 4.0 °C. What is the heat of reaction in this experiment? [Specific heat capacity of solution = 4.2 J g–1 °C–1; relative atomic mass of Mg = 24] A –0.42 J mol–1 B –0.84 J mol–1 C –4.20 kJ mol–1 D –8.4 kJ mol–1 21 A student wanted to determine the heat of displacement for the reaction Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s) Which of the following measurements is not required for the experiment? A Temperature of CuSO4 solution B Volume of CuSO4 solution C Concentration of CuSO4 solution D Mass of magnesium reacted 22 Consider the following reactions. I NaOH(s) → NaOH(aq) ’09 II NH Cl(s) → NH Cl(aq) 4 4 III NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g) IV Mg(s) + FeSO4(aq) → MgSO4(aq) + Fe(s) Identify the energy-releasing reactions. A I and II B II and III C III and IV D I and IV Thermochemistry

4.4

Heat of Neutralisation

23 Which of the following neutralisation reactions releases ’04 the least heat? A 1 mol of HCl(aq) + 1 mol NH3(aq) B 1 mol of CH3COOH + 1 mol of NH3(aq) C 1 mol of CH3COOH + 1 mol of KOH(aq) D 1 mol of HCl(aq) + 1 mol of NaOH(aq) 24 When 40 cm3 of 0.1 mol dm–3 hydrochloric acid is mixed with 10 cm3 of 0.1 mol dm–3 sodium hydroxide solution, the rise in temperature is t °C. What is the rise in temperature when 20 cm3 of 0.1 mol dm–3 hydrochloric acid reacts with 30 cm3 of 0.1 mol dm–3 sodium hydroxide solution? 1 A — t °C C 2t °C 2 B t °C D 4t °C 25 When 50 cm3 of 1.5 mol dm–3 HCl is added to 100 cm3 of 1.5 mol dm–3 sodium hydroxide solution, x kJ of heat is produced. When 45 cm3 of sulphuric acid is added to 100 cm3 of 1.5 mol dm–3 sodium hydroxide, the same amount of heat is produced. What is the concentration of sulphuric acid? A 0.2 mol dm–3 B 0.4 mol dm–3 C 0.8 mol dm–3 D 1.5 mol dm–3 26

A 4.6 °C B 6.8 °C

27 The heat of neutralisation between hydrochloric acid and sodium hydroxide is –57 kJ mol–1. What is the heat energy released, when 500 cm3 of 2 mol dm–3 sulphuric acid reacts with excess potassium hydroxide? A –0.5 3 57 kJ B –57 kJ C –2 3 57 kJ D –4 3 57 kJ 28 The information that must be obtained in an experiment to determine the heat of neutralisation are I volumes of acids and alkalis. II concentrations of acids and alkalis. III maximum rise in temperature. IV degree of dissociation of acids and alkalis. A I and III only B II and IV only C I, II and III only D I, II, III and IV 29 Which of the following reactions will produce the same amount of heat of neutralisation? Reaction I

494

Reaction II

A HNO3(aq) + H2SO4(aq) + NaOH(aq) KOH(aq) B HNO3(aq) + HNO3(aq) + NaOH(aq) NH3(aq) C HCl(aq) + NH3(aq)

CH3COOH(aq) + NH3(aq)

D HCl(aq) + NaOH(aq)

CH3COOH(aq) + NaOH(aq)

4.5 Based on the energy level diagram given above, what is the temperature rise when 25 cm3 of 1 mol dm–3 H2SO4 is mixed with 50 cm3 of 1 mol dm–3 sodium hydroxide solution? [Specific heat capacity of solution = 4.2 J g–1 °C–1; density of solution = 1 g dm–3]

C 9.1 °C D 13.5 °C

Heat of Combustion

30 The combustion of 3.6 g of fuel X increases the temperature of 500 cm3 of water by 22.8 °C. What is the temperature rise if the combustion of 7.2 g of fuel X is used to heat up 1.0 dm3 of water? A 11.4 °C B 14.2 °C C 22.8 °C D 28.4 °C

32 The diagram shows the arrangement of apparatus used for determining the heat of combustion of methanol.

Which of the following steps must be taken to ensure that the experimental results obtained are accurate? I Replace the glass beaker with a metal container. II Stir the water continuously. III Remove the wire gauze. IV Place the spirit lamp closer to the water. A III and IV only B I, II and IV only C II, III and IV only D I, II, III and IV

33 C4H9OH(l) + 6O2(g) → 4CO2(g) + 5H2O(l); ΔH = –2670 kJ mol–1 Based on the thermochemical equation given above, what is the mass of butanol needed for combustion in order to produce 80 kJ of heat? [Relative atomic mass: C, 12; H, 1; O, 16] A 2.22 g B 4.44 g C 6.66 g D 8.88 g 34 The heat of combustion of alcohols increases in the order: ethanol < propanol < butanol Which of the following statements explain the trend? I The number of carbon atoms per molecule increases. II The number of hydrogen atoms per molecule increases. III The number of oxygen atoms per molecule increases IV The chemical bonds in the molecule become stronger. A I and II only B III and IV only C II and III only D I, II and IV only 35 The heat of combustion of propane is –2220 kJ mol–1. C3H8(g) + 5O2(g) →

1 B – —  2220 kJ 4 1 C – ___  2220 kJ 3 D – 4  3  2220 kJ 36 The heat of combustion of a fuel, X is –273 kJ mol–1. What is the maximum temperature rise when 100 cm3 of water is heated by the combustion of 0.80 g of X? [Relative molecular mass of X = 40; specific heat capacity of water = 4.2 J g–1 °C–1] A 6 °C C 13 °C B 9 °C D 15 °C 37 The equation for the combustion of ethanol in excess oxygen is ’07 shown below. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l); ΔH = –1371 kJ mol–1 Which of the energy level diagrams represents the reaction? A

B

C

3CO2(g) + 4H2O(l); ΔH = –2220 kJ mol–1 How much heat energy is evolved when 11 g of carbon dioxide is produced from the complete combustion of propane? 1 1 A – —  —  2220 kJ 4 3

495

D



Thermochemistry

4

31 The results of an experiment to determine the heat change for ’06 the combustion of propanol, C3H7OH, are shown below. Volume of water in the copper container = 300 cm3 Initial temperature of water = 28.5 °C Final temperature of water = 69.0 °C What is the heat released by the combustion of propanol? [Specific heat capacity of water = 4.2 J g–1 °C–1; assume that 1 cm3 of water is equal to 1 g] A 25.20 kJ B 34.65 kJ C 51.03 kJ D 86.94 kJ

Structured Questions 1 Two experiments are carried out to determine the heat of displacement of copper by zinc. Experiment 1 A heating element is immersed in copper(II) sulphate solution and 1.0 kJ of heat energy is supplied. Experiment II 3.27 g of zinc powder is added to 200 cm3 of copper(II) sulphate solution (in excess). In both these experiments, • the copper(II) sulphate solution is placed in a polystyrene beaker, • the initial temperature of copper(II) sulphate is 30.4 °C. Diagram 1 shows the thermometer readings for the maximum temperature reached in Experiments I and II.

Diagram 2

Which of these polystyrene beakers should be used? Why? [2 marks]

4

2 The apparatus set-up as shown below is used to determine the heat of combustion of ethanol. The copper container contains 100 cm3 of water.

The results of the experiment are shown below.

Diagram 1

(a) What is meant by heat of displacement of copper by zinc? [1 mark] (b) Explain why polystyrene beaker is used in Experiments I and II. [1 mark] (c) Calculate the rise in temperatures in Experiments I and II. [2 marks] (d) (i) State two observations in Experiment II.

Mass of ethanol + lamp before combustion (g)

150.38

Mass of ethanol + lamp after combustion (g)

149.83

Initial temperature of water (°C)

30.0

Highest temperature of water (°C)

61.5

[2 marks]

(ii) State one precaution that must be taken while carrying out Experiment II. [1 mark]

(a) What is meant by heat of combustion of ethanol? [1 mark]

(e) (i) Calculate the heat energy change in Experiment II. (ii) Calculate the number of moles of copper metal formed in Experiment II. (iii) With reference to your answers in (i) and (ii), calculate the heat of displacement of copper by zinc. [Specific heat capacity of solution: 4.2 J g–1 °C–1; density of solution: 1.0 g cm–3; relative atomic mass: Zn, 65.4] [3 marks]

(b) Write a chemical equation for the complete combustion of ethanol. [1 mark]



(c) (i) Calculate the number of moles of ethanol used in the experiment. [2 marks] (ii) Calculate the heat released in the experiment. [1 mark] [Specific heat capacity of water = 4.2 J g–1 °C–1; density of water = 1 g cm–3] (iii) Calculate the heat of combustion of ethanol. [1 mark] (d) The theoretical heat of combustion of ethanol is –1371 kJ mol–1. Give two reasons why the heat of combustion obtained from this experiment is different from the theoretical value. [2 marks]

(f) Experiment II can be carried out by using polystyrene beaker A or polystyrene beaker B (Diagram 2). Thermochemistry

496

3 (a) Experiment I was carried out to determine the heat of neutralisation between sulphuric acid and potassium hydroxide. The experimental result shows that the heat of neutralisation is –57 kJ mol–1. (i) What is meant by heat of neutralisation? (ii) Draw the energy level diagram for the reaction.

[1 mark] [1 mark]

(b) In Experiment II, 50 cm3 of 1 mol dm–3 nitric acid is added to 50 cm3 of 1 mol dm–3 ammonia solution in a polystyrene cup. The initial temperatures of the acid and alkali are measured. The mixture is stirred slowly and the maximum temperature of the mixture is recorded. The results of the experiment are shown below. Initial temperature of nitric acid = 28.2 °C Initial temperature of ammonia solution = 28.2 °C The maximum temperature = 34.2 °C

[The specific heat of solution = 4.2 J g–1 °C–1; Density of solution = 1 g cm–3] (i) State two precautions that must be taken when carrying out the experiment. (ii) Calculate the number of moles of water molecules produced. (iii) Calculate the heat of neutralisation between nitric acid and ammonia solution.

[2 marks] [2 marks] [3 marks]

(c) Explain why the heat of neutralisation obtained in Experiment I is different from that in Experiment II. (d) Predict the temperature change if Experiment II is repeated by adding 100 cm of hydrochloric acid to 100 cm3 of 1 mol dm–3 ammonia solution. Explain your answer.

[2 marks]

3

4

[1 mark]

4 (a) Diagram 3 shows the apparatus set-up for Experiments I, II and III and the readings of the initial temperature and the highest or lowest temperature in each experiment.

Diagram 3



(i) Identify the exothermic and endothermic reaction in these reactions. (ii) Three substances are used for this experiment: ammonium sulphate, sodium hydrogen carbonate and sodium hydroxide. Match these substances with X, Y and Z. (iii) Write the equation for Experiment III. (iv) Draw an energy level diagram for the reaction in Experiment I.

[2 marks] [2 marks] [1 mark] [2 marks]

(b) When 25.0 cm3 of 0.5 mol dm–3 silver nitrate solution is poured into 25.0 cm3 of 0.5 mol dm–3 of sodium chloride solution, the temperature increases by 3 °C. (i) Calculate the heat change in this experiment. [Specific heat capacity of solution is 4.2 J g–1 °C–1; Density of solution is 1.0 g cm–3] (ii) Calculate the heat of precipitation of silver chloride for this experiment. (iii) The experiment is repeated using 25.0 cm3 of 0.5 mol dm–3 silver nitrate solution to react with 25.0 cm3 of 0.5 mol dm–3 hydrochloric acid. Predict the temperature change in this experiment. Give a reason for your prediction.

497

[1 mark] [2 marks]

[2 marks]

Thermochemistry

Essay Questions (i) What is the fuel value of propanol. (ii) State two main factors that must be considered when choosing a fuel. [3 marks]

1 (a) Describe one experiment to determine the heat of precipitation of lead(II) chloride. In your answer, show the workings in your calculation, assumptions made and the energy level diagram for the precipitation of lead(II) chloride.

2 (a) Describe one experiment to determine the heat of neutralisation between dilute sulphuric acid and sodium hydroxide solution. Show the workings in your calculation. [12 marks]

[12 marks]



(b) The heats of combustion of methanol and ethanol are –715 kJ mol–1 and –1371 kJ mol–1 respectively. State two similarities between the combustion of methanol and the combustion of ethanol. Explain why the heat of combustion of ethanol is different from the heat of combustion of methanol. [5 marks]

(b) Explain why the heat of neutralisation between hydrochloric acid and sodium hydroxide is the same as the heat of neutralisation between nitric acid and potassium hydroxide solution. [4 marks] (c) The reaction between nitrogen and hydrogen to form ammonia is an exothermic reaction. Explain this observation in terms of bond breaking and bond forming. [4 marks]

4

(c) The heat of combustion of propanol is –2010 kJ mol–1.

Experiments 1 Diagram 1 shows two experiments carried out to determine the heat of neutralisation.

(b) Construct a table to tabulate the results of Experiments I and II. [3 marks]

’06

(c) State three observations in Experiment I other than the change in temperature. [3 marks]

Experiment I: Reaction between 25 cm3 of methanoic acid, HCOOH, 2.0 mol dm–3 and 25 cm3 of sodium hydroxide solution, NaOH, 2.0 mol dm–3.

(d) State one hypothesis for both experiments. [3 marks] (e) State the constant variables in this experiment. [3 marks]

45

(f) Based on the temperatures in Experiment I, predict the rise in temperature in Experiment II. [3 marks] (g) (i) Define heat of neutralisation. [3 marks] (ii) Calculate the heat of neutralisation for the reaction between methanoic acid and sodium hydroxide. [3 marks] (Specific heat capacity of solution = 4.2 J g–1 °C–1; density of solution = 1.0 g cm–3)

Diagram 1

Initial temperature of the mixture = … … … °C Highest temperature of the mixture = … … … °C Increase in temperature = … … … °C Experiment II: Reaction between 25 cm3 of hydrochloric acid, HCl, 2.0 mol dm–3 and 25 cm3 of sodium hydroxide solution, NaOH, 2.0 mol dm–3.

2 A student bought two canisters of fuels (brand X and brand Y) for camping. It is found that 200 g of fuel of brand X can boil 200 cm3 of water in 12 minutes, whereas 200 g of fuel of brand Y can boil 200 cm3 of water in 20 minutes. With reference to the above observation, you are asked to plan an experiment to compare the fuel values (heat of combustion expressed in kJ g–1) of fuel X and fuel Y.

Initial temperature of the mixture = T1 °C Highest temperature of the mixture = T2 °C Increase in temperature = T3 °C



(a) Write the initial and the highest temperature of the solution and the rise of temperature for Experiment I. [3 marks] Thermochemistry

498

Your planning should include the following aspects: (a) Aim of experiment (b) All the variables (c) Statement of the hypothesis (d) List of substances and apparatus (e) Procedure of the experiment (f) Tabulation of data [17 marks]

FORM 5 THEME: Production and Management of Manufactured Chemicals

CHAPTER

5

Chemicals for Consumers

SPM Topical Analysis 2008

Year 1

Paper

2 A

Section Number of questions

3

2009

–

3

B

1

C

1

2 A

–

–

2

2010

–

B –

3

1

C –

–

1

2011

2

3

A

B

C

–

1 — 2

–

–

1

2

2

3

A

B

C

–

1

–

–

ONCEPT MAP CHEMICALS FOR CONSUMERS

Cleansing agents

Soap Palm oil + NaOH ↓ Soap + Glycerol

Food additives

Detergent Preparation of sodium alkyl sulphate R – O – SO3H + NaOH ↓ ROSO3–Na+ + H2O

Cleansing action of soap and detergent • The ‘head’ (hydrophilic) dissolves in water • The ‘tail’ (hydrophobic) dissolves in greasy dirt • Wetting action • Emulsification Effectiveness of soap and detergent • Soap is not effective in hard water or acidic water • Detergents are effective in both hard water and acidic water

• • • •

Preservative: sodium nitrite Antioxidant: ascorbic acid Flavouring agent: MSG Stabiliser and thickening agent: gelatin and acacia gum • Dyes: azo compound and triphenyl compound

Effects on health • Allergy • Cancer causing • Brain damage • Hyperactivity Additives in detergents • Builders (sodium tripolyphosphate): to soften hard water • Whitening agent (sodium perborate): for making clothes white • Biological enzyme: decomposes protein stains such as blood and food

Medicine

Traditional medicine • derived from plants and animals

Modern medicine

• Analgesic: aspirin, paracetamol and codeine • Antibiotic: penicillin; streptomycin • Psychotherapeutic medicine – stimulant – antipsychotic – antidepressant

Side effects Aspirin: causes stomach bleeding Penicillin: allergy Psychotherapeutic medicine: addiction

5.1 Soaps

1 Cleansing agents are chemical substances used SPM to remove grease and dirt. ’04/P2 2 There are two types of cleansing agents: (a) Soaps (b) Detergents

Example of soap

C17H33COOH Oleic acid

C17H33COO–K+ Potassium oleate

C17H35COOH Stearic acid

C17H35COO–K+ Potassium stearate

2 Soap is produced by the reaction between sodium hydroxide or potassium hydroxide with animal fats or vegetable oils. This reaction is known as saponification. 3 Fats and vegetable oils are large, naturally occurring ester molecules. When fats or oils are boiled with concentrated alkalis, such as sodium hydroxide or potassium hydroxide, saponification occurs and the ester molecules are broken down into soap and glycerol.

The History of Soap Manufacturing

5

Fatty acid

Soaps and Detergents

1 Soaps have been used for more than 3000 years. It was recorded that the Babylonians were making soap around 2800 B.C. 2 In ancient times, soaps were made from ashes of plants which contain sodium carbonate and potassium carbonate. The ashes were boiled with lime (calcium oxide) to produce caustic potash (potassium hydroxide). Caustic potash is then boiled with animal fats to produce soap. boil (a) Ash + lime ⎯⎯→ caustic potash (K2CO3) (CaO) (KOH)

Fats and oils (natural ester)

Glycerol (an alcohol with three –OH groups)

boil (b) Caustic potash + animal fats ⎯⎯→ soap

SPM

1 (a) Soap is a chemical used as a cleansing agent to remove grease and dirt. (b) Soaps are sodium or potassium salts of long chain fatty acids with 12 to 18 carbon atoms per molecule. The general formula of soap is RCOO–Na+ or RCOO–K+ (Table 5.1).

C15H31COOH Palmitic acid

C15H31COO–Na+ Sodium palmitate

Chemicals for Consumers

Soap (RCOONa or RCOOK)

glyceryl tristearate (in fats and oils)



Table 5.1 Examples of soap

Example of soap

+

O i CH2 – C – O – C17H35 ⎮ O concentrated ⎮ i NaOH CH – C – O – C17H35 + 3Na OH ⎯⎯⎯⎯⎯→ boil ⎮ O ⎮ i CH2 – C – O – C17H35

’08/P1, ’09/P1, ’10/P1

Fatty acid

NaOH(aq) or KOH(aq) (alkali)

4 Saponification is the alkaline hydrolysis of esters using alkali solutions. From a chemical aspect, soaps are sodium salts or potassium salts of long chain fatty acids (with 12 to 18 carbon atoms per molecule). 5 Glyceryl tristearates are naturally occurring esters commonly found in animal fats and vegetable oils. When the ester is boiled with concentrated sodium hydroxide solution, saponification (alkaline hydrolysis) occurs and a mixture of sodium stearate (soap) and glycerol is obtained.

3 In 1816, the French chemist Michel Chevreul (1786 – 1889) discovered that animal fats are composed of fatty acids and glycerol. This discovery contributed to the rapid development of the soap and candle industry. 4 In 1861, the Belgian chemist Ernest Solvay (1838–1922) discovered the process to make soda (sodium carbonate) from common salt (sodium chloride) and calcium carbonate. This process is known as the Solvay Process which produces sodium carbonate which is used for making glass, soaps and detergents. Preparation of Soap by Saponification

+



CH2 – OH ⎮ CH – OH + 3C17H35COO–Na+ ⎮ soap ⎮ (sodium stearate) CH2 – OH glycerol

500

6 The soap produced can be precipitated by adding common salt (sodium chloride) to the reaction mixture. Sodium chloride reduces the solubility of soap in aqueous solution and causes the soap to be precipitated out. 7 The properties of soap depend on (a) the type of alkali used for saponification, (b) the type of animal fats or vegetable oils used.

8 Soaps produced from sodium hydroxide are hard whereas soaps produced from potassium hydroxide are soft. 9 Animal fats (tallow) from cows and vegetable oils (such as palm oil or olive oil) are used for making soap. 10 Activity 5.1 describes the method for making soap by the saponification process.

Apparatus

Beaker, measuring cylinder, glass rod, Bunsen burner, filter funnel, filter paper, wire gauze, spatula and tripod stand.

Materials

5 mol dm–3 sodium hydroxide solution, sodium chloride solution, palm oil, olive oil, corn oil and peanut oil.

Procedure (A) Preparation of soap 1 Using a small (100 cm3) measuring cylinder, 10 cm3 of palm oil is measured out into a beaker. 2 Using a large (250 cm3) measuring cylinder, 50 cm3 of concentrated (5 mol dm–3) sodium hydroxide solution is measured out and poured into the palm oil in the beaker. 3 The mixture is heated until boiling (Figure 5.1 (a)). The solution is stirred with a glass rod throughout the experiment. 4 The boiling is continued for about 5 minutes until the layer of oil disappears. 5 The Bunsen flame is then turned off and the reaction mixture is left to cool.

5

To prepare soap

(B) To separate soap from the reaction mixture 1 Three to four spatula of sodium chloride is dissolved in about 100 cm3 of water. 2 The salt solution is then added to the soap solution obtained in Section A. The mixture is boiled again for about 5 minutes (Figure 5.1 (b)).

3 When the mixture is cooled, a white precipitate is formed, which floats on the surface of the mixture. 4 The mixture is filtered (Figure 5.1 (c)) and washed with distilled water to remove excess alkali and salt. 5 A small amount of the residue on the filter paper is examined by (a) rubbing it on the hand, (b) shaking it with water in a boiling tube. The observations are recorded. 6 The experiment is repeated using olive oil, corn oil and peanut oil. 501

Chemicals for Consumers

Activity 5.1

Figure 5.1 Preparation of soap

Results Test

Observation

(a) Rubbed on the hand

Soapy (slippery) feel

(b) Shaken with water

A lot of foam is produced

Discussion 1 The white residue on the filter paper is soap because it has a slippery feel and produces foam when shaken with water. 2 The soap produced in Section A can be precipitated from the solution by adding sodium chloride and

boiling the solution again. This process is called salting out. 3 The salting out process occurs because sodium chloride reduces the solubility of soap in water and causes it to be precipitated. 4 The precautionary step in this experiment is to control the flame of the Bunsen burner. This is because boiling a mixture of palm oil with sodium hydroxide solution produces a lot of froth. Conclusion Soaps can be prepared from vegetable oils and concentrated sodium hydroxide solution by saponification.

5

The Structure of Soap Molecule

‘tail’. The ‘head’ is negatively-charged and the ‘tail’ is a long hydrocarbon chain. O  3 The ‘head’ contains the –C–O– ion which dissolves readily in water (hydrophilic) but does not dissolve in oil. Conversely, the ‘tail’ contains a long hydrocarbon chain which is insoluble in water (hydrophobic) but dissolves readily in oil. 4 Soaps made from palmitic acid are known as sodium palmitate. Figure 5.2 shows the structure of the palmitate ion (C15H31COO–).

1 When soap is dissolved in water, it will dissociate and produce sodium ions and carboxylate ions (RCOO–). For example, sodium stearate (soap) dissolves in water to form sodium ions and stearate ions. C17H35COONa(s) + water → C17H35COO–(aq) sodium stearate stearate ions + Na+(aq) 2 The stearate ions take part in the removal of dirt but the sodium ions do not. The stearate ion consists of two parts: the ‘head’ and the SPM

’04/P1 ’05/P2

Figure 5.2 The structure of a palmitate ion

5 Figure 5.3 (a) shows the molecular model of the palmitate ion and Figure 5.3 (b) shows the simple representation of the structure of the palmitate ion.

Figure 5.3 The palmitate ion Chemicals for Consumers

502

Detergents

3 The equations for the preparation of sodium lauryl sulphate is shown below. Step 1 Preparation of lauryl hydrogen sulphate CH3(CH2)10CH2O H + H2SO4 ⎯→ lauryl alcohol ( HO SO3H) CH3(CH2)10CH2OSO3H + H2O

SPM

’08/P1

1 Detergents are synthetic cleansing agents made from hydrocarbons obtained from petroleum fractions. Thus, detergents are petrochemicals. 2 Detergents can be classified into three main types, depending on the charge on the detergent ion. (a) Anionic detergents where the head of the detergent particle contains a negativelycharged ion.

lauryl hydrogen sulphate

Step 2 Preparation of sodium lauryl sulphate CH3(CH2)10CH2OSO3 H + Na OH ⎯→ lauryl hydrogen sulphate

negativelycharged ion Example: R – O – SO3–Na+ (sodium alkyl sulphate)

CH3(CH2)10CH2OSO3–Na+ + H2O sodium lauryl sulphate

(b) Cationic detergents where the head of the detergent particle contains a positivelycharged ion. positivelycharged ion

Example: R – N(CH3)3+Br–

5

(c) Non-ionic detergents Example: R– O– CH2CH2OH Preparation of Detergent 1 The detergent, sodium alkyl sulphate can be prepared from alcohols with chain lengths of 12 to 18 carbon atoms in two steps. Step 1 Reaction with concentrated sulphuric acid O O   R–O– H+H–O –S–O–H→ R–O–S–O–H   + H2O O O long chain alcohol

concentrated sulphuric acid

Toothpaste The main ingredients in toothpaste are sodium dodecyl sulphate (detergent), calcium carbonate or aluminium hydroxide (as abrasives to rub away the food and bacteria), glycerol (sweetener), cellulose gum (thickener) and mint (for taste).

The Structure of Detergent Molecule

alkyl hydrogen sulphate

When a detergent is dissolved in water, it dissociates to form sodium ions and detergent ions. The detergent ions have the same basic structure as the soap ions, that is, it consists of two parts: (a) The ‘head’ is the sulphate group (–OSO3–) which is negatively-charged and hydrophilic (dissolves readily in water but not in oils and grease). (b) The ‘tail’ is the long hydrocarbon chain, which is neutral and hydrophobic (dissolves readily in oils and grease but not in water).

Step 2 Neutralisation with sodium hydroxide solution O O   R – O – S – O – H + NaOH → R – O – S – O–Na+   + H2O O O alkyl hydrogen sulphate

sodium alkyl sulphate

2 An example of a long chain alcohol is lauryl alcohol (dodecan-1-ol), CH3(CH2)10CH2OH. The deter­gent prepared from dodecan-1-ol is called sodium dodecyl sulphate (IUPAC name) or sodium lauryl sulphate (common name), CH3(CH2)10CH2O-SO3–Na+.

alkyl sulphate ion

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Chemicals for Consumers

The Cleansing Action of Soap and Detergent The cleansing action of soap or detergent depends on their chemical bonding and structures. (a) The ionic ‘head’ (negatively-charged) is soluble in water (hydrophilic) but insoluble in oil. (b) The long hydrocarbon ‘tail’ (neutral) is insoluble in water (hydrophobic) but soluble in oil. Step 1 Action of soap on dirt • When soap or detergent is mixed with water, it lowers the surface tension of water and wets the dirty surface. • The negatively-charged ‘heads’ of soap or detergent ions dissolve in water (hydrophilic), • The hydrocarbon ‘tails’ of soap or detergent ions dissolve in the layer of grease (hydrophobic).

5

The arrangement of soap or detergent ions in the greasy layer

Step 2 Dirt being surrounded by soap/ detergent ions If the water is agitated slightly, the grease begins to lift off the surface. On agitation, grease begins to lift off the surface

Step 3 Lifting dirt from cloth On further agitation during washing, the greasy dirt is lifted from the surface since the density of grease is less than water. Greasy dirt particles are lifted from the layer of cloth

Step 4 Emulsifying dirt in water (a) When the water is shaken, the grease will be emulsified when it breaks into smaller droplets. (b) These greasy droplets repel one another be­cause they carry the same charge. As a result, the grease is suspended in the solution. (c) When the cloth is rinsed with water, the droplets will be carried away.

Emulsification of greasy dirt in water as droplets Chemicals for Consumers

504

Part Y is soluble in water. The structural formula represents a soap ion (a carboxylate). Since the formula contains 16 carbon atoms, it is called a palmitate. Answer C

• Soaps and detergents can act as emulsifying agents to emulsify oil and grease. • The process of emulsification breaks large drops of grease into smaller droplets that float in water.

Additives in Detergents 1 Modern detergents used for washing clothes usually contain a few types of additives to (a) increase their cleaning power, (b) make them attractive and saleable. 2 Only about 20% of the substances in a detergent are cleansing agents (sodium alkyl sulphate or sodium alkylbenzene sulphonate). The other substances are additives. The examples of additives and their functions are described on pages 508 to 509.

The cleaning process becomes more efficient if the water containing the soap or detergent solution is stirred vigorously or the washing is done at a higher temperature. Heating and stirring help to loosen the dirt particles from the material being washed.

5

• When soap or detergent is added to clean water, the ‘head’ of the soap or detergent ion dissolves in water while the ‘tail’ sticks out from the surface of water.

• Figure 5.4 shows the action of brighteners. The brighteners absorb the invisible ultraviolet light and reradiate it as blue light.

The ‘heads’ of the soap or detergent ions will weaken the forces of attraction between water molecules and lower the surface tension of water. When the surface tension is lowered, the water molecules will spread out and wet the cloth or plates to be cleaned. Hence, soaps act as wetting agents. • The dirt particles surrounded by soap or detergent ions are called micelles.

1

Figure 5.4 Action of brighteners • Brighteners makes clothes appear whiter and brighter because the blue light can hide any yellowing on the fabric.

’03 ’04

Part of the structural formula of a cleansing agent is shown below. CH3(CH2)14COO–

The Effectiveness of Soaps and Detergents as Cleansing Agents

X Y Which of the following statements is true? A Part X represents alkylbenzene. B Part Y is soluble in grease. C The name of the structural formula is palmitate. D The structural formula represents a detergent ion.

Advantages of Soaps

SPM

’11/P1

SPM

’11/P2

1 Soaps are effective cleansing agents in soft water, that is, water that does not contain Mg2+ and Ca2+ ions. 2 Soaps do not cause pollution problems to the environ­ment. This is because soaps are made from chemicals found in animals and plants. This means that soaps are biodegradable.

Comments Part X represents an alkyl group and not alkylbenzene. 505

Chemicals for Consumers

Disadvantages of Soaps 2R – O – SO3–(aq) + Ca2+(aq) → (R – O – SO3)2Ca(aq)

1 Soaps are ineffective in hard water, that is, water that contains magnesium and calcium salts. 2 In hard water, soaps will react with Mg2+ and Ca2+ ions to form a precipitate called scum. Thus, soaps do not lather in hard water. 3 Scum is a grey solid that is insoluble in water. It consists of magnesium stearate and calcium stearate.

alkyl sulphate ion (detergent)

Hence, scum is not formed and the detergents are still active in hard water and lathers easily. 3 Detergents are synthetic cleansing agents. This means that the structure of the hydrocarbon chain can be modified to produce detergents with specific properties. Nowadays, different types of detergents have been synthesised for specific uses such as shampoos and dish cleaners. 4 Furthermore, detergents are also effective in acidic water because H+ ions in acidic water do not combine with detergent ions.

Mg2+(aq) + 2C17H35COO–(aq) → (C17H35COO)2Mg(s)

stearate ion (soap ion)

magnesium stearate (scum)

Ca2+(aq) + 2C17H35COO–(aq) → (C17H35COO)2Ca(s)

5



stearate ion (soap ion)

calcium stearate (scum)

4 Soaps are also not effective in acidic water, for example, rainwater containing dissolved acids. H+ ions from acids will react with soap ions to produce fatty acid molecules of large molecular size that are insoluble in water.

Disadvantages of Detergents 1 Most detergents have branched hydrocarbon chains and are non-biodegradable. As a result, non­­ -biodegradable detergents cause water pollution. 2 Phosphates in detergents act as fertilisers and promotes the growth of water plants and algae. When the plants die and decay, they use up the oxygen dissolved in water. This will decrease the oxygen content in water and kill fishes and other aquatic organisms. 3 Detergents produce a lot of foam in water. The layer of foam that covers the water surface will prevent oxygen from dissolving in water. This condition will cause fish and other aquatic life to die from oxygen starvation. 4 Additives such as sodium hypochlorite (bleaching agent) releases chlorine gas in water that is acidic. Chlorine gas is highly toxic and kills aquatic life.

C17H35COO–(aq) + H+(aq) → C17H35COOH(s) stearate ion (soap ion)

stearic acid

5 Stearic acids and other fatty acids do not act as cleansing agents because they exist mainly as molecules and do not have anionic hydrophilic ends (‘heads’) that dissolve in water. Advantages of Detergents 1 Detergents are cleansing agents that are effective in soft water as well as hard water. This is because detergents do not form scum with Mg2+ ions and Ca2+ ions found in hard water. 2 The detergent ions (R – O – SO3–) react with Mg2+ and Ca2+ ions in hard water. However, the magnesium salts and calcium salts which are formed are soluble in water.

The cleansing agents for window glass or mirrors are volatile liquids. The chemicals usually used in the cleansing agents are isopropyl alcohol (propan-2-ol). Ammonia is added to enhance the cleansing power.

2R – O – SO (aq) + Mg (aq) → (R – O – SO3)2Mg(aq) – 3

alkyl sulphate ion (detergent)

Chemicals for Consumers

calcium alkyl sulphate (dissolves in water)

2+

magnesium alkyl sulphate (dissolves in water)

506

2

’06

The following chemical equations show the reactions between alkyl sulphate ions, ROSO3– and stearate ions, CH3(CH2)16COO– with magnesium ions, Mg2+ in hard water. R represents long hydrocarbon chain. 2ROSO3–(aq) + Mg2+(aq) → (ROSO3)2Mg(aq)

Comments The equation shows that alkyl sulphate ions do not form a precipitate with Mg2+ ions but stearate ions form a precipitate with Mg2+ ions. As CH3(CH2)16COO– (aq) reacts with Mg2+(aq) to form a precipitate of (CH3(CH2)16COO)2Mg, the concentration of CH3(CH2)16COO–(aq) decreases. Answer C

5

2CH3(CH2)16COO–(aq) + Mg2+(aq) → (CH3(CH2)16COO)2Mg(s) What is the effect of adding magnesium ions to a solution containing alkyl sulphate ions or stearate ions? A The concentration of alkyl sulphate ions decreases. B The concentration of alkyl sulphate ions increases.

C The concentration of stearate ions decreases. D The concentration of stearate ions remains unchanged.

Cleansing agents Soaps • Soaps are cleansing agents made from animal fats or vegetable oils.

History of soap making • In ancient times, soaps were made from ashes and soda. • Ernest Solvay and Michel Chevreul are the chemists that contributed to the development of the soap industry in the past.

Detergents • Detergents are synthetic cleansing agents made from petroleum products.

Preparation of soap by saponification Fats or oils + NaOH(aq) → soap + glycerol ∆ Soap is RCOONa (R is a long hydrocarbon chain)

Preparation of detergent from petroleum fractions • Sodium alkyl sulphate ROSO3H + NaOH(aq) → ROSO3Na + H2O

Cleansing action of soaps and detergents • The ‘head’ dissolves in water. Effectiveness of the cleansing action of soaps and detergents • Soaps are ineffective in hard water or acidic water because of the formation of precipitates that are insoluble in water. • Soaps react with Mg2+ and Ca2+ in hard water to form scum. • Detergents are effective in both hard water and acidic water because no scum is formed. • Detergents cause water pollution.

–COO – (in soaps) and OSO3– (in detergents) • The ‘tail’ (long hydrocarbon chain) dissolves in oil. • Agitation of water causes grease to lift off the surface.

Additives in detergents • Builders: soften water • Whitening agents: remove coloured stains • Biological enzymes: remove protein stains • Brighteners: make clothes look brighter and whiter • Drying agent: to keep the detergent dry • Perfume: to make clothes smell fresh • Stabilisers • Antiseptics

507

Chemicals for Consumers

5 5 Chemicals for Consumers

508

5 5 509

Chemicals for Consumers

3

’07

(a) Name the type of reaction that has occured in the following reaction. Vegetable oil + NaOH → soap + compound X (b) State the name of compound X. (c) What happens when sodium chloride is added to the soap mixture?

Solution (a) Saponification (alkaline hydrolysis) (b) Glycerol (c) Soap is precipitated out.

5.1 1 (a) What is meant by soap? (b) State the raw materials needed in the preparation of soap. (c) A naturally occurring ester has the structural formula as shown below.

3 Complete the flowchart to compare the advantages and disadvantages of soap and detergent. Soap

5

O  CH2 – O – C – C17H35 O  CH – O – C – C15H31 O  CH2 – O – C – C17H35

Advantages (a) (b)

Disadvantage

Write an equation to show the reactions involved when soap is prepared from this ester.

Detergent

2 Sodium alkyl sulphate acts as a detergent and has the formula shown below. C12H25 – O – SO3–Na+

Advantages

(a) State the characteristics of sodium alkyl sulphate that enable it to act as a detergent. (b) Using sodium alkyl sulphate as an example, describe the action of detergents on the greasy stains found on a piece of cloth.

5.2

(a) (b)



(a) making food last longer by preventing the growth of microorganism, (b) making food last longer by preventing the oxidation of fats and oils by oxygen in air, (c) making food taste or smell better, (d) improving the texture of food and to prevent food from becoming liquid, (e) restoring the colour of food destroyed during food processing, and (f) adding colouring to food so as to make the food look fresher, more interesting or more appetising.

Uses of Food Additives

Types of Additives and Examples 1 Food preservatives have been used since ancient times. Ancient civilisation used salt to preserve meat and fish, herbs and spices to improve the flavour of food. 2 Food additives are chemicals that are added to food in small quantities for specific purposes such as Chemicals for Consumers

Disadvantages

510

3 There are six types of food additives as shown below. Preservatives Stabilisers

Drying fruits and meat is one of the oldest methods of food preservation. Drying is effective because water is necessary for the growth of microorganisms. Salt and sugar can also be used for food preservation because the dissolved salt and sugar will cause water to flow by osmosis from the microorganisms to its surroundings. The dehydration process stops the growth of microorganisms.

Antioxidants Type of food additives

Thickeners

Flavouring agents

Dyes

Functions of Food Additives

SPM

’09/P1, ’10/P2, ’11/P1, P2

1 Preservatives are chemicals that are added to food to retard or to prevent the growth of microorganisms such as bacteria, mould or fungus so that the food can be stored for a longer period. 2 In ancient times, food additives from natural sources such as salt, sugar and vinegar were used to preserve food and to make the food taste better. 3 Nowadays, synthetic preservatives are used. Table 5.2 shows the types of preservatives commonly used. Many of the preservatives are organic acids and salts of organic acids. Table 5.2 Types of preservatives and their uses

Preservative

Molecular formula

Uses

Sodium nitrite Sodium nitrate

NaNO2 NaNO3

• To preserve meat, sausage, cheese and dried fish • To prevent food poisoning in canned foods • To maintain the natural colour of meat and to make them look fresh

Benzoic acid Sodium benzoate

C6H5COOH C6H5COONa

• To preserve sauces (oyster, tomato or chili), fruit juice, jam and margarine

Sulphur dioxide Sodium sulphite

SO2 Na2SO3

• Used as bleaches and antioxidants to prevent browning in fruit juices • Maintain the colour and freshness of vegetables • To prevent the growth of microorganisms

Antioxidants 1 Antioxidants are chemicals that are added to food to prevent the oxidation of fats and oils by oxygen in the air. 2 Food containing fats or oils are oxidised and become rancid when exposed to air. This makes the food unfit for consumption. The rancid products are volatile organic compounds with

foul odours (for example, butanoic acid, C3H7COOH). 3 Antioxidants are added to fats, oils, cakes, sausages, biscuits and fried food to slow down the oxidation process. 4 Ascorbic acid, citric acid and sodium citrate are examples of antioxidants.

511

Chemicals for Consumers

5

Preservatives

Flavouring Agents 1 There are two types of flavouring agents: artificial flavours and flavour enhancers. They are added to food to make them taste better. 2 Flavour enhancers have little or no taste of their own. They are chemicals that are added to food to bring out the flavours or to enhance the tastes of food. An example of a flavour enhancer is monosodium glutamate (MSG). 3 Artificial flavours include sweeteners and other flavours such as peppermint or vanilla. Aspartame and saccharin are examples of artificial sweeteners. 4 Both aspartame and saccharin can be used as a substitute for sugar to enhance the sweetness

in food and drink. However, the use of saccharin is banned in many countries because it is a suspected carcinogen. 5 Many esters have fruity odours and tastes and are used as artificial flavours. Table 5.3 shows some examples of esters that are used in making drinks. Table 5.3 Uses of esters as flavouring agents

Ester

Benzyl ethanoate

Octyl ethanoate

Ethyl butanoate

Flavour

Strawberry

Orange

Pineapple

5

Stabilisers and Thickening Agents 1 Stabilisers and thickening agents improve the texture and the blending of food. 2 Stabilisers are chemicals that are used to enable oil and water in the food to mix together properly in order to form an emulsion. Examples of stabilisers are gelatin, acacia gum, lecithin and pectin. 3 Stabilisers are added to improve the texture of food. For example, stabilisers are added to ice cream and peanut butter to keep them smooth and creamy. 4 Without stabilisers, ice crystals would form in ice cream, particles of chocolate would settle out

of chocolate milk, oil and vinegar in salad dressing will separate as soon as mixing is stopped. 5 Thickening agents are chemicals that are added to food to thicken the liquid and to prevent the food from liquefying. Thickening agents (also called thickeners) absorb water and thicken the liquid in food to produce a jelly-like structure. 6 Most thickening agents are natural carbohydrates. Gelatin and pectin are added to help jams and jellies set.

Dyes 1 Dyes (colouring agents) are chemicals that are added to food to give them colour so as to improve their appearance. 2 Some food are naturally coloured, but the colour is lost during food processing. The food industry uses synthetic food colours to (a) restore the colour of food lost during food processing, (b) enhance natural colours, so as to increase the attractiveness of food, (c) give colour to food that do not have colour. 3 Some dyes are natural plant pigments while others are synthetically prepared. The synthetic colours used in food are azo and triphenyl compounds. Both these compounds are organic compounds. 4 The synthetic dye, brilliant blue, is an example of triphenyl compound. The synthetic dye, tartrazine and sunset yellow are examples of azo compounds. 5 Azo compounds are organic compounds containing the diazo group, – N = N –, and are usually yellow, red, brown or black in colour. Triphenyl compounds are organic compounds containing three phenyl groups, –C6H5, and are usually green, blue or purple in colour. Chemicals for Consumers

512

• Pectin, gelatin and acasia gum can be used as both stabilisers and thickeners. • Some food are hygroscopic, that is, they absorb moisture from the air and become lumpy. Anticaking agents such as calcium silicate and magnesium silicate are added to table salt and baking powder to keep them dry. • Other food additives include – acidity regulators (to control pH of food) – antifoaming agents – bulking agents (like starch to increase food bulk) – humectants (prevent food from drying out)

Brain damage Excessive intake of nitrites for a prolonged period of time can cause brain damage. In this condition, the supply of oxygen to the brain is disrupted and this causes brain damage.

Naturally occurring colouring agents can be extracted from the flowers, leaves or roots of plants. For example, beta-carotene, the orange-red substance in carrots and a variety of plants is an example of natural food colour. Beta-carotene is added to butter and margarine.

Hyperactivity SPM

1 Food additives such as tartrazine can cause hyperactivity. 2 Children who are hyperactive become very active, find it difficult to relax or sleep and are very restless.

’10/P2

1 The types of food additives allowed and the quantity permitted are controlled by the 1983 Food Act and the 1985 Food Regulation. 2 The permissible quantity depends on the type of food and the food additives. For example, benzoic acid added must not exceed 800 mg per kg in cordial drinks, whereas sodium nitrite must not exceed 100 mg per kg in meat products. 3 The excessive intake of food additives for a prolonged period of time will ruin our health. The side effects arising from taking food additives are allergy, cancer, brain damage and hyperactivity.

Reading the Food Label 1 You must read the label on the food package to identify (a) brand name, (b) the net weight, (c) the halal symbol for Muslim consumers, (d) the nutrient content (for example, carbohy­­­­­­­­ drates, proteins and fats, minerals and vitamins), (e) food additives used, (f) expiry date, (g) address of the manufacturer. 2 Food additives listed in the food labels are usually represented by the code number – E. This code number shows that the food additives have been approved. 3 Figure 5.5 shows a typical food label for an orange drink.

Allergy 1 Food additives such as sodium sulphite (preservative), BHA and BHT (antioxidants), MSG (flavouring) and some food colours (e.g. Yellow No. 5) can cause allergic reactions in some people. 2 The symptoms of MSG allergy are giddiness, chest pain and difficulty in breathing. This condition is called the ‘Chinese restaurant syndrome’. 3 The presence of sodium nitrate or sodium nitrite in food can cause ‘blue baby’ syndrome that is fatal for babies. This syndrome is due to the lack of oxygen in the blood. Hence, the use of nitrate and nitrite is not allowed in baby food.

ascorbic acid

sodium benzoate

expiry date

XYZ Orange drink Ingredients: Water, sugar, Use before citric acid, E330 31-1-2009 Stabilisers: E466, preservative E211 Sweetener: aspartame Colouring agents: E110, E102, vitamins A and D

Cancer 1 Chemicals that cause cancer are called carcinogens. Sodium nitrite (a preservative) is a potent carcinogen. 2 The nitrite reacts with the amines in food to produce nitrosamine which can cause cancer.

yellow-orange tartrazine (dye) (dye)

Figure 5.5 A typical food label

513

Chemicals for Consumers

5

Effects of Food Additives on Health

Disadvantages

The Rationale for Using Food Additives

1 Eating food additives such as preservatives, antioxidants and flavour enhancers in excess quantities over a long period of time is detrimental to health. These additives have little nutritional value. 2 Some of the synthetic food additives can cause allergy, cancer and hyperactivity.

5

Advantages 1 To prevent food spoilage (a) Oxidation and microorganisms (bacteria, fungi) are the main causes in the decomposition of food. In a hot climate, meat and fish rot easily. The use of preservatives is an effective way to prevent food spoilage. (b) If preservatives are not used, food spoilage might drastically reduce the food supply, making for costlier food. (c) Few deaths are associated with the use of food additives. However, many people have died due to food poisoning caused by bacterial toxins. 2 For medical reasons (a) Aspartame and sorbitol are used to make food and drink sweet without using sugar. These food additives are particularly useful as artificial sweeteners for diabetic patients. (b) Artificial sweeteners give the sweet taste but without adding calories to the food. Thus, they can be used to reduce obesity.

Life Without Food Additives Imagine life without food additives. The following situations will occur. 1 Food spoilage will drastically reduce the food supply. This will result in food shortages around the world. 2 Diseases will flourish again due to deficiencies of vitamins and minerals in our food. 3 Loss of appetite will be common particularly among the younger generation because our food will not look, taste and smell good and this will affect their appetites.

Uses of food additives • To make food last longer (to extend the shelf life) • To make food taste better and look better

Types of food additives • Preservatives: sodium nitrite, sodium benzoate • Antioxidant: ascorbic acid • Flavouring: MSG, aspartame • Stabiliser and thickener: gelatin and acacia gum • Dye: azo compounds, triphenyl compounds

Chemicals for Consumers

Functions of food additives • Preservative: to inhibit or prevent the growth of moulds, yeasts and bacteria that spoil food • Antioxidants: to prevent spoilage of food due to oxidation by oxygen • Flavouring agents: to enhance the taste of food • Stabilisers and thickeners: to improve the texture of food • Colouring agents: to restore the colour or to enhance the natural colour of food

514

Effects on health • Allergy: MSG • Carcinogenic: NaNO2 • Brain damage • Hyperactivity

2 Traditional medicine are medicine derived from natural sources such as plants and animals without being processed chemically.

5.2 1 (a) What is meant by food additives? (b) State two methods of food preservation which ’03 are used in our daily lives. Explain how these methods work. 2 State the function and application of (a) acacia gum, (b) aspartame in the food industry.

The source and uses of some animals and animal parts as medicine are shown below. • Sea horses – For treating respiratory disorders such as asthma and skin ailments • Soft-shelled turtles – Improve blood circulation • Tiger bones – For curing joint pain and stiffness, back pain and rheumatism • Bear bile (from bear gall bladder) – To treat illnesses ranging from liver cancer to haemorrhoids (common name: piles) to conjunctivitis The use of animals or animal parts as medicine should be banned. This is because killing animals or removing animal parts is a cruel act. Furthermore, many of these animals are endangered and should be protected.

3 Explain the advantages and disadvantages of using artificial colourings as food additives.

5.3

Medicine

5

Sources and Uses of Traditional Medicine

3 Medicine obtained from plants are known as herbal medicine. The sources and uses of some herbal medicine are shown in Table 5.4.

1 Medicine is a substance used to prevent or cure diseases or to reduce pain and suffering due to illnesses.

Table 5.4 Some examples of herbal medicine SPM

’06/P2

Plant

Part of the plant used

Uses

Garlic

Corm

• For preventing flu or asthma attacks • For reducing high blood pressure

Ginger

Rhizome (horizontal underground stem)

• For treating stomach pain due to wind in the stomach • For supplying heat energy to keep the body warm • For preventing flu attack

Aloe vera

Leaves

• For treating itchy skin • For treating burns (scalding) on the skin

Lemon (lime)

Fruits

• For treating boils or abscesses on the skin • For preventing flu attack • For treating skin diseases

Quinine

Bark of Cinchona tree

• For treating malaria • For preventing muscle cramps

Ginseng

Roots

• As a tonic to improve the overall health of human beings • For increasing energy, endurance and reducing fatigue

Lemon grass

Stem/leaves

• Has antibacterial and antifungal properties • For treating coughs

Tongkat ali

Roots

• As a tonic for after birth and general health 515

Chemicals for Consumers

Modern Medicine 1 Modern medicine can be classified as follows based on their effects on the human body: (a) Analgesics (b) Antibiotics (c) Psychotherapeutic drugs 2 Some examples of modern medicine are shown in Table 5.5.

Modern drugs have a trade name and a generic name. For example, the analgesic aspirin (generic name) is sold under different brand names such as Caprin and Disprin. Similarly, paracetamol (generic name) is sold under the trade name of Panadol.

Function of Each Type of Modern Drugs

Table 5.5 Examples of modern drugs

Type of modern drug

Aspirin, paracetamol, codeine

(b) Antibiotics

Penicillin, streptomycin

(c) Psychotherapeutic drugs

Stimulant, antidepressant and antipsychotic

5

Analgesics

Example

(a) Analgesics

SPM

’04/P1 ’04/P2 ’05/P1

1 Analgesics are medicine that relieve pain. Examples of analgesics are aspirin, paracetamol and codeine. Analgesics are sometimes called painkillers. 2 Aspirin and paracetamol are mild painkillers whereas codeine is a powerful painkiller. 3 Analgesics relieve pain but do not cure the disease. Table 5.6 shows the chemical aspect and functions of some analgesics.

Table 5.6 Types of analgesics and their functions

Type of analgesic Aspirin

Chemical aspect IUPAC name: Acetyl salicylic acid Functional groups: a carboxylic acid group and an ester group carboxylic group COOH O  O – C – CH3 ester group

Function • Relief pain and has anti-inflammatory action • Used to – reduce fever – relieve headaches, muscle aches and joint aches – treat arthritis, a disease caused by inflammation of the joints – act as an anticoagulant. It prevents the clotting of blood and reduce the risk of heart attacks and strokes

Thus, aspirin is acidic in nature Paracetamol SPM

’07/P1, ’09/P1

Structural formula: H O |  – N – C – CH3 HO –

• Similar to aspirin in its effects but it does not reduce inflammation • Reduces or relieves flu symptoms such as fever, bone aches and runny nose

Thus, unlike aspirin, paracetamol is neutral in nature Codeine

Chemicals for Consumers

Codeine is an organic compound that contains the elements carbon, hydrogen, oxygen and nitrogen

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• Used to relieve minor to moderate pain. It is more powerful than aspirin and paracetamol but less powerful than morphine. Codeine and morphine are narcotic drugs • Also used in cough mixtures for suppressing coughs

Antibiotics: Antibacterial Medicine

(c) Penicillin is only effective on certain bacteria. For example, it cannot be used to treat tuberculosis. 5 Streptomycin is the antibiotic that is effective in treating tuberculosis.

1 Antibiotics are chemicals that destroy or prevent the growth of infectious microorganisms. 2 Two examples of antibiotics are penicillin and streptomycin. 3 Antibiotics are used to treat diseases caused by bacteria and are not effective against diseases caused by viral infections such as influenza, measles or small pox. 4 Penicillin (a) Penicillin is derived from the mould Penicillium chrysogenum. (b) Penicillin is used to treat diseases, caused by bacteria, such as pneumonia, gonorrhoea and syphilis.

Psychotherapeutic Medicine 1 Psychotherapeutic medicine are a group of drugs that change the emotions and behaviour of the patient and are used for treating mental or emotional illnesses. 2 Table 5.7 shows the types and functions of psychotherapeutic drugs.

Type of psychotherapeutic drug

Functions

Examples

Comments

Caffeine

• It is found in coffee, tea and Cola drinks • It is a weak, naturally occurring stimulant

Amphetamine

• A strong synthetic stimulant • It increases the heart and respiration rates as well as blood pressure

• To alleviate depression • To relieve anxiety or tension • To make a person feel calm and sleepy

• Tranquiliser • Barbiturate

• They are substances that depress the central nervous system and cause drowsiness

• To treat mental illness such as schizophrenia (madness)

• Lithium carbonate (Li2CO3) • Chloropromazine • Haloperidol

• Mental (psychotic) patients have extreme mood swings. Their mood changes rapidly from high spirits to deep depression • Antipsychotic medicine do not cure mental illness but it can reduce some of the symptoms to help the person live a more normal life

Stimulants These are drugs that stimulate (excite) the activity of the brain and the central nervous system

• To maintain or increase alertness • To counteract normal fatigue • To elevate mood

Antidepressants These are drugs that increase the brain’s level of neurotransmitters and thus improve mood Antipsychotic medicine

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Chemicals for Consumers

5

Table 5.7 Types and functions of psychotherapeutic drugs

4

Side Effects of Traditional Medicine

’07

1 It is generally believed that traditional medicine have little side effects compared to modern medicine. 2 However, taking high doses of quinine for a prolonged period may cause hearing loss. German health officials reported 40 cases of liver damage which were linked to the herbal medicine containing kava-kava. 3 While the use of traditional medicine is rising globally, health experts have insufficient data about how it affects patients. 4 The World Health Organisation (WHO) hopes to set up a global monitoring system to monitor the adverse side effects of traditional medicine.

Which of the following can relieve joint and muscle aches? I Paracetamol II Barbiturate III Aspirin IV Streptomycin A I and II only C II and IV only B I and III only D III and IV only Comments Analgesics such as aspirin, paracetamol and codeine are used to relieved pain. Answer B

5

Side Effects of Modern Medicine 1 Table 5.8 shows the side effects of modern medicine. Table 5.8 Side effects of modern medicine

Type of modern drug Aspirin Amphetamine

Codeine Penicillin Streptomycin Stimulant Antidepressant Antipsychotic drug

Side effects • Can cause bleeding in the stomach because aspirin is very acidic • Can cause allergic reactions, skin rashes and asthmatic attacks • People who abuse amphetamines are excitable and talkative • Psychologically addictive and can cause heart attack • Can cause anxiety, sleeplessness, aggressive behaviour and decrease appetite • Can cause enlarged pupils, heavy perspiration and trembling hands • Can cause addiction • Can cause allergic reactions • Can cause death to people who are allergic to it • Can cause nausea, vomiting, dizziness, rashes and fever • Can cause loss of hearing following long-term use • Can cause addiction • Can cause addiction • Can cause headaches, grogginess and loss of appetite • Can cause dry mouth, blurred vision, urinary retention, constipation • Can cause tremor and restlessness • Sedation (make people calmer or make people sleepy)

2 Correct ways of using medicine In taking any medicine, we should know why the medicine is prescribed, how the medicine should be used, what special precautions should be followed, what special diet should be followed, what are the side effects and what storage conditions are needed. In addition, we should note the following points: (a) Self-medication Do not prescribe medicine for yourself (self-medication) or for other people. Chemicals for Consumers

Discuss with your doctor and listen to him concerning the medicine to be taken. (b) Follow the instructions given Follow the instructions given by your doctor or pharmacist concerning the dosage and method of taking the medicine. Dosage: Never take larger or more frequent doses and do not take the drug for longer than directed. Method of taking: Different drugs are taken in different ways. For example, medicine 518

that are acidic (such as aspirin) must be taken after food. If taken on an empty stomach, the medicine will damage the stomach lining. Conversely, antacid tablets are usually taken before food. (c) Medicine for adults and children Medicine for adults should not be given to children and vice versa. (d) Use of antibiotics Complete the whole course of antibiotic treatment given by your doctor. Do not stop taking the antibiotic just because you are already feeling well. If not, the remaining bacteria will mutate and develop resistance to the antibiotic and render it ineffective in the future. (e) Side effects Visit the doctor immediately if there are symptoms of allergy or other side effects of the drugs. (f) Expiry date Like food, medicine also have expiry dates. Do not take medicine after their expiry dates.

5.3 1 (a) State the difference in the effect of using aspirin and paracetamol as medicine. (b) Give a simple chemical test that can be used to distinguish aspirin from paracetamol. 2 State two similarities and one difference between penicillin and streptomycin. 3 State (a) the use of aspirin, (b) the side effects of aspirin.

Medicine

Addiction to certain medicine causes psychological dependence, physical dependence and tolerance. • Psy­­­chological dependence: an uncontrollable desire for the medicine. • Physical dependence: acute withdrawal symptoms such as convulsions. • Tolerance for the medicine is shown by the increasing dosage required to produce the same effect.

5

• Types and functions of medicine • Side effects of modern and traditional medicines • Correct usage of modern and traditional medicines

Modern medicine

Analgesics • Aspirin • Paracetamol • Codeine

Antibiotics • Penicillin • Streptomycin

Psychotherapeutic drugs • Stimulants • Antidepressants • Antipsychotics

’05

Which of the following statements about aspirin are correct? I It is an analgesic. II It destroys bacteria. III It can relieve headache. IV It calms down the emotions of the patient. A I and III only C I, II and III only B II and IV only D II, III and IV only

5.4

Appreciating the Existence of Chemicals

1 Since the last 100 years, thousands of new chemicals are synthesised. These new chemicals include synthetic polymers, composite materials, antibiotics, detergents and modern medicine. These chemicals improve the quality of life. 2 Modern chemical substances have brought enormous benefits to mankind. However, the chemicals have side effects on life and

Comments Aspirin is an analgesic and is used to relieve pain. Answer A

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Chemicals for Consumers

5

Traditional medicine • Derived from plants and animals

5

the environment. We must practise proper management of chemicals towards a better life, hygiene and health. 3 Modern living depends on chemical substances. Try to imagine the world without chemicals. For example, without petroleum, the transport system will break down; without modern medicine, diseases will spread and without soap and detergent, the world will become more dirty and less hygienic.

In recent years, food irradiation is used in food preservation. It is well known that gamma rays can kill insects, parasitic worms (such as trichinae in pork) and bacteria such as Salmonella sp. and E. coli (in beef) . In food irradiation, the food is exposed to highenergy gamma radiation. Irradiation can extend the shelf lives of food for weeks or even months.

1 Soaps are the sodium or potassium salts of long chain fatty acids. 2 Detergents are the sodium salts of alkyl hydrogen sulphate. 3 Soap is prepared by the saponification (alkaline hydrolysis) of vegetable oils or animal fats.

11 Food additives that are commonly used are preservatives, antioxidants, flavouring agents, stabilisers, thickening agents and dyes. 12 Preservatives (sodium nitrite and sodium benzoate) are used to retard the growth of bacteria, mould or fungus. 13 Antioxidants (ascorbic acid and sodium citrate) are used to prevent the oxidation of fats and oils by air. 14 Flavouring agents such as monosodium glutamate (MSG) enhance the taste of food. 15 Stabilisers and thickening agents (gelatin and acacia gum) improve the texture and the blending of food. 16 Dyes (azo compound and triphenyl compound) are colouring agents added to food to give them colour so as to improve their appearance. 17 Traditional medicine are medicine derived from natural sources such as plants and animals without being processed chemically. 18 Garlic is used for reducing high blood pressure and ginger is used to remove wind in the stomach. 19 Modern medicine can be classified as follows: • Analgesics • Psychotherapeutic drugs • Antibiotics 20 Analgesics are used to relief pain. Examples of analgesics are aspirin, paracetamol and codeine. 21 Antibiotics are used to treat diseases caused by bacteria. Examples of antibiotics are penicillin and streptomycin. 22 Psychotherapeutic medicine are drugs used for treating mental or emotional illness. 23 Examples of psychotherapeutic medicine: • Stimulants (for example, amphetamine) • Antidepressants (for example, barbiturate) • Antipsychotic medicine – to treat mental illness 24 Most medicine have side effects. For example, • aspirin can cause bleeding in the stomach, • amphetamine, codeine, stimulants and antide­ pressants can cause addiction.

boil

Vegetable oil + NaOH(aq) ⎯→ glycerol + soap

4 The structure of a soap ion or a detergent ion consists of a hydrophobic part and a hydrophilic part. The hydrophobic part is the long chain hydrocarbon which is soluble in grease but insoluble in water. 5 For a soap ion, the hydrophilic part is –COO –. For the detergent ion, the hydrophilic part is –OSO3–. The hydrophilic part is negatively-charged and is soluble in water but insoluble in grease. 6 The following additives are added to detergents to increase their cleaning power. • Builders • Whitening agents • Biological enzymes • Brighteners 7 Builders – to remove Mg2+ and Ca2+ ions in hard water Whitening agents – to remove coloured stains Biological enzymes – to remove protein stains Brighteners – make fabrics appear whiter and brighter 8 Soaps do not cause water pollution because they are biodegradable. However, they are not suitable for use in hard water (or acidic water) because they form scum and their cleansing action is reduced. 9 Detergents cause water pollution because they are non-biodegradable. However, they are suitable for use in hard water (or acidic water) because they do not form scum. 10 Food additives are chemicals that are added to food in small quantities for specific purposes.

Chemicals for Consumers

520

5 Multiple-choice Questions Soaps and Detergents

1 Which of the following is produced when a mixture of palm oil and concentrated sodium hydroxide solution (in excess) is boiled? A Fatty acids only B Salt of the fatty acids only C A mixture of glycerol and fatty acids D A mixture of glycerol and salt of the fatty acids 2 Which of the following is an ester? A Palm oil B Soap C Sodium palmitate D Glycerol 3 Which of the following is correctly matched with the manufacturing process? Chemical

Manufacturing process

A

Soap

Neutralisation of vegetable oils

B

Soap

Alkaline hydrolysis of vegetable oils

C

Detergent Neutralisation of long chain alcohol

D Margarine Oxidation of unsaturated fats 4 In the soap industry, sodium chloride is used to A prevent the formation of scum. B decrease the surface tension of water. C produce a softer soap. D decrease the solubility of soap in soap solution. 5 The following equation shows a chemical equation. ’09 Fat + NaOH → Glycerol + sodium salt

Which statement about the reaction is correct? A It is a saponification reaction. B It is an esterification reaction. C It is used to make detergent. D The product glycerol is a cleaning agent.

’03

Which of the following statements about the carbon compound is true? A It is a detergent ion. B Part X and part Y are both soluble in water. C Part X and part Y are both soluble in grease. D Part X is soluble in grease and part Y is soluble in water.

6 Which of the following ions reacts with soap solution to form scum? A Na+ C Cl– 2+ B Ca D SO42– 7 Which of the following chemicals is not needed in the preparation of the detergent, sodium alkyl sulphate? A Sodium hydroxide B Concentrated sulphuric acid C Alkylbenzene sulphonic acid D Alcohol with large molecular size 8 Which of the following statements about detergent is true? A Detergents are covalent compounds. B Detergents are made from petroleum. C Detergents contain the –COO– group in its molecules. D Detergents are less effective than soap in hard water. 9 The hydrophilic group in the detergent, sodium dodecyl (lauryl) sulphate is A –OSO3– B –OSO42– C –COO – D CH3(CH2)10CH2– 10 Which of the following substances acts as a whitening agent? A Sodium borate B Sodium perborate C Sodium tripolyphosphate D Biological enzyme 11 The diagram shows the structure of a carbon compound.

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12 Soaps and detergents A are salts of fatty acids. B contain the same functional group. C form scum with strongly acidic solution. D contain the hydrophobic part that is soluble in oil and grease. 13 Which of the following statements are true regarding soaps? I Soaps are biodegradable. II Soaps lather readily in hard water. III Soaps are ineffective in acidic water. IV Soaps are prepared from sulphonic acid. A I and III only B II and IV only C I, III and IV only D I, II, III, and IV 14 A sodium compound has the following formula: ’08

O  R – O – S – O –Na+  O



R is a long hydrocarbon chain. Which of the following statements are true regarding this compound? I The compound is used as a detergent. Chemicals for Consumers

5

5.1



II The molecule has a part that is insoluble in water. III The compound is an example of sodium alkyl sulphate. IV The compound can break large droplets of grease into smaller droplets of grease. A I and III only B II and IV only C I, II and III only D I, II, III and IV

5

15 Amylase and lipase are additives added to detergent. What is the function of these additives? A They act as drying agents. B They act as bleaching agents C They remove protein stains. D They make clothes look brighter. 16 Which of the following molecules act as cleansing agents? I CH3(CH2)3COONa II CH3(CH2)16CH3 III CH3(CH2)16OSO3Na IV CH3(CH2)15COONa A I and II only B III and IV only C II, III and IV only D I, II, III and IV 17 A section of the anion of a cleaning agent is shown below. ’04

CH3(CH2)14COO – Which of the following is true about the ion? A It is glycerol. B It is a detergent ion. C It is palmitate ion. D It is stearate ion.

18 When a concentrated sodium chloride solution is added to solution Q, a white precipitate is formed. Solution Q most likely contains I sodium stearate II sodium palmitate III lead(II) nitrate IV methyl ethanoate A I and II only B III and IV only C I, II, and III only D I, II, and IV only

Chemicals for Consumers

19 Which of the following is true about soap or detergent? ’06 A Soap forms scum in soft water. B Detergent forms scum in hard water. C Scum decreases the effectiveness of soap as a cleansing agent. D The presence of sulphate ion in detergent forms scum.

5.3

Medicine

24 A student has a headache. Which of the following medicine ’05 will relieve his headache? A Streptomycin C Aspirin B Amphetamine D Insulin 25 Which of the following modern medicine is matched correctly with its function? ’09

Medicine

5.2

Uses of Food Additives

20 Which of the following are the functions of food additives? ’09 I To enhance the flavour of food II To increase the rate of metabolism III To decrease the rate of oxidation of food IV To make food more digestible A I and II only B I and III only C II and IV only D I, III and IV only 21 Which of the following food additives is used to stop the activity of microorganisms in food? A Ascorbic acid B Benzoic acid C Citric acid D Triphenyl compounds 22 Which of the following food additives is used to improve the texture of ice cream? A Pectin B Citric acid C Sodium nitrate D Aspartame 23 Which food additive is matched correctly with its function? ’11

Food additive

Function

A

Thickener

Tartrazine

B

Flavouring

Acasia gum

C

Antioxidant

Ascorbic acid

D

Preservative

Aspartame

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Function

A

Aspirin

Antibiotic

B

Insulin

Antibiotic

C

Paracetamol

Analgesic

D

Streptomycin

Analgesic

26 Which of the following belongs to the class of medicine called psychotherapeutic drugs? A Codeine C Penicillin B Aspirin D Stimulant 27 A patient is experiencing depression and has difficulty in sleeping. Which of the following medicine is suitable for treating this patient? A Barbiturate B Codeine C Lithium carbonate D Paracetamol 28 Which of the following medicine have antibacterial properties? I Barbiturate II Penicillin III Paracetamol IV Streptomycin A I and III only B II and IV only C III and IV only D I, II and III only 29 Which of the following modern medicine can cause addiction? A Streptomycin C Codeine B Paracetamol D Aspirin 30 A patient complained of pain in the stomach after taking aspirin. ’08 Which substance will help to relieve the pain? A Ammonia B Ethanoic acid C Sodium chloride D Magnesium hydroxide

Structured Questions 1 The following equation shows the reaction between palm oil and concentrated potassium hydroxide solution. ’05 boil Palm oil + potassium hydroxide ⎯⎯→ potassium palmitate + glycerol (a) What is the name of this reaction? (b) (i) Which of the chemicals in the above equation is a soap? (ii) Name the homologous series to which palm oil belongs. (iii) Complete the anion part of the palmitate ion. CH2 CH2 CH2 CH2 CH2 CH2 CH2

[1 mark] [1 mark] [1 mark] [1 mark]

CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2

5

(c) Diagram 1 shows part of the washing action of a detergent particle, sodium alkyl sulphate on a cloth stained with grease.

Diagram 1



(i) Write the anion part of sodium alkyl sulphate. (ii) State the part of sodium alkyl sulphate that is soluble in grease.



(iii) Identify the particles represented by + and



[1 mark] [1 mark]

– in Diagram 1.

[2 marks]

(iv) Based on Diagram 1, explain the cleansing action of detergents.

[3 marks]

(d) (i) State what happens to the grease particles on further agitation. (ii) Complete Diagram 2 to illustrate your statement in (d)(i).

[1 mark] [1 mark]

Diagram 2

2 A student carries out four experiments to study the effectiveness of two cleansing agents, X and Y, on an oil-stained cloth in hard and soft water. Table I shows the arrangement of apparatus, the type of water used and the observations for Experiments I, II, III and IV. Experiment

I

II

III

IV

Oily stains disappear

Oily stains disappear

Oily stains remain

Oily stains disappear

Arrangement of apparatus

Observation

Table 1

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Chemicals for Consumers

(a) Based on the observation shown above, deduce which of the cleansing agents, X or Y, is (i) a detergent, (ii) a soap? [2 marks]

(iii) How is ginger used to treat this illness? 4 (a) State three reasons why food additives are used in the food industry. [3 marks]

(b) Describe the observation (if any) when solutions X and Y are separately shaken with hard water.

(b) Pineapple jelly contains gelatin. What is the role of gelatin in the jelly?

[2 marks]

[1 mark]

(c) Modern medicines are classified based on their effects on the human body. Explain the terms analgesics and psychotherapeutic drugs. [2 marks]

(c) Predict the observation if hard water in Experiment III is replaced by unpolluted river water. [1 mark] (d) What inference can be drawn regarding the effectiveness of X and Y as cleansing agents? [2 marks]

(d) Complete the table to show the type of medicine to which the following medicines belong to.

(e) State the variables involved in this experiment. (i) Manipulated variable (ii) Responding variable (iii) Constant variable [3 marks]

Name of medicine

Type of medicine

Aspirin Barbiturate

3 (a) Toothpaste contains the following chemicals:

Amphetamine

Sodium lauryl sulphate: CH3(CH2)11OSO3Na Calcium pyrophosphate: Ca2P2O7 Magnesium hydroxide: Mg(OH)2 Silicon(IV) oxide: SiO2 Mint oil Cellulose gum Saccharin

5

[1 mark]



Codeine

[4 marks]

(e) State one side effect of taking (i) aspirin, (ii) barbiturate for a prolonged period. [2 marks] 5 Aspirin has the following structural formula:

(i) What is the purpose of using sodium lauryl sulphate, mint oil and saccharin in toothpaste? [3 marks] (ii) In the food industry, cellulose gum is used as a thickening agent. What is meant by thickening agent? [1 mark] (iii) Suggest a reason why magnesium hydroxide is used in toothpaste. [1 mark] (iv) State the names of chemicals needed for the preparation of sodium lauryl sulphate. [2 marks] (v) Write the equations for the reactions. [2 marks]

’04

(a) What is the molecular formula of aspirin? [1 mark]

(b) Ginger can be used as a traditional medicine. Diagram 3 shows a ginger plant.

(b) Write the formulae of the ions produced when aspirin is added to water. [1 mark]

’06

(c) (i) State two observations for the reaction that occurs when aspirin is added to sodium carbonate solution. [2 marks] (ii) Name one important application of aspirin. [1 mark] (d) State one side effect of aspirin.

[1 mark]

(e) What is the correct way of taking aspirin? [1 mark] (f) Give the name of a medicine that can be used as a substitute for aspirin. [1 mark]

Diagram 3

(i) What illness can be cured by using (ii) Which of the parts W, X, Y or Z, is treat the illness in (i)? Chemicals for Consumers

(g) An aspirin pill contains 0.32 g of aspirin. Calculate the volume of 0.050 mol dm–3 of sodium carbonate needed to neutralise the aspirin pill. [3 marks] (Relative atomic mass: H, 1; C, 12; O, 16)

ginger? [1 mark]

used to [1 mark]

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Essay Questions 1 (a) Describe one experiment to prepare a sample of soap using coconut oil in the school laboratory. [8 marks]

(d) State three wrong methods of taking medicine and the problems that may arise. [6 marks] 3 (a) What is meant by food additives?

(b) Describe four types of additives in liquid detergents. Give one example of the additives and state its function. [7 marks]

’08

(c) Describe the differences between soaps and detergents in terms of their effectiveness as cleansing agents and their effects on the environment. [5 marks] 2 (a) What is meant by (i) medicine and (ii) traditional medicine? [3 marks]

[3 marks]

(b) A label on a box of ice-cream states that it contains the following food additives: • Azo compounds • Citric acid • Benzyl ethanoate • Sodium benzoate • Sucrose State the function of each of the food additives used. [12 marks]

(b) Give two examples of traditional medicine derived from plants and state their functions. [2 marks]

(c) What are the advantages and disadvantages of using food additives. [5 marks]

5

(c) Describe the types of modern medicine and their examples. State the function of each type of modern medicine. [9 marks]

Experiment 1 You are asked to devise an experiment to compare the effectiveness of the cleansing action of soaps and detergents in soft water and hard water. Your explanation must contain the following: (a) Aim of experiment (b) All the variables (c) Statement of the hypothesis (d) Lists of materials and apparatus (e) Procedure of the experiment (f) Tabulation of data [17 marks]

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Chemicals for Consumers

PAPER 1

Time: 1 hour 15 minutes

(50 marks)

Instructions: Question 1 to Question 50 are followed by four options A, B, C and D. Choose the best option for each question. 1 The chemical used to coagulate latex into solid rubber is A ammonia B methanoic acid C sodium chloride D calcium carbonate 2 The element 147 X I has 5 valence electrons. II is located in Period 3 of the Periodic Table. III forms a negatively charged ion of charge –3. IV has 7 protons and 14 neutrons. A I and III only B II and IV only C I, II and III only D I, III and IV only 3 The electronic configuration of the ion X2+ is 2.8.18.18.8. The ion X2+ has 81 neutrons. Determine the nucleon number of the atom of element X. A 133 C 136 B 135 D 137 4 The diagram shows the structural formula of the compound acrolein. H2C == CH — C == O | H Which statements below are true concerning acrolein? [Relative atomic mass: H, 1; C, 12; O, 16; Avogadro number = 6 3 1023 mol-1] I Acrolein is soluble in ethanol. II Acrolein has a low melting point. III Acrolein can undergo addition polymerisation.

I V 1.12 gram of acrolein contains 1.2 3 1023 molecules. A I and III only B II and III only C I, II and III only D I, II, III and IV 5 Graphs I and II show the heating of two samples of naphthalene. graph I X

Y

time (s) temperature (°C) graph II 80

P

Q

time (s)

Why is the region XY of graph I shorter than region PQ of graph II? A Sample II used is purer than sample I. B Sample I is heated earlier than sample II. C Sample I is heated more strongly than sample II. D Sample I is heated using a water bath whereas sample II is heated directly. 6 Which compound in the table below is correctly matched with its relative formula mass? [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Na, 23; P, 31; S, 32; Ca, 40; Fe, 56; Cu, 64]

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Relative formula mass

I CH3C6H2(NO2)3

227

II HOOCC6H4COOCH3

180

III FeSO4.7H2O

278

IV Cu(NH3)4SO4

228

A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV

temperature (°C)

80

Compound

7 Which of the elements below are inert? I: 2W III: 8Y II: 4 X IV: 10 Z A I and III only B I and IV only C II and III only D II and IV only 8 Brilliant blue is used as a food colouring. Its molecular formula is C47H50N3O7S2 . Calculate the number of moles of brilliant blue in 2.08 gram of the compound. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; S, 32] A 0.0025 mol B 0.0030 mol C 0.0050 mol D 0.0060 mol 9 Which of the following hydrocarbons burns with the most soot? [Relative atomic mass: H, 1; C, 12] A C2H2 C C7H14 B C5H12 D C9H20 10 The table shows the proton numbers of four elements P, Q, R and S.

Element

Number Number of of protons neutrons

P

3

Q

8

8

R

9

10

S

11

12

4

Which of the following pairs of elements react to form a compound with the relative formula mass of 54? A P and Q B P and R C R and S D Q and R 11 The chemical that can be used to differentiate dilute sulphuric acid and dilute hydrochloric acid solutions is I sodium carbonate solution II silver nitrate solution III lead(II) nitrate solution IV barium nitrate solution A I and III only B II and IV only C I, II and IV only D II, III and IV only 12 20.00 cm3 of sulphuric acid neutralises 25.0 cm3 of 2 mol dm–3 sodium hydroxide solution. Calculate the concentration of the sulphuric acid solution in mol dm–3. A 0.40 mol dm–3 B 0.80 mol dm–3 C 1.25 mol dm–3 D 2.50 mol dm–3 13 As we go down Group 1, A reactivity decreases B electropositivity decreases C melting point decreases D density decreases 14 Which of the following statements is true about halogens? A Chlorine is more reactive than fluorine. B Iodine reacts with iron to produce iron(III) iodide. C Bromine exists as gas at room temperature and pressure. D Bromine is more electronegative than chlorine.

15 Which of the following metallic oxides is amphoteric? A Aluminium oxide, tin(II) oxide and lead(II) oxide B Aluminium oxide, lead(II) oxide and silicon(IV) oxide C Magnesium oxide, tin(II) oxide and lead(II) oxide D Magnesium oxide, aluminium oxide and tin(II) oxide 16 Concentrated sodium chloride solution is electrolysed using carbon electrodes. At the anode, gas X is produced first followed by gas Y. What is gas X and gas Y? Gas X

Gas Y

A

Oxygen

Chlorine

B

Oxygen

Hydrogen

C

Chlorine

Hydrogen

D

Chlorine

Oxygen

17 Potassium iodide solution of concentration 1 mol dm–3 is electrolysed using carbon electrodes. Which of the following halfequations represent the reactions occurring at both electrodes?

19 Streptomycin is an antibiotic. Its molecular formula is C21H39N7O12. Calculate the number of streptomycin molecules contained in 1.162 gram of the compound. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Avogadro number = 6 3 1023 mol–1] A 1.2 3 1021 C 1.5 3 1022 B 1.5 3 1021 D 1.5 3 1023 20 Magnesium sulphate salt is used as laxative. 10 g of magnesium oxide is reacted with excess dilute sulphuric acid solution. Calculate the maximum mass of magnesium sulphate salt formed. [Relative atomic mass: O, 16; Mg, 24; S, 32] A 12 g C 24 g B 28 g D 30 g 21 X and Y form a covalent compound with the formula YX3. Which of the following is the likely proton numbers of X and Y? Proton number of atom X

Proton number of atom Y

Cathode

A

13

9

4OH– → 2H2O + O2 + 4e–

2H+ + 2e– → H2

B

17

13

C

15

17

B

2I– → I2 + 2e–

2H+ + 2e– → H2

D

17

15

C

2H+ + 2e– → H2

2I– → I2 + 2e–

D

2H+ + 2e– → H2

4OH– → 2H2O + O2 + 4e–

Anode A

18 Calculate the volume of oxygen gas needed for complete combustion of 3.2 g of methane at room temperature and pressure. [Relative atomic mass: H, 1; C, 12; 1 mol of gas occupies 24 dm3 at r.t.p.] A 2.4 dm3 B 4.8 dm3 C 9.6 dm3 D 12.0 dm3

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22 The diagram below shows a chemical cell. A

zinc

iron

sodium chloride(aq)

How can the voltage of the cell be increased? I Increase the concentration of sodium chloride solution II Replace the iron electrode with copper electrode III Replace the sodium chloride solution with iron(II) sulphate solution

time volume of CO2 (cm3)

IV Replace the zinc electrode with magnesium electrode A I and III only B II and IV only C I, II and IV only D II, III and IV only 23 Which of the following is a composite material? A Fibreglass B Borosilicate glass C Lead glass D Soda glass

II

C

I Sodium iodide solution II Potassium sulphate solution I III Hydrochloric acid solution IV Ammonia solution A I, II and III only B I, II and IV only C II, III and IV only D I, II, III and IV volume of CO2 (cm3)

I

II

II

D

volume of CO2 (cm3)

I

time

I

29 The diagram shows some 50 cents coins.

volume of CO2 (cm3) volume of CO2 (cm3)

time time

volume of CO2 (cm3)

II

I

II time 31 The diagram below shows a graph obtained when hydrogen peroxide decomposed to oxygen when manganese(IV) oxide is added into the solution.

time

24 The bite of fire ants contains the compound with the molecular formula HCOOH. Which of the following chemicals can be used to counter the effect of this chemical? A Ethanoic acid B Ethyl ethanoate C Sodium bicarbonate D Ethanol 25 10.0 gram of sodium hydroxide is dissolved in V cm3 of distilled water. The solution obtained has a concentration of 0.5 mol dm–3. Calculate the value of V. [Relative atomic mass: H, 1; O, 16; Na, 23] A 200 cm3 C 500 cm3 B 250 cm3 D 1000 cm3 26 Which of the solutions below has the lowest pH value? A 25 cm3 of 1.0 mol dm–3 sulphuric acid B 25 cm3 of 1.5 mol dm–3 nitric acid C 25 cm3 of 2.0 mol dm–3 ethanoic acid D 45 cm3 of 1.0 mol dm–3 hydrochloric acid



The composition of elements in the coins are A 10% tin and 90% copper B 30% zinc and 70% copper C 50% tin and 50% lead D 25% nickel and 75% copper

30 The table shows the volumes and concentrations of hydrochloric acid added to 10 g of marble (in excess) in experiments I and II.

I 27 nCHCH3CH2 → -(CHCH3CH2)n The reaction I represented by the equation above is A cracking B polymerisation volume II C addition reaction of CO2 (cm3) D reduction reaction

time

B

volume ofvolume of (cm3) CO 2 3 CO 2 (cm )

I

I

I

volume of CO2 (cm3)

I

90

150 time (s)

From the graph we can conclude that I the rate of decomposition of hydrogen peroxide decreases with time. Experiment Volume of Concentration II the average rate of hydrochloric of hydrochloric decomposition of hydrogen acid (cm3) acid V (mol dm–3) peroxide is —— cm3 s—1. 90 I 50 1.0 III the rate of reaction at time II 100 0.5 150 second is zero. IV all the hydrogen peroxide Which of the following graphs of had decomposed at 90 total volumes of carbon dioxide seconds. collected against time for both A I, II and III only experiments I and II is correct? B I, III and IV only A volume C II, III and IV only volume II I of of 3 3 II IV CO2 (cm ) CO2 (cm )D I, II, III and I

28 Which of the following reagents can be used to differentiate lead(II) nitrate and aluminium nitrate solutions?

volume of oxygen gas V (cm3)

CO2 (cm )

II II

time

time time

volume 528 II of CO2 (cm3)

I

32 2SO2(g) + O2(g) → 2SO3(g) The rate of the above reaction above can be increased by A adding vanadium powder. time B cooling the reaction vessel. C increasing the pressure. volume D immediately removing the II Itrioxide formed. of sulphur 3 33 Which of the following compounds does sulphur have the highest oxidation state? A H2S C H2S2O7 time B SO2 D H2SO3

34 W, X, Y and Z are four metals. Consider the reactions below involving these metals: X + Z sulphate → Z + X sulphate W + Z sulphate → Z + W sulphate W + X sulphate → No reaction Y + Z sulphate → No reaction Arrange the metals W, X, Y and Z in decreasing order of electropositivity. A X, W, Z, Y C W, X, Z, Y B Y, Z, W, X D Y, Z, X, W

P

Alcohol

Alkane



38 The diagram below shows three iron nails wrapped with metals P, Q and R and then left in agar-agar solution added with a few drops of potassium hexacyanoferrate(III) for 1 day. agar-agar solution + a few drops of potassium hexacyanoferrate(III) I

35



III M x+ ion accepts electron during the reaction. IV M x+ ion releases electron during the reaction. A I and III only C II and III only B I and IV only D II and IV only

Alkene S

III

Chloroalkane

An alcohol can be converted to chloroalkane via three processes P, Q and R. Name the processes P, Q and R. P

II

Q

Q

R

A

Fermentation Hydrogenation Substitution

B

Fermentation Hydrolysis

C

Dehydration

Hydrogenation Hydrolysis

D

Dehydration

Hydrogenation Substitution

iron nail + P

iron nail + Q

The results of the experiment are tabulated below: Test tube

Observation

I

Large amount of dark blue precipitate formed

II

No dark blue precipitate formed

III

Small amount of dark blue precipitate formed

Substitution

36 The IUPAC name for the compound shown below is C2H5 CH3CHCH2CHCH3 C3H7

A 2,4-dimethyloctane B 3,5-dimethyloctane C 2-ethyl-4-methylheptane D 2-ethyl-4-propylpentane 37 Mx+ → My+ The element M has two oxidation numbers. The conversion of Mx+ to My+ is a reduction reaction if I the value of x is more than y. II the value of y is more than x.

iron nail + R

Arrange the metals P, Q and R in decreasing order of electropositivity. A P, R, Q C Q, R, P B P, Q, R D Q, P, R 39 The heat of combustion of an alkane X is 2020 kJ mol–1. When 11 g of the alkane X is burned, 505 kJ of heat is released. Determine the molecular formula of alkane X. [Relative atomic mass: H, 1; C, 12] A C2H6 C C4H10 B C3H8 D C5H12 40 The initial temperature of 100 cm3 dilute hydrochloric acid solution is x °C. When 1.3 g of zinc powder is added to the

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acid, the heat energy released increases the temperature of the acid to y oC. Determine the heat of reaction between zinc and hydrochloric acid. [Relative atomic mass: Zn, 65; Specific heat capacity of water = 4.2 J/g oC] 1.3 A ———— 3 100 3 4.2 3 65 (y – x) kJ/mol 1.3 100 B ———— 3 —————————— 3 4.2 3 65 1 000 (y – x) kJ/mol 65 C ———— 3 100 3 4.2 3 1.3 (y – x) kJ/mol 65 100 D ———— 3 —————————— 3 4.2 3 1.3 1 000 (y – x) kJ/mol

41 Which of the following statements is not true concerning soap? A Soap does not pollute river water because it is biodegradable. B Cleansing action of soap is not affected in acidic water but the cleansing action of soap is lower in hard water. C Soap is produced by alkaline hydrolysis of fat molecules. D Part of the soap molecule dissolves in greasy stain and a part of the molecule dissolves in water. 42 The diagram shows the content of a box of ice cream. Contents : Vanillin, Lecithin, Sugar, Full cream milk Manufactured date : Sept 2012 What is the function of vanillin and lecithin? Vanillin

Lecithin

A

Emulsifier

Colouring

B

Emulsifier

Flavouring

C

Flavouring

Colouring

D

Flavouring

Emulsifier

43 Which of the following medicine is correctly categorised? Analgesic

Antibiotic

A

Streptomycin

Colouring

Valium

B

Aspirin

Flavouring

Codeine

C

Paracetamol

Streptomycin

Ketamine

D

Aspirin

Insulin

Barbiturate Which of the following experiments will produce graph Q?

44 The table shows the proton numbers of four elements. Element

W

Proton number

Psychotherapeutic

9

X 11

Y

Z

17

45 When a mixture of bromine water, potassium iodide and carbon tetrachloride is shaken, I the iodide ions will be oxidised. II redox reaction will take place. III iodide ions act as reducing agent. IV the carbon tetrachloride layer will turn brown. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV 46 The diagram shows graph P obtained when 5 grams (in excess) of granulated zinc is added to 50 cm3 of 0.5 mol dm–3 hydrochloric acid at 30 oC. Q

Manipulated variable

I

5 grams of zinc powder added to 50 cm3 of 0.5 mol dm–3 HCl at 30 °C

II

5 grams of granulated zinc added to 25 cm3 of 1.0 mol dm–3 HCl at 30 °C

19

Which of the following pairs of elements reacts most vigorously? A W and Z B X and Y C X and Z D W and X

total volume of H2 (cm3)

Experiment

III

IV

5 grams of granulated zinc added to 50 cm3 of 1.0 mol dm–3 HCl at 30 °C

47 Consider the reaction: phosphoric potassium acid dichromate(VI)

C3H6 + steam ⎯⎯→ X ⎯⎯⎯→ Y What is the molecular formula of X and Y? X

P

time (s)

49 The diagram shows an apparatus set-up of an experiment used to prepare gas X. mineral wool aluminium soaked oxide in ethanol

5 grams of granulated zinc added to 50 cm3 of 0.5 mol dm–3 HCl at 40 °C

A I and III only B II and IV only C I, II and III only D I, III and IV only



48 Which of the statements below are true? I Vulcanised rubber is more elastic than natural rubber. II Vulcanised rubber is less stretchable than natural rubber. III Vulcanised rubber is harder than natural rubber IV The melting point of vulcanised rubber is higher than natural rubber. A I, II and III only B I, III and IV only C II, III and IV only D I, II, III and IV

Y

A

C3H7OH

C3H7COOH

B

C3H7OH

C2H5COOH

C

C3H6(OH)2

C3H7COOH

D

C3H6(OH)2

C2H5COOH

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heat

gas X

water

Which of the statements below is true about gas X? A Rekindle a glowing splint B Decolourises brown colour of bromine water C Produces a ‘pop’ sound when tested with a burning splint D Changes colour of acidified potassium dichromate from orange to green

acidified potassium dichromate(VI) butanol

50 C3H7OH ⎯⎯⎯→ A ⎯⎯→ B The flowchart above shows the conversion of C3H7OH to products A and B. What of the following is the molecular formula of product B? A C2H5COOC3H7 B C2H5COOC4H9 C C3H7COOC3H7 D C3H7COOC4H9



PAPER 2

Time: 2 hour 30 minutes

Section A (60 marks) Answer all questions in this section. The time suggested to complete this section is 90 minutes. 1 Table 1 shows a list of elements labelled P, Q, …… V. The letters are not the actual symbols of the elements. Element

7 3

P

12 6

Q

14 7

R

S

16 8

T

27 13

U

35 17

(b) Two drops of phenolphthalein is added into the sodium hydroxide solution.

phenolphthalein

V

40 20

Electronic configuration Table 1

nitric acid

(a) Write the electronic configuration of the elements in the spaces provided. [1 mark] (b) (i) Complete Table 2 below by writing the formula of the compound formed between a pair of elements given.

(ii) State whether the compound formed is ionic or covalent.



20

20

(iii) Determine the relative formula mass of the compound.

Pair of elements

Formula of compound

Ionic or covalent

Relative formula mass

(d) Diagram shows the final burette reading.

20

30

Q and S

(c) A burette is filled with nitric acid of concentration 2.0 mol dm–3. The initial burette reading is 0.00 cm3. The sodium hydroxide in the conical flask is then titrated with the nitric acid until the neutral point.

30

P and R T and U

30

R and U V and R Q and U

Diagram 1

T and S

(a) Name the apparatus P used in measuring 25.0 cm3 of the sodium hydroxide solution. [1 mark]

Table 2

[7 marks]

(c) Hydrogen (H) and R form a covalent compound. Draw the Lewis structure of this compound. [2 marks] 2 Diagram 1 shows the steps involved in preparing a salt.

P sodium hydroxide

(b) State the colour change of the phenolphthalein indicator at the neutral point. [1 mark] (c) (i) State the volume of nitric acid needed to neutralise the sodium hydroxide solution. [1 mark]

(a) 25.0 cm3 of sodium hydroxide solution of concentration 2.0 mol dm–3 is measured using apparatus P and then transferred into a conical flask.

(ii) The experiment is repeated by adding this volume of nitric acid in (c)(i) above to 25.0 cm3 sodium hydroxide solution of concentration 2.0 mol dm–3 without addition of phenolphthalein indicator. Why was the experiment repeated without the addition of phenolphthalein indicator? Write the chemical formula of this compound. [1 mark]

531

(d) (i) Write a balanced equation neutralisation reaction.

for

the

0.2 g of lithium was added into 200 cm3 of water in a conical flask at room temperature and pressure. The gas collected was recorded at 20-second intervals. The result was tabulated below:

[1 mark]



(ii) Calculate the maximum mass of salt formed. [Relative atomic mass: N, 14; O, 16; Na, 23] [2 marks]

(iii) State one use of this salt.

[1 mark]

(e) The salt formed remained in the solution. Describe how the salt is removed from the solution. [2 marks]

Time (s)

0

20

Gas syringe reading (cm3)

0

250 310 355 360 360

40

60

80

100

(a) Write a chemical equation for the reaction [1 mark] between lithium metal and water.

3 The flowchart below shows the steps involved in the production of ammonia in industry.

(b) Sketch a graph of volume of gas evolved against time. [3 marks]

Unreacted nitrogen and hydrogen

volume of gas (cm3)

Nitrogen Compression chamber

Catalyst chamber

Hydrogen

Ammonia

Liquid ammonia

(a) How are (i) nitrogen (ii) hydrogen obtained? [2 marks]

time (s)

(b) To what pressure is the mixture of nitrogen and hydrogen compressed? [1 mark]

(c) (i) State how the reaction rate changes with time. [1 mark] (ii) Explain your answer. [1 mark]

(c) Name the catalyst used in the catalyst chamber. [1 mark] (d) To what temperature is the catalyst heated? [1 mark]

(d) Calculate the number of moles of gas collected. [1 mol of gas occupies a volume of 24 dm3 at room temperature and pressure] [2 marks]

(e) After all these conditions are attained, the nitrogen and hydrogen react to form ammonia. Write a equation for the reaction. [1 mark] (f) How is the liquid ammonia removed from the reaction chamber? [1 mark] (g) State two uses of ammonia.

[2 marks]

4 Diagram 2 shows the apparatus set-up used to study the rate of reaction between lithium metal and water.

hydrogen gas

burette

conical flask lithium

(e) The experiment was repeated as stated below: Experiment II : 0.2 g of lithium was added to 200 cm3 of water at 40 °C. Experiment III : 0.1 g of lithium was added to 200 cm3 of water at 30 °C. Sketch the shape of the graphs for each of these experiments on the same axes as graph (b) above. [2 marks]

5 Diagram 3 shows the apparatus set-up to study the reactivity of metals P, Q, R and S with oxygen.

Diagram 2

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potassium chlorate, KCIO3

glass wool

(d) Identify the metal (i) P (ii) S [2 marks]

Metal P

(e) (i) Identify metal Q : heat

(ii) Name the product formed when metal Q was used in the reaction. [1 mark]



(iii) Write a chemical equation for this reaction.

heat

[1 mark]



(f) Suggest another experiment which can be carried out to determine the position of the metals in the reactivity series. [1 mark]

Diagram 3

The P metal powder was heated strongly first. Then the potassium chlorate was heated. On heating, the potassium chlorate decomposes to give oxygen gas. The reactivity of the reaction between the hot metal and oxygen gas was recorded. The experiment was repeated by replacing P metal powder with Q, R and S metal powders. The result of the experiment is shown in the table below: Results Metals

Observation

6 Table 3 below shows four categories of medicine.

Cold

P

Burns brightly

Yellow

White

Q

Glows dimly

Black

Black

R

Burns very brightly

White

White

S

Glows brightly

Brown

Yellow

Category P Q R Hormone



Example Streptomycin Valium Aspirin Insulin Table 3

(a) Name the category of the following medicine (i) Streptomycin (iii) Aspirin (ii) Valium [2 marks]

Colour of product formed when… Hot

[1 mark]



(b) State the function of each category of medicine. (i) P (ii) Q (iii) R [3 marks] (c) (i) Name a disease that can be treated with streptomycin. [1 mark]

(a) What is the function of the glass wool in the experiment? [1 mark]

(ii) A patient must complete the dosage of streptomycin prescribed by the doctor. Explain why. [1 mark]

(d) (i) Aspirin cannot be prescribed to small children. Explain why. [2 marks]

(b) Write a balanced chemical equation for the decomposition of potassium chlorate when heated. [1 mark]



(c) Arrange the metals P, Q, R and S in decreasing order of reactivity. [2 marks]

(ii) Name an alternative medicine to aspirin for small children. [1 mark]

(e) Valium cannot be used for prolonged period. State a reason why. [1 mark]

Section B (20 marks) Answer any one question from this section. The time suggested to complete this section is 30 minutes.

7 (a) Table 4 shows the products formed at the carbon electrodes when dilute and concentrated sodium chloride are electrolysed. Electrolyte

Electrodes

[10 marks]

(b) Electroplating can be used to coat an iron chain with a thin coat of silver. Explain how you carry out the process. [7 marks]

Products at Anode

Cathode

Dilute sodium chloride solution

Carbon

Gas P

Gas Q

Concentrated sodium chloride solution

Carbon

Gas R

Gas S

8 (a) Starting with magnesium oxide, describe how you can prepare a sample of magnesium carbonate salt. [10 marks] (b) You are given four solutions each containing sulphate, nitrate, carbonate and chloride ions. Describe experiments on how you can identify the ions present in each solution. [10 marks]

Table 4



(ii) Using half-equations, explain how these products are formed at the electrodes.

(i) Name the gases P, Q, R and S released at the electrodes. [3 marks]

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Section C (20 marks) Answer any one question from this section. The time suggested to complete this section is 30 minutes. (c) Describe an experiment to determine the heat of precipitation of silver chloride. Your answer should include: (i) Procedure of experiment (ii) Tabulation of result (iii) Calculation [8 marks]

9 (a) With the aid of an energy level diagram, explain the meaning of (i) exothermic reaction (ii) endothermic reaction [6 marks] (b) Table 5 shows the heat of displacement Displacement reaction

Fe(s) + CuSO4(aq) → Cu(s) + FeSO4(aq) 2Al(s) + 3ZnSO4(aq) → 3Zn(s) + Al2(SO4)3(aq) Mg(s) + 2AgNO3(aq) → 2Ag(s) + Mg(NO3)2(aq)



10 (a) Using pentane as an example, define the term isomerism. [5 marks]

Heat of displacement (kJ mol–1)

(b) With an example, explain the meaning of substitution reaction. [6 marks] (c) X = C6H12 Y = C6H14 X and Y are two organic compounds. (i) Write the general formula of each compound. [2 marks] (ii) Describe an experiment on how you can differentiate both compounds. [5 marks] (iii) X can be converted to Y. Name the reaction of this conversion.

ΔH1 ΔH2 ΔH3

Table 5



Arrange the values of the heat of displacements of the three reactions above in increasing order. [2 marks] Explain your answer above. [4 marks]

Paper

3

[1 mark]



Write a chemical equation for the reaction. [1 mark]

Time: 1 hour 30 minutes

(50 marks)

Instructions: Answer all questions. below the water level. A thermometer is placed in the boiling tube.

1 Diagram 1 shows the apparatus set-up to determine the melting point of a substance X.

(d) The water bath is heated with a slow fire. thermometer

(e) After some time the stopwatch is started and the temperature of the substance X is recorded at 30-second intervals.

retort stand

boiling tube

(f) The diagram below shows the temperature readings of the substance at 30-second intervals.

beaker substance X water heat

80 80

80 80

80 80

70 70

70 70

70 70

tripod stand

Diagram 1

(a) 3 spatulas of substance X are added into a boiling tube. (b) A 500 ml beaker is filled with water until it is 3 about —— -full and placed on a tripod stand. 4 (c) The boiling tube containing substance X is then clamped in the water bath with the substance X

Time: 0 minute Temperature =

534

80 80

1 Time: —— minute 2 Temperature = 80 80

80 80

80 80

80 80

80

80

80

80 80

(a) 80Record the temperature80reading in the spaces provided. [3 marks]

80 80

80 80

80 80 (b) Construct a table to tabulate the results of the experiment above. [3 marks]

(c) Plot a graph of temperature against time. [3 marks] (d) Explain why a water bath is used to heat substance X. [3 marks] 70 70

70 70

70

70

70 70 Time: 1 minute Temperature =

70

70 70

70 the temperature from (e) 70What can you say about

1 70 70 time 1 minute to 2— minutes? 2 Explain your answer.

70 1 70 Time: 1—— minutes 2 Temperature =

(f) 90 (i) What is the melting90point of substance X? [3 marks]

80

80 80

80

80

80 80

80

80 80

80

80

70 70

70

70

70 70

70

70 70

Time: 2 minutes Temperature = 90 80

90 90

80

90

70

80 80

80 70

70 70

glass rod glass rod90 90

90

90 90

80

70

80

80

nitric acid nitric acid 80 80

80 80 70

Time: 3 minutes Temperature =

80

2 Diagram 2 below shows the apparatus used in an 80 80 experiment carried out in the school laboratory to study the property of oxide of elements in Period 3 of the Periodic Table.

70 70

90

70 70

90

70

(ii) If the melting point of substance X from another laboratory is determined and is 80 found to be lower than this value, what can you say about the sample? [3 marks]

magnesium oxide magnesium oxide glass rod glass rod

magnesium oxide magnesium oxide

80

80 80

70

70

80

1 Time: 2—— minutes 2 Temperature =

80 80 80



80

70

[3 marks]

after stirring after stirring

sodium hydroxide sodium hydroxide

MgO MgO

after stirring after stirring

Experiment I Observation :

1 Time: 3—— minutes 2 Temperature =

glass rod glass rod

90

silicon(IV) oxide silicon(IV) oxide

SiO2 SiO2 after sodium hydroxide stirring sodium hydroxide after stirring

80

Time: 4 minutes Temperature =

glass rod glass rod

sodium hydroxide sodium hydroxide

Experiment II Observation : aluminium oxide glass rod glass rod

535

aluminium oxide

silicon(IV) oxide silicon(IV) oxide

glass rod glass rod

after stirring after stirring

aluminium oxide aluminium oxide

stirring

hydroxide

aluminium oxide glass rod

stirring

(b) Record your observations in the spaces provided.

aluminium oxide



glass rod

[3 marks]

(c) What conclusions can you make from the result of the experiment above? [3 marks] (d) Name another oxide that has the same property as (i) magnesium oxide : (ii) silicon(IV) oxide : (iii) aluminium oxide : [3 marks]

nitric acid

after stirring

sodium hydroxide

Experiment III Observation :

after stirring

Diagram 2

The oxides of the elements are added into dilute nitric acid and dilute sodium hydroxide. The mixtures are stirred. The results are shown in Diagram 2. (a) State three variables that must be kept constant in the experiment. [3 marks]

3 “Cleansing power of soap is lower in hard water whereas the cleansing power of detergent is unaffected in hard water”. Describe an experiment to prove the above statement. Your planning should include the following: (a) Problem statement (b) Hypothesis (c) Variables of the experiment (d) Apparatus and chemicals used in experiment (e) Procedure of experiment (f) Tabulation of result [17 marks]

536

Answers 1

Introduction to Chemistry

Self Assess 1.1 1 Field

Chemicals used

Agriculture Fertilisers and pesticides Medicine Antibiotics and analgesic drugs Food preservatives and Food processing food flavourings Self Assess 1.2 1 (a) Table salt is soluble in water but not in kerosene. (b) (i) Manipulated variable: Solvents used, that is, water and kerosene (ii) Responding variable: Solubility of table salt (iii) Constant variable: Volume of solvent (c) Apparatus: Two test tubes, glass rod, 10 cm3 measuring cylinder, electronic balance, filter paper and spatula. Materials: Water, kerosene and table salt. (d) (i) 2 grams of table salt is weighed onto a filter paper. (ii) 5 cm3 of water is measured using a measuring cylinder and then poured into a test tube. (iii) The salt is added to the water and the mixture is stirred using a glass rod. (iv) The procedure above is repeated by adding 2 grams of salt to kerosene and stirred. (e) Solvent Water Solubility of salt Soluble

Structured Questions 1 (a) (i) Concentration of hydrochloric acid solutions (ii) Time taken for the 5 cm length of magnesium ribbon to dissolve (iii) The length of magnesium ribbon 3 (b) The higher the concentration of the hydrochloric acid solution, the shorter the time taken for the magnesium ribbon to dissolve. 1 (c) (i)

4

5 (ii) 0.425 mol dm–3 1 2 (a) Volume of water 1 (b) (i) Temperature of water 3 (c)

SPM Exam Practice 1 Multiple-choice Questions 1 B 2 D 3 C 4 A 6 B 7 C 8 D 9 A 11 A 12 C 13 A 14 B 16 C 17 A 18 B 19 B 21 D 22 C 23 A 24 D 26 D 27 A 28 D

5 A 10 D 15 C 20 B 25 B

(d) (i) The solubility of the salts increases with the increase in the temperature of the water. (ii) The solubility of salt Q is higher than that of salt P. 2 3 (a) The higher the temperature of the water, the greater the mass of sugar that can dissolve in it. 1 (b) (i) Temperature of water (ii) Mass of sugar that dissolves (iii) Volume of water 3

Temperature of Initial mass of beaker Final mass of beaker Mass of sugar dissolved (g) P and contents (g) P and contents (g) water (°C) 92.50 87.50 77.50 62.50 42.50

30 40 50 60 70

Kerosene Not soluble

Self Assess 1.3 1 (a) Add concentrated acid to the water. (b) Make sure the mouth of the test tube is not directed at any student. (c) Carry out the experiment in a fume cupboard.

(ii) Volume of water (iii) Mass of salt that dissolved in 100 cm3 of water 3 (c)

87.50 77.50 62.50 42.50 17.50

5 10 15 20 25 2

(d)

3 (e) 12.5 gram 1 4 (a) Food preservatives (i.e. potassium nitrate), antioxidant (i.e. ascorbic acid), food colouring (i.e.

537

tartrazine) and food enhancer (i.e. monosodium glutamate). 4 (b) Pharmacist, dentist, agriculturist, chemical engineer, biochemist and geologist. 6 (c) The government can collect tax from the export of these chemicals. Jobs are created by these industries. The country saves on foreign exchange if the chemicals needed are produced locally. 3 Answers

F O R M 4

CHAPTER 1

Form 4

(d) Cooperate with fellow classmates when carrying out an experiment. Be honest in recording the results obtained in the experiment. Do not waste chemicals and only take the required amount needed for the experiment. Always follow safety procedures while carrying out an experiment. Clean the apparatus after carrying out an experiment. 5 (e) Urea as nitrogenous fertiliser to enhance growth. Paraquat as a herbicide to kill the weeds which compete with the crop for nutrients from the ground. 2

F O R M 4

Essay Questions 1 (a) Apparatus Two test tubes, two rubber stoppers, test tube rack, sandpaper, 100 cm3 beaker, Bunsen burner, tripod stand and wire gauze. Materials Water, iron nails, cooking oil. 3 (b) (i) Water with air and water without air (ii) Rusting of iron nail (iii) Presence of water in both sets of experiments 3 (c) Procedure

(b) (i) Making an observation, making an inference, making a hypothesis, identifying variables, planning the procedure of the experiment, carrying out an experiment, collecting data, interpreting data and making a conclusion. 8 (ii) An inference is an initial opinion based on an observation. A hypothesis is a statement that relates a manipulated variable to a responding variable. An experiment has to be carried out to prove the validity of the hypothesis. 4 (iii) In graphical form, a pie chart or tabulation. 6 Experiments 1 (a) To determine the effect of the volume of water on the solubility of salt 3 (b) The greater the volume of water, the greater the amount of salt that dissolves. 3 (c) Manipulated variable Volume of water

Responding variable Mass of salt that dissolved Constant variable Temperature of water 3 (d) Apparatus 100 cm3 measuring cylinder, electronic balance, 100 cm3 beaker and spatula. Materials Water and common salt. 3 (e) Procedure (i) Initial mass of beaker A and salt is taken (x g). (ii) 50 cm3 of water is measured using a measuring cylinder and poured into a 100 cm3 beaker. Salt is added to 50 cm3 of water a little at a time while stirring the mixture until no more salt can further dissolve. (iii) The final mass of beaker A and salt is taken (y g). The mass of salt that dissolved is (x – y) gram. (iv) The experiment is repeated by dissolving the salt in 30 cm3, 40 cm3, 60 cm3 and 3 70 cm3 of water.

(f) Tabulation of data

CHAPTER 1 & 2



(i) Two iron nails are cleaned with sandpaper. (ii) A 100 cm3 beaker is filled with water. The water is heated until boiling to expel the air from the water. (iii) The boiled water is then poured into test tube A. (iv) Cooking oil is poured onto the water in test tube A to prevent air from dissolving in the water. (v) Tap water is poured into test tube B. (vi) An iron nail is then dropped into each of the tubes above and left for three days. 10 (d) Results Test tube

A

Observation Iron nail does after 3 days not rust

Volume of water (cm3)

Initial mass of beaker A and contents (g)

Final mass of beaker A and contents (g)

Mass of salt dissolved (g)

30 40 50 60 70

x

y

(x –y)

2 (a) Manipulated variable Concentration of NaOH solution Responding variable pH value Constant variable Volume of solution 3 (b) The higher the concentration of sodium hydroxide solution, the higher the pH value. 3 (c)

2

B Iron nail rusts

4 2 (a) Scientific method is a systematic approach taken when carrying out scientific research. 2 Answers

3 (d) (i) pH = 13.55 (ii) 0.25 mol dm–3 3 3 (a) (i) Mass of catalyst (ii) Time taken to collect 50 cm3 of oxygen gas (iii) Volume and concentration of hydrogen peroxide solution 3 (b) The greater the quantity of catalyst added, the shorter the time taken to collect 50 cm3 of oxygen gas. 3

3

538

The Structure of the Atom

Self Assess 2.1 1 (a) Molecules (b) lons (c) Atoms (d) lons (e) Atoms (f) Molecules

Self Assess 2.2 1 (a) Proton number = 92 Nucleon number = 92 + 143 = 235 Nuclear symbol is 235 U. 92 (b) It has 53 protons and 53 electrons. Nucleon number = proton number + number of neutrons 127 = 53 + number of neutrons Number of neutrons = 127 – 53 = 74 The 53 protons and 74 neutrons are located in the nucleus of the atom, whereas the electrons are arranged in shells surrounding the nucleus. 2 (a) B and E are atoms of the same element because they have the same proton number. (Note: Atoms of different elements have different proton numbers) (b) Al is the symbol for aluminium. It has 13 protons, 13 electrons and 14 neutrons (27 – 13 = 14). (Note: In a neutral atom, the number of protons is equal to the number of electrons) Self Assess 2.3 1 Isotopes are atoms with the same number of protons but different number of neutrons. Each has 92 protons. The number of neutrons in each isotope are 142, 143 and 146 respectively. (Note: The number of neutrons are obtained by deducting the proton number from the nucleon number) 2 127 B and 131 E are isotopes because 53 53 they have the same proton number but different nucleon numbers.

3 Carbon-14, 146C. Radioisotope is an isotope that is unstable and will decay by emitting radioactive rays. Carbon-14 is used in dating the age of archaeological artifacts. Self Assess 2.4 1

2 (a) (i) Five valence electrons (ii) 15 protons. (Number of protons is equal to the total number of electrons) (b) Nucleon number = Number of protons + Number of neutrons = 15 + 16 = 31 X (c) 31 15 SPM Exam Practice 2 Multiple-choice Questions 1 C 2 C 3 B 4 D 6 B 7 B 8 A 9 C 11 B 12 B 13 D 14 A 16 C 17 B 18 C 19 A 21 D 22 C 23 A 24 B 26 B 27 D 28 A 29 B 31 C 32 A 33 C 34 A 36 B 37 B 38 D 39 A

5 B 10 A 15 C 20 D 25 C 30 D 35 D 40 C

Structured Questions 1 (a) (i) Atom Proton number Nucleon number 12 6

C

6

12

14 6

C

6

14

2 12 (ii) C has six neutrons while 146C 6 has eight neutrons. 1 (b)

2 (c) In carbon dating: To determine the age of archaeological specimens. 1 (d) 4 valence electrons 1 2 (a) (i) Solid (ii) Mixture of solid and liquid (iii) Liquid (iv) Mixture of liquid and gas (v) Gas 3

539

(b) t3 minute 1 (c) 115 °C 1 (d) Heat is absorbed to overcome the forces of attraction between the solid particles. Heat absorbed is equal to the heat supplied in heating. 1 (e)

3 (a) V: 2.4 W: 2.6 Z: 2.8.6 (b) Four (c)

X: 2.6

2 Y: 2.8.1 2 1

F O R M

2

(d) Three electron-filled shells. 1 (e) Nucleon number is the sum of the number of protons and the number of neutrons. 1 (f) 23 – 11 = 12 neutrons. 1 (g) (i) Isotopes are atoms with the same proton number but different nucleon numbers. 2 (ii) W and X. 1 4 (a) Bromine and naphthalene. 1 (b) Iron 1 (c) (i) Iron 1 (ii)

1 (d) Sodium ions and chloride ions 1 (e) (i) T1 1 (ii) The heat is absorbed to overcome the forces of attraction between the naphthalene molecules. 2 (iii) The speed of movement of the particles increases during heating from R to S. 1 Answers

4

CHAPTER 2

2 (a) T2 (b) Mixture of solid and liquid naphthalene. (c) The heat is absorbed to overcome the forces of attraction between the naphthalene molecules. (d)

Essay Questions 1 (a)

F O R M 4



Solid

Liquid

Gas

Particle arrangement

Particles are closely packed.

The particles are still closely packed but there are more empty spaces between the particles.

The particles are very far apart.

2

Forces of attraction

The forces of attraction between the particles are strong.

The forces of attraction between the particles are weaker than in a solid but stronger than in a gas.

The forces of attraction between the particles are very weak.

2

Kinetic energy

The kinetic energy of the particles is low.

The kinetic energy of the particles is high on average.

The kinetic energy of the particles is very high.

2

Compressibility

Low

Low

High

2

(b) Apparatus Boiling tube, retort stand and clamp, tripod stand, Bunsen burner, wire gauze, thermometer (0 – 110 °C), 500 cm3 beaker, 250 cm3 conical flask, test tube holder and stopwatch. 1 Material X and Y, water 1 Procedure



(vi) The experiment is repeated by replacing substance X with substance Y. 6 Tabulation of results Substance X Time (s)

0 30 60 90 120 …

Temp (°C) Substance Y Time (s)

0 30 60 90 120 …

CHAPTER 2

Temp (°C)









Answers

(i) Three spatulas of X powder are put into a boiling tube. (ii) A 500 cm3 beaker is filled with water until it is about –34 full and placed on a tripod stand. (iii) The boiling tube containing X powder is then clamped in the water bath with the level of X powder below the water level. (iv) The water bath is heated until it reaches a temperature of about 55 °C. Then the water is heated with low flame. (v) The stopwatch is started and the temperature of the substance X is recorded every 30 seconds. The result is then tabulated.

1 A graph of temperature against time is plotted for both substances. The melting point of the substance is obtained from the region where the temperature remains constant at T. 1 If the melting point of the substance is 65 °C, then the substance is naphthol. If its melting point is 80 °C, then the substance is naphthalene. 1

1 2 (a) (i) Proton number is the number of protons in the nucleus of an atom. 1

540

(ii) Nucleon number is the total number of protons and neutrons in the nucleus of an atom. 1 (iii) Valence electrons are the electrons in the outermost electron shell. 1 (b) (i) Isotopes are atoms with the same proton number but different nucleon numbers. 3 16 O and 188O are isotopes of (ii) 8 oxygen. 2 (iii) Six uses of isotopes: 1 Medicine (a) Cobalt-60 is used to kill malignant cancer cells in radiotherapy. (b) Some medicine, surgical gloves, bandages, plastic hypodermic syringes which cannot be sterilised by boiling are sterilised using gamma radiation. 2 2 Agriculture (a) Radioactive 14CO2 is used to trace the path of carbon during photosynthesis. (b) Radioactive 32PO43– ions are used to determine the rate of absorption of phosphorus by the plant. 2 3 Industry (a) Beta radiation is used to control the thickness of paper, plastic, metals and rubber made in industry. (b) Radioisotope is used to detect leaks in pipes carrying gas. 2 4 Archaeology The activity of the carbon-14 still remaining in an archaeological specimen can be used to estimate the age of the specimen. 2 5 Food preservation (a) Fungus, bacteria and larvae that cause the rotting of food can be killed by irradiating the food with radiation. (b) Budding in potatoes and onions is slowed down by irradiation, thus extending their shelf-life. 2 6 Electricity generation The nuclear energy released when uranium-235 decays spontaneously is used to heat water into steam which then drives turbines of the generator to produce electricity. 2

Temp (°C)

Initial

30

60

90

120

150

180

210

70.0

78.0

80.0

80.0

80.0

82.0

85.0

95.0

3

1 (b) Let the number of carbon atoms that has the same mass as one Mo atom be n. 12n = 96



Materials Pure naphthalene, a mixture of naphthalene with some acetamide and water. 3 (d) Procedure (i) 3 spatulas of naphthalene powder are put into a boiling tube. (ii) A 500 cm3 beaker is filled with water



3

(d) 80 °C 3 (e) A mixture of solid and liquid naphthalene. 3 (f) The heat energy absorbed is used to overcome the forces of attraction between the naphthalene molecules. Hence, the temperature remains constant during the melting process. 3 (g)

3 2 (a) Aim To show that the presence of acetamide as an impurity will lower the melting point of naphthalene. 3 (b) Manipulated variable Pure naphthalene and a mixture of naphthalene with some acetamide. Responding variable Melting point of substance Constant variable Atmospheric pressure, constant stirring during heating. 3 (c) Apparatus 500 cm3 beaker, two boiling tubes, thermometer, stopwatch, retort stand and clamp, tripod stand, wire gauze.

3

until it is about — full 4

and is then placed on a tripod stand. (iii) The boiling tube containing naphthalene is then clamped in the water bath with the level of naphthalene powder below the water level. (iv) The water bath is heated until it reaches a temperature of about 65°C. Then the water is heated with a low flame. (v) The stopwatch is started and the temperature of the naphthalene is recorded at 30-second intervals until the temperature reaches 90 °C. The results are tabulated. (vi) The experiment is repeated by replacing the pure naphthalene with a mixture of naphthalene with acetamide. 3

96



12

(c) Let relative atomic mass of phosphorus = m Mass of five aluminium atoms = mass of six lithium atoms + mass of three phosphorus atoms (5  27) = (6  7) + 3m 3m = 135 – 42 93 m=— — — = 31

3

(c)



n = — — —= 8

(b)

3

(d) (i) Relative atomic mass of Th = (6  7) + (2  39) + (2  56) = 232 (ii) Assume that the mass of n W atoms = mass of two X atoms + mass of one Y atom + mass of one Z atom 7n = (2  39) + 56 + 195 n = 47 atoms 2 (a) Relative molecular mass of C17H35COONa = 17(12) + 35(1) + 12 + 2(16) + 23 = 204 + 35 +12 +32 +23 = 306 (b) Relative formula mass of Cu(NH3)4SO4 = 64 + 4(14 +3) + 32 + 4(16) = 64 + 68 +32 +64 = 228

0

1

–2–

1

1 1–– 2

3 Assume the relative atomic mass of the element X is a. Relative molecular mass of X2B4O7 = 202 2a + 4(11) + 7(16) = 202 a = 23 The relative atomic mass of X is 23. Thus, X is sodium atom.

2

1 2– –– 2

3

1 3–– 2

4

1 4–– 2

5

1 2– –– 2

3

1 3–– 2

4

1 4–– 2

5

Temp (°C)

Mixture of naphthalene and acetamide Time (min)

0

1

–2–

1

1 1–– 2

Temp (°C)

3

2

3

Chemical Formulae and Equations

Self Assess 3.1 1 (a) The relative atomic mass of Pt atom = 5  relative atomic mass of K atom = 5  39 = 195

541

4

There are four ammonia, NH3 molecules

(e) Tabulation of results Pure naphthalene Time (min)

F O R M

Answers

CHAPTER 2 & 3

Experiments 1 (a) Time (s)

Self Assess 3.2 1 1 (a) —6—  6 

10

23

= 1  1023 copper atoms (b) 0.0625  6  1023 = 3.75  1022 water molecules (c) 1.3  6  1023 = 7.8  1023 sodium ions 2 (a) 0.012 mol of C2H6 contains 0.012  6  1023 molecules. One molecule of C2H6 has eight atoms (two carbon and six hydrogen atoms). Therefore, the number of atoms = 8  0.012  6  1023 atoms = 5.76  1022 atoms (b) 1.1 mol of SO3 contains 1.1  6 1023 molecules. One molecule of SO3 has four atoms (one sulphur and three oxygen atoms).

F O R M 4

Therefore, the number of atoms = 4  1.1  6  1023 atoms = 2.64  1024 atoms 3 (a) 1 mol of sodium ions contains 6  1023 ions. Therefore, 6  1022 sodium ions 6 1022 ions = ———————— ——————  1 mol-ion 6 1023 ions = 0.1 mol-ion (b) 1 mol of H2S contains 6  1023 molecules Therefore 1.8 10 H2S 1.8 1024 molecules 1 mol= ————————————————–  6 1023 molecules molecule = 3 mol-molecules

CHAPTER 3

24

Self Assess 3.3 1 (a) 1 mol of He = 4 g 1.25 mol of He = 1.25  4 g =5g (b) 1 mol of Co = 59 g 2 – – mol 5

of Co =

–25– 

59 g

= 23.6 g (c) 1 mol of CuSO4.5H2O = 64 + 32 + 4(16) + 5(18) g = 250 g 0.15 mol of CuSO4.5H2O = 0.15  250 g = 37.5 g (d) 1 mol of KMnO4 = 39 + 55 + 4(16) g = 158 g 0.05 mol of KMnO4 = 0.05  158 g = 7.9 g 2 (a) 1 mol of Fe = 56 g 2.8 2.8 g of Fe = ——— 56

Answers

 1 mol

(b) 1 mol of C10H14N2 = 10(12) + 14 + 2(14) g = 162 g 4.05 4.05 g of C10H14N2 = ——— 162

mol

= 0.025 mol (c) 1 mol of (NH4)3PO4 = 3(14 + 4) + 31 + 4(16) = 149 g 1.49 1.49 g of (NH4)3PO4 = ——— 149

mol

= 0.01 mol (d) 1 mol of C2H5OH = 2(12) + 6 + 16 g = 46 g 2.3 2.3 g of C2H5OH = —— 46

mol

= 0.05 mol 3 (a) 1 mol contains 6  1023 atoms. 3  1023 titanium atoms 3 1023 atom = ——————————————— 1 mol 6 1023 atom = 0.5 mol 1 mol of Ti = 48 g 0.5 mol of Ti = 0.5  48 g = 24 g (b) 1.2  1024 of argon atoms contain 1.2  1024 = ————————23 —— mol 6 10 = 2 mol 1 mol of argon = 40 g 2 mol of argon = 2  40 g = 80 g (c) 7.5  1022 molecules of C12H16O14 7.5  1022 = ————————23 —— mol 6 10 = 0.125 mol 1 mol of C12H16O14 = 12(12) + 16 + 14(16) g = 384 g 0.125 mol of C12H16O14 = 0.125  384 g = 48 g 4 (a) 1 mol of S = 32 g 4 4 g of S = —— 32

mol

= 0.125 mol 1 mol of S contains 6  1023 atoms. 0.125 mol of S contains 0.125  6  1023 atoms = 7.5  1022 atoms (b) 1 mol of Cd = 112 g 2.24 2.24 g of Cd = ——— 112

mol

= 0.02 mol 1 mol of Cd contains 6  1023 atoms. 0.02 mol of Cd contains 0.02  6  1023 atoms = 1.2  1022 atoms (c) 1 mol of C6H12O6 = 180 g 36 36 g of C6H12O6 = ——— 180

= 0.05 mol

mol

= 0.2 mol

542

1 mol of C6H12O6 contains 6  1023 molecules. 0.2 mol of C6H12O6 contains 0.2  6  1023 molecules = 1.2  1023 molecules 5 (a) The molecule consist of 10 carbon atoms, 16 hydrogen atoms and one oxygen atom. Hence its molecular formula is C10H16O. Thus, the mass of 1 mol of C10H16O = 10(12) + 16 + 16 g = 152 g (b) 1 mol of C10H16O = 152 g 0.02 mol of C10H16O = 0.02  152 g = 3.04 g (c) 1 mol of C10H16O =152 g Hence, 7.6 g of C10H16O = —7.6 —– mol 152



= 0.05 mol 1 mol of C10H16O contains 6  1023 molecules. 0.05 mol of C10H16O contains 0.05  6  1023 = 3  1022 molecules. (d) 7.5  1022 molecules 7.5  1022 = ——————————  1 mol 6 1023 = 0.125 mol 1 mol of C10H16O = 152 g 0.125 mol of C10H16O = 0.125  152 g = 19 g

Self Assess 3.4 1 1 mol of gas occupies a volume of 24 dm3 at room temperature. 0.55 mol of gas occupies a volume of 0.55  24 dm3 = 13.2 dm3 2 1 mol of gas occupies a volume of 22 400 cm3 at s.t.p. The number of mol of gas in 672 cm3 672 is ———— mol 22 400

= 0.03 mol

3 1 mol of C2H4 = 28 g 1.4 1.4 g of C2H4 = — –— mol = 0.05 mol 28 1 mol of gas occupies a volume of 24 dm3 at room temperature 0.05 mol of gas occupies a volume of 0.05  24 dm3 = 1.2 dm3 4 (a) 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p. The number of mol of gas in 16.8 16.8 dm3 of gas is —––— 22.4

mol

= 0.75 mol 1 mol of CH4 = 16 g 0.75 mol of CH4 = 0.75  16 g = 12 g (b) 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p. The number of moles of gas in 6720 6720 cm3 of gas is — ———— 22 400

= 0.3 mol

mol

4 Assume that the relative atomic mass of V is a. Element

3.6 5 (a) 3.6 dm3 of gas = —–— mol 24

Mass

= 0.15 mol The number of molecules present is 0.15  6  1023 = 9  1022 molecules

Number of moles

= 0.05 mol The number of molecules present is 0.05  6  1023 = 3  1022 molecules

Percentage mass

Si

H

87.5 %

12.5 %

3.125 ——————— 3.125

=1



12.5 —————— 3.125

=4

Its empirical formula is SiH4. 3 Mass of oxygen that combines with tin is (7.55 – 5.95) g = 1.60 g Element Mass

Sn

O

5.95 g

1.6 g

Number of 5.95 ————— moles 119 = 0.05

1.6 ——— 16

Simplest ratio

0.1 ————— 0.05

0.05 ————— 0.05

=1

= 0.1

=2

Its empirical formula is SnO2.

—— = 0.5

16

5

——— = —— 0.5 5 10.2  5 a = ————––––––— = 51 2  0.5 Mass Number of moles

N

H

87.5 %

12.5 %

87.5 ————— 14

= 6.25 Simplest ratio

6.25 ————— 6.25

=1

12.5 ————— 1

= 12.5 12.5 ————— 6.25

=2

(a) Empirical formula is NH2. (b) Let its molecular formula be (NH2)n

(14 + 2)n = 32 16n = 32 n=2

Its molecular formula is N2H4.

6

Element

Fe

Mass

xg

Number of moles

x ————— 56

Simplest ratio

Number of 87.5 12.5 ————— ————— moles 28 1 = 3.125 = 12.5 Simplest ratio

2

8

10.2 — — — a 2

The volume occupied by 0.015 mol of gas is 0.015  22.4 dm3 = 0.336 dm3 at s.t.p.

Element

10.2 ————— a

5 Element

Self Assess 3.5 1 (a) AgNO3 (b) Na2S2O3 (c) (NH4)3PO4 (d) Ca(OH)2 (e) MgCO3 (f) Zn3P2 (g) Fe(OH)3 (h) Al2O3 (i) CrCl3 (j) CuSO4 (k) NiCl (l) Mg3N2 2 Percentage by weight of hydrogen = (100 – 87.5)% = 12.5 %

O 8g

Simplest ratio

1200 (b) 1200 cm3 of gas = ————— mol 24000

6 9  1021 molecules of CO 9  1021 = ————————— = 0.015 mol 6 1023

V 10.2 g

O 5.04 g 5.04 ————— 16

= 0.315

2

3

C

H

80 %

20 %

x —56 — — 2 —————— = —— 0.315 3 x = 11.76 g

7

Element Percentage by weight

Number of 80 20 ——— = 20 ——— = 6.67 moles 1 12 Simplest ratio



6.67 ————— 6.67

=1

20 ————— 6.67

=3

(a) Its empirical formula is CH3. (b) 1 mol of gas has mass of 22.4 ——––—  6 g = 30 g 4.48 Therefore, its relative molecular mass is 30. (c) Assume its molecular formula is (CH3)n.

543

Relative molecular mass of gas is [12 + 3]n = 30 n = 2 Therefore, its molecular formula is C2H6. Self Assess 3.6 1 (a) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) (b) Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) (c) CuCO3(s) + 2HNO3(aq) → Cu(NO3)2(aq) + CO2(g) + H2O(l) (d) 2HCl(aq) + Na2S2O3(aq) → 2NaCl(aq) + SO2(g) + S(s) + H2O(l) (e) 2Pb(NO3)2(s) → heat

2PbO(s) + 4NO2(g) + O2(g) 2 (a) 2SO2(g) + O2(g) → 2SO3(g) (b) 2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l) (c) 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) (d) Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq) (e) ZnCO3(s) → ZnO(s) + CO2(g) heat

(f) PbO2(s) + 2H2(g) → Pb(s) + 2H2O(l) 3 (a) Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) (b) (i) 1 mol of Fe (56 g) produces 1 mol of FeCl2 (127 g) Hence, 2.8 g of Fe produce 2.8 ——–— 127 g = 6.35 g of FeCl2 56 (ii) 1 mol of Fe (56 g) produces 1 mol of H2 (24 dm3) Hence, 2.8 g of Fe produce 2.8 ——–— 24 dm3 56 =1.2 dm3 of H2. 4 (a) 1 mol of CaC2(40 + 12 + 12) g produces 1 mol of C2H2 (24 dm3) at room temperature and pressure. Hence, 4.8 g of CaC2 4.8 produce ———  24 dm3 = 1.8 dm3 64 of C2H2 gas. (b) 1 mol of CaC2(40 + 12 + 12) g produces 1 mol of Ca(OH)2 (40 + 17 + 17) g. Hence, 4.8 g of CaC2 produce 4.8 ——–— 74 g 64 = 5.55 g of Ca(OH)2 5 (a) 1 mol of CH4(16 g) produces 3 mol (3  24 dm3) of H2. Hence, 60 dm3 of H2 is produced from 60 dm3 H2 = ——————————————  16 g 3  24 dm3 H2 = 13.33 g of CH4 Answers

F O R M 4

CHAPTER 3

1 mol of CO = 28 g 0.3 mol of CO = 0.3  28 g = 8.4 g

(b) 1 mol of CO (6  1023 molecules) is produced together with 3 mol of H2. Hence, the number of CO molecules produced together with 60 dm3 of hydrogen is 60 dm3 H2 = ———————————3———  6  1023 3  24 dm H2 = 5  1023 molecules 6 (a) 3Zn(s) + 2P(s) → Zn3P2 (b) 2 mol of P react with excess zinc to produce 1 mol of Zn3P2. 2  31 g of P react with excess zinc to produce 257 g of Zn3P2. Therefore, 51.4 kg of Zn3P2 is produced from 51.4 kg Zn3P2 —————————————  2  31 kg of P 257 kg Zn3P2 = 12.4 kg of phosphorus

F O R M

CHAPTER 3

4

Self Assess 3.7 1 (a) HBr (b) Zn(OH)2 (c) KNO3 (d) Na2SO4 (e) AgNO3 (f) MgCI2 2 (a) 1 mol of copper(II) oxide reacts with 1 mol of sulphuric acid to form 1 mol of copper(II) sulphate and 1 mol of water. (b) 2 mol of magnesium nitrate decom­poses to form 2 mol of magnesium oxide, 4 mol of nitrogen dioxide and 1 mol of oxygen gas when heated. (c) 1 mol of hydrogen gas react with 1 mol of lead(II) oxide to form 1 mol of lead metal and 1 mol of water. (d) 1 mol of calcium carbonate reacts with 2 mol of hydrochloric acid to form 1 mol of calcium chloride, 1 mol of carbon dioxide and 1 mol of water.

SPM Exam Practice 3 Multiple-choice Questions 1 C 2 A 3 B 4 D 6 C 7 B 8 A 9 A 11 B 12 D 13 D 14 B 16 A 17 C 18 C 19 B 21 D 22 C 23 D 24 D 26 B 27 B 28 D 29 C 31 C 32 D 33 C 34 B 36 C 37 A 38 C 39 D

5 D 10 C 15 A 20 A 25 A 30 B 35 A 40 B

Structured Questions 1 (a) A hydrocarbon is a compound that contains the elements carbon and hydrogen only. 1 Answers

(b) Element Percentage mass Number of moles

C

H

82.76 %

100 – 82.76 = 17.24 % 17.24 ————— 1

82.76 —————— 12

= 17.24

= 6.9 Simplest ratio

17.24 ————— 6.9

6.9 ———— 6.9

=1 =132 =2

= 2.5 = 2.5 3 2 =5

The empirical formula is C2H5 . 2 (c) 1.2 dm3 of gas has a mass of 2.9 g. Hence, 1 mol of gas (24 dm3) 24 has a mass of ———  2.9 g = 58 g 1.2 The relative molecular mass of X is 58. 2 (d) Let the molecular formula be (C2H5)n. (24 + 5)n = 58 58 n = ——— = 2 29 Hence the molecular formula is C4H10. 2 (e) 2C4H10 + 13O2 → 8CO2 + 10H2O 2 (f) (i) 1 mol of C4H10 (58 g) produces 5 mol of water (5  18 g). 11.6 g of C4H10 produce 11.6 ————  5  18 g 58 = 18 g of water 1 (ii) 1 mol of C4H10 (58 g) produces 4 mol of CO2 (4  6  1023 molecules). 11.6 g of C4H10 produce 11.6 ————  4  6  1023 58 = 4.8  1023 molecules 1 2 (a) The air in the combustion tube must be displaced before lighting the hydrogen gas. 1 (b) The heating, cooling and weighing is repeated until a constant mass is obtained. 1 (c) (i) Zinc and hydrochloric acid 1 (ii) Zn + 2HCI → ZnCI2 + H2 1 (iii) Anhydrous calcium chloride 1 (d) (i) Element Mass Number of moles

M

O

5.5 g

3.2 g

5.5 ———— 55

Simplest ratio

= 0.1 1

544

3.2 ———— 16

= 0.2 2

Empirical formula is MO2. 2 (ii) 2H2 + MO2 → M + 2H2O 2 (e) No. Magnesium is more reactive than hydrogen. Thus, hydrogen gas cannot reduce magnesium oxide. 2 3 (a) Cation 1 (b) Pb(NO3)2 1 (c) (i) Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) Yellow 2 (ii) 1 mol of lead nitrate reacts with 2 mol of potassium iodide to form 1 mol of lead(II) iodide and 2 mol of potassium nitrate. 1 (iii) Lead(II) iodide 1 (iv) 2 mol of KI produce 1 mol of PbI2(461 g). Hence 0.04 mol of KI will 0.04

produce ———  461 g

2

= 9.22 g of PbI2 2 4 (a) (i) 2Mg + O2 → 2MgO 2 (ii) 2 mol of Mg produce 2 mol of MgO. 1 mol of Mg (24 g) produce 1 mol of MgO (40 g). Hence 3 g of Mg produce 3

——  40 g

24

= 5 g of MgO 2 (iii) Used as an antacid medicine 1 (iv) MgO + 2HNO3 → Mg(NO3)2 + H2O 2 (b) (i) X : Copper(II) oxide Gas P : Carbon dioxide 2 (ii) CuCO3 → CuO + CO2 1 heat (iii) (a) 1 mol of CuCO3 (124 g) produces 1 mol of CuO (80 g). 6.2 g of CuCO3 produce 6.2 ———  124

80 g

= 4 g of CuO 1 (b) 1 mol CuCO3 (124 g) produces 1 mol of CO2 (24 dm3). 6.2 g of CuCO3 produce 6.2 ———  124

24 dm3

= 1.2 dm3 of CO2 1 5 (a) 2NaHCO3 → heat Na2CO3 +CO2 + H2O 2 (b) The gas bubbles is passed into limewater. If CO2 is present, the limewater turns cloudy. 2 (c) (i) 2 mol of NaHCO3 (168 g) form 1 mol of CO2 (24 dm3) at room temperature. Hence, 8.4 g of NaHCO3



8.4 form — ——  24 dm3 168

= 1.2 dm3 of CO2 3 (ii) 2 mol of NaHCO3 (168 g) form 1 mol of Na2CO3 (106 g). Hence, 8.4 g of NaHCO3 8.4 form — ——  106 168

= 5.3 g of Na2CO3 3 (d) (i) NaHCO3 + HNO3 → NaNO3 + CO2 + H2O 2 (ii) 1 mol of NaHCO3 (84 g) produces 1 mol of NaNO3 (85 g). Hence, 17 g of NaNO3 is produced from

17 g NaNO3 — —————————— 84 85 g of NaNO3



Essay Question 1 (a) (i) The empirical formula of a compound shows the simplest ratio of the atoms of the elements that combine to form the compound. 2 (ii) The molecular formula of a compound shows the actual numbers of atoms of the elements that combine to form the compound. 2 For example, the molecular formula of glucose is C6H12O6 and its empirical formula is CH2O. (b) By reducing a sample of lead oxide with hydrogen gas. 2 (c) No. Magnesium is more reactive than hydrogen. Thus, hydrogen gas cannot reduce magnesium oxide. 3 (d) (i) Procedure • A combustion tube with an empty porcelain dish is weighed and its mass recorded. • A spatula of lead oxide is put into the porcelain dish. The combustion tube and its contents is weighed again. The mass is recorded. • Dry hydrogen gas is passed th­rough the tube to expel all the air before the excess hydrogen is lit and the lead oxide heated. • After the lead oxide is reduced to lead, a stream of hydrogen gas is passed

(iii) The magnesium ribbon is coiled and then placed in the crucible. The crucible, lid and the mag­ne­sium ribbon is weighed again and its weight recorded. (iv) The crucible is heated until the magnesium ribbon starts to burn. (v) The lid is lifted and closed occasionally during the burning of magnesium. (vi) After combustion, the crucible and its contents is cooled and weighed again. Its weight is recorded. (vii) The heating, cooling and weighing is repeated until a constant weight is obtained. 5 (e) Tabulation of data

Mass of combustion tube + x gram empty porcelain dish Mass of combustion tube + y gram porcelain dish + lead oxide

g

= 16.8 g of NaHCO3 3 (iii) As a food preservative 2 (e) (i) Sodium hydroxide 1 (ii) NaOH + HNO3 → NaNO3 + H2O 2

through the tube to cool the lead and to prevent O2 from air mixing with the hot lead. • The tube with the lead metal formed is weighed and its mass recorded. • The heating, cooling and weighing is repeated until a constant mass is obtained. 4 (ii) Tabulation of data

Mass of combustion tube + z gram porcelain dish + lead metal

3 (iii) Calculation Mass of lead formed = (z – x) gram Mass of oxygen that combines with lead = (y – z) gram Pb Mass

Mass of empty crucible + lid

Mass of crucible + lid + magne­sium oxide (after heating)

O

3

(z – x) gram (y – z) gram

Number of moles

z—x ————— 207

y—z ————— 16

Simplest ratio

a

b

Empirical formula is PbaOb. 4 Experiment 1 (a) Statement of the problem What is the empirical formula of magnesium oxide? 3 (b) Variables Manipulated variable Mass of magnesium ribbon used. Responding variable Mass of magnesium oxide formed. Constant variable Excess supply of oxygen. 3 (c) List of substances and apparatus Materials Magnesium ribbon about 20 cm long, oxygen Apparatus Sandpaper, crucible with lid, tripod stand, Bunsen burner, clay pipe triangle, electronic balance and tongs. 3 (d) Procedure (i) An empty crucible with its lid is weighed and its weight recorded. (ii) A 20 cm length of magnesium ribbon is polished with sandpaper.

545

F O R M

Mass of crucible + lid + magnesium (before heating)

4

4

Periodic Table of Elements

Self Assess 4.1 1 Valence electrons are electrons in the outermost shell of an atom. (a) One valence electron (b) Two valence electrons (c) Five valence electrons (d) Seven valence electrons 2 Element

W

X

Y

Z

Electron 2.8.5 2.1 2.8.8.3 2.8.18. arrangement 32.18.7 Group

15

1

13

17

Period

3

2

4

6

3 Arsenic has a proton number of 33; it has 33 electrons. The electron arrangement of arsenic is 2.8.18.5. Arsenic belongs to Group 15 of the Periodic Table because it has five valence electrons. Arsenic belongs to Period 4 of the Periodic Table because it has four electron shells filled with electrons. Nitrogen (2.5) and phosphorus (2.8.5) have the same chemical properties as arsenic because each element has five valence electrons. Answers

CHAPTER 3 & 4



4 (a)

Element

A

B

C

D

E

F

G

H

J

K

Proton number 2

9

13

19

18

16

7

20

17

6

Electron arrangement

2 2.7 2.8.3 2.8.8.1 2.8.8 2.8.6 2.5 2.8.8.2 2.8.7 2.4

Group number 18 17

13

1

18

16

15

2

17

14

Period number

3

4

3

3

2

4

3

2

1

2

(b) B and J have the same chemical properties because each element has seven valence electrons.

F O R M

CHAPTER 4

4

Self Assess 4.2 1 Element P has a duplet (2) electron arrangement and element S (2.8.8) has an octet electron arrangement. These elements do not need to accept, donate or share electrons with other elements. Therefore, they exist as monatomic gases. 2 Argon: To fill electric light bulbs so as to increase their lifespan. Neon: Fill advertising bulbs. It gives a red light. 3 The melting points and boiling points increase as we go down the group because the sizes of the atoms increase. As the sizes of the atoms increase, the van der Waals forces of attraction become stronger. 4 The electronic configuration of neon is 2.8. Neon has attained an octet in its electron arrangement. It does not need to share, donate or accept electrons from other elements to form a compound. Self Assess 4.3 1 (a) Pentium has a silvery surface; it is soft and can be cut by a penknife and it is a conductor of electricity. Pentium reacts with oxygen to form pentium oxide. This oxide dissolves in water, forming pentium hydroxide solution. 4Pn(s) + O2(g) → 2Pn2O(s) Pn2O(s) + H2O(I) → 2PnOH(aq) Pentium metal reacts with cold water, forming pentium hydroxide solution and hydrogen gas. 2Pn(s) + 2H2O(l) → 2PnOH(aq) + H2(g) Pentium burns in chlorine gas, forming pentium chloride salt.

2 Francium, caesium, rubidium, potassium, sodium, lithium. 3 (a) Decreases (c) Increases (b) Increases (d) Increases 4 The burning sodium metal reacts with the brown bromine gas to produce white sodium bromide salt. 2Na(s) + Br2(g) → 2NaBr(s) Self Assess 4.4 1 Chlorine dissolves in water to form hy­drochloric acid and hypochlorous(I) acid. CI2 + H2O → HCI + HOCI These acids ionise to give hydrogen ions, H+ resulting in the solution becoming acidic. The hypochlorous acid (HOCI) formed has a bleaching property. 2 (a) Test both solutions with starch solution in separate test tubes. Iodine solution will form a dark blue precipitate with starch solution. Bromine solution does not show any reaction. (b) See experiment 4.6 (Reaction of halogen with iron wool) 3 (a) Yes, it shows similar chemical properties. The electron arrangement of iodine is 2.8.18.18.7 and the electron arrangement of chlorine is 2.8.7. Both elements have 7 valence electrons. During a chemical reaction, both iodine and chlorine will accept one electron to attain an octet electron arrangement. (b) (i) Higher (ii) Higher (c) (i) I2(s) + 2NaOH(aq) → NaI(aq) + NaOI(aq) + H2O(I) (ii) 2Fe(s) + 3I2(s) → 2FeI3(s) 4 This can be explained by atomic sizes.

2Pn(s) + CI2(g) → 2PnCI(s) (b) Pentium is more reactive than sodium because the atomic size of Pn is larger than Na. The force of attraction between the nucleus and the valence electron in pentium atom is weaker and thus the electron is more easily transferred to other elements during a chemical reaction. Answers

3

Element

Proton number

Electron arrangement

Chlorine

17

2.8.7

Bromine

35

2.8.18.7

Iodine

53

2.8.18.18.7

• Halogens have 7 valence electrons. During a chemical reaction all halogens will accept one electron so that it can attain a stable octet electron arrangement.

1

— CI2 + e—

2

→ CI—

2.8.7 2.8.8 1

— Br2 + e—

→ Br—

2.8.18.7

2.8.18.8

2 1



— I2 + e —

2

→ I—

2.8.18.18.7 2.8.18.18.8 • As we go down the group, the atomic size increases. The force of attraction between the nucleus and electrons become weaker. • The elements down the group has a lower tendency to attract an electron to form a halide ion. • Therefore, the elements down the group is less reactive. 5 Compound Uses Silver bromide (AgBr)

To make photo­ graphic films

Potassium iodide (KI)

Added to iodised salt to prevent goitre

Silver chloride (AgCI)

To make photo­ chromic glass

Dichlorodiphenyl- An insecticide trichloroethane, used to kill CCI3CH(C6H4CI)2 mosquito larvae Tin(II) fluoride, SnF2

Added to fluoridated toothpaste

Self Assess 4.5 1 (a) Decreases (b) Increases (c) The oxides change from basic to amphoteric to acidic. 2 (a) The electron arrangement of X is 2.1 and Y is 2.7. Both elements have two electron shells. Thus, they belong to Period 2 of the Periodic Table. (b) Y is more electronegative. Y has 7 valence electrons. Thus, it has a tendency to accept an electron during a reaction to form a negative ion. It is more electronegative. Element X has one valence electron. It will donate the valence electron during a reaction to form a positive ion. It is electropositive.

Na2O

MgO

AI2O3

SiO2

P4O10

SO2 or SO3

CI2O7

Basic

Basic

Amphoteric

Acidic

Acidic

Acidic

Acidic

546

Self Assess 4.6 1 To each solution, add sodium hydroxide reagent. The solution containing Cu2+ ion will form a blue precipitate. Cu2+(aq) + 2OH—(aq) → Cu(OH)2(s) blue precipitate The solution containing the blue food colou­ring does not form any blue precipitate. 2 The alloy is strong, light and can withstand high temperatures caused by combustion of the jet fuel. SPM Exam Pracitce 4 Multiple-choice Questions 1 D 2 A 3 D 4 C 6 B 7 A 8 C 9 B 11 A 12 C 13 C 14 D 16 B 17 A 18 B 19 D 21 A 22 D 23 A 24 B 26 B 27 C 28 D 29 B 31 D 32 D 33 A 34 D 36 C 37 C 38 D 39 A

5 D 10 D 15 A 20 C 25 A 30 C 35 B 40 B

Structured Questions 1 (a) Liquid 1 (b) Br2(I) + H2O(I) → HBr(aq) + HOBr(aq) hydrobromic hypobromous(I) acid acid 1 (c)

2 (d) Seven 1 (e) Bromine is more reactive. Each halogen atom has seven valence electrons. During reactions, a halogen will accept one electron to attain the stable octet in its electron arrangement. A bromine atom has a smaller atomic radius than an iodine atom. Thus, bromine has a greater tendency to accept an electron than iodine. 3 2 (a) It is a soft metal and is a conductor of electricity. 2 (b) Store in paraffin oil. 1

(c) (i) 2Rb(s) + 2H2O(I) → 2RbOH(aq) + H2(g) (ii) 2Rb(s) + CI2(g) → 2RbCI(s) 2 (d) Each Group 1 element has one valence electron. This electron will be released in a chemical reaction. The atomic radius of rubidium is larger and the valence electron is further from the nucleus compared to potassium. The electrostatic force of attraction between the nucleus and the valence electron is weaker. Hence, the valence electron of rubidium is more easily released compared to potassium. 2 (e) (i) RbNO3 (ii) Rb2SO4 2 (f) White 1 3 (a) E, R, C, G, or H 2 (b) (i) E+ (ii) Q— 2 (c) G 1 (d) B. It is used to fill airships or meteorological balloons. 2 (e) Q 1 (f) R is more reactive. Both E and R donate one valence electron each during chemical reac­tions. The atomic radius of R is larger than the atomic radius of E. The valence electron in R is further from the nucleus and the force of attraction is weaker. Thus, the element R can donate its valence electron more easily. 2 4 (a) R and T or P and U 1 (b) (i) Group 1 2 (ii) Period 4 (Note that the electron arrangement of U is 2.8.8.1) (c) P is used in the hydrogenation of oil to produce margarine. (P is hydrogen) 1 (d)

2 (e) (i) U (ii) 2U(s) + 2H2O(I) → 2UOH(aq) + H2(g) 2 (f) U+ 1 (g) (i) Electronegativity is a measurement of the tendency of an element to attract electrons towards the nucleus. 1 (ii) R is more electronegative. 1 T is below R in the same group (Group 17). Atom R has less filled electron shells than T. The atomic radius of R is less than that of T. The force

547

of attraction of the nucleus of atom R towards an electron is stronger than atom T. Thus, R is more electronegative. 2 5 (a) Good conductors of electricity and high melting points 2 (b) Blue 1 (c) (i) Iron powder (ii) Nickel powder 2 (d) They have more than one oxidation number in their compounds. They form complex ions. 2 (e) Aqueous ammonia solution or sodium hydroxide 1 6 (a) Na, Mg or AI 1 (b) The atomic radius becomes smaller. The charge of the nucleus increases by one unit from one element to the next element across the period but the number of filled electron shells is the same. Thus, the force of attraction between the nucleus and the valence electrons becomes stronger. 3 (c) Argon (Ar). Argon has attained an octet in its electron arrangement. It does not need to donate, receive or share electrons with other elements. 2 (d) Na2O, MgO, AI2O3, SiO2, 3 P4O10, SO2, CI2O7 (e) (i) Na2O (ii) Na2O(s) + H2O(I) → 2NaOH(aq) sodium hydroxide 2 (f) (i) Sulfur dioxide (ii) SO2(g) + H2O(I) → H2SO3(aq) sulphurous acid 2 (g) Amphoteric oxide 1 (h) Silicon is used as a semiconductor. 1 Essay Questions 1 (a) — Each Group 1 element has one valence electron. 1 — During chemical reactions, a Group 1 element will donate its valence electron to attain the stable octet in its electron arrangement. 1 — The reactivity of Group 1 elements depends on the tendency of the elements to donate their valence electrons. 1 — When descending Group 1 from lithium to potassium, the atomic size becomes larger. 1 — The force of attraction between the protons in the nucleus and the valence electron becomes weaker. 1 Answers

F O R M 4

CHAPTER 4

4 (a) The melting point of silicon is higher than that of sulphur. (b) Silicon is a weak conductor of electricity but iron is a good conductor of electricity.



— The elements lower in Group 1 loses its valence electron more easily. 1 — Therefore, reactivity increases down Group 1. 1 (b) Physical properties Rubidium is a soft metal, it is a good conductor of electricity and heat. 3 Chemical properties (i) Rubidium reacts with cold water to form an alkaline solution.



2Rb(s) + 2H2O(I) → 2RbOH(aq) + H2(g) 2 (ii) Rubidium reacts with chlorine to form rubidium chloride salt.

2Rb(s) + CI2(g) → 2RbCI(s) 2 (iii) Rubidium burns in the air to form rubidium oxide. This oxide dissolves in water to form rubidium hydroxide solution. 1

F O R M

CHAPTER 4

4

4Rb(s) + O2(g) → 2Rb2O(s) 1 Rb2O(s) + H2O(I) → 2RbOH(aq) 1 (c) The number of protons in sodium atom is 23 – 12 = 11. The number of protons in chlorine atom is 35 – 18 = 17. The number of electrons is equal to the number of protons. 1 Therefore, the electron arrangement of Na is 2.8.1 and the electron arrangement of CI is 2.8.7. 1 As both atoms have three filled electron shells, they belong to Period 3 of the Periodic Table. 1 2 (a) (i) — Each halogen atom has seven valence electrons. 1 — During a chemical reaction, a halogen atom will accept one electron so that it can attain a stable octet in its electron arrangement. 1 — The reactivity of Group 17 elements depend on their tendency to accept an electron. 1 — The atomic size of Group 17 elements increase down the group as the number of filled electron shell increases. 1 — The force of attraction between the nucleus and the valence electron becomes weaker when descending the group. 2 — The lower elements in the group has a lower tendency to accept an electron to form an ion. 1 Answers

(ii) Each element in Period 3 has three filled electron shells. 1 The proton number increases by one unit from one element to the next across Period 3. 1 As the number of protons in the nucleus increases, the electrostastic force of attraction between the nucleus and the valence electrons become stronger. 2 The valence electrons are attracted closer to the nucleus, causing the atomic radius to decrease. 1 (b) — Transition elements have variable oxidation numbers in their compounds. 1 For example, iron forms two oxidation states: Fe2+ and Fe3+ 1 — Transition elements form coloured compounds in aqueous solutions. 1 For example, aqueous Fe2+ solution is green while an aqueous Fe3+ solution is brown. 1 — Transition elements or their compounds have catalytic properties. 1 For example, iron powder is used as a catalyst in the Haber process to manufacture ammonia. 1 — Transition elements can form complex ions. 1 For example, copper(II) ions form the dark blue tetraammine copper(II), Cu(NH3)42+ ion. 1

3 (a) Helium: used to fill airships. Neon: used to fill advertising bulbs. Argon: used to fill electric bulbs. 5 (b) Helium has attained a duplet in its electron arrangement. The other noble gases have attained the octet in their electron arrangement. 2 Therefore, they do not donate, accept or share electrons with other elements, thus they exist as monatoms. 3 (c) (i) — A piece of sodium is heated in a gas jar spoon until it starts to burn. 1 — The ignited sodium in the gas jar spoon is placed in a gas jar of chlorine gas. 1 — The sodium burns brightly with a yellowish flame. A white solid is obtained. 2 2Na(s) + CI2(g) → 2NaCI(s) 1 (ii) — Iron wool inside a combustion tube is heated strongly. 1 — Chlorine gas is passed through the heated iron wool. 1 — The iron wool glows forming brown iron(III) chloride. 2 2Fe(s) + 3CI2(g) → 2FeCI3(s) 1

4 (a) (Period 3 is used as an example) Sodium Magnesium Aluminium Metal

Metal

Metal

Silicon

Phosphorus

Sulphur

Argon

Semi-metal Non-metal Non-metal Non-metal Non-metal

(b) The electronegativity increases from sodium to argon in Period 3. 1 Across Period 3, the number of protons increases by one for each element. 1 The force of attraction of the nucleus increases as the number of protons increases. 1 The atomic radius decreases across Period 3. 1 This causes an increase in the force of attraction between the nucleus and electrons. 1 (c) Na2O, MgO, AI2O3, SiO2, P4O10, SO2, CI2O7 2 Experiment: Reaction of oxides of Period 3 with 2 mol dm—3 nitric acid and 2 mol dm—3 sodium hydroxide solutions

548

Chlorine

5

Procedure 1 A little sodium oxide powder is added to two separate test tubes. 2 5 cm3 of nitric acid and 5 cm3 of sodium hydroxide are added separately to the two different test tubes. 3 The contents in each test tube are heated slowly while being stirred with glass rods. 4 The solubility of sodium oxide in the two solutions is recorded. 5 The experiment is repeated by replacing the sodium oxide with magnesium oxide, aluminium oxide, silicon(IV) oxide and phosphorus(V) oxide. 3

(b) Atoms P and Q will take part in chemical bonding. This is because they have less than 8 electrons in the outermost shell. 3 Ionic bond and covalent bond.

Results Oxides

Solubility in NaOH

Solubility in HNO3

Inference

Na2O

Not soluble

Dissolves in nitric acid Sodium oxide is a base forming a colourless because it reacts with solution an acid

MgO

Not soluble

Dissolves in nitric acid Magnesium oxide is a forming a colourless base because it reacts solution with an acid

AI2O3

Dissolves in sodium hydroxide forming a colourless solution

Dissolves in nitric acid Aluminium oxide is ampho­­ forming a colourless teric because it reacts solution with both acid and alkali

SiO2

Dissolves in sodium hydroxide forming a colourless solution

Not soluble

Dissolves in sodium hydroxide forming a colourless solution

Not soluble

P4O10

Silicon(IV) oxide is acidic because it reacts with an alkali Phosphorus(V) oxide is acidic because it reacts with an alkali

Self Assess 5.2 1 (a) Ca2+ (b) P3— (c) S2— (d) K+ (e) N3— 2 (a) P + e— → P– 2.7 2.8 Q → Q2+ + 2e— 2.8.8.2 2.8.8

(b)

Thus, we can conclude that the oxides change from basic to amphoteric and then to acidic. 5 (i) The way to manipulate a variable: Pass the different halogen gases over the heated iron wool

(ii) Responding variable: The glow of iron wool

(ii) What to observe in the responding variable: Brightness of flame or glow

(iii) Constant variable: Quantity of iron wool

(iii) The way to maintain the constant variable: Use the same amount of iron wool

(b) The higher the position of a halogen is in Group 17, the more reactive it is with heated iron wool. 1 (c) Chlorine, bromine, iodine 1 (d) The reactivity of the halogens in Group 17 decreases down the group. 1 (e) Least reactive, it will only glow faintly with heated iron wool. 1 2 (a) Problem statement What is the trend of the reactivity of Group 1 metals when descending the group from lithium to potassium? (b) Hypothesis The reactivity of Group 1 metals increases down the group. (c) Materials Small pieces of lithium, sodium and potassium metals, basin filled with water, filter paper and blue litmus paper. Apparatus Penknife, basin and tongs. (d) Procedure 1 A piece of lithium metal is removed from the bottle. A small piece of the metal is cut using a penknife. A piece of filter paper is used to absorb the paraffin oil from the metal.



6



2 The lithium metal is then dropped into a basin of water carefully using a pair of tongs. 3 The observation is recorded. 4 The solution formed in the basin is tested with a piece of blue litmus paper. 5 The experiment is repeated using small pieces of sodium and potassium metal. (e) Tabulation of data Metal

Observation

Lithium Sodium Potassium 15 5

Chemical Bonding

Self Assess 5.1 1 Neon: 2.8 Argon: 2.8.8 Both are stable because each of them has 8 valence electrons corresponding to the stable octet electron arrangement. 2 (a) Atom R is chemically inert. This is because atom R has 8 electrons in the outermost shell.

549

F O R M

3 (a) E+, F2+, G2—, H — (b) (i) E2G (ii) EH (iii) FG (iv) FH2

4

Self Assess 5.3 1 (a) H

(b) H



(c) H



H

(d)

CHAPTER 4 & 5

Experiments 1 (a) (i) Manipulated variable: Types of halogens

2 (a) (i) VZ4 (ii) VY2 (iii) XZ3 (b) (i)

(ii)

(iii) (c) (i) Covalent compound (ii) Covalent compound (iii) Covalent compound 3 (a)

(b)



(c)

Answers

F O R M

CHAPTER 5

4

Self Assess 5.4 1 (a) Aluminium oxide, copper(II) chloride/silicon dioxide/ ammonium nitrate (b) Ammonium nitrate, copper(II) chloride (c) Ethanol (d) Silicon dioxide (e) Ethanol, tetrachloromethane (f) Aluminium oxide 2 (a) R and S (c) Q and S (b) Q and S 3 Carbon dioxide is a covalent compound consisting of simple covalent molecules with weak intermolecular forces of attraction between the molecules. Little energy is required to overcome these weak intermolecular forces of attraction. Hence, it exists as a gas. Magnesium chloride is an ionic compound with strong ionic bonds holding the ions together. High energy is required to separate the strong bonds. It has a high melting point and hence it exists as a solid. Magnesium chloride can conduct electricity in the molten state. The ionic bonds between ions are overcome in the molten state. The ions are free to move to conduct electricity. SPM Exam Practice 5 Multiple-choice Questions 1 B 2 C 3 B 4 B 6 C 7 A 8 C 9 C 11 B 12 D 13 D 14 A 16 D 17 B 18 B 19 A 21 C 22 D 23 A 24 C 26 A 27 D 28 B 29 A 31 D 32 D 33 A 34 C 36 D Structured Questions 1 (a) (i) Ionic bond (ii)

5 D 10 B 15 C 20 C 25 A 30 B 35 B

1

2 (iii) High melting point and boiling point/soluble in water/conducts electricity in liquid state or in aqueous solution. 1 (b) (i) 2.8 1 (ii)



2 (c) No, because U has a stable octet in the electronic configuration. U does not gain, lose or share electrons with other elements. 2 (d) 2V + 2H2O → 2VOH + H2 1 Answers

2 (a) P: 2.8.1, Q: 2.6 2 (b) P is in Group 1, Q is in Group 16 2 (c) (i) P → P+ + e– (ii) Q + 2e– → Q2– 2 1 (d) (i) P2Q (ii) Ionic bond 1 (iii) 1

(e) Low melting point and boiling point/does not conduct electricity in any state. 1 3 (a) (i) Covalent bond 1 (ii) Ionic bond 1 (b) The weak intermolecular forces require little energy to overcome it. 2 (c) Any ionic compound, example: sodium chloride, magnesium chloride 1 (d) R 1 (e) There are freely moving ions in the liquid state. The ions are not able to move freely in the solid state because they are held together by the strong electrostatic force of attraction. 2 (f) Particles in R are atoms. Particles in S are molecules. R: any metal, examples: copper/ aluminium S: macromolecules , examples: diamond/silicon dioxide. 2 4 (a) 19 1 (b) (i) Covalent bond 1 (ii) WY4 1 (iii) 88 (12 + 4 3 19) 1 (c) W and X. They have the same number of protons but different number of neutrons. 2 (d) (i) Y and Z 1 (ii) They have the same number of valence electrons. 1 1 (e) (i) FeZ3 (ii) High melting point and boiling point/soluble in water/conducts electricity in liquid state or in aqueous solution. 1 1 5 (a) PQ4 (b) P is in Group 14, Q is in Group 17 2 (c) Covalent bond 1 (d) It has low melting point and boiling point, it is insoluble in water but soluble in organic solvents and it cannot conduct electricity in any state. 3 (e) (i)

550

2



(ii) Element Q has a low melting point because the intermolecular forces of attraction between molecules (van der Waals forces) are weak. 1

Essay Questions 1 (a) Element

C

Na

Cl

No. of protons

6

11

17

No. of neutrons

6

12

18

Electron arrangement No. of valence electrons

2.4 2.8.1 2.8.7 4

1

7

4 (b) • One carbon atom forms 4 covalent bonds with 4 chlorine atoms to form a CCl4 molecule. • One carbon atom shares 4 valence electrons with 4 chlorine atoms while each chlorine atom shares 1 valence electron with the carbon atom. • The sharing of 4 pairs of electrons enables carbon and chlorine to achieve the stable octet electron arrangement as in the noble gases. • The electron arrangement of CCl4 is as follows:

• A sodium atom donates 1 valence electron to form a sodium ion with the stable octet electron arrangement as in the noble gases: Na → Na+ + e– 2.8.1 2.8 • A chlorine atom accepts 1 electron to form a chloride ion with the stable octet electron arrangement as in the noble gases: Cl + e– → Cl– 2.8.7 2.8.8 • The sodium ions and the chloride ions are attracted by the strong electrostatic force of attraction. • The electron arrangement of sodium chloride is as follows:

8



CCl4

NaCl

Low melting and boiling points

High melting and boiling points

Insoluble in water

Soluble in water

Soluble in organic solvents

Insoluble in organic solvents

Does not conduct electricity in any state

Conducts electricity in the liquid and aqueous state

4 2 (a) (i) — Two elements that form an ionic compound are W and Z. 1 — The electron arrangement of atom W is 2.8.1 and the electron arrangement of atom Z is 2.8.7. 1 + 1 — To achieve the stable octet electron arrangement as in the noble gas: 1 — An atom W donates one electron to form a W+ ion. 1 — An atom Z accepts one electron to form a Z– ion. 1 — The positive ions and negative ions formed are attracted by the strong electrostatic force of attraction. 1

— An ionic compound has high boiling and melting points but a covalent compound has low melting and boiling points. 1+1 — An ionic compound is soluble in water but insoluble in organic solvents whereas a covalent compound is soluble in organic solvents but insoluble in water. 1+1 — An ionic compound conducts electricity in the liquid state or in an aqueous solution but a covalent compound does not conduct electricity in any state. 1+1 3 (a) — A single covalent bond is formed from the sharing of one pair of electrons. 1 — Example: a chlorine molecule consists of a single covalent bond formed between 2 chlorine atoms. 1 — A chlorine atom has 7 valence electrons and requires 1 more electron to achieve the stable octet electron arrangement. 1 — Two chlorine atoms share 1 pair of electrons. 1 — Diagram showing the electron arrangement of a chlorine molecule:

1 — A double covalent bond is formed from the sharing of 2 pairs of electrons. 1 — For example, an oxygen molecule consists of a double covalent bond formed between 2 oxygen atoms. 1 — An oxygen atom has 6 valence electrons and requires 2 electrons to achieve the stable octet electron arrangement. 1 — Two oxygen atoms share 2 pairs of electrons between them. 1 — Diagram showing the electron arrangement of an oxygen molecule:

1 (ii) — Two elements that form a covalent compound are Y and Z. 1 — The electron arrangement of atom Y is 2.4 and the electron arrangement of atom Z is 2.8.7. 1 — One atom of Y contributes 4 electrons to be shared with 4 atoms of Z. 1 — Four pairs of shared electrons form 4 single covalent bonds. 1

1 — A triple covalent bond is formed from the sharing of 3 pairs of electrons. 1 — For example, a nitrogen molecule consists of a triple covalent bond between 2 nitrogen atoms. 1 — A nitrogen atom has 5 valence electrons and requires 3 electrons to achieve the stable octet electron arrangement. 1

2 (b) Three differences between an ionic compound and a covalent compound:

551



— Two nitrogen atoms share 3 pairs of electrons between them. 1 — Diagram showing the electron arrangement of a nitrogen molecule:

1 (b) An ionic compound consists of ions that carry charges. 1 • In the liquid state, heat energy has overcome the ionic bond; the ions are free to move. 1 • In aqueous solution, water molecules separate the ions so that the ions are free to move. 1 • A covalent compound consists of molecules. 1 • There are no freely moving ions and therefore it does not conduct electricity. 1 Experiments 1 (a) Chemical Physical compound state

Observation when shaken with Water

Potassium chloride

Solid

Acetone

Forms a Does not colourless dissolve in solution acetone

3 (b) X is soluble in water but is insoluble in acetone. Y is insoluble in water but is soluble in acetone. Z is soluble in both water and acetone. 3 (c) (i) X and potassium chloride (ii) Y and Z 3 (d) X is potassium nitrate because potassium nitrate is a solid ionic compound, soluble in water and insoluble in acetone. 3 (e) Z is ethanol because ethanol is a liquid and is soluble in both water and acetone. 3 2 (a) Problem statement Do ethanol and aqueous sodium nitrate solution conduct electricity? (b) Hypothesis Ethanol does not conduct electricity whereas aqueous sodium nitrate solution conducts electricity. (c) Variables Manipulated variable Types of compounds. Responding variable Electrical conductivity. Constant variable Physical state of compound. (d) Substances Ethanol and aqueous sodium nitrate solution. Answers

F O R M 4

CHAPTER 5

(c)

Apparatus Beaker, measuring cylinder, graphite rods, batteries, light bulb, switch and connecting wires. (e) Procedure (i) 30 cm3 of ethanol is measured by a measuring cylinder and is poured into a beaker. (ii) Two graphite rods are immersed in ethanol and the circuit is completed by connecting to batteries, light bulb and a switch. (iii) The switch is turned on and the bulb is checked if it lights up. (iv) The experiment is repeated using aqueous sodium nitrate solution to replace ethanol. (f) Tabulation of data Chemical compound

F O R M 4

Ethanol

Bulb does not light up when circuit is completed.

Aqueous sodium nitrate solution

Bulb lights up when circuit is completed.



6

CHAPTER 5 & 6

Observation

17

Electrochemistry

Self Assess 6.1 1 (a) (i) An electrolyte is a chemical substance that conducts electricity in the molten state or in aqueous solution and undergoes chemical changes. Examples: molten lead(II) chloride, aqueous potassium iodide solution (molten or aqueous solution of any other ionic compounds). (ii) A non-electrolyte is a chemical compound that does not conduct electricity in any state. Examples: naphthalene, glucose (or any covalent compounds). (b) A conductor can conduct electric current. An electrolyte can conduct electric current as well as be decomposed by electricity. 2 Electrolytes: molten zinc oxide, aqueous zinc chloride solution, aqueous ethanoic acid solution and molten sodium chloride. Non-electrolytes: molten sulphur, solid zinc oxide, molten zinc metal, solid zinc metal, acetone and aqueous glucose solution. 3 Solid magnesium chloride consists of positive ions and negative ions held together by strong ionic bonds. The ions Answers

are not free to move and hence cannot conduct electricity. In the molten state, heat energy has overcome the ionic bonds. The ions are freely moving and hence able to conduct electricity.

Self Assess 6.2 1 Name of compound

2

Ions that move to

Ions produced

anode

cathode

–

—

2—

2—

(a) Zinc chloride

Zn , Cl

CI

Zn2+

(b) Magnesium oxide

Mg , O

O

Mg2+

2+

2+

Half-equation at

Name of compound

anode

Products formed at

cathode

2O2— → O2 + 4e— Ca2+ + 2e— → Ca

(a) Calcium oxide (b) Aluminium iodide

2I— → I2 + 2e—

anode

cathode

Oxygen gas

Calcium metal

AI3+ + 3e— → AI Iodine vapour

Aluminium metal

3 (a) (i) Shiny grey metal is deposited (ii) Greenish-yellow gas is evolved (b) (i) Lead metal (ii) Chlorine gas (c) (i) Pb2+ + 2e— → Pb (ii) 2CI— → CI2 + 2e— Self Assess 6.3 1 Ions discharged Ions discharged Ions present Aqueous solution at the cathode at the anode (a) Aqueous nitric acid

H+, NO3—, OH—

H+

OH—

(b) Silver nitrate

Ag+, NO3—, H+, OH—

Ag+

OH—

(c) Very dilute copper(II) chloride

Cu2+, H+, CI—, OH—

Cu2+

OH—

2 (a) Cell I: copper(II) ions, hydrogen ions, sulphate ions, hydroxide ions Cell II: silver ions, hydrogen ions, nitrate ions, hydroxide ions (b) (i) Colourless gas bubbles are formed at electrode P. A brown deposit is formed at electrode Q. (ii) The blue colour becomes paler. The concentration of copper(II) ions decreases. (iii) Electrode R becomes thinner. Electrode S becomes thicker. (iv) Electrode P: 4OH— → 2H2O + O2 + 4e— Electrode R: Ag → Ag+ + e— (c) (i) Electrode Q: Positions of ions in the electrochemical series. (ii) Electrode R: Types of electrodes used. Self Assess 6.4 1 Three uses of electrolysis: (a) To extract reactive metals from their ores (b) To purify impure metals (c) To electroplate a metal object with another metal 2 (a) Silver (b) The iron key is deposited with a layer of shiny silver metal. The silver electrode becomes thinner.

552

(c) (i) Ag+ + e— → Ag (ii) Ag → Ag+ + e— (d) No change. The concentration of silver ions remains constant because the rate of formation of silver ions at the anode is the same as the rate of discharge of silver ions at the cathode. (e) The iron key is polished with a piece of sandpaper. Electrolysis is carried out using a low current over a longer period of time. Self Assess 6.5 1 (a)

(b) Iron metal is the negative terminal (c) At the negative terminal: Fe → Fe2+ + 2e— At the positive terminal: Cu2+ + 2e— → Cu (d) The iron metal corrodes while the copper metal thickens.

Advantages

Disadvantages

Dry cells

• Cheap • No spillage occurs • Produces moderately stable current and voltage • Can be carried around easily • Can be made into different sizes

• Cannot last long • Cannot be recharged • Zinc metal will dissolve and the electrolyte leaks, corroding electrical instruments • Current and voltage produced is low

Alkaline cells

• Can last longer than a dry cell • Produces a higher and more stable voltage

• Not rechargeable • More expensive than a dry cell • If leakage occurs, the electrolyte is more corrosive

Type of cell

3 Lead-acid accumulator At the negative terminal: Lead metal releases electrons. Pb → Pb2+ + 2e— Electrons that are released flow to the positive terminal through the external circuit, producing electric current. At the positive terminal: Electrons are received by lead(II) oxide. PbO2 + 4H+ + 2e— → Pb2+ + 2H2O Self Assess 6.6 1 (a) X, Z, Y, W (b) (i) W becomes the negative terminal. (ii) 0.7 volt [(1.2 – 0.9) + 0.4] (c) (i) Y will displace metal Z and Y ions are produced. (ii) No change will occur. (iii) W will displace metal X and W ions are produced. 2 (a) No change occur. (b) Displacement reaction occurs: copper will displace silver from silver nitrate. Copper metal dissolves, silver metal deposits and a blue colour solution is formed. Cu + 2Ag+ → 2Ag + Cu2+ (c) Displacement reaction occurs: magnesium will displace silver from silver nitrate. Magnesium metal dissolves and silver metal deposit is formed. Mg + 2Ag+ → 2Ag + Mg2+ SPM Exam Practice 6 Multiple-choice Questions 1 C 2 C 3 B 4 C 6 D 7 A 8 D 9 A 11 C 12 C 13 D 14 B 16 D 17 D 18 A 19 A 21 A 22 C 23 B 24 A 26 C 27 D 28 D 29 C 31 C 32 C 33 C 34 B 36 C 37 B 38 C 39 C

5 B 10 C 15 D 20 D 25 A 30 A 35 B 40 C

Structured Questions 1 (a) Naphthalene does not contain freely moving ions. 1 (b) In lead(II) bromide solid, the ions are held in a lattice by strong bonds and are not free to move. In the molten form, the ions are free to move when the bonds holding them are overcome. 2

(c) At the anode: brown gas is evolved. At the cathode: grey metal is deposited. 2 (d) At the cathode: Pb2+ + 2e— → Pb At the anode: 2Br– → Br2 + 2e— 2 (e) Zinc metal will form at the cathode and chlorine gas will form at the anode. 2 2 (a) H+, Cl–, OH– 1 (b) Carbon/graphite/platinum 1 (c) (i) Gas X is oxygen gas, gas Y is hydrogen gas. 2 (ii) 4OH—→ 2H2O + O2 + 4e— 1 (d) (i) H+ ions and OH– ions are discharged at the cathode and anode respectively in electrolysis. The effect is equal to the removal of water which causes the concentration of hydrochloric acid to increase. 2 (ii) Chlorine gas is collected at the anode. This is because the concentration of Cl– ions is higher than OH– ions, hence Cl– ions are selectively discharged at the anode. 2 (iii) 2Cl— → Cl2 + 2e— 1 3 (a) In cell A, chemical energy is converted into electrical energy. In cell B, electrical energy is converted into chemical energy. 2 (b) (i) Mg → Mg2+ + 2e— 1 (ii) Cu2+ + 2e— → Cu 1 (c) (i) Carbon electrode Q 1 (ii) Q is connected to the negative terminal (magnesium electrode) in cell A. 1 (iii) Electrons flow from the magnesium electrode through the external circuit to electrode Q. 1 (d) (i) Copper metal, Cu2+ + 2e— → Cu 2 (ii) Oxygen gas, 4OH— → 2H2O + O2 + 4e— 2 (e) Electrons flow from the copper electrode to the silver electrode. 1 (f) Electrode P will dissolve forming copper(II) ions. Copper metal will deposit on electrode Q. 1

553

4 (a) Cu2+, SO42–, H+, OH– 1 (b) Electrons flow from the zinc plate to the copper plate through the external circuit. 1 (c) (i) The cathode increases in mass. 1 (ii) Cu2+ + 2e– → Cu 1 (iii) The colour intensity remains the same. 1 (iv) Oxygen gas 1 (d) (i) The zinc plate becomes thinner. 1 (ii) Zn → Zn2+ + 2e– 1 (iii) Zn + Cu2+ → Zn2+ + Cu 1 (iv) The cell voltage increases. 1 5 (a) Positive terminal is copper metal. Negative terminal is lead metal. 2 (b) (i) Pb → Pb2+ + 2e— 1 (ii) Cu2+ + 2e— → Cu 1 (c) Pb + Cu2+ → Pb2+ + Cu 1 (d) To complete the circuit for the ions to flow. 1 (e) Higher than 0.57 V. 1 6 (a) Electrode L and electrode S. 2 (b) Cu2+, SO42–, H+, OH– 2 (c) (i) Cu → Cu2+ + 2e– (ii) Cu2+ + 2e– → Cu 2 (d) (i) Copper metal (ii) Oxygen gas 2 (e) Intensity of blue colour in beaker I is constant. Intensity of blue colour in beaker II decreases. 2 (f) (i) Rheostat. (ii) To control and to ensure that the magnitude of electric current remains constant. 2 Essay Questions 1 (a) Electrolysis is a process involving the decomposition of an ionic compound (electrolyte) in the molten form or in aqueous solution by an electric current. 2 (b) Magnesium chloride is an ionic compound. In the solid state, the ions are held by strong electrostatic force of attraction. Hence, the ions are not free to move and subsequently cannot conduct electric current. When it melts, the ions are free to move and can conduct electric current. 4 (c) The arrangement of apparatus for the experiment to be carried out is shown in the diagram below.

Answers

F O R M 4

CHAPTER 6

2

Magnesium chloride is heated until it melts and the circuit is completed to allow the flow of electric current. Shiny grey solids will deposit at the cathode and greenish-yellow gas will evolve at the anode. In terms of the ionic theory, molten magnesium chloride consists of freely moving magnesium ions, Mg2+ and chloride ions, Cl—. MgCl2 → Mg2+ + 2Cl— Magnesium ions, Mg2+ move to the cathode to accept electrons and form magnesium metal. Mg2+ + 2e– → Mg

F O R M

CHAPTER 6

4

Chloride ions move to the anode and donate electrons to form chlorine atoms. Two atoms of chlorine combine to form a chlorine molecule. 2Cl— → Cl2 + 2e— 14 2 (a) (i) In the electrolysis of aqueous copper(II) chloride solution, the ions present are Cu2+ ions and Cl— ions from CuCl2, H+ ions and OH–­ ions from water. If carbon is used as the anode and cathode, 1 hydroxide ions, OH— ions are discharged at the anode and oxygen gas is released. 1 Copper is deposited at the cathode. Anode: 4OH— → 2H2O + O2 + 4e— Cathode: Cu2+ + 2e— → Cu 1 If copper metal is used as the cathode and anode, copper anode dissolves to form copper(II) ions. 1 Anode: Cu → Cu2+ + 2e— Cathode: Cu2+ + 2e— → Cu 1 When different substances are used as the anode, different products will be produced at the anode, even though copper metal is deposited at the cathode in both cases. 1 (ii) In the electrolysis of copper(II) chloride aqueous solution using carbon as the anode and cathode, the ions present are Cu2+ ion and Cl— ion from CuCl2, H+ ions and OH– ions from water. 1 If dilute copper(II) chloride solution is used, hydroxide ions are discharged at the anode producing oxygen gas. 1 Answers

Anode: 4OH— → 2H2O + O2 + 4e— 1 If concentrated copper(II) solution is used, chloride ions are discharged at the anode and chlorine gas is produced. 1 This is because in concentrated copper(II) chloride solution, the concentration of chloride ions is higher than that of 1 hydroxide ions. Hence, chloride ions are discharged. Anode: 2Cl— → Cl2 + 2e — 1 (b) Electroplating of an iron spoon with copper in electrolysis: In the experiment, copper(II) sulphate solution is used as the electrolyte. The iron spoon is 1 used as the cathode and 1 copper metal is used as the anode. Copper anode will dissolve and 1 copper metal will be deposited on the surface of the iron spoon. 1

2 Anode: Cu → Cu2+ + 2e— 1 Cathode: Cu2+ + 2e– → Cu 1 3 (a) An electrolytic cell is a circuit arrangement consisting of electrodes, connected by wires to batteries, immersed in an electrolyte. 1 The passage of an electric current produces a chemical reaction. 1 A voltaic cell is a circuit arrangement that consists of different metals as electrodes immersed in an electrolyte. A 1 chemical reaction takes place and chemical energy is converted into electrical energy. 1 (b)

2 • 30 cm3 of P(NO3)2 solution and 30 cm3 of Q(NO3)2 1 solution are placed into different beakers. 1 • Metal P is dipped in P(NO3)2 solution and metal Q is dipped in Q(NO3)2 solution. 1 • A salt bridge is used to complete the circuit by allowing the ions to flow. 1

554

• P releases electrons to form P ions, P2+. P → P2+ + 2e— 2 • Q ions receive electrons to form metal Q. Q2+ + 2e— → Q 2 • Electrons flow from metal P to metal Q. 1 • Deflection of the voltmeter needle indicates the flow of an electric current. 1 (c) Displacement reaction will occur. P will displace Q from Q(NO3)2 solution. Metal P will dissolve to form P2+ ions. Metal Q will be deposited. P + Q2+ → P2+ + Q 4 Experiments 1 (a) (i) Anode: oxygen gas Cathode: hydrogen gas (ii) Positions of ions in the electrochemical series. 3 (b) (i) Iodine (ii) Add a little starch solution, a dark blue colour will be formed. (iii) Concentration of ions in the electrolyte 3 (c) (i) Anode: 2I­— → I2 + 2e— Cathode: 2H+ + 2e— → H2 (ii) 4OH— → O2 + 2H2O + 4e— 3 (d) (i) Copper metal (ii) Bromine gas (iii) Copper(II) bromide 3 2 (a) Aim of experiment To investigate the effect of the concentration of an electrolyte on the types of products formed in electrolysis. 2 (b) Hypothesis Ions with higher concentration will be selectively discharged during electrolysis. 3 (c) Variables • Manipulated variable Concentration of aqueous copper(II) chloride solutions 1 • Responding variable Products of electrolysis 1 • Constant variables Types of electrodes and electrolytes used 1 (d) Substances 1.0 mol dm–3 copper(II) chloride solution and 0.001 mol dm–3 copper(II) chloride solution. Apparatus Batteries, carbon electrodes, rheostat, ammeter, connecting wires with crocodile clips, switch and test tubes. 3 (e) Procedure 1 1.0 mol dm–3 aqueous copper(II) chloride solution is

Chemical substance

Observation Observation at the anode at the cathode

Copper(II) Greenishchloride yellow gas is solution of 1.0 evolved mol dm–3

Brown deposit is formed

Copper(II) Colourless Brown deposit chloride gas is evolved is formed solution of 0.001 mol dm–3 7

3

Acids and Bases

Self Assess 7.1 1 (a) (i) An acid is a chemical compound that produces hydrogen ions when it dissolves in water. (ii) A base is a chemical compound that neutralises an acid to produce salt and water. (iii) An alkali is a chemical compound that produces hydroxide ions when it dissolves in water. (b) An acid changes moist blue litmus paper to red. An alkali changes moist red litmus paper to blue. 2 Uses Acids or bases To make antacid

Mg(OH)2

To make fertiliser

H2SO4, HNO3, NH3, NaOH

To make soap

NaOH

To neutralise acidity in soil

CaO, Ca(OH)2

3 Q = MgCO3 R = HNO3 X = HCl Y = KOH Z = NH3 4 (a) H2SO4 + MgO → MgSO4 + H2O (b) 6HNO3 + 2Al → 2Al(NO­3)3 + 3H2 (c) 2HCl + CaCO3 → CaCl2 + CO2+ H2O (d) CH3COOH + NaOH → CH3COONa + H2O (e) KOH + NH4Cl → NH3 + KCl + H2O

5 Hydrochloric acid dissociates to H+ ions in aqueous solution. H+ ions react with magnesium to produce hydrogen gas: 2HCl + Mg → MgCl2 + H2 Hydrogen chloride dissolved in methylbenzene exists as molecules and is not acidic. H+ ions are not produced to react with magnesium. Self Assess 7.2 1 (a) (i) E (ii) A (b) (i) E (ii) A (iii) D/F (iv) C (c) (i) B (ii) E (iii) C (iv) A 2 A strong acid is a chemical substance that dissociates completely into hydrogen ions in water. An example is hydrochloric acid, which undergoes complete dissociation in water to form hydrogen ions, H+ and chloride ions, Cl—: HCl → H+ + Cl— A weak acid is a chemical substance that dissociates partially into hydrogen ions in water. An example is ethanoic acid, which undergoes partial dissociation in water to form hydrogen ions, H+ and ethanoate ions, CH3COO—: CH3COO— + H+ CH3COOH pH value of the strong acid is lower than that of the weak acid of the same concentration. 3 (a) Propanoic acid, ethanoic acid, methanoic acid (b) pH of methanoic acid < 3, pH of propanoic acid > 3 Self Assess 7.3 Mass 1 (a) Concentration = — ————— Volume

mass ∴46.4 g dm—3 = ——————— 3 2.0 dm

Mass of KOH = 2.0  46.4 g = 92.8 g (b) Relative molecular mass of KOH = 39 + 16 + 1 = 56 Number of moles = — MV — — — — — 1000 Number of moles of KOH 100 ——— = 0.2 = 2.0  — 1000 Mass of KOH = 0.2  56 = 11.2 g Mass 2 (a) Concentration = — — ——— Volume

2.12 g = ————— ————————3 = —2.12 ————g—3 (500/1000) dm

0.5 dm

= 4.24 g dm—3 (b) Relative molecular mass of Na2CO3 = (23  2) + 12 + (16  3) = 106 4.24 Molarity = — —— = 0.04 mol dm—3 106 3 (a) Relative molecular mass of NaOH = 23 + 16 + 1 = 40 8.0 Molarity = —40 —— g

dm—3

= 0.2 mol dm—3 (b) M1V1 = M2V2 M final  (100 + 100) = 0.2 100 M final = 0.1 mol dm—3

555

(c) Number of moles of NaOH in 100 cm3 of 0.2 mol dm—3 NaOH 0.2  100 = ——— ————— 1000

= 0.02

Number of moles of NaOH in 20 cm3 of 2.0 mol dm—3 NaOH 2.0  20 = —— ———— = 1000

0.04

Total number of moles of NaOH = 0.02 + 0.04 = 0.06 Total volume = 100 + 20 = 120 cm3 Molarity of new solution ———— 0.06 = ————— (120 / 1000)

= 0.5 mol dm—3 Self Assess 7.4 1 (a) MgCl2, H2 (b) ZnSO4, 2H2O (c) CH3COOH (d) Ca(OH)2, H2SO4 2 (a) 2HNO3 + Mg(OH)2 → Mg(NO3)2 + 2H2O (b) From the equation, 1 mol of Mg(OH)2 reacts with 2 mol of HNO3. Hence, 0.001 mol of Mg(OH)2 will react with 0.002 mol of HNO3. 0.002 mol

M V M  10 =— ——— = — ———— 1000 1000

4

Concentration of HNO3,

M = 0.002



F O R M

1000 — — —— 10

= 0.2 mol dm—3 MAVA M 3 ———V— — B B

2

= —1—

1.25  VA ∴ ———————— 1.0  25.0

2

= —1—

Volume of HCl, V = 2  25.0  —1.0 —— A 1.25 = 40 cm3 4 H2SO4 + 2NH3 → (NH4)2SO4 MAVA —M—— — — V B B

CHAPTER 6 & 7

placed in an electrolytic cell with carbon electrodes. 2 Two test tubes filled with copper(II) chloride solutions are inverted on top of the anode and cathode. 3 The switch is turned on and electrolysis is carried out. 4 Observations at the anode and cathode are recorded. 5 The experiment is repeated using 0.001 mol dm–3 copper(II) chloride solution. 3 (f) Tabulation of data

1

= —2—

1 2.0  25.0 MB  VB ∴ MA = ——— ———— —— = ———————— V 2 15.0  2 A



= 1.67 mol dm—3

SPM Exam Practice 7 Multiple-choice Questions 1 A 2 D 3 C 4 B 6 D 7 B 8 A 9 C 11 C 12 D 13 D 14 D 16 B 17 A 18 A 19 C 21 B 22 B 23 D 24 D 26 B 27 C 28 D 29 B 31 C 32 A 33 B 34 A 36 C 37 B 38 C 39 C

5 C 10 B 15 D 20 C 25 D 30 B 35 B 40 B

Structured Questions 1 (a) To prevent the solution from being sucked up. 1 (b) (i) No noticeable change in 1 beaker A. Effervescence occurs in beaker B. 1 Answers

F O R M

CHAPTER 7

4

(ii) Hydrogen chloride in methylbenzene does not have any acidic properties. In 1 beaker B, hydrogen chloride in water produces H+ ions which react with magnesium to produce hydrogen gas. 1 (c) (i) Hydrogen chloride molecules 1 (ii) Hydrogen ions and chloride ions 1 (d) (i) Carbon dioxide gas. Deliver the gas produced into limewater. Limewater will turn cloudy. 1 (ii) In the presence of water, hydrogen chloride dissociates to H+ ions that will react with sodium carbonate solution to produce carbon dioxide gas. 1 (iii) 2H+­(aq) + CO32—(aq) → CO2(g) + H2O(l) 1 (iv) Hydrogen chloride can only act as an acid in the presence of water. 1 2 (a) (i) A standard solution is a solution with a known concentration. 1 (ii) Molarity of the solution gives the number of moles of sodium hydroxide dissolved in 1 dm3 of the solution. 1 (b) Number of moles of NaOH 0.5 3 100 = ——————————— = 0.05 1000 1 Mass of NaOH = 0.05 3 40 = 2 g 1 (c) (i) A measuring cylinder is unsuitable because it cannot measure the volume of water accurately. 1 (ii) A volumetric flask 1 (d) Parameter I: mass of sodium hydroxide 1 Parameter II: volume of the solution 1 (e) Water is added carefully (drop by drop) so that the level of the solution does not exceed the graduation mark of the volumetric flask. 1 The volumetric flask is stoppered to prevent the evaporation of water which can change the concentration of the solution prepared. 1 3 (a) 1 mol of acid reacts with 2 mol of OH— ions (or 1 mol of acid produces 2 mol of H+ ions) 1 (b) H2A(aq) + 2KOH(aq) → K2 A(aq) + 2H2O(l) 1 (c) From pink colour to colourless. 1 (d) (i) 26.05 cm3, 26.15 cm3, 26.10 cm3 1

(e) Concentration of H2A



0.5  25.0 = ———————— = 2  26.1

0.24 mol dm

—3

2

(f)

2 4 (a) (i) H+ ion. 1 (ii) Mg + 2H+ → Mg2+ + H2 1 (b) (i) Experiment I. 1 (ii) Sulphuric acid is a strong acid whereas ethanoic acid is a weak acid. Sulphuric 1 acid undergoes complete dissociation to produce more hydrogen ions whereas 1 ethanoic acid undergoes partial ionisation in the presence of water. 1 (c) pH value of sulphuric acid is lower than that of ethanoic acid. 1 The concentration of hydrogen ions is higher in sulphuric acid than in ethanoic acid. 1 (d) pH value will be higher. 1 Addition of water will dilute the sulphuric acid. This will lower the hydrogen ion concentration and hence increase the pH value. 1 5 (a) pH value is 7 (see graph). 2

(b) Volume of nitric acid is 20.0 cm3 (see graph). 2 (c) HNO3 + NaOH → NaNO3 + H2O 1 (d) From yellow to orange 1 (e) Concentration of HNO3 0.5  25.0 = ——— ————— 20.0



= 0.625 mol dm—3 2

(f)

(26.05 + 26.15 + 26.10) cm3

(ii) ——————————3——————————— = 26.10 cm3 Answers

1



2

556

Essay Questions 1 (a) Neutralisation is a reaction between an acid and a base to produce salt and water only. 1 For example, sulphuric acid reacts with sodium hydroxide to 1 produce the salt, sodium sulphate and water only. 1 H2SO4 + 2NaOH → Na2SO4 + 2H2O 1 (b) Sodium hydroxide is a strong alkali that undergoes complete 1 dissociation in aqueous solution. 1 Ammonia is a weak alkali that 1 undergoes partial dissociation 1 only. The concentration of hydroxide ion in sodium hydroxide is thus higher than that in ammonia. 1 Hence the pH value of sodium hydroxide is higher than that of ammonia. 1 (c) Molar mass of KOH = 39 + 16 + 1 = 56 g mol–1 1 250 cm3 of 1.0 mol dm—3 KOH 1.0  250 contains ——1000 ————— mol

KOH

which is 0.25  56 = 14.0 g of KOH. 1 • Weigh exactly 14.0 g of KOH accurately in a weighing bottle. 1 • Dissolve 14.0 g of KOH in a little water in a beaker and transfer 1 the contents into a 250 cm3 volumetric flask. 1 • Rinse the beaker with distilled water and transfer all the contents into the volumetric flask. 1 • Distilled water is added to the volumetric flask until the graduation mark. The solution produced is 1.0 mol dm—3 potassium hydroxide. 1 • To prepare a 0.1 mol dm—3 potassium hydroxide solution, 25.0 cm3 of 1.0 mol dm—3 1 potassium hydroxide solution is transferred to a 250 cm3 1 volumetric flask using a 25.0 cm3 pipette. • Distilled water is added to the volumetric flask until the 1 graduation mark. 2 (a) Chemical formula of magnesium hydroxide is Mg(OH)2. 1 Magnesium hydroxide is a base 1 which can neutralise the excess hydrochloric acid in the stomach. 1 Another chemical found in antacid is aluminium hydroxide. 1 (b) (i) Solution X is an acid whereas solution Y is an alkali. 1 • Solution X is sour in taste. 1

(c)

Concentration of H+ ions (mol dm—3) pH value

0.100

0.060

0.040

0.025

0.015

0.010

1.0

1.2

1.4

1.6

1.8

2.0



3

557

(d)

3 (e) The pH value of a 0.020 mol dm—3 nitric acid is 1.69 (accept 1.68 1.70). 3 2 (a) Problem statement How to determine the volume of sulphuric acid required to neutralise 25.0 cm3 of 0.5 mol dm—3 potassium hydroxide solution? (b) Variables Manipulated variable Volume of potassium hydroxide solution used Responding variable Volume of sulphuric acid required Constant variables Concentrations of sulphuric acid and potassium hydroxide used (c) Statement of hypothesis The volume of sulphuric acid required in neutralisation depends on the volume of potassium hydroxide used. (d) Apparatus 25 cm3 pipette, pipette filler, 50 cm3 burette, retort stand and clamp, conical flasks, filter funnel, dropper and white tile. Materials Sulphuric acid, 0.5 mol dm—3 potassium hydroxide solution and methyl orange. (e) Procedure 1 A 25 cm3 pipette is cleaned with distilled water and rinsed with a little potassium hydroxide solution. 2 Using a pipette, 25 cm3 of 0.5 mol dm—3 potassium hydroxide is transferred to a clean conical flask. Three drops of methyl orange indicator are added to the alkali and the colour of the solution is noted. 3 A 50 cm3 burette is cleaned with distilled water and is rinsed with a little sulphuric acid. 4 The burette is then filled with sulphuric acid and is clamped to a retort stand. The initial burette reading is recorded. Answers

F O R M 4

CHAPTER 7

• It can react with a base to ions, H+ which is responsible for its acidic properties. 1 produce salt and water. 1 • CH3COOH(aq) • It can react with a reactive H+(aq) + CH3COO—(aq) 1 metal to produce a salt • H+ ions change blue litmus and hydrogen gas. 1 to red and react with reactive • It can react with a metal metals or carbonates or bases. carbonate to produce a 1 salt, carbon dioxide gas (b) (i) Test with magnesium ribbon. and water. 1 • About 3 cm of magnesium • Solution Y is bitter in taste ribbon is added to about and it feels soapy. 1 5 cm3 of acidic solution • It can react with an acid to in a test tube. 1 produce salt and water. 1 • The gas evolved is tested • When it is heated with by placing a glowing ammonium salts, ammonia wooden splint near 1 gas is produced. 1 the mouth of the test tube. • It forms a metal hydroxide If a ‘pop’ sound is when it is added to an produced, indicating the aqueous salt solution. 1 presence of hydrogen gas, An example of solution X is the solution in the test 1 hydrochloric acid. An example tube is proven to be acidic. of solution Y is sodium • Mg + 2H+ → Mg2+ + H2 1 hydroxide solution. 1 (ii) Tests with sodium carbonate (ii) When solution X and solution • A little sodium carbonate Y are mixed together, powder is added to about neutralisation reaction takes 1 5 cm3 of acidic solution in place, salt and water is a test tube. 1 produced. The solution 1 • The gas evolved is tested produced has a pH value of 7. by passing the gas into lime 1 water. If the limewater 1 The ionic equation is: turns milky, indicating the H+ + OH— → H2O 1 presence of carbon dioxide (iii) M1V1 = M2V2 gas, the solution in the test tube is proven to be acidic. M final  (80 + 20) 1 1 = 0.1  20 • CO32— + 2H+ → CO2 + H2O M final = 0.02 mol dm—3 1 1 3 (a) • Pure and dry acid, for (iii) Test with copper(II) oxide example, glacial ethanoic acid • A little copper(II) oxide does not show any acidic powder is added to about property. 1 5 cm3 of acidic solution in • Glacial ethanoic acid exists as a test tube and the mixture molecules. 1 is heated. 1 • Hydrogen ions are not present • If the black copper(II) to show any acidic properties. oxide powder dissolves and 1 a blue solution is formed, • Glacial ethanoic acid does not the solution in the test tube change blue litmus to red. is proven to be acidic. 2 It also does not react with • CuO + 2H+ → Cu2++ H2O 1 reactive metals or carbonates or bases. 1 Experiments • When water is added to 1 (a) Manipulated variable glacial ethanoic acid, aqueous Concentration of nitric acid solutions. ethanoic acid produced shows Responding variable pH values acidic properties. 1 Constant variable Type of acid • In the presence of water, used 3 aqueous ethanoic acid (b) The higher the concentration of H+ ions, the lower the pH value. 3 dissociates to form hydrogen

5 The conical flask containing 25 cm3 of potassium hydroxide is placed below the burette. A piece of white tile is placed below the conical flask for clearer observation of the colour change. 6 Sulphuric acid is added slowly from the burette to the potassium hydroxide solution in the conical flask while swirling the flask gently. 7 Titration is stopped when methyl orange changes from yellow colour to orange colour. The final burette reading is recorded. 8 Steps 1 to 7 are repeated until accurate titration values are obtained, that is, until the difference in the volumes of sulphuric acid used in two experiments is less than 0.10 cm3. (f) Tabulation of data









F O R M 4

Rough

Accurate 1

2

3

Final burette reading (cm3) Initial burette reading (cm3) Volume of sulphuric acid used (cm3) 17

CHAPTER 7 & 8

8

Salts

Self Assess 8.1 1 (a) Neutralisation reaction between NaOH and H2SO4. (b) Neutralisation reaction between NH3 and H2SO4. (c) Reaction between excess Al powder/ Al2O3/ Al2(CO3)3 and H2SO4. (d) Reaction between excess Pb powder / PbO/ PbCO3 and HNO3. (e) Reaction between excess Zn powder/ ZnO/ ZnCO3 and HCl. (f) Precipitation between Pb(NO3)2 and Na2SO4/ K2SO4/ H2SO4. (g) Precipitation between AgNO3 and NaCl /KCl/ HCl. 2 Magnesium powder (or magnesium carbonate or magnesium oxide) and nitric acid. Mg + 2HNO3 → Mg(NO3)2 + H2 3 (a) Mole ratio of Ca2+: SO42— = 1 : 1 (b) Mole ratio of Al3+: OH— = 1 : 3 4 (a) 2Ag+ + CrO42— → Ag2CrO4 (b) 0.3 mol of Pb2+ ions react with 0.6 mol of Br — ions. Answers



0.3 Hence, — — mol of Pb2+ ions 0.3 0.6 reacts with —— mol of Br – ions. 0.3

1 mol of Pb2+ ions reacts with 2 mol of Br — ions. Pb2+ + 2Br — → PbBr2 5 Number of moles of Ba2+ ions 5 = 0.2  — ——— = 0.001 1000

Number of moles of CrO42— ions = 0.1  1000 —10 —— = 0.001 0.001 mol of Ba2+ ions react with 0.001 mol of CrO42— ions. Hence, 1 mol of Ba2+ ions react with 1 mol of CrO42— ions. Ba2+ + CrO42— → BaCrO4



5 Add aqueous ammonia a little at a time until in excess to the salt solution. If a white precipitate that dissolves in excess aqueous ammonia is formed, the salt is zinc nitrate. If no precipitate is formed, the salt is calcium nitrate. Add potassium iodide solution to the salt solution. If a yellow precipitate is formed, the salt is lead(II) nitrate. 6 Test

Inference

1

Zn , Al or Pb2+ ions may be present.

2

Zn2+ ions may be present.

3

NO2 gas and O2 gas are evolved. NO3— ions may be present. The residue may be ZnO, Zn2+ ion may be present.

4

SO42— ions are not present.

2+

0.5  10 —— ———— 1000

6 Number of moles of Ag + ions = = 0.005 Number of moles of CO32— ions 0.5  5 = —— ———— = 0.0025 1000 0.005 mol Ag + react with 0.0025 mol of CO32—. Hence, 2 mol of Ag + react with 1 mol of CO32— . 2Ag + + CO32— → Ag2CO3

7 (a) 3MgO + 2H3PO4 → Mg3(PO4)2 + 3H2O (b) 3 mol of MgO produce 1 mol of Mg3(PO4)2. Hence, 3  1.2 = 3.6 mol of MgO is used to produce 1.2 mol of Mg3(PO4)2. Self Assess 8.2 1 (a) PbCl2, ZnCO3, CuCO3 (b) PbCl2, ZnSO4, ZnCO3, Na2CO3, Al2(SO4)3, Mg(NO3)2 (c) CuCl2, Cu(NO3)2 (d) ZnSO4, Al2(SO4)3 (e) ZnSO4, Al2(SO4)3 (f) Al2(SO4)3, Mg(NO3)2 2 (a) Iron(III) carbonate (b) Fe2(CO3)3 + 6HNO3 → 2Fe(NO3)3 + 3CO2 + 3H2O (c) Y decomposes to iron(III) oxide and carbon dioxide gas is evolved. (d) A blood red colour solution is formed. (e) Fe3+ + 3OH— → Fe(OH)3 3 P contains chloride ions, Q contains sulphate ions and R contains carbonate ions. 4 Add barium chloride solution to each type of the acid. If a white precipitate is formed, sulphuric acid is present. Add silver nitrate solution to each type of the acid. If a white precipitate is formed, hydrochloric acid is present. If there is no precipitate formed with either barium chloride solution or silver nitrate solution, nitric acid is present.

558

3+

Salt Z contains Zn2+ ions and NO3— ions. SPM Exam Practice 8 Multiple-choice Questions 1 C 2 B 3 B 4 B 6 C 7 C 8 B 9 C 11 D 12 D 13 B 14 B 16 C 17 C 18 A 19 C 21 C 22 A 23 A 24 A 26 C 27 D 28 C 29 B

5 D 10 C 15 D 20 D 25 A 30 D

Structured Questions 1 (a) (i) Nitric acid 1 (ii) PbO + 2HNO3 → Pb(NO3)2 + H2O 1 (b) (i) Gas B is nitrogen dioxide. 1 Solid D is lead(II) oxide. 1 (ii) Gas C will rekindle a glowing wooden splint. 1 (c) (i) Lead(II) iodide. Yellow 1 (ii) Pb2+ + 2I– → PbI2 2 (d) (i) Lead(II) hydroxide 1 (ii) The white precipitate formed dissolves in excess aqueous sodium hydroxide to produce a colourless solution. 1 (e) Lead(II) carbonate or lead metal 1 2 (a) MgO + 2HNO3 → Mg(NO3)2 + H2O 1 (b) Number of moles of HNO3 2.0  50



= ——————— = 0.1

1000

1

0.1 mol of HNO3 will produce 0.05 mol of Mg(NO3)2. 1 Mass of the salt, Mg(NO3)2 produced = 0.05  {24 + 2(14+16  3)} = 7.4 g 1 (c) (i) A white precipitate is formed, 1 insoluble in excess aqueous sodium hydroxide. 1



0.06 mol of HCl will produce 0.03 mol of salt P. 1 (ii) ZnCl2 + Na2CO3 → ZnCO3 + 2NaCl 0.03 mol P will produce 0.03 mol of ZnCO3 1

Mass of ZnCO3 produced = 0.03  125 = 3.75 g 1 (e) Heat zinc carbonate strongly, zinc carbonate will decompose to zinc oxide. 1 5 (a) Filtration 1 (b) Ammonia gas, ammonium ion 2 (c) Barium sulphate, sulphate ion 2 (d) Zinc ion 1 (e) Metal Q is zinc, gas T is hydrogen gas. 2 Essay Questions 1 (a) (i) Soluble salt

Insoluble salt

Sodium sulphate Lead(II) chloride Magnesium nitrate

3

(ii) To prepare lead(II) chloride, the reactants are lead(II) nitrate solution and sodium chloride / potassium chloride solution. 3 (b) • The unsaturated salt solution is evaporated until it is saturated. 1 • This is tested by placing a drop of the solution on a piece of glass plate. If crystals are formed, then the solution is saturated. 1 • The saturated salt solution is cooled to allow crystallisation to occur. 1 • The salt crystals formed are then filtered. 1

diagram 1 , label 1 (c) (i) The anion present is nitrate ion. 1 Chemical test: Add a little dilute sulphuric acid and iron(II) sulphate solution to about 5 cm3 of salt X solution in a test tube, 1 followed by concentrated sulphuric acid along the sides of the test tube. 1 A brown ring will be formed, thus confirming the presence of nitrate ions. 1 (ii) The 3 cations present are Al3+ ions, Pb2+ ions and Zn2+ ions. 3 Add potassium iodide solution to a little salt X solution in a test tube. 1 The formation of yellow precipitate will confirm the presence of Pb2+ ions. 1 2 (a) A salt is an ionic compound formed by replacing the hydrogen ion of an acid by a metal ion or ammonium ion. 2 (b) • Add lead metal powder a little at a time to 30 cm3 of dilute nitric acid, 1 • until the lead metal powder is in excess, with a little remaining undissolved. 1 • The reaction that occurs: Pb + 2HNO3 → H2 + Pb(NO3)2 1 • Filter off the excess undissolved lead metal powder, keep the filtrate. 1 • 30 cm3 of sodium carbonate solution is added to the filtrate. 1

559

• The reaction that occurs: Na2CO3 + Pb(NO3)2 → PbCO3 + 2NaNO3 1 • The precipitate formed is lead(II) carbonate salt. • Filter the mixture to obtain the precipitate. 1 • The residue is rinsed with distilled water and dried between filter papers. 1 (c) • Add ammonia solution to 2 cm3 of solution X in a test tube, a little at a time until the ammonia solution is in excess. 1 • Repeat the test on solution Y followed by solution Z to replace solution X. 1 • If a white precipitate that does not dissolve in excess ammonia is formed, the solution is magnesium sulphate. 2 • If a white precipitate that dissolves in excess ammonia is formed, the solution is zinc sulphate or zinc nitrate. 2 • Add a little barium nitrate solution to 2 cm3 of the solutions in separate test tubes. 1 • If the solution produces a white precipitate that dissolves in excess ammonia and also a white precipitate with barium nitrate solution, the solution is zinc sulphate. 1 • If the solution produces a white precipitate that dissolves in 1 excess ammonia but does not produce any precipitate with barium nitrate solution, the solution is zinc nitrate. 1 Experiment 1 (a) 50.00 cm3 burette 3 (b) Manipulated variable Volume of potassium chromate(VI) solution Responding variable Height of precipitate Constant variable Volumes of barium nitrate solution and size of test tubes 3 (c)

3 (d) (i) 5.0 cm3 (ii) Number of moles of Ba2+ ion 0.5  5.0 = —————————— = 0.0025 3 1000 Answers

F O R M 4

CHAPTER 8

(ii) Magnesium nitrate 1 (d) (i) Gas E is nitrogen dioxide. Solid C is magnesium oxide. 2 (ii) Gas D will rekindle a glowing wooden splint. 1 3 (a) (i) Carbon dioxide 1 (ii) Carbonate ion, CO32– 1 (b) Zinc ion, Zn2+ 1 (c) ZnCO3 → ZnO + CO2 1 (d) Yellow when hot, white when cold. 1 (e) (i) ZnO + 2HCl → H2O + ZnCl2 1 (ii) Residue B dissolves in HCl effervescence occurs 1 (f) (i) Zinc(II) hydroxide 1 (ii) Zn2+ + 2OH– → Zn(OH)2 1 (g) A white precipitate is formed which is soluble in excess sodium hydroxide solution. 1 4 (a) ZnO + 2HCl → ZnCl2 + H2O 1 (b) Hydrochloric acid is heated in a beaker. Add zinc oxide to hydrochloric acid a little at a time until a little of solid zinc oxide is in excess. Filter the mixture and 1 the filtrate is salt solution P. 1 (c) (i) Sodium carbonate solution 1 (ii) Double decomposition 1 (iii) Zn2+ + CO32— → ZnCO3 1 (d) (i) Number of moles of HCl 30 = 2.0  ————— = 0.06 1 1000 ZnO + 2HCl → ZnCl2 + H2O

(e) From the formula BaCrO4, 0.0025 mol of Ba2+ ions will precipitate with 0.0025 mol of CrO42– ions. Molarity of CrO42– ion 1000 = 0.0025 mol  ————— dm–3 5 = 0.5 mol dm–3 3 (f) All the barium ions have been precipitated. The chromate ions are in excess. 3 9

F O R M

CHAPTER 8 & 9

4

Self Assess 9.3 1 (a) An alloy is a mixture of two or more elements with a fixed composition of which the major component is a metal. (b) (i) Carbon (ii) Chromium and nickel (c) (i)

Manufactured Substances in Industry

Self Assess 9.1 1 (a) (i) Ammonium sulphate, 2NH3­ + H2SO4 → (NH4)2SO4 (ii) Potassium sulphate, 2KOH + H2SO4 → K2SO4 + 2H2O (b) H2SO4 + Ba(OH)2 → BaSO4 + 2H2O 2 To make fertilisers, detergents and white paint pigment 3 (a) To absorb water (b) (i) Carbon (ii) C6H12O6 → 6C + 6H2O 4 (a) Contact process (b) Gas X is sulphur dioxide, gas Y is sulphur trioxide and liquid Z is oleum. (c) Step 1 : S + O2 → SO2 Step 2 : 2SO2 + O2 2SO3 (d) Temperature of 450 °C – 550 °C, pressure of one atmosphere and vanadium(V) oxide as the catalyst. (e) A lot of heat will be evolved when gas Y dissolves in water, this vaporises the sulphuric acid produced as acid mist. Self Assess 9.2 1 (a) Nitrogen gas and hydrogen gas (b) Iron powder (c) Temperature of 450 °C – 550 °C; pressure of 200 – 500 atmospheres (d) N2 + 3H2 2NH3 (e) To make fertilisers and nitric acid 2 (a) Gas A is nitrogen gas, gas B is ammonia, acid C is nitric acid (b) Haber process (c) Air (d) White fumes are formed. (e) NH3 + HNO3 → NH4NO3 (f) Relative molecular mass of NH4NO3 = 14 + 4(1) + 14 + 3(16) = 80 1 mol of NH4NO3 consists of 2 mol-atoms of nitrogen. Percentage of nitrogen in 1 mol of 2(14) NH4NO3 = ———— 80

Answers

= 35%

 100%

• Reduce, reuse and recycle • Produce biodegradable polymers Self Assess 9.5 1 (a) Silicon dioxide or silica (b) Sand (c) Fused glass, soda lime glass, borosilicate glass and lead glass 2 (a) Silicon dioxide (b) Similarities: Hard, brittle, inert to chemicals, withstand compression, poor conductors of heat and insulators of electricity. Differences: Glass



2 3

(ii)

(d) Stainless steel does not rust, whereas iron rusts. (a) Tin, lead and antimony (b) Harder, stronger and more shiny when polished (c) To make decorative ornaments (a) Copper alloys are stronger, harder and more resistant to tarnish than pure copper. (b) Bronze and brass (c) Pure copper metal contains atoms of the same size arranged in a regular and organised manner. Pure copper is soft and ductile because the orderly arrangement of atoms enables the layers of atoms to slide over each other easily when an external force is applied. There are empty spaces in the structures of pure copper due to imperfections. Pure copper is malleable because groups of metal atoms may glide into new positions in these empty spaces when pressed.

Self Assess 9.4 1 (a) A polymer is a large molecule made up of many smaller and identical repeating units, joined together by covalents bonds. (b) Cotton and silk (c) Nylon and Terylene 2 Ethene

Polythene

Chloroethene

Polychloroethene or PVC

Propene

Polypropene

Styrene

Polystyrene

3 Nonbiodegradable and produces poisonous gas when burnt

560

Ceramics

• Transparent

• Opaque

• Becomes soft when heated

• Withstand heating

• Impermeable

• Usually porous unless glazed

(c)

• To make lenses • To make laboratory apparatus • To make light bulbs • To make mirrors and window panes • To make cooking utensils (d) In making cement, in making bricks 3 Differences

Soda lime glass

Borosilicate glass

(a) Composition Sodium Boron oxide oxide, calcium oxide (b) Property

Does not withstand heating

Withstand heating

Self Assess 9.6 1 (a) Composite material is a structural material made by combining two or more materials with different properties into a complex mixture. (b) Composite materials Reinforced concrete

Uses Construction of framework of high rise buildings

Superconductor To make stronger and lighter electromagnets Fibre optic

Transmit data in telecommunications

Fibreglass

To make boats, fishing rods and containers

Photochromic glass

To make light sensitive optical lens and camera lens

ultraviolet 1 AgBr → Ag + — Br2 2 When the ultraviolet ray intensity decreases, silver atoms and bromine gas recombine to form silver bromide.

1 Ag + — Br2 → AgBr 2

SPM Exam Practice 9 Multiple-choice Questions 1 B 2 D 3 B 4 B 6 B 7 D 8 D 9 C 11 D 12 A 13 A 14 B 16 D 17 D 18 D 19 A 21 D 22 C 23 B 24 D 26 A 27 B 28 D 29 C 31 A 32 C 33 C 34 B 36 D 37 D

5 C 10 C 15 D 20 A 25 B 30 B 35 B

Structured Questions 1 (a) (i) Sulphur dioxide 1 (ii) S + O2 → SO2 1 (b) (i) Sulphur trioxide 1 (ii) Temperature of 450°C – 550°C, one atmosphere pressure and in the presence of vanadium(V) oxide as a catalyst. 2 (c) (i) Oleum 1 (ii) Gas Z (or sulphur trioxide) is dissolved in concentrated sulphuric acid. 1 (d) A lot of heat will be evolved when sulphur trioxide gas is dissolved in water. This will vaporise the sulphuric acid produced into fine droplets as acid mist. 1

(e) To make fertilisers and to make paint pigments/ detergent/ as an electrolyte in lead-acid accumulator 2 2 (a) Process P is the Contact process. 1 Process Q is the Haber process. 1 (b) Temperature of 450 – 550 °C 1 Pressure of 200 – 500 atmospheres. 1 Presence of iron powder as catalyst 1 (c) (i) Ammonium sulphate 1 (ii) H2SO4 + 2NH3 → (NH4)2 SO4 1 (d) 0.4 mol of NH3 is required to react with 0.2 mol of sulphuric acid. 1 Mass of ammonia = 0.4 3 17 = 6.8 g 1 (e) Urea 1 3 (a) Temperature of 450 °C – 550 °C, 1 pressure of 200 – 500 1 atmospheres and in the presence of iron powder as a catalyst. 1 (b) (i) Ostwald process 1 (ii) Platinum 1 (c) Physical property: Pungent smell/ very soluble in water. 1 Chemical property: Dissolves in water to produce an alkaline solution/forms white fumes with hydrogen chloride gas. 1 (d) Nitric acid 1 1 NH3­ + HNO3 → NH­4NO3 (e) (i) Potassium hydroxide 1 (ii) KOH + HNO3 → KNO3 + H2O 1 4 (a) (i) Silicon dioxide 1 (ii) Sand 1 (iii) Boron oxide 1 (b) (i) Polypropene 1 (ii) Lighter / does not rust 1 (c) (i) Magnesium metal 1 (ii) Magnesium atoms disrupt the orderly arrangement of aluminium atoms. Hence 1 layers of aluminium atoms are prevented from sliding over each other easily. 1 (d) Z is reinforced concrete. 1 (e) (i) T is a composite material. 1 (ii) It is sensitive to light intensity (darkens when light intensity is high, becomes clear when light intensity is low). 1 5 (a) An alloy is a mixture of two or more elements in which the major component is a metal. 1 (b) (i) Zinc 1 (ii) Tin 1 (c) Copper atoms 1

561

(d) (i)

1 (ii)

1 (e) (i) Aluminium, copper, magnesium 1 (ii) Stronger, harder and does not tarnish easily compared to aluminium. 1 (iii) In the making of aircraft body 1 6 (a) Polymerisation 1 (b) (i) Ethene 1 (ii) To make plastic bag / containers / toys 1 H H   (c) (i) H – C = C – Cl 1 (ii) Polyvinyl chloride is lighter 1 and does not rust. 1 (iii) It is not biodegradable. It 1 produces poisonous gas when burnt. 1 (d) Petroleum 1 Essay Questions 1 (a) A polymer is a big molecule formed by many small molecules (monomers) joined together by covalent bonds. For example, in polythene, many molecules of ethene are joined by covalent bonds to form the big molecule called polythene. 4 (b) (i) Three examples of natural polymers are: Natural rubber: to make car tyres Protein: as a class of food Carbohydrate: as a class of food 3 (ii) Three examples of synthetic polymers are: Polythene: to make plastic bags or bottles Polyvinyl chloride (PVC): to make water pipes Polystyrene: to make packaging materials 3 (c) Sulphuric acid is manufactured by the Contact process in industry. 1 Answers

F O R M 4

CHAPTER 9

2 (a) Fibreglass is a composite material made by embedding glass fibre filaments in plastic. Fibreglass is lighter, stronger and tougher than glass and is not brittle. It is more resilient and flexible than plastic and is not flammable. (b) Reinforced concrete is a composite material made by adding a concrete mixture of cement, water, sand and small stones into a frame of steel bars or wire netting. It is strong, hard, can withstand compression and pressure. It does not rust like iron. 3 Lenses of spectacles made of photochromic glass will darken when the light intensity is high and becomes clear when the light intensity is low. This eliminates the necessity for a separate pair of sunglasses. Photochromic glass is produced when a dispersion of silver chloride, AgCl or silver bromide, AgBr is added to normal glass. When exposed to ultraviolet light, AgCl or AgBr decomposes to form silver and halogen atoms. The fine silver deposited in the glass is black and the glass is darkened.

F O R M

CHAPTER 9

4

• Sulphur is burned in air to produce sulphur dioxide gas. S + O2 → SO2 2 • Sulphur dioxide gas is converted to sulphur trioxide gas by excess air in the presence of vanadium (V) oxide as a catalyst, 450 – 550 °C temperature and one atmosphere pressure. 2SO2 + O2 2SO3 3 • Sulphur trioxide gas is dissolved in concentrated sulphuric acid to produce oleum. SO3 + H2SO4 → H2S2O7 2 • Oleum is diluted with an equal volume of water to produce concentrated sulphuric acid. H2S2O7 + H2O → 2H2SO4 2 2 (a) A composite material is a structural material formed by combining two or more materials with different properties into a complex mixture. • An example of composite material is reinforced concrete made from concrete and steel. • Concrete is hard but brittle, with low tensile strength and can crack under the action of bending forces. • Steel is strong with high tensile strength but can corrode and is expensive. • Reinforced concrete is formed by adding a concrete mixture of cement, water, sand and small stones into a frame of steel bars or steel wire netting and allowed to set. The structural material formed on setting combines the compressive strength of concrete and the tensile strength of steel. It can withstand very high pressures and can support heavy loads. • Reinforced concrete is used in the construction of framework for highways, bridges and high rise building. 5 • Another example of composite material is fibreglass made from plastic and glass. • Plastic is light, flexible and elastic but is weak and inflammable. • Glass is strong but is heavy, brittle and non-flexible. • Fibreglass is formed by embedding glass filaments in polyester resin (a type of plastic). The new material formed is strong, tough, resilient, flexible, impermeable to water and is not inflammable (does not catch fire easily). • Fibreglass is used in the making of boats, containers, pipes and fishing rods. 5 Answers













(b) An example of an alloy is brass, a mixture of copper and zinc. To compare the hardness of brass and copper, the following experiment is carried out. Materials Copper block, brass block, ball bearing, 1 kg weight, metre rule, retort stand with clamp, cellophane tape and thread. Procedure 1 A metre rule is clamped to a retort stand and a piece of copper block is placed on the base of the retort stand. 2 A steel ball bearing is placed on the copper block and a piece of cellophane tape is used to hold the ball bearing in place. 3 A 1 kg weight is hung at a height of 50 cm above the copper block. The weight is dropped onto the ball bearing placed on the copper block. 4 The diameter of the dent made by the ball bearing is measured to the nearest 0.5 mm. 5 The experiment is repeated three times using different areas of the surface of the copper block. 6 The average diameter of the dent is calculated. 7 The whole experiment from steps 1 to 6 is repeated using a piece of brass block in place of the copper block.

To compare the hardness of an alloy with a pure metal Results Metal block

Diameter of the dent (mm) I

II

III

Average

Copper Brass If the average diameter of the dents made by the steel ball bearing on the copper block is bigger than that on the brass block, the steel bearing has been pressed deeper onto the surface of copper metal than that on brass. This will show that, brass, a type of alloy, is harder than pure copper. 10

562

Experiments 1 (a) Brass : 3.0 mm Copper : 4.0 mm 3 (b) Type of Diameter of Block depression (mm) Brass

3.0

Copper

4.0

3 (c) Alloy is a mixture of two or more elements with a fixed composition in which the major component is a metal. 3 (d) The smaller the diameter of the depression, the harder the metal. 3 (e) Brass is harder than copper / copper is softer than brass. 3 (f) Mass of weight used, height of weight from which it is dropped, size of steel ball bearing. 3 2 (a) Problem statement How to compare the rate of rusting of an iron nail and a stainless steel nail? (b) Variables • Manipulated variable Types of nails (iron nail and stainless steel nail). • Responding variable Rate of rusting. • Constant variable Size of nails, duration for rusting and conditions of experiment (temperature, supply of water and air). (c) Hypothesis Stainless steel can resist rusting better than iron. (d) Materials Iron nail, stainless steel nail, sandpaper, 5% jelly solution and potassium hexacyanoferrate(III) solution. (e) Procedure 1 Two test tubes are half-filled with jelly solution. 2 1 cm3 of potassium hexacyanoferrate(III) solution is added to each test tube. 3 An iron nail and a stainless steel nail are polished with sandpaper and placed in the two test tubes separately. 4 The two test tubes are allowed to stand for 2 days after which they are examined to see if any colour change has taken place. (f) Tabulation of data Test tube

Type of nail

1

Iron nail

2

Stainless steel nail

Observation

17

1

Rate of Reaction

Self Assess 1.1 1 (a) Reactions I and IV. (b) Reactions II and III. 2 (a) • Place zinc in dilute sulphuric acid. • Using a stopwatch, record the time taken for all the zinc to dissolve. (b) Let the time taken for zinc to dissolve completely be t s. Mass of zinc dissolved = 2.0 g Units of reaction rate are g s–1. (c) Assumptions: Sulphuric acid is in excess. All the zinc (2.0 g) dissolves completely in sulphuric acid. 3 Method 1: Measure the total volume of hydrogen given off at regular time intervals. Plot the graph of volume of hydrogen released against time. Method 2: Measure the mass of the conical flask and its contents (mixture of iron and sulphuric acid) at regular time intervals. Calculate the loss of mass at regular intervals. Plot the graph of loss of mass (of the conical flask and its contents) against time. (Comments: The loss of mass of the conical flask and its contents is equal to the mass of hydrogen released.) 4 (a) 5.6 minutes (Comments: The time at which the curve becomes completely flat is the time at which the reaction is complete). (b) The slope of the graph at 1 minute is less steep than the slope at 2 minutes. Hence, the rate of reaction at 1 minute is lower than the rate at 2 minutes. (c) It is not a normal behaviour. Normally, the curve is steepest at the start because the reaction is most vigorous at the start. But as the reaction proceeds, it becomes less vigorous and the graph becomes less steep. This unusual behaviour is due to the fact that the magnesium ribbon is not cleaned and is therefore covered with a layer of magnesium oxide. The acid must first dissolve the magnesium oxide layer before it can react with the magnesium ribbon.

(d) From the graph, the total volume of H2 produced at 1 minute = 4 cm3. ∴ The average rate of reaction between magnesium and hydrochloric acid volume of hydrogen (cm3) =— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — time taken (minutes) 4 =— — — = 4 cm3 min–1 1 The average rate of reaction between hydrochloric acid and another metal = 12 cm3 min–1. Thus, the average rate of reaction between hydrochloric acid and another metal is higher. 5 Average rate of reaction between nickel and hydrochloric acid volume of hydrogen (cm3) =— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — time taken (s) 60 =— — — — — — — = 0.5 cm3 s–1 120 Average rate of reaction between zinc and hydrochloric acid 45 =— — — — — = 0.8 cm3 s–1 56 The reaction between zinc and hydrochloric acid is a faster reaction. Hence, zinc is a more reactive metal than nickel. 6 (a) The reaction stops at 70 s. total volume of H2 Average rate = — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — time 36 =— — — — — = 0.51 cm3 s–1 70 (b) After 0.5 minutes, amount reacted = 0.0030 3 0.5 3 60 = 0.09 mol Mass of Mg reacted = 0.09 3 24 = 2.16 g Mass of Mg unreacted = 4.0 – 2.16 = 1.84 g 144 7 (a) Time = 144 s = — — — — — — — = 2.4 min. 60 Since the volume of O2 gas collected at 2.4 minutes is not given in the table, we need to plot a graph of volume of O2 evolved against time and use the graph to determine this value.

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From the graph: Volume of O2 = 62 cm3 Average rate of reaction 62 =— — — — — — — = 0.43 cm3 s–1 144 or, Average rate of reaction 62 =— — — — — — — = 25.8 cm3 min–1 2.4 (b) Time taken for the overall reaction = 4 minutes (from the table given in the question) = 240 s Average rate of the overall reaction 74 =— — — — — — — = 0.31 cm3 s–1 240 (c) The average rate of reaction between 1 minute and 3 minutes (69 – 32) 37 =— — — — — — — — — — — — — — — — — — — —= — — — — — — cm3 min–1 (3 – 1) 2 = 18.5 cm3 min–1 (d) Time taken = 150 s 150 =— — — — — — — = 2.5 minutes 60 Rate of reaction at 2.5 minutes = gradient of the curve at 2.5 minutes 78 – 50 =— — — — — — — — — — — — — — — — — = 13.3 cm3 min–1, or 3.6 – 1.5 13.3 =— — — — — — — = 0.22 cm3 s–1 60 Self Assess 1.2 1 The rate of reaction of zinc powder with dilute sulphuric acid can be increased by (a) increasing the temperature of sulphuric acid, (b) increasing the concentration of sulphuric acid, (c) using a catalyst (for example, copper(II) sulphate solution). 2 (a) At 6.6 seconds (the point when the graph becomes flat) (b) The reaction stops because all the hydrochloric acid has reacted. (Comment: As stated in the question, excess calcium carbonate is used in this experiment). (c)

Answers

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CHAPTER 1

Form 5

Comments: At a lower temperature, the rate of reaction is lower. Hence, the curve is less steep. Since the reaction is slower, it takes a longer time to complete. The total volume of CO2 collected at the end is the same because the same amounts of CaCO3 and HCl are used. (d) Constant variables: Concentration and volume of hydrochloric acid, mass of calcium carbonate. 3 (a)

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5

(b) Step 1: Calculate the total number of moles of oxygen released. Experiment I: Number of moles of H2O2 used MV 1.0  100 =— — — — — — — — — —= — — — — — — — — — — — — — — — — — — — — — — — 1000 1000 = 0.1 From the equation, 1 1.0 mol of H2O2 produces — mol 2 of O2. ∴Total number of moles of O2 released 1 = —  0.1 2 = 0.05 Experiment II: Number of moles of H2O2 used MV 0.2  300 =— — — — — — — — —= — — — — — — — — — — — — — — — — — — — — — 1000 1000 = 0.06 Total number of moles of O2 released 1 = —  0.06 2 = 0.03 Step 2: Predict which reaction has a higher rate of reaction. Since the concentration of hydrogen peroxide used in Experiment I is higher than the concentration of hydrogen peroxide in Experiment II, the rate of reaction in Experiment I is higher. Thus, the graph in Experiment I is steeper than the graph in Experiment II. Answers

4 (a) Hypothesis: The higher the temperature of the reaction, the faster the reaction, that is, the higher the rate of release of carbon dioxide. Comment: The experimental results in the table below confirm the hypothesis. The decrease in mass corresponds to the mass of carbon dioxide released. Experiment

Temperature (°C)

Decrease in mass (g)

I

28

II

35

III

40

270.35 – 270.04 = 0.31 271.42 –270.01 = 1.41 268.20 –266.00 = 2.20

(b) Constant variable: Size and mass of marble, concentration and volume of hydrochloric acid. (c) There is a loss of mass due to the release of carbon dioxide gas into the atmosphere. The loss of mass = mass of carbon dioxide released = 270.35 – 270.04 = 0.31 g (d) Different beakers are used for the experiments. Although the beakers are the same size, their masses are different. 5 (a)

(b) (i) Reaction II is faster than reaction I because it is carried out at a higher temperature. The higher the temperature, the higher the rate of reaction. (ii) Reaction III is a faster reaction than reaction II because the catalyst, copper(II) sulphate, increases the rate of reaction. (c) This shows that the catalyst is the Cu2+ ion. Both the Na+ and SO42– ions are not the catalyst for the reaction.

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(d) (i) A blue precipitate will be formed, insoluble in excess sodium hydroxide. (ii) Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s) (e) A catalyst remains chemically unchanged at the end of the reaction. Self Assess 1.3 1 Only I and II will increase the rate of reaction. Explanation: I: Effect of concentration on reaction rate When the concentration of sulphuric acid increases, the number of reacting particles per unit volume also increases. As a result, the particles are closer together and they collide more frequently. Hence, the frequency of effective collisions increases and this causes the rate of reaction to increase. II: Effect of temperature on reaction rate When the temperature is increased, the particles gain energy and they move faster and collide more frequently. At a higher temperature, the number of reactant particles having the activation energy increases. As a result, there are more effective collisions and the rate of reaction increases. III: An increase in the volume of sodium thiosulphate solution will not increase the rate of reaction. IV: The sodium hydroxide solution added will neutralise the sulphuric acid. Hence, the concentration of sulphuric acid decreases and this will decrease the rate of reaction. 2 (a) Number of moles of H2O2 used 0.2 3 30 =— — — — — — — — — — — — — — — — — — = 0.0060 1000 2H2O2(aq) → 2H2O(l) + O2(g) Number of moles of O2 produced 0.0060 =— — — — — — — — — — — — — = 0.0030 2 Volume of O2 produced = 0.0030 3 24 000 = 72 cm3 (b) (i)

(ii) • The graph is steep at the start of reaction because the rate of reaction is the highest. • As the reaction proceeds, the concentration of hydrogen peroxide decreases. As a result, the graph becomes less steep because the rate of reaction decreases. • The graph becomes flat when the reaction stops, that is, when all the hydrogen peroxide has decomposed. The total volume of oxygen gas produced is 72 cm3. (c) (i) It acts as a catalyst. (ii) Manganese(IV) oxide lowers the activation energy of the reaction by providing an alternative reaction route. Thus the effective rate of collision between hydrogen peroxide molecules increases and this causes the rate of decomposition of hydrogen peroxide to increase. 3 (a) (i) The rate of reaction at the 1st minute is higher than the rate of reaction at the 2nd minute. (ii) When the reaction continues, the total surface area of calcium carbonate decreases, and the concentration of hydrochloric acid also decreases. As a result, the effective collisions that cause the reaction to occur decrease. Thus, the reaction rate decreases. (b) (i)



(ii) When the experiment is carried out at a lower temperature, • the reactant particles move slower, • the number of reactant particles that possess

the activation energy decreases. Consequently, the frequency of effective collisions decreases and the rate of reaction also decreases. 4 Negative catalysts reduce the rate of reaction by increasing the activation energy of the reaction. In the presence of a negative catalyst, the reaction occurs via a pathway with a higher activation energy. Ea = activation energy without a catalyst Ea’ = activation energy with a negative catalyst As a result, fewer particles have sufficient energy to overcome the higher activation energy (Ea’). Hence, the frequency of effective collisions decreases and this causes the rate of reaction to decrease.

SPM Exam Practice 1 Multiple-choice Questions 1 C 2 C 3 D 4 A 6 A 7 C 8 C 9 C 11 D 12 C 13 C 14 A 16 D 17 D 18 C 19 A 21 D 22 A 23 B 24 B 26 A 27 A 28 A 29 C 31 A 32 C 33 A 34 D 36 D

(iii) Both reactions have stopped. The reaction for experiment II was completed earlier than experiment I. (iv) The mass of zinc, the concentration and volume of sulphuric acid used are the same for both experiments. 4 (e) The smaller the particle size, the larger the total surface area exposed for reaction and the higher the rate of reaction. 1 (f)

2 (a) (i) C3H8 + 5O2 → 3CO2 + 4H2O 1 mol of propane needs 5 mol of O2 for complete reaction. Rate of consumption of O2 = 0.20 3 5 = 1.0 mol s–1 1 (ii) 1 mol of propane produces 3 mol of CO2. Rate of production of CO2 = 0.2 3 3 = 0.6 mol s–1 1 (b) (i) Mg(s) + 2H+(aq) →

F O R M 5

Mg2+(aq) + H2(g) 1 5 D 10 D 15 A 20 D 25 C 30 B 35 A

Structured Questions 1 (a) The size (total surface area) of reactant. 1 (b) Temperature/concentration of acid/mass of zinc (any two variables). 1 (c) (i) Hydrogen (ii) The volume of hydrogen can be measured easily. 2 (d) (i) Experiment II. (ii) The gradient of curve II is steeper than the gradient of curve I. Alternative answer: It takes a shorter time for reaction II to complete.

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(ii) The time, t1, is shorter because HCl is a strong acid but CH3COOH is a weak acid. 1 A weak acid dissociates only partially in water but a strong acid dissociates completely in water. Hence the concentration of H+ ions in Experiment I is higher. 1 The higher the concentration, the faster the reaction, the shorter the time needed to dissolve magnesium completely. 1 (c) (i) Curve X: Experiment II Curve Y: Experiment III Curve Z: Experiment I 3 (ii) The amount (in moles) of HCl used in Experiment II (curve X) is half the amount (in moles) of HCl used in Experiment I (curve Z). 1

Answers

CHAPTER 1



F O R M

3 (a) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) 1 (b) Number of moles of magnesium 0.12 = ————— = 0.005 24 From the equation, volume of hydrogen liberated = 0.005 3 24 = 0.12 dm3 = 120 cm3 2 (c) Maximum volume of hydrogen = 120 cm3. CuSO4(aq) acts as a catalyst. A catalyst increases the rate of reaction but does not affect the yield of the products. 2 (d) Volume of hydrogen produced in Experiment II = 120 cm3 120 Rate of reaction = ———— 32 1 = 3.75 cm3 s–1 (e) (i) Experiment III has the highest rate of reaction. It takes the shortest time to complete the reaction. 1 (ii)



1 (c) Rate of reaction ∝ — — — — — — — — — — — — — — — — —— — — time taken The time taken for Experiment III is longer than Experiment II because the reaction rate in Experiment III is lower than in Experiment II. This is because hydrogen peroxide used in Experiment III has a lower concentration after dilution with water. When the concentration of a reactant decreases, the number of particles per unit volume decreases. Therefore, the frequency of effective collisions decreases. This causes a lower rate of reaction. 3 (d) Increase the temperature of the hydrogen peroxide solution. 1 Fe2O3 (e) 2H2O2(aq) → 2H2O(l)+O2(g)

The time taken for experiment II is shorter than Experiment I. Hence, reaction rate in Experiment II is higher than in Experiment I. This shows that iron(III) oxide acts as a catalyst in Experiment II. A catalyst provides an alternative reaction pathway that has a lower activation energy than the original reaction. As a result, a larger number of reactant particles have sufficient energy to overcome the lower activation energy. Thus the frequency of effective collisions increases and the rate of reaction increases. 3 Answers

CHAPTER 1

Experiment

Concentration (mol dm–3)

Time (s)

A C E

0.15 0.10 0.05

65 85 105

The higher the concentration of the acid, the shorter the time taken for a fixed mass of sulphur to be precipitated. 1 As rate of reaction ∝ — — — — — — — —, time this shows that the higher the concentration of the acid, the higher the rate of reaction. 1 (c) (i) Experiments B and C or Experiments D and E. 1 (ii) Comparing Experiments B and C, the acid concentration is fixed at 0.10 mol dm–3, but the temperature varies from 30 °C to 20 °C. Comparing Experiments D and E, the acid concentration is fixed at 0.05 mol dm–3 but the temperature varies from 30 °C to 20 °C. 1 (iii) Comparing Experiments B and C or Experiments D and E show that the higher the temperature, the shorter the time taken for the cross ‘X’ to disappear from view. 1 Since rate of reaction ∝ — — — — — — —, time this shows that the higher the temperature, the higher the rate of reaction. 1 (d) At a higher temperature, the H+ and thiosulphate (S2O32–) ions have greater kinetic energy and move faster. The frequency of effective collisions between the ions increases. Hence, the rate of reaction increases. 2

1 (f)

5

2 4 (a) Average rate of reaction 30 =— — — — — 35 = 0.857 cm3 s–1 2 1 (b) Rate of reaction ∝ — — — — — — — — — — — — — — — — —— — — time taken

(iii)

2 5 (a) (i) Na2S2O3 + H2SO4 → S + Na2SO4 + SO2 + H2O (ii) S2O32- + 2H+ → S + SO2 + H2O 2 (b) (i) Experiments A, C and E. 1 (ii) In these three experiments, the temperature was kept constant at 20 °C but the acid concentration varies from 0.15 mol dm–3 to 0.05 mol dm–3. 1 6 (a) Experiment

1

2

3

4

Volume of hydrogen peroxide (cm3)

40

40

40

80

Volume of water (cm3)

40

40

40

0

Temperature (°C)

30

30

32

30

Mass of MnO2 used (g)

1.0

0

1.0

1.0

• Comparing Experiments 1 and 2 Constant variable: Concentration and temperature of hydrogen peroxide Manipulated variable: Mass of catalyst Explanation Decomposition of hydrogen peroxide does not occur because a catalyst is not used.

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Essay Questions 1 (a) (i) Average rate of reaction 480 = ———— 100 = 4.8 cm3 s–1 2 (ii) The gradient of graph I at any time is steeper than that of graph II. Hence, the rate of reaction for Experiment I is greater than that of Experiment II. 2 This is because 1 mol of H2SO4 dissociates in water to form 2 mol of H+ ions but 1 mol of HCl dissociates to form only 1 mol of H+ ions. Thus the concentration of H+ ions in Experiment I is higher. 2 At a higher concentration, the frequency of effective collisions between H+ ions and iron atoms is greater. Hence the rate of reaction is higher. 2 (b)







(ii) A rubber stopper with a delivery tube is immediately inserted into the conical flask and the stopwatch is started simultaneously. (iii) The total volume of oxygen evolved is determined from the burette readings at halfminute intervals. (iv) The experiment is repeated using 0.5 g of copper(II) oxide instead of 0.5 g of lead(IV) oxide. Tabulation of data Experiment I: The decomposition of hydrogen peroxide in the presence of PbO2

Time (minute)

1 1 1 1 1— 2— 3— 0 — 2 1 2 2 2 3 2

Burette reading (cm3)

50

Volume of 0 O2 (cm3) Experiment II: The decomposition of hydrogen peroxide in the presence of CuO



Experiment To investigate the effectiveness of lead(IV) oxide and copper(II) oxide on the rate of decomposition of hydrogen peroxide. Hypothesis Lead(IV) oxide is a more effective catalyst than copper(II) oxide. Variables Manipulated variable Type of catalyst used. Responding variable Volume of oxygen liberated at particular time intervals. Constant variables Volume and concentration of hydrogen peroxide, temperature of experiment and mass of catalyst used. Procedure (i) Using a measuring cylinder, 25 cm3 of 0.2 mol dm–3 hydrogen peroxide is measured out and poured into a conical flask. 0.5 g of lead(IV) oxide is then added to the conical flask.

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Time (minute)

1 1 1 1 1— 2— 3— 0 — 2 1 2 2 2 3 2

Burette reading (cm3)

50

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Volume of 0 O2 (cm3) Based on the experimental results obtained, two graphs of total volume of oxygen against time for the decomposition of hydrogen peroxide are plotted on the same axes.

Graph I shows the effect of decomposition of hydrogen graph II shows the effect of decomposition of hydrogen

PbO2 on the peroxide while CuO on the peroxide.

Conclusion The gradient of graph I is steeper than the gradient of graph II at any particular time. This shows that the rate of reaction Answers

CHAPTER 1

• Comparing Experiments 1 and 3 Constant variable: Concentration of hydrogen peroxide and mass of catalyst Manipulated variable: Temperature of hydrogen peroxide Explanation The total volume of oxygen released is the same for both Experiments 1 and 3. Thus, the concentration and volume of hydrogen peroxide are the same for both Experiments 1 and 3. The volume of oxygen released (Table 4) shows that the rate of reaction in Experiment 3 is higher than Experiment 1 because a higher temperature is used in Experiment 3. • Comparing Experiments 1 and 4 Constant variable: Mass of catalyst and temperature of reaction Manipulated variable: Concentration of hydrogen peroxide Explanation The total volume of oxygen released in Experiment 4 is twice the total volume of oxygen released in Experiment 1. This implies that the concentration of hydrogen peroxide used in Experiment 4 is twice the concentration of hydrogen peroxide in Experiment 1. The volume of oxygen released (Table 4) shows that the rate of reaction in Experiment 4 is higher than Experiment 1 because a higher concentration of hydrogen peroxide is used in Experiment 4. 5 (b) (i) The volume of oxygen collected is less than 18 cm3 because of the lower rate of reaction. (ii) The total volume of oxygen collected remains the same (that is, 38 cm3) because a catalyst does not affect the yield of the product obtained when the reaction is complete. 2 (c) (i) This means that the mass and chemical composition of the catalyst are the same before and after the reaction. (ii) The catalyst is weighed before the experiment. After the reaction, the reaction mixture is filtered. The catalyst is collected and dried. It is then weighed again. The mass of the catalyst remains the same after the reaction. 3

in Experiment I is higher than the rate of reaction in Experiment II. Thus, lead(IV) oxide is more effective as a catalyst for the decomposition of hydrogen peroxide. The hypothesis is accepted. 12 2 (a) The dissolving of iron in dilute hydrochloric acid is a chemical reaction. When iron dissolves in hydrochloric acid, the following reaction occurs:

refrigerator. The souring of milk is a decomposition reaction caused by bacteria. If the milk is kept at a low temperature in the fridge, the rate of decomposition is reduced and it will remain fresh for a longer period of time. 4 (d) Hydrogen peroxide decomposes slowly at room temperature to give water and oxygen.

Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)

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5

Iron powder has a much larger surface area than an equal mass of coarse iron filings. With a larger surface area, the frequency of effective collisions increases. Hence, even at room temperature, the reaction rate of iron powder with dilute hydrochloric acid is much greater than the rate of coarse iron filings with acid. This explains why iron powder dissolves readily in cold dilute hydrochloric acid at room temperature. In the case of coarse iron filings, the rate of reaction is low at room temperature. Hence, a higher temperature is required to speed up the reaction. When a substance is heated, the particles in the substance absorb energy. This causes the number of reactant particles having the activation energy to increase. As a result, the frequency of effective collisions increases, and the rate of iron dissolving in dilute hydrochloric acid increases. 7 (b) Coal is a solid fuel and thus flammable. But big pieces of coal are safe because they do not catch fire easily. In contrast, coal dust is highly explosive because it is in the finely divided state. Explosions are caused by very rapid chemical reactions. In the coal mine, the air is filled with coal dust which forms an explosive mixture with air. It only needs a spark to trigger off an explosion because the reaction between finely divided coal and oxygen is a very fast reaction. 6 (c) A reaction can be made to go slower by decreasing the temperature. Conversely, a reaction can be made to go faster by increasing the temperature. Milk will turn sour very quickly if it is exposed to the air at room temperature, but it will keep for several days in a Answers

2H2O2(aq) → 2H2O(l) + O2(g)

The human blood contains enzymes. Enzymes are biological catalysts. The enzyme in the blood catalyses the decomposition of hydrogen peroxide. Hence, adding a drop of blood suddenly speeds up the decomposition of hydrogen peroxide to produce bubbles of oxygen gas. Thus, a vigorous effervescence occurs. 3 3 (a) (i) The rate of reaction is defined as the amount of a reactant used up or the amount of a product obtained per unit time. 2 (ii) The collision theory states that for a chemical reaction to occur, the reactant particles must collide with each other. At high pressures, the nitrogen and hydrogen molecules are packed closer together. This means that the number of gaseous molecules per unit volume is increased. Consequently, the frequency of effective collisions increases and the rate of reaction also increases. 5 (b)



Hypothesis The rate of reaction between hydrochloric acid and sodium thiosulphate is proportional to the concentration of sodium thiosulphate used.

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Variables Manipulated variable Concentration of sodium thiosulphate solution. Responding variable Time taken for the cross 'X' to disappear from view. Constant variables The total volume of solution, the concentration of hydrochloric acid and the temperature of the experiment. Chemicals Sodium thiosulphate solution (0.2 mol dm–3), hydrochloric acid (1 mol dm–3) and distilled water. Apparatus Measuring cylinders, conical flask, stopwatch and white paper marked with a cross ‘X’. Procedure (i) A white piece of paper with an ink cross ‘X’ on it, 0.2 mol dm–3 sodium thiosulphate solution and 1 mol dm–3 hydrochloric acid are prepared. (ii) Using a measuring cylinder, 50 cm3 of sodium thiosulphate solution is measured out and poured into a conical flask. (iii) The conical flask is placed on the white piece of paper marked with ‘X’. (iv) Using a small measuring cylinder, 5 cm3 of 1 mol dm–3 hydrochloric acid is measured out. (v) Hydrochloric acid is quickly poured into the sodium thiosulphate solution and the mixture is shaken. (vi) The stopwatch is then started simultaneously. (vii) The cross ‘X’ is viewed from the top of the conical flask. The solution becomes cloudy because sulphur is precipitated. (viii) The time is taken as soon as the cross disappears from view. (ix) Steps (ii) to (viii) are repeated using different volumes of sodium thiosulphate solution which are diluted with distilled water as shown in the following table. Calculations The concentration of sodium thiosulphate solution after mixing

volume of Na2S2O3 (aq) = 0.2  ———————————————— mol dm–3 50

1

2

3

4

Volume of sodium thiosulphate (cm3)

50

40

30

20

10

Volume of water (cm3)

0

10

20

30

40

Total volume of solution (50 cm3) Concentration of Na2S2O3(aq) after mixing

5

50

50

50

50

50

0.20

0.16

0.12

0.08

0.04

Time taken (s) 1 ———— (s–1) Time Results 1 A graph of ————— against time concentration of sodium thiosulphate solution is plotted.

Conclusion 1 Rate ∝ ——————————— time taken Experiment

Time taken (s)

Based on the graph obtained, the rate of reaction between sodium thiosulphate solution and hydrochloric acid is directly proportional to the concentration of sodium thiosulphate used. The hypothesis is accepted. 10 (c) 20 cm3 of ethanedioic acid is added to 5 cm3 of potassium manganate(VII) solution. The time taken for the purple colour of potassium manganate(VII) to disappear (that is, for decolourisation to occur) is recorded. The experiment is repeated by adding 20 cm3 of ethanedioic acid to 5 cm3 of potassium manganate(VII) solution containing one or two drops of (i) manganese(II) sulphate solution, (ii) sodium sulphate solution. All other conditions are kept unchanged.

H2C2O4 + KMnO4

H2C2O4 + KMnO4 + MnSO4(aq)

H2C2O4 + KMnO4 + Na2SO4

t1

t2

t3

The results of the experiment show that t1 = t3 but t2 is very much less than t1 or t3. This proves that the manganese(II) ion, Mn2+ and not the sulphate ion, SO42–, increases the rate of reaction. Thus, sulphate ion does not act as a catalyst. 3 4 (a) (i) Experiment I: Sulphuric acid Experiment II: Hydrochloric acid (ii) Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) (iii)

Explanation • The energy profile diagram shows that the energy content of the products

(ZnCl2 + H2) is lower than the energy content of the reactants (Zn + 2HCl). Hence, the reaction is exothermic and the temperature of the reaction increases. 2 • The energy profile diagram also shows that the activation energy for a catalysed reaction is lower than that of an uncatalysed reaction. 1 • A catalyst lowers the activation energy by providing an alternative reaction pathway for the reaction. 1 • As a result, more reacting particles possess sufficient

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energy to overcome the lower activation energy required for effective collisions. 1 • Hence, the frequency of effective collisions increases and the rate of reaction increases. 2 (b) (i) Combustion of charcoal The rate of reaction between charcoal and oxygen in the air depends on the area of contact between them. A large block of charcoal will not catch fire easily and thus will burn slowly. If the block of charcoal is cut into smaller pieces, the total surface area of the charcoal increases and the rate of burning is higher. The larger the total surface area, that is, the smaller the size of the charcoal pieces, the higher the rate of combustion. 3 (ii) Cooking food in a pressure cooker The pressure cooker is a cooking utensil that allows the pressure inside it to become greater than the atmospheric pressure. Pressure cookers are used to speed up cooking. In the pressure cooker, the high pressure causes the water to boil at a temperature above 100 °C or the cooking oil to boil above its boiling point. Furthermore, an increase in pressure causes the number of water vapour or cooking oil molecules that are in contact with the food to increase. Thus, food cooks faster in a pressure cooker than at normal atmospheric pressure. Faster cooking preserves the essential vitamins and other nutrients in the food. 3 Experiments 1 (a) t1 = 52 s t4 = 36 s t2 = 46 s t5 = 32 s t3 = 38 s 3 (b) Manipulated variable The temperature of sodium thiosulphate Responding variable Time taken for the ‘X’ sign to disappear Constant variable The concentrations and volumes of both sodium thiosulphate solution and dilute sulphuric acid 3 Answers

F O R M 5

CHAPTER 1

Experiment

(c) Temperature (°C)

25

30

35

40

45

Time (s)

52

46

38

36

32

1 ————— (s–1) Time

(c) The volume of gas evolved decreases as the reaction proceeds. Thus it can be concluded that rate of reaction decreases as the reaction proceeds. 3

0.019 0.022 0.026 0.028 0.031

2 (d) (i)

2

Carbon Compounds

Self Assess 2.1 1 Organic compounds Rubber, sugar, vinegar, polyvinylchloride, urea

F O R M

CHAPTER 1 & 2

5 4 1 (ii) The rate of reaction  ————— . time The graph shows that the rate of reaction increases as the tem­perature of sodium thiosulphate solution increases. 3 1 (e) At 50 °C, ————— = 0.0345 s–1 time Time = 29 s 3 (f) The higher the temperature, the higher the rate of reaction. Conversely, the lower the temperature, the lower the rate of reaction. 3 (g) Meat and fish decay (turn bad) rapidly at room temperature. The lower the temperature, the lower the rate of decay. Hence, meat and fish are kept in refrigerators to prevent them from turning bad. 3 2 (a) Rate of reaction is the volume of gas evolved per second. Rate of reaction volume of gas evolved (cm3) =— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — time taken (s) 3 (b) At 30 s Burette reading = 28.10 cm3 At 40 s Burette reading = 24.70 cm3 3 Answers

Inorganic compounds Limestone, carbon dioxide, calcium carbonate, sand, sodium chloride, ammonium sulphate

2 (a) Hydrocarbons are organic compounds that contain the elements carbon and hydrogen only. (b) Three sources of hydrocarbon are petroleum, natural gas and coal. 3 Two main products formed are carbon dioxide and water. Rubber is an organic compound containing carbon and hydrogen. When rubber is burnt in excess air, the carbon combines with oxygen to form carbon dioxide and the hydrogen combines with oxygen to form water. Self Assess 2.2 1 (a) Propane, C3H8 (b) Pentane, C5H12 (c) Hexane, C6H14 2 (a) C8H18 (b) 2C8H18 + 25O2 → 18H2O + 16CO2 3 (a) The presence of ultraviolet light or sunlight (b) Substitution reaction (c) C2H6 + Cl2 → C2H5Cl + HCl 4 The general formula of a saturated hydrocarbon is CnH2n+2. CnH2n+2 = 58 12n + 2n+2 = 58 56 n = ——— = 4 14 X is C4H10, butane. Self Assess 2.3 1 (a) CnH2n, where n = 2, 3, 4… (b) (i) C3H6 (ii) C5H10 (iii) C7H14 2 (a) Both have low boiling points and both are insoluble in water but soluble in organic solvents. (b) Both will undergo complete combustion in excess oxygen to produce carbon dioxide and water.

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3 (a) C3H6 + Cl2 → C3H6Cl2 (b) C3H6 + H2O → C3H7OH (c) C3H6 + H2 → C3H8 (d) 2C3H6 + 9O2 → 6CO2 + 6H2O (e) C3H6 + H2O + [O] → C3H6(OH)2 Self Assess 2.4 1 (a) 2-methylbut-2-ene (b) 3,4-dimethylpent-1-ene (c) 2,4-dimethylpent-2-ene 2 C5H10 has five isomers.

H H H H H | | | | | H—C=C—C—C—C—H | | | H H H pent-1-ene

H H H H H | | | | | H—C—C=C—C—C—H | | | H H H pent-2-ene H | H—C—H H H H | | | H—C=C—C—C—H | | H H 2-methylbut-1-ene H | H H H — C — H H | | | | H  C = C  C  C  H | | H H 3-methylbut-1-ene H | H—C—H H H H | | | H—C—C=C—C—H | | H H 2-methylbut-2-ene 3

Cl H | | H—C—C—H | | Cl H 1,1-dichloroethane and



H H | | H—C—C—H | | Cl Cl 1,2-dichloroethane 1,2-dichloroethane is formed when ethene reacts with chlorine in addition reaction.



H H H H | | | | H—C—C—C—C—H | | | | H O H H | H butan-2-ol H | H—C—H H H | | H—C—C—C—H | | | O H H | H 2-methylpropan-1-ol H | H—C—H H H | | H—C—C—C—H | | | H O H | H 2-methylpropan-2-ol (d) C4H9OH + 2[O] → C3H7COOH + H2O 2 (a) Compound P is ethanol, C2H5OH. Compound Q is ethanoic acid, CH3COOH. Compound R and compound S are carbon dioxide and water. Compound T is ethene, C2H4. 3 (a) Dehydration, W is C3H7OH. (b) Hydration, X is C3H6. (c) Oxidation, Y is C3H7COOH. (d) Combustion, Z is C4H9OH. Self Assess 2.6 1 (a) (CH2O)n = 60 (12 + 2 + 16)n = 60 60 n = ——— = 2 30 Molecular formula of X is C2H4O2 or CH3COOH. Structural formula of X is H O | i H—C—C—O—H | H

(b) The functional group of X is the carboxyl group, –COOH. The general formula is CnH2n+1 COOH, where n = 0, 1, 2, 3… (c) 2CH3COOH + CaCO3 → (CH3COO)2Ca + CO2 + H2O Effervescence occurs and a gas that turns limewater cloudy is produced. 2 (a) Pentanoic acid, C4H9COOH (b) C4H9COOH + NaOH → C4H9COONa + H2O 2C4H9COOH + Mg → (C4H9COO)2Mg + H2 3 (a) Compound Q is propanoic acid, C2H5COOH. Compound R is propyl propanoate, C2H5COOC3H7. (b) Reflux the reagents in the presence of concentrated sulphuric acid as a catalyst. (c) Step I is oxidation. Step II is esterification. Self Assess 2.7 1 (a) Methyl methanoate, methanol and methanoic acid (b) Propyl ethanoate, propanol and ethanoic acid (c) Methyl butanoate, methanol and butanoic acid 2 (a) Methyl propanoate, O i CH3 — CH2 — C — O — CH3 (b) Propyl ethanoate, O i CH3 — C — O — CH2 — CH2 — CH3 (c) Propyl methanoate, O i H — C — O — CH2 — CH2 — CH3 3 (a) CH3OH + C3H7COOH → C3H7COOCH3 + H2O (b) Reflux methanol and butanoic acid in the presence of concentrated sulphuric acid as a catalyst. (c) Has a sweet fruity smell, insoluble in water. Self Assess 2.8 1 Similarity: (a) Both molecular structures have the carboxylate group, –COO– as the functional group. (b) Both are insouble in water but soluble in organic solvents. Difference: (a) Fats do not have carbon-carbon double bond, C = C in their structures whereas oils have. (b) Fats have higher melting points / exist as solids at room

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temperature whereas oils have lower melting points / exist as liquids at room temperature. 2 (a) Saturated fatty acids are longchained carboxylic acids that do not contain carbon-carbon double bonds. Unsaturated fatty acids are long-chained carboxylic acids that contain carbon-carbon double bonds. (b) Shake 5 cm3 of each type of oil with 1 cm3 of bromine dissolved in trichloromethane (or acidified potassium manganate(VII)). Oleic acid will decolourise the brown colour of bromine (or the purple colour of potassium manganate(VII)) whereas there is no noticeable change in stearic acid. 3 (a) Hydrogenation (b) Heat with nickel as a catalyst (c) Palm oil has a lower melting point than margarine. Self Assess 2.9 1 Natural polymer

Monomer

Protein

Amino acid

Rubber

Isoprene

Carbohydrate

Glucose

Starch

Glucose

2 (a) Acids such as methanoic acid or ethanoic acid; the liquid latex is changed to solid rubber. (b) Sulphur of sulphur compounds; the rubber becomes harder, stronger, more elastic and more resistant towards oxidation and heat. 3 The organic acids produced by bacteria present in latex neutralise the negative charges on the membranes of the latex colloidal particles. Coagulation occurs when the membranes of the latex particles break during collision. To prevent this, aqueous ammonia is added to neutralise the organic acids produced by the bacteria.

SPM Exam Practice 2 Multiple-choice Questions 1 B 2 A 3 A 4 A 6 D 7 B 8 A 9 D 11 C 12 C 13 B 14 D 16 C 17 D 18 C 19 D 21 B 22 C 23 B 24 A 26 B 27 C 28 C 29 B 31 A 32 C 33 D 34 D

5 B 10 B 15 C 20 A 25 A 30 D 35 C

Answers

F O R M 5

CHAPTER 2

Self Assess 2.5 1 (a) The hydroxyl (–OH) group (b) CnH2n+1OH where n = 1, 2, 3… (c) C4H9OH has four isomers. H H H H | | | | H—C—C—C—C—H | | | | O H H H | H butan-1-ol

F O R M

CHAPTER 2

5

Structured Questions 1 (a) Alkane, CnH2n+2, n = 1, 2, 3… 1 + 1 (b) Propane, C3H8 1 Butane, C4H10 1 H | (c) H — C — H | H 1 (d) (i) Gas 1 (ii) Liquid 1 1 (e) Hexane, C6H14 (f) 2C2H6 + 7O2 → 4CO2 + 6H2O 1 2 (a) Alkene 1 (b) Hydrogenation 1 H H H | | | (c) nH — C = C — C — H | H propene H CH3 polymerisation | | → – C — C  | | H H n polypropene 1 (d) (i) Heat propene vapour with steam over phosphoric(V) acid as a catalyst 1 + 1 H H H | | | (ii) H — C — C — C — H | | | O H H | H 1 (e) (i) Purple colour changes to colourless 1 (ii) C3H6 + H2O + [O] → C3H6(OH)2 1 (f) The quantity of soot produced by the combustion of propene is more than that by propane. This is because the percentage of carbon by weight in propene is more than in propane. 1 + 1 3 (a) For complete reaction between ethanol and ethanoic acid to take place / to prevent the vapour of ethanol and ethanoic acid to escape into the atmosphere. 1 (b) As a catalyst (or as a dehydrating agent) 1 (c) (i) Esterification 1 (ii) CH3COOH + C2H5OH → CH3COOC2H5 + H2O 1 (d) (i) Methyl ethanoate 1 O i (ii) CH3 — C — O — CH3 1 (iii) Has a sweet / fragrant / fruity smell 1 Answers

(e) (i) C2H4 + H2O → C2H5OH 1 (ii) C2H5OH + 2[O] → CH3COOH + H2O 1 (iii) Acidified potassium dichromate(VI) / potassium manganate(VII) 1 4 (a) (i) Fermentation 1 (ii) H H | | H — C — C — H | | O H | H 1 1 (b) (i) Oxidation (ii) C2H5OH + 2[O] → CH3COOH + H2O 1 (c) Add magnesium powder / magnesium carbonate powder to liquid A and liquid B in two separate test tubes. Effervescence occurs / gas bubbles form in liquid B but no noticeable change occurs in liquid A. 1 + 1 (d) (i) Ester 1 (ii) Ethyl ethanoate 1 (iii) As a solvent 1 (e) Hydration of ethene 1 5 (a) Process I is coagulation. Process II is vulcanisation. 1 + 1 (b) Ethanoic acid / methanoic acid 1 (c) Vulcanised rubber 1 (d) Compound X is stronger / harder / more elastic / more resistant to heat / more resistant to oxidation than natural rubber 1 + 1 (any 2) (e) Heat natural rubber with sulphur 1 (f) (i) Aqueous ammonia 1 (ii) The hydroxide ions of aqueous ammonia neutralise the organic acids produced by the bacteria in latex. 1 (g) To make tyres / shoe soles / mattress / balloons 1 Essay Questions H H H H | | | | 1 (a) H — C — C = C — C — H | | H H but-2-ene H | H—C—H H H | | H—C=C—C—H | H 2-methylpropene

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1 1

1 1

(b) (i) Element

C

H

O

Mass in 100 g

40

6.7

53.3

Mol

40 6.7 53.3 ——— ——— ————— 12 1 16 = 3.3 = 6.7 = 3.3 1 Mol ratio

1

2

1

1

The empirical formula is CH2O. 1 (CH2O)n = 60 (12 + 2 + 16)n = 60 n = 2 1 The molecular formula of carbon compound X is C2H4O2. 1 (ii) The structural formula of X is H O | i H—C—C—O—H | H 1 (iii) The name is ethanoic acid. 1 (iv) The general formula is CnH2n+1COOH, n = 0, 1, 2… 1 (c) Structures Similarity: Both margarine and palm oil are esters / they have –COO– as the functional group 1 Differences: Margarine is saturated / has covalent single bonds 1 Palm oil is unsaturated / has covalent double bonds 1 Physical properties Similarity: Both are insoluble in water / non-conductors of electricity 1 Differences: Margarine is a solid / has a higher melting point 1 Palm oil is a liquid / has a lower melting point 1 Palm oil can be converted to margarine by hydrogenation. 1 Hydrogen gas and palm oil is passed over heated nickel at 200 °C 1 2 (a) To prepare ethanoic acid from ethanol Materials Ethanol, concentrated sulphuric acid, potassium dichromate(VI) 1 Apparatus Boiling tubes, delivery tube with stopper, test tube, Bunsen burner, test tube holder, blue litmus paper 1

6 Equation C2H5OH + CH3COOH → CH3COOC2H5 + H2O 1

ethanoic acid

water

heat

Diagram + label 1 + 1 Procedure 1 A boiling tube is filled with about 5 cm3 of potassium TC A2/1 dichromate(VI) solution and 2 cm3 of concentrated sulphuric acid. About 3 cm3 of ethanol is added to the tube. 1 2 A piece of tiny porcelain is put in to promote smooth boiling. The boiling tube is closed with a stopper connected with a delivery tube. The end of the delivery tube is placed in a test tube cooled by cold water in a beaker (or The content in the boiling tube is heated and the distillate is collected in the test tube cooled by water). 1 3 Observation The potassium dichromate(VI) changes from orange to green 1 4 Test The distillate changes blue litmus to red, indicating an acid is produced. 1 5 Equation C2H5OH + 2[O] → CH3COOH + H2O 1 6 The carboxylic acid produced is ethanoic acid. 1 (b) Ester prepared Ethyl ethanoate 1 Alcohol used Ethanol 1 Carboxylic acid used Ethanoic acid 1 Apparatus Beaker, boiling tube, Bunsen burner, test tube holder 1 Procedure 1 About 3 cm3 of ethanol and 3 cm3 of ethanoic acid are put in a boiling tube. 1 2 About 1 cm3 of concentrated sulphuric acid is added to the mixture. 1 3 A piece of tiny porcelain is put in to promote smooth boiling. 1 4 The contents of the boiling tube are heated to boiling for a few minutes. 1 5 The contents are then poured into a beaker half-full of water. 1

Experiment 1 (a) Statement of the problem How to differentiate between hexane, a saturated hydrocarbon and hexene, an unsaturated hydrocarbon? 3 (b) All the variables Manipulated variable Types of hydrocarbon, hexane and hexene Responding variable Reactions towards addition reaction / bromine water / acidified potassium manganate(VII) Constant variables Quantity of hydrocarbon and quantity of reagent 3 (c) Hypothesis Hexene undergoes addition reaction whereas hexane does not undergo addition reaction / Hexene decolourises the brown colour of bromine water whereas hexane does not. 3 (d) List of materials and apparatus Hexene, hexane, acidified potassium manganate(VII) solution, bromine water, test tubes. 3 (e) Procedure 1 5 cm3 of hexene is put in a test tube. 2 1 cm3 of acidified potassium manganate(VII) solution is added to hexene. 3 The mixture is then shaken and any colour changed is recorded. 4 The experiment is repeated using hexane to replace hexene. 5 Steps 1 to 4 are repeated using bromine water to replace the acidified potassium manganate(VII) solution. 3 (f) Tabulation of data Hydrocarbon Observation Hexene Hexane

3

3

Oxidation and Reduction

Self Assess 3.1 1 (a) Oxidised. Oxidation number of Fe increases. (b) Oxidised. Oxidation number of Zn increases.

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(c) Reduced. Loss of oxygen. (d) Oxidised. Loss of hydrogen or gain of oxygen. 2 (a) (i) Calcium (ii) Chlorine (b) Oxidising agent: chlorine Reducing agent: calcium 3 (a) Compound

Oxidation number

(i) CaCrO4

+6

(ii) HNO2

+3

(iii) NaClO3

+5

(iv) Cr2(SO4)3

+3

(v) FeCl3.6H2O

+3

(vi) K2MnO4

+6

(b) Ion (i) SO42–

(ii) CuCl43–

(iii) ClO

–

(iv) NO3–

Oxidation number +6

F O R M

+1 +1 +5

4 (a) (i) Cu2SO4 (ii) MnCl2 (iii) NiSO4 (b) (i) Chromium(III) chloride (ii) Cobalt(II) oxide (iii) Iron(III) sulphate (c) (i) Fe: +3 Al: +3 (ii) Iron(III) oxide Aluminium oxide (iii) Iron shows variable oxidation states. The Roman numeral III refers to the oxidation number of iron in the compound. Aluminium has only one oxidation state, that is, +3. Hence, the Roman numeral figure, III, is not shown in the names of aluminium compounds. 5 (a) Magnesium is oxidised. (b) (i) Loss of electrons to form Mg2+ ion in the compound, MgO. Mg(s) → Mg2+(s) + 2e—

loss of electrons (ii) Increase in oxidation number of magnesium from 0 in Mg to +2 in MgO. 6 (a) Zinc powder dissolves. The brown solution of Fe3+(aq) changes to the green solution of Fe2+(aq). Answers

5

CHAPTER 2 & 3

potassium dichromate + concentrated sulphuric acid + ethanol

(b) Zn(s) → Zn2+(aq) + 2e– Fe3+(aq) + e– → Fe2+(aq) (c) Zinc is oxidised because it has donated electrons to form Zn2+ ions. 7 (a) (i), (ii) and (v) are redox reactions. (b) Reaction (i) Reaction (ii) Reaction (v) Cl2 + 2e– → 2Cl– Fe → Fe3+ + 3e– Fe3+ + e– → Fe2+ 2Br– → Br2 + 2e– O2 + 4e– → 2O2– Sn2+ → Sn4+ (c) Oxidising agent Reducing agent (decrease in oxidation number) (increase in oxidation number) (i)

FeCl3 (Fe: +3 to +2)

SnCl2 (Sn: +2 to +4)

(ii)

Pb(NO3)2 (Pb: +2 to 0)

Zn (Zn: 0 to +2)

(iii)

Cl2 (Cl: 0 to –1)

KBr (Br: –1 to 0)

(iv)

H2O2 (O: –1 to –2)

KI (I: –1 to 0)

8 (a) Oxidising agent: potassium dichromate(VI) Reducing agent: iron(II) sulphate (b) At the negative terminal Fe2+ → Fe3+ + e–

CHAPTER 3

5

Self Assess 3.2 1 (a) A brown substance is formed on the surface of iron. (b) Iron reacts with oxygen in the air and water to form rust, Fe2O3. xH2O. 2 (a) 4Fe(s) + 3O2(g) + 2H2O(l) → 2Fe2O3.H2O(s) (b) A layer of grease can prevent iron from having contact with air (oxygen) and water. Without oxygen and water, rusting of iron cannot occur. 3 (a) Galvanised iron refers to iron that has been coated with a layer of zinc to protect it from corrosion. (b) Zinc is more electropositive than iron. Hence, zinc donates electrons more readily. When galvanised iron is exposed to air and water, zinc will corrode first. Zn(s) → Zn2+(aq) + 2e– The electrons donated by zinc will move to iron. It is, therefore, difficult for iron(II) ions to form. Hence, rusting of iron does not occur. 4 (a) (i) 2H+(aq) + 2e– → H2(g) (ii) Mg(s) → Mg2+(aq) + 2e– (b) Magnesium is more electropositive than zinc This means that magnesium donates electrons more readily than zinc to form Mg2+ ions. Hence, magnesium prevents the corrosion of zinc.

Mixture

Observation

Inference

Metal P + oxide of metal Q

Glows

P is more reactive than Q

Metal Q + oxide of metal R

Glows

Q is more reactive than R

Metal Q + oxide of metal T

Does not glow

T is more reactive than Q

Metal P + oxide of metal T

Does not glow

T is more reactive than P

Answers

(b) Metal Q does not react with the oxide of metal P because metal Q is less reactive than metal P. 2 (a) Metal X is less reactive than carbon but more reactive than hydrogen. Thus, X is zinc. C > Zn > H → reactivity decreases (b) (i) 2Al(s) + 3ZnO(s) → 3Zn(s) + Al2O3(s) (ii) Oxidising agent : ZnO Reducing agent : Al 3 Method 1: Reduction of hot lead(II) oxide by carbon in powdered form

Method 2: Reduction of lead(II) oxide by hydrogen gas

(c) At the negative terminal, oxidation process occurs. Fe2+ is oxidised to Fe3+. The oxidation number of iron increases from +2 to +3. Electrons are donated to the electrode. At the positive terminal, reduction process occurs. Cr2O72– is reduced to Cr3+. The oxidation number of chromium decreases from +6 to +3. Electrons are accepted from the electrode. (d) At the negative electrode The colour changes from green (Fe2+) to brown (Fe3+). At the positive electrode The colour changes from orange (Cr2O72–) to green (Cr3+).

Self Assess 3.3 1

(a) T > P > Q > R → reactivity decreases

2PbO(s) + C(s) → 2Pb(s) + CO2(g)

At the positive terminal Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O

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574

PbO(s) + H2(g) → Pb(s) + H2O(l) Method 3: Electrolysis of molten lead(II) oxide Molten lead is produced at the cathode Pb2+(l) + 2e– → Pb(l) … reduction Method 4: Displacement of lead from its salt solution Step 1: Prepare lead(II) nitrate solution by dissolving lead(II) oxide in dilute nitric acid. PbO(s) + 2HNO3(aq) → Pb(NO3)2(aq) + H2O(l) Step 2: Add zinc powder to lead(II) nitrate produced Zn(s) + Pb(NO3)2(aq) → Pb(s) + Zn(NO3)2(aq) Self Assess 3.4 1 (a) Electrolysis of concentrated potassium iodide solution At the anode (carbon) Anions present: I– and OH– Oxidation of iodide ions to iodine occurs because of its high concentration. Iodide ions are oxidised to form iodine and electrons are released. 2I–(aq) → I2(aq) + 2e– At the cathode (carbon) Cations present: K+ and H+ Reduction of H+ ions to hydrogen occurs. H+ ions accept electrons and are preferentially discharged to form hydrogen gas. 2H+(aq) + 2e– → H2(g)

SPM Exam Practice 3 Multiple-choice Questions 1 B 2 D 3 C 4 A 6 C 7 D 8 B 9 C 11 B 12 A 13 A 14 B 16 D 17 C 18 A 19 C 21 B 22 B 23 A 24 B 26 D 27 A 28 C 29 B 31 B 32 A 33 D 34 A 36 A 37 B 38 A 39 B

5 C 10 B 15 C 20 A 25 D 30 D 35 D 40 D

Structured Questions 1 (a) Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) 1 (b) Iron is less electropositive than magnesium. Hence there is no reaction between iron and magnesium chloride solution. Iron is more electropositive than copper. Hence it displaces copper from copper(II) nitrate solution. 2 (c) (i) Copper metal 1 (ii) A brown precipitate of copper is formed. The solution changes from blue to green. 2 (d) (i) Cu2+ ion (ii) Fe 2 (e) Cu2+ oxidises Fe to Fe2+. Hence it is an oxidising agent. Fe reduces Cu2+ to Cu. Hence it is reducing agent. 2 2 (a) (i) From +2 to 0. 1 (ii) Ammonia is oxidised. The oxidation number of nitrogen increases from –3 to 0. 1 (b) (i) An oxidising agent is a substance that brings about the oxidation of another substance but is itself reduced. 1 (ii) Oxidising agent Potassium manganate(VII) solution 1 Substance reduced The manganate(VII) ion, MnO4– 1 (iii) Electrode Y 1 (iv) Fe2+(aq) → Fe3+(aq) + e– 1 (v) Potassium nitrate solution 1 (c) At electrode X: Green to brown 1 At electrode Y: Purple to colourless 1 3 (a) (i)

Some pieces of zinc are put into the round-bottomed flask. By means of a thistle funnel, dilute sulphuric acid is added to the zinc. Hydrogen gas released is passed through a U-tube containing a drying agent such as anhydrous calcium chloride. 3

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(ii) A mixture of hydrogen and air explodes easily when burnt. In order to avoid an explosion, all the air in the apparatus must be removed before hydrogen gas is ignited. 1 (iii) Melting point: 0 °C Boiling point: 100 °C 1 (iv) The oxide of X is iron(III) oxide. 1 Reason: Iron(III) oxide is brown and can be reduced to iron which is grey in colour. (v) The oxide of Y is zinc oxide. (vi) Y > H > X 1 (vii) Hydrogen reduces the oxide of X into metal X. 1 This means that hydrogen is more reactive than the metal X. Hydrogen does not react with the oxide of Y. Hence, hydrogen is less reactive than Y. 2 (b) (i) To produce oxygen for the combustion of magnesium, carbon and zinc. 1 (ii) Element

Observation

F O R M 5

Flame/glow Residue (if any)

Magnesium Very bright flame

White powder

Zinc

The glow spreads

The residue is yellow when hot and white when cold

Carbon

Bright flame No residue

3 (iii) The powdered element (magnesium, carbon and zinc) is first heated strongly. When the powdered element has become very hot, potassium manganate(VII) is then heated strongly. The oxygen released is allowed to pass over the elements. 2 4 (a) The presence of water and oxygen. 1 (b)

3 (c) The electrons flow to the edge of the water droplet. Oxygen accepts Answers

CHAPTER 3

(b) Electrolysis of copper(II) sulphate solution At the anode (copper) Anions present: SO42– and OH– ions Both OH– and SO42– ions are not discharged. The copper anode dissolves because oxidation of copper to Cu2+ ions occurs. Cu(s) → Cu2+(aq) + 2e– At the cathode (copper) Cations present: Cu2+ and H+ Reduction of Cu2+ ions to Cu occurs. Cu2+ ions accept electrons and are reduced to form copper. Cu2+(aq) + 2e– → Cu(s) 2 (a) (i) The diagram shows that electrode X is an anode because it is connected to the positive terminal of the battery. Anions present: OH– and SO42– ions Oxidation occurs. OH– ions are oxidised to oxygen. 4OH–(aq) → O2(g) + 2H2O(l) + 4e– Electrode Y is the cathode because it is connected to the negative terminal of the battery. Cations present: H+ and Fe2+ ions Reduction occurs. Hydrogen ions are reduced to hydrogen gas. 2H+(aq) + 2e– → H2(g) Comment: Hydrogen is below iron in the electrochemical series. Hence, hydrogen ions are preferentially discharged. (ii) The electrons are transferred from electrode X to electrode Y through the external circuit. (b) In electrolysis, electrical energy is converted into chemical energy. 3 (a) Oxidising agent: NiO2 Reducing agent: Cd (b) (i) Cd + 2OH– → Cd(OH)2 + 2e– (ii) NiO2 + 2H2O + 2e– → Ni(OH)2 + 2OH– (c) From Cd electrode to NiO2 electrode.

F O R M

CHAPTER 3

5

these electrons and is reduced to hydroxide ions. O2(g) + 2H2O(l) + 4e– → 4OH– (aq) Fe2+ and OH– combine to form Fe(OH)2. Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) The iron(II) hydroxide produced is oxidised by oxygen to form Fe2O3. xH2O(rust). 3 5 (a) Metal X is magnesium. This is a displacement reaction. Copper and lead are less reactive than iron. They will not displace iron from its salt solution. Potassium will react vigorously with water and hence cannot be used. 3 (b) (i) Magnesium: Changes from 0 to +2. Iron: Changes from +3 to +2. 2 (ii) Mg → Mg2+ + 2e– Fe3+ + e– → Fe2+ 2 (c) Iron(III) chloride acts as the oxidising agent in this reaction. It oxidises Mg to Mg2+ and is itself reduced to Fe2+. 2 6 (a) (i) Oxidation and reduction 1 (ii) In cell P, electrical energy is coverted to chemical energy. In cell Q, chemical energy is converted to electrical energy. 1 (b) (i) 2Cl–(aq) → Cl2(g) + 2e– (ii) 2H+(aq) + 2e– → H2(g) 2 (c) (i) 2NaCl(aq) + 2H2O(l) → 2NaOH(aq) + H2(g) + Cl2(g) (ii) Cl– ions are oxidised to chlorine and H+ ions are reduced to hydrogen. 2 (d) (i) The copper plate is coated with a brown deposit of copper metal. (ii) The zinc plate becomes thinner. 2 (e) (i) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) 1 (ii) Zinc is oxidised to Zn2+ ions. Cu2+ ions are reduced to copper. 2 Essay Questions 1 (a) Iron can be extracted from haematite by reduction with carbon. Haematite is the iron ore that contains iron(III) oxide. Iron ore, coke (carbon) and limestone (calcium carbonate) are added at the top of the blast furnace. A blast of hot air is blown into the furnace from the bottom. Coke: It is oxidised to carbon monoxide. Answers

The coke burns in the hot air and is oxidised to carbon dioxide. C(s) + O2(g) → CO2(g) Carbon dioxide then ascends to the top part of the furnace and is reduced by coke to carbon monoxide. C(s) + CO2(g) → 2CO(g) Iron ore: Reduced to iron. The iron ore is reduced by coke and carbon monoxide to produce molten iron. Fe2O3(s) + 3C(s) → 2Fe(l) + 3CO(g) Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g) Molten iron flows to the bottom of the furnace and is collected. Limestone: To remove impurities Limestone decomposes to calcium oxide and carbon dioxide. CaCO3(s) → CaO(s) + CO2(g) The calcium oxide produced reacts with the impurities such as sand to form calcium silicate (slag). CaO + SiO2 → CaSiO3 sand slag (silica) (calcium silicate) The molten slag floats on top of the molten iron. The two layers are run off from separate tap holes and allowed to solidify. 6 (b) Metals: reducing agents Reducing agents are electron donors. Metals are reducing agents because they tend to donate electrons to form stable metal ions (cations) with the noble gas electron arrangement. For example, Na → Na+ + e– (2.8.1) (2.8) Mg → Mg2+ + 2e– (2.8.2) (2.8) Al → Al3+ + 3e– (2.8.3) (2.8) Na+, Mg2+ and Al3+ ions have the same stable electron arrangement as a neon atom (2.8). Non-metals: oxidising agents Oxidising agents are acceptors of electrons. Non-metals are oxidising agents because they accept electrons readily to form stable anions (negative ions) with the stable electron arrangement of noble gases. For example, O2 + 4e– → 2O2– (2.6) (2.8) Cl2 + 2e– → 2Cl– (2.8.7) (2.8.8)

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Oxide (O2–) ion has the same stable electron arrangement as neon (2.8). Chloride (Cl–) ion has the same stable electron arrangement as argon (2.8.8). 10 (c) Galvanising is the coating of iron or steel with zinc for protection against rusting. Even if the zinc coating is scratched, the iron below it does not rust. Rusting occurs when iron donates electrons to form Fe2+ ions. Fe(s) → Fe2+(aq) + 2e– In galvanised iron, zinc is corroded and not iron. This is because zinc is more electropositive than iron. When zinc corrodes, zinc ions are formed and electrons are released. Zn(s) → Zn2+(aq) + 2e– The electrons released will prevent iron from forming Fe2+ ions and thus prevent the rusting of iron. 4 2 (a) Experiment 1 Apparatus Wire gauze, tripod stand and Bunsen burner. Materials Magnesium powder, copper(II) oxide and asbestos paper. Procedure 1 Magnesium powder is added to copper(II) oxide and mixed uniformly. 2 The mixture is placed on an asbestos paper. 3 The asbestos paper with the mixture is placed on a wire gauze over a tripod stand. 4 The mixture of powdered magnesium and copper(II) oxide is then heated strongly.

Observation A glow appears and spreads across the surface of the mixture. Conclusion Magnesium is more reactive with oxygen than copper.

Experiment 2 Apparatus Spatula and test tube. Materials Magnesium powder and copper(II) sulphate solution. Procedure A spatula of magnesium powder is added to copper(II) sulphate solution in a test tube. The mixture is shaken gently.



Observation 1 The magnesium powder dissolves. 2 The blue colour of copper(II) sulphate changes to colourless. 3 A reddish-brown precipitate is produced. Conclusion A more reactive metal will displace a less reactive metal from its salt. Hence, the experiment shows that magnesium is more reactive than copper. Discussion 1 The brown precipitate is copper. 2 Magnesium displaces copper from copper(II) sulphate solution. Mg(s) + CuSO4(aq) → Cu(s) + MgSO4(aq) 10 (b) The reaction of iron(II) sulphate as an oxidising agent When excess zinc powder is added to iron(II) sulphate solution, zinc dissolves to form zinc sulphate. The green colour of iron(II) sulphate slowly disappears

as iron is precipitated. When excess sodium hydroxide is added to the reaction mixture, a white precipitate is formed. The precipitate dissolves in excess sodium hydroxide solution to form a colourless solution. This shows the presence of Zn2+ ions. In this reaction, iron(II) sulphate acts as the oxidising agent. Iron(II) sulphate oxidises zinc to Zn2+ ions and is itself reduced to iron (Fe). Zn(s) → Zn2+(aq) + 2e– …oxidation Fe2+(aq) + 2e– → Fe(s) …reduction The overall redox reaction between zinc and iron(II) sulphate can be represented by the equation: FeSO4(aq) + Zn(s) → Fe(s) + ZnSO4(aq) oxidising reducing agent agent The reaction of iron(II) sulphate as a reducing agent When chlorine water is added to iron(II) sulphate solution, the green colour of iron(II) sulphate changes to a brown colour. When sodium hydroxide solution is added to the reaction mixture, a brown precipitate is formed, which is insoluble in excess sodium hydroxide. This shows the presence of Fe3+(aq) ions. In this reaction, iron(II) sulphate acts as the reducing agent. It reduces chlorine to chloride ions and is itself oxidised to Fe3+ ions. Cl2(aq) + 2e– → 2Cl–(aq) … reduction 2Fe2+(aq) → 2Fe3+(aq) + 2e– … oxidation The overall redox reaction between chlorine and iron(II) sulphate can be represented by the ionic equation: 2Fe2+(aq) + Cl2(aq) → reducing oxidising agent agent 2Fe3+(aq) + 2Cl–(aq) 10 Experiments 1 (a) Copper(II) oxide reacts vigorously with carbon. (b) (i) If carbon is more reactive than the metal, a reaction will occur and the mixture will glow. Conversely, if carbon is less reactive than the metal, the mixture will not glow. 3

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(ii)

Metals more reactive Metals less reactive than carbon than carbon Magnesium Sodium

Copper Lead

2 (c) (i) The purpose is to produce oxygen gas from potassium manganate(VII). 1 (ii) Metal

Observation

Iron

• Moderately bright flame • Bright glow

Lead

• Faint flame • Moderately bright glow

Magnesium

• Very bright flame • Very bright glow

3 (d) The more reactive the metal, the brighter the flame or the glow is on the metal. 3 (e) Magnesium

Iron

Lead

Copper

3 (f) Metal

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Metal

(i) Manipulated (i) Use different variable metals for the experiment (ii) Responding variable

(ii) Observe the brightness of the flame or the glow on the metal

(iii) Constant variable

(iii) Use the same mass of potassium manganate(VII) and the same mass of each metal

3 2 (a) Problem statement Is steel more resistant to rusting than iron? (b) Hypothesis Iron rusts more readily than steel. (c) Manipulated variable Steel and iron nails Responding variable Colour change in the gelatin solution Constant variables Conditions of experiment (d) Apparatus Test tubes Materials Iron nail, steel nail, gelatin, sandpaper and potassium hexacyanoferrate(III). Answers

CHAPTER 3

Discussion 1 The experiment shows that magnesium reacts with copper(II) oxide. 2 Magnesium reduces copper(II) oxide to copper and is itself oxidised to magnesium oxide. Mg(s) + CuO(s) → Cu(s) + MgO(s)

(e) Procedure 1 The iron nail and steel nail are cleaned with sandpaper. 2 They are then placed in test tubes A and B respectively. 3 A solution of gelatin in hot water is prepared. A few drops of potassium hexacyano­ ferrate(III) are added to the hot gelatin solution. 4 The mixture is stirred and poured into test tubes A and B. 5 The test tubes are set aside for three days and examined. (f) Tabulation of results Test tube

Intensity of Inference on blue colour rate of rusting

A (iron nail) B (steel nail) 17

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CHAPTER 3 & 4

5

4

Thermochemistry

Self Assess 4.1 1 (a) Endothermic (b) Exothermic (c) Endothermic (d) Exothermic 2 (a) The temperature of the solution increases until the maximum temperature is reached. The temperature then decreases until room temperature is reached. (b)

3 (a) Exothermic (b) Characteristic

Reaction in the hot pack

(i) Temperature change

Temperature increases

(ii) Energy content of reactants and products

Energy content of products is higher than energy content of the reactants

(iii) Energy involved in bond breaking and bond forming

Energy absorbed in bond breaking is less than energy released in bond forming

Answers

4 (a) The formation of covalent bonds between nitrogen and oxygen to form NO2 molecule. (b) The breaking of covalent bonds in nitrogen molecules and oxygen molecules. (c) Total energy content in nitrogen and oxygen is less than the total energy content in NO2 molecule. The reaction between nitrogen and oxygen to form nitrogen dioxide is endothermic, that is, heat energy is absorbed when the reaction occurs. 5 (a) Number of moles of anhydrous copper(II) sulphate needed 16.5 = ————— 66 1 mol → 66 kJ = 0.25 ? mol → 16.5 kJ Relative molecular mass of CuSO4 = 64 + 32 + (4  16) = 160 Mass of CuSO4 needed = 0.25  160 = 40 g (b) Number of moles of nitrogen used 12 = ——— 24 = 0.5 Number of moles of oxygen used 24 = ——— 24 = 1.0 0.5 mol of N2 will react with 0.5 mol of O2 to form 1.0 mol of NO2. Heat absorbed 1 = ——— 3 66 2.0 = 33 kJ ∆H = +33 kJ Self Assess 4.2 1 (a) (i) 2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq) (ii) Ag+(aq) + Cl–(aq) → AgCl(s) (b) Step 1: Calculate the number of moles of AgCl precipitated Number of moles of AgNO3 used 0.1  200 = ——————————— 1 000 = 0.02 Number of moles of CaCl2 used 0.1  100 = ——————————— 1 000 = 0.01 Number of moles of Cl– ions produced = 2  0.01 = 0.02

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Ag+(aq) + Cl–(aq) → AgCl(s) 0.02 mol of Ag+ ions on mixing with 0.02 mol of Cl– ions will produce 0.02 mol of AgCl. Step 2: Calculate the heat released when 0.02 mol of AgCl is precipitated The precipitation of 1.0 mol of AgCl releases 66 kJ of heat. Heat change for the precipitation of 0.02 mol of AgCl = –0.02  66 = –1.32 kJ 2 (a) M2+(aq) + CO32–(aq) → MCO3(s) (b) Heat change (mcθ) = (100 + 100) 3 4.2 3 (30.5 – 29.5) = 840 J = 0.84 kJ (c) Number of moles of M2+ MV = ————— 1000 0.5 3 100 = ——————————— 1000 = 0.05 Number of moles of CO32– MV 1.0 3 100 = ————— = ———————————  1000 1000 = 0.10 Number of moles of MCO3 precipated = 0.05 ∆H of precipitation –0.84 = —————— 0.05 = –16.8 kJ mol–1 (d)

3 Experiment 1 Number of moles of M2+ ions 0.5  100 = —————————— 1000 = 0.05 Number of moles of Cl– ions 1.0  100 = —————————— 1000 = 0.1 Number of moles of MCl2 formed = 0.05 Total volume of solution = 200 cm3 (a) Experiment 2 Number of moles of M2+ ions 0.5  200 = —————————— 1000 = 0.1

Self Assess 4.3 1 (a) (i) Magnesium powder, lead nitrate solution. (ii) Magnesium is used in excess to ensure all the lead metal is displaced from lead nitrate. (b) (i)



(ii) Measuring cylinder, electronic balance. (c) (i) Maximum increase in temperature, volume and concentration of lead nitrate solution. (ii) Mass of zinc added.

2 Step1: Calculate the number of moles of copper displaced Number of moles of copper displaced mass = ————————————————————— relative atomic mass 2.56 = ————— = 0.04 64 Step 2: Calculate the heat energy released From the thermochemical equation given, the displacement of 1.0 mol of copper releases 210 kJ of heat energy. Heat change when 0.04 mol of copper is formed = –210  0.04 = –8.4 kJ 3 Step 1: Calculate the number of moles of copper displaced Number of moles of Cu2+ used 0.50  50 = ——————————— = 0.025 1000 Number of moles of iron used mass = ————————————————————— relative atomic mass 0.7 = ———— 56 = 0.0125 Fe(s) + Cu2+(aq) → Fe2+(aq) +Cu(s) 0.0125 0.0125 (in excess) Number of moles of Cu displaced = 0.0125 Step 2: Calculate the heat energy released in the experiment The heat of displacement = –150 kJ mol–1 Displacement of 0.0125 mol of Cu releases 0.0125  150 = 1.875 kJ of heat energy. Step 3: Calculate the maximum temperature reached ∆H (in J) = mcθ From step 2, 1.875  1000 = 50  4.2  θ ∆T = 8.9 °C Maximum temperature reached = 30 + 8.9 = 38.9 °C 4 (a) As the mass of magnesium added increases, more iron is displaced from iron(II) chloride solution. Hence more heat energy is released and the rise in temperature increases. (b) All the iron present in iron(II) chloride is displaced on the addition of 0.3 g of magnesium. (c) Step 1: Calculate the heat released in the experiment Heat released (mcθ) = 50  4.2  12 = 2520 J = 2.52 kJ

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Step 2: Calculate the number of moles of iron displaced Number of moles of Fe2+ used 0.25  50 = ——————————— 1000 = 0.0125 Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s) 0.0125 mol of Fe2+ reacts with excess magnesium to form 0.0125 mol of Fe. Step 3: Calculate the heat of displacement of iron by magnesium From steps (1) and (2), displacement of 0.0125 mol of Fe releases 2.52 kJ of heat energy. Heat released for the formation of 1.0 mol of Fe 1 = 2.52  ——————— 0.0125 = 201.6 kJ That is, heat of displacement of iron by magnesium = –201.6 kJ mol–1 (d) Number of moles of iron displaced 252 = —————— = 1.25 201.6 Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s) Number of moles of Mg needed = 1.25 Relative atomic mass of magnesium = 24.3 Mass of magnesium needed = 1.25  24.3 = 30.4 g Self Assess 4.4 1 (a) ∆H represents the heat energy evolved when 1.0 mol of H2SO4 reacts with 2.0 mol of NaOH. That is, when 2 mol of H+ ions react with 2 mol of OH– ions to form 2 mol of water molecules, 114 kJ of heat energy is released. (b) Heat of neutralisation. –114 = ————— = –57 kJ mol–1 2 (c) Heat change (mcθ) = (50 + 50) 3 4.2 3 6.5 J = 2730 J = 2.73 kJ H+(aq) + OH–(aq) → H2O(l); ∆Hneut = –57 kJ mol–1 Number of moles of OH– ion Heat change = ———————————— ∆Hneut 2.73 = ————— 57 = 0.045 Answers

F O R M 5

CHAPTER 4

Number of moles of Cl– ions 1.0  200 = —————————— 1000 = 0.2 M2+(aq) + 2Cl–(aq) → MCl2(s) 0.1 mol of M2+ ions react with 0.2 mol of Cl– ions to form 0.1 mol of MCl2. Total volume = 400 cm3 Rise in temperature is also 10 °C. (Comment: The total number of moles of MCl2 increases two times (from 0.05 mol to 0.1 mol), but the total volume of solution also increases 2 times. Hence, the temperature remains unchanged). (b) Experiment 3 Number of moles of M2+ ions 1.0  100 = ——————————— = 0.10 1000 Number of moles of Cl– ions 2.0  100 = ——————————— = 0.2 1000 Number of moles of MCl2 formed = 0.10 Total volume = 200 cm3 Rise in temperature = 2  10 °C = 20 °C (Comment: Number of moles of MCl2 precipitated is increased two times (from 0.05 mol to 0.10 mol), but the total volume of solution does not change. Thus, the temperature of the solution is increased two times).

MV 0.045 = —————— 1000 M 3 50 = ———————— 1000 M = 0.9 Concentration of NaOH = 0.9 mol dm–3 2 Number of moles of acid/alkali Reaction

Concentration  Volume (cm3) (mol dm–3) = ——————————————————————————————— 1000 (a) Reactions I and IV Reason Same number of moles of water molecules produced from strong acid and strong base reaction. Number of moles of OH– ions

Number of moles of H+ ions

I 1.0 3 50 ————————— = 0.05 HNO3 + NaOH 1000

Number of moles of H2O formed

1 3 25 ————————— = 0.025 1000

0.025

II HCl + NaOH

1.0 3 50 ————————— = 0.05 1000

1.0 3 50 ————————— = 0.05 1000

0.05

III CH3COOH + NH3

1.0 3 50 ————————— = 0.05 1000

1.0 3 25 ————————— = 0.025 1000

0.025

0.025

1.0 3 50 1.0 3 25 IV ————————— 3 2 = 0.10 ————————— = 0.025 H2SO4 + KOH 1000 1000

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CHAPTER 4

5

V CH3COOH + NaOH

1.0 3 25 ————————— = 0.025 1000

2.0 3 25 —————————— = 0.05 1000

(b) Reaction II Reason The largest number of moles of water molecules produced from strong acid and strong base reaction. (c) Reaction III Reason 0.025 mol of water produced from weak acid and weak base reaction. 3 Experiment 1 Number of moles of CH3COOH 1.0  100 = ——————————— = 0.1 1000 Number of moles of NaOH 1.0  100 = ——————————— = 0.1 1000 CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) 0.1 mol of CH3COOH reacts with 0.1 mol of NaOH to form 0.1 mol of water. Experiment 2 Number of moles of CH3COOH 0.8  100 = ——————————— = 0.08 1000 Number of moles of NaOH 0.8  100 = ——————————— = 0.08 1000 CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) 0.08 mol of CH3COOH reacts with 0.08 mol of NaOH to form 0.08 mol of water. The total volume of solution for the two experiments are the same. Thus, Answers

0.025

the formation of 0.08 mol of water will increase the temperature by 0.08 —————  5.5 = 4.4 °C 0.1 4 (a) Number of moles of NaOH used 1.0  50 = —————————— = 0.05 1000 0.05 mol of OH– ions will react with 0.05 mol of H+ ions to form 0.05 mol of H2O molecules. Formation of 0.05 mol of water releases 2.65 kJ of heat. Heat of neutralisation 1 = –2.65  ————— 0.05 = –53 kJ mol–1 (b) For weak acid/strong alkali reaction, the heat of neutralisation is less than –57 kJ mol–1. Hence, the acid, HA, is a weak acid. (c) The calculation is based on the assumption that the apparatus is well insulated and there is no heat loss to the surroundings. Self Assess 4.5 1 (a) Mass of pentane used. The volume of water heated by pentane. The initial temperature and the final temperature of water after heating. (b) The molecular formula of butane is CH3CH2CH2CH3.

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The molecular formula of propane is CH3CH2CH3. 1 H2(g) + — O2(g) → H2O(l); 2 ∆H = –286 kJ mol–1 C(s) + O2(g) → CO2(g); ∆H = –394 kJ mol–1 One molecule of butane contains one CH2 group more than one molecule of propane. Expected heat of combustion of butane = heat of combustion of propane + heat of combustion of carbon + heat of combustion of hydrogen = (–2220) + (–286) + (–394) = –2900 kJ mol–1 2 (a) H H H H H H       H – C – C – C – OH; H – C – C – C – H       H H H H OH H propan-1-ol propan-2-ol (b) The temperature rise is 18.2 °C because propan-1-ol and propan2-ol are isomeric, that is, containing the same number of carbon, hydrogen and oxygen atoms. The heat energy released depends on the molecular formula and not on the structural formula of the isomer. (c) Increase in temperature 2.28 = —————  18.2 1.14 = 36.4 °C (d) The mass of propan-1-ol decreases by two times, that is, from 1.14 g to 0.57 g, but the mass of water also decreases by two times, that is, from 500 cm3 to 250 cm3. The rise in temperature is the same, that is, 18.2 °C. (e) Step 1: Calculate the heat change when 1.14 g of propan-1-ol is burnt Heat change (mcθ) = –500  4.2  18.2 J = –38 220 J = –38.2 kJ Step 2: Calculate the number of moles of propan-1-ol used Relative molecular mass of propan-1-ol = (3  12) + 8 + 16 = 60 Number of moles of propan-1-ol used 1.14 = ————— = 0.019 60

Step 4: Calculate the value of x In the equation given, ∆H = –x kJ refers to 2 mol of propanol burnt. Hence, x = 2  2011 = 4022 3 Number of moles of heptane used 1104 1 mol → 5 520 kJ = —————— 2 mol → 1 104 kJ 5520 = 0.2 Relative molecular mass of heptane (C7H16) = (7  12) + (16  1) = 100 Mass of heptane used = 0.2  100 = 20.0 g 4 (a) Coal as fuel. Advantages: Cost per gram is cheap. Can be stored easily and safely. Disadvantages: Low fuel value. A lot of residue is produced after combustion. (b) Natural gas as fuel Advantages: High fuel value. Products of combustion do not pollute the air. Disadvantages: Explodes easily if not handled properly. It is a fossil fuel which is nonrenewable. SPM Exam Practice 4 Multiple-choice Questions 1 B 2 C 3 A 4 A 6 A 7 D 8 C 9 D 11 C 12 A 13 D 14 A 16 A 17 B 18 D 19 B 21 D 22 D 23 B 24 C 26 C 27 C 28 C 29 A 31 C 32 D 33 A 34 A 36 C 37 C

5 D 10 A 15 A 20 C 25 C 30 C 35 A

Structured Questions 1 (a) The heat of displacement is the heat evolved when 1 mol of copper is displaced by zinc from copper(II) sulphate solution. 1 (b) The polystyrene beaker is an insulator of heat. The polystyrene beaker is used to reduce the loss of heat to the surroundings. 1 (c) The highest temperature reached in Experiment 1 = 32.8 °C. Increase in temperature in Experiment 1 = 32.8 – 30.4 = 2.4 °C. The highest temperature obtained in Experiment II = 51.2 °C



Increase in temperature in Experiment II = 51.2 – 30.4 = 20.8 °C. 2 (d) (i) The blue colour of copper(II) sulphate becomes colourless. A brown precipitate of copper is produced. 2 (ii) The solution must be stirred continuously to obtain a uniform temperature. 1 (e) (i) Experiment I shows that the release of 1.0 kJ of heat energy increases the temperature by 2.4 °C. Energy change in Experiment II 20.8 = 1.0  —————— 2.4 = 8.67 kJ (ii) Relative atomic mass of zinc = 65.4 Number of moles of zinc used 3.27 = ———— 65.4 = 0.05 Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) From the equation, 1 mol of Zn displaces 1 mol of copper metal. 0.05 mol of Zn displaces 0.05 mol of Cu. (iii) Displacement of 0.05 mol of Cu releases 8.67 kJ of heat. Heat change in the displacement of 1 mol of Cu 1 = –8.67  ———— 0.05 = –173.4 kJ Heat of displacement of Cu by Zn = –173.4 kJ mol–1. 3 (f) The polystyrene container A should be used. If polystyrene A is used, the surface area of the solution exposed to the atmosphere is reduced. This will reduce the loss of heat to the surroundings. 2 2 (a) The heat of combustion of ethanol is the heat released when 1 mol of ethanol is burnt completely in excess oxygen. 1 (b) C2H5OH + 3O2 → 2CO2 + 3H2O 1 (c) (i) Mass of of ethanol used = 150.38 – 149.83 = 0.55 g Relative molecular mass of C2H5OH = 46 Number of moles of ethanol used 0.55 = ————— = 0.012 2 46

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(ii) Heat released = mcθ = 100  4.2  (61.5 – 30.0) = 13 230 J = 13.23 kJ 1 (iii) ∆H of combustion 13.23 = ————— 0.012 = 1102 kJ mol–1 1 (d) The heat of combustion obtained from the experiment is less than the theoretical value because • of the loss of heat to the surroundings, • the combustion of ethanol is incomplete (or, some of the ethanol escaped due to evaporation). NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g) 3 (a) (i) Heat of neutralisation is the heat energy released when 1 mol of H+ ions react with 1 mol of OH– ions to form 1 mol of water. 1 (ii)

F O R M 5

(b) (i) • The acid must be added quickly to the alkali to prevent heat loss to the surroundings. • The mixture must be stirred continuously to ensure that the temperature is constant. (ii) Number of moles of nitric MV acid = ————— 1000 1.0 3 50 = ————————— = 0.05 1000 Number of moles of ammonia MV 1.0 3 50 = ————— = ————————— = 0.05 1000 1000 HNO3(aq) + NH4+(aq) + OH–(aq) ammonia solution → NH4NO3(aq) + H2O(l) From the equation, the number of moles of water molecules formed = 0.05 (iii) Temperature change = 34.2 – 28.2 = 6 °C Answers

CHAPTER 4

Step 3: Calculate the heat of combustion of propan-1-ol The combustion of 0.019 mol of propan-1-ol releases 38.2 kJ of heat. Heat change when 1 mol of propan-1-ol is burnt 1 = –38.2  —————— = –2011 kJ 0.019

F O R M

CHAPTER 4

5

∆H = mcθ = –(50 + 50) 3 4.2 3 6 J = –2520 J = –2.52 kJ Formation of 0.05 mol of water molecules releases 2.52 kJ of heat. ∴ Heat evolved in the formation of 1.0 mol of water molecules 1 = 2.52 3 ————— 0.05 = 50.4 kJ ∴ Heat of neutralisation between HNO3(aq) and NH3(aq) = –50.4 kJ mol–1 (c) In Experiment I, sulphuric acid is a strong acid and potassium hydroxide is a strong alkali. Strong acids and strong alkalis are fully dissociated in aqueous solution. In Experiment II, nitric acid is a strong acid but ammonia solution is a weak alkali. A weak alkali is only partially dissociated in aqueous solution. Part of the heat energy released during neutralisation is absorbed to produce hydroxide ions. Consequently, heat of neutralisation of nitric acid and ammonia solution is less than –57 kJ mol–1. (d) Both nitric acid and hydrochloric acid are strong acids. When the volume of acid and alkali is increased by two (from 50 cm3 to 100 cm3), the number of moles of water molecules produced is also increased by two. This means that the heat released is increased by two. However, the total volume of solution also increased by two (from 100 cm3 to 200 cm3), hence the rise in temperature remains the same, that is, 6 °C. 4 (a) (i) Exothermic reaction: Experiment I. Endothermic reactions: Experiments II and III. 2 (ii) X is sodium hydroxide. Y is ammonium sulphate. Z is sodium hydrogen carbonate. 2 (iii) NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g) 1 Answers



(iv)

(b) (i) Heat change = mcθ = (25 + 25)  4.2  3 = 630 J = 0.63 kJ 1 (ii) Step 1: Calculate the number of moles of AgCl precipitated Number of moles of AgNO3 used 0.5  25 = ————————— = 0.0125 1000 Number of moles of NaCl used 0.5  25 = —————————— = 0.0125 1000 Ag+(aq) + Cl–(aq) → AgCl(s) Number of moles of AgCl precipitated = 0.0125 Step 2: Calculate the heat of precipitation of silver chloride Precipitation of 0.0125 mol of AgCl releases 0.63 kJ of heat. Precipitation of 1.0 mol of AgCl releases 1 0.63  ———————— 0.0125 = 50.4 kJ of heat Heat of precipitation of AgCl is –50.4 kJ mol–1. 2 (iii) Rise in temperature = 3.0 °C The volume of the reaction mixture remains constant, that is, 50.0 cm3. The concentrations of Ag+ ions and Cl– ions remain constant. Hence, the temperature rise remains constant, that is, 3.0 °C. 2 Essay Questions 1 (a) Procedure Using a measuring cylinder, 25 cm3 of 0.5 mol dm–3 sodium chloride solution is measured into a plastic cup. The initial temperature of sodium chloride solution is measured and recorded. Using another measuring cylinder, 25 cm3 of 0.5 mol dm–3 lead(II) nitrate is measured out. The initial temperature of lead(II) nitrate is measured and recorded. Lead(II) nitrate solution is poured quickly into the sodium chloride solution.

582

The mixture is stirred and the highest temperature is recorded. Results Initial temperature of sodium chloride solution = t1 °C Initial temperature of lead(II) nitrate solution = t2 °C Maximum temperature of the reaction mixture = t3 °C Rise in temperature t1 + t2 = t3 – (——————— ) = t °C 2 Assumptions Specific heat capacity for solution = 4.2 J g–1 °C–1 Density of solution = 1.0 g cm–3 Calculation Heat change = mcθ = –(25 + 25)  4.2  t J –y = –y J = —————— kJ 1000 Number of moles of Pb2+ ions used 0.50  25 = ——————————— 1000 = 0.0125 Number of moles of Cl– ions used 0.50  25 = ——————————— = 0.0125 1000 Pb2+(aq) + 2Cl–(aq) → PbCl2(s) The equation shows that 2 mol of Cl– ions react with 1 mol of Pb2+ ions to produce 1 mol of PbCl2. 0.0125 mol of Cl– ions reacts with 0.00625 mol of Pb2+ ions to form 0.00625 mol of PbCl2. Heat evolved on the precipitation of 0.00625 mol of PbCl2 y = —————— kJ 1000 Heat evolved on the precipitation of 1.0 mol of PbCl2 y 1 = ——————  ————————— kJ 1000 0.00625 Heat of precipitation of lead(II) chloride –y 1 = ——————  ————————— kJ mol–1 1000 0.00625 –y = —————— kJ mol–1 6.25 Energy level diagram

12

hydroxide solution. The mixture is stirred and the highest temperature is recorded. Results Initial temperature of sodium hydroxide = t1 °C Initial temperature of sulphuric acid = t2 °C Maximum temperature of reaction mixture = t3 °C Average initial temperature 1 = —  (t1 + t2) °C 2 Rise in temperature 1 = t3– —  (t1 + t2) °C = t °C 2 Assumptions Specific heat capacity of solution = 4.2 J g-1 °C–1 Density of solution = 1.0 g cm–3 Calculation Total amount of heat energy released = mcθ = (100 + 100)  4.2  t J y = y J = ————— kJ 1000 Number of moles of H2SO4 used 0.5  100 = ——————————— = 0.05 1000 Number of moles of NaOH used 1.0  100 = ——————————— = 0.1 1000 H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) From the equation, 1 mol of H2SO4 reacts with 2 mol of NaOH to produce 2 mol of water. 0.05 mol of H2SO4 reacts with 0.1 mol of NaOH to form 0.1 mol of water. Formation of 0.1 mol of water y releases —————— kJ of heat energy. 1000 Heat energy released when 1 mol of water is formed y 1 = —————  ——— kJ 1000 0.1 Heat of neutralisation for the reaction between sulphuric acid and sodium hydroxide –y 1 = —————  ——— kJ mol–1 1000 0.1 –y = ————— kJ mol–1 100 12 (b) Hydrochloric acid and nitric acid are strong acids that dissociate completely in aqueous solution. HCl(aq) → H+(aq) + Cl–(aq) HNO3(aq) → H+(aq) + NO3–(aq)

583

Sodium hydroxide and potassium hydroxide are strong alkalis that dissociate completely in aqueous solution. NaOH(aq) → Na+(aq) + OH–(aq) KOH(aq) → K+(aq) + OH–(aq) In the reaction between 1 mol of hydrochloric acid and 1 mol of sodium hydroxide or 1 mol of nitric acid and 1 mol of potassium hydroxide, the change that occurs is the combination of 1 mol of H+ ions with 1 mol of OH– ions to form 1 mol of water molecules. H+(aq) + OH–(aq) → H2O(l) Heat of neutralisation is the heat released when 1 mol of H+ ions react with 1 mol of OH– ions to form 1 mol of water. Hence, the heats of neutralisation for all reactions between a strong acid and a strong base are the same, that is, –57 kJ mol–1. 4 (c) When N2 and H2 react, the bonds between N2 and H2 molecules are broken and new N – H bonds in NH3 are formed. Bond breaking absorbs heat and bond forming releases heat. In this reaction, the heat released in bond forming is greater than the heat absorbed in bond breaking. Hence the reaction is exothermic. 4 Experiments 1 (a) Initial temperature of the mixture = 29 °C Highest temperature of the mixture = 41 °C Increase in temperature = 12 °C 3 (b) Experiment

Experiment I Experiment II (HCOOH + (HCl + NaOH) NaOH)

Initial temperature of the mixture (°C) Highest temperature of the mixture (°C) Increase in temperature (°C) 3 (c) The polystyrene cup gets warmer when reaction occurs. Answers

F O R M 5

CHAPTER 4

3 (b) CH3OH(l) + — O2(g) → 2 CO2(g) + 2H2O(l); ∆H= –715 kJ mol–1 CH3CH2OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l); ∆H = –1371 kJ mol–1 Similarities in combustion Both the alcohols burn in excess air to form carbon dioxide and water. Both the alcohols release a large amount of heat energy when burnt in excess air. Difference in heat of combustion The heat of combustion of ethanol is higher than the heat of combustion of methanol. This is because a molecule of ethanol contains one carbon atom and two hydrogen atoms (that is, the –CH2 group) more than a molecule of methanol. Hence, the combustion of 1 mol of ethanol produces 1 mol of CO2 and 1 mol of H2O more than the combustion of 1 mol of methanol. C(s) + O2(g) → CO2(g); ∆H = –x kJ mol–1 1 H2(g) + — O2(g) → H2O(l); 2 ∆H = –y kJ mol–1 Hence, the heat released on the combustion of 1 mol of ethanol increases by (1371 – 715) = 656 kJ. That is, (x + y) = 656 5 (c) (i) Relative molecular mass of propanol (C3H7OH) = 60 Fuel value Heat of combustion = ———————————————————————— Relative molecular mass 2010 = ————— 60 = 33.5 kJ g–1 (ii) The factors are high fuel value and no pollution problem (or low fuel price). 3 2 (a) Experiment • Using a measuring cylinder, 100 cm3 of 1.0 mol dm–3 sodium hydroxide solution is measured into a plastic cup. The initial temperature of the alkali is determined and recorded. • Using another measuring cylinder, 100 cm3 of 0.5 mol dm–3 sulphuric acid is measured out and its initial temperature determined. • The sulphuric acid is poured quickly into the sodium

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CHAPTER 4 & 5

5

The mercury level in the thermometer rises. The smell of methanoic acid disappears. 3 (d) The heat of neutralisation between methanoic acid and sodium hydroxide is lower than the heat of neutralisation between hydrochloric acid and sodium hydroxide. 3 (e) Concentrations of acids and sodium hydroxide solution. Volume of acids and sodium hydroxide solution. Polystyrene cup 3 (f) In Experiment I, the reaction is between a weak acid and a strong alkali. In Experiment II, the reaction is between a strong acid and a strong alkali. Therefore, the rise in temperature in Experiment II is higher than 12 °C. 3 (g) (i) The heat of neutralisation is the heat released when 1.0 mol of H+ ions react with 1.0 mol of OH– ions to produce 1.0 mol of water molecules. 3 (ii) ∆H = mcθ = –50 g 3 4.2 J g–1 °C–1 3 12 °C = – 2520 J Number of moles of methanoic acid used 2.0 3 25 = ——————————— = 0.05 1000 Number of moles of NaOH used 2.0 3 25 = ——————————— = 0.05 1000 CH3COOH + NaOH → CH3COONa + H2O Number of moles of water molecules formed = 0.05 Heat of neutralisation – 2520 = ——————— = – 50.4 kJ mol–1 0.05 2 (a) Aim To compare the fuel values of fuel X and Y. (b) Variables Manipulated variable Type of fuel. Responding variable Heat of combustion. Constant variable Volume of water and the metal container. (c) Hypothesis Fuel X has a higher heat of combustion than fuel Y. (d) Substances Water, canisters X and Y Apparatus Screen, thermometer, metal Answers

calorimeter, tripod stand, measuring cylinder (e) Procedure

1 200 cm3 of water is measured out using a measuring cylinder and poured into a metal calorimeter. 2 The initial temperature of water is read and recorded. 3 The metal calorimeter is put on a tripod stand. 4 A canister containing fuel of brand X is weighed and its mass recorded. 5 The canister with fuel of brand X is then placed below the metal calorimeter and lit. The flame is protected from air movement by putting a screen around it as shown in the diagram. 6 The water in the metal calorimeter is stirred continuously with a thermometer. 7 The flame from the burning of fuel X is extinguished when the temperature of water has gone up by 30 °C. 8 The canister containing fuel of brand X is weighed again and its mass recorded. 9 The experiment is repeated using canister of brand Y to replace canister of brand X. (f) Tabulation of data Experimental results Type of fuel

Brand X Brand Y

Initial mass of canister (g) Final mass of canister (g) Mass of fuel used (g) Mass of water in metal calorimeter (g) Initial temperature of water (°C) Final temperature of water (°C) Temperature rise (°C) 17

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5

Chemicals for Consumers

Self Assess 5.1 1 (a) Soaps are sodium or potassium salts of long chain fatty acids that contain 12 to 18 carbon atoms per molecule. (b) Animal fats or vegetable oils, concentrated sodium hydroxide or potassium hydroxide, water and sodium chloride. (c) Naturally occurring ester + 3NaOH ↓ 2C17H35COONa + C15H31COONa soap + CH2OH  CHOH  CH2OH glycerol 2 (a) The detergent ion has two parts. (i) The ‘tail’ is a long hydrocarbon chain, C12H25, that is insoluble in water but dissolves readily in grease. (ii) The ‘head’ is a negativelycharged ion, –O – SO3–, that dissolves readily in water but does not dissolve in grease. (b) When the detergent is dissolved in water, the detergent molecule dissociates into the detergent ion (negatively-charged) and the sodium ion (positively-charged). C12H25 – O – SO3–Na+ → C12H25 – O – SO3– + Na+ detergent ion • When detergent ions come into contact with grease-stained cloth, – the ‘head’ dissolves in water because it is hydrophilic, – the ‘tail’ dissolves in the greasy stain because it is hydrophobic. • On agitation, the greasy stain is lifted from the cloth. • Finally, the stain is broken up to form tiny oily drops, which are then suspended in the solution. • When rinsed with water, the cloth becomes clean, because the oily droplets are removed by water. 3 Soap Advantages (a) Effective in soft water (b) Biodegradable, do not cause pollution

Self Assess 5.2 1 (a) Food additives are chemical substances added in small quantities to food for specific purposes, such as preserving food or enhancing the flavour or smell of food. (b) Any method that prevents the growth of microorganisms or retards oxidation is an effective preservation process. Method 1: By adding preservative By adding a preservative such as sodium nitrate or benzoic acid, the growth of microorganisms such as bacteria or mould is retarded. As a result, the food can be kept longer. Method 2: By adding antioxidant By adding an antioxidant such as ascorbic acid, the oxidation of fats and oils by atmospheric oxygen can be prevented. In this way, the food does not go rancid. 2 (a) Acacia gum acts as a stabiliser to improve the texture of food so that the different ingredients (for example, oil and water) can blend (mix together) properly. For example, acacia gum is added to ice cream so that the oil and water ingredients can mix properly. In this way, tiny ice crystals do not form in the ice cream. (b) Aspartame is an artificial sweetener. It is used to enhance the sweetness of food. 3 Advantages Artificial colourings • restore colours lost during food processing, • enhance natural colours of food, and • make food look more interesting or appetising. Disadvantages • The food may not be fresh and the dyes are used to make the food look fresh. • The dyes used may not be the permitted dyes.

Eating food that are not fresh can cause food poisoning and eating food containing unapproved dyes may affect the health. Self Assess 5.3 1 (a) Both aspirin and paracetamol are analgesics, that is, medicine that relieves pains. However, aspirin is strongly acidic and can cause bleeding in the stomach. In contrast, paracetamol is neutral in nature and does not cause stomach bleeding. (b) Add aspirin and paracetamol separately to two solutions of sodium carbonate. For aspirin, effervescence occurs and the gas evolved turns limewater cloudy. This is because aspirin contains –COOH group which is acidic. For paracetamol, no effervescence is seen. This is because paracetamol does not contain the – COOH group. 2 Similarities • Both penicillin and streptomycin are organic compounds. • Both penicillin and streptomycin are antibiotics. They are used to kill bacteria so that illnesses and infections caused by bacteria can be cured. Differences Penicillin is used to treat infectious diseases such as pneumonia and gonorrhaea. Streptomycin is used to treat tuberculosis. 3 (a) Aspirin is used as a an analgesic (painkiller) to relieve pain. (b) Aspirin can cause stomach bleeding. SPM Exam Practice 5 Multiple-choice Questions 1 D 2 A 3 B 4 D 6 B 7 C 8 B 9 A 11 D 12 D 13 A 14 D 16 B 17 C 18 C 19 C 21 B 22 A 23 C 24 C 26 D 27 A 28 B 29 C

5 A 10 B 15 C 20 B 25 C 30 D

Structured Questions 1 (a) Saponification 1 (b) (i) Potassium palmitate 1 (ii) Ester 1 (iii) O  C – O– 1 O  (c) (i) – O – S – O–  O 1

585

(ii) The long hydrocarbon chain (the alkyl group). 1 (iii) ⊕ represents Na+ ions and represents the anion part of the detergent ion. 2 (iv) When the detergent particles come into contact with grease, the negativelycharged ‘heads’ of the detergent ions dissolve in water and the hydrocarbon ‘tails’ dissolve in the grease (Diagram 1 (a)). On agitation, grease particles begin to separate from the cloth (Diagram 1 (b)). 3 (d) (i) On further agitation, each large drop of grease particle is completely separated from the cloth and broken into small droplets and suspended in water. 1 (ii) –

–

–

+

–

–

+

– – –

–

–

–

+

– – +

+

– – –

– –

+

– –

F O R M

–

5

2 (a) (i) Cleansing agent Y is a detergent. (ii) Cleansing agent X is a soap. 2 (b) Solution X produces little lather. Solution Y produces a lot of lather. 2 (c) The oily stain disappears and lather is formed. 1 (d) Y is a more effective cleansing agent because Y can remove oily stains in both hard water and soft water. In contrast, the cleansing agent X can only remove oily stains in soft water. 2 (e) (i) Manipulated variable Type of cleansing agent and type of water. (ii) Responding variable Disappearance of oily stains. (iii) Constant variable Amount of cleansing agent, amount of oil stains and volume of water. 3 3 (a) (i) Sodium lauryl sulphate: To act as cleansing agent Mint oil: To provide a fragrant flavour Saccharin: To make the toothpaste taste sweet 3 (ii) A thickening agent is a food additive to absorb water and Answers

CHAPTER 5

Disadvantage Forms scum in hard water Detergent Advantages (a) Effective in both soft water and hard water (b) Different types of detergent can be synthesied for specific uses Disadvantages (a) Non-biodegradable (b) Phosphates in detergents cause water pollution

prevent food from becoming liquid. 1 (iii) Tooth decay is caused by bacteria that converts sugars to acids. Magnesium hydroxide is used to neutralise the acid and thus prevents tooth decay. 1 (iv) Lauryl alcohol (dodecan-1-ol), concentrated sulphuric acid and sodium hydroxide. 2 (v) CH3(CH2)10CH2OH + H2SO4 → CH3(CH2)10CH2OSO3H + H2O CH3(CH2)10CH2OSO3H + NaOH → CH3(CH2)10CH2OSO3–Na+ + H2O 2 (b) (i) Wind in the stomach 1 (ii) X 1 (iii) Boil part X in water. Drink the ginger water obtained. 1 4 (a) • To prevent the growth of microorganisms • To prevent the oxidation of fats and oils in food • To enhance the taste of food 3 (b) Gelatin acts as a stabiliser/ thickener. 1 (c) Analgesics are medicines that relief pain. Psychotherapeutic medicines are medicines that change the mood and behaviour of the patient. 2 (d) Name of Type of medicine medicine

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CHAPTER 5

5

Aspirin

Analgesic

Barbiturate

Antidepressant

Amphetamine

Stimulant

Codeine

Analgesic

4 (e) (i) Can cause bleeding in the stomach (ii) Can cause addiction 2 5 (a) C9H8O4 1 (b) Aspirin molecule dissociates in water to form H+ ion and the following ion: 1

(c) (i) Aspirin dissolves slowly. Effervescence occurs as –COOH reacts with Na2CO3 and a colourless gas is released. 2 (ii) It is used as an analgesic to relieve pain. 1 (d) Aspirin is an acid. It irritates the stomach to cause gastric problems. In severe cases, it can cause bleeding of the stomach lining. 1 (e) Aspirin should be taken after food to reduce irritation of the stomach wall. 1 (f) Paracetamol 1 (g) The formula of aspirin is C9H8O4. Relative molecular mass = (9  12) + (8  1) + (4  16) = 180 Number of moles of aspirin 0.32 = ————— 180 = 0.00178 1 mol of aspirin contains one –COOH group. Let formula of aspirin be RCOOH. 2RCOOH + Na2CO3 → 2RCOONa + H2O + CO2 1 1 mol of aspirin reacts with — mol 2 of Na2CO3. Number of moles of Na2CO3 needed 1 = —  0.00178 2 = 0.00089 0.05  V 0.00089 = ——————————— 1 000 V = 17.8 cm3 3 Essay Questions 1 (a) Soaps can be prepared in the laboratory by boiling a mixture of coconut oil and concentrated sodium hydroxide solution.

The following steps are taken in the preparation of soap. (A) Saponification of coconut oil 1 10 cm3 of coconut oil is poured into a beaker. 2 50 cm3 of 5.0 mol dm–3 sodium hydroxide solution is added to the coconut oil. Answers

586

3 The mixture is boiled and stirred continuously until the layer of oil disappears. (B) Precipitation of soap 1 100 cm3 of water is added to the reaction mixture followed by three spatulas of sodium chloride. 2 The mixture is boiled for 5 minutes with constant stirring and then set aside to cool. 3 When cooled, a white precipitate of soap is produced. 4 The solution is filtered to obtain soap. The soap is then washed with distilled water and dried. 8 (b) Four types of additives in liquid detergent are builder, whitening agent, biological enzyme and brightening agent. Builder: sodium tripolyphosphate Function: It is used to soften hard water. In the presence of builders, Ca2+ and Mg2+ ions are removed. Whitening agent: sodium perborate Function: Whitening agents remove the colour of dirty stains (such as coffee or tea stains) by the oxidation process. When the stain is oxidised, the colour of the stain disappears. Biological enzyme: amylase Function: To break down protein stains such as blood stains or food stains Brightening agent: Blancophor R Function: It is used to make white fabrics whiter and coloured fabrics brighter. 7 (c) Soaps act as good and effective cleansing agents in soft water. However, soaps are not effective in hard water. This is because the Mg2+ and Ca2+ ions in hard water react with soap ions to form scum. Soaps are also not effective in acidic water because the H+ ions from the acid react with soap ions to form a white precipitate of long chain fatty acid. The advantage of soaps is that they do not cause pollution because they are biodegradable. They can be decomposed by bacteria. Detergents are effective cleansing agents in both hard and soft water because they do not form precipitates with hard or soft

Antidepressants make a person feel calm and more cheerful. Example: Prozac Antipsychotic medicine: Antipsychotic medicine are medicine used to reduce the symptoms of mental illness such as fear, hallucination and hearing of voices. Example: Chloropromazin 8 (d) Wrong methods of taking medicine include • not taking the dosage directed by the doctor. If a diabetic patient takes a higher dosage of insulin than recommended, the patient may suffer or die from insulin shock. • not taking the medicine at the proper time. If aspirin or other acidic drugs are taken on an empty stomach, the patient may suffer from gastritis (irritation of stomach wall). • not completing the whole course of treatment. If antibiotics given by the doctor are not taken until the course is completed, the bacteria will develop resistance towards the antibiotic and cause the antibiotic to be ineffective. 6 3 (a) Food additives are chemical substances added in small quantities to food for specific purposes, such as preventing food spoilage, or enhancing the flavour, smell and colour of food. 3 (b) • Azo compound is used as a dye to give colour to food. • Citric acid is used as an antioxidant to prevent the oxidation of fats and oils by oxygen in the air. In this way, food does not become rancid. • Benzyl ethanoate is an ester and has a fruity smell. It is used as artificial flavour. • Sodium benzoate is used as a preservative to stop the growth of microorganisms. • Sucrose is used as an artificial sweetener to give a sweet taste to food. 12 (c) Advantages • Reduce food spoilage • Make food taste better, smell better and look more attractive

587

• Aspartame (artificial sweetener) is a good substitute for sugar for diabetic patients Disadvantages • Artificial food additives, when consumed in large quantities, may harm our health • Some food additives can cause allergy, cancer, brain damage and hyperactivity 5

Experiment 1 (a) Aim of experiment To compare the effectiveness of the cleansing action of a soap and a detergent in soft water and hard water (b) All the variables Manipulative variable Type of cleansing agent and type of water Responding variable Effect of the cleansing agent on the oily stain Constant variables Amount of the cleansing agent and oil stain, volume of water, soft water and hard water used. (c) Hypothesis (i) Both soap and detergent are effective in soft water. (ii) Soap is ineffective but detergent is effective in hard water. (d) Materials Soap solution, detergent solution, 1 mol dm–3 magnesium sulphate solution, four pieces of cloth of the same size and stained with the same volume of oil. Apparatus Beaker, measuring cylinder and glass rod. (e) Procedure 1 Four beakers are labelled as A, B, C and D. 2 Beakers A and C are filled with the cleansing agent and distilled water. Beakers B and D are filled with the cleansing agent, distilled water and magnesium sulphate. 3 A piece of oil-stained cloth is placed in each of the beakers and the solution is stirred. 4 The observation of the effect of soap and detergent on the oily stain for each beaker is recorded. Answers

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CHAPTER 5

water or acidic water. However, detergents cause water pollution. For example, the phosphates in detergents act as nutrients for water plants and algae and speed up their growth. When the water plants and algae die, they are decomposed by bacteria which use up dissolved oxygen. Consequently, fish and aquatic life will die because of lack of oxygen. 5 2 (a) (i) Medicine are substances used for treating illnesses, preventing illnesses or reducing pain or suffering. (ii) Traditional medicine is medicine derived directly from plants (leaves, roots or barks of the plants) or animals without being chemically processed. 3 (b) Two examples of traditional medicine derived from plants: Garlic and quinine. Garlic — reduces high blood pressure — prevents attack of common colds or influenza Quinine – for treating malaria 2 (c) Analgesics: Medicine to relieve pain. Examples: Aspirin, paracetamol and codeine. Antibiotics: Medicine to destroy or prevent the growth of infectious microorganisms. Examples: Penicillin and streptomycin. Psychotherapeutic medicine: For treating mental or emotional illness. Types of psychotherapeutic medicine: Stimulant, antidepressant and antipsychotic medicine. Stimulants: Stimulants are chemical substances that can stimulate the activities of the brain and the central nervous system and hence increase alertness and physical activity. Examples of stimulants are caffeine and amphetamine. Stimulants stimulate the heart to beat faster and make a person more alert and less tired. Antidepressants: Antidepressants are medicine that decrease the activities of the brain and the central nervous system.

Beaker

Chemicals used

A

50 cm3 of soap solution + 50 cm3 distilled water

B

50 cm3 of soap solution + 20 cm3 of MgSO4(aq) + 30 cm3 distilled water

C

50 cm3 of detergent solution + 50 cm3 distilled water

D

50 cm3 of detergent solution + 20 cm3 of MgSO4(aq) + 30 cm3 distilled water

(f) Tabulation of data Experiment

Solution

Observation on oily stain

A

Soap + soft water

B

Soap + hard water

C

Detergent + soft water

D

Detergent + hard water

Inference

(c) Fine iron powder (d) 450 °C (e) N2(g) + 3H2(g) → 2NH3(g) (f) The ammonia gas is cooled to liquefy it (g) To make nitric acid and urea 4 (a) 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) (b) volume of gas (cm3) I

360

17 SPM Model Test

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SPM Model Test

4 & 5

Paper 1 1 B 2 A 11 B 12 C 21 D 22 B 31 D 32 C 41 B 42 D

3 D 13 C 23 A 33 C 43 C

4 C 14 B 24 C 34 A 44 A

5 C 15 A 25 C 35 D 45 A

6 D 16 D 26 A 36 B 46 C

7 B 17 B 27 B 37 A 47 B

8 A 18 C 28 A 38 C 48 D

9 A 19 A 29 D 39 B 49 B

10 D 20 D 30 B 40 D 50 B

Paper 2 1 (a) P = 2.1; Q = 2.4; R = 2.5; S = 2.6; T = 2.8.3; U = 2.8.7; V = 2.8.8.2 (b) Pair of elements Formula of compound

Ionic or covalent

Relative formula mass

Q and S

QS2

Covalent

12 + 2(16) = 44

P and R

P3R

Ionic

3(7) + 14 = 35

T and U

TU3

Ionic

27 + 3(35) = 132

R and U

RU3

Covalent

14 + 3(35) = 119

V and R

V3R2

Ionic

3(40) + 2(14) = 148

Q and U

QU4

Covalent

12 + 4(35) = 152

T and S

T2S3

Ionic

2(27) + 3(16) = 102

Note: Metals and non-metals form ionic compounds. Non-metallic and non-metallic elements form covalent compounds. (c) o xx o Hx x R H ox H 2 (a) Pipette (b) From pink to colourless (c) (i) 24.00 cm3 (ii) So that the salt obtained is not contaminated with the phenolphthalein indicator (d) (i) NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(l) (ii) Number of moles of sodium hydroxide neutralised, n 25 3 2 = ——————– mol 1000 = 0.05 mol 0.05 mol NaOH will produce 0.05 mol of NaNO3 salt. Answers

1 mol of NaNO3 salt = (23 + 14 + 48) g = 85 g 0.05 mol of NaNO3 salt = 0.05 x 85 g = 4.25 g (iii) Used as food preservative (e) The mixture is heated in an evaporating dish to expel excess water so as to saturate the solution. When the saturated solution is cooled, sodium nitrate salt will form. 3 (a) (i) From fractional distillation of liquid air (ii) From reaction of methane with steam (b) 200 atmosphere

588

time (s)

80

(c) (i) Rate decreases with time (ii) The quantity of lithium decreases with time. 360 (d) Number of moles = —————— 24000 = 0.015 mol (e) volume of gas (cm3) 360

II

I

III

180

time (s)

5 (a) To prevent the KClO3 from mixing with the metal powders (b) 2KClO3 → 2KCl + 3O2 heat (c) R, P, S, Q (d) (i) Zinc (ii) Lead Note: ZnO is yellow when hot and white when cold; PbO is brown when hot and yellow when cold (e) (i) Copper (ii) Copper(II) oxide (iii) 2Cu + O2 → 2CuO Note: Copper(II) oxide is black (f) Carry out displacement reactions. A more electropositive metal can displace a less electropositive metal from its salt solution. 6 (a) (i) Antibiotic (ii) Psychotherapeutic medicine (iii) Analgesic (b) (i) To kill pathogenic bacteria (ii) Treat psychotic patients (iii) Relieve pain and fever (c) (i) Tuberculosis (ii) If the dosage is not completed, the surviving bacteria will continue to

Section B 7 (a) (i) Gas P = Oxygen ; Gas Q = Hydrogen ; Gas R = Chlorine ; Gas S = Hydrogen 3 (ii) Dilute sodium chloride solution The ions present in dilute NaCl solution are Na+, H+, Cl– and OH–. 1 The Cl– and OH– ions move to the anode. The hydroxide ions are selected for discharge to produce oxygen gas because it is lower in the electrochemical series. 1 4OH– → 2H2O + O2 + 4e– 1 The Na+ and H+ ions move to the cathode. 1 The H+ ions are selected for discharge to produce hydrogen gas because it is lower in the electrochemical series. 1 2H+ + 2e– → H2 1 Concentrated sodium chloride solution The Cl– and OH– ions move to the anode. The chloride ions are selected for discharge to produce chlorine gas because its concentration is higher. 1 1 2Cl– → Cl2 + 2e– The Na+ and H+ ions move to the cathode. 1 The H+ ions are selected for discharge to produce hydrogen gas because it is lower in the electrochemical series. 2H+ + 2e– → H2 1 (b) The iron chain is connected to the negative terminal of the cell. 1 A piece of silver metal is connected to the positive terminal of the cell. 1 The electrolyte used is aqueous silver nitrate solution. The switch is closed and current is passed through the solution. 1

At anode Silver anode ionises. Each silver atom releases one electron to form silver ion. 1 Ag → Ag+ + e– 1 At cathode The silver ions are then attracted to the cathode. At the cathode, the silver ion will discharge by accepting one electron and form silver atom. 1 Ag+ + e– → Ag 1 The silver atoms will deposit on the iron chain. 8 (a) 25 cm3 of nitric acid is poured into a 100 ml beaker. The solution is heated. 1 Magnesium oxide powder is added into the acid solution while stirring the mixture with a glass rod until some magnesium oxide remains undissolved. 2 The reaction taking place can be represented by the chemical equation: 2HNO3(aq) + MgO(s) → Mg(NO3)2(aq) + H2O(l) 1 The mixture is filtered to remove the excess magnesium oxide powder. 1 The filtrate contains the soluble magnesium nitrate salt. Now excess sodium carbonate solution is poured into the filtrate and stirred. 1 Double decomposition takes place and insoluble magnesium carbonate salt is formed. 1 The reaction can be represented by the chemical equation: Mg(NO3)2(aq) + Na2CO3(aq) → MgCO3(s) + 2NaNO3(aq) 1 The mixture is filtered to separate the insoluble magnesium carbonate salt. 1 The salt is rinsed with distilled water and then dried by pressing with dry filter paper. 1 (b) Each solution is poured into four different test tubes. Dilute nitric acid is added into each solution. 1 The solution which releases gas which turns limewater cloudy, contains carbonate ions. 1 Carbonate ions react with hydrogen ions from acid to release carbon dioxide gas according to the equation: 2H+ + CO32– → CO2 + H2O 1 The other three solutions do not show any reaction. Now barium nitrate solution is added to the three remaining

589













solutions. 1 The mixture which produces a white precipitate contains sulphate ions. 1 Barium ions react with sulphate ions to form insoluble white barium sulphate salt according to the equation: Ba2+(aq) + SO42–(aq) → BaSO4(s) 1 The othe two solutions either contain nitrate or chloride ions. Each of the two remaining solutions is poured into separate test tubes. Dilute nitric acid is added followed by silver nitrate solution. 1 The mixture which produces a white precipitate is the one which contains chloride ions. 1 Silver ions react with chloride ions to produce insoluble white silver chloride salt according to the equation: Ag+(aq) + Cl–(aq) → AgCl(s) 1 To confirm the presence of nitrate ions in the remaining solution, the following procedure is performed: Dilute sulphuric acid is added to the solution, followed by iron(II) sulphate solution and then concentrated sulphuric acid is added. 1 A brown ring formed confirms the presence of nitrate ions.

Section C 9 (a) (i) An exothermic reaction is one that releases heat energy to the surroundings. 1 Thus the temperature of the surrounding increases. 1 In the exothermic reaction, the energy content of products is lower than the energy content of the reactants. 1 energy

reactants

∆H = –x kJ mol-1 products



(ii) An endothermic reaction is one that absorbs heat energy from the surroundings. 1 Thus the temperature of the surrounding decreases. 1

Answers

F O R M 4 & 5

SPM Model Test

multiply and become resistant to the antibiotic. A higher dosage of the antibiotic is needed to kill off the mutant bacteria. (d) (i) It is acidic and can cause intestinal wall bleeding. (ii) Paracetamol (e) It can cause dependency / addiction.

In the endothermic reaction, the energy content of products is higher than the energy content of the reactants. 1

H

products

energy

∆H = +y kJ

H

H

H

H

H

C

C

C

C

C

H

H

H

H

H



mol-1

H H H

C

C

H

pentane

reactants

H

(b) (Lowest) ∆H2, ∆H1, ∆H3 (Highest) The reactivity series of metals is shown below:

F O R M

H

H

H

H

H

C

C

C

C

C

H

H

H

2

H

Mg Al ∆H2 Zn H H H H H Fe Sn ∆H3 H C C C C C H H ∆H1 Pb H H H H H Cu Hg Ag The further apart the two metals, the higher the heat of displacement. The distance between the metals Mg/Ag > Fe/Cu > Al/Zn. Thus the value of ∆H2 < ∆H1 < ∆H3. (c)

H

SPM Model Test

C

C

H

H

H

H



2-methylbutane

H

CH3 H

C

C

H

CH3 H

H H

C

C

C

H

H

H

H

1 1 1 1

3 (b) When a mixture of methane and chlorine gas is exposed to ultraviolet light, substitution reaction will take place. The hydrogen atoms in methane will be substituted by the chlorine atoms. 2 CH4 + Cl2 → CH3Cl + HCl

Procedure (i) The initial temperatures of 25.0 cm3 of 0.5 mol dm–3 silver nitrate solution 1 and 25.0 cm3 of 0.5 mol dm–3 of sodium chloride were recorded. (ii) The solutions are then mixed in a polystyrene cup and stirred. 1 (iii) The highest temperature attained by the mixture was recorded. 1 Tabulation Initial temperature of AgNO3 solution = a °C Initial temperature of NaCl solution = b °C Highest temperature of mixture = c °C

2

Calculation a+b Average initial temperature of reactants is —————— = t °C 2 Rise in temperature of mixture = (c – t) = T °C Heat released in the experiment = 50 3 4.2 3 T joule = 210T joule = 0.21T kJ 0.5 3 25 Number of moles of silver chloride formed is ———————— = 0.0125 mol 1000 0.21T The heat released when 1 mol of AgCl is formed is ————— kJ 0.0125 = 16.8T kJ Heat of precipitation of silver chloride is, ∆H = –16.8T kJ mol–1 10 (a) Isomerism is a phenomena whereby there exists organic molecules with the same molecular formula with different structural formulae. For example, pentane has three isomers as shown:

H H



AgNO3(aq)

590

C

C

AgCl

Answers

C

CH3 H

polystyrene cup



CH3 H

H

thermometer

NaCI(aq)

H

H

thermometer

4 & 5

H H

H

1

H

2,2-dimethylpropane



methane



chloromethane

chloromethane



dichloromethane



trichloromethane tetrachloromethane

CH3Cl + Cl2 → CH2Cl2 + HCl dichloromethane

CH2Cl2 + Cl2 → CHCl3 + HCl trichloromethane

CHCl3 + Cl2 → CCl4 + HCl

4 (c) (i) X = CnH2n Y = CnH2n + 2 2 (ii) X and Y solutions are poured into two different test tubes. Bromine water is added into each solution. The mixture is shaken. 3 Observation

Solution 1

C

Observation

X

Brown colour of bromine water is decolourised

Y

Brown colour of bromine water is not decolourised

1



2

2

(iii) Hydrogenation C6H12 + H2 → C6H14

2

H

CH

C

C

H

CH

0

1 — 2

1

Temperature (°C)

73.0

75.0

77.0

1 1— 2 77.0

2 77.0

1 2— 2 77.0

3 79.0

(Note : All temperature readings must be in 1 decimal place) (c) temperature (ºC)

1 3— 2 82.0

4 85.0 6

90 80 70

0

1

2

3

4

time (min) 3 (d) To ensure even heating. 3 (e) The temperature remains constant. Heat supplied by the Bunsen burner is used to overcome the force of attraction between the molecules. 3 (f) (i) 77 °C 3 (ii) The substance from the laboratory is contaminated / impure. 3

2 (a) Mass of oxides added to the acid and alkali solutions Concentrations and volumes of the nitric and sodium hydroxide solutions 3 (b) Experiment I : Magnesium oxide is soluble in nitric acid solution but insoluble in sodium hydroxide solution. Experiment II : Silicon(IV) oxide is soluble in sodium hydroxide solution but insoluble in nitric acid solution. Experiment III : Aluminium oxide is soluble in both nitric acid and sodium hydroxide solutions. 3

(c) Magnesium oxide is basic, silicon(IV) oxide is acidic and aluminium oxide is amphoteric. 3 (d) (i) Copper(II) oxide (ii) Phosphorus pentoxide (iii) Tin(II) oxide / lead(II) oxide 3 3 (a) Problem statement Is the cleansing action of soap and detergent affected by hard water? (b) Hypothesis Cleansing power of soap is lower in hard water whereas the cleansing power of detergent is unaffected in hard water. (c) Variables Manipulated variable Addition of soap and detergent to hard water

591

Beaker

Observation



A (Soap added) Greasy spots remain on fabric

B (Detergent added) Greasy spots removed from fabric 17

Answers

F O R M 4 & 5

SPM Model Test

Time (min)

Responding variable Removal of greasy spots Controlled variable Mass of soap and detergent added (d) Apparatus Two 500 ml beakers, glass rod, fabric stained with five spots of grease, weighing balance and measuring cylinder Chemicals 5 g of soap and 5 g of detergent powder, magnesium sulphate salt and distilled water (e) Procedure • Two portions of 400 ml of distilled water were poured into two beakers labelled A and B. • Two spatulas of MgSO4 salt were added into each beaker and stirred until it dissolved. (Dissolution of MgSO4 in water produces hard water) • 5 g of soap was added into beaker A and 5 g of detergent powder was added into beaker B. • Two pieces of fabric stained with five greasy spots are dropped into each beaker and the mixtures were stirred with a glass rod for 10 minutes. • The fabrics were removed and observation recorded. (f) Tabulation

Paper 3 1 (a) & (b)

Glossary Form 4 Chapter 1: Introduction to Chemistry Hypothesis A statement that relates a manipulated variable with a responding variable. Constant variable The factor that is kept constant throughout an experiment. Manipulated variable The factor that is changed during the experiment. Responding variable The variable that responds to the change due to the manipulated variable. Variable A factor that affects the result of an experiment. Chapter 2: The Structure of the Atom Compound A compound is made up of two or more elements bonded together. Element A substance that cannot be split into simpler substances by a chemical reaction. Ion A charged particle. The positively-charged ion is called cation whereas the negativelycharged ion is called anion. Isotopes Isotopes are atoms having the same number of protons but different number of neutrons. Nucleon number The total number of protons and neutrons in the nucleus of an atom. Chapter 3: Chemical Formulae and Equations Chemical formulae A concise way of writing the exact number of atoms of elements that make up a compound. Empirical formula A chemical formula which gives the composition of elements in a molecule in their lowest relative proportions. Mole A mole is the amount of substance which contains the same number of particles as there are in 12 grams of carbon-12 isotope. Molecular formula A formula which gives the actual number of atoms of each element present in a compound. Glossary

Relative atomic mass The number of times an atom of an element is heavier than one-twelfth of the mass of a carbon-12 atom. Relative molecular mass The number of times a molecule of a compound is heavier than one-twelfth of the mass of a carbon-12 atom. Chapter 4: Periodic Table of Elements Amphoteric oxide An oxide which react with both acid and base to form salt and water. Catalyst A chemical which alter the rate of a reaction. Electronegativity A measure of the tendency of an element to gain electron and form a negative ion. Electropositivity A measure of the tendency of an element to lose electron and form a positive ion. Group The horizontal row of elements in the Periodic Table. Inert Unreactive Period The vertical column of elements in the Periodic Table. Metalloids Elements which show metallic and nonmetallic properties. Chapter 5: Chemical Bonds Covalent bond A chemical bond formed from the sharing of electrons between two atoms. Duplet electron arrangement A stable electron arrangement in which the only electron shell is occupied with two electrons. Ionic bond A chemical bond formed by the transfer of electrons from a metal atom to a nonmetal atom. Octet electron arrangement A stable electron arrangement in which the outermost electron shell is occupied with eight electrons. Volatile compound A compound with low boiling point that can be changed to the vapour state easily. Volatility The ability of a compound to be changed to the vapour state.

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Chapter 6: Electrochemistry Anion Negatively-charged ion Anode An electrode connected to the positive terminal of a battery or power supply in an electrolytic cell. Cathode An electrode connected to the negative terminal of a battery or power supply in an electrolytic cell. Cation Positively-charged ion. Electrochemical series An arrangement of metals based on the tendency of each atom to donate electrons. Electrolysis A process in which an electrolyte is decomposed by the passage of an electric current. Electrolyte A chemical compound which conducts electricity in the molten state or in aqueous solution and undergoes chemical changes. Electrolytic cell A cell that consists of two electrodes connected to a battery and immersed in an electrolyte. Voltaic cell A cell that produces electrical energy from chemical energy. Chapter 7: Acids and Bases Acid A chemical compound that ionises in water to produce hydrogen ions, H+. Alkali A chemical compound that ionises in water to produce hydroxide ions, OH–. Base A chemical compound that can neutralise an acid to produce salt and water. Concentration The number of grams or moles of solute that is present in 1 dm3 of solution. Dilution A process of diluting a concentrated solution by adding a solvent such as water to obtain a dilute solution. End point The point in a titration process when an acid-base indicator changes colour to indicate the end of titration. Molarity The number of moles of solute that is present in 1 dm3 of solution.

Neutralisation A reaction in which an acid reacts with a base to produce salt and water only. pH scale A set of numbers which measures the degree of acidity or alkalinity of a solution. pH value A value that measures the concentration of hydrogen ions, H+ in a solution. Standard solution A solution of known concentration. Strong acid A chemical that dissociates completely to form hydrogen ions, H+ in water. Strong alkali A chemical that dissociates completely to form hydroxide ions, OH– in water. Weak acid A chemical that dissociates partially to form hydrogen ions, H+ in water. Weak alkali A chemical that dissociates partially to form hydroxide ions, OH– in water. Titration A quantitative analysis that involves the gradual addition of a chemical solution from a burette to another chemical solution of known quantity. Chapter 8: Salts Crystallisation A process in which salt crystals are formed from saturated salt solution. Double decomposition A reaction in which two aqueous solutions react to produce an insoluble salt as a precipitate and another soluble salt. Ionic equation An equation that shows the ions that take part in a chemical reaction. Precipitate An insoluble solid produced from a solution during a chemical reaction. Precipitation A reaction used in the preparation of an insoluble salt. Qualitative analysis of salts A technique used to determine the cations and anions of a salt. Recrystallisation A process used to purify a salt by the repeat crystallisation process. Salt A compound formed when the hydrogen ion of an acid is replaced by a metal ion or ammonium ion, NH 4+. Chapter 9: Manufactured Substances in Industry Acid rain Rain that has a pH value between 2.5 to 5.0 when acidic gases dissolve in rain water. Alloy A mixture of two or more elements with

a certain composition in which the major component is a metal. Catalyst A substance that changes the rate of a chemical reaction. Composite material A structural material formed by combining two or more materials with different physical properties, producing a complex mixture with more superior properties. Monomer A small molecule that make up the repeating unit of a polymer. Polymer A large molecule made up of many smaller and identical repeating units joined together by covalent bonds. Polymerisation A chemical process in which monomers are joined together to form big molecules known as polymers.

Form 5 Chapter 1: Rate of Reaction Activation energy The minimum energy that the reactant molecules must possess at the time of collision in order for a chemical reaction to occur. Catalyst A substance than changes the rate of a chemical reaction but is itself chemically unchanged at the end of the reaction. Contact process The industrial process of making sulphuric acid from sulphur dioxide and oxygen. Effective collision The collision that is successful in producing a chemical reaction. Haber process The industrial process of making ammonia from nitrogen and hydrogen. Inhibitor A negative catalyst that decreases the rate of a chemical reaction. Rate of reaction Change in amount or concentration of a reactant or a product per unit time. Chapter 2: Carbon Compounds Addition reaction A reaction in which an unsaturated organic compound combines with another element of compound to form a single new compound which is saturated. Alcohol An organic compound that contains the hydroxyl (–OH) group and have the general formula, CnH2n+1OH. Alkane A family of saturated hydrocarbons that have the general formula, CnH2n+2.

593

Alkene A family of unsaturated hydrocarbons that have the general formula, CnH2n. Amino acids Organic compounds containing the amino group (–NH2) and the carboxylic acid group (–COOH). Carbohydrates Organic compounds that have the empirical formula, CH2O, and the general formula, Cn(H2O)n. Carboxylic acid An organic acid that has the general formula, CnH2n+1COOH, where n = 0, 1, 2, 3… Dehydration A chemical reaction in which a water molecule is removed from the organic compound. Diol A saturated alcohol with two hydroxyl (–OH) groups on adjacent carbon atoms. Esterification The reaction between a carboxylic acid and an alcohol to produce ester and water. Ester An organic compound produced from the reaction between a carboxylic acid and an alcohol. Fermentation The chemical process in which microorganisms such as yeast act on carbohydrates to produce ethanol and carbon dioxide. Functional group An atom or a group of atoms that determines the characteristic properties of an organic compound. Glycerol An alcohol that contains three hydroxyl (–OH) groups per molecule. Halogenation The addition reaction between an alkene and a halogen (chlorine or bromine). Homologous series A family of organic compounds that have the same functional group and similar chemical properties. Hydration The reaction in which a water molecule is added across the carbon-carbon double bond. Hydrocarbons Organic compounds that contain carbon and hydrogen only. Hydrogenation The addition of a hydrogen molecule across a carbon-carbon double bond. Isomers Organic compounds which have the same molecular formula but different structural formulae. Oils and Fats Naturally occurring esters produced from glycerol and carboxylic acids with long hydrocarbon chains. Glossary

Organic compounds Compounds containing carbon atoms joined to other atoms by covalent bonds. Polymer A large molecule made up of many smaller and identical repeating units. Proteins Polymers formed by linking together different amino acids. Refining The separation of petroleum into useful fractions by fractional distillation. Saponification The alkaline hydrolysis of esters using alkali. Saturated fats Fats that contain saturated fatty (carboxylic) acids. Saturated hydrocarbons Hydrocarbons that contain only carboncarbon single bonds (– C – C –). Starch Naturally occurring polymer that is produced from glucose monomers by condensation polymerisation. Substitution reaction A reaction in which an atom (or a group of atoms) in an organic molecule is replaced by another atom (or group of atoms). Unsaturated fats Fats that contain fatty (carboxylic) acids with at least one carbon-carbon double bond. Unsaturated hydrocarbons Hydrocarbons containing carbon-carbon double bonds (C = C) or carbon-carbon triple bonds (– C ≡ C –). Vulcanisation The process of hardening rubber by heating it with sulphur or sulphur compounds. Chapter 3: Oxidation and Reduction Accumulator A chemical cell that can be recharged by passing a direct current through it. Anodising The electrolytic process of depositing a thin layer of aluminium oxide on the surface of aluminium. Corrosion A redox reaction in which a metal is oxidised spontaneously at room condition to produce metal ions. Daniell cell A chemical cell made up of a zinc plate dipped into zinc sulphate solution and a copper plate dipped into copper(II) sulphate solution. Galvanising The coating of iron or steel with zinc for protection from rusting. Oxidation The reaction in which a substance gains

Glossary

oxygen, loses hydrogen, loses electrons or increases in oxidation number. Oxidation number An arbitrary charge assigned to an element according to a set of rules. Oxidising agent A substance that brings about the oxidation of another substance but is itself reduced. Plating A process carried out to coat the surface of metal objects with a thin layer of another metal. Reactivity series A list of elements arranged according to their chemical reactivity with oxygen. Redox reaction A reaction in which both oxidation and reduction take place simultaneously. Reducing agent A substance that brings about the reduction of another substance but is itself oxidised. Reduction The reaction in which a substance loses oxygen, gains hydrogen, accepts electrons or decreases in oxidation number. Rusting A redox reaction that occurs slowly between iron and oxygen to form hydrated iron(III) oxide. Chapter 4: Thermochemistry Bond energy Energy needed to break one mole of covalent bonds. Endothermic reactions Reactions that absorb heat energy from the surroundings. Exothermic reactions Reactions that release heat energy to the surroundings. Fuels Chemical substances that can burn easily in air to produce heat energy. Heat of combustion Heat released when one mole of a substance is burnt completely in excess oxygen. Heat of displacement Heat released when one mole of a metal is displaced by more electropositive metals from an aqueous solution of its salt solution Heat of neutralisation Heat released when one mole of H+ ions react with one mole of OH– ions to produce one mole of water. Heat of precipitation Heat released when one mole of a precipitate is formed from its ions. Heat of reaction Heat energy absorbed or released in a reaction when the number of moles

594

of the reactants (as shown in the thermochemical equation) react to form the products. Chapter 5: Chemicals for Consumers Analgesics Medicines that relieve pain. Antibiotics Chemicals that destroy or prevent the growth of infectious microorganisms. Antidepressants Medicines that can increase the brain’s level of neurotransmitters, thus improving mood. Antioxidants Food additives used to prevent the oxidation of fats and oils by oxygen in the air. Antipsychotic drugs Medicines used for treating psychotic patients, such as schizophrenics. Colouring agents Food additives that restore or enhance natural colours of food or give colour to food which do not normally have it. Detergents Synthetic cleaning agents produced from petroleum products. Psychotherapeutic medicine A group of drugs for treating mental or emotional illnesses. Soaps Cleaning agents made from animal fats or vegetable oils by saponification. Stabilisers Food additives that are used to enable oil and water in the food to mix properly. Stimulants Drugs that stimulate (excite) the activity of the brain and central nervous system. Thickeners Food additives used to thicken liquids and to prevent food from becoming liquid. Flavour enhancers Food additives that bring out the flavours in food. Food additives Chemicals added in small quantities to food to preserve the food, to give or enhance the colours, taste or smell of food. Food preservatives Chemicals added in small quantities to food in order to slow down or to prevent the growth of microorganisms such as bacteria or fungi that cause food spoilage. Traditional medicines Drugs obtained from natural resources such as animals, plants and minerals without being processed chemically. Tranquilisers Medicines that calm the central nervous system and make people calmer when they are very anxious or agitated.