PDF CHEMISTRY F4

149 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Confirmatory Tests for Cations LS 6.11.1 Sodium hydroxide solution/ammonia solution Add an excess of sodium hydroxide solution/ammonia solution Precipitate dissolves (Precipitate is soluble in excess of sodium hydroxide solution/ ammonia solution) Precipitate does not dissolve (Precipitate is insoluble in excess of sodium hydroxide solution/ammonia solution) White/coloured precipitate (Insoluble metal hydroxide) Salt solution A Using sodium hydroxide, NaOH: 1 2 cm3 of calcium nitrate solution is poured into a test tube. 2 A few drops of sodium hydroxide solution are added into the test tube using dropper. 3 The test tube is shaken. 4 Observation on whether the precipitate is produced and its colour is recorded. 5 If a precipitate is formed, sodium hydroxide solution is added continuously until no further changes occurred. 6 The test tube is shaken. 7 Observation on whether the precipitate dissolved in excess sodium hydroxide solution is recorded. 8 Steps 1 – 7 are repeated by replacing calcium nitrate solution with aluminium nitrate, copper(II) sulphate, iron(II) sulphate, iron(III) sulphate, lead(II) nitrate, magnesium nitrate, zinc nitrate and ammonium chloride solutions. B Using ammonia solution, NH3: 1 2 cm3 of calcium nitrate solution is poured into a test tube. 2 A few drops of ammonia solution are added into the test tube using dropper. 3 The test tube is shaken. 4 Observation on whether the precipitate is produced and its colour is recorded. 5 If a precipitate is formed, ammonia solution is added continuously until no further changes occurred. 6 The test tube is shaken. 7 Observation on whether the precipitate dissolved in excess ammonia solution is recorded. 8 Steps 1 – 7 are repeated by replacing calcium nitrate solution with aluminium nitrate, copper(II) sulphate, iron(II) sulphate, iron(III) sulphate, lead(II) nitrate, magnesium nitrate, zinc nitrate and ammonium chloride solutions. 06 U6b Chemistry F4(p132-160)csy2p.indd 149 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 150 UNIT 6 MODULE • Chemistry FORM 4 Results: Cations Sodium hydroxide solution Ammonia solution Confirmatory test with other reagent small amount excess small amount excess Ca2+ White precipitate Insoluble in excess No change No change – Mg2+ White precipitate Insoluble in excess White precipitate Insoluble in excess – Zn2+ White precipitate Soluble in excess White precipitate Soluble in excess – Al3+ White precipitate Soluble in excess White precipitate Insoluble in excess Refer to page 149 Pb2+ White precipitate Soluble in excess White precipitate Insoluble in excess Refer to page 149 Fe2+ Green precipitate Insoluble in excess Green precipitate Insoluble in excess Add a few drops of potassium hexacyanoferrate(III) solution, dark blue precipitate is formed Fe3+ Brown precipitate Insoluble in excess Brown precipitate Insoluble in excess • Add a few drops of potassium hexacyanoferrate(II) solution, dark blue precipitate is formed • Add a few drops of potassium thiocyanate solution, blood red colouration is formed Cu2+ Blue precipitate Insoluble in excess Blue precipitate Soluble in excess – NH4 + No change. A gas that changes red litmus paper to blue is released when heated. No change No change Add a few drops of Nessler’s reagent, brown precipitate is formed (a) Reaction with small amount until excess of sodium hydroxide solution: (refer to the above table) Add a little sodium hydroxide solution No precipitate Pungent smell, moist red litmus paper turn to blue Heat Add excess sodium hydroxide solution Precipitate formed Coloured precipitate White precipitate Insoluble Soluble Solution contains: Ca2+, Mg2+, Al3+, Zn2+, Pb2+, Fe2+, Fe3+, Cu2+, NH4 + NH4 + NH4 + Cu2+ (blue), Fe2+ (green), Fe3+ (brown) Pb2+, Al3+, Zn2+, Ca2+, Mg2+ Zn2+, Al3+, Pb2+ Ca2+, Mg2+ 06 U6b Chemistry F4(p132-160)csy2p.indd 150 21/12/2022 3:54 PM

151 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. (b) Reaction with small amount until excess of ammonia solution: Add a little ammonia solution Solution contains: Ca2+, Mg2+, Al3+, Zn2+, Pb2+, Fe2+, Fe3+, Cu2+ Ca2+ Cu2+ Zn2+ Fe2+, Fe3+ Mg2+, Al3+, Pb2+ Add excess aqueous ammonia Add excess aqueous ammonia Coloured precipitate White precipitate Precipitate formed No precipitate Cu2+ (blue), Fe2+ (green), Fe3+ (brown) Pb2+, Al3+, Zn2+, Mg2+ Insoluble Insoluble Soluble Soluble (c) Conclusion of the confirmatory test for colourless white cations: (i) Zn2+ : White precipitate that soluble in excess of sodium hydroxide and ammonia solution (ii) Mg2+ : White precipitate that insoluble in excess of sodium hydroxide and ammonia solution (iii) Al3+ : White precipitate that soluble in excess of sodium hydroxide and Pb2+ insoluble in excess ammonia solution (iv) Ca2+ : White precipitate that insoluble in excess of sodium hydroxide and no precipitate with ammonia solution (v) NH4 + : No precipitate with sodium hydroxide solution and pungent smell released when heated, gas released changes red litmus paper to blue (d) Conclusion of the confirmatory test for coloured cations. (i) Cu2+ : Blue precipitate that insoluble in excess of sodium hydroxide solution and soluble in excess ammonia solution (ii) Fe2+ : Green precipitate that insoluble in excess of sodium hydroxide and ammonia solution (iii) Fe3+ : Brown precipitate that insoluble in excess of sodium hydroxide and ammonia solution 06 U6b Chemistry F4(p132-160)csy2p.indd 151 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 152 UNIT 6 MODULE • Chemistry FORM 4 (e) All cations can be identified with confirmatory test using sodium hydroxide solution and ammonia solution except Al3+ and Pb2+. (f) To differentiate between Al3+ and Pb2+: • Al3+ and Pb2+ are differentiated by double decomposition reaction. An aqueous solution containing SO4 2–/ Cl– / I– anion is used to detect the presence of Al3+ and Pb2+. • Precipitate is formed when solution containing SO4 2–/ Cl– / I– added to Pb2+. • No precipitate when solution containing SO4 2–/ Cl– / I– added to Al3+. (g) Write the ionic equations for the formation of precipitates: Pb Al3+ 2+ No change No change No change White precipitate White precipitate Yellow precipitate Add sodium sulphate solution Add sodium chloride solution Add potassium iodide solution Pb2+ + SO4 2– PbSO4 Pb2+ + 2Cl– PbCl2 Pb2+ + 2I– PbI2 Al3+ Al3+ Pb2+ Al3+ and Pb2+ Pb2+ Confirmatory Tests for Anions LS 6.11.1 List the anions. CO3 2–, Cl– , SO4 2–, NO3 – What are the essential observation? The observation could be one of the following: • Colour of precipitate • Gas released 06 U6b Chemistry F4(p132-160)csy2p.indd 152 21/12/2022 3:54 PM

153 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 2 cm3 of solution that contains anion Xn– is poured into a test tube. 1 4 cm3 of dilute hydrochloric acid is added into the test tube. 2 The gas given off is passed through limewater. Observation: • Brown ring is formed. Conclusion: Nitrate ion present. 1 2 cm3 of dilute sulphuric acid is added into the test tube followed by 2 cm3 of iron(II) sulphate solution. The test tube is shaken. 2 The test tube is slanted and held with a test tube holder. 3 A few drops of concentrated sulphuric acid are dropped along the wall of the test tube and is held upright. 1 Dilute nitric acid is added into the test tube until no further changes. 2 2 cm3 of silver nitrate solution is added into the test tube. Observation: • A white precipitate is formed. Inference: • The precipitate is barium sulphate . Conclusion: Sulphate ion present. Ionic equation: Ba2+ + SO42– → BaSO4 1 Dilute hydrochloric acid/nitric acid is added into the test tube until no further changes 2 2 cm3 of barium chloride/barium nitrate solution is added into the test tube. Observation: • Effervescence occurs. • Limewater turns chalky. Inference: • The gas is carbon dioxide . Conclusion: Carbonate ion present. Ionic equation: CO32– + 2H+ → 2H O + CO 2 2 Acid Sodium carbonate Limewater Confirmatory Tests for Anions Brown ring White precipitate Observation: • A white precipitate is formed. Inference: • The precipitate is silver chloride . Conclusion: Chloride ion present. Ionic equation: Ag+ + Cl– → AgCl White precipitate LS 6.11.1 Confirmatory Tests for Anions 06 U6b Chemistry F4(p132-160)csy2p.indd 153 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 154 UNIT 6 MODULE • Chemistry FORM 4 1 50 cm3 of 1 mol dm–3 sodium hydroxide solution is neutralised by 25 cm3 of sulphuric acid. Calculate the concentration of sulphuric acid in mol dm–3 and g dm–3. [RAM: H = 1, S = 32, O = 16] M = 1 mol dm–3 M = ? v = 50 cm3 v = 25 cm3 2NaOH + H2SO4 Na2SO4 + 2H2O Number of moles of NaOH = 1 × 50 1 000 = 0.05 mol From the equation, 2 mol NaOH : 1 mol H2SO4 0.05 mol NaOH : 0.025 mol H2SO4 Concentration of H2SO4 = n mol v dm3 = 0.025 mol ( 25 1 000 ) dm3 = 1 mol dm–3 Concentration of H2SO4 = 1 mol dm–3 × (2 × 1 + 32 + 16 × 4) g mol–1 = 98 g dm–3 2 Calculate the volume of 2 mol dm–3 sodium hydroxide needed to neutralise 100 cm3 of 1 mol dm–3 hydrochloric acid. M = 2 mol dm–3 M = 1 mol dm–3 v = ? cm3 v = 100 cm3 NaOH + HCl NaCl + H2O Number of moles of HCl = 1 × 100 1 000 = 0.1 mol From the equation, 1 mol HCl : 1 mol NaOH 0.1 mol HCl : 0.1 mol NaOH Volume of NaOH = n mol M mol dm–3 = 0.1 mol 2 mol dm–3 = 0.05 dm3 = 50 cm3 3 Experiment I 1 mol dm–3 of nitric acid is used to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution. Experiment II 1 mol dm–3 of sulphuric acid is used to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution. Compare the volume of acids needed to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution in Experiment I and Experiment II. Explain your answer. Subjective Questions PL3 PL3 PL4 ENRICHMENT EXERCISE 06 U6b Chemistry F4(p132-160)csy2p.indd 154 21/12/2022 3:54 PM

155 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Answers: Experiment Experiment I Experiment I Balanced equation NaOH + HNO3 NaNO3 + H2O 2NaOH + H2SO4 Na2SO4 + 2H2O Calculation Mol of NaOH = 1 × 100 1 000 = 0.1 mol From the equation, 1 mol NaOH : 1 mol HNO3 0.1 mol NaOH : 0.1 mol HNO3 Mol of HNO3 = Mv 1 000 M= Concentration of HNO3 v = Volume of HNO3 in cm3 1 mol dm–3 × v 1 000 = 0.1 mol v = 100 cm3 Mol of NaOH = 1 × 100 1 000 = 0.1 mol From the equation, 2 mol NaOH : 1 mol H2SO4 0.1 mol NaOH : 0.05 mol H2SO4 Mol of H2SO4 = Mv 1 000 M = Concentration of H2SO4 v = Volume of H2SO4 in cm3 1 mol dm–3 × v 1 000 = 0.05 mol v = 50 cm3 Comparison and explanation • The volume of acid needed in Experiment I is doubled of Experiment II. • Sulphuric acid is diprotic acid while nitric acid is monoprotic . • One mol of sulphuric acid ionises to two mol of H+ ions, one mol of nitric acid ionises to one mol of H+ ions. • The number of H+ ions in the same volume and concentration of both acids is doubled in sulphuric acid compared to hydrochloric acid. 4 The diagram below shows the apparatus set-up for the titration of potassium hydroxide solution with sulphuric acid. 0.5 mol dm–3 sulphuric acid 50 cm3 of 1 mol dm–3 potassium hydroxide solution + methyl orange 0.5 mol dm–3 sulphuric acid is titrated to 50 cm3 of 1 mol dm–3 potassium hydroxide solution and methyl orange is used as indicator. (a) (i) Name the reaction between sulphuric acid and potassium hydroxide. Neutralisation (ii) Name the salt formed in the reaction. Potassium sulphate (b) Suggest an apparatus that can be used to measure 25.0 cm3 of potassium hydroxide solution accurately. Pipette PL1 PL1 PL2 06 U6b Chemistry F4(p132-160)csy2p.indd 155 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 156 UNIT 6 MODULE • Chemistry FORM 4 (c) What is the colour of methyl orange (i) in potassium hydroxide solution? Yellow (ii) in sulphuric acid? Red (iii) at the end point of the titration? Orange (d) (i) Write a balanced equation for the reaction that occurs. 2KOH + H2SO4 K2SO4 + 2H2O (ii) Calculate the volume of the 0.1 mol dm–3 sulphuric acid needed to completely react with 50 cm3 of 0.1 mol dm–3 potassium hydroxide. Number of moles of KOH = 0.1 × 50 1 000 = 0.005 mol From the equation, 2 mol KOH : 1 mol H2SO4 0.005 mol KOH : 0.0025 mol H2SO4 Volume of H2SO4 = n mol M mol dm–3 = 0.0025 mol 0.1 mol dm–3 = 0.025 dm3 = 25 cm3 (e) (i) The experiment is repeated with 0.1 mol dm–3 hydrochloric acid to replace sulphuric acid. Predict the volume of hydrochloric acid needed to neutralise 50.0 cm3 potassium hydroxide solution. 50 cm3 // double the volume of sulphuric acid (ii) Explain your answer in 4(e)(i). • Hydrochloric acid is a monoprotic acid whereas sulphuric acid is a diprotic acid. • The same volume and concentration of both acids, hydrochloric acid contains half the number of mole of H+ ions as in sulphuric acid. 5 (a) Substance A is white in colour. When A is strongly heated, brown gas, B and gas C are released. Gas C lighted a glowing wooden splinter. Residue D which is yellow in colour when hot and white when cold is formed. (i) Name substances A, B, C and D. A: Zinc nitrate C: Oxygen B: Nitrogen dioxide D: Zinc oxide (ii) Write the chemical equation when substance A is heated. 2Zn(NO3)2 2ZnO + 4NO2 + O2 (b) A colourless solution E gives the following results when a few series of tests are conducted. S1 – Add sodium hydroxide solution, a white precipitate is formed. The precipitate is soluble in excess sodium hydroxide solution. S2 – Add ammonia solution, a white precipitate is formed. The precipitate is insoluble in excess ammonia solution. S3 – Add potassium iodide solution, a yellow precipitate F, is formed. PL1 PL3 PL3 PL4 PL4 PL5 06 U6b Chemistry F4(p132-160)csy2p.indd 156 21/12/2022 3:54 PM

157 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. (i) What are the possible cations present in substance E as a result of S1 test? Pb2+, Al3+, Zn2+ (ii) What are the possible cations present in solution E as a result from S1 and S2 tests? Pb2+, Al3+ (iii) What is the ion present in E after S3 test has been done? Write an ionic equation for the formation of substance F. Ion present: Pb2+ Ionic equation: Pb2+ + 2I– PbI2 6 The table below shows the colour of five solutions labelled A, B, C, D and E added with small amount until excess of sodium hydroxide solution and ammonia solution. Solution Colour With sodium hydroxide solution With ammonia solution A Blue Blue precipitate insoluble in excess Blue precipitate soluble in excess B Colourless White precipitate soluble in excess White precipitate soluble in excess C Light green Dirty green precipitate Dirty green precipitate D Colourless White precipitate soluble in excess White precipitate insoluble in excess E Colourless White precipitate insoluble in excess White precipitate insoluble in excess (a) State the cations present in: A: Cu2+ B: Zn2+ C: Fe2+ E: Mg2+ (b) State another test to identify C. Add potassium hexacyanoferrate(III) solution, dark blue precipitate formed (c) What are the possible cations present in solution D? Al3+, Pb2+ (d) Describe briefly a test that can differentiate the cations present in solution D. • Add a few drops of potassium iodide/sodium chloride/sodium sulphate solution into 1 cm3 of solution D. • Yellow/white precipitate formed, lead(II) ion/Pb2+ present/No precipitate, aluminium ion, Al3+ present. 7 The diagram below shows the flow chart of changes that took place beginning from solid M. Solid M is a zinc salt. When solid M is heated strongly, it decomposes into solid Q which is yellow when hot and white when cold. + + Solid M Solution S Carbon dioxide gas Water Zinc metal + Magnesium nitrate solution Solid Q + carbon dioxide gas Reaction I Reaction II Heat Add dilute nitric acid Reaction III + Magnesium PL4 PL4 PL4 PL4 06 U6b Chemistry F4(p132-160)csy2p.indd 157 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 158 UNIT 6 MODULE • Chemistry FORM 4 (a) (i) State one chemical test for carbon dioxide gas. Passed the gas through limewater, limewater turns chalky (ii) Draw a diagram of the apparatus set-up to carry out reaction I. Limewater Heat Solid M (b) Name solids M and Q. M: Zinc carbonate Q: Zinc oxide (c) State the observations made when excess ammonia solution is added to solution S. White precipitate, soluble in excess of ammonia solution. (d) (i) Write the chemical equation for reaction II. ZnCO3 + 2HNO3 Zn(NO3)2 + H2O + CO2 (ii) For reaction II, calculate the volume of carbon dioxide gas released at room condition if 12.5 g solid M decomposes completely. [Relative atomic mass: C = 12, O = 16, Zn = 65, 1 mole of gas occupies 24 dm3 at room condition] Number of moles of solid M = 12.5 125 = 0.1 mol From the equation, 1 mol M : 1 mol CO2 0.1 mol M : 0.1 mol CO2 Volume of CO2 = 0.1 mol × 24 dm3 mol–1 = 2.4 dm3 (e) Name reaction III. Displacement reaction (f) Describe a chemical test to determine the presence of anion in the magnesium nitrate solution. • About 2 cm3 of magnesium nitrate solution is poured into a test tube. • 2 cm3 of dilute sulphuric acid is added to the solution followed by 2 cm3 of iron(II) sulphate solution. The mixture is shaken . • The test tube is slanted and held with a test tube holder. • A few drops of concentrated sulphuric acid are dropped along the wall of the test tube and is held upright. • A brown ring is formed between two layers. Anion present is nitrate ion. PL3 PL3 PL2 PL2 PL4 PL3 PL2 06 U6b Chemistry F4(p132-160)csy2p.indd 158 21/12/2022 3:54 PM

159 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 8 You are given zinc chloride crystals. Describe how you would conduct a chemical test in the laboratory to identify the ions present in zinc chloride crystals. • Dissolve half spatula of zinc chloride crystals in 10 cm3 of distilled water . • The solution is poured in three test tubes. • Add a few drops of sodium hydroxide solution to zinc chloride solution until excess . A white precipitate soluble in excess of sodium hydroxide solution formed. • Add a few drops of ammonia solution to another zinc chloride solution until excess . A white precipitate soluble in excess of ammonia solution formed. Ions present are zinc ions . • About 2 cm3 of dilute nitric acid is added to 2 cm3 of zinc chloride solution followed by 2 cm3 of silver nitrate solution. White precipitate formed. Ions present are chloride ions. 9 Drinking water may contain toxic component. In order to obtain healthier drinking water, qualitative analysis must be carried out. A scientist has done several tests of a drinking water and the observations are shown in the table below: Tests Observations Test 1: Add sodium hydroxide solution into the water until excess White precipitate soluble is formed Test 2: Add ammonia solution into the water until excess White precipitate insoluble is formed Test 3: Add potassium iodide solution into the water Yellow precipitate is formed (a) From table above, what is the ions presence in the water based on the three tests? Pb2+ ion. (b) What is the purpose of doing test 3? • Results in test 1 and 2 are the same result with test for aluminium ion, Al3+ and lead(II) ion, Pb2+. • Test 3 is to differentiate wether lead(II) ion, Pb2+ or aluminium ion, Al3+ are present. (c) If sodium chloride solution is added into the water, state the observation. Explain your answer. • White precipitate if formed. • Lead(II) ion, Pb2+ reacts with chloride ions, Cl– to form lead (II) chloride, PbCl2. (d) How to remove ion in 9(a) in drinking water? • Add potassium chloride/potassium sulphate/potassium carbonate solutions until excess. • Filter the mixture and dispose the precipitate. PL6 PL2 PL3 PL3 PL4 06 U6b Chemistry F4(p132-160)csy2p.indd 159 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 160 UNIT 6 MODULE • Chemistry FORM 4 1 Which of the following substances changes red litmus paper to blue when dissolved in water? A Sulphur dioxide C Lithium oxide B Carbon dioxide D Sodium carbonate 2 The table below shows the pH value of four solutions which have the same concentration. Solution pH value P 2 Q 7 R 12 S 13 Which of the following solutions has the highest concentration of hydroxide ion? A P C R B Q D S 3 Which of the following pairs of reactants would result in a reaction? A Sulphuric acid and copper(II) sulphate solution B Nitric acid and magnesium oxide C Hydrochloric acid and sodium nitrate solution D Ethanoic acid and sodium sulphate solution 4 Which of the following reactions will not produce any gas? A Copper metal with sulphuric acid B Zinc metal with hydrochloric acid C Ammonium chloride with calcium hydroxide D Sodium carbonate with hydrochloric acid 5 The table below shows the concentration of hydrochloric acid and ethanoic acid. Acid Concentration / mol dm–3 Hydrochloric acid 0.1 Ethanoic acid 0.1 Which of the following statements is true about both acids? A Both are strong acids B Both acids are strong electrolyte C The pH value of both acid are equal D Hydrochloric acid needs more sodium hydroxide to achieve neutralisation compare to ethanoic acid 6 The molarity of sodium hydroxide solution is 0.5 mol dm–3. What is the concentration of the solution in g dm–3? [Relative atomic mass: H = 1, O = 16, Na = 23] A 20 C 80 B 40 D 120 7 Which of the following is a salt? A Lead(II) oxide B Calcium hydroxide C Barium sulphate D Tetrachloromethane 8 Which of the following salts is soluble in water? A Iron(II) sulphate C Calcium carbonate B Silver chloride D Lead(II) bromide 9 Which of the following salts can be prepared by precipitation reaction? A Copper(II) chloride B Lead(II) nitrate C Barium sulphate D Zinc sulphate 10 Which pair of substances represented by the following formulae can react to produce salt? I HNO3(aq) + NaOH(aq) II HCl(aq) + NaCl(aq) III H2SO4(aq) + MgSO4(aq) IV H2CO3(aq) + KOH(aq) A I and II only C I, II and IV only B I and IV only D I, II, III and IV 11 If 0.2 mol of calcium carbonate is heated until no further change, what is the mass of calcium oxide produced? [Relative atomic mass of C = 12, O = 16, Ca = 40] A 5.6 g C 16.8 g B 11.2 g D 22.4 g 12 Which of the following reactions will produce copper(II) chloride? I Copper and hydrochloric acid II Copper(II) oxide and hydrochloric acid III Copper(II) carbonate and hydrochloric acid IV Copper(II) sulphate and sodium chloride A I and II only B II and III only C III and IV only D I, II, III and IV Objective Questions SPM PRACTICE PL3 PL2 PL4 PL4 PL3 PL3 PL3 PL3 PL3 PL2 PL3 PL4 06 U6b Chemistry F4(p132-160)csy2p.indd 160 21/12/2022 3:54 PM

MODULE • Chemistry FORM 4 UNIT 7 161 © Nilam Publication Sdn. Bhd. RATE OF REACTION Particles possess activation energy Particles collide at the correct orientation Collision Theory Effective collision Frequency of effective collision Factors affect rate of reaction Rate of reaction increases Application Rate of reaction at any given time Average rate of reaction Daily activities Industry processes Concentration The higher the concentration of solution, the higher the number of particles per unit volume Catalyst The presence of catalyst lower the activation energy of the reaction Measurement of the change in quantity of reactant or product per unit time Examples in Meaning Experiments on the effect of Measuring rate of reaction Can be explained using Related to Happen if Cause Related to The smaller the size of solid reactant, the larger total surface area expose to collision Size The higher the temperature, the higher the kinetic energy of particles Temperature Concept Map 7 UNIT RATE OF REACTION 07 U7 Chemistry F4(p161-195)csy2p.indd 161 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 162 State the meaning of rate of reaction. LS 7.1.2 • The rate of reaction is a measurement of the change in quantity of reactant or product per unit time. State the relationship between rate of reaction and time. (a) The rate of reaction is high if the reaction occurs fast within a short period of time. (b) The rate of reaction is low if the reaction occurs slowly within a long period of time. (c) The rate of reaction is inversely proportional to time: Rate of reaction ∝ 1 Time taken Give examples of fast reactions. LS 7.1.1 • Reaction of marble chip with hydrochloric acid. • Reaction of magnesium with sulphuric acid. • Reaction of potassium with water. • Burning of fuel. Give examples of slow reactions. LS 7.1.1 • Rusting of iron in the air. • Photosynthesis. • Fermentation of fruit juice to form alcohol. How to determine rate of reaction? LS 7.1.4 • Rate of reaction can be determined by calculating the rate of chemical change or measured quantity in a chemical change per unit time. Rate of reaction = Change in quantity of reactant/product Time taken for the change to occur How to identify the change in quantity of reactant/product for measuring rate of reaction? Give example. LS 7.1.3 • The change in amount of reactant or product in any reaction which is chosen for the purpose of measuring rate of reaction must be observable and measurable . Examples: (a) Decrease in the mass of reactant. (b) Increase in the mass of product. (c) Increase in volume of gas released. (d) Formation of precipitate as a product. What are the possible unit for the rate of reaction? • Units for the rate of reaction depends on the unit for the reactant or product of the reaction. • The possible units are: (a) g s–1 or g min–1 for increase in mass of product or decrease in mass of reactant (b) cm3 s–1 or cm3 min–1 for increase of volume gas released (c) s–1 or min–1 for fixed amount of precipitate formed in different experiments 7.1 DETERMINE RATE OF REACTION CS 7.1 PERFORMANCE LEVEL (PL) Mastered Not mastered PL1 PL2 Recall knowledge and basic skills about rate of reaction. Understand and explain rate of reaction. 07 U7 Chemistry F4(p161-195)csy2p.indd 162 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 163 © Nilam Publication Sdn. Bhd. How to measure the observable changes when a reaction produce gas? Chemical reaction Observable changes Method of measuring the observable changes Reaction between magnesium and hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Decrease in mass of magnesium 100 g Magnesium Hydrochloric acid Reading from the balance is recorded in every 30 seconds. Increase in volume of hydrogen Method I Hydrochloric acid Magnesium Water Hydrogen gas is collected by water displacement in a burette. The volume of hydrogen gas collected is recorded every 30 seconds. Method II Hydrochloric acid Magnesium The rate of reaction is measured by the volume of gas collected in the gas syringe per unit time. * This apparatus set-up can also be used to measure the increase in volume of other gases that are insoluble for example oxygen, hydrogen and carbon dioxide. How to measure the observable change when a reaction produces precipitate? Chemical reaction Observable changes Method of measuring the observable changes Reaction between sodium thiousulphate and hydrochloric acid: Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g) + S(s) Formation of sulphur as a precipitate * Volume of sulphur dioxide gas, SO2 cannot be measured by water displacement because sulphur dioxide is soluble in water. Sodium thiosulphate solution + hydrochloric acid Amount of solid sulphur formed is measured by the time taken for the mark ‘X’ placed under the conical flask can no longer be seen. LS 7.1.3 LS 7.1.3 07 U7 Chemistry F4(p161-195)csy2p.indd 163 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 164 Rate of reaction is measured by volume of gas released in every 30 seconds by water displacement in a burette. Example Water Hydrochloric acid Calcium carbonate Sketch of graph: Time / s Volume of carbon dioxide gas / cm3 D x D y t1 The rate of reaction at t1 second = The gradient of tangent to the curve at t1 s = D y cm3 D x s Average rate of reaction within certain period of time: Rate of reaction is measured by volume of gas released in every 30 seconds by water displacement in a burette. Example Water Hydrochloric acid Calcium carbonate Sketch of graph: Time / s 1 V1 V2 5 3 7 2 6 4 8 Volume of carbon dioxide gas / cm3 Average rate of reaction in the fourth minute = (V2 – V1) cm3 (4 – 3) s = y cm3 s–1 Rate of reaction is measured by time taken for the formation of precipitate. Example Sodium thiosulphate solution + hydrochloric acid Sketch of graph: (a) Time / s Concentration of sodium thiosulphate solution / mol dm–3 t1 M1 Rate of reaction for sodium thiosulphate solution with concentration M1 mol dm–3 = 1 t1 s = x s–1 (b) Time / s Temperature of sodium thiosulphate solution / °C t1 T1 Rate of reaction for sodium thiosulphate solution at temperature T1 °C = 1 t1 s = y s–1 Average rate of reaction from 0 second: Rate of reaction is measured by volume of gas released in every 30 seconds by water displacement in a burette. Example Water Hydrochloric acid Calcium carbonate Sketch of graph: Time / s 1 V 5 3 7 2 6 4 8 Volume of carbon dioxide gas / cm3 Average rate of reaction in the first 4 minutes = (V – 0) cm3 (4 – 0) s = x cm3 s–1 Rate of reaction at any given time/temperature/concentration Average rate of reaction Measurement of Rate of Reaction LS 7.1.4 How to determine the rate of reaction from the graph? 07 U7 Chemistry F4(p161-195)csy2p.indd 164 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 165 © Nilam Publication Sdn. Bhd. 1 An experiment is carried out to determine the rate of reaction of 20 cm3 of 0.5 mol dm–3 hydrochloric acid with excess calcium carbonate. The results are shown below. Time / s 0 15 30 45 60 75 90 105 120 135 150 165 Volume of CO2 / cm3 0.00 10.00 16.00 22.00 27.00 31.50 36.00 39.50 42.00 44.00 44.00 44.00 (a) (i) Write a chemical equation for the above reaction. CaCO3 + 2HCl → CaCl2 + H2O + CO2 (ii) State the observable and measurable changes in the experiment. Increase in volume of carbon dioxide/decrease in mass of calcium carbonate (iii) State the meaning of the rate of reaction for the above reaction. Change in volume of carbon dioxide gas in one second/change in mass of calcium carbonate in one second. (iv) Draw an apparatus set-up to measure rate of reaction in the given reaction. Calcium carbonate Water Hydrochloric acid LS 7.1.5 PL3 PL3 PL3 PL3 Exercise PERFORMANCE LEVEL (PL) Mastered Not mastered PL3 Apply knowledge on the rate of reaction to explain the natural occurrences or phenomena and be able to carry out simple tasks. 07 U7 Chemistry F4(p161-195)csy2p.indd 165 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 166 (b) Draw the graph of the volume of carbon dioxide gas collected against time. Time / s 30 60 90 120 150 180 0 10 20 30 40 50 Volume of CO2 / cm3 PL4 07 U7 Chemistry F4(p161-195)csy2p.indd 166 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 167 © Nilam Publication Sdn. Bhd. (c) From the graph, determine: (i) the average rate of reaction in the first minute. = Total volume of carbon dioxide gas collected in the first minute Time taken for the change to occur = 27 60 = 0.45 cm3 s–1 (ii) the average rate of reaction in the second minute = Total volume of carbon dioxide gas collected between first minute and the second minute Time taken for the change to occur = 42 – 27 60 = 0.25 cm3 s–1 (iii) the time when the reaction has completed 135 s (iv) the average rate of reaction for overall reaction = Total volume of carbon dioxide collected Time taken for the change to occur = 44 135 = 0.326 cm3 s–1 (v) the rate of reaction at 30 seconds = the gradient of the graph at 30 seconds = 0.405 ± 0.1 cm3 s–1 (vi) the rate of reaction at 105 seconds = the gradient of the graph at 105 seconds = 0.217 ± 0.1 cm3 s–1 (d) Compare the rate of reaction at 30 seconds and 105 seconds. Explain your answer. Rate of reaction at 30 seconds is higher than at 105 seconds because the concentration of hydrochloric acid decreases as time increases. PL3 PL3 PL3 PL3 PL3 PL3 PL4 07 U7 Chemistry F4(p161-195)csy2p.indd 167 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 168 2 Excess of zinc powder is added to 50 cm3 of 1 mol dm–3 hydrochloric acid. The volume of hydrogen gas collected and time taken are recorded. Complete the following table. Sketch a curve for volume of hydrogen gas collected against time for the reaction between excess of zinc powder with 50 cm3 of 1 mol dm–3 hydrochloric acid. The tangents on the curve at t1, t2 and t3 are shown. Time / min Volume of hydrogen / cm3 t1 V 0 t2 t3 Tangent on the curve at t1, t2 and t3 respectively Write the balanced equation for the reaction. Zn + 2HCl → ZnCl2 + H2 Calculate the volume of hydrogen gas collected in the experiment at room conditions. From the equation, 2 mol of HCl : 1 mol H2 0.05 mol HCl : 0.025 mol H2 Volume of H2 = 0.025 mol × 24 dm3 mol–1 = 0.6 dm3 = 600 cm3 Compare the gradient of the curve at t1 and t2. Explain your answer. • The gradient of tangent on the curve at t2 is lower than t1. • The rate of reaction at t2 is lower than at t1. • The rate of reaction decreases as the time increases because mass of zinc and concentration of hydrochloric acid decreases . What is the gradient at t3? Explain your answer. • The gradient of tangent on the curve at t3 is zero , the rate of reaction at t3 is zero . • The reaction is completed at t3. • All hydrochloric acid has reacted because zinc powder used is in excess . • At t3, maximum volume of hydrogen gas is collected. The maximum volume of hydrogen gas collected is 600 cm3 . Sketch a curve for mass of zinc against time. Time / s Mass of zinc / g LS 7.1.3 LS 7.1.4 PL3 PL4 07 U7 Chemistry F4(p161-195)csy2p.indd 168 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 169 © Nilam Publication Sdn. Bhd. Sketch a curve for concentration of hydrochloric acid against time. Time / s Concentration of hydrochloric acid / mol dm–3 3 The diagram below shows a conical flask containing calcium carbonate powder and hydrochloric acid. The experiment is set up to study the rate of reaction by measuring the mass loss of the reacting conical flask. 100 g Calcium carbonate Cotton wool Hydrochloric acid Electronic balance (a) (i) Write a balanced chemical equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3 + HCl → CaCl2 + H2O + CO2 (ii) What could have caused the mass loss of the conical flask in the reaction? The carbon dioxide gas released from the reaction between hydrochloric acid and calcium carbonate (b) What is the function of cotton wool? To prevent any liquid from escaping from the flask. (c) Describe how this set up of apparatus can be used to measure the rate of reaction. • The gas release causes the mass of conical flask decreases as the reaction progresses • The reading of electronic balance are taken at regular intervals of time • The rate reaction is calculated by the total amount of mass loss of the conical flask per unit time taken (d) The sketch of graph below shows the results of the experiment when the reading is plotted. Time / s Mass of conical flask / g A B C 80 State and describe the rate of reaction in the following points shown in the graph. (i) Point A The gradient of the graph is the steepest. The rate of reaction is the highest at the start. PL3 PL3 PL3 PL4 07 U7 Chemistry F4(p161-195)csy2p.indd 169 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 170 (ii) Point B The gradient of the graph becomes less steep. The rate of reaction decreases. (iii) Point C The graph is a horizontal straight line. The reaction has stopped. (e) State one factor that affect the rate of reaction from point A to B. Concentration of hydrochloric acid. (f) Sketch a graph to show the decrease in mass against time on the graph below. Time / s Decrease in mass / g 80 (g) The chemical equation below shows a neutralization reaction. NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Can the rate of reaction for the reaction measured by the mass loss of the reacting conical flask? Explain your answer. • Cannot. • The reaction does not produce any gas. • No gas escaping from the flask to cause the decrease in mass in the content of conical flask. 1 The rate of reaction is affected by: (i) Size of solid reactant (ii) Concentration of solution (for the reactant used in the form of solution) (iii) Temperature of solution at which the reaction occurs (iv) Presence of catalyst (for a particular reaction) (v) Pressure of gas reactant 7.2 FACTORS AFFECTING THE RATE OF REACTIONS CS 7.2 PL5 PERFORMANCE LEVEL (PL) Mastered Not mastered PL4 Analyse knowledge on the rate of reaction in the context of problem solving the natural occurrences or phenomena. Quiz Rate of Reaction 07 U7 Chemistry F4(p161-195)csy2p.indd 170 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 171 © Nilam Publication Sdn. Bhd. Hydrochloric acid Water Calcium carbonate Problem statement: How does the size of calcium carbonate chips affect the rate of its reaction with dilute hydrochloric acid? Hypothesis: The smaller the size of calcium carbonate, the higher the rate of reaction. Manipulated variable: The size of calcium carbonate Responding variable: The rate of reaction Fixed variables: Mass of calcium carbonate chips, volume and concentration of hydrochloric acid, the temperature of the reaction mixture. Apparatus: Conical flask, basin, delivery tube and rubber stopper, retort stand and clamp, measuring cylinder, burette, stop watch and weighing balance. Planning of Experiments to Study the Factor that Affect the Rate of Reaction Sodium thiosulphate solution + Sulphuric acid Problem statement: How does the concentration of sodium thiousulphate solution affect the rate of its reaction with sulphuric acid? Hypothesis: When the concentration of sodium thiosulphate solution increases, the rate of its reaction with sulphuric acid increases. Manipulated variable: Concentration of sodium thiosulphate solution Responding variable: The rate of reaction Fixed variables: Volume of sodium thiosulphate solution, volume and concentration of dilute sulphuric acid, temperature of mixture, size of conical flask. Apparatus: 100 cm3 conical flask, 50 cm3 and 5 cm3 measuring cylinder, stop watch, white paper with mark ‘X’. X Sodium thiosulphate solution + sulphuric acid 43 42 41 40 39 38 37 36 35 34 Sodium thiosulphate solution Conical flask Paper Heat ‘X’ mark Problem statement: How does the temperature of sodium thiousulphate solution affect the rate of its reaction with sulphuric acid? Hypothesis: When the temperature of sodium thiosulphate solution increases, the rate of its reaction with sulphuric acid increases. Manipulated variable: Temperature of sodium thiosulphate solution Responding variable: The rate of reaction Fixed variables: Volume and concentration of sodium thiosulphate solution, volume and concentration of dilute sulphuric acid, size of conical flask. Apparatus: 100 cm3 conical flask, 50 cm3 and 5 cm3 measuring cylinder, stop watch, thermometer. Problem statement: How does manganese(IV) oxide affect the rate of decomposition of hydrogen peroxide? Hypothesis: Manganese(IV) oxide increases the rate of decomposition of hydrogen peroxide. Manipulated variable: The presence of manganese(IV) oxide Responding variable: The rate of reaction Fixed variables: Volume and concentration of hydrogen peroxide solution Apparatus: Test tubes, measuring cylinder, retort stands and clamp, filter funnel, spatula, electronic balance, beaker. Size of solid reactant Catalyst Concentration of solution Temperature LS 7.2.1 Manganese(IV) oxide Hydrogen peroxide solution Glowing wooden splinter B A 07 U7 Chemistry F4(p161-195)csy2p.indd 171 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 172 Materials: 0.2 mol dm–3 sodium thiosulphate solution, 1 mol dm–3 sulphuric acid, distilled water. Procedures: 1 50 cm3 of sodium thiosulphate solution is measured and poured into a conical flask. 2 The temperature of this solution is measured using thermometer and recorded. 3 The conical flask is placed on top of a piece of paper with a mark ‘X’ at the centre. 4 5 cm3 of sulphuric acid is measured and poured quickly and carefully into the conical flask. Swirl the conical flask at the same time start the stop watch. 5 The mark ‘X’ as shown in the above diagram is observed. 6 The stop watch is stopped immediately when the mark ‘X’ is no longer visible. 7 The time taken for the mark ‘X’ is no longer visible is recorded. 8 Steps 1 to 7 are repeated using same volume and concentration of sodium thiosulphate solution but heated gently to a higher temperature of 35 °C, 40 °C, 45 °C, 50 °C and 55 °C. All other conditions remain unchanged. 9 A graph of temperature against time is plotted and temperature against 1 time are plotted. Tabulation of data: Temperature of Na2S O2 3, solution/ °C 55 50 45 40 35 30 Time taken for ‘X’ to disappear / s 1 time / s–1 Materials: Large and small calcium carbonate chips, 0.2 mol dm–3 of hydrochloric acid. Procedures: 1 A basin is filled with water until half full. 2 A burette full with water is inverted into the basin. 3 It is then clamped vertically using retort stand. 4 5 g of large calcium carbonate chips is weighed and put inside the conical flask as shown in the diagram. 5 50 cm 3 of hydrochloric acid 0.2 mol dm–3 is measured and poured into the conical flask. 6 The conical flask is closed immediately with a stopper fitted with delivery tube as shown in the diagram. 7 A stop watch is started immediately. 8 The volume of gas released is recorded for every 30 seconds. 9 Steps 1 to 8 are repeated using 5 g small calcium carbonate chips to replace 5 g of large calcium carbonate chips. 10 The graphs of volume of carbon dioxide against time for the two experiments are plotted on the same axes. Tabulation of data: A Large calcium carbonate chips Time / s Burette reading / cm 3 Volume of gas / cm3 B Small calcium carbonate chips Time / s Burette reading / cm 3 Volume of gas / cm3 Materials: 0.2 mol dm–3 sodium thiosulphate solution, 1 mol dm–3 sulphuric acid, distilled water. Procedures: 1 50 cm3 of sodium thiosulphate solution is measured and poured into a conical flask. 2 The conical flask is placed on top of a piece of paper with a mark ‘X’ at the centre. 3 5 cm3 of sulphuric acid 1 mol dm–3 is measured and poured quickly and carefully into the conical flask. Swirl the conical flask at the same time start the stop watch. 4 The mark ‘X’ as shown in the above diagram is observed. 5 The stop watch is stopped immediately when the mark ‘X’ is no longer visible. 6 The time taken for the mark ‘X’ is no longer visible is recorded. 7 Steps 1 to 6 are repeated using different volumes of sodium thiosulphate solution with different volumes of distilled water as shown in the table. 8 Graphs of concentration of sodium thiosulphate against time and concentration of sodium thiosulphate against 1 time are plotted. Tabulation of data: Volume of Na2S O2 3 / cm3 50 40 30 20 10 Volume of water / cm3 0 10 20 30 40 Concentration of Na2S O2 3 solution / mol dm –3 Time taken for ‘X’ to disappear / s 1 time / s–1 Materials: 20-volume of hydrogen peroxide solution, manganese (IV) oxide powder, filter paper, glowing wooden splinter Procedures: 1 Two test tubes are labelled A and B respectively. 2 5 cm3 of 20-volume of hydrogen peroxide solution is measured and poured separately into test tubes A and B. 3 1 g of manganese(IV) oxide powder is weighed and added to test tube B. 4 A glowing wooden splinter is quickly placed at the mouth of each test tube. Observe the changes. 5 At the end of the reaction, the mixture in test tube B is filtered to separate the manganese(IV) oxide powder and the residue is rinsed with distilled water. 6 Manganese(IV) oxide powder is pressed between to filter papers to dry it. Dry manganese(IV) is weighed again. 7 All observations are recorded. Tabulation of data: Test tubes Observations A B Mass of manganese(IV) oxide before reaction = ......g Mass of manganese(IV) oxide after reaction = ......g 07 U7 Chemistry F4(p161-195)csy2p.indd 172 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 173 © Nilam Publication Sdn. Bhd. LS 7.2.1 2 The curve for graph of amount of product formed against time in a chemical reaction consists of two parts: (a) The maximum quantity of product: It depends on the number of mol of reactants react in the chemical reaction. (b) The gradient of the curve: It depends on the factors that affect the rate of reaction. Factors Effects on the gradient of the curve The size of solid reactant • The smaller the size of a solid reactant, the higher the rate of reaction, the greater is the gradient of the curve. The concentration of solution • The higher the concentration of a solution, the higher the rate of reaction, the greater is the gradient of the curve. The temperature of reaction mixture • The higher the temperature of a solution, the higher the rate of reaction, the greater is the gradient of the curve. The presence of catalyst • The presence of catalyst in certain chemical reaction increases the rate of reaction, the gradient of the curve becomes greater . • An increase in the quantity of catalyst chemical reaction increases the rate of reaction, the gradient of the curve becomes greater . Time (minute/ second) Quantity of product (g/mol/cm3) V (a) The maximum quantity of product (b) The gradient of the curve Extra 07 U7 Chemistry F4(p161-195)csy2p.indd 173 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 174 (a) Experiments Number of mol of reactant/Quantity of product/Factor Sketch of the graph Experiment I: Excess of zinc powder + 100 cm3 of 1.0 mol dm–3 sulphuric acid at 40°C Experiment II: Excess of zinc powder + 50 cm3 of 1.0 mol dm–3 sulphuric acid at 30°C Balanced equation: Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) Observable changes to measure rate of reaction: Volume of hydrogen gas collected in every 30 seconds by water displacement in the burette. 1 Zinc is in excess in experiments I and II, the volume of hydrogen gas collected is not affected by the quantity of zinc. 2 Volume of H2 collected depends on number of mol of H2SO4 (a) Number of mol of H2SO4 in experiment I = 100 × 1 1 000 = 0.1 mol (b) Number of mol of H2SO4 in experiment II = 50 × 1 1 000 = 0.05 mol ⇒ The maximum volume of hydrogen gas collected in experiment I is double of experiment II . 3 Compare rate of reaction: Experiment Type of zinc Concentration of H2SO4 Temperature I Powder 1.0 mol dm–3 40°C II Powder 1.0 mol dm–3 30°C ⇒ The rate of reaction in experiments I and II is not affected by size of zinc and concentration of sulphuric acid. ⇒ Initial rate of reaction in experiment I is higher than experiment II because the temperature of reaction mixture in experiment I higher than experiment II, the gradient of the curve for experiment I is greater than experiment II. Volume of hydrogen / cm3 Experiment I Experiment II Time / s Exercise: Sketch the graph of volume of gas produced against time for following experiments. LS 7.2.1 07 U7 Chemistry F4(p161-195)csy2p.indd 174 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 175 © Nilam Publication Sdn. Bhd. (b) Experiments Number of mol of reactant/Quantity of product/Factor Sketch of the graph Experiment I: Excess calcium carbonate chips and 25 cm3 of 1.0 mol dm–3 hydrochloric acid Experiment II: Excess calcium carbonate chips and 25 cm3 of 0.5 mol dm–3 hydrochloric acid Experiment III: Excess calcium carbonate chips and 100 cm3 of 0.5 mol dm–3 hydrochloric acid Balanced equation: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H O(l) + CO 2 2(g) Observable changes to measure rate of reaction: Volume of carbon dioxide collected in every 30 seconds by water displacement in the burette. 1 Calcium carbonate is in excess in experiment I, II and III, the volume of carbon dioxide gas collected is not affected by the quantity of calcium carbonate. 2 Volume of CO2 collected depends on the number of mol of HCl (a) Number of mol of HCl in experiment I = 25 × 1.0 1 000 = 0.025 mol (b)Number of mol of HCl in experiment II = 25 × 0.5 1 000 = 0.0125 mol (c) Number of mol of HCl in experiment III = 100 × 0.5 1 000 = 0.05 mol ⇒ The maximum volume of carbon dioxide gas in experiment III is double of experiment I . ⇒ The maximum volume of carbon dioxide gas in experiment I is double of experiment II . 3 Compare rate of reaction: Experiment Type of CaCO3 Concentration of HCl I Chips 1.0 mol dm–3 II Chips 0.5 mol dm–3 III Chips 0.5 mol dm–3 ⇒ The rate of reaction in experiments I, II and III is not affected by size of calcium carbonate. ⇒ Initial rate of reaction in experiment I is higher than experiment II because the concentration of HCl in experiment I is higher than experiment II, the gradient of the curve for experiment I is greater than experiment II. ⇒ Initial rate of reaction in experiment II is equal to experiment III because the concentration of HCl in experiments II and III is the same , the gradient of the curve in experiments II and III is the same . Volume of carbon dioxide / cm3 Experiment III Experiment I Experiment II Time / s 07 U7 Chemistry F4(p161-195)csy2p.indd 175 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 176 (c) Experiments Number of mol of reactant/Quantity of product/Factor Sketch of the graph Experiment I: Excess of magnesium powder + 100 cm3 of 1.0 mol dm–3 hydrochloric acid Experiment II: Excess of magnesium ribbon + 200 cm3 of 1.0 mol dm–3 hydrochloric acid Balanced equation: Mg(s) + 2HCl(aq) → MgCl2(aq) Observable changes to measure rate of reaction: Volume of hydrogen gas collected in every 30 seconds by water displacement in the burette. 1 Magnesium is in excess in experiments I and II, the volume of hydrogen gas collected is not affected by the quantity of magnesium. 2 Volume of H2 collected depends on the number of mol of HCl. (a) Number of mol of HCl in experiment I = 100 × 1 1 000 = 0.1 mol (b)Number of mol of HCl in experiment II = 200 × 1 1 000 = 0.2 mol ⇒ The maximum volume of hydrogen gas collected in experiment II is double of experiment I . 3 Compare rate of reaction: Experiments Types of Mg Concentration of HCl I Powder 1.0 mol dm–3 II Ribbon 1.0 mol dm–3 4 Initial rate of reaction in experiment I is higher than experiment II because the total surface area of magnesium powder in experiment I is higher than magnesium ribbon in experiment II, the gradient of the curve for experiment I is greater than experiment II. Volume of hydrogen / cm3 Experiment I Experiment II Time / s 07 U7 Chemistry F4(p161-195)csy2p.indd 176 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 177 © Nilam Publication Sdn. Bhd. (d) Experiments Number of mol of reactant/Quantity of product/Factor Sketch of the graph Experiment I: Excess of zinc granules and 100 cm3 of 1.0 mol dm-3 sulphuric acid + 5 cm3 of copper(II) sulphate solution Experiment II: Excess of zinc granules and 100 cm3 of 0.5 mol dm–3 sulphuric acid + 5 cm3 of copper(II) sulphate solution Experiment III: Excess of zinc granules and 100 cm3 of 0.5 mol dm–3 sulphuric acid Balanced equation: Experiments I and II: Zn + H2SO4 → ZnSO4 + H2 Experiment III: Zn + H2SO4 → ZnSO4 + H2 Observable changes to measure rate of reaction: Volume of hydrogen gas collected in every 30 seconds by water displacement in the burette. 1 Zinc is in excess in experiments I, II and III, the volume of hydrogen gas collected is not affected by the quantity of zinc. 2 Volume of H2 depends on the number of mol of H2SO4. (a) Number of mol of H2SO4 in experiment I = 100 × 1 1 000 = 0.1 mol (b)Number of mol of H2SO4 in experiment II = 100 × 0.5 1 000 = 0.05 mol (c) Number of mol of H2SO4 in experiment III = 100 × 0.5 1 000 = 0.05 mol ⇒ The maximum volume of hydrogen gas collected in experiment I is double of experiment II . ⇒ The maximum volume of hydrogen gas collected in experiment II is equal to experiment III. 3 Compare rate of reaction: Experiments Types of zinc Concentration of H2SO4 Presence of catalyst I Granules 1.0 mol dm–3 Copper(II) sulphate II Granules 0.5 mol dm–3 Copper(II) sulphate III Granules 0.5 mol dm–3 – ⇒ Initial rate of reaction in experiment I is higher than experiment II because the concentration of H2SO4 in experiment I higher than experiment II, the gradient of the curve for experiment I is greater than experiment II. ⇒ Initial rate of reaction in experiment II is higher than experiment III because the catalyst copper(II) sulphate is present in experiment II. The gradient of the curve in experiment II is greater than experiment III. Volume of hydrogen / cm3 Experiment I Experiment III Experiment II Time / s 07 U7 Chemistry F4(p161-195)csy2p.indd 177 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 178 APPLICATION OF FACTORS THAT AFFECT THE RATE OF REACTION IN DAILY LIFE 7.3 CS 7.3 PERFORMANCE LEVEL (PL) Mastered Not mastered PL5 Evaluate knowledge on rate of reaction in the context of problem solving and decision-making to carry out a task. Give examples of the application of factor of size that affect the rate of reaction in daily activities. LS 7.3.1 (a) Burning of charcoal • When food is cooked with smaller pieces of charcoal, the food cooked faster . • The smaller pieces of charcoal have a larger total exposed surface area. • Hence, smaller pieces charcoal burns faster to produce more heat. (b) Cooking of smaller pieces of food • The total surface area on a smaller cut pieces of food is larger . • The food can absorb more heat. • Hence, the time taken for the food to be cooked is shorter . Give examples of the application of factor of temperature that affect the rate of reaction in daily activities. LS 7.3.1 (a) Storage of food in a refrigerator • When the food is kept in refrigerator, the food lasts longer. • The lower temperature in the refrigerator slows down the activity of the bacteria . • The bacteria produces less toxin , hence the rate of decomposition of food is lower . (b) Cooking food in a pressure cooker • The high pressure in pressure cooker increases the boiling point of water to a temperature above 100 ºC. • The kinetic energy of the particles in the food is increase/higher . • Hence, time taken for the food to be cooked is shorter . • Thus, the food cooked faster at a higher temperature in a pressure cooker. Give examples of the application of factor of concentration that affect the rate of reaction in daily activities. LS 7.3.1 (a) Statue, bridge or building made from iron deteriorate more rapidly than in less polluted air. • In a polluted atmosphere, the concentration of sulphur dioxide is high. • Sulphur dioxide dissolve in rain water to form high concentration of acid rain. • High concentration of acid corrodes building, monuments and statues made from marble (calcium carbonate) faster because calcium carbonate react with acid to produce salt, water and carbon dioxide. • Iron rust faster with the high concentration of acid rain. (b) Iron structures facing seafront rust faster. • Sea breeze contains high concentration of dissolved salts. • The high concentration dissolved salts becomes better electrolyte. • An electrolyte will increase the electrical conductivity of water. Hence, the rate corrosion of iron increases. Give examples of factor of catalyst in industrial processes and daily life. LS 7.3.1 (a) Haber process is an industrial process to manufacture ammonia gas . • The optimum conditions to run the process at a higher rate are: (i) The temperature is 400 ºC – 500 ºC. (ii)The pressure is 200 – 300 atm. (iii) The catalyst is iron , Fe. 07 U7 Chemistry F4(p161-195)csy2p.indd 178 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 179 © Nilam Publication Sdn. Bhd. What are the two processes that occur during a chemical reaction? 1 Chemical bonds in the reactants are broken. 2 New bonds in the products are formed. How can a reaction take place according to the collision theory? • A chemical reaction can only occur when a reacting particles collide with one another with certain amount of kinetic energy. What is activation energy, Ea? • It is the minimum amount of energy needed for the reacting particles to react. Are all chemical reactions have the same activation energy, Ea? LS 7.4.2 • No. • Different chemical reactions have different activation energy. Define effective collision. LS 7.4.1 • The collisions that lead to a chemical reaction and result in the formation of products. What happens when the energy of collision between particles less than activation energy, Ea? LS 7.4.2 • The reactants particles just bounce off each other and no reaction occur. Reactant A Reactant B No new products formed Bounce Below Ea Collision ⇒ Collision energy of particles < activation energy. ⇒ The chemical bonds in the reactants are not broken. ⇒ No reaction. ⇒ The collision is an ineffective collision. 7.4 THE COLLISION THEORY CS 7.4 Chemical equation: N2(g) + 3H2(g) 2NH3(g) Fe, 200 – 300 atm 400 ºC – 500 ºC (b) Contact process is an industrial process to manufacture sulphuric acid. The optimum conditions to run the process at a higher rate are: (i) The temperature is 450 ºC. (ii) The pressure is 1 atm. (iii)The catalyst is vanadium(V) oxide , V2O5. Chemical equation: 2SO2 + O2 2SO3 V2O5, 450 ºC 1 atm (c) Ostwald process is an industrial process to manufacture nitric acid . The catalyst used for this process is platinum . (d) A catalytic converter is a device that uses platinum as a catalyst to convert three harmful compounds in car exhaust into harmless compounds. In a catalytic converter, the catalyst converts: • carbon monoxide into carbon dioxide. • hydrocarbons into carbon dioxide and water. • nitrogen oxides back into nitrogen and oxygen. 07 U7 Chemistry F4(p161-195)csy2p.indd 179 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 180 What happens when the collision between particles does not occur at the correct orientation? LS 7.4.2 • The reactants particles just bounce off each other and no reaction occur. Reactant A Reactant B No new products formed Collision Wrong orientation Bounce ⇒ The chemical bonds in the reactants are not broken. No reaction. ⇒ The collision is an ineffective collision. What happens when the energy of collision between the reacting particles have equal or more than the activation energy, Ea and collide at correct orientation? LS 7.4.2 Reactant A Reactant B New products are formed Achieves Ea Correct orientation Collision ⇒ Collision energy of particles ≥ activation energy. ⇒ The chemical bonds in the reactants are broken. ⇒ Reaction occurs. ⇒ The collision is an effective collision. What are the two conditions for a reaction to occur? LS 7.4.2 (a) The colliding particle must have enough energy i.e equal or more than a minimum amount of energy known as *activation energy, Ea. * The activation energy differs in different chemical reaction. The lower the activation energy, the higher the rate of reaction. (b) The colliding particles must also have the right orientation of collisions. How to show the energy changes of reacting particles and the activation energy, Ea in a reaction? LS 7.4.3 Collision Theory Part 2 • The energy changes of the reacting particles and the activation energy of a reaction is shown in an energy profile diagram. (a)* Exothermic reaction Reaction path Reactants Products Energy Ea (b) * Endothermic reaction Reaction path Reactants Products Energy Ea Ea – The minimum energy needed for the reacting particles to react. *Exothermic and endothermic reactions will studied in Form 5, Thermochemistry. Collision Theory Part 1 07 U7 Chemistry F4(p161-195)csy2p.indd 180 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 181 © Nilam Publication Sdn. Bhd. How to relate the frequency of effective collision with the rate of reaction? LS 7.4.2 • The effective collisions will result in chemical reaction. • When frequency of effective collision increases, the rate of reaction will also increase. Remark: Frequency of collision is the number of collisions in one second. How to change the frequency of effective collision? (i) The frequency of collision between particles (number of collisions per unit time). Or (ii) The activation energy of the chemical reaction. State factors that can change the frequency of effective collision. LS 7.4.2 (i) Size of solid reactant. (ii) Concentration of reacting solution. (iii) Temperature of reaction mixture. (iv) The presence of catalyst. Remark: Factors Frequency of collision or activation energy Frequency of effective collision Rate of reaction Affect Affect Affect 1 Size of Solid Reactant How does the change in size of solid reactant affect rate of reaction? • The smaller the size of solid reactant, the larger total surface area exposed to collision that allows more collision between reacting particles. Explain how this factor affects the rate of reaction. • When the size of solid reactant is larger, the total surface area at which reaction occur is smaller. • When the solid reactant is broken into smaller size, the total surface area at which reaction occur becomes larger. When the size of solid substance B is large, only outer layer of reacting particles B can react with reacting particles A When size of solid substance B is broken into smaller size, more reacting particles B can react with reacting particles A When the size substance B decreases Particle of substance A Particle of substance B • This will increase the frequency of collision between particles. • Thus, frequency of effective collisions between particles increases which leads to an increase of rate of reaction. Relationship Between Frequency of Effective Collision with Factors Affecting Rate of Reaction 07 U7 Chemistry F4(p161-195)csy2p.indd 181 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 182 2 Concentration How does the concentration of solution affect rate of reaction? • Concentration is the number of solutes dissolved in a given volume of solution. • A higher concentration means that the number of particles of solute per unit volume is also higher, and vice versa. Explain how this factor affects the rate of reaction. • When the concentration of solution is higher, the number of particles per unit volume increases. • This will increase the frequency of collision between particles. • Thus, frequency of effective collisions between particles increases. At a lower concentration, the number of particles of A and B in a unit volume is lower, thus the frequency of collision between particles is low. At a higher concentration, the number of particles of A and B in a unit volume is higher, thus the frequency of collision between particles is high. When the concentration increases Particle of substance A Particle of substance B • This leads to an increase of rate of reaction. Explain why the monoprotic acid and diprotic acid have different rate of reaction when the concentration of the acids is the same. • For diprotic acid, 1 mol dm–3 of the acid ionises to 2 mol dm–3 of hidrogen ions, H+ Example: H2SO4 2H+ + SO4 2– 1 mol dm–3 2 mol dm–3 • For monoprotic acid, 1 mol dm–3 of the acid ionises to 1 mol dm–3 of hidrogen ions, H+ Example: HCl H+ + Cl– 1 mol dm–3 1 mol dm–3 • Hence, when the same concentration of sulphuric acid and hydrochloric acid are used, the concentration of hydrogen ions in sulphuric acid is double . • The rate of reaction using sulphuric acid is higher than hydrochloric acid. 3 Temperature How does the temperature of solution affect rate of reaction? • The higher the temperature of a substance, the higher the kinetic energy of the reacting particles and vice versa. Explain how this factor affects the rate of reaction. • When the temperature increases, the kinetic energy of reacting particles increases and the particles move faster. • This will increase the frequency of collision between particles. • Thus, frequency of effective collisions between particles increases. • This leads to an increase of rate of reaction. Factor: Concentration 07 U7 Chemistry F4(p161-195)csy2p.indd 182 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 183 © Nilam Publication Sdn. Bhd. 4 Catalyst Define catalyst. • It is a substance that change the rate of reaction without itself being chemically changed at the end of the reaction. How does catalyst affect the rate of reaction? • Catalyst provides alternative pathway of a reaction at lower activation energy for the reacting particles to react. Explain how this factor affects the rate of reaction. (i) In a reaction, if the reacting particles collide with energy lower than the activation energy, there will be no reaction. (ii) When catalyst is added to the same reaction, its provides alternative pathway with a lower activation energy for the reacting particles to react. (iii) With this lower activation energy, more reacting particles will have enough energy to react when they collide. (iv) This will increase the frequency of effective collisions between particles which leads to an increase of rate of reaction. – Ea : The activation energy without the presence of a catalyst. – Ea′ : The lower activation energy in the presence of a catalyst. Reaction path Products Reactants Uncatalysed reaction pathway Energy Catalysed reaction pathway Ea′ Ea State the characteristics of catalyst. (i) A catalyst does not change the amount of products of a reaction. (ii) Catalyst is specific to a particular reaction, different chemical reactions need different catalyst. (iii) Catalyst remains unchanged chemically during a reaction. (iv) Catalyst may undergo physical change during a reaction. (v) Only a small amount of catalyst needed to achieve big increase in rate of reaction. (vi) More amount of catalyst used can further increase the rate of reaction. (vii) Catalyst in powder form can further increase the rate of reaction. Factor: Catalyst 07 U7 Chemistry F4(p161-195)csy2p.indd 183 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 184 The frequency of collision between particles increases. The rate of reaction increases . Size of Reactant The smaller the size of reactant, the total surface area exposed to collision is larger. Concentration of Reactant The higher the concentration of reactants, the number of particles in a unit volume is higher. Temperature of Reaction Mixture The higher the temperature, the kinetic energy of reacting particles is higher. The reacting particles move faster. Catalyst Catalyst provides an alternative path of reaction which needs lower activation energy (Ea′). More colliding particles achieve activation energy . *State the particles that collide based on ionic equation. The frequency of effective collision between* particles increases. Conclusion Steps to explain factors that affect the rate of reaction based on collision theory: LS 7.2.1 LS 7.4.1 Rate of reaction can be increased by increasing temperature. You are required to conduct an experiment to study the factor of temperature on rate of reaction between sodium thiosulphate with hydrochloric acid. You are provided with 0.1 M sodium thiosulphate solution labelled with S1, 0.5 mol dm–3 hydrochloric acid labelled with S2, distilled water, 100 cm3 conical flask, 50 cm3 and 5 cm3 measuring cylinder, stopwatch, thermometer, Bunsen burner, paper with mark ‘X’, tripod stand, wire gauze, glass rod Diagram 1 shows the apparatus set up for this experiment. X Sodium thiosulphate solution + hydrochloric acid 34 35 36 37 38 39 40 41 42 43 Sodium thiosulphate solution Conical flask Paper Heat ‘X’ mark Diagram 1 Experiment: Factor of Temperature on the Rate of Reaction SPM K3 S P O T S P M K 3 07 U7 Chemistry F4(p161-195)csy2p.indd 184 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 185 © Nilam Publication Sdn. Bhd. 1 Conduct the experiment by using the procedures below: (i) By using a measuring cylinder, measure 50 cm3 of S1. Pour the Solution S1 into 100 cm3 conical flask. (ii) Place the conical flask on the tripod stand with wire gauze. Solution S1 heated until temperature reach 30 °C. (iii)Place the conical flask on the paper with ‘X’ mark. (iv) Add 5 cm3 of S2 into the conical flask. Stir the mixture using glass rod and start immediately the stop watch. (v) Record the time taken for the ‘X’ mark disappears from sight. (vi) Steps (i) – (v) are repeated using the same volume and concentration of sodium thiosulphate solution but heated gently to a higher temperature of 35 °C, 40 °C, 45 °C, 50 °C and 55 °C. All other conditions remain unchanged. 2 (a) Construct a table to record the time taken for ‘X’ marks disappear from sight at different temperature. Include the value of 1 time in the table. Temperature of Na2S2O3 / °C 30 35 40 45 50 55 Time taken for ‘X’ to disappear / s 1 time / s–1 [4 marks] (b) Write the chemical equation for the reaction between sodium thiosulphate with hydrochloric acid. Na2S2O3 + 2HCl→2NaCl + S + SO2 + H2O [2 marks] (c) Plot a graph of temperature of sodium thiosulphate solution against 1 time . Temperature of sodium thiosulphate solution / °C 1 time / s–1 [3 marks] (d) Based on your plotted graph, (i) what is represent by 1 time ? Rate of reaction [1 mark] (ii) does the rate of reaction increase, decrease or remain the same with temperature? Rate of reaction increases when temperature increases [1 mark] (e) How does the change of kinetic energy of particles of reactant when temperature is increases? Increases [1 mark] (f) Explain the effect of temperature on rate of reaction by using Collison Theory. • At high temperature, the kinetic energy of thiosulphate ion and hydrogen ion increases. • The thiosulphate ion and hydrogen ion move rapidly and collide with each other. • The frequency of collision increases. • The frequency of effective collision also increase and rate of reaction is higher. [3 marks] S P O T S P M K 3 07 U7 Chemistry F4(p161-195)csy2p.indd 185 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 186• Analyse the condition for both experiments given in the form of description/balanced equation/diagram to identify factor that affects the rate of reaction. Steps to Compare And Explain Factor that Affects Rate of Reaction Between Any Two Experiments 1 Compare factor that affects the rate of reaction in both experiments. Examples: (a) Factor size of solid reactant (for reaction between different size of calcium carbonate and hydrochloric acid) The size of calcium carbonate chips in experiment I is smaller than experiment II (b) Factor concentration (for the reaction between different concentration of hydrochloric acid with zinc) The concentration of hydrochloric acid in experiment I is higher than experiment II (c) Factor temperature (for the reaction between different temperature of hydrochloric acid and zinc) The temperature of reaction mixture in experiment I is higher than experiment II (d) Factor catalyst (for the reaction between zinc and sulphuric acid) Copper(II) sulphate present as a catalyst for the reaction between zinc and sulphuric acid in experiment I but not in experiment II 4 Compare frequency of effective collision between *particles in both experiments. (a) Factor size solid reactant The frequency of effective collision between calcium carbonate and hydrogen ion in experiment I is higher than experiment II (b) Factor concentration The frequency of effective collision between hydrogen ions and zinc atoms in experiment I is higher than experiment II (c) Factor temperature The frequency of effective collision between hydrogen ions and zinc atoms in experiment I is higher than experiment II (d) Factor catalyst The frequency of effective collision between hydrogen ions and zinc atoms in experiment I is higher than experiment II 5 Compare rate of reaction in both experiments Rate of reaction in experiment I is higher than experiment II. 2 Compare how the factor affects the rate of reaction in both experiments. (a) Factor size solid reactant The total surface area of calcium carbonate chips in experiment I is larger than experiment II (b) Factor concentration The number of hydrogen ions per unit volume in experiment I is higher than experiment II (c) Factor temperature The kinetic energy of hydrogen ion in experiment I is higher than experiment II (d) Factor catalyst • Copper(II) sulphate lower the activation energy for the reaction between zinc and sulphuric acid in experiment I • More colliding particles achieve the activation energy *State the particles that collide in the reaction based on the ionic equation **Catalyst does not increase the frequency of collision between particles 3 Compare frequency of collision between number of *particles in both experiments. (a) Factor size solid reactant The frequency of collision between calcium carbonate and hydrogen ion in experiment I is higher than experiment II (b) Factor concentration The frequency of collision between hydrogen ions and zinc atoms in experiment I is higher than experiment II (c) Factor temperature The frequency of collision between hydrogen ions and zinc atoms in experiment I is higher than experiment II (d) **Factor catalyst 07 U7 Chemistry F4(p161-195)csy2p.indd 186 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 187 © Nilam Publication Sdn. Bhd. 1 Complete the following table. Chemical equation Ionic equation *Particles that collide in the reaction (i) Mg + 2HCl → MgCl2 + H2 Mg + 2H+ → Mg2+ + H2 Magnesium atom and hydrogen ion (ii) Mg + 2HNO3 → Mg(NO3)2 + H2 Mg + 2H+ → Mg2+ + H2 Magnesium atom and hydrogen ion (iii) Zn + H2SO4 → ZnSO4 + H2 Zn + 2H+ → Zn2+ + H2 Zinc atom and hydrogen ion (iv) Zn + 2CH3COOH → (CH3COO)2Zn + H2 Zn + 2H+ → Zn2+ + H2 Zinc atom and hydrogen ion (v) CaCO3 + 2HCl → CaCl2 + H2O + CO2 CaCO3 + 2H+ → H2O + CO2 + Ca2+ Calcium carbonate and hydrogen ion (vi) 2H2O2 → 2H2O + O2 – Hydrogen peroxide molecules (vii) Na2S2O3 + H2SO4 → Na2SO4 + H2O + SO2 + S S2O3 2– + 2H+ → H2O + SO2 + S Thiosulphate ion and hydrogen ion 2 Compare rate of reaction in Experiment I and Experiment II. Explain based on collision theory. Experiment I Experiment II Reactants 20 cm3 of 0.5 mol dm–3 hydrochloric acid + excess of calcium carbonate powder at 30°C 20 cm3 of 0.5 mol dm–3 hydrochloric acid + excess of calcium carbonate chips at 30°C Balanced equation CaCO3 + 2HCl → CaCl2 + H2O + CO2 Ionic equation CaCO3 + 2H+ → H2O + CO2 + Ca2+ Calculate volume of carbon dioxide gas released in both experiments at room conditions. 1 mol of gas occupies 24 dm3 at room conditions Number of moles of HCl = 20 × 0.5 1 000 = 0.01 mol From the equation 2 mol HCl : 1 mol CO2 0.01 mol HCl : 0.005 mol CO2 Volume of CO2 = 0.005 mol × 24 dm3 mol–1 = 0.12 dm3 = 120 cm3 Factor that affects rate of reaction • Size of calcium carbonate in experiment I is smaller than experiment II. Compare how the factor affects rate of reaction • The total surface area of calcium carbonate exposed to collision in experiment I is higher than experiment II. Compare the frequency of collision between * particles • The frequency of collisions between calcium carbonate and hydrogen ions in experiment I is higher than experiment II. PL3 PL4 Exercise PERFORMANCE LEVEL (PL) Mastered Not mastered PL6 Invent by applying the knowledge on rate of reaction in the context of problem solving and decision-making or when carrying out an activity/task in new situations in a creatively and innovatively; giving due considerations to social/economic/ cultural values of the community. 07 U7 Chemistry F4(p161-195)csy2p.indd 187 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 188 Experiment I Experiment II Compare the frequency of effective collisions between *particles • The frequency of effective collisions between calcium carbonate and hydrogen ions in experiment I is higher than experiment II. Compare rate of reaction • Rate of reaction in experiment I is higher than experiment II. Sketch of the graph Volume of carbon dioxide gas / cm3 Experiment I Experiment II Time / s * particles – State the type of particles (atom/ion/molecule) that collide based on the ionic equation for the reaction. 3 Complete the following table to compare and explain the rate of reaction in Experiment I and Experiment II based on collision theory. Experiment I Experiment II Reactants 20 cm3 of 0.5 mol dm–3 hydrochloric acid + excess of magnesium powder at 30°C 20 cm3 of 0.5 mol dm–3 sulphuric acid + excess of magnesium powder at 30°C Balanced equation Mg + 2HCl → MgCl2 + H2 Mg + H2SO4 → MgSO4 + H2 Ionic equation Mg + 2H+ → Mg2+ + H2 Mg + 2H+ → Mg2+ + H2 Calculate volume of gas released in each experiment at room conditions Number of moles of HCl = 20 × 0.5 1 000 = 0.01 mol From the equation 2 mol HCl : 1 mol H2 0.01 mol HCl : 0.005 mol H2 Volume of H2 = 0.005 mol × 24 dm3 mol–1 = 0.12 dm3 = 120 cm3 Number of moles of H2SO4 = 20 × 0.5 1 000 = 0.01 mol From the equation 1 mol H2SO4 : 1 mol H2 0.01 mol H2SO4 : 0.001 mol H2 Volume of H2 = 0.01 mol × 24 dm3 mol–1 = 0.24 dm3 = 240 cm3 Ionisation equation of acid and concentration of H+ ion HCl → H+ + Cl– 1 mol of hydrochloric acid ionise to 1 mol H+ 1 mol dm–3 hydrochloric acid ionise to 1 mol dm–3 H+ H2SO4 → 2H+ + SO4 2– 1 mol of sulphuric acid ionise to 2 mol H+ 1 mol dm–3 sulphuric acid ionise to 2 mol dm–3 H+ Compare the concentration of reactant • Concentration of hydrogen ion, H+ in experiment II is double of experiment I. Compare how the factor affects rate of reaction • The number of hydrogen ion per unit volume in experiment II is double of experiment I. PL4 07 U7 Chemistry F4(p161-195)csy2p.indd 188 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 189 © Nilam Publication Sdn. Bhd. Experiment I Experiment II Compare the frequency of collision between *particles • Frequency of collisions between hydrogen ions and magnesium atoms in experiment II is higher than experiment I. Compare the frequency of effective collisions between *particles • Frequency of effective collisions between hydrogen ions and magnesium atoms in experiment II is higher than experiment I. Compare rate of reaction • Rate of reaction in experiment II is higher than experiment I. Sketch of the graph Volume of hydrogen gas / cm3 Experiment I 240 Experiment II 120 Time / s Compare the gradient and amount of product for the curve in both experiments. Explain. (i) The gradient of the curve for experiment II is greater than experiment I because the rate of reaction in experiment II is higher than experiment I. (ii) The volume of hydrogen gas collected in experiment II is double of experiment I. Sulphuric acid is diprotic acid while hydrochloric acid is monoprotic acid. One mol of sulphuric acid ionises to two mol of H+ ions, one mol of hydrochloric acid ionises to one mol of H+ ions. (iii) The number of H+ ions in the same volume and concentration of both acids is double in sulphuric acid. * particles – State the type of particles (atom/ion/molecule) that collide based on the ionic equation for the reaction. 4 Complete the following table to compare and explain how does copper(II) sulphate solution affect the rate of reaction. Experiment I Experiment II Water Excess of zinc granules 50 cm3 of 1.0 mol dm–3 hydrochloric acid Water Excess of zinc granules 50 cm3 of 1.0 mol dm–3 hydrochloric acid + Copper(II) sulphate solution Balanced equation Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Zn(s) + 2HCl(aq) CuSO4 ZnCl2(aq) + H2(g) PL4 07 U7 Chemistry F4(p161-195)csy2p.indd 189 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 190 Experiment I Experiment II Ionic equation Zn + 2H+ → Zn2+ + H2 Calculate volume of hydrogen gas released in experiments I and II at room conditions. 1 mol of gas occupies 24 dm3 at room conditions Number of moles of HCl = 50 × 1 1 000 = 0.05 mol From the equation 2 mol HCl : 1 mol H2 0.05 mol HCl : 0.025 mol H2 Volume of H2 = 0.025 mol H2 × 24 dm3 mol–1 = 0.6 dm3 = 600 cm3 Compare the factor that affects rate of reaction • Reaction between zinc and hydrochloric acid in experiment I is without catalyst while copper(II) sulphate added in experiment II acts as a catalyst. Compare how the factor affects rate of reaction • Copper(II) sulphate lowers the activation energy for the reaction between zinc and hydrochloric acid in experiment II. Compare the frequency of effective collisions between *particles • Frequency of effective collisions between zinc atom and hydrogen ion in experiment II is higher than experiment I. Compare rate of reaction • Rate of reaction in experiment II is higher than experiment I. Sketch of the graph for experiment I and experiment II Volume of hydrogen gas / cm3 Experiment I Experiment II Time / s Compare the gradient and amount of product for the curve in both experiments. Explain. 1 The gradient of the curve for experiment II is greater than experiment I because the rate of reaction in reaction experiment II is higher than experiment I. 2 The volume of hydrogen gas released in experiment I is equal to experiment II. The volume and concentration of hydrochloric acid in experiments I and II are similar. Copper(II) sulphate in experiment II that is used as a catalyst does not affect the total volume of hydrogen gas produced. 07 U7 Chemistry F4(p161-195)csy2p.indd 190 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 191 © Nilam Publication Sdn. Bhd. 1 The table below shows the data for three experiments that have been carried out to determine the effect of catalyst on the decomposition of hydrogen peroxide to water and oxygen. Volume of H2O2 / cm3 Concentration of H2O2 / mol dm–3 Mass of MnO2 / g I 20 1.0 – II 20 1.0 1.0 III 20 1.0 2.0 (a) Write the chemical equation for the decomposition of hydrogen peroxide. 2H2O2 → 2H2O + O2 (b) Draw the set-up of apparatus for the experiment. Water H2O2 Manganese(IV) oxide (c) Between experiments I, II and III, which has the highest rate of reaction? Explain your answer. Experiment III. The amount of catalyst used in experiment III is more than experiment II. (d) Explain how does manganese(IV) oxide affect the rate of decomposition of hydrogen peroxide by using collision theory. • Manganese(IV) oxide provides an alternative path with lower activation energy for the decomposition of hydrogen peroxide. • The frequency of effective collisions between hydrogen peroxide molecules increases. • The rate of decomposition of hydrogen peroxide increases. (e) Calculate the volume of oxygen gas released in experiment II at room temperature and pressure. Number of moles of H2O2 = 20 × 1 1 000 = 0.02 mol From the equation, 2 mol H2O2 : 1 mol O2 0.02 mol H2O2 : 0.01 mol O2 Volume of O2 = 0.01 mol × 24 dm3 mol–1 = 0.24 dm3 = 240 cm3 Subjective Questions PL3 PL3 PL4 PL4 PL3 ENRICHMENT EXERCISE Quiz 07 U7 Chemistry F4(p161-195)csy2p.indd 191 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 192 (f) Sketch the graph of volume of oxygen gas against time for experiments I, II and III. Time / s Volume of oxygen gas / cm3 III II I (g) State one factor other than concentration and catalyst that can affect the rate of decomposition of hydrogen peroxide. Temperature of hydrogen peroxide. 2 The diagram on the right shows a sketch of curve I and curve II for volume of carbon dioxide gas collected against time. Curve I is for the reaction in Experiment I between 2.0 g marble chips added to 50 cm3 of 1 mol dm–3 hydrochloric acid at room temperature. (a) Write a chemical equation to show the reaction between marble chips and hydrochloric acid. CaCO3 + 2HCl → CaCl2 + H2O + CO2 (b) State two ways to obtain curve II for the reaction of 50 cm3 of 1 mol dm–3 hydrochloric acid at room temperature. (i) 2.0 g marble powder is used to replace 2.0 g marble chips (ii) The experiment is carried out at 40 °C (higher than room temperature) (c) Explain each of your answer in 2(b). (i) 2.0 g marble powder in Experiment II has a larger total surface area than 2.0 g marble chips in Experiment I, hence initial rate of reaction of Experiment II is higher than Experiment I. Maximum volume of carbon dioxide gas collected in Experiment I and II are the same because the quantity of marble and hydrochloric acid are the same in both experiments. (ii) Experiment II is conducted at higher temperature compared to Experiment I, hence intial rate of reaction Experiment II is higher than Experiment I. Maximum volume of carbon dioxide gas collected in Experiment I and II are the same because the quantity of marble and hydrochloric acid are the same in both experiments. (d) Which of the two reactants in Experiment I is in excess? Explain your answer. [Relative atomic mass: C = 12, O = 16, Ca = 40] Number of moles CaCO3 = 2 g 100 g mol–1 = 0.02 mol Number of moles HCl = 50 × 1 1 000 = 0.05 mol From the equation: 1 mol of CaCO3 react with 2 mol of HCl 0.02 mol of CaCO3 react with 0.04 mol of HCl → Hydrochloric acid is in excess PL4 PL2 PL3 PL3 PL4 PL4 Time / s Volume of carbon dioxide gas / cm3 II I 07 U7 Chemistry F4(p161-195)csy2p.indd 192 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 193 © Nilam Publication Sdn. Bhd. 3 The diagram below shows the set up of apparatus to study the effect of concentration on the rate of reaction. Hydrochloric acid Magnesium ribbon In each experiment, x g magnesium ribbon is allowed to react with excess y mol dm–3 hydrochloric acid. The table shows the mixing of different volumes of hydrochloric acid y mol dm–3 and water in experiments A, B and C. Experiment Volume of hydrochloric acid y mol dm–3 / cm3 Volume of water / cm3 A 50 350 B 400 0 C 200 200 The result of these experiments are plotted in the same graph shown below. Time / s Volume of hydrogen gas / cm 20 50 100 Curve I Curve II Curve III 0 50 80 (a) What is meant by concentration? It is the amount of solute in a given volume of solution (b) Which of the experiments are represented by: (i) Curve I: B (ii) Curve II: C (iii) Curve III: A (c) Why is the volume of hydrogen gas the same in all experiments? • Magnesium is the limiting factor. • The volume of hydrogen gas released is limited by the number of mol of magnesium present. (d) By using collision theory, compare rate of reaction in experiment A and experiment B. Explain your answer. • Rate of reaction of experiment B is higher than experiment A • Concentration of hydrochloric acid in experiment B is higher than experiment A • Number of hydrogen ions per unit volume in experiment B is higher than A • Frequency of collision between hydrogen ion and magnesium atom in experiment B is higher than A • Frequency of effective collision between hydrogen ion and magnesium atom in experiment B is higher than A (e) (i) Write a balanced chemical equation for the reaction between magnesium and hydrchloric acid. Mg + 2HCl → MgCl2 + H2 PL1 PL5 PL2 PL4 PL3 07 U7 Chemistry F4(p161-195)csy2p.indd 193 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 © Nilam Publication Sdn. Bhd. 194 (ii) Calculate the mass of magnesium that reacts in all three experiments. [Relative atomic mass : Mg = 24, 1 mol of gas occupy 24 dm3 at room conditions] Volume of hydrogen = 100 cm3 = 0.1 dm3 Mol of hydrogen gas = 0.1 dm3 24 dm3 mol–1 = 0.00417 mol From the equation: 1 mol of H2 : 1 mol of Mg 0.00417 mol H2 : 0.00417 mol of Mg Mass of Mg = 0.00417 mol × 24 g mol–1 = 0.1 g 4 In an experiment, marble chips is added with vinegar as shown in diagram below. Table below shows the decreasing of the mass of marble for every 10 s. 180.00g Cotton wool Vinegar Marble Chips Time (s) 0 10 20 30 40 Mass (g) 180.00 68.00 44.00 40.00 40.00 (a) Write chemical equation for the reaction between marble and vinegar. CaCO3 + 2CH3COOH → (CH3COO)2Ca + CO2 + H2O (b) The mass of marble is decreasing with time. Explain your answer. Concentration of vinegar decreases (c) Plot graph mass against time in the graph provided below. 160 180 120 140 100 80 60 40 20 10 20 30 40 Time (s) Mass of marble (g) (c) (d) (d) The experiment is repeated by using marble powder to replace marble chips but the concentration of vinegar remain the same. Predict and sketch the curve in the same graph in 5(c) above. Topical Excercise PL3 PL4 PL5 PL5 07 U7 Chemistry F4(p161-195)csy2p.indd 194 21/12/2022 3:40 PM

MODULE • Chemistry FORM 4 UNIT 7 195 © Nilam Publication Sdn. Bhd. 1 The reaction between excess of magnesium with sulphuric acid is represented by the following equation. Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g) Which of the following represents the graph of mass of magnesium against time? A Time / s Mass / g C Time / s Mass / g B Time / s Mass / g D Time / s Mass / g 2 The table below shows the total volume of carbon dioxide gas collected at regular intervals in a reaction. Time (s) 0 30 60 90 120 150 180 210 Volume of gas (cm3 ) 0 2.0 3.7 5.2 6.4 7.3 8.6 8.6 What is the average rate of reaction in this experiment? A 0.041 cm3 s–1 B 0.048 cm3 s–1 C 0.049 cm3 s–1 D 0.053 cm3 s–1 3 The table below shows the volume of hydrogen gas obtained at regular intervals for the reaction between zinc and sulphuric acid. Time (min) 0 1.0 2.0 3.0 4.0 5.0 Volume of hydrogen gas (cm3 ) 0 5.5 10.0 14.0 17.5 20.0 What is the average rate of reaction from the second minute to the fourth minute? A 17.5 cm3 min–1 B 12.5 cm3 min–1 C 3.75 cm3 min–1 D 4.17 cm3 min–1 4 Which of the following will increase the frequency of collision? I Use a reactant with a smaller size II Remove the product of the reaction III Increase the temperature of the reaction IV Increase the concentration of the reactant A IV only B II and IV only C II and III only D I, III and IV only 5 The diagram below shows the graph of the volume of oxygen gas against time for the reaction between zinc and sulphuric acid. Curve X is obtained when excess granulated zinc is reacted with 25 cm3 of 1 mol dm–3 sulphuric acid. Time / s Volume / cm3 Y X Which of the following must be done to produce curve Y? A Replace granulated zinc with zinc powder. B Add a few drops of copper(II) sulphate solution. C Replace 25 cm3 of 1 mol dm–3 sulphuric acid with 25 cm3 of 2 mol dm–3 sulphuric acid. D Replace 25 cm3 of 1 mol dm–3 sulphuric acid with 50 cm3 of 1 mol dm–3 sulphuric acid. 6 Which of the following is the function of a catalyst in a chemical reaction? A To increase the kinetic energy of the particles of the reactants. B To provide a bigger total surface area for the reaction. C To increase the frequency of collision between the particles of the reactants. D To provide an alternative pathway with a lower activation energy. Objective Questions SPM PRACTICE PL3 PL3 PL3 PL3 PL4 PL4 07 U7 Chemistry F4(p161-195)csy2p.indd 195 21/12/2022 3:40 PM

Concept Map MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 196 UNIT 8 8 UNIT Component Component Purpose of alloying General Properties General Properties Type of ceramic Component Composition, properties and uses of alloy Compositions, properties and uses of different types of glass Properties Example, properties and uses MANUFACTURED SUBSTANCE IN INDUSTRY • Alloy is a mixture of two or more elements with a certain fixed/ specific/ composition. • The major component in the mixture is a metal. • From sand: Silicon and Oxygen Traditional ceramic From clay: Aluminosilicate (Al2O3.2SiO2. 2H2O) Advanced ceramic Non-organic substance such as oxide, carbide and nitride • An inorganic non-metallic solid. • It is made up of either metal or semi metal compounds. Example, properties and uses • Combination of two or more nonhomogeneous materials. • To increase hardness and strength • To resist corrosion • To improve appearance • Hard but brittle • Chemically inert • Transparent • Good electric and heat insulator • Noncompressible • Hard and strong • Chemically inert • High melting and boiling points • Good electric and heat insulator • Breakable • Matrix materials and reinforce materials Alloy: • Bronze • Brass • Steel • Stainless steel • Duralumin • Pewter Type of glass • Fused glass • Borosilicate glass • Soda-lime glass • Lead glass • Composite materials have properties that are superior to those of the original components • Superconductor • Reinforced concrete • Fiberglass • Photochromic glass Alloy Glass Ceramic Composite Material Meaning Major component Meaning Meaning MANUFACTURED SUBSTANCES IN INDUSTRY 08 U8 Chemistry F4(p196-206)csy2p.indd 196 21/12/2022 3:41 PM

MODULE • Chemistry FORM 4 197 UNIT 8 © Nilam Publication Sdn. Bhd. 8.1 ALLOY AND ITS IMPORTANCE CS 8.1 PERFORMANCE LEVEL (PL) Mastered Not mastered PL1 Recall knowledge and basic skills about manufactured substances in industry. What is the meaning of alloy? LS 8.1.1 • Alloy is a mixture of two or more elements with a certain fixed/specific composition. • The major component in the mixture is a metal . State the relationship between arrangement of atom in pure metal with its ductility and malleability. LS 8.1.2 Pure metal is easily ductile Force Pure metals • Pure metals are formed from one type of atom . • The atoms in pure metal have the same size . • Atoms of the same size are arranged in layers . • When a force is applied to a pure metal, the atomic layers slide against each other easily. • Therefore, pure metals are ductile and easily pulled into fine wires. Pure metals is easily malleable Force • There is empty space between atoms in pure metal. When force is applied, the layer of metal atoms will slide and fill the space empty to new position. • Therefore, metals are malleable and its shape can be easily changed. Draw the arrangement of atoms in (a) Bronze (90% copper and 10% tin) (b) Steel (99% iron and 1% of carbon) [Relative atomic mass: Cu = 64, Sn = 119, Fe = 56, C = 12] (a) Bronze (90% Cu & 10% Sn) Tin Copper (b) Steel (99% Fe & 1% C) Carbon Iron Explain why an alloy is stronger than its pure metal in terms of the arrangement of atoms in metals and alloys. LS 8.1.2 • Atoms of other element added to the pure metal to make an alloy consists of atoms different in size. • These atoms disrupt the orderly arrangement of atoms in pure metal. • This makes it difficult for the layers of atoms in alloy to slide over each other when force is applied. Compare the properties of an alloy and a pure metal. LS 8.1.3 Property Alloy Pure metal Surface Shiny surface Dull surface Resistant to corrosion Resistant to corrosion Easily corrode Hardness and strength Hard and strong Less hard and strong 08 U8 Chemistry F4(p196-206)csy2p.indd 197 21/12/2022 3:41 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 198 UNIT 8 Flow chart shows the composition, properties and uses of some alloys. Major component Type of alloy Type of alloy Type of alloy Type of alloy ALLOY COPPER IRON ALUMINIUM BRONZE (90% Cu, 10% Sn) • Hard and strong, does not corrode, (shiny surface) • Uses: Building statue or monuments, medals, swords and artistic materials. BRASS (70% Cu, 30% Zn) • Hard and strong. • Uses: Musical instrument and kitchenware. DURALUMIN (93% Al, 3% Cu, 3% Mg, 1% Mn) • Light and strong. • Uses: Building body of aeroplane and bullet train. CUPRONICKEL (75% Cu, 25% Ni) • Shiny, hard and does not corrode. • Uses: Making coins STEEL (98% Fe, 2% C) • Hard and strong. • Uses: Construction of building and bridge and railway tracks. STAINLESS STEEL (73% Fe, 18% Cr, 8% Ni, 1% C) • Shiny, strong and does not rust. • Uses: Making cutlery and surgical instrument. TIN PEWTER (95% Sn, 3.5% Sb, 1.5% Cu) • Luster, shiny and strong. • Uses: Making souvenirs. Experiment to compare the hardness of alloy and pure metal. LS 8.1.2 (a) Aim: To compare the hardness of brass and copper (b) Hypothesis: Brass is harder than copper (c) Manipulated variable: Copper and brass blocks (d) Responding variable: Hardness of the copper and brass blocks (e) Fixed variable: 1 kg weight (f) Apparatus: Retort stand and clamp, 1 kg weight, string, metre ruler Materials: Steel ball, copper block, brass block PERFORMANCE LEVEL (PL) Mastered Not mastered PL2 PL3 Understand and explain about manufactured substances in industry. Apply knowledge on manufactured substances in industry to explain the natural occurrences or phenomena and be able to carry out simple tasks. 08 U8 Chemistry F4(p196-206)csy2p.indd 198 21/12/2022 3:41 PM