Suggested Teaching Scheme Suggested Teaching Scheme The following suggested teaching schemes are for teachers reference only. Teachers may revise them based on the time-tabling arrangement of their own schools. Scheme 1: Chemistry to be studied in Secondary 3, 4, 5 and 6 In many schools, the Chemistry curriculum is studied in Secondary 3, 4, 5 and 6. Although the distribution of periods varies from school to school, the total number of periods for the curriculum is generally around 416. A possible distribution of periods is as follows: A possible distribution of periods for S3, S4, S5 and S6 S3 S4 S5 S6 Number of teaching weeks per year 28 28 28 16 Number of periods per week 2 5 5 5 Total number of periods per year 56 140 140 80 Total number of periods for the curriculum 416 Suggested teaching scheme for the curriculum Level S3 (56 periods) S4 (140 periods) S5 (140 periods) S6 (80 periods) Only 2 out of 3 Content Suggested number of period(s) Topic 1 Planet Earth 12 Topic 2 Microscopic World I 44 Revision on laboratory safety 1 Topic 3 Metals 39 Topic 4 Acids and Bases 45 Topic 5 Redox Reactions, Chemical Cells and Electrolysis 41 Topic 6 Microscopic World II 14 Revision on laboratory safety 1 Topic 7 Fossil Fuels and Carbon Compounds 32 Topic 8 Chemistry of Carbon Compounds 45 Topic 9 Chemical Reactions and Energy 13 Topic 10 Rate of Reaction 16 Topic 11 Chemical Equilibrium 18 Topic 12 Patterns in the Chemical World 15 Revision on laboratory safety 1 Topic 13 Industrial Chemistry 39 Topic 14 Materials Chemistry 39 Topic 15 Analytical Chemistry 40 Schools taking investigative study need to allocate an extra of 30 periods for the curriculum.
Topic 10 Rate of Reaction Scheme 2: Chemistry to be studied in Secondary 4, 5 and 6 In some schools, the Chemistry curriculum is studied in Secondary 4, 5 and 6. The total number of periods for the curriculum is generally around 360. A possible distribution of periods is as follows: A possible distribution of periods for S4, S5 and S6 S4 S5 S6 Number of teaching weeks per year 28 28 16 Number of periods per week 5 5 5 Total number of periods per year 140 140 80 Total number of periods for the curriculum 360 Suggested teaching scheme for the curriculum Level S4 (140 periods) S5 (140 periods) S6 (80 periods) Only 2 out of 3 Content Suggested number of period(s) Topic 1 Planet Earth 8 Topic 2 Microscopic World I 31 Topic 3 Metals 32 Topic 4 Acids and Bases 36 Topic 5 Redox Reactions, Chemical Cells and Electrolysis 33 Revision on laboratory safety 1 Topic 6 Microscopic World II 13 Topic 7 Fossil Fuels and Carbon Compounds 29 Topic 8 Chemistry of Carbon Compounds 41 Topic 9 Chemical Reactions and Energy 12 Topic 10 Rate of Reaction 15 Topic 11 Chemical Equilibrium 16 Topic 12 Patterns in the Chemical World 13 Revision on laboratory safety 1 Topic 13 Industrial Chemistry 39 Topic 14 Materials Chemistry 39 Topic 15 Analytical Chemistry 40 Schools taking investigative study need to allocate an extra of 30 periods for the curriculum. 2
Suggested Teaching Scheme Suggested number of periods for Topic 10 Chemistry for Total number of periods Suggested number of periods for each unit S3 S6 (Scheme 1) 16 Unit 36 An introduction to rate of reaction Unit 37 Factors affecting the rate of a reaction Unit 38 Gas volume calculations 5 7 4 S4 S6 (Scheme 2) 15 Unit 36 An introduction to rate of reaction Unit 37 Factors affecting the rate of a reaction Unit 38 Gas volume calculations 5 6 4 3
Topic 10 Rate of Reaction Teaching Plan Rate of reaction is a fundamental concept in the study of chemistry and in daily life. This topic aims to help students build up concepts related to rate of reaction. In Unit 36, students will learn different methods that can be used to follow the progress of a reaction, such as methods using a variety of physical properties of the reaction mixture and titrimetric analysis. In Unit 37, students will learn factors affecting rate of reaction, including concentration, surface area, temperature and the presence of catalyst. They will design and perform experiments to study the effects of these factors on the rate of a reaction. Students will learn to explain the effect of changes in concentration, surface area and temperature on reaction rate in terms of the behaviour of reactant particles. The important role of catalysts in both research and chemical industries will be discussed as well. Students can find a further study of catalysis in Topic 3 Industrial Chemistry. Unit 38 includes a study of molar volume of gases at room temperature and pressure, and calculations involving molar volume of gases. Organization of the topic Rate of Reaction Unit 36 An introduction to rate of reaction Unit 38 Gas volume calculations Unit 37 Factors affecting the rate of a reaction 4
Teaching Plan Unit 36 An introduction to rate of reaction Section Key point(s) Suggested task(s) for students Total number of period = 1 (Scheme 1), total number of period = 1 (Scheme 2) Remark 36.1 Fast and slow reactions 36.2 The rate of a reaction Examples of fast and slow reactions Determining the rate of a reaction by measuring the change in concentration (or amount) of a reactant per unit time, or the change in concentration (or amount) of a product per unit time Total number of period = 1 (Scheme 1), total number of period = 1 (Scheme 2) 36.3 Instantaneous rate of reaction Determining the instantaneous rate of reaction by measuring the slope of the tangent to a concentration-time curve at a particular time Practice 36.1 Total number of period = 1 (Scheme 1), total number of period = 1 (Scheme 2) 36.4 Methods for following the progress of a reaction 36.5 Following the progress of a reaction by measuring the change in volume of a gaseous product 36.6 Following the progress of a reaction by measuring the change in mass of the reaction mixture Methods using a variety of physical properties of the reaction mixture Titrimetric analysis Following the progress of the reaction between magnesium and dilute hydrochloric acid by measuring the volume of hydrogen produced Following the progress of the reaction between calcium carbonate and dilute hydrochloric acid by measuring the loss in mass of the reaction mixture Activity 36.1 Following the progress of a reaction by measuring the change in volume of a gaseous product. 5 Continued on next page
Topic 10 Rate of Reaction Section Key point(s) Suggested task(s) for students Remark Total number of period = 1 (Scheme 1), total number of period = 1 (Scheme 2) 36.7 Following the progress of a reaction by measuring the change in pressure of the reaction mixture 36.8 Following the progress of a reaction by measuring the change in colour intensity of the reaction mixture 36.9 Following the progress of a reaction by measuring the change in turbidity of the reaction mixture Following the progress of the reaction between magnesium and dilute hydrochloric acid using a pressure sensor Following the progress of the oxidation of oxalate ions by permanganate ions using a colorimeter Basic components of a colorimeter Following the progress of the reaction between sodium thiosulphate solution and dilute sulphuric acid by measuring the time to reach an opaque stage Practice 36.2 Practice 36.3 The following website contains an animation demonstrating how to follow the progress of the reaction between sodium thiosulphate solution and hydrochloric acid via the light transmittance of the reaction mixture: http://www.kscience. co.uk/animations/thio_ ict.htm (accessed July 2014) Total number of period = 1 (Scheme 1), total number of period = 1 (Scheme 2) 36.10 Following the progress of a reaction using titrimetric analysis Following the progress of the alkaline hydrolysis of ethyl ethanoate Common quenching techniques Advantages and disadvantages of using titrimetric analysis Discussion 6
Teaching Plan Unit 37 Factors affecting the rate of a reaction Section Key point(s) Suggested task(s) for students Remark Total number of periods = 2 (Scheme 1), total number of periods = 2 (Scheme 2) 37.1 Factors affecting the rate of a reaction Concentration Surface area Temperature Catalyst 37.2 Studying the effect of change in concentration of a reactant on the rate of a reaction Effect of varying the concentration of permanganate ions on the rate of its reaction with oxalate ions in an acidic solution Activity 37.1 Investigating the effect of varying the concentration of hydrochloric acid on the rate of its reaction with magnesium Activity 37.2 Investigating the effect of varying the concentration of hydroxide ion on the rate of its reaction with phenolphthalein Refer to the following website for a simulation of the progress of the catalytic decomposition of hydrogen peroxide solution: http://www.chem. iastate.edu/group/ Greenbowe/sections/ projectfolder/flashfiles/ kinetics2/rxnrate01.html (accessed July 2014) Practice 37.1 Total number of periods = 3 (Scheme 1), total number of periods = 2 (Scheme 2) 37.3 Studying the effect of change in surface area of a solid reactant on the rate of a reaction Comparing the rate of reaction of dilute hydrochloric acid with powdered calcium carbonate / calcium carbonate lumps Activity 37.3 Investigating the effect of varying the surface area of marble chips on the rate of their reaction with dilute hydrochloric acid Do you know Powders in factories Practice 37.2 Continued on next page 7
Topic 10 Rate of Reaction Section Key point(s) Suggested task(s) for students Remark 37.4 Studying the effect of change in temperature on the rate of a reaction Effect of varying the temperature on the rate of a reaction Activity 37.4 Investigating the effect of varying the temperature on the rate of the reaction between sodium thiosulphate solution and dilute sulphuric acid Investigating the effect of varying the temperature on the rate of the reaction between sodium thiosulphate solution and dilute sulphuric acid Activity 37.5 Investigating the effect of varying the temperature on the rate of the reaction between ingredients of Alka Seltzer tablet Investigating the effect of varying the temperature on the rate of the reaction between ingredients of Alka Seltzer tablet Practice 37.3 37.5 Studying how the presence of a catalyst affects the rate of a reaction What a catalyst is Positive and negative catalysts Activity 37.6 Catalyzing the decomposition of hydrogen peroxide in solution Do you know Catalysts and the ozone layer Total number of periods = 2 (Scheme 1), total number of periods = 2 (Scheme 2) 37.6 Reaction rate and effective collisions 37.7 Why does reaction rate increase with the concentration of reactants? Why does reaction rate increase with the concentration of reactants? Why does reaction rate increase with the surface area of a solid reactants? Why does reaction rate increase with temperature? Industrial catalysts Catalytic converters in car exhaust systems Hydrogenation of unsaturated vegetable oils Practice 37.4 The following website contains an animation demonstrating the correct and incorrect orientation of reactant particles in a reaction: http://www.mhhe. com/physsci/chemistry/ essentialchemistry/flash/ collis.swf (accessed July 204) Reaction rate and effective collisions Effect of concentration Continued on next page 8
Teaching Plan Section Key point(s) Suggested task(s) for students Remark 37.8 Why does reaction rate increase with the surface area of a solid reactant? Reaction rate and effective collisions Effect of surface area 37.9 Why does reaction rate increase with the temperature? Practice 37.5 Discussion Reaction rate and effective collisions Effect of temperature 37.0 Applications of catalysts Do you know Shrimp shells help make biodiesel 37. Enzymes Yeast and fermentation Uses of enzymes in industries Chemistry Magazine Saved by a very fast chemical reaction Do you know Biological washing powders 9
Topic 10 Rate of Reaction Unit 38 Gas volume calculations Section Key point(s) Suggested task(s) for students Total number of periods = 2 (Scheme 1), total number of period = 1 (Scheme 2) Remark 38.1 The relationship between gas volume and moles: Avogadro s Law 38.2 Molar volume of a gas Equal volumes of gases at the same temperature and pressure contain equal numbers of particles Calculations involving mass, number of moles and volume of a gas Practice 38.1 Activity 38.1 Determining the molar volume of carbon dioxide Practice 38.2 Total number of periods = 2 (Scheme 1), total number of periods = 2 (Scheme 2) 38.3 Calculations from chemical equations 38.4 Gas volume gas volume calculations from chemical equations Steps for calculating the quantities of reactants or products in a reaction Calculations involving masses and gas volumes Calculations involving gas volumes Practice 38.3 Practice 38.4 Do you know Density of a gas Do you know From hypothesis to law 10
Teaching Notes Teaching Notes Unit 36 An introduction to rate of reaction N1 page 3 Deducing the chemical equation of a reaction based on changes in concentration of reactant(s) and product(s) with time More difficult examination questions may ask students to deduce the chemical equation of a reaction based on curves showing changes in concentration of reactant(s) and product(s) with time. X(g) and Y(g) give Z(g) when reacted. Suppose a mixture of X(g) and Y(g) is allowed to react in a closed container of volume 1 dm 3. Changes in concentration of the species with time are shown below. 11
Topic 10 Rate of Reaction From the curves, we can see that 2 moles of X(g) react with 1 mole of Y(g) to give 1 mole of Z(g). Hence the chemical equation of the reaction is 2X(g) + Y(g) Z(g) If the mixture is allowed to react in a closed container of volume 2 dm 3, the time required for the reaction to complete will be longer as the concentrations of reactants become smaller in a larger container. N3 page 10 Saturating the hydrochloric acid with carbon dioxide The hydrochloric acid should be saturated with carbon dioxide. Otherwise, some of the carbon dioxide produced in the reaction will dissolve in the acid and affect the experimental results. Saturate the hydrochloric acid with carbon dioxide by adding a suitable amount of marble chips to the acid. N4 page 12 Following the progress of decomposition of hydrogen peroxide solution The decomposition of hydrogen peroxide solution gives oxygen and water. 2H 2 O 2 (aq) O 2 (g) + 2H 2 O(l) The progress of the decomposition can be followed by using a pressure sensor connected to a data-logger interface and a computer as oxygen is evolved. 12
Teaching Notes The rate of decomposition at A is high as the concentration of hydrogen peroxide solution is high. The rate of decomposition at B decreases as the concentration of hydrogen peroxide solution decreases during the decomposition. The decomposition stops at C because all the hydrogen peroxide has been used up. N7 page 14 Following the progress of the reaction between iodine and propanone Other than by using a colorimeter, the progress of the reaction can also be followed by the methods below: measuring the concentration of iodine by titrating with standard sodium thiosulphate solution; measuring the concentration of H + (aq) ions by titrating with standard alkali; measuring the concentration of I (aq) ions by titrating with standard silver nitrate solution; measuring the acidity of the reaction mixture by using a ph meter; measuring the electrical conductivity of the reaction mixture as there are two ions on the right side of the equation. N8 page 14 Following the progress of the reaction between oxalate ions and permanganate ions Other than by using a colorimeter, the progress of the reaction can also be followed by the methods below: measuring the loss of CO 2 (g) by weighing; measuring the volume of CO 2 (g) produced by gas collection; measuring the concentration of H + (aq) ions by titrating with a standard alkali; measuring the ph of the reaction mixture by using a ph meter. 13
Topic 10 Rate of Reaction N10 page 18 Ethyl ethanoate can be hydrolyzed in the presence of dilute hydrochloric acid to form ethanoic acid and ethanol. HCl(aq) CH 3 COOC 2 H 5 (l) + H 2 O(l) CH 3 COOH(aq) + C 2 H 5 OH(aq) ethyl ethanoate water ethanoic acid ethanol We can follow the progress of the reaction by first withdrawing small samples of the reaction mixture at regular time intervals; then titrating the total amount of acid in the sample (both the ethanoic acid formed and the hydrochloric acid used as catalyst) with a standard sodium hydroxide solution. During the hydrolysis, each molecule of ethyl ethanoate hydrolyzed produces one molecule of ethanoic acid. As hydrochloric acid is a catalyst in the reaction, its amount remains the same. Hence the increase in total amount of acid in the sample is a direct measure of the amount of ethyl ethanoate that has undergone hydrolysis. Suppose V o, V t and V are the volumes of alkali consumed to neutralize the acid in a withdrawn sample at the beginning, at time t and at the end respectively. V o is the volume of alkali required to neutralize the hydrochloric acid used originally. V is the volume of alkali required to neutralize the hydrochloric acid and the ethanoic acid formed upon complete hydrolysis. Hence (V V o ) is proportional to the concentration of ethyl ethanoate at the beginning of the reaction. V t is the volume of alkali required to neutralize the hydrochloric acid and the ethanoic acid formed at time t. Hence (V V t ) is proportional to the concentration of unreacted ethyl ethanoate at time t. Suppose a student investigated the rate of hydrolysis at 25 C and 35 C. The results obtained are shown in the following graph. (V V t ) is proportional to the concentration of unreacted ethyl ethanoate at time t. From the curves, it can be deduced that the concentration of unreacted ethyl ethanoate falls more rapidly at 35 C, i.e. the rate of hydrolysis of ethyl ethanoate is higher. 14
Teaching Notes Unit 37 Factors affecting the rate of a reaction N1 page 31 Comparing reaction of acids of different concentrations Examination questions often ask students to compare the reaction of acids of different concentrations. Examples: Comparing the reactions of 100 cm 3 of 1 mol dm 3 HCl(aq) and 50 cm 3 of 2 mol dm 3 HCl(aq) with 1 g of magnesium. they react with magnesium at different rates; they give the same amount of hydrogen gas with 1 g of magnesium (magnesium being the limiting reagent and limits the amount of hydrogen produced); the same amount of magnesium chloride is produced in both cases, but the magnesium chloride solutions produced are of different concentrations because the volumes of the two mixtures are different. Comparing the reactions of 100 cm 3 of 1 mol dm 3 HCl(aq) and 150 cm 3 of 0.8 mol dm 3 HCl(aq) with excess zinc granules. they react with zinc granules at different rates; they give different amount of hydrogen gas with excess zinc granules (the acids being the limiting reagents and limit the amount of hydrogen produced). Comparing the reactions of 100 cm 3 of 1 mol dm 3 HCl(aq) and 100 cm 3 of 1 mol dm 3 H 2 SO 4 (aq) with magnesium. the acids contain different concentration of hydrogen ions; they react with magnesium at different rates. 15
Topic 10 Rate of Reaction N2 page 37 Curve showing the effect of change in temperature on the rate of the reaction between sodium thiosulphate solution and dilute sulphuric acid N3 page 39 Characteristics of catalysts The mass of a catalyst remains unchanged at the end of the reaction. It does not change the amount of product formed in a reaction. Catalysts do not undergo any permanent chemical changes, though sometimes they may be changed physically. For example, the surface of a solid catalyst may crumble or become roughened. This suggests that the catalyst is taking some parts in the reaction, but is being regenerated. Only small amounts of catalyst are usually needed. N4 page 42 Energy profiles Energy profile for an exothermic reaction Plots like that shown below are a useful way of picturing how the potential energy changes as a reaction proceeds. The curve line is the energy pathway for a pair of colliding molecules, A B and C. It is called the energy profile for the reaction. 16
Teaching Notes In going from reactants (A B + C) to products (A + B C), the highest point on the pathway corresponds to an arrangement of atoms where old bonds are breaking and new bonds are starting to form. For the reaction of A B and C, it would be an arrangement in which the bond in A B is breaking and a new bond between B and C is forming. The highly energetic and unstable species that exist briefly at the point of maximum potential energy is called the transition state or activated complex. The activation energy for the forward reaction is represented by the difference in potential energy of the activated complex and the reactants. Notice that the products have a lower potential energy than the reactants. The difference between the potential energy of the products and that of the reactants represents the enthalpy change of the reaction ( H). Energy profile for an endothermic reaction Energy profiles will be discussed in Topic 13 Industrial Chemistry. N5 page 49 Manufacture of nitric acid The manufacture of nitric acid from ammonia involves three stages: Stage A Catalytic oxidation of ammonia to nitrogen monoxide (NO) 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) The conditions used are 900 C, about 10 atmospheres, and a platinum-rhodium catalyst. Stage B Oxidation of nitrogen monoxide (NO) to nitrogen dioxide (NO 2 ); dinitrogen tetroxide (N 2 O 4 ) is formed upon cooling the nitrogen dioxide 2NO(g) + O 2 (g) 2NO 2 (g) 2NO 2 (g) N 2 O 4 (g) Low temperatures (about 40 C) and 7 12 atmospheres are used. 17
Topic 10 Rate of Reaction Stage C Reaction of dinitrogen tetroxide with water to form nitric acid 3N 2 O 4 (g) + 2H 2 O(l) cold water 4HNO 3 (aq) + 2NO(g) This will be discussed in Topic 13 Industrial Chemistry. Unit 38 Gas volume calculations N1 page 73 Finding the volume of one mole of oxygen gas 1 Put some potassium permanganate crystals in a dry test tube and find the total mass. 2 Connect the tube to a syringe as shown below. 3 Gently heat the crystals until 100 cm 3 of oxygen gas are produced. 4 Cool the apparatus down to room temperature. 5 Note the volume of the oxygen gas collected. 6 Find the mass of the test tube and the contents again. Experimental data obtained: Volume of oxygen gas produced = 100 cm 3 = 0.100 dm 3 Loss in mass of the test tube and contents = mass of oxygen collected = 0.1333 g Since the mass of 1 mole of oxygen gas (O 2 ) = 32.0 g \ volume occupied by 1 mole of oxygen gas = 0.100 dm 3 x = 24.0 dm 3 32.0 g 0.1333 g The volume that is occupied by 1 mole of oxygen is 24.0 dm 3 at room temperature and pressure. This is the molar volume of oxygen. 18
Suggested Answers Suggested Answers page 1 1 Surface area of the fuel / presence of catalyst 2 2NaHCO 3 (s or aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + 2H 2 O(l) + 2CO 2 (g) 3 One mole Unit 36 An introduction to rate of reaction Practice P36.1 page 7 1 X: blue line Y: red line Z: green line 2 a) Average rate of reaction = b) Initial rate of reaction = P36.2 page 12 a) i) 2NaOCl(aq) 2NaCl(aq) + O 2 (g) 46 cm 3 2.0 min = 23 cm3 min 1 100 cm 3 2.5 min = 40 cm3 min 1 ii) A gas is evolved in the decomposition. b) i) Point A The tangent to the curve at point A is the steepest. c) The pressure no longer changed. d) Any one of the following: Measure the volume of gas evolved. Measure the loss in mass of the solution. 19
Topic 10 Rate of Reaction P36.3 page 17 a) Measure the volume of gas evolved. / Measure the pressure of the reaction mixture in a closed reaction vessel. Hydrogen gas is produced in the reaction. b) Measure the light absorbance of the reaction mixture using a colorimeter. When the reaction proceeds, the blue colour of the reaction mixture becomes less intense as the concentration of Cu 2+ (aq) ions falls. Concentration of Cu 2+ (aq) ions absorbance of reaction mixture c) Measure the pressure if the reaction mixture in a closed reaction vessel. 2 moles of gases react to produce 1 mole of gas. d) Any one of the following: Use an electrical conductivity meter. The electrical conductivity of the reaction mixture increases as the ammonia produced forms ammonium ions in water. Use a ph meter. The ph of the reaction mixture changes as ammonia is produced. Measure the volume of gas collected. Carbon dioxide gas is produced in the reaction. Withdraw samples at regular time intervals and titrate with standard hydrochloric acid. Ammonia is produced in the reaction. Discussion page 19 1 Major differences between titrimetric analysis and methods based on changes in physical properties for following the progress of a reaction: Titrimetric analysis can be applied to most reactions involves withdrawing samples from the reaction mixture cannot monitor concentration change continuously requires simple laboratory apparatus only takes time, and hence inappropriate for rapid reactions Methods based on changes in physical properties a method using a certain physical property can be applied to certain reactions only no need to withdraw samples from the reaction mixture can monitor concentration change continuously often requires specific apparatus readings can be taken rapidly, and hence more appropriate for rapid reactions 2 Withdrawing samples from the reaction mixture disturbs the reaction mixture. 20
Suggested Answers Unit Exercise pages 23 29 1 a) measuring the change in mass b) measuring the change in turbidity c) any reaction in which a gas is formed d) e) alkaline hydrolysis of an ester (e.g. ethyl ethanoate) 2 a) Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) b) Hydrogen gas escapes. c) Point A represents the most rapid reaction because the tangent to the curve at point A is the steepest. 3 A In Reaction W, the greatest volume of gas (22 cm 3 ) were produced in the first minute. 4 D In Reaction Z, only 12 cm 3 of gas were produced with five minutes. A total of 40 cm 3 of gas would be produced. Thus, the reaction did NOT finish within five minutes. 5 A 6 D 7 D The mass of the beaker and contents decreased as the reaction proceeded. The gaseous product (carbon dioxide) escaped from the reaction mixture. (0.050 0.200) mol dm 3 8 C Average reaction rate during the first 20 s = 20 s = 0.0075 mol dm 3 s 1 9 C 10 B 11 a) Measure the volume of gas evolved. / Pass the gas into limewater and measure the turbidity of the limewater. Carbon dioxide gas is produced in the reaction. 21
Topic 10 Rate of Reaction b) Any one of the following: Measure the light absorbance of the reaction mixture using a colorimeter. When the reaction proceeds, the brown colour of the reaction mixture becomes more intense as the concentration of iodine increases. Use a ph meter The ph of the reaction mixture changes as hydrogen ions are consumed. Withdraw samples at regular time intervals and titrate with standard sodium thiosulphate solution. The concentration of iodine increases as the reaction proceeds. c) Measure the pressure of the reaction mixture in a closed reaction vessel. mole of gas decomposes to produce 2 moles of gases. d) Withdraw sample at regular time intervals and titrate with standard sodium hydroxide solution. When the hydrolysis is complete, carry out a final titration to find out the amount of alkali needed to neutralize the acid catalyst present. Ethanoic acid is produced in the reaction. 12 a) By measuring the pressure of the reaction mixture in a closed vessel because 4 moles of gas react to give 3 moles of gas and the pressure decreases as the reaction proceeds. (0.100 1.20) mol dm 3 b) i) Average rate of reaction = 20.0 min = 0.0550 mol dm 3 min 1 ii) Draw a tangent to the curve at time = 0. iii) Determine the slope of the tangent. 22
Suggested Answers 13 a) 56 cm 3 b) 36 38 seconds c) The acid has been used up. 14 a) b) The volume of gas produced increases until levels off. The rate of production is high at the start, and then decreases. 15 a) The time taken for the solution to become colourless. b) Rate of reaction 1 time 16 a) 2MnO 4 (aq) + 5C 2 O 4 2 (aq) + 16H + (aq) 2Mn 2+ (aq) + 10CO 2 (g) + 8H 2 O(l) b) When the reaction proceeds, the purple colour of the reaction mixture becomes less intense as the concentration of MnO 4 (aq) ions falls. Concentration of MnO 4 (aq) ions absorbance of reaction mixture c) i) Average rate of consumption of MnO 4 (aq) ions ii) (0.40 1.0) mol dm 3 = 20 s = 0.030 mol dm 3 s 1 Initial rate of consumption of MnO 4 (aq) ions (0.16 1.0) mol dm 3 = 19 s = 0.044 mol dm 3 s 1 23
Topic 10 Rate of Reaction (159.68 159.99) g 17 a) Average rate of reaction = 20 min = 0.0155 g min 1 b) Draw a tangent to the curve at the particular time. Determine the slope of the tangent. c) Measure the volume of gas evolved. d) As the change in the mass is very small in this experiment, the use of a data-logger can give more accurate results. 24
Suggested Answers Unit 37 Factors affecting the rate of a reaction Practice P37.1 page 33 a) Rate of the reaction 1 t b) i) To keep the total volume of each sample constant. Thus, the concentration of sodium thiosulphate in the sample is directly proportional to the volume of thiosulphate solution used. c) ii) To ensure that the only variable is the change in the concentration of sodium thiosulphate in the sample. It can be concluded that the rate of the reaction increases when the concentration of sodium thiosulphate in the sample is increased. P37.2 page 36 a) All the zinc was used up and dissolved in the acid. b) i) t 1 is less than 200 s. During the reaction between zinc and the acids, zinc would react with hydrogen ions in the acids. Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) H 2 SO 4 (aq) is a strong dibasic acid while HCl(aq) is a strong monobasic acid. Thus, 1 mol dm 3 H 2 SO 4 (aq) has a higher concentration of hydrogen ions than 1 mol dm 3 HCl(aq). The reaction rate between zinc and 1 mol dm 3 H 2 SO 4 (aq) is thus higher and the reaction takes less time to complete. 25
Topic 10 Rate of Reaction ii) t 2 is more than 200 s. The surface area of 1 g of zinc granules is less than that of 1 g of zinc powder. The reaction rate between zinc granules and 1 mol dm 3 HCl(aq) is thus lower and the reaction takes more time to complete. P37.3 page 38 a) b) c) d) 26
Suggested Answers P37.4 page 44 a) b) The rate of reaction increases with the concentration of OH (aq) ions. Increasing the concentration of OH (aq) ions means increasing the number of reactant particles per unit volume. The particles are more crowded. The chance of collision increases and the number of effective collisions increases. P37.5 page 48 a) From the curves, 1 mole of X(g) reacts with 2 moles of Y(g) to give 1 mole of Z(g) X(g) + 2Y(g) Z(g) b) The time would be longer. In a larger container, the concentrations of the reactants become smaller. The chance of collision decreases and hence the rate of the reaction decreases. Discussion page 48 The rate of a reaction depends on how many effective collisions between reacting particles there are in a unit volume per unit time. Only those colliding particles with an energy equal to or greater than the activation energy can react. When the temperature increases, the reacting particles have more kinetic energy and collide more often. More importantly, a larger portion of the particles have an energy equal to or greater than the activation energy. 27
Topic 10 Rate of Reaction Chemistry Magazine page 53 Saved by a very fast chemical reaction 1 2NaN 3 (s) 2Na(s) + 3N 2 (g) 2 Any one of the following: The Na(s) produced will react with water vapour in the air / water from the skin to give corrosive NaOH(aq) which causes skin burns. The reaction between Na(s) and water is highly exothermic. 3 Airbags had been designed based on car occupants of average size, so the force and direction of impact of the bag itself could hurt children or small adults. Car manuals say young children should not be put in front seats where they might be injured or killed by an inflating airbag. Unit Exercise pages 58 69 1 a) activation energy b) effective c) concentration of reactant d) surface area of solid reactant e) temperature f) enzymes 2 Rate increases / remains the same / decreases Using powdered calcium carbonate Using an acid of higher concentration Adding water to the reaction mixture Adding more acid of the same concentration Raising the temperature of the acid Rate increases Rate increases Rate decreases Rate remains the same Rate increases 3 a) Cutting whole potatoes into chips increases the surface area of potatoes, thus increases the cooking rate. b) Low temperatures slow down reactions that lead to the spoiling of food. Thus storing food products in refrigerators helps keep food from spoiling. 4 A The reaction rate in Experiment 1 was the highest. Thus, the smallest marble chips were used. 28
Suggested Answers 5 D The volume of gas in Experiment 4 was lower than the other three. Thus, some carbon dioxide leaked from the gas syringe. 6 C Option C The rate of decomposition in Experiment I was higher because a higher concentration of ammonium nitrite solution was used. The volume of nitrogen collected in Experiment I was higher because a greater amount of ammonium nitrite was used. 7 C Using a more concentrated hydrochloric acid increases the frequency of collision. However, the average energy of collisions does NOT change because the temperature remains the same. 8 A When the temperature is lowered, both the frequency and energy of particle collisions decrease. 9 D Option D A catalyst can increase the rate of the reaction. A catalyst does not change the amount of product formed in the reaction. 10 B 11 B 12 a) The temperature rised. b) Repeat the experiment to obtain similar results. c) The contact surface area between magnesium and the acid increases. There is a greater chance for collision and the number of effective collisions increases. d) Dilute the acid / cooler temperature of acid / fold magnesium ribbon 13 a) The concentration of propanone in each sample is directly proportional to the volume of propanone solution used. b) When the reaction proceeds, the brown colour of the reaction mixture becomes less intense as the concentration of iodine falls. c) Concentration of I 2 (aq) absorbance of reaction mixture. 29
Topic 10 Rate of Reaction 14 a) 18 seconds b) It was because the number of acid particles per unit volume increases. There is a grater chance for collision and the number of effective collisions increases. c) The contact surface area between the tablet and hydrochloric acid increases. There is a greater chance for collision and the number of effective collisions increases. d) CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + H 2 O(l) + CO 2 (g) 15 a) Volume of 0.1 mol dm 3 Na 2 S 2 O 3 (aq) (cm 3 ) Volume of water (cm 3 ) Volume of 0.1 mol dm 3 HCl(aq) (cm 3 ) Reaction time (s) 100 0 5 20 80 20 5 25 60 40 5 33 40 60 5 50 20 80 5 100 b) Mark a cross on a piece of paper. Put a beaker containing some sodium thiosulphate solution on top of the paper. Add dilute hydrochloric acid to the beaker. Start the top watch at the same time. The cross gets fainter as the precipitate forms. Stop the stop watch when the cross can no longer be seen from above. c) The relative rate of reaction is inversely proportional to the time taken for enough sulphur to form to make the reaction mixture opaque. d) The rate of reaction increases with the concentration of sodium thiosulphate solution. Increasing the concentration of sodium thiosulphate solution means increasing the number of particles per unit volume. The particles are more crowded. The chance of collision increases and the number of effective collisions increases. 16 a) The rate of the reaction is the highest at the start. It slows down as the reaction proceeds and then stops finally. b) The change: Acid of lower concentration is used. Explanation: The rate of reaction is lower; less product is formed; the reaction ends sooner. 30
Suggested Answers 17 a) Mg(s) + 2CH 3 COOH(aq) Mg(CH 3 COO) 2 (aq) + H 2 (g) b) i) 55 cm 3 ii) Hydrochloric acid is a strong acid while ethanoic acid is a weak acid. Dilute hydrochloric acid has a higher concentration of hydrogen ions than dilute ethanoic acid of the same concentration. There is more collision between particles in a unit volume per unit time for hydrochloric acid and calcium carbonate, and thus more effective collisions. 18 a) Draw a tangent to the curve at time = 0. Determine the slope of the tangent. b) During the course of the reaction, only the concentration of Br 2 (aq) is changing. The concentration of HCOOH(aq) would be effectively constant. c) The rate of reaction increases with the concentration of bromine. Increasing the concentration of bromine means increasing the number of reactant particles per unit volume. The particles are more crowded. The chance of collision increases and the number of effective collisions increases. 19 a) i) 4.5 minutes ii) b) The contact surface area between zinc and sulphuric acid increases. There is a greater chance for collision and the number of effective collisions increases. 31
Topic 10 Rate of Reaction 74 cm 3 20 a) Rate of reaction during the first 30 seconds = 30 s = 2.5 cm 3 s 1 b) The rate of a reaction depends on how many effective collisions between reacting particles there are in a unit volume per unit time. Only those colliding particles with an energy equal to a greater than the activation energy can react. At a higher temperature, the reacting particles have more kinetic energy and collide more often. More importantly, a larger portion of the particles have an energy equal to a greater than the activation energy. 21 a) b) c) d) The concentration of hydrogen peroxide decreases. The rate of decomposition decreases. e) i) 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) ii) Hydrogen bromide is regenerated in Step 2. 32
Suggested Answers 22 a) b) c) The time required will be longer. In a larger container, concentrations of the reactants become smaller, and hence the rate of reaction decreases. d) Colliding molecules will undergo reaction if they possess an energy equal to or greater than the activation energy, and collide in the right orientation. 23 24 a) Heat is released during the reaction. b) i) Speed up the reaction. ii) 2CO + O 2 2CO 2 33
Topic 10 Rate of Reaction iii) Catalyst is not used up in the reaction. iv) Different reactions require different catalysts. 25 Answers for the HKCEE question are not provided. 26 Unit 38 Gas volume calculations Practice P38.1 page 73 Gas Mass of 1 mole of gas Mass of 1 dm 3 of gas Molar volume N 2 28.0 g 1.166 g 24.0 dm 3 O 2 32.0 g 1.333 g 24.0 dm 3 CO 2 44.0 g 1.833 g 24.0 dm 3 NH 3 17.0 g 0.708 g 24.0 dm 3 P38.2 page 76 6.80 g 1 Number of moles of NH 3 = 17.0 g mol 1 = 0.400 mol Volume of NH 3 (at room temperature and pressure) = number of moles of NH 3 x molar volume of gas (at room temperature and pressure) = 0.400 mol x 24.0 dm 3 mol 1 = 9.60 dm 3 \ 6.80 g of ammonium occupy 9.60 dm 3. 2 Number of moles of Cl 2 = = volume of Cl 2 (at room temperature and pressure) molar volume of gas (at room temperature and pressure) 672 cm 3 24 000 cm 3 mol 1 = 0.0280 mol \ 672 cm 3 of chlorine contain 0.0280 mole of the gas. 34 3 Number of moles of gas = volume of gas (at room temperature and pressure) molar volume of gas (at room temperature and pressure) = 600 cm 3 24 000 cm 3 mol 1 = 0.0250 mol
Suggested Answers Mass of gas = (78.98 77.53) g = 1.45 g 1.45 g Molar mass of gas = 0.0250 mol = 58.0 g mol 1 P38.3 page 80 1 Number of moles of Ag 2 O = 11.6 g 231.8 g mol 1 = 0.0500 mol According to the equation, 2 moles of Ag 2 O give 1 mole of O 2 upon complete decomposition. 0.0500 i.e. number of moles of O 2 formed = mol 2 = 0.0250 mol Volume of O 2 formed (at room temperature and pressure) = number of moles of O 2 x molar volume of gas (at room temperature and pressure) = 0.0250 mol x 24.0 dm 3 mol 1 = 0.600 dm 3 \ the volume of O 2 formed is 0.600 dm 3. 2 a) No more gas was produced. b) Calcium carbonate and nitric acid react according to the following equation: CaCO 3 (s) + 2HNO 3 (aq) Ca(NO 3 ) 2 (aq) + H 2 O(l) + CO 2 (g) Number of moles of gas collected (at room temperature and pressure) volume of gas (at room temperature and pressure) = molar volume of gas (at room temperature and pressure) 84.0 cm 3 = 24 000 cm 3 mol 1 = 0.00350 mol According to the equation, 1 mole of CaCO 3 reacts with HNO 3 to produce 1 mole of CO 2. i.e. number of moles of CaCO 3 = 0.00350 mol Mass of CaCO 3 in the sample = 0.00350 mol x 100.1 g mol 1 = 0.350 g 0.350 g Percentage by mass of CaCO 3 in the sample = 0.380 g x 100% = 92.1% c) The sample of calcite does not contain other carbonates that give carbon dioxide with nitric acid. 35
Topic 10 Rate of Reaction P38.4 page 83 1 a) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l)... (1) C 2 H 6 (g) + 7 2 O 2(g) 2CO 2 (g) + 3H 2 O(l)... (2) b) According to equation (1), 1 mole of CH 4 requires 2 moles of O 2 for complete combustion. i.e. 24 000 cm 3 of CH 4 require 2 x 24 000 cm 3 of O 2 for complete combustion. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) 24 000 cm 3 2 x 24 000 cm 3 00 cm 3? cm 3 Volume of O 2 required = 100 cm 3 x = 200 cm 3 2 x 24 000 cm 3 24 000 cm 3 According to equation (2), 1 mole of C 2 H 6 requires 7 2 moles of O 2 for complete combustion. i.e. 24 000 cm 3 of C 2 H 6 require 7 2 x 24 000 cm3 of O 2 for complete combustion. 7 C 2 H 6 (g) + 2 O 2(g) 2CO 2 (g) + 3H 2 O(l) 24 000 cm 3 7 x 24 000 cm3 2 00 cm 3? cm 3 Volume of O 2 required = 100 cm 3 x = 350 cm 3 7 2 x 24 000 cm3 24 000 cm 3 Total volume of O 2 required = (200 + 350) cm 3 2 X(g) + 3Y(g) 2Z(g) 40 cm 3 90 cm 3 = 550 cm 3 According to the equation, 1 mole of X reacts with 3 moles of Y to produce 2 moles of Z. \ 30 cm 3 of X react with 90 cm 3 of Y to produce 60 cm 3 of Z, i.e. 10 cm 3 of X would remain. \ volume of the resultant gaseous mixture = (10 + 60) cm 3 = 70 cm 3 Unit Exercise pages 86 90 1 a) volume of gas molar volume of gas b) mass of gas (in gram); volume of gas at room temperature and pressure (in dm 3 ) 36
Suggested Answers 2 Formula of gas Mass of 1 mole of gas Mass of gas in sample CH 4 16.0 g 4.00 g NO 2 NH 3 46.0 g 17.0 g 0.125 mol x 46.0 g mol 1 = 5.75 g 1.25 mol x 17.0 g mol 1 = 21.3 g Cl 2 71.0 g 3.55 g Number of moles of gas in sample Volume of gas sample at room temperature and pressure 4.0 g 16.0 g mol = 0.25 mol 0.25 mol x 24.0 dm 3 mol 1 1 = 6.0 dm 3 3.00 dm 3 24.0 dm 3 mol 1 = 0.125 mol 1.25 mol 3.00 dm 3 1.25 mol x 24.0 dm 3 mol 1 = 30.0 dm 3 3.55 g 71.0 g mol = 0.05 mol 0.05 mol x 24.0 dm 3 mol 1 1 = 1.20 dm 3 3 D Option Sample Number of moles of gas A B C D ethane oxygen fluorine neon 1.0 g 1 = 0.033 mol 30.0 g mol 1.0 g 1 = 0.031 mol 32.0 g mol 1.0 g 1 = 0.026 mol 38.0 g mol 1.0 g 1 = 0.050 mol 20.0 g mol \ 1 gram of neon contains the greatest number of mole of gas and thus occupies the greatest volume at room temperature and pressure. 4 B 1 dm 3 of HCl(g) and 1 dm 3 of CH 4 (g) contain the same number of molecules. One molecule of HCl(g) contains 2 atoms while one molecule of CH 4 (g) contains 5 atoms. Thus, if 1 dm 3 of HCl(g) contains x atoms, then 1 dm 3 of CH 4 (g) contains 5 2 x atoms. 5 C 2C 8 H 18 (l) + 25O 2 (g) 16CO 2 (g) + 18H 2 O(l) 0.0 mol According to the equation, 2 moles of C 8 H 18 require 25 moles of O 2 for complete combustion. \ 10.0 moles of C 8 H 18 required 125 moles of O 2 for complete combustion. Volume of O 2 required = 125 mol x 24.0 dm 3 mol 1 = 3 000 dm 3 \ 3 000 dm 3 of oxygen are required for complete combustion. 37
Topic 10 Rate of Reaction 6 C 7 B 8 C 9 B Let C n H 2n+2 be the formula of the hydrocarbon. C n H 2n+2 + n + n + 1 2 O 2 nco 2 + (n+1)h 2 O 50 cm 3 350 cm 3 According to the equation, 1 mole of C n H 2n+2 gives n moles of CO 2 upon complete combustion. i.e. 50 cm 3 350 cm 3 = 1 n n = 7 \ the formula of the hydrocarbon is C 7 H 16. 10 B 11 a) Density of CO 2 = = mass (g) volume (dm 3 ) 0.7488 g ( 400.0 1 000 ) dm3 = 1.872 g dm 3 b) Molar mass of CO 2 = 44.0 g mol 1 Volume of one mole of CO 2 = c) Number of moles of gas G = = molar mass density 44.0 g mol 1 1.872 g dm 3 = 23.5 dm 3 mol 1 volume of gas G (under the experimental conditions) molar volume of gas (under the experimental conditions) = ( 200.0 1 000 ) dm3 23.5 dm 3 mol 1 = 8.51 x 10 3 mol 0.5448 g Molar mass of gas G = 8.51 x 10 3 mol = 64.0 g mol 1 38
Suggested Answers 0.535 g 12 Number of moles of NH 4 Cl = 53.5 g mol 1 = 0.0100 mol According to the equation, 2 moles of NH 4 Cl give 2 moles of NH 3 when reacted with Ca(OH) 2. i.e. number of moles of NH 3 formed = 0.0100 mol Volume of NH 3 formed (at room temperature and pressure) = number of moles of NH 3 x molar volume of gas (at room temperature and pressure) = 0.0100 mol x 24.0 dm 3 mol 1 = 0.240 dm 3 5.29 g 13 Number of moles of Sr(NO 3 ) 2 = 211.6 g mol 1 = 0.0250 mol According to the equation, 2 moles of Sr(NO 3 ) 2 give 5 moles of gas upon decomposition. i.e. number of moles of gas = 0.0250 x 5 2 mol = 0.0625 mol Volume of gas obtained (at room temperature and pressure) = number of moles of gas x molar volume of gas (at room temperature and pressure) = 0.0625 mol x 24.0 dm 3 mol 1 = 1.50 dm 3 \ the volume of gas obtained is 1.50 dm 3. 14 a) 2NaHCO 3 (s) Na 2 CO 3 (s) + CO 2 (g) + H 2 O(l) 54.6 g b) Number of moles of NaHCO 3 = 84.0 g mol 1 = 0.650 mol According to the equation, 2 moles of NaHCO 3 give 1 mole of CO 2 upon decomposition. i.e. number of moles of CO 2 formed = 0.650 mol 2 = 0.325 mol Volume of CO 2 formed (at room temperature and pressure) = number of moles of CO 2 x molar volume of gas (at room temperature and pressure) = 0.325 mol x 24.0 dm 3 mol 1 = 7.80 dm 3 \ the volume of carbon dioxide formed is 7.80 dm 3. 39
Topic 10 Rate of Reaction 15 a) Number of moles of N 2 = = volume of N 2 molar volume of gas 108 dm 3 24.0 dm 3 mol 1 = 4.50 mol b) According to the equation, 2 moles of NaN 3 produce 3 moles of N 2 upon decomposition. i.e. number of moles of NaN 3 required = 4.50 x 2 3 mol = 3.00 mol Mass of NaN 3 required = 3.00 mol x 65.0 g mol 1 16 Number of moles of CO 2 = = = 195 g volume of CO 2 (at room temperature and pressure) molar volume of gas (at room temperature and pressure) 480 cm 3 24 000 cm 3 mol 1 = 0.0200 mol According to the equation, 1 mole of MCO 3 decomposes to give 1 mole of CO 2. i.e. number of moles of MCO 3 = 0.0200 mol 2.50 g Formula mass of MCO 3 = 0.0200 mol = 125 g mol 1 Let x be the relative mass of metal M. x + 12.0 + 3 x 16.0 = 125 x = 65 \ M is zinc. 17 Number of moles of CO 2 = = volume of CO 2 (at room temperature and pressure) molar volume of gas (at room temperature and pressure) 3.60 dm 3 24.0 dm 3 mol 1 = 0.150 mol i.e. number of moles of oxygen atoms in the oxide = 2 x 0.150 mol = 0.300 mol 38.1 g Number of moles of M in the oxide = 63.5 g mol 1 = 0.600 mol Number of moles of M : number of moles of O = 0.600 : 0.300 = 2 : 1 \ the empirical formula of the oxide is M 2 O. 40
Suggested Answers 18 a) i) Number of moles of CO 2 = = volume of CO 2 molar volume of gas 58.4 cm 3 24 000 cm 3 mol 1 = 0.00243 mol ii) According to the equation, 1 mole of NaHCO 3 reacts with hydrochloric acid to give 1 mole of CO 2. i.e. number of moles of NaHCO 3 present = 0.00243 mol Mass of NaHCO 3 present = 0.00243 mol x 84.0 g mol 1 = 0.204 g 0.204 g iii) Percentage purity of NaHCO 3 = 0.227 g x 100% = 89.9% b) Any one of the following: Carbon dioxide is soluble in water. Carbon dioxide reacts with water. / Carbon dioxide forms carbonic acid with water. 19 20 21 a) MgCO 3 (s) + 2HCl(aq) MgCl 2 (aq) + CO 2 (g) + H 2 O(l) b) MgCO 3 (s) + 2HCl(aq) MgCl 2 (aq) + CO 2 (g) + H 2 O(l) 3.88 g 2.00 mol dm 3? cm 3 50.0 cm 3 3.88 g Number of moles of MgCO 3 present = 84.3 g mol 1 = 0.0460 mol Number of moles of HCl present = 2.00 mol dm 3 x = 0.100 mol 50.0 1 000 dm3 According to the equation, 1 mole of MgCO 3 reacts with 2 moles of HCl to produce 1 mole of CO 2. HCl is in excess in this case. The amount of MgCO 3 limits the amount of CO 2 liberated. Number of moles of CO 2 liberated = 0.0460 mol Volume of CO 2 liberated (at room temperature and pressure) = number of moles of CO 2 x molar volume of gas (at room temperature and pressure) = 0.0460 mol x 24.0 dm 3 mol 1 = 1.10 dm 3 \ 1.10 dm 3 of carbon dioxide gas are liberated. 41
Topic 10 Rate of Reaction 22 a) Let C x H y be the molecular formula of the two isomer compounds. The compounds undergo complete combustion according to the following equation: C x H y (g) + ( y x + 4 ) O y 2(g) xco 2 (g) + 2 H 2O(l) 0.0150 mol 080 cm 3 0.810 g Number of moles of CO 2 = \ x 1 = 0.0450 mol 0.0150 mol x = 3 Number of moles of H 2 O = \ y 2 1 = 0.0450 mol = 0.0150 mol y = 6 volume of CO 2 (at room temperature and pressure) molar volume of gas (at room temperature and pressure) 1 080 cm 3 24 000 cm 3 mol 1 = 0.0450 mol 0.810 g 18.0 g mol 1 = 0.0450 mol \ the molecular formula of the two isomeric compounds is C 3 H 6. Structures of the two isomeric compounds: CH 2 H 2 C CH 2 CH 3 CH=CH 2 b) Any one of the following: Mix each compound with aqueous bromine. Only CH 3 CH=CH 2 can turn the yellow-brown aqueous bromine colourless quickly. Mix each compound with cold acidified dilute potassium permanganate solution. Only CH 3 CH=CH 2 can turn the purple permanganate solution colourless quickly. 0.150 g 23 a) Number of moles of NO = 30.0 g mol 1 = 0.00500 mol 112 Volume of 1 mole of gas = 0.00500 cm3 = 22 400 cm 3 42
Suggested Answers b) 2NO(g) + O 2 (g) 2NO 2 (g) 200 cm 3? cm 3 According to the equation, 2 moles of NO give 2 moles of NO 2 when reacted with oxygen. i.e. 2 x 24 000 cm 3 of NO give 2 x 24 000 cm 3 of NO 2. 2NO(g) + O 2 (g) 2NO 2 (g) 2 x 24 000 cm 3 2 x 24 000 cm 3 200 cm 3? cm 3 Volume of NO 2 formed = 200 cm 3 x = 200 cm 3 2 x 24 000 cm 3 2 x 24 000 cm 3 c) Number of moles of NO = = volume of NO (at room temperature and pressure) molar volume of gas (at room temperature and pressure) 480 cm 3 24 000 cm 3 mol 1 = 0.0200 mol Number of molecules in 480 cm 3 of NO = 0.0200 L 24 a) Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) b) i) As the reaction proceeds, the reactants are consumed gradually, i.e. the concentrations decreases. The chance of collision decreases and the number of effective collisions decreases. Hence the rate of the reaction decreases. The reaction stops when zinc is used up. 0.981 g ii) Number of moles of Zn = 65.4 g mol 1 = 0.0150 mol According to the equation, 1 mole of Zn gives 1 mole of H 2 when reacted with hydrochloric acid. i.e. number of moles of H 2 = 0.0150 mol Volume of H 2 (at room temperature and pressure) = number of moles of H 2 x molar volume of gas (at room temperature and pressure) = 0.0150 mol x 24 000 cm 3 mol 1 = 360 cm 3 c) i) During the reaction between zinc and the acids, zinc would react with hydrogen ions in the acids. Hydrochloric acid is a strong acid that completely dissociates in water while ethanoic acid is a weak acid that only partially dissociates in water. Therefore hydrochloric acid has a higher concentration of hydrogen ions than ethanoic acid. There is more collisions between reactant particles in a unit volume per unit time for hydrochloric acid and thus more effective collisions. ii) The same volume of gas would be given off as zinc was the limiting reagent. 43