Suggested answers to in-text activities and unit-end exercises

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1 Topic 7 Unit 30 Suggested answers to in-text activities and unit-end exercises In-text activities Checkpoint (page 70) a) NO(g) + 1 O (g) Enthalpy (kj mol 1 ) NO (g) ΔH = +34 kj mol 1 1 N (g) + O (g) ΔH = +90 kj mol 1 ΔH = 56 kj mol 1 NO (g) route 1 route b) route 1 1 N (g) + O (g) ΔH = +34 kj mol 1 NO (g) ΔH x =? ΔH y =? NO(g) + 1 O (g) route ΔH x = +90 kj mol 1 ΔH y = 56 kj mol 1 47

2 Checkpoint (page 78) 1 The following data is given in the question: (1) N O(g) N (g) + 1 O (g) ΔH 1 = 8 kj mol 1 () NO(g) + 1 O (g) NO (g) ΔH = 57 kj mol 1 (3) N (g) + O (g) NO(g) ΔH 3 = +181 kj mol 1 Looking at the target equation, we need 1 mole of N O(g) as the reactant. So, keep equation (1) as it is. We need 1 mole of NO (g) as the reactant. Equation () has 1 mole of NO (g), but it is on the product side. So, reverse the equation, giving equation (). By combining equations (1), (3) and (), followed by collecting like items, we can obtain the target equation. (1) N O(g) N (g) + 1 O (g) ΔH 1 = 8 kj mol 1 () NO (g) NO(g) + 1 O (g) ΔH = +57 kj mol 1 (3) N (g) + O (g) NO(g) ΔH 3 = +181 kj mol 1 N O(g) + NO (g) 3NO(g) ΔH 4 = +156 kj mol 1 the enthalpy change for the reaction is +156 kj mol 1. a) Rhombic sulphur is more stable. b) The following data is given in the question: (1) S(rhombic) + O (g) SO (g) ΔH = 96.0 kj mol 1 () S(monoclinic) + O (g) SO (g) ΔH = 96.4 kj mol 1 By combining equations (1) and (), followed by collecting like items, we can obtain the target equation. (1) S(rhombic) + O (g) SO (g) ΔH = 96.0 kj mol 1 () SO (g) S(monoclinic) + O (g) ΔH = kj mol 1 S(rhombic) S(monoclinic) ΔH = +0.4 kj mol 1 the enthalpy change for the process is +0.4 kj mol 1. 48

3 3 a) Zn(s) + S(s) + 3 O (g) ΔH O f [ZnS(s)] ZnS(s) + 3 O (g) ΔH O = [( 348) + ( 97)] kj mol 1 ΔH O = 441 kj mol 1 b) By Hess s Law ZnO(s) + SO (g) c) ΔH o f[zns(s)] = [( 348) + ( 97) ( 441)] kj mol 1 = 04 kj mol 1 the enthalpy change of formation of zinc sulphide is 04 kj mol 1. Enthalpy (kj mol 1 ) Zn(s) + S(s) + 3 O (g) ΔH O f [ZnS(s)] = 04 kj mol 1 ZnS(s) + 3 O (g) ΔH O = [( 348) + ( 97)] kj mol 1 = 645 kj mol 1 ΔH O = 441 kj mol 1 ZnO(s) + SO (g) Checkpoint (page 86) a) i) Amount of heat released = 80 g x 4.18 J g 1 K 1 x 9.6 K = 3 00 J 8.0 g Number of moles of CuSO 4 (s) added = g mol 1 = mol 3 00 J Enthalpy change of solution of CuSO 4 (s) = mol = J mol 1 = 64.0 kj mol 1 ii) Amount of heat taken in = 80 g x 4.18 J g 1 K 1 x.0 K = 670 J 1.0 g Number of moles of CuSO 4 5H O(s) added = 49.6 g mol 1 = mol 49

4 +670 J Enthalpy change of solution of CuSO 4 5H O(s) = mol = J mol 1 = kj mol 1 b) i) ΔH r = ΔH 1 ΔH ii) Enthalpy change of the process ΔH r = ΔH 1 ΔH = [( 64.0) (+14.0)] kj mol 1 = 78.0 kj mol 1 the enthalpy change of the process is 78.0 kj mol 1. Checkpoint (page 93) 1 a) ΔH represents the enthalpy change of formation of C H 4 (g). ΔH 3 represents the enthalpy change of combustion of C H 4 (g). b) ΔH 1 represents the enthalpy change of combustion of C(graphite) and the enthalpy change of combustion of H (g). c) ΔH o f[c H 4 (g)] = x ΔH o c[c(graphite)] + x ΔH o c[h (g)] ΔH o c[c H 4 (g)] = [( 394) + ( 86) ( 1 411)] kj mol 1 = +51 kj mol 1 the standard enthalpy change of formation of ethene is +51 kj mol 1. d) C(graphite) + H (g) + 3O (g) C H 4 (g) + 3O (g) ΔH O f [C H 4 (g)] = +51 kj Enthalpy (kj) x ΔH O c [C(graphite)] + x ΔH O c [H (g)] = [( 394) + ( 86)] kj = kj ΔH O c [C H 4 (g)] = kj CO (g) + H O(l) e) Carbon and hydrogen do not react to produce ethene under normal conditions. ΔH o f[c 1 H O 11 (s)] refers to the standard enthalpy change of the following process: 1C(graphite) + 11H (g) + 11 O (g) C 1 H O 11 (s) ΔH o f[compound] = ΔH o c[constituent elements] ΔH o c[compound] ΔH o f[c 1 H O 11 (s)] = 1 x ΔH o c[c(graphite)] + 11 x ΔH o c[h (g)] ΔH o c[c 1 H O 11 (s)] = [1( 394) + 11( 86) ( 5 640)] kj mol 1 = 30 kj mol 1 50 the standard enthalpy change of formation of sucrose is 30 kj mol 1.

5 Checkpoint (page 99) 1 a) ΔH O r (NH 4 ) Cr O 7 (s) N (g) + 4H O(g) + Cr O 3 (s) ΔH O f [(NH 4 ) Cr O 7 (s)] 4 x ΔH O f [H O(g)] + ΔH O f [Cr O 3 (s)] N (g) + 4H (g) + Cr(s) + 7 O (g) b) ΔH o f[n (g)] = 0 c) ΔH o r = ΔH o f[products] ΔH o f[reactant] = 4 x ΔH o f[h O(g)] + ΔH o f[cr O 3 (s)] ΔH o f[(nh 4 ) Cr O 7 (s)] = [4( 4) + ( 1 140) ( 1 810)] kj mol 1 = 98 kj mol 1 the standard enthaply change of the reaction is 98 kj mol 1. a) Amount of heat released when 1.00 g of butan-1-ol was burnt = 00.0 g x 4.18 J g 1 K 1 x (44.0.0) K = J = 18.4 kj 1.00 g Number of moles of butan-1-ol burnt = 74.0 g mol 1 = mol 18.4 kj Enthalpy change of combustion of butan-1-ol = mol = kj mol 1 the enthalpy change of combustion of butan-1-ol is kj mol 1. b) ΔH o c[c 4 H 9 OH(l)] refers to the standard enthalpy change of the following process: C 4 H 9 OH(l) + 6O (g) 4CO (g) + 5H O(l) ΔH o r = ΔH o f[products] ΔH o r[reactants] ΔH o c[c 4 H 9 OH(l)] = 4 x ΔH o f[co (g)] + 5 x ΔH o f[h O(l)] ΔH o f[c 4 H 9 OH(l)] = [4( 394) + 5( 86) ( 37)] kj mol 1 = 680 kj mol 1 the standard enthaply change of combustion of butan-1-ol is 680 kj mol 1. 51

6 c) Any two of the following: Heat loss to the surroundings. Incomplete combustion of butan-1-ol. Heat capacities of the container and the thermometer are ignored. Experiment not carried out under standard conditions. Loss of butan-1-ol / water via evaporation. The thermometer not precise enough. STSE Connections (page 101) 1 ΔH o c[c 8 H 18 (l)] refers to the standard enthalpy change of the following process: C 8 H 18 (l) + 5 O (g) 8CO (g) + 9H O(l) ΔH o c[c 8 H 18 (l)] = ΔH o f[products] ΔH o f[reactants] = [8( 394) + 9( 86) ( 50)] kj mol 1 = kj mol 1 ΔH o f[c H 5 OH(l)] refers to the standard enthalpy change of the following process: C H 5 OH(l) + 3O (g) CO (g) + 3H O(l) ΔH o c[c H 5 OH(l)] = ΔH o f[products] ΔH o f[reactants] = [( 394) + 3( 86) ( 78)] kj mol 1 = kj mol 1 Advantages and disadvantages of using ethanol as a fuel for cars Advantages Ethanol can be produced from many different types of food crops. Ethanol can burn directly as a fuel and does not require other additives to improve its efficiency. Ethanol burns to form mainly carbon dioxide and water. It is a cleaner fuel than other fossil fuels. Ethanol is a renewable resource. Ethanol can be easily transported to various places. Distilleries are built for ethanol production. This creates jobs for people. Cars can be easily converted to run on pure ethanol. Disadvantages It requires a large area of land to produce enough food crops to meet the demand. It may lead to diversion of investment from food production, resulting in increased food prices. The toxic effluent from distilleries causes water pollution. O Ethanal (CH 3 C H) is also produced during combustion. Without adequate pollution control, it can harm vegetation, irritate the skin and eyes, and damage the lung at high concentrations. Fuels are required to produce ethanol in distilleries. Combustion of 1 kg of ethanol can give only 60% of the energy that can be released from 1 kg of petrol. It may not be possible to obtain cheap ethanol in every part of the world. 5

7 Unit-end exercises (pages ) Answers for the HKCEE (Paper 1) and HKALE questions are not provided. 1 A possible concept map is shown below: HC CH(g) + H (g) Enthalpy (kj mol 1 ) Step 1 Step CH =CH (g) + H (g) CH 3 CH 3 (g) 53

8 3 a) C 8 H 18 (l) + 5 O (g) 8CO (g) + 9H O(l) b) ΔH O f [C 8 H 18 (l)] 5 8C(graphite) + 9H (g) + O (g) C 8 H 18 (l) + 5 O (g) 8CO (g) + 9H O(l) c) ΔH o f[compound] = ΔH o c[constituent elements] ΔH o c[compound] ΔH o f[c 8 H 18 (l)] = 8 x ΔH o c[c(graphite)] + 9 x ΔH o c[h (g)] ΔH o c[c 8 H 18 (l)] = [8( 394) + 9( 86) ( 5 470)] kj mol 1 = 56 kj mol 1 4 a) ΔH O r CH 4 (g) + H O(g) CO (g) + 4H (g) C(graphite) + 4H (g) + O (g) b) ΔH o f[h (g)] = 0 c) ΔH o r = ΔH o f[products] ΔH o f[reactants] 5 a) None ΔH o r = ΔH o f[co (g)] ΔH o f[ch 4 (g)] x ΔH o f[h O(g)] = [( 394) ( 74.8) ( 4)] kj mol 1 = +165 kj mol 1 ΔH o r = ΔH o f[products] ΔH o f[reactants] = 3 x ΔH o f[cao(s)] ΔH o f[fe O 3 (s)] Both ΔH o f[cao(s)] and ΔH o f[fe O 3 (s)] are given in the question. b) ΔH o r = ΔH o f[products] ΔH o f[reactants] = ΔH o f[cao(s)] ΔH o f[cuo(s)] the value of ΔH o f[cuo(s)] is required for the calculations. c) ΔH o r = ΔH o f[products] ΔH o f[reactants] = ΔH o f[fe O 3 (s)] 3 x ΔH o f[cuo(s)] the value of ΔH o f[cuo(s)] is required for the calculations. 6 B ΔH o r = ΔH o f[products] ΔH o f[reactants] = 3 x ΔH o f[co(g)] ΔH o f[fe O 3 (s)] = [3( 110) ( 8)] kj = +49 kj 54

9 7 C ΔH o r = ΔH o f[products] ΔH o f[reactants] = 3 x ΔH o f[mgo(s)] + ΔH o f[kcl(s)] ΔH o f[kclo 3 (s)] = [3( 60) + ( 437) ( 391)] kj = 1 85 kj 8 B ΔH o r = ΔH o f[products] ΔH o f[reactants] = ΔH o f[ca(oh) (s)] x ΔH o f[h O(l)] the extra information required is the enthalpy change of formation of H O(l), i.e. the enthalpy change of combustion of H (g). 9 A The following data is given in the question: (1) H (g) + 1 O (g) H O(l) ΔH = q () S(s) + O (g) SO (g) ΔH = r (3) H S(g) + 3 O (g) H O(l) + SO (g) ΔH = s Looking at the target equation, we need 1 mole of S(s) and 1 mole of H (g) as the reactants. So, keep equations (1) and () as they are. We need 1 mole of H S(g) as the product. Equation (3) has 1 mole of H S(g), but it is on the reactant side. So, we need to reverse the equation, giving equation (3). By combining equations (1), () and (3), followed by collecting like items, we can obtain the target equation. (1) H (g) + 1 O (g) H O(l) ΔH = q () S(s) + O (g) SO (g) ΔH = r (3) H O(l) + SO (g) H S(g) + 3 O (g) ΔH = s S(s) + H (g) H S(g) ΔH = p = q + r s 10 A (1) CuSO 4 (s) + 5H O(l) CuSO 4 5H O(s) 11 The above reaction cannot be properly carried out. Some CuSO 4 (s) may dissolve but may not be completely hydrated. () C(s) + H (g) CH 4 (g) Carbon and hydrogen do not react to produce methane under normal conditions. (3) C 6 H 1 O 6 (s) + 6O (g) 6CO (g) + 6H O(l) The enthalpy change of combustion of glucose can be measured directly: burn a fixed mass of glucose and measure the heat evolved. 55

10 1 a) 3C(graphite) + H (g) CH 3 C CH(g) b) 3C(graphite) + H (g) + 4O (g) ΔH O f [CH 3 C CH(g)] CH 3 C CH(g) + 4O (g) 3 x ΔH O c [C(graphite)] + x ΔH O c [H (g)] ΔH O c [CH 3 C CH(g)] 3CO (g) + H O(l) c) No ΔH o f[ch 3 C CH(g)] = 3 x ΔH o c[c(graphite)] + x ΔH o c[h (g)] ΔH o c[ch 3 C CH(g)] = [3( 394) + ( 86) ( 1 938)] kj mol 1 = +184 kj mol 1 the enthalpy change of formation of propyne is +184 kj mol 1. Carbon and hydrogen do not react to produce propyne under normal conditions. 13 a) N (g) + 5 O (g) N O 5 (g) b) Looking at equation (I), we need to eliminate the NO(g) on the product side. So, multiply equation (II) by, giving equation (II). By combining equation (I), (II) and (III), followed by collecting like terms, we can obtain the equation for the formation of N O 5 (g) from its constituent elements. (I) N (g) + O (g) NO(g) ΔH = +180 kj (II) NO(g) + O (g) NO (g) ΔH = 114 kj (III) NO (g) + 1 O (g) N O 5 (g) ΔH = 55 kj N (g) + 5 O (g) N O 5 (g) By Hess s Law ΔH o f[n O 5 (g)] = [ ( 114) + ( 55)] kj mol 1 = +11 kj mol 1 the enthalpy change of formation of dinitrogen pentoxide is +11 kj mol 1. 56

11 14 a) C 6 H 1 O 6 (s) + 6O (g) 6CO (g) + 6H O(l) b) C 6 H 1 O 6 (s) + 6O (g) ΔH O c [C 6 H 1 O 6 (s)] 6CO (g) + 6H O(l) ΔH O f [C 6 H 1 O 6 (s)] 6 x ΔH O f [CO (g)] + 6 x ΔH O f [H O(l)] 6C(graphite) + 6H (g) + 9O (g) c) ΔH o c[c 6 H 1 O 6 (s)] = 6 x ΔH o f[co (g)] + 6 x ΔH o f[h O(l)] ΔH o f[c 6 H 1 O 6 (s)] = [6( 394) + 6( 86) ( 1 74)] kj mol 1 = 806 kj mol 1 Enthalpy (kj) 6C(graphite) + 6H (g) + 9O (g) ΔH O f [C 6 H 1 O 6 (s)] C 6 H 1 O 6 (s) + 6O (g) ΔH O c [C 6 H 1 O 6 (s)] 6 x ΔH O f [CO (g)] + 6 x ΔH O f [H O(l)] 6CO (g) + 6H O(l) 15 a) NaHCO 3 (s) ΔH O r Na CO 3 (s) + CO (g) + H O(l) x ΔH O f [NaHCO 3 (s)] ΔH O f [Na CO 3 (s)] + ΔH O f [CO (g)] + ΔH O f [H O(l)] Na(s) + C(graphite) + H (g) + 3O (g) ΔH o r = ΔH o f[na CO 3 (s)] + ΔH o f[co (g)] + ΔH o f[h O(l)] x ΔH o f[nahco 3 (s)] = [( 1 131) + ( 394) + ( 86) ( 951)] kj = +91 kj the standard enthaply change accompanying the thermal decomposition of moles of NaHCO 3 is +91 kj. 57

12 b) Na(s) + C(graphite) + H (g) + 3O (g) Enthalpy (kj) x ΔH O f [NaHCO 3 (s)] ΔH O f [Na CO 3 (s)] +ΔH O f [CO (g)] +ΔH O f [H O(l)] Na CO 3 (s) + CO (g) + H O(l) NaHCO 3 (s) ΔH O r 16 a) C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O(l) b) The standard enthalpy change of combustion of a substance is the enthalpy change when one mole of the substance is completely burnt in oxygen under standard conditions. 17 c) i) Amount of heat produced = 00 g x 4.18 J g 1 K 1 x ( ) K = J = 4.1 kj 1.00 g ii) Number of moles of C 3 H 8 burnt = 44.0 g mol 1 = 0.07 mol 4.1 kj iii) Enthalpy change of combustion of C 3 H 8 = 0.07 mol = kj mol 1 the enthalpy change of combustion of C 3 H 8 is kj mol 1. d) i) ΔH o f[compound] = ΔH o c[constituent elements] ΔH o c[compound] ΔH o f[c 3 H 8 (g)] = 3 x ΔH o c[c(graphite)] + 4 x ΔH o c[h (g)] ΔH o c[c 3 H 8 (g)] = [3( 394) + 4( 86) ( 19)] kj mol 1 = 107 kj mol 1 the enthalpy change of formation of propane is 107 kj mol 1. ii) Carbon and hydrogen do not react to produce propane under normal conditions. 58

13 18 a) Amount of heat released = 600 J K 1 x 8.60 K = J 0.43 g Number of moles of Mg reacted = 4.3 g mol 1 = mol J Enthalpy change of the reaction = mol = J mol 1 = 516 kj mol 1 b) ΔH o f[mg(oh) (s)] refers to the enthalpy change of the following process: Mg(s) + O (g) + H (g) Mg(OH) (s) ΔH values of the following processes are known: (1) Mg(s) + HCl(aq) MgCl (aq) + H (g) ΔH = 516 kj mol 1 () Mg(OH) (s) + HCl(aq) MgCl (aq) + H O(l) ΔH = 163 kj mol 1 (3) H (g) + 1 O (g) H O(l) ΔH = 86 kj mol 1 Looking at the target equation, we need 1 mole of Mg(OH) (s) as the product. Equation () has 1 mole of Mg(OH) (s), but it is on the reactant side. So, reverse the equation, giving equation (). We need moles of H (g) as the reactant, 1 mole for the target equation and 1 mole for elimination with the H (g) in equation (1). So, multiply equation (3) by, giving equation (3). By combining equations (1), () and (3), followed by collecting like items, we can obtain the equation for the formation of Mg(OH) (s) from its constituent elements. (1) Mg(s) + HCl(aq) MgCl (aq) + H (g) ΔH = 516 kj () MgCl (aq) + H O(l) Mg(OH) (s) + HCl(aq) ΔH = +163 kj (3) H (g) + O (g) H O(l) ΔH = 57 kj Mg(s) + H (g) + O (g) Mg(OH) (s) ΔH o f[mg(oh) (s)] = [( 516) + (+163) + ( 57)] kj mol 1 = 95 kj mol 1 the enthalpy change of formation of Mg(OH) is 95 kj mol a) i) Amount of heat produced by the reaction of the first sample = 5.0 g x 4.18 J g 1 K 1 x 16.5 K = 1 70 J Amount of heat produced by the reaction of the second sample = 5.0 g x 4.18 J g 1 K 1 x 5.5 K = 660 J 59

14 1.00 g ii) Number of moles of Ca(OH) (s) used = 74.1 g mol 1 = mol 1 70 J iii) Enthalpy change for the reaction of the first sample = mol = J mol 1 = 17 kj mol J Enthalpy change for the reaction of the second sample = mol = J mol 1 = 197 kj mol 1 b) i) ΔH r = enthalpy change for the reaction of the first sample enthalpy change for the reaction of the second sample = [( 17) ( 197)] kj mol 1 = +70 kj mol 1 ii) Any two of the following: Using a glass beaker with no lid is likely to lead to heat loss. The glass beaker has a significant heat capacity. It is difficult to know whether Ca(OH) was fully decomposed. The thermometer may not be precise enough. iii) Any one of the following: It is difficult to measure the temperatures of solids. High temperatures are involved (above the boiling points of mercury / ethanol), so laboratory thermometers cannot be used. It is difficult to know whether Ca(OH) has fully decomposed. Given the high temperatures involved, it is impossible to use a thermometer to measure the heat taken in by Ca(OH). 0 a) i) Amount of heat change = 104 g x 4.09 J g 1 K 1 x ( ) K = J = 11.1 kj 4.00 g ii) Number of moles of AlCl 3 reacted = g mol 1 = mol 11.1 kj Enthalpy change when 1 mole of AlCl 3 reacted = mol = 370 kj mol 1 60 b) Record the temperature of the water every half minute for 1 minutes. At precisely the third minute, add AlCl 3 (s) to the water. Stir the mixture and record the temperature for an additional 6 minutes. Plot the temperature against time. Join the points before the addition of AlCl 3 (s) using a straight line and extrapolate to the third minute (the time at which AlCl 3 (s) is added). Label the temperature at the third minute as T i. Join the points after the addition of AlCl 3 (s) using a straight line and extrapolate back to the third minute. Label the temperature at the third minute as T f. The separation of the lines at the third minute corresponds to the maximum temperature change for the process, i.e. maximum temperature rise ΔT = T f T i

15 1 a) 3Al(s) + 3NH 4 ClO 4 (s) Al O 3 (s) + AlCl 3 (s) + 3NO(g) + 6H O(g) ΔH o r = ΔH o f[products] ΔH o f[reactants] = ΔH o f[al O 3 (s)] + ΔH o f[alcl 3 (s)] + 3 x ΔH o f[no(g)] + 6 x ΔH o f[h O(g)] 3 x ΔH o f[nh 4 ClO 4 (s)] = [( 1 676) + ( 706) + 3(+90.) + 6( 4) 3( 95)] kj mol 1 = 680 kj mol kj Enthalpy change per gram of Al(s) / NH 4 ClO 4 (s) fuel = [3 x x 117.5] g = 6.18 kj g 1 H (g) + 1 O (g) H O(g) ΔH o r = ΔH o f[products] ΔH o f[reactants] = 4 kj mol 1 Enthalpy change per gram of hydrogen-oxygen fuel = b) The hydrogen-oxygen fuel is more environmentally friendly. 4 kj [ x 3.0] g = 13.4 kj g 1 Any one of the following reasons: The hydrogen-oxygen fuel produces H O(g) only. The Al(s) / NH 4 ClO 4 (s) fuel produces NO(g), which is an air pollutant and precursor to acid rain. The Al(s) / NH 4 ClO 4 (s) fuel produces solid aluminium compounds, which are particulates. The Al(s) / NH 4 ClO 4 (s) fuel uses Al(s), the production of which causes environmental damage. 61