2. New Way Chemistry for Hong Kong A-Level Book 11 Energetics 6.1What is Energetics? 6.2Enthalpy Changes Related to Breaking and Forming of Bonds 6.3Standard Enthalpy Changes 6.4Experimental Determination of Enthalpy Changes by Calorimetry 6.5Hess’s Law 6.6Calculations involving Standard Enthalpy Changes of Reactions 6

3. New Way Chemistry for Hong Kong A-Level Book 12 What is energetics? Energetics is the study of energy changes associated with chemical reactions. Thermochemistry is the study of heat changes associated with chemical reactions. Some terms Enthalpy(H) = heat content in a substance Enthalpy change(  H) = heat content of products - heat content of reactants = H p - H r 6.1 What is energetics? (SB p.136)

4. New Way Chemistry for Hong Kong A-Level Book 13 Law of conservation of energy 6.1 What is energetics? (SB p.136) The law of conservation of energy states that energy can neither be created nor destroyed.

5. New Way Chemistry for Hong Kong A-Level Book 14 Internal energy and enthalpy e.g. Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) 6.1 What is energetics? (SB p.137)

6. New Way Chemistry for Hong Kong A-Level Book 15 Internal energy and enthalpy 6.1 What is energetics? (SB p.138) (Heat change at constant volume) Enthalpy change Heat change at constant pressure = Change in internal energy Work done on the surroundings +

7. New Way Chemistry for Hong Kong A-Level Book 16 Exothermic and endothermic reactions An exothermic reaction is a reaction that releases heat energy to the surroundings. (  H = -ve) 6.1 What is energetics? (SB p.138)

8. New Way Chemistry for Hong Kong A-Level Book 17 An endothermic reaction is a reaction that absorbs heat energy from the surroundings. (  H = +ve) 6.1 What is energetics? (SB p.139) Exothermic and endothermic reactions Check Point 6-1 Check Point 6-1

9. New Way Chemistry for Hong Kong A-Level Book 18 6.2 Energy Changes Related to Breaking and Forming of Bonds

10. New Way Chemistry for Hong Kong A-Level Book 19 Enthalpy changes related to breaking and forming of bonds e.g. CH 4 + 2O 2  CO 2 + 2H 2 O 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)

11. New Way Chemistry for Hong Kong A-Level Book 110 Enthalpy changes related to breaking and forming of bonds 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140) In an exothermic reaction, the energy required in breaking the bonds in the reactants is less than the energy released in forming the bonds in the products (products contain stronger bonds).

12. New Way Chemistry for Hong Kong A-Level Book 111 Enthalpy changes related to breaking and forming of bonds 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)

13. New Way Chemistry for Hong Kong A-Level Book 112 Enthalpy changes related to breaking and forming of bonds In an endothermic reaction, the energy required in breaking the bonds in the reactants is more than the energy released in forming the bonds in the products (reactants contain stronger bonds). 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140) Check Point 6-2 Check Point 6-2

14. New Way Chemistry for Hong Kong A-Level Book 113 6.3 Standard Enthalpy Changes

15. New Way Chemistry for Hong Kong A-Level Book 114 Standard enthalpy changes CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ mol -1 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H = -890 kJ mol -1 6.3 Standard enthalpy changes (SB p.141)

16. New Way Chemistry for Hong Kong A-Level Book 115 Standard enthalpy changes As enthalpy changes depend on temperature and pressure, it is necessary to define standard states and conditions: 1.elements or compounds in their normal physical states; 2.a pressure of 1 atm (101325 Nm -2 ); and 3.a temperature of 25 o C (298 K) Enthalpy change under standard conditions denoted by symbol:  H ø 6.3 Standard enthalpy changes (SB p.141)

17. New Way Chemistry for Hong Kong A-Level Book 116 Standard enthalpy changes of neutralization Standard enthalpy change of neutralization (  H neut ) is the enthalpy change when one mole of water is formed from the neutralization of an acid by an alkali under standard conditions. ø e.g. H + (aq) + OH - (aq)  H 2 O(l)  H neut = -57.3 kJ mol -1 ø 6.3 Standard enthalpy changes (SB p.142)

18. New Way Chemistry for Hong Kong A-Level Book 117 H + (aq) + OH - (aq)  H 2 O(l) -57.1 -57.2 -52.2 -68.6 NaOH KOH NH 3 NaOH HCl HF  H neu AlkaliAcid ø Standard enthalpy changes of neutralization 6.3 Standard enthalpy changes (SB p.142)

19. New Way Chemistry for Hong Kong A-Level Book 118 6.3 Standard enthalpy changes (SB p.142) Standard enthalpy changes of neutralization Enthalpy level diagram for the neutralization of a strong acid and a strong alkali

20. New Way Chemistry for Hong Kong A-Level Book 119 Standard enthalpy change of solution (  H soln ) is the enthalpy change when one mole of a solute is completely dissolved in a sufficiently large volume of solvent to form an infinitely dilute solution under standard conditions. ø 6.3 Standard enthalpy changes (SB p.142) Standard enthalpy change of solution

21. New Way Chemistry for Hong Kong A-Level Book 120 e.g. NaCl(s) + water  Na + (aq)+Cl - (aq)  H soln = +3.9 kJ mol -1 ø 6.3 Standard enthalpy changes (SB p.143) Standard enthalpy change of solution Enthalpy level diagram for the dissolution of NaCl

22. New Way Chemistry for Hong Kong A-Level Book 121 6.3 Standard enthalpy changes (SB p.143) Standard enthalpy change of solution e.g. LiCl(s) + water  Li + (aq) + Cl - (aq)  H soln = -37.2 kJ mol -1 ø Enthalpy level diagram for the dissolution of LiCl in water

23. New Way Chemistry for Hong Kong A-Level Book 122 -44.7 +3.9 -57.8 +20.0 NaOH NaCl KOH KBr  H soln (kJ mol -1 ) Salt ø Standard enthalpy change of solution 6.3 Standard enthalpy changes (SB p.143)

24. New Way Chemistry for Hong Kong A-Level Book 123 Standard enthalpy change of formation (  H f ) is the enthalpy change of the reaction when one mole of the compound in its standard state is formed from its constituent elements under standard conditions. ø 6.3 Standard enthalpy changes (SB p.144) Standard enthalpy change of formation

25. New Way Chemistry for Hong Kong A-Level Book 124 Standard enthalpy change of formation of NaCl is -411 kJ mol -1. e.g. 2Na(s) + Cl 2 (g)  2NaCl(s)  H = -822 kJ mol -1 ø Na(s) + ½Cl 2 (g)  NaCl(s)  H f = -411 kJ mol -1 ø 6.3 Standard enthalpy changes (SB p.144) Standard enthalpy change of formation

26. New Way Chemistry for Hong Kong A-Level Book 125 The enthalpy change of formation of an element is always zero. N 2 (g)  N 2 (g) What is  H f [N 2 (g)] ? ø  H f [N 2 (g)] = 0 ø 6.3 Standard enthalpy changes (SB p.144) Standard enthalpy change of formation

27. New Way Chemistry for Hong Kong A-Level Book 126 e.g. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l)  H 1 = -2220 kJ 2C 3 H 8 (g) + 10O 2 (g)  6CO 2 (g) + 8H 2 O(l)  H 2 = ?  H 2 = -4440 kJ  It is more convenient to report enthalpy changes per mole of the main reactant reacted/product formed. 6.3 Standard enthalpy changes (SB p.146) Standard enthalpy change of combustion

28. New Way Chemistry for Hong Kong A-Level Book 127 Standard enthalpy change of combustion (  H c ) of a substance is the enthalpy change when one mole of the substance burns completely under standard conditions. ø e.g. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l) 1 mole  H c = -2220 kJ mol -1 ø 6.3 Standard enthalpy changes (SB p.146) Standard enthalpy change of combustion

29. New Way Chemistry for Hong Kong A-Level Book 128 -285.8 -395.4 -393.5 -283.0 -890.4 H 2 (g) C (diamond) C (graphite) CO(g) CH 4 (g)  H c (kJ mol -1 ) Substance ø 6.3 Standard enthalpy changes (SB p.147) Standard enthalpy change of combustion Check Point 6-3 Check Point 6-3

30. New Way Chemistry for Hong Kong A-Level Book 129 6.4 Experimental Determination of Enthalpy Changes by Calorimetry

31. New Way Chemistry for Hong Kong A-Level Book 130 Experimental determination of enthalpy changes by calorimetry Calorimeter = a container used for measuring the temperature change of solution 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.148)

32. New Way Chemistry for Hong Kong A-Level Book 131 Determination of enthalpy change of neutralization 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)

33. New Way Chemistry for Hong Kong A-Level Book 132 Heat evolved = (m 1 c 1 + m 2 c 2 ) ΔT wherem 1 is the mass of the solution, m 2 is the mass of calorimeter, c 1 is the specific heat capacity of the solution, c 2 is the specific heat capacity of calorimeter, ΔT is the temperature change of the reaction 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149) Example 6-4A Example 6-4A

34. New Way Chemistry for Hong Kong A-Level Book 133 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.150) Determination of enthalpy change of combustion The Philip Harris calorimeter used for determining the enthalpy change of combustion of a liquid fuel

35. New Way Chemistry for Hong Kong A-Level Book 134 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151) Determination of enthalpy change of combustion A simple apparatus used to determine the enthalpy change of combustion of ethanol

36. New Way Chemistry for Hong Kong A-Level Book 135 Heat evolved = (m 1 c 1 + m 2 c 2 ) ΔT Wherem 1 is the mass of water in the calorimeter, m 2 is the mass of the calorimeter, c 1 is the specific heat capacity of the water, c 2 is the specific heat capacity of calorimeter, ΔT is the temperature change of the reaction 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151) Example 6-4B Example 6-4B

37. New Way Chemistry for Hong Kong A-Level Book 136 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.152) Determination of enthalpy change of solution By measuring the temperature change when a known mass of solute is added to a known volume of solvent in a calorimeter Heat change = (m 1 c 1 + m 2 c 2 )  T Example 6-4C Example 6-4C

38. New Way Chemistry for Hong Kong A-Level Book 137 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153) Determination of enthalpy change of formation The enthalpy change of formation of a substance can be quite high Found out by applying Hess’s law of constant heat summation Check Point 6-4 Check Point 6-4

39. New Way Chemistry for Hong Kong A-Level Book 138 6.5 Hess’s Law

40. New Way Chemistry for Hong Kong A-Level Book 139 Hess’s Law A + BC + D Route 1 H1H1 E H2H2 H3H3 Route 2  H 1 =  H 2 +  H 3 Hess’s law of constant heat summation states that the total enthalpy change accompanying a chemical reaction is independent of the route by which the chemical reaction takes place. 6.5 Hess’s law (SB p.153)

41. New Way Chemistry for Hong Kong A-Level Book 140 6.5 Hess’s law (SB p.154) Enthalpy level diagram Relate substances together in terms of enthalpy changes of reactions Enthalpy level diagram for the oxidation of C(graphite) to CO 2 (g)

42. New Way Chemistry for Hong Kong A-Level Book 141 6.5 Hess’s law (SB p.155) Enthalpy cycle (Born-Haber cycle) Relate the various equations involved in a reaction Enthalpy cycle for the oxidation of C(graphite) to CO 2 (g)

43. New Way Chemistry for Hong Kong A-Level Book 142 Importance of Hess’s law The enthalpy change of some chemical reactions cannot be determined directly because: the reactions cannot be performed in the laboratory the reaction rates are too slow the reactions may involve the formation of side products But the enthalpy change of such reactions can be determined indirectly by applying Hess’s Law. 6.5 Hess’s law (SB p.155)

44. New Way Chemistry for Hong Kong A-Level Book 143 Enthalpy change of formation of CO(g) = -393.5 - (-283.0 ) = -110.5 kJ mol -1 Given:  H f [CO 2 (g)] = -393.5 kJ mol -1 ;  H c [CO(g)] = -283.0 kJ mol -1 ø ø H2H2 + ½O 2 (g) CO 2 (g) H1H1 + ½O 2 (g) C(graphite) + ½O 2 (g)CO(g)  H f [CO(g)] ø  H f [CO(g)] +  H 2 =  H 1 ø  H f [CO(g)] =  H 1 -  H 2 ø 6.5 Hess’s law (SB p.153)

45. New Way Chemistry for Hong Kong A-Level Book 144 6.5 Hess’s law (SB p.153) Enthalpy change of formation of CaCO 3 (s) Ca(s) + C(graphite) + O 2 CaCO 3 (s) CaO(s) + CO 2 (g) H1H1 H2H2  H f [CaCO 3 (s)] ø  H f [CaCO 3 (s)] =  H 1 +  H 2 = -1028.5 kJ mol -1 + (-178.0) kJ mol -1 = -1206.5 kJ mol -1 ø

46. New Way Chemistry for Hong Kong A-Level Book 145 Enthalpy change of hydration of MgSO 4 (s) aq MgSO 4 (s) + 7H 2 O(l) MgSO 4 ·7H 2 O(s) Mg 2+ (aq) + SO 4 2- (aq) + 7H 2 O(l) ΔH ΔH 2 aq ΔH 1 ø ΔH = enthalpy of hydration of MgSO 4 (s) ΔH 1 = molar enthalpy change of solution of anhydrous magnesium sulphate(VI) ΔH 2 = molar enthalpy change of solution of magnesium sulphate(VI)-7-water ΔH = ΔH 1 - ΔH 2 ø ø 6.5 Hess’s law (SB p.153) Check Point 6-5 Check Point 6-5

47. New Way Chemistry for Hong Kong A-Level Book 146 6.6 Calculations involving Standard Enthalpy Changes of Reactions

48. New Way Chemistry for Hong Kong A-Level Book 147 Calculation of standard enthalpy change of reaction from standard enthalpy changes of formation  H reaction =   H f [products] -   H f [reactants] ø ø 6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)

49. New Way Chemistry for Hong Kong A-Level Book 148 6.6 Calculations involving standard enthalpy changes of reactions (SB p.159) Example 6-6A Example 6-6A Example 6-6B Example 6-6B Example 6-6D Example 6-6D Example 6-6C Example 6-6C

50. New Way Chemistry for Hong Kong A-Level Book 149 6.6 Calculations involving standard enthalpy changes of reactions (SB p.162) Calculation of standard enthalpy change of formation from standard enthalpy changes of combustion  H f =   H c [products] -   H c [reactants] ø ø ø

51. New Way Chemistry for Hong Kong A-Level Book 150 6.6 Calculations involving standard enthalpy changes of reactions (SB p.162) Example 6-6E Example 6-6E Example 6-6F Example 6-6F Check Point 6-6 Check Point 6-6

52. New Way Chemistry for Hong Kong A-Level Book 151 6.7 Entropy Change 6.7 Entropy change (SB p.164)

53. New Way Chemistry for Hong Kong A-Level Book 152 6.7 Entropy change (SB p.164) Entropy A process is said to be spontaneous If no external forces are required to keep the process going The process may be physical change or a chemical change Example of spontaneous physical change: cooling of hot water Example of spontaneous chemcial change: burning of wood once the fire is started

54. New Way Chemistry for Hong Kong A-Level Book 153 6.7 Entropy change (SB p.164) Entropy Exothermicity is the reason for the spontaneity of a process Some spontaneous changes are endothermic Examples: Melting of ice, dissolution of ammonium nitrate in water

55. New Way Chemistry for Hong Kong A-Level Book 154 6.7 Entropy change (SB p.164) Melting of ice Dissolution of ammonium nitrate in water

56. New Way Chemistry for Hong Kong A-Level Book 155 6.7 Entropy change (SB p.165) Entropy Entropy is a measure of the randomness or the degree of disorder of a system Solid Liquid Gas Entropy increases

57. New Way Chemistry for Hong Kong A-Level Book 156 6.7 Entropy change (SB p.166) Entropy change Entropy change means the change in the degree of disorder of a system  S = S final - S initial

58. New Way Chemistry for Hong Kong A-Level Book 157 6.7 Entropy change (SB p.166) Positive entropy Increase in entropy Final state has a larger entropy that the initial state Example: Ice (less entropy)  Water (more entropy)  S = S water – S ice = +ve

59. New Way Chemistry for Hong Kong A-Level Book 158 6.7 Entropy change (SB p.166) Negative entropy Decrease in entropy Initial state has a larger entropy that the final state Example: Water (more entropy)  Ice (less entropy)  S = S ice – S water = -ve Check Point 6-7 Check Point 6-7

60. New Way Chemistry for Hong Kong A-Level Book 159 6.8 Free Energy Change

61. New Way Chemistry for Hong Kong A-Level Book 160 6.8 Free energy change (SB p.168) Free energy change Entropy is temperature dependent At a higher temp, the entropy of the system is higher At a lower temp, the entropy of the system is lower

62. New Way Chemistry for Hong Kong A-Level Book 161 6.8 Free energy change (SB p.168) Free energy Another driving force for a process Called free energy (G) G = H – TS where H is the enthalpy T is Kelvin temperature S is the entropy

63. New Way Chemistry for Hong Kong A-Level Book 162 6.8 Free energy change (SB p.168) Free energy change  G =  H – T  S At a given temp, there are two driving forces for a process to occur Overall enthalpy of the system tends to be low Overall entropy of the system tends to be high

64. New Way Chemistry for Hong Kong A-Level Book 163 6.8 Free energy change (SB p.168) Significance of the equation Process favoured by:  H = -ve;  S = +ve Process not favoured by:  H = +ve;  S = -ve

65. New Way Chemistry for Hong Kong A-Level Book 164 6.8 Free energy change (SB p.168) Significance of the equation A process is spontaneous or favourable when  G is negative A process is not spontaneous or favourable as indicated when  G is positive, but is spontaneous in the opposite direction

66. New Way Chemistry for Hong Kong A-Level Book 165 6.8 Free energy change (SB p.169) How  H and  S affect the spontaneity of a process HH SS  G =  H - T  S Result -ve+ve-veProcess is spontansous at all temepratures +ve-ve+veProcess is not spontaneous at any temperature (reverse process is spontaneous at all temperatures)

67. New Way Chemistry for Hong Kong A-Level Book 166 6.8 Free energy change (SB p.170) Effects of relative magnitudes of  H and  S on the spontaneity of a process HH SS Condition  G =  H - T  S Result +ve At high temp, T  S >  H  G = -ve Process is spontaneous at high temp +ve At low temp,  H > T  S  G = +ve Process is not spontaneous -ve At high temp, T  S >  H  G = +ve Process is not spontaneous -ve At low temp,  H > T  S  G = -ve Process is spontaneous at low temp

68. New Way Chemistry for Hong Kong A-Level Book 167 6.8 Free energy change (SB p.170) Check Point 6-8 Check Point 6-8

69. New Way Chemistry for Hong Kong A-Level Book 168 The END

70. New Way Chemistry for Hong Kong A-Level Book 169 State whether the following processes are exothermic or endothermic. (a) Melting of ice. (b) Dissolution of table salt. (c) Condensation of steam. Back Answer 6.1 What is energetics? (SB p.140) (a) Endothermic (b) Endothermic (c) Exothermic

71. New Way Chemistry for Hong Kong A-Level Book 170 (a)State the difference between exothermic and endothermic reactions with respect to (i)the sign of  H; (ii)the heat change with the surroundings; (iii) the total enthalpy of reactants and products. Answer (a)(i) Exothermic reactions:  H = -ve; endothermic reactions:  H = +ve (ii)Heat is given out to the surroundings in exothermic reactions whereas heat is taken in from the surroundings in endothermic reactions. (iii)In exothermic reactions, the total enthalpy of products is less than that of the reactants. In endothermic reactions, the total enthalpy is greater than that of the reactants. 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)

72. New Way Chemistry for Hong Kong A-Level Book 171 (b)Draw an enthalpy level diagram for a reaction which is (i) endothermic, having a large activation energy. (ii)exothermic, having a small activation energy. Answer 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)

73. New Way Chemistry for Hong Kong A-Level Book 172 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141) (b)(i)

74. New Way Chemistry for Hong Kong A-Level Book 173 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141) (ii) Back

75. New Way Chemistry for Hong Kong A-Level Book 174 (a)Why must the condition “burnt completely in oxygen” be emphasized in the definition of standard enthalpy change of combustion? Answer 6.3 Standard enthalpy changes (SB p.147) (a)If the substance is not completely burnt in excess oxygen, other products such as C(s) and CO(g) may be formed. The enthalpy change of combustion measured will not be accurate.

76. New Way Chemistry for Hong Kong A-Level Book 175 (b) The enthalpy change of the following reaction under standard conditions is –566.0 kJ. 2CO(g) + O 2 (g)  2CO 2 (g) What is the standard enthalpy change of combustion of carbon monoxide? Answer 6.3 Standard enthalpy changes (SB p.147) (b)Standard enthalpy change of combustion of CO =  (-566.0) kJ = -283.0 kJ

77. New Way Chemistry for Hong Kong A-Level Book 176 (c)What terms may be given for the enthalpy change of the following reaction? N 2 (g) + O 2 (g)  NO 2 (g) Answer 6.3 Standard enthalpy changes (SB p.147) (c) Enthalpy change of combustion of nitrogen or enthalpy change of formation of nitrogen dioxide. Back

78. New Way Chemistry for Hong Kong A-Level Book 177 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149) Determine the enthalpy change of neutralization of 25 cm 3 of 1.25 M hydrochloric acid and 25 cm 3 of 1.25 M sodium hydroxide solution using the following data: Mass of calorimeter = 100 g Initial temperature of acid = 15.5 o C (288.5 K) Initial temperature of alkali = 15.5 o C (288.5 K) Final temperature of the reaction mixture = 21.6 o C (294.6 K) The specific heat capacities of water and calorimeter are 4200 J kg -1 K -1 and 800 J kg -1 K -1 respectively. Answer

79. New Way Chemistry for Hong Kong A-Level Book 178 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149) Assume that the density of the reaction mixture is the same as that of water, i.e. 1 g cm -3. Mass of the reaction mixture = (25 + 25) cm 3  1 g cm -3 = 50 g = 0.05 kg Heat given out = (m 1 c 1 + m 2 c 2 )  T = (0.05 kg  4200 J kg -1 K -1 + 0.1 kg  800 J kg -1 K -1 )  (294.6 – 288.5) K = 1769 J H + (aq) + OH - (aq)  H 2 O(l) Number of moles of HCl = 1.25 mol dm -3  25  10 -3 dm 3 = 0.03125 mol Number of moles of NaOH = 1.25 mol dm -3  25  10 -3 dm 3 = 0.03125 mol Number of moles of H 2 O formed = 0.03125 mol

80. New Way Chemistry for Hong Kong A-Level Book 179 Back 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149) Heat given out per mole of H 2 O formed = = 56608 J mol -1 The enthalpy change of neutralization is –56.6 kJ mol -1.

81. New Way Chemistry for Hong Kong A-Level Book 180 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151) Determine the enthalpy change of combustion of ethanol using the following data: Mass of spirit lamp before experiment = 45.24 g Mass of spirit lamp after experiment = 44.46 g Mass of water in copper calorimeter = 50 g Mass of copper calorimeter without water = 380 g Initial temperature of water = 18.5 o C (291.5 K) Final temperature of water = 39.4 o C (312.4 K) The specific heat capacities of water and copper calorimeter are 4200 J kg -1 K -1 and 2100 J kg -1 K -1 respectively. Answer

82. New Way Chemistry for Hong Kong A-Level Book 181 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151) Heat evolved by the combustion of ethanol = Heat absorbed by the copper calorimeter = (m 1 c 1 + m 2 c 2 )  T = (0.05 kg  4200 J kg -1 K -1 + 0.38 kg  2100 J kg -1 K -1 )  (312.4 – 291.5)K = 21067 J C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(l) Mass of ethanol burnt = (45.24 – 44.46) g = 0.78 g Number of moles of ethanol burnt = = 0.017 mol

83. New Way Chemistry for Hong Kong A-Level Book 182 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151) Heat given out per mole of ethanol = = 1239235 J mol -1 = 1239 kJ mol -1 The enthalpy change of combustion of ethanol is –1239 kJ mol -1. There was heat loss by the system to the surroundings, and incomplete combustion of ethanol might occur. Also, the experiment was not carried out under standard conditions. Therefore, the experimentally determined value (-1239 kJ mol -1 ) is less than the theoretical value of the standard enthalpy change of combustion of ethanol (-1371 kJ mol -1 ). Back

84. New Way Chemistry for Hong Kong A-Level Book 183 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.152) 0.02 mol of anhydrous ammonium chloride was added to 45 g of water in a polystyrene cup to determine the enthalpy change of solution of anhydrous ammonium chloride. It is found that there was a temperature drop from 24.5 o C to 23.0 o C in the solution. Given that the specific heat capacity of water is 4200 J kg -1 K -1 and NH 4 Cl(s) + aq  NH 4 Cl(aq) Calculate the enthalpy change of solution of anhydrous ammonium chloride. (Neglect the specific heat capacity of the polystyrene cup.) Answer

85. New Way Chemistry for Hong Kong A-Level Book 184 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.152) Heat absorbed = m 1 c 1  T (  c 2  0) = 0.045 kg  4200 J kg -1 K -1  (297.5 – 296) K = 283.5 J (0.284 kJ) Heat absorbed per mole of ammonium chloride = = 14.2 kJ mol -1 The enthalpy change of solution of anhydrous ammonium chloride is +14.2 kJ mol -1. Back

86. New Way Chemistry for Hong Kong A-Level Book 185 (a) A student tried to determine the enthalpy change of neutralization by putting 25.0 cm 3 of 1.0 M HNO 3 in a polystyrene cup and adding 25.0 cm 3 of 1.0 M NH 3 into it. The temperature rise recorded was 3.11 o C. Given that the mass of the polystyrene cup is 250 g, the specific heat capacities of water and the polystyrene cup are 4200 J kg -1 K -1 and 800 J kg -1 K -1 respectively. Determine the enthalpy change of neutralization of nitric acid and aqueous ammonia. (Density of water = 1 g cm -3 ) Answer 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)

87. New Way Chemistry for Hong Kong A-Level Book 186 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153) (a)Heat evolved = m 1 c 1  T + m 2 c 2  T = 0.050 kg  4200 J kg -1 K -1  3.11 K + 0.25 kg  800 J kg -1 K -1  3.11 K = (653.1 + 622) J = 1275.1 J No. of moles of HNO 3 used = 1.0 M  25  10 -3 m 3 = 0.025 mol

88. New Way Chemistry for Hong Kong A-Level Book 187 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153) (a)No. of moles of NH 3 used = 1.0 M  25  10 -3 dm 3 = 0.025 mol No. of moles of H 2 O formed = 0.025 mol Heat evolved per mole of H 2 O formed = = 51.004 kJ mol -1 The enthalpy change of neutralization of nitric acid and aqueous ammonia is –51.004 kJ mol -1.

89. New Way Chemistry for Hong Kong A-Level Book 188 (b)When 0.05 mol of silver nitrate was added to 50 g of water in a polystyrene cup, a temperature drop of 5.2 o C was recorded. Assuming that there was no heat absorption by the polystyrene cup, calculate the enthalpy change of solution of silver nitrate. (Specific heat capacity of water = 4200 J kg -1 K -1 ) Answer 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)

90. New Way Chemistry for Hong Kong A-Level Book 189 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153) (b)Energy absorbed = mc  T = 0.05 kg  4200 J kg -1 K -1  5.2 K = 1092 J No. of moles of AgNO 3 used = 0.05 mol Energy absorbed per mole of AgNO 3 used = = 21.84 kJ mol -1 The enthalpy change of solution of silver nitrate is +21.84 kJ mol -1.

91. New Way Chemistry for Hong Kong A-Level Book 190 (c) A student used a calorimeter as shown in Fig. 6-15 to determine the enthalpy change of combustion of methanol. In the experiment, 1.60 g of methanol was used and 50 g of water was heated up, raising the temperature by 33.2 o C. Given that the specific heat capacities of water and copper calorimeter are 4200 J kg -1 K -1 and 2100 J kg -1 K -1 respectively and the mass of the calorimeter is 400 g, calculate the enthalpy change of combustion of methanol. Answer 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)

92. New Way Chemistry for Hong Kong A-Level Book 191 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153) (c)Heat evolved = m 1 c 1  T + m 2 c 2  T = 0.05 kg  4200 J kg -1 K -1  33.2 K + 0.4 kg  2100 J kg -1 K -1  33.2 K = (6972 + 27888) J = 34860 J No. of moles of methanol used = = 0.05 mol Heat evolved per mole of methanol used = = 697.2 kJ mol -1 The enthalpy change of combustion of methanol is –697.2 kJ mol -1. Back

93. New Way Chemistry for Hong Kong A-Level Book 192 6.5 Hess’s law (SB p.158) (a)Given the following thermochemical equation: 2H 2 (g) + O 2 (g)  2H 2 O(l) (i)Is the reaction endothermic or exothermic? (ii)What is the enthalpy change for the following reactions? (1) 2H 2 O(l)  2H 2 (g) + O 2 (g) (2) H 2 (g) + O 2 (g)  H 2 O(l) (iii) If the enthalpy change for the reaction H 2 O(l)  H 2 O(g) is +41.1 kJ mol -1, calculate the  H for 2H 2 (g) + O 2 (g)  2H 2 O(g). Answer

94. New Way Chemistry for Hong Kong A-Level Book 193 6.5 Hess’s law (SB p.158) (a)(i) Exothermic (ii) (1) +571.6 kJ mol -1 (2) –285.8 kJ mol -1 (iii)  H = [-571.6 + 2  (+41.1)] kJ mol -1 = -489.4 kJ mol -1

95. New Way Chemistry for Hong Kong A-Level Book 194 6.5 Hess’s law (SB p.158) (b)Given the following information about the enthalpy change of combustion of allotropes of carbon:  H c [C(graphite)] = -393.5 kJ mol -1  H c [C(diamond)] = -395.4 kJ mol -1 (i) Which allotrope of carbon is more stable? (ii) What is the enthalpy change for the following process? C(graphite)  C(diamond) ø ø Answer

96. New Way Chemistry for Hong Kong A-Level Book 195 6.5 Hess’s law (SB p.158) (b)(i) Graphite (ii)  H = [-393.5 – (-395.4)] kJ mol -1 = +1.9 kJ mol -1

97. New Way Chemistry for Hong Kong A-Level Book 196 6.5 Hess’s law (SB p.158) (c)The formation of ethyne (C 2 H 2 (g) can be represented by the following equation: 2C(graphite) + H 2 (g)  C 2 H 2 (g) (i) Draw an enthalpy level diagram relating the above equation to carbon dioxide and water. (ii) Calculate the standard enthalpy change of formation of ethyne. (Given:  H c [C(graphite)] = -393.5 kJ mol -1 ;  H c [H 2 (g)] = -285.8 kJ mol -1 ;  H c [C 2 H 2 (g)] = -1299 kJ mol -1 ) ø ø ø Answer

98. New Way Chemistry for Hong Kong A-Level Book 197 6.5 Hess’s law (SB p.158) (c)(i) (ii)  H f [C 2 H 2 (g)] =  H c [C(graphite)]  2 +  H c [H 2 (g)] –  H c [C 2 H 2 (g)] = [(-393.5)  2 + (-285.8) – (-1299)] kJ mol -1 = +226.2 kJ mol -1 ø ø ø ø Back

99. New Way Chemistry for Hong Kong A-Level Book 198 6.6 Calculations involving standard enthalpy changes of reactions (SB p.159) Given the following information, find the standard enthalpy change of the reaction: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H f [C 2 H 4 (g)] = +52.3 kJ mol -1  H f [C 2 H 6 (g)] = -84.6 kJ mol -1 ø ø Answer

100. New Way Chemistry for Hong Kong A-Level Book 199 6.6 Calculations involving standard enthalpy changes of reactions (SB p.159) Note:  H 1 =  [  H f (reactants)] =  H f [C 2 H 4 (g)] +  H f [H 2 (g)]  H 2 =  [  H f (products)] =  H f [C 2 H 6 (g)] Applying Hess’s law,  H 1 +  H =  H 2  H =  H 2 -  H 1 =  H f [C 2 H 6 (g)] – (  H f [C 2 H 4 (g)] +  H f [H 2 (g)]) = [-84.6 – (+52.3 + 0)] kJ mol -1 =-136.9 kJ mol -1 The standard enthalpy change of the reaction is –136.9 kJ mol -1. ø ø ø ø ø ø ø ø ø ø Back

101. New Way Chemistry for Hong Kong A-Level Book 1100 6.6 Calculations involving standard enthalpy changes of reactions (SB p.160) Given the following information, find the standard enthalpy change of the reaction: 6PbO(s) + O 2 (g)  2Pb 3 O 4 (s)  H f [PbO(g)] = -220.0 kJ mol -1  H f [Pb 3 O 4 (g)] = -737.5 kJ mol -1 ø ø Answer

102. New Way Chemistry for Hong Kong A-Level Book 1101 6.6 Calculations involving standard enthalpy changes of reactions (SB p.160) Note:  H 1 =  [  H f (reactants)] = 6   H f [PbO(s)] +  H f [O 2 (g)]  H 2 =  [  H f (products)] = 2   H f [Pb 3 O 4 (s)] Applying Hess’s law,  H 1 +  H =  H 2  H =  H 2 -  H 1 = 2   H f [Pb 3 O 4 (s)] – (6   H f [PbO(s)] +  H f [O 2 (g)]) = [2  (-737.5) – 6  (-222.0) – 0] kJ mol -1 =-155.0 kJ mol -1 The standard enthalpy change of the reaction is –155.0 kJ mol -1. ø ø ø ø ø ø ø ø ø ø Back

103. New Way Chemistry for Hong Kong A-Level Book 1102 6.6 Calculations involving standard enthalpy changes of reactions (SB p.160) Given the following information, find the standard enthalpy change of the reaction: Fe 2 O 3 (s) + 3CO(g)  2Fe(s) + 3CO 2 (g)  H f [Fe 2 O 3 (s)] = -822.0 kJ mol -1  H f [CO(g)] = -110.5 kJ mol -1  H f [CO 2 (g)] = -393.5 kJ mol -1 ø ø ø Answer

104. New Way Chemistry for Hong Kong A-Level Book 1103 6.6 Calculations involving standard enthalpy changes of reactions (SB p.160) Note:  H 1 =  [  H f (reactants)] =  H f [Fe 2 O 3 (s)] + 2   H f [CO(g)]  H 2 =  [  H f (products)] = 2   H f [Fe(s)] + 3   H f [CO 2 (g)] Applying Hess’s law,  H 1 +  H =  H 2  H =  H 2 -  H 1 = 2   H f [Fe(s)] + 3   H f [CO 2 (g)] -  H f [Fe2O3(s)] - 3   H f [CO(g)] = [2  (0) + 3  (-393.5) –(-822.0) – 3  (-110.5)] kJ mol -1 =-27.0 kJ mol -1 The standard enthalpy change of the reaction is –27.0 kJ mol -1. ø ø ø ø ø ø ø ø ø ø ø ø Back

105. New Way Chemistry for Hong Kong A-Level Book 1104 6.6 Calculations involving standard enthalpy changes of reactions (SB p.161) Given the following information, find the standard enthalpy change of the reaction: 4CH 3 · NH · NH 2 (l) + 5N 2 O 4 (l)  4CO 2 (g) + 12H 2 O(l) + 9N 2 (g)  H f [CH 3 · NH · NH 2 (l)] = +53 kJ mol -1  H f [N 2 O 4 (l)] = -20 kJ mol -1  H f [CO 2 (g)] = -393.5 kJ mol -1  H f [H 2 O(l)] = -285.8 kJ mol -1 ø ø ø ø Answer

106. New Way Chemistry for Hong Kong A-Level Book 1105 6.6 Calculations involving standard enthalpy changes of reactions (SB p.161) Note:  H 1 =  [  H f (reactants)] = 4   H f [CH 3 ·NH ·NH 2 (l)] + 5   H f [N 2 O 4 (l)]  H 2 =  [  H f (products)] = 4   H f [CO 2 (g)] + 12   H f [H 2 O(l)] + 9   H f [N 2 (g)] Applying Hess’s law,  H 1 +  H =  H 2  H =  H 2 -  H 1 = (4   H f [CO 2 (g)] + 12   H f [H 2 O(l)] + 9   H f [N 2 (g)] – (3   H f [CH 3 ·NH ·NH 2 (l)] + 5   H f [N 2 O 4 (l)] = [4  (-393.5) + 12  (-285.8) + 9  (0) – 4  (+53) – 5  (-20)] kJ mol -1 =- 5115.6 kJ mol -1 The standard enthalpy change of the reaction is –5115.6 kJ mol -1. ø ø ø ø ø ø ø ø ø Back

107. New Way Chemistry for Hong Kong A-Level Book 1106 6.6 Calculations involving standard enthalpy changes of reactions (SB p.162) Given the following information, find the standard enthalpy change of formation of methane gas. C(graphite) + O 2 (g)  CO 2 (g)  H c [C(graphite)] = -393.5 kJ mol -1 H 2 (g) + O 2 (g)  H 2 O(l)  H c [H 2 (g)] = -285.8 kJ mol -1 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H c [CH 4 (g)] = -20 kJ mol -1 ø ø ø Answer

108. New Way Chemistry for Hong Kong A-Level Book 1107 6.6 Calculations involving standard enthalpy changes of reactions (SB p.162) Direct measurement of ΔH f [CH 4 (g)] is impossible because carbon(graphite) and hydrogen do not combine directly, and methane does not decompose directly to form carbon(graphite) and hydrogen. Since methane contain carbon and hydrogen only, they can be related to carbon dioxide and water by the combustion of methane and its constituent elements as shown in the diagram below. ø

109. New Way Chemistry for Hong Kong A-Level Book 1108 6.6 Calculations involving standard enthalpy changes of reactions (SB p.163) Note:  H 1 =  H c [C(graphite)]  H 2 = 2   H c [H 2 (g)]  H 3 =  H c [CH 4 (g)] Applying Hess’s law,  H f [CH 4 (g)] +  H 3 =  H 1 +  H 2  H f [CH 4 (g)] =  H 1 +  H 2 -  H 3 =  H c [C(graphite)] + 2   H c [H 2 (g)] -  H c [CH 4 (g)] = [-393.5 + 2  (-285.8) –(-890.4)] kJ mol -1 = -74.7 kJ mol -1 The standard enthalpy change of formation of methane gas is –74.7 kJ mol -1. ø ø ø ø ø ø ø Back

110. New Way Chemistry for Hong Kong A-Level Book 1109 6.6 Calculations involving standard enthalpy changes of reactions (SB p.163) Given the following information, find the standard enthalpy change of formation of methanol. C(graphite) + O 2 (g)  CO 2 (g)  H c [C(graphite)] = -393.5 kJ mol -1 H 2 (g) + O 2 (g)  H 2 O(l)  H c [H 2 (g)] = -285.8 kJ mol -1 C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H c [C 2 H 5 OH(l)] = -1371 kJ mol -1 ø ø ø Answer

111. New Way Chemistry for Hong Kong A-Level Book 1110 6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)

112. New Way Chemistry for Hong Kong A-Level Book 1111 6.6 Calculations involving standard enthalpy changes of reactions (SB p.163) Note:  H 1 = 2   H c [C(graphite)]  H 2 = 3   H c [H 2 (g)]  H 3 =  H c [C 2 H 5 OH(l)] Applying Hess’s law,  H f [C 2 H 5 OH(l)] +  H 3 =  H 1 +  H 2  H f [C 2 H 5 OH(l)] =  H 1 +  H 2 -  H 3 = 2   H c [C(graphite)] + 3   H c [H 2 (g)] -  H c [C 2 H 5 OH(l)] = [2  (-393.5) + 3  (-285.8) –(-1371)] kJ mol -1 = -273.4 kJ mol -1 The standard enthalpy change of formation of ethanol is –273.4 kJ mol -1. ø ø ø ø ø ø ø Back

113. New Way Chemistry for Hong Kong A-Level Book 1112 6.6 Calculations involving standard enthalpy changes of reactions (SB p.164) (a) Find the standard enthalpy change of formation of butane gas (C 4 H 10 (g)). Given:  H c [C(graphite)] = -393.5 kJ mol -1  H c [H 2 (g)] = -285.8 kJ mol -1  H c [C 4 H 10 (g)] = -2877 kJ mol -1 ø ø ø Answer

114. New Way Chemistry for Hong Kong A-Level Book 1113 6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)  H f [C 4 H 10 (g)] =  H c [C(graphite)]  4 +  H c [H 2 (g)]  5 -  H c [C 4 H 10 (g)] = [(-393.5)  4 + (-285.8)  5 – (-2877)] kJ mol -1 = -126 kJ mol -1 ø ø ø ø

115. New Way Chemistry for Hong Kong A-Level Book 1114 6.6 Calculations involving standard enthalpy changes of reactions (SB p.164) (b) Find the standard enthalpy change of the reaction: Br 2 (l) + C 2 H 4 (g)  C 2 H 4 Br 2 (l) Given:  H f [C 2 H 4 (g)] = +52.3 kJ mol -1  H f [C 2 H 4 Br 2 (l)] = -80.7 kJ mol -1 ø ø Answer

116. New Way Chemistry for Hong Kong A-Level Book 1115 6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)  H =  [  H f (products)] -  [  H f (reactants)] = [-80.7 – (+52.3) – 0)] kJ mol -1 = -133 kJ mol -1 ø ø ø Back

117. New Way Chemistry for Hong Kong A-Level Book 1116 Predict whether the following changes or reactions involve an increase or a decrease in entropy. Dissolving salt in water to form salt solution Condensation of steam on a cold mirror Complete combustion of carbon Complete combustion of carbon monoxide Oxidation of sulphur dioxide to sulphur trioxide Answer 6.7 Entropy change (SB p.167) (a) Increase (b) Decrease (c) Increase (d) Decrease (e) Decrease Back

118. New Way Chemistry for Hong Kong A-Level Book 1117 6.8 Free energy change (SB p.170) In the process of changing of ice to water, at what temperature do you think  G equals 0? Back  G equals 0 means that neither the forward nor the reverse process is spontaneous. The system is therefore in equilibrium. Melting point of ice is 0 o C (273 K) at which the process of changing ice to water and the process of water turning to ice are at equilibrium. At 0 o C,  G of the processes equals 0. Answer

119. New Way Chemistry for Hong Kong A-Level Book 1118 (a)At what temperatures is the following process spontaneous at 1 atmosphere? Water  Steam (b)What are the two driving forces that determine the spontaneity of a process? Answer 6.8 Free energy change (SB p.170) (a)100 o C (b) Enthalpy and entropy

120. New Way Chemistry for Hong Kong A-Level Book 1119 (c)State whether each of the following cases is spontaneous at all temperatures, not spontaneous at any temperature, spontaneous at high temperatures or spontaneous at low temperatures. (i) positive  S and positive  H (ii)positive  S and negative  H (iii) negative  S and positive  H (iv) negative  S and negative  H Answer Back (i)Spontaneous at high temperatures (ii)Spontaneous at all temperatures (iii)Not spontaneous at any temperature (iv)Spontaneous at low temperatures 6.8 Free energy change (SB p.170)