1. 1 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

2. 2 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. 6.1 What is energetics? (SB p.136)

3. 3 U = kinetic energy + potential energy Internal Energy (U)

4. 4 Energy translational rotational vibrational heat  T (K)Kinetic Energy Potential Energy Relative position among particles Bond breaking  P.E.  Bond forming  P.E. 

5. 5 H – H(g)  H(g) + H(g) P.E.  H(g) + H(g)  H – H(g) P.E.  Na(g)  Na + (g) + e  P.E.  Bond breaking : - Bond forming : - Ionization : -

6. 6 Reaction coordinate Internalenergy bond breaking 4H(g) + 2O(g) bond forming  U = U 2 – U 1 = -(y-x) kJ 2H 2 (g) + O 2 (g) U1U1 2H 2 O(l) U2U2 Q.1

7. 7 Internal energy and enthalpy 6.1 What is energetics? (SB p.137) H = U + PV enthalpy Internal energy

8. 8 Internal energy and enthalpy e.g. Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) 6.1 What is energetics? (SB p.137) q v =  U = -473 kJ mol  1 q p =  H = -470 kJ mol  1 Mg

9. 9 q v = q p – 3 = q p – w = q p - P  V Work done against the surroundings q v =  U = -473 kJ mol  1 q p =  H = -470 kJ mol  1 P  V  (Nm  2 )(m 3 ) = Nm Force  displacement

10. 10 Internal energy and enthalpy 6.1 What is energetics? (SB p.138)  H =  U +  PV =  U + P  V at constant P q p = q v + P  V Heat change at fixed P Heat change at fixed V Work done

11. 11 Internal energy and enthalpy 6.1 What is energetics? (SB p.138) q p = q v + P  V On expansion, P  V > 0 Work done by the system against the surroundings System gives out less energy to the surroundings q p is less negative than q v (less exothermic)

12. 12 Internal energy and enthalpy 6.1 What is energetics? (SB p.138) q p = q v + P  V On contraction, P  V < 0 Work done by the surroundings against the system System gives out more energy to the surroundings q p is more negative than q v (more exothermic)

13. 13  H is more easily measured than  H as most reactions happen in open vessels. i.e. at constant pressure. The absolute values of H and H cannot be measured.

14. 14 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)

15. 15 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

16. 16 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, but can be exchanged between a system and its surroundings The law of conservation of energy states that energy can neither be created nor destroyed, but can be exchanged between a system and its surroundings

17. 17 Exothermic : - P.E. of the system  K.E. of the surroundings Endothermic : - K.E. of the surroundings  P.E. of the system

18. 18 Enthalpy changes related to breaking and forming of bonds CH 4 + 2O 2  CO 2 + 2H 2 O 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)

19. 19 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140) In an exothermic reaction, E absorbed to break bonds < E released as bonds are formed. In an exothermic reaction, E absorbed to break bonds < E released as bonds are formed.

20. 20 Enthalpy changes related to breaking and forming of bonds 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140) N 2 (g) + 2O 2 (g)  2NO 2 (g)

21. 21 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140) In an endothermic reaction, E absorbed to break bonds > E released as bonds are formed. In an endothermic reaction, E absorbed to break bonds > E released as bonds are formed. Check Point 6-2 Check Point 6-2

22. 22 For non-gaseous reactions, P  V  0  H =  U + P  V   U For gaseous reactions,  H =  U + P  V =  U + (  n)RT (PV = nRT)

23. 23 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  U =  H – (  n)RT =  885 kJ mol  1 Given : R = 8.314 J K  1 mol  1, T = 298 K Q.2

24. 24 Reaction coordinate Enthalpy bond breaking C(g) + 4H(g) + 4O(g) bond forming  H =  890 kJ mol  1 CH 4 (g) + 2O 2 (g) H1H1 CO 2 (g) + 2H 2 O(l) H2H2

25. 25 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  890  H / kJ mol  1 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  802 At 298K At 373K 2H 2 O(l)  2H 2 O(g) +88 kJ 373K

26. 26 Reaction coordinate Enthalpy CH 4 (g) + 2O 2 (g) C(g) + 4H(g) + 4O(g) CO 2 (g) + 2H 2 O(l) 298 K  H =  890 kJ mol  1 H1H1 H2H2

27. 27 Reaction coordinate Enthalpy CH 4 (g) + 2O 2 (g) C(g) + 4H(g) + 4O(g) CO 2 (g) + 2H 2 O(l) 373 K  H =  890 kJ mol  1 Assume constant  H H1’H1’ H2’H2’

28. 28 Reaction coordinate Enthalpy CH 4 (g) + 2O 2 (g) C(g) + 4H(g) + 4O(g) CO 2 (g) + 2H 2 O(l) 373 K  H =  890 kJ mol  1 In fact,  H depends on T

29. 29 Reaction coordinate Enthalpy CH 4 (g) + 2O 2 (g) C(g) + 4H(g) + 4O(g) CO 2 (g) + 2H 2 O(l) CO 2 (g) + 2H 2 O(g)  H =  802 kJ mol  1  H = +88 kJ mol  1 373 K

30. 30 6.3 Standard Enthalpy Changes

31. 31 Standard enthalpy changes CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ mol -1 at 373 K CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H = -890 kJ mol -1 at 298 K 6.3 Standard enthalpy changes (SB p.141)

32. 32 Standard enthalpy changes As enthalpy changes depend on temperature and pressure, it is necessary to define standard 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)

33. 33 Standard enthalpy change of reaction The enthalpy change when the molar quantities of reactants as stated in the equation react under standard conditions. 2H 2 (g) + O 2 (g)  2H 2 O(l)  H =  572 kJ mol  1  per mole of O 2

34. 34 Standard enthalpy change of reaction 2H 2 (g) + O 2 (g)  2H 2 O(l)  H =  572 kJ mol  1  per mole of O 2  H =  286 kJ mol  1 H 2 (g) + O 2 (g)  H 2 O(l)  per mole of H 2 or H 2 O HH  depends on the equation 4H 2 (g) + 2O 2 (g)  4H 2 O(l)  H =  1144 kJ 

35. 35 Standard enthalpy change of formation HfHf  The enthalpy change when one mole of the substance is formed from its elements under standard conditions. H 2 (g) + O 2 (g)  H 2 O(l)  H f [H 2 O] =  286 kJ mol  1  Q.3 O 2 (g)  O 2 (g)  H f [O 2 ] = 0 kJ mol  1   H f [element] = 0 kJ mol  1 

36. 36 C(graphite)  C(diamond)  H f [diamond] = +1.9 kJ mol  1  Most stable allotrope

37. 37 Q.4 (i)C(graphite) + O 2 (g)  CO 2 (g) (ii)C(graphite) + 2H 2 (g)  CH 4 (g) (iii)Mg(s) + O 2 (g)  MgO(s) (v)2C(graphite) + 2H 2 (g) + O 2 (g)  CH 3 COOH(l) (iv)Na(s) + H 2 (g) + C(graphite) + O 2 (g)  NaHCO 3 (s)

38. 38 H 2 (g) + O 2 (g)  H 2 O(l) q v =  U =  140.3 kJ per g of H 2 Q.5  n = 0 – 0.496 – 0.248 = -0.744 mol

39. 39  H =  U +  nRT = -142.1 kJ Heat released for the formation of 0.496 mol of water Molar  H f [H 2 O] = Q.5

40. 40 Standard enthalpy change of combustion HcHc  The enthalpy change when one mole of the substance undergoes complete combustion under standard conditions. C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H c [C 2 H 5 OH(l)] = -1368 kJ mol  1 

41. 41 -395.4 -393.5 C (diamond) C (graphite)  H c (kJ mol -1 ) Substance ø 6.3 Standard enthalpy changes (SB p.147) Reaction coordinate Enthalpy C(diamond) + O 2 (g) CO 2 (g)  395.4 C(graphite) + O 2 (g)  393.5 1.9  H f [diamond]  = +1.9 kJ mol  1

42. 42 Q.6 HH  =  H f [CO(g)]    H c [graphite]  HH  = 2   H c [H 2 (g)]  (b)2H 2 (g) + O 2 (g)  2H 2 O(l) (a)C(graphite) + O 2 (g)  CO(g) Incomplete combustion = 2   H f [H 2 O(l)] 

43. 43 Q.6 (c)C(graphite) + O 2 (g)  CO 2 (g) HH  =  H c [graphite]  =  H f [CO 2 (g)]  (d)CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l) HH  =  H c [CH 4 (g)]    H f [CO 2 (g)]   2   H f [H 2 O(l)]  Not formed from elements Check Point 6-3 Check Point 6-3

44. 44 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)

45. 45 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

46. 46 -57.1 -57.2 -52.2 -68.6 NaOH KOH NH 3 NaOH HCl HF  H neu AlkaliAcid ø 6.3 Standard enthalpy changes (SB p.142) NH 3 (aq) + HCl(aq)  NH 4 Cl (aq) H + (aq) + OH - (aq) + Cl  (aq)  H 2 O(l) + Cl  (aq)  H 2 = -57.3 NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH  (aq)  H 1 > 0 ø  H neu =  H 1 +  H 2 =  52.2 kJ mol  1

47. 47 -57.1 -57.2 -52.2 -68.6 NaOH KOH NH 3 NaOH HCl HF  H neu AlkaliAcid ø 6.3 Standard enthalpy changes (SB p.142) HF(aq) + NaOH(aq)  NaF(aq) + H 2 O(l) H + (aq) + OH - (aq) + Na + (aq)  H 2 O(l) + Na + (aq)  H 2 = -57.3 HF(aq) H + (aq) + F  (aq)  H 1 < 0 ø  H neu =  H 1 +  H 2 =  68.6 kJ mol  1

48. 48 Standard enthalpy change of solution (  H soln ) is the enthalpy change when one mole of a solute is dissolved in a specified number of moles of solvent (e.g. water) under standard conditions. ø 6.3 Standard enthalpy changes (SB p.142) Standard enthalpy change of solution NaCl(s) + 10H 2 O(l)  Na + (aq) + Cl - (aq)  H soln [NaCl(s)]= +2.008 kJ mol -1 ø NaCl(aq) dilution  H > 0

49. 49 Standard enthalpy change of solution (  H soln ) is the enthalpy change when one mole of a solute is dissolved to form an infinitely dilute solution under standard conditions. ø 6.3 Standard enthalpy changes (SB p.142) Standard enthalpy change of solution NaCl(s) + water  Na + (aq) + Cl - (aq) ø  H soln = +4.98 kJ mol -1 concentration  0

50. 50 e.g. NaCl(s) + water  Na + (aq) + Cl - (aq)  H soln = +4.98 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 + 4.98 kJ mol  1

51. 51 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

52. 52  35.5  43.1  73.0  74.0  37.2 +4.98 +21.0 +22.6 NH 3 NaOH HCl H 2 SO 4 LiCl NaCl NaNO 3 NH 4 Cl  H soln (kJ mol -1 )Salt ø Standard enthalpy change of solution 6.3 Standard enthalpy changes (SB p.143)

53. 53 6.4 Experimental Determination of Enthalpy Changes by Calorimetry

54. 54 Experimental determination of enthalpy changes by calorimetry Calorimeter is any set-up used for the determination of  H. By temperature measurement. 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.148)  H = q p = (m 1 c 1 + m 2 c 2 )  T

55. 55  H = q p = (m 1 c 1 + m 2 c 2 )  T where m 1 is the mass of the reaction mixture, m 2 is the mass of the calorimeter, c 1 is the specific heat capacity of the reaction mixture, c 2 is the specific heat capacity of the calorimeter,  T is the temperature change of the reaction mixture. 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.148)

56. 56 Determination of enthalpy change of neutralization 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)

57. 57 If the reaction is fast enough, T 1  T 2 T0T0 T1T1 t1t1 T2T2 t2t2  H = (m 1 c 1 + m 2 c 2 )(T 1 – T 0 )  H  (m 1 c 1 + m 2 c 2 )(T 2 – T 0 ) Check Point 6-4(a) Check Point 6-4(a)

58. 58 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

59. 59 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

60. 60 Heat evolved = (m 1 c 1 + m 2 c 2 ) ΔT Where m 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 the calorimeter, ΔT is the temperature change of the reaction 6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)

61. 61 Q.7(Example) Check Point 6-4(c) Check Point 6-4(c) heat given out

62. 62 6.5 Hess’s Law

63. 63 Hess’s Law 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 and depends only on the difference between the total enthalpy of the reactants and that of the products. 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 and depends only on the difference between the total enthalpy of the reactants and that of the products. 6.5 Hess’s law (SB p.153)

64. 64 Hess’s Law A(H A )B(H B ) Route 1 H1H1 D H4H4 H5H5 Route 3  H 1 = H B – H A 6.5 Hess’s law (SB p.153) =  H 2 +  H 3 =  H 4 +  H 5 C H2H2 H3H3 Route 2

65. 65 Importance of Hess’s law The enthalpy change of some chemical reactions cannot be determined directly because: But the enthalpy change of such reactions can be determined indirectly by applying Hess’s Law. 6.5 Hess’s law (SB p.155) the reactions cannot be performed/controlled in the laboratory the reaction rates are too slow the reactions may involve the formation of side products

66. 66 Enthalpy change of formation of CO(g) C(graphite) + ½O 2 (g)CO(g)  H f [CO(g)] ø 6.5 Hess’s law (SB p.153) due to further oxidation of CO to CO 2 The reaction cannot be controlled.  H f [CO(g)] ø cannot be determined directly

67. 67 Enthalpy change of formation of CO(g) = -393 - (-283 ) = -110 kJ mol -1  H c [graphite] = -393 kJ mol -1 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)  H c [CO(g)] = -283.0 kJ mol -1 ø

68. 68 6.5 Hess’s law (SB p.155) Enthalpy cycle (Born-Haber cycle) Relate the various equations involved in a reaction 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

69. 69 Steps for drawing Born-Haber cycle C(graphite) + ½O 2 (g)CO(g)  H f [CO(g)] ø 6.5 Hess’s law (SB p.153) 1. Give the equation for the change being considered.

70. 70 Steps for drawing Born-Haber cycle H2H2 + ½O 2 (g) CO 2 (g) H1H1 + ½O 2 (g) 6.5 Hess’s law (SB p.153) 2. Complete the cycle by giving the equations for the combustion reactions of reactants and products. C(graphite) + ½O 2 (g)CO(g)  H f [CO(g)] ø  H f [CO(g)]=  H c [graphite] -  H c [CO(g)] ø ø ø

71. 71 Steps for drawing Born-Haber cycle H2H2 + ½O 2 (g) CO 2 (g) H1H1 + ½O 2 (g) 6.5 Hess’s law (SB p.153) 2. Complete the cycle by giving the equations for the combustion reactions of reactants and products. C(graphite) + ½O 2 (g)CO(g)  H f [CO(g)] ø  H f [CO(g)]=  H c [reactant] -  H c [product] ø ø ø

72. 72 Calculation of standard enthalpy change of formation from standard enthalpy changes of combustion  H f =   H c [reactants] -   H c [product] ø ø ø 6B

73. 73 Q.8  H f [C 4 H 10 (g)] = 4  H c [C(graphite)] + 5  H c [H 2 (g)] -  H c [C 4 H 10 (g)] ø ø øø 4C(graphite) + 5H 2 (g)C 4 H 10 (g)  H f [C 4 H 10 (g)] ø 4CO 2 (g) 4  H c [graphite] ø + 4O 2 (g) 5  H c [H 2 (g)] ø + 5H 2 O(l) + 2.5 O 2 (g)  H c [C 4 H 10 (g)] ø + 6.5 O 2 (g) = [4(-393)+ 5(-286) – (  2877)] kJ mol  1 =  125 kJ mol  1

74. 74 Q.8Method B : By addition and/or subtraction of equations with known HcHc  (1) C(graphite) + O 2 (g)  CO 2 (g)  393  H c /kJ mol  1  (2) H 2 (g) + O 2 (g)  H 2 O(l)  286 (3) C 4 H 10 (g) + O 2 (g)  4CO 2 (g) + 5H 2 O(l)  2877 Overall reaction : 4  (1) + 5  (2) – (3) 4C(graphite) + 5H 2 (g)C 4 H 10 (g)  H f [C 4 H 10 (g)] ø  H f [C 4 H 10 (g)] = [4(-393) + 5(-286)  (  2877)] kJ mol  1  =  125 kJ mol  1

75. 75 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)

76. 76 Enthalpy / kJ mol  1 Steps for drawing enthalpy level diagram 1. Draw the enthalpy level of elements. C(graphite) + O 2 (g)

77. 77 Enthalpy / kJ mol  1 Steps for drawing enthalpy level diagram 2. Enthalpies of elements are arbitrarily taken as zero. C(graphite) + O 2 (g)

78. 78 Enthalpy / kJ mol  1 Steps for drawing enthalpy level diagram 3. Higher enthalpy levels are drawn above that of elements C(graphite) + O 2 (g)

79. 79 Enthalpy / kJ mol  1 Steps for drawing enthalpy level diagram C(graphite) + O 2 (g) 4. Lower enthalpy levels are drawn below that of elements Route 1 CO 2 (g)  H c [graphite] =  393 kJ 

80. 80 Enthalpy / kJ mol  1 Steps for drawing enthalpy level diagram C(graphite) + O 2 (g) 4. Lower enthalpy levels are drawn below that of elements Route 1 CO 2 (g)  H c [graphite] =  393 kJ  CO(g) + O 2 (g)  H c [CO(g)] =  283 kJ   H f [CO(g)] =  110 kJ  Route 2

81. 81 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 cycle 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

82. 82 6.5 Hess’s law (SB p.158) 2C(graphite) + H 2 (g)  C 2 H 2 (g) + H 2 O(l) HfHf  2CO 2 (g) 2  H c [graphite]  + 2O 2 (g)  H c [H 2 (g)]  + 0.5O 2 (g)  H c [C 2 H 2 (g)]  + 2.5O 2 (g) By Hess’s law, HfHf  +  H c [C 2 H 2 (g)] =  2  H c [graphite]  +  H c [H 2 (g)]  HfHf  -  H c [C 2 H 2 (g)]  = 2  H c [graphite]  +  H c [H 2 (g)]  = [2(  393.5) + (  285.8) –(  1299)] kJ mol  1 = +226.2 kJ mol  1

83. 83 Enthalpy / kJ mol  1 2C(graphite) + H 2 (g) + 2.5O 2 (g) (iii). Draw an enthalpy level diagram for the reaction using the enthalpy changes in (ii) Route 1 2CO 2 (g) + H 2 O(l) Route 2  H c [H 2 (g)]   H f [C 2 H 2 (g)]  C 2 H 2 (g) + 2.5O 2 (g) 2CO 2 (g) + H 2 (g) + 0.5O 2 (g)  2  H c [graphite]  H c [C 2 H 2 (g)] 

84. 84 6.6 Calculations involving Standard Enthalpy Changes of Reactions

85. 85 Calculation of standard enthalpy change of reaction from standard enthalpy changes of formation 6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)  H from  H f  

86. 86 Q.9 NH 3 (g) + HCl(g) NH 4 Cl(s) + 0.5H 2 (g) + 0.5Cl 2 (g) 0.5N 2 (g) + 1.5H 2 (g)  H f [NH 4 Cl(s)]   H f [NH 3 (g)]   H f [HCl(g)]  HH  By Hess’s law,  H f [NH 3 (g)]   H f [HCl(g)]  HH  + +=  H f [NH 4 Cl(s)]    H f [NH 3 (g)]  HH  - -=  H f [HCl(g)]  = [  314 –(  46) –(  92)] kJ mol  1 =  176 kJ mol  1

87. 87 Q.9 NH 3 (g) + HCl(g) NH 4 Cl(s) + 0.5H 2 (g) + 0.5Cl 2 (g) 0.5N 2 (g) + 1.5H 2 (g)  H f [NH 4 Cl(s)]   H f [NH 3 (g)]   H f [HCl(g)]  HH  By Hess’s law,  H f [NH 3 (g)]   H f [HCl(g)]  HH  + +=  H f [NH 4 Cl(s)]    H f [NH 3 (g)]  HH  - -=  H f [HCl(g)]   H reaction =   H f [products] -   H f [reactants] ø ø

88. 88 Q.10 4CH 3 NHNH 2 (l) + 5N 2 O 4 (l) 4CO 2 (g) + 9N 2 (g) + 12H 2 O(l) 4  H f [CO 2 (g)]  4C(graphite) + 12H 2 (g) + 4N 2 (g) 4  H f [CH 3 NHNH 2 (l)]  + 5N 2 (g) + 10O 2 (g) 5  H f [N 2 O 4 (l)]  HH  By Hess’s law, = [4(  393) + 12(  286) - 4(+53) – 5(  20)] kJ =  5116 kJ 12  H f [H 2 O(l)]  HH  Highly exothermic/ignites spontaneously Used as rockel fuel in Apollo 11

89. 89 Q.11 C 3 H 6 (g) + H 2 (g) C 3 H 8 (g) HH  (1)C 3 H 6 (g) + 4.5O 2 (g)  3CO 2 (g) + 3H 2 O(l)  2058  H /kJ mol  1  (2)H 2 (g) + 0.5O 2 (g)  H 2 O(l)  286 (3)C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l)  2220 Overall reaction : (1) + (2) – (3) = [(  2058) + (  286) – (  2220)] kJ mol  1 =  124 kJ mol  1 HH 

90. 90 Q.11 C 3 H 6 (g) + H 2 (g) C 3 H 8 (g) HH  = [(  2058) + (  286) – (  2220)] kJ mol  1 =  124 kJ mol  1 HH   H c [C 3 H 8 (g)]  3H 2 O(l) + 3CO 2 (g)  H c [C 3 H 6 (g)]   H f [H 2 O(l)]  + H 2 O(l) HH  + (  2220) = (  2058) + (  286) By Hess’s law,

91. 91 Q.11 C 3 H 6 (g) + H 2 (g) C 3 H 8 (g) HH   H c [C 3 H 8 (g)]  3H 2 O(l) + 3CO 2 (g)  H c [C 3 H 6 (g)]   H c [H 2 (g)]  + H 2 O(l)  H f =   H c [reactants] -   H c [product] ø ø ø  H rx =   H c [reactants] -   H c [products] ø ø ø

92. 92 Relative stability of compounds and  H f   H f indicates the energetic stability of the compound with respect to its elements  HfHf  < 0  energetically more stable than its elements HfHf  > 0  energetically less stable than its elements HfHf  Elements  Compound

93. 93 H 2 (g) + O 2 (g)  H 2 O 2 (l)  H f [H 2 O 2 (l)] =  188 kJ mol  1  H 2 O 2 (l)  H 2 O(l) + 0.5O 2 (g)  H rx =  98 kJ mol  1  H 2 O(l) + 0.5O 2 (g) Enthalpy / kJ mol  1 H 2 (g) + O 2 (g) H 2 O 2 (l)  H rx =  98 kJ   H f [H 2 O 2 (l)] =  188 kJ 

94. 94 Enthalpy / kJ mol  1 H 2 (g) + O 2 (g) H 2 O(l) + 0.5O 2 (g) H 2 O 2 (l)  H rx =  98 kJ  H 2 O 2 is energetically stable w.r.t. H 2 and O 2 H 2 O 2 is energetically unstable w.r.t. H 2 O and 0.5O 2  H f [H 2 O 2 (l)] =  188 kJ 

95. 95 Enthalpy / kJ mol  1 H 2 (g) + O 2 (g) H 2 O(l) + 0.5O 2 (g)  H c [H 2 (g) ] =  286 kJ H 2 O 2 (l)  H rx =  98 kJ  H 2 and O 2 are energetically unstable w.r.t. H 2 O 2 and H 2 O  H f [H 2 O 2 (l)] =  188 kJ 

96. 96 C(graphite) + O 2 (g) Enthalpy / kJ mol  1 C(diamond) + O 2 (g) CO 2 (g)  H c [diamond] =  395 kJ  H =  2 kJ  C(diamond)  C(graphite)  H =  2 kJ mol  1   H c [graphite] =  393 kJ Energetically unstable

97. 97 C(graphite) + O 2 (g) Enthalpy / kJ mol  1 C(diamond) + O 2 (g) CO 2 (g)  H c [diamond] =  395 kJ  H =  2 kJ  C(diamond)  C(graphite)  H =  2 kJ mol  1   H c [graphite] =  393 kJ Kinetically stable The rate of conversion is extremely low

98. 98 C(graphite) + O 2 (g) Enthalpy / kJ mol  1 C(diamond) + O 2 (g) CO 2 (g)  H c [diamond] =  395 kJ  H c [graphite] =  393 kJ C(g) + O(g) + O(g) bond breaking bond forming

99. 99 C(graphite) + O 2 (g) Enthalpy / kJ mol  1 C(diamond) + O 2 (g) CO 2 (g)  H c [diamond] =  395 kJ  H =  2 kJ   H c [graphite] =  393 kJ Rate of reaction depends on the ease of bond breaking in reactants (e.g. diamond) The minimum energy required for a reaction to start is known as the activation energy, E a

100. 100 C(graphite) + O 2 (g) Enthalpy / kJ mol  1 C(diamond) + O 2 (g) CO 2 (g)  H c [diamond] =  395 kJ  H =  2 kJ  C(diamond)  C(graphite)  H =  2 kJ mol  1   H c [graphite] =  393 kJ The extremely low rate is due to high E a

101. 101 C(graphite) + O 2 (g) Enthalpy / kJ mol  1 C(diamond) + O 2 (g) CO 2 (g)  H c [diamond] =  395 kJ  H =  2 kJ  C(diamond)  C(graphite)  H =  2 kJ mol  1   H c [graphite] =  393 kJ E a is always > 0 as bond breaking is endothermic

102. 102 E a tells how fast a reaction can proceed. HfHf  tells how far a reaction can proceed Rate of reaction / kinetics / kinetic stability Equilibrium / energetics / energetic stability

103. 103  Energetic stability HfHf  E a  kinetic stability (rate of reaction) Higher E a  higher kinetic stability of reactants w.r.t. products HfHf  > 0  lower energetic stability w.r.t. its elements < 0  higher energetic stability w.r.t. its elements HfHf  Lower E a  lower kinetic stability of reactants w.r.t. products  lower rate to give products  higher rate to give products

104. 104 diamond graphite Diamond  energetically unstable w.r.t. graphite  kinetically stable w.r.t. graphite Graphite  stable w.r.t. diamond energetically and kinetically

105. 105 diamond graphite extremely slow diamond graphite

106. 106 6.7 Spontaneity of Changes 6.7 Entropy change (SB p.164)

107. 107 Spontaneity : The state of being spontaneous Spontaneous : - self-generated - natural - happening without external influence internal external boundary

108. 108 6.7 Entropy change (SB p.164) A process is said to be spontaneous If no external “forces” are required to keep the process going, although external “forces” may be required to get the process started (E a ). The process may be a physical change or a chemical change

109. 109 6.7 Entropy change (SB p.164) Spontaneous physical change: E.g. condensation of steam at 25°C Spontaneous chemcial change: E.g. burning of wood once the fire has started Exothermic  spontaneous ? Endothermic  not spontaneous ? Q.12

110. 110 No Dissolution of NH 4 Cl in water at 25°C No Melting of ice at 25°C Yes Condensation of steam at 25°C Yes Burning of CO at 25°C Spontaneous ? (Yes / No) Exothermic ? (Yes / No) Change

111. 111 6.7 Entropy change (SB p.164) Exothermicity is NOT the only reason for the spontaneity of a process Some spontaneous changes are endothermic E.g.Melting of ice Dissolution of NH 4 Cl in water

112. 112 6.7 Entropy change (SB p.164) Melting of ice Dissolution of ammonium nitrate in water

113. 113 6.7 Entropy change (SB p.165) Entropy Entropy is a measure of the randomness or the degree of disorder (freedom) of a system Solid Liquid Gas Entropy increases

114. 114 6.7 Entropy change (SB p.166) Entropy change (  S) Entropy change means the change in the degree of disorder of a system  S = S final - S initial

115. 115 6.7 Entropy change (SB p.166) Positive entropy change (  S > 0) Increase in entropy S final > S initial Example: Ice (low entropy)  Water (high entropy)  S = S water – S ice = +ve

116. 116 6.7 Entropy change (SB p.166) Negative entropy change (  S < 0) Decrease in entropy S initial > S fianl Example: Water (high entropy)  Ice (low entropy)  S = S ice – S water = -ve Q.13

117. 117 Changes SS CO(g) + O 2 (g)  CO 2 (g) H 2 O(g)  H 2 O(l) H 2 O(s)  H 2 O(l) NH 4 Cl(s)  NH 4 Cl(aq)

118. 118 Changes SS 2CO(g) + O 2 (g)  2CO 2 (g)-ve H 2 O(g)  H 2 O(l)-ve H 2 O(s)  H 2 O(l)+ve NH 4 Cl(s)  NH 4 Cl(aq)+ve

119. 119 Changes SS C(s) + O 2 (g)  CO 2 (g) SO 2 (g) + O 2 (g)  SO 3 (g) Diffusion of a drop of ink in water diamond  graphite

120. 120 Changes SS C(s) + O 2 (g)  CO 2 (g)+ve 2SO 2 (g) + O 2 (g)  2SO 3 (g)-ve Diffusion of a drop of ink in water +ve diamond  graphite+ve 2 2

121. 121 Consider an isolated system which has no exchange of energy and matter with its surroundings

122. 122  S = S 2 – S 1 > 0 S1S1 S2S2 S2S2 S1S1  S = S 1 – S 2 < 0 Which one is spontaneous ?

123. 123 Spontaneous,  S = S 2 – S 1 > 0 S1S1 S2S2 S2S2 S1S1 Not spontaneous,  S = S 1 – S 2 < 0

124. 124 Free adiabatic expansion of an ideal gas into a vacuum. A molecular statistical interpretation Slit is open vacuum Probability

125. 125 Free adiabatic expansion of an ideal gas into a vacuum. A molecular statistical interpretation Slit is open vacuum Probability 1 mole

126. 126 A spontaneous process taking place in an isolated system is always associated with an increase in entropy (I.e.  S > 0) In closed system, The spontaneity of a process depends on both  H and  S. The driving force of a process is a balance of  H and  S. 6B

127. 127 6.7 Entropy change (SB p.166) Ice (less entropy)  Water (more entropy)  H is +ve  not favourable  S is +ve  favourable Considering both  S &  H, the process is spontaneous Q.14

128. 128 Changes HH SS Spontaneous (Yes / No) H 2 O(g)  H 2 O(l) at 25°C H 2 O(g)  H 2 O(l) at 110°C H 2 O(s)  H 2 O(l) at 25°C H 2 O(s)  H 2 O(l) at -10°C

129. 129 Changes HH SS Spontaneous (Yes / No) H 2 O(g)  H 2 O(l) at 25°C -ve Yes H 2 O(g)  H 2 O(l) at 110°C -ve No H 2 O(s)  H 2 O(l) at 25°C +ve Yes H 2 O(s)  H 2 O(l) at -10°C +ve No

130. 130 Changes HH SS Spontaneous (Yes / No) H 2 O(g)  H 2 O(l) at 25°C -ve Yes H 2 O(g)  H 2 O(l) at 110°C -ve No H 2 O(s)  H 2 O(l) at 25°C +ve Yes H 2 O(s)  H 2 O(l) at -10°C +ve No Favourable

131. 131 Changes HH SS Spontaneous (Yes / No) H 2 O(g)  H 2 O(l) at 25°C -ve Yes H 2 O(g)  H 2 O(l) at 110°C -ve No H 2 O(s)  H 2 O(l) at 25°C +ve Yes H 2 O(s)  H 2 O(l) at -10°C +ve No Not favourable

132. 132 Changes HH SS Spontaneous (Yes / No) H 2 O(g)  H 2 O(l) at 25°C -ve Yes H 2 O(g)  H 2 O(l) at 110°C -ve No H 2 O(s)  H 2 O(l) at 25°C +ve Yes H 2 O(s)  H 2 O(l) at -10°C +ve No Spontaneity depends on temperature

133. 133 Changes HH SS Spontaneous (Yes / No) H 2 O(g)  H 2 O(l) at 25°C -ve Yes H 2 O(g)  H 2 O(l) at 110°C -ve No H 2 O(s)  H 2 O(l) at 25°C +ve Yes H 2 O(s)  H 2 O(l) at -10°C +ve No Favourable

134. 134 Changes HH SS Spontaneous (Yes / No) H 2 O(g)  H 2 O(l) at 25°C -ve Yes H 2 O(g)  H 2 O(l) at 110°C -ve No H 2 O(s)  H 2 O(l) at 25°C +ve Yes H 2 O(s)  H 2 O(l) at -10°C +ve No Not favourable

135. 135 Changes HH SS Spontaneous (Yes / No) H 2 O(g)  H 2 O(l) at 25°C -ve Yes H 2 O(g)  H 2 O(l) at 110°C -ve No H 2 O(s)  H 2 O(l) at 25°C +ve Yes H 2 O(s)  H 2 O(l) at -10°C +ve No Spontaneity depends on temperature

136. 136 Spontaneity of a process depends on  H,  S & T  G =  H –T  S G is the (Gibbs’) free energy J. Willard Gibbs (1839 – 1903)

137. 137  G =  H –T  S Spontaneity depends on  G ‘Free’ means the energy free for work

138. 138 A spontaneous process is always associated with a decrease in the free energy of the system.  G < 0  spontaneous process  G > 0  not spontaneous process Q.15

139. 139 HH SST GGResults +ve high +ve low -ve high -ve low  G =  H –T  S

140. 140 Spontaneous-velow-ve Not spontaneus+vehigh-ve Not spontaneous+velow+ve Spontaneous-vehigh+ve Results GGT SS HH  G =  H –T  S

141. 141 HH SST GGResults -ve+vehigh -ve+velow +ve-vehigh +ve-velow  G =  H –T  S

142. 142 low-ve+ve Not spontaneous+ve high-ve+ve low+ve-ve Spontaneous-ve high+ve-ve Results GGT SS HH  G =  H –T  S

143. 143  G =  H –T  S Q.16

144. 144 Diamond Graphite  G =  H –T  S < 0  H < 0  S > 0 The process is spontaneous, although activation energy is required to start the conversion.

145. 145 Spontaneous  S = S 2 – S 1 > 0 S1S1 S2S2 The entropy of a system can be considered as a measure of the availability of the system to do work. Before expansion, the system is available to do work. After expansion, the system is not available to do work.

146. 146 Spontaneous  S = S 2 – S 1 > 0 S1S1 S2S2 The lower the entropy of a system(before expansion), the more available is the system to do work. Thus, entropy is considered as a measure of the useless energy of a system.

147. 147  G =  H –T  S G = H – TS H = G + TS Useless energyTotal energy Useful energy

148. 148 H = G + TS Useless energyTotal energy Useful energy If the universe is an isolated system H is a constant and  H is always zero

149. 149 H = G + TS  H =  G +  TS = 0 cosmic background radiation = 4K Useful energy Useless energy =  G + T  S = 0  S always > 0,  G always < 0, S always ,useless energy always  G always ,useful energy always 

150. 150 In an isolated system, entropy will only increase with time, it will not decrease with time. The second law of thermodynamics

151. 151 If the universe is an isolated system,  S universe =  S system +  S surroundings > 0 the total entropy (randomness) of the universe will tend to increase to a maximum; the total free energy of the universe will tend to decrease to a minimum.

152. 152 As time increases, the universe will always become more disordered. Entropy is considered as a measure of time. Entropy can be used to distinguish between future and past.

153. 153 Time can only proceed in one direction that results in an increase in the total entropy of the universe. This is known as the arrow of time.

154. 154 The history of the universe minimum entropy, maximum free energy (singularity) expanding maximum entropy, minimum free energy Big bang Big chill H = G +TS T  0 K T  1.4  10 32 K

155. 155 Planck’s units Planck’s length Planck’s time 1.416785(71) × 10 32 K Planck’s temperature h = Planck’s constant G = gravitational constant c = speed of light in vacuum k = Boltzmann constant

156. 156 Planck’s units 1.416785(71) × 10 32 K Planck’s temperature Absolute hot beyond which all physical laws break down Planck’s length Planck’s time

157. 157 Planck’s units 1.416785(71) × 10 32 K Planck’s temperature Planck’s length Planck’s time 1.616252(81) × 10 −35 m Physical significance not yet known

158. 158 Planck’s units 1.416785(71) × 10 32 K Planck’s temperature Planck’s length Planck’s time 1.616252(81) × 10 −35 m the diameter of proton

159. 159 Planck’s units 1.416785(71) × 10 32 K Planck’s temperature Planck’s length Planck’s time 1.616252(81) × 10 −35 m = 7.3  10 -37 m  22 times of Mr Chio’s wavelength

160. 160 1.616252(81) × 10 −35 m Planck’s units 1.416785(71) × 10 32 K Planck’s temperature It is the time required for light to travel, in a vacuum, a distance of 1 Planck length. 5.39124(27) × 10 −44 s Planck’s length Planck’s time

161. 161 1.616252(81) × 10 −35 m Planck’s units 1.416785(71) × 10 32 K Planck’s temperature 10 -15 s  femtosecond( 飛秒 ) 10 -18 s  attosecond( 阿托秒 ) 5.39124(27) × 10 −44 s Planck’s length Planck’s time Time taken for light to travel the length of 3 H atoms

162. 162 Q.17(a) The drop in temperature of the system is accompanied by the rise in temperature of its surroundings.  S system < 0  S surroundings > 0  S universe =  S system +  S surroundings > 0

163. 163 Q.17(b)  S system < 0  S surroundings > 0  S universe =  S system +  S surroundings > 0 The drop in entropy of the system is at the cost of the rise in entropy of its surroundings.

164. 164 6.8 Free energy change (SB p.170) Check Point 6-8 Check Point 6-8

165. 165 The END

166. 166 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

167. 167 (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)

168. 168 (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)

169. 169 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141) (b)(i)

170. 170 6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141) (ii) Back

171. 171 (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.

172. 172 (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

173. 173 (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) ½ Enthalpy change of combustion of nitrogen or enthalpy change of formation of nitrogen dioxide. Back

174. 174 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

175. 175 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

176. 176 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.

177. 177 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

178. 178 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

179. 179 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

180. 180 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

181. 181 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

182. 182 (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)

183. 183 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 dm 3 = 0.025 mol

184. 184 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 Back 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.

185. 185 (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)

186. 186 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. Back

187. 187 (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)

188. 188 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 = 50 g  4.18 J g -1 K -1  33.2 K + 400g  2.10 J g -1 K -1  33.2 K = (6939 + 27888) J = 34827 J No. of moles of methanol used = = 0.05 mol Heat evolved per mole of methanol used = = 696.5 kJ mol -1 The enthalpy change of combustion of methanol is –696.5 kJ mol -1.

189. 189 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

190. 190 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

191. 191 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

192. 192 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

193. 193 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

194. 194 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

195. 195 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

196. 196 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

197. 197 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

198. 198 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

199. 199 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

200. 200 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

201. 201 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

202. 202 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

203. 203 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

204. 204 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

205. 205 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

206. 206 6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)

207. 207 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

208. 208 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

209. 209 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

210. 210 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

211. 211 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

212. 212 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

213. 213 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

214. 214 (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

215. 215 (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)

216. 216 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

217. 217 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 change 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)