1. General reaction: Enthalpy change Consider the reaction a A + b B c C + d D = c (C) + d (D) - a (A) - b (B) 61

2. General reaction: Enthalpy change Consider the reaction a A + b B c C + d D = c (C) + d (D) - a (A) - b (B) For the case of CO 2, the result would read = (CO 2 ) - { (C (graphite) ) + (O 2 )} can be measured by experiment. 62

3. We can find (CO 2 ) if (C (graphite) ) and (O 2 ) can be determined. 63

4. We can find (CO 2 ) if (C (graphite) ) and (O 2 ) can be determined. Unfortunately, there is no way to determine (C (graphite) ) and (O 2 ). In fact, there is no way to determine for any element. 64

5. We can find (CO 2 ) if (C (graphite) ) and (O 2 ) can be determined. Unfortunately, there is no way to determine (C (graphite) ) and (O 2 ). In fact, there is no way to determine for any element. A simple way out of the difficulty is to arbitrarily define the standard enthalpy of formation of any element in its stable form as zero. 65

6. Therefore, (C (graphite) ) = 0 (O 2 ) = 0 (both results at 1 bar, 25 o C). 66

7. Therefore, (C (graphite) ) = 0 (O 2 ) = 0 (both results at 1 bar, 25 o C). Hence, we may write: = (CO 2 ) 67

8. Therefore, (C (graphite) ) = 0 (O 2 ) = 0 (both results at 1 bar, 25 o C). Hence, we may write: = (CO 2 ) Tables of are put together in this way. 68

9. Example: The metabolism of glucose in our bodies can be represented by the equation: C 6 H 12 O 6 + 6 O 2(g) 6 CO 2(g) + 6 H 2 O (l). Calculate the standard enthalpy for the reaction. 69

10. Example: The metabolism of glucose in our bodies can be represented by the equation: C 6 H 12 O 6 + 6 O 2(g) 6 CO 2(g) + 6 H 2 O (l). Calculate the standard enthalpy for the reaction. The standard enthalpies of formation of C 6 H 12 O 6(s), CO 2(g), and H 2 O (l), are -1274.5 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively. 70

11. Example: The metabolism of glucose in our bodies can be represented by the equation: C 6 H 12 O 6 + 6 O 2(g) 6 CO 2(g) + 6 H 2 O (l). Calculate the standard enthalpy for the reaction. The standard enthalpies of formation of C 6 H 12 O 6(s), CO 2(g), and H 2 O (l), are -1274.5 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively. Note, by definition (O 2 ) = 0. 71

12. Hence: = {6 + 6 } -{ + 6 } 72

13. Hence: = {6 + 6 } -{ + 6 } = {6 mol x (-393.5 kJ/mol) + 6 mol x (-285.84 kJ/mol)} - { 1 mol (-1273.3 kJ/mol) } = -2803 kJ 73

14. Hess’s Law Suppose we wish to determine. 74

15. Hess’s Law Suppose we wish to determine. In principle, this quantity can be obtained by measuring the heat given off by the following reaction 75

16. Hess’s Law Suppose we wish to determine. In principle, this quantity can be obtained by measuring the heat given off by the following reaction C (graphite) + ½ O 2(g) CO (g) 76

17. Hess’s Law Suppose we wish to determine. In principle, this quantity can be obtained by measuring the heat given off by the following reaction C (graphite) + ½ O 2(g) CO (g) What is the major problem with this approach? 77

18. Can avoid the problem in the following manner: (a) CO (g) + ½ O 2(g) CO 2(g) = -283.0 kJ (b) C (graphite) + O 2(g) CO 2(g) = -393.5 kJ 78

19. Can avoid the problem in the following manner: (a) CO (g) + ½ O 2(g) CO 2(g) = -283.0 kJ (b) C (graphite) + O 2(g) CO 2(g) = -393.5 kJ If we reverse reaction (a) and add (b) CO 2(g) CO (g) + ½ O 2(g) = 283.0 kJ 79

20. Can avoid the problem in the following manner: (a) CO (g) + ½ O 2(g) CO 2(g) = -283.0 kJ (b) C (graphite) + O 2(g) CO 2(g) = -393.5 kJ If we reverse reaction (a) and add (b) CO 2(g) CO (g) + ½ O 2(g) = 283.0 kJ C (graphite) + O 2(g) CO 2(g) = -393.5 kJ 80

21. Can avoid the problem in the following manner: (a) CO (g) + ½ O 2(g) CO 2(g) = -283.0 kJ (b) C (graphite) + O 2(g) CO 2(g) = -393.5 kJ If we reverse reaction (a) and add (b) CO 2(g) CO (g) + ½ O 2(g) = 283.0 kJ C (graphite) + O 2(g) CO 2(g) = -393.5 kJ 81

22. Can avoid the problem in the following manner: (a) CO (g) + ½ O 2(g) CO 2(g) = -283.0 kJ (b) C (graphite) + O 2(g) CO 2(g) = -393.5 kJ If we reverse reaction (a) and add (b) CO 2(g) CO (g) + ½ O 2(g) = 283.0 kJ C (graphite) + ½ O 2(g) CO 2(g) = -393.5 kJ 82

23. Can avoid the problem in the following manner: (a) CO (g) + ½ O 2(g) CO 2(g) = -283.0 kJ (b) C (graphite) + O 2(g) CO 2(g) = -393.5 kJ If we reverse reaction (a) and add (b) CO 2(g) CO (g) + ½ O 2(g) = 283.0 kJ C (graphite) + O 2(g) CO 2(g) = -393.5 kJ C (graphite) + ½ O 2(g) CO (g) = -110.5 kJ 83

24. Hence = = -110.5 kJ/mol 84

25. The preceding is an illustration of Hess’s Law. Hess’s Law: The overall enthalpy change in a reaction is equal to the sum of the enthalpy changes for the individual steps. 85

26. Hess’s Law energy diagram 86 enthalpy

27. Specific Heat (Specific Heat Capacity) Specific heat: The specific heat of a substance is the amount of heat required to raise the temperature of 1 g of that substance by 1 o C. 87

28. Specific Heat (Specific Heat Capacity) Specific heat: The specific heat of a substance is the amount of heat required to raise the temperature of 1 g of that substance by 1 o C. 88

29. Specific Heat (Specific Heat Capacity) Specific heat: The specific heat of a substance is the amount of heat required to raise the temperature of 1 g of that substance by 1 o C. or (The symbol s is also used in place of c). 89

30. Specific Heat (Specific Heat Capacity) Specific heat: The specific heat of a substance is the amount of heat required to raise the temperature of 1 g of that substance by 1 o C. or (The symbol s is also used in place of c). A common (non SI) unit is. At 25 o C the value is 0.998 cal g -1 o C -1 for water. (i.e. approximately 1 cal g -1 o C -1 ). 90

31. Heat Capacity Heat Capacity: This is defined by 91

32. Heat Capacity Heat Capacity: This is defined by Water has a relatively high heat capacity. 92

33. Heat Capacity Heat Capacity: This is defined by Water has a relatively high heat capacity. The units of C are cal o C -1. The conversion to SI units is 1 cal o C -1 = 4.184 J K -1 93

34. The human body has a high percentage (by mass) of water, and thus, a relatively high heat capacity. 94

35. The human body has a high percentage (by mass) of water, and thus, a relatively high heat capacity. If C is high, then for modest changes in heat q, the change in temperature of the body is small. Humans are sensitive to even small changes in body temperature. 95

36. Calorimetry Heat changes are measured in calorimetry. There are two principal types of heat that can be measured. 96

37. Calorimetry Heat changes are measured in calorimetry. There are two principal types of heat that can be measured. (1) Heat changes measured under constant pressure conditions. Symbol q p This quantity is equal to the enthalpy change: 97

38. Calorimetry Heat changes are measured in calorimetry. There are two principal types of heat that can be measured. (1) Heat changes measured under constant pressure conditions. Symbol q p This quantity is equal to the enthalpy change: (2) Heat changes measured under constant volume conditions. Symbol q V This quantity is equal to the change in the internal energy: 98

39. Which quantity is measured, q p or q V, will depend on the type of calorimeter used to make the experimental measurements. 99

40. Which quantity is measured, q p or q V, will depend on the type of calorimeter used to make the experimental measurements. A device which is open to the atmosphere will give q p, i.e. the enthalpy change. 100

41. Which quantity is measured, q p or q V, will depend on the type of calorimeter used to make the experimental measurements. A device which is open to the atmosphere will give q p, i.e. the enthalpy change. A device that is a sealed container, that is one having a fixed volume, will yield q V, i.e. the internal energy change. 101

42. Which quantity is measured, q p or q V, will depend on the type of calorimeter used to make the experimental measurements. A device which is open to the atmosphere will give q p, i.e. the enthalpy change. A device that is a sealed container, that is one having a fixed volume, will yield q V, i.e. the internal energy change. The enthalpy and the internal energy are connected by the formula: H = E + p V 102

43. 103

44. 104

45. Calorimetry Heats of combustion are usually measured in a calorimeter. 105

46. Calorimetry Heats of combustion are usually measured in a calorimeter. A known mass of the sample is placed in a steel container which is filled under pressure with oxygen gas. The sample is ignited electrically, and the temperature rise in the surroundings indicates the amount of heat given off. 106

47. Calorimetry Heats of combustion are usually measured in a calorimeter. A known mass of the sample is placed in a steel container which is filled under pressure with oxygen gas. The sample is ignited electrically, and the temperature rise in the surroundings indicates the amount of heat given off. The surroundings are a known mass of water plus the actual calorimeter device. The assumption is usually made that no heat is lost to the outside world. 107

48. The heat capacity of the calorimeter is determined using a compound for which the heat of combustion is accurately known. 108

49. Example: If a given amount of a standard compound releases 2.000 x 10 4 J, and that raises the temperature of 1000.0 g of water in the calorimeter by 2.00 o C, determine the heat taken up by the calorimeter. For water, c = 4.18 Jg -1 o C -1. 109

50. Example: If a given amount of a standard compound releases 2.000 x 10 4 J, and that raises the temperature of 1000.0 g of water in the calorimeter by 2.00 o C, determine the heat taken up by the calorimeter. For water, c = 4.18 Jg -1 o C -1. = (1000.0 g) (4.18 Jg -1 o C -1 )(2.00 o C) = 8.36 kJ 110

51. Now: heat released = heat gained + heat gained by sample by water by calorimeter 111

52. Now: heat released = heat gained + heat gained by sample by water by calorimeter Therefore: heat gained = 2.000 x 10 4 J - 0.836 X 10 4 J by calorimeter = 1.164 x 10 4 J 112

53. The heat capacity of the calorimeter is: 1.164 x 10 4 J = 2.00 o C = 5.82 x 10 3 J o C -1 113

54. Sample Calculation A 0.569 g sample of benzoic acid is burned in a constant volume bomb calorimeter. The heat capacity of the calorimeter is 3.36 x 10 3 J/ o C. If the temperature rise registered in the 500.0 g of water in the calorimeter is 2.05 o C, calculate the heat given off by the benzoic acid in kJ/mol. 114

55. Sample Calculation A 0.569 g sample of benzoic acid is burned in a constant volume bomb calorimeter. The heat capacity of the calorimeter is 3.36 x 10 3 J/ o C. If the temperature rise registered in the 500.0 g of water in the calorimeter is 2.05 o C, calculate the heat given off by the benzoic acid in kJ/mol. The chemical reaction is : C 6 H 5 COOH + 15/2 O 2 7 CO 2 + 3 H 2 O 115

56. 116 The heat gained by the water: = 500.0 g x 4.18 Jg -1 o C -1 x 2.05 o C = 4.28 x 10 3 J

57. 117 The heat gained by the water: = 500.0 g x 4.18 Jg -1 o C -1 x 2.05 o C = 4.28 x 10 3 J The heat gained by the calorimeter: = 3.36 x 10 3 J o C -1 x 2.05 o C = 6.89 x 10 3 J

58. The total heat gained by the calorimeter + water contents = 4.28 x 10 3 J + 6.89 x 10 3 J = 1.117 x 10 4 J 118

59. The total heat gained by the calorimeter + water contents = 4.28 x 10 3 J + 6.89 x 10 3 J = 1.117 x 10 4 J So we can write the following equivalence statement: 0.569 g benzoic acid 1.117 x 10 4 J 119

60. The total heat gained by the calorimeter + water contents = 4.28 x 10 3 J + 6.89 x 10 3 J = 1.117 x 10 4 J So we can write the following equivalence statement: 0.569 g benzoic acid 1.117 x 10 4 J The molar mass of benzoic acid = 122 g/mol 120