1. Thermochemistry

2. Thermochemistry
Important Terminology:

3. Thermochemistry
Important Terminology:
Thermodynamics …

4. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...

5. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy …

6. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...

7. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ...

8. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ...

9. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2

10. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ...

11. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ...

12. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh

13. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ...

14. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ...

15. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...

16. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ...

17. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance

18. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d

19. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ...

20. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ...

21. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same

22. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same
Universe ...

23. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same
Universe ... Composed of the system and the surroundings

24. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same
Universe ... Composed of the system and the surroundings ...

25. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same
Universe ... Composed of the system and the surroundings ...
System ... That which we are focusing upon ...

26. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same
Universe ... Composed of the system and the surroundings ...
System ... That which we are focusing upon ...
Surroundings ...

27. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same
Universe ... Composed of the system and the surroundings ...
System ... That which we are focusing upon ...
Surroundings ... Everything else in the universe outside the system under study ...

28. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same
Universe ... Composed of the system and the surroundings ...
System ... That which we are focusing upon ...
Surroundings ... Everything else in the universe outside the system under study ...
Exothermic ...

29. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same
Universe ... Composed of the system and the surroundings ...
System ... That which we are focusing upon ...
Surroundings ... Everything else in the universe outside the system under study ...
Exothermic ... Energy (as heat) that flows out of the system ...

30. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same
Universe ... Composed of the system and the surroundings ...
System ... That which we are focusing upon ...
Surroundings ... Everything else in the universe outside the system under study ...
Exothermic ... Energy (as heat) that flows out of the system ...
Endothermic ...

31. Thermochemistry Important Terminology:
Thermodynamics … The study of energy an its interconversions ...
Energy … The capacity to produce work or heat ...
Kinetic Energy ... The energy of motion ... KE = 1/2mv2
Potential Energy ... Energy that can be converted to useful work ... PE = mgh
Heat ... Involves transfer of energy between two objects ...
Work ... Force times distance ... W = F x d
State Function ... Independent of pathway ... difference in the value is the same
Universe ... Composed of the system and the surroundings ...
System ... That which we are focusing upon ...
Surroundings ... Everything else in the universe outside the system under study ...
Exothermic ... Energy (as heat) that flows out of the system ...
Endothermic ... Energy (as heat) that flows into the system ...

32. Thermochemistry
Work:

33. Thermochemistry
Work:
Force

34. Force = mass x acceleration =
Thermochemistry
Work:
Force = mass x acceleration =

35. Force = mass x acceleration = ma =
Thermochemistry
Work:
Force = mass x acceleration = ma =

36. Force = mass x acceleration = ma = kg x m/s2 =
Thermochemistry
Work:
Force = mass x acceleration = ma = kg x m/s2 =

37. Force = mass x acceleration = ma = kg x m/s2 = N
Thermochemistry
Work:
Force = mass x acceleration = ma = kg x m/s2 = N

38. Force = mass x acceleration = ma = kg x m/s2 = N
Thermochemistry
Work:
Force = mass x acceleration = ma = kg x m/s2 = N
Work =

39. Force = mass x acceleration = ma = kg x m/s2 = N
Thermochemistry
Work:
Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance

40. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d =

41. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m =

42. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2

43. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m

44. Force = mass x acceleration = ma = kg x m/s2 = N
Thermochemistry
Work:
Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P =

45. Force = mass x acceleration = ma = kg x m/s2 = N
Thermochemistry
Work:
Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A

46. Force = mass x acceleration = ma = kg x m/s2 = N
Thermochemistry
Work:
Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2

47. Force = mass x acceleration = ma = kg x m/s2 = N
Thermochemistry
Work:
Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2

48. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa

49. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V =

50. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2

51. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2 = N x m

52. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2 = N x m = J

53. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2 = N x m = J
So:

54. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2 = N x m = J
So:
W =

55. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2 = N x m = J
So:
W = P x ∆V

56. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2 = N x m = J
So:
W = P x V = Constant

57. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2 = N x m = J
So:
W = P x V = Constant

58. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2 = N x m = J
So:
W = P x V = Constant

59. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2 = N x m = J
So:
W = P x V = Constant

60. Thermochemistry Work: Force = mass x acceleration = ma = kg x m/s2 = N
Work = Force x distance = F x d = kg x m/s2 x m = kg x m2/s2 = N x m = J
P = F/A = kg m/s2/m2 = kg/ms2 = Pa
P x V = kg/ms2 x m3 = kg m2/s2 = N x m = J
So:
W = P x V = Constant

61. Thermochemistry
Sign Conventions …

62. Thermochemistry
Sign Conventions …
System expansion ...

63. Thermochemistry Sign Conventions …
System expansion work done on surroundings …

64. Thermochemistry Sign Conventions …
System expansion work done on surroundings … - work done on system ...

65. Thermochemistry Sign Conventions …
System expansion work done on surroundings … - work done on system ...
System contraction ...

66. Thermochemistry Sign Conventions …
System expansion work done on surroundings … - work done on system ...
System contraction work done on system work done on surroundings ...

67. Thermochemistry Sign Conventions …
System expansion work done on surroundings … - work done on system ...
System contraction work done on system work done on surroundings ...
Work =

68. Thermochemistry Sign Conventions …
System expansion work done on surroundings … - work done on system ...
System contraction work done on system work done on surroundings ...
Work = -P∆V

69. Thermochemistry
Known:

70. Thermochemistry
Known: Unknown:

71. Thermochemistry
Known: Unknown:
W = ? L atm

72. Thermochemistry
Known: Unknown:
W = ? L atm
W = ? J

73. Thermochemistry
Known: Unknown:
Vi = 75 L W = ? L atm
W = ? J

74. Thermochemistry Known: Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J

75. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J

76. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:

77. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ...

78. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work

79. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work
W = ? L atm ||

80. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work
W = ? L atm || -P∆V =

81. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work
W = ? L atm || -P∆V = -P(Vf – Vi) = -(6.0 atm) (30 L – 75 L) =

82. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work
W = ? L atm || -P∆V = -P(Vf – Vi) = -(6.0 atm) (30 L – 75 L) = -(6.0 atm)(-45 L)

83. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work
W = ? L atm || -P∆V = -P(Vf – Vi) = -(6.0 atm) (30 L – 75 L) = -(6.0 atm)(-45 L)
= 270 L atm

84. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work
W = ? L atm || -P∆V = -P(Vf – Vi) = -(6.0 atm) (30 L – 75 L) = -(6.0 atm)(-45 L)
= 270 L atm
W = ? J ||

85. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work
W = ? L atm || -P∆V = -P(Vf – Vi) = -(6.0 atm) (30 L – 75 L) = -(6.0 atm)(-45 L)
= 270 L atm
W = ? J || (270 L atm) (101.3 J/1 L atm)

86. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work
W = ? L atm || -P∆V = -P(Vf – Vi) = -(6.0 atm) (30 L – 75 L) = -(6.0 atm)(-45 L)
= 270 L atm
W = ? J || (270 L atm) (101.3 J/1 L atm) =

87. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work
W = ? L atm || -P∆V = -P(Vf – Vi) = -(6.0 atm) (30 L – 75 L) = -(6.0 atm)(-45 L)
= 270 L atm
W = ? J || (270 L atm) (101.3 J/1 L atm) = x 104 J

88. Thermochemistry Known: @ Pressure = k Unknown: Vi = 75 L W = ? L atm
Vf = 30 L W = ? J
1 L atm = J
Solution:
1st: Calculate work done on the system ... system compression is + work
W = ? L atm || -P∆V = -P(Vf – Vi) = -(6.0 atm) (30 L – 75 L) = -(6.0 atm)(-45 L)
= 270 L atm
W = ? J || (270 L atm) (101.3 J/1 L atm) = x 104 J

89. Thermochemistry
Law of Conservation of Energy …

90. Thermochemistry
Law of Conservation of Energy … energy can neither be created nor destroyed, merely transformed from one form to another ...

91. Thermochemistry
Law of Conservation of Energy … energy can neither be created nor destroyed, merely transformed from one form to another ...
TE = PE + KE

92. The Energy of the Universe is constant
Thermochemistry
Law of Conservation of Energy … energy can neither be created nor destroyed, merely transformed from one form to another ...
TE = PE + KE
The Energy of the Universe is constant
∆Euniv = ∆Esys + ∆Esurr

93. Thermochemistry
Law of Conservation of Energy … energy can neither be created nor destroyed, merely transformed from one form to another ...
TE = PE + KE
The Energy of the Universe is constant
∆Euniv = ∆Esys + ∆Esurr = 0

94. Thermochemistry
Law of Conservation of Energy … energy can neither be created nor destroyed, merely transformed from one form to another ...
TE = PE + KE
The Energy of the Universe is constant
∆Euniv = ∆Esys + ∆Esurr = 0
Chemical Bonds are made and broken ...
Chemical Bonds are made and broken ... Energy is converted between the PE (stored in bonds) ...

95. Thermochemistry
Law of Conservation of Energy … energy can neither be created nor destroyed, merely transformed from one form to another ...
TE = PE + KE
The Energy of the Universe is constant
∆Euniv = ∆Esys + ∆Esurr = 0
Chemical Bonds are made and broken ...
Chemical Bonds are made and broken ... Energy is converted between the PE (stored in bonds) and KE, i.e., thermal energy as heat …

96. Thermochemistry
Change in Internal Energy of a System …

97. Thermochemistry
Change in Internal Energy of a System …
∆Esys =

98. Thermochemistry
Change in Internal Energy of a System …
∆Esys = q + W

99. Thermochemistry Change in Internal Energy of a System … ∆Esys = q + W

100. Thermochemistry Change in Internal Energy of a System … ∆Esys = q + W
∆Esys = (-) means the system loses energy

101. Thermochemistry Change in Internal Energy of a System … ∆Esys = q + W
∆Esys = (-) means the system loses energy
∆Esys = (+) means the system gaines energy

102. Thermochemistry
Known: Unknown:
∆Esys = ? J

103. Thermochemistry Known: Unknown:
W = J (work done by system) ∆Esys = ? J

104. Thermochemistry Known: Unknown:
W = J (work done by system) ∆Esys = ? J
q = J (heat loss by system)

105. Thermochemistry Known: Unknown:
W = J (work done by system) ∆Esys = ? J
q = J (heat loss by system)
Solution:
1st: Calculate ∆Esys ...

106. Thermochemistry Known: Unknown:
W = J (work done by system) ∆Esys = ? J
q = J (heat loss by system)
Solution:
1st: Calculate ∆Esys ...
∆Esys = ? J ||

107. Thermochemistry Known: Unknown:
W = J (work done by system) ∆Esys = ? J
q = J (heat loss by system)
Solution:
1st: Calculate ∆Esys ...
∆Esys = ? J || q + W =

108. Thermochemistry Known: Unknown:
W = J (work done by system) ∆Esys = ? J
q = J (heat loss by system)
Solution:
1st: Calculate ∆Esys ...
∆Esys = ? J || q + W = J + (-38.9 J) =

109. Thermochemistry Known: Unknown:
W = J (work done by system) ∆Esys = ? J
q = J (heat loss by system)
Solution:
1st: Calculate ∆Esys ...
∆Esys = ? J || q + W = J + (-38.9 J) = J

110. Thermochemistry Known: Unknown:
W = J (work done by system) ∆Esys = ? J
q = J (heat loss by system)
Solution:
1st: Calculate ∆Esys ...
∆Esys = ? J || q + W = J + (-38.9 J) = J ... system lost J of energy

111. Thermochemistry Known: Unknown: Piston compressed … ∆Esys = ? J
Vi = 8.3 L

112. Thermochemistry Known: Unknown: Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L

113. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L

114. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J

115. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:

116. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? ||

117. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W =

118. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V =

119. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V = (+350 J) – (1.9 atm) (2.8 L – 8.3 L)

120. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V = (+350 J) – (1.9 atm) (2.8 L – 8.3 L) =

121. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V = (+350 J) – (1.9 atm) (2.8 L – 8.3 L) =
= +350 J L atm =

122. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V = (+350 J) – (1.9 atm) (2.8 L – 8.3 L) =
= +350 J L atm = +350 J + (10.45 L atm)(101.3 J/1 L atm) =

123. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V = (+350 J) – (1.9 atm) (2.8 L – 8.3 L) =
= +350 J L atm = +350 J + (10.45 L atm)(101.3 J/1 L atm) =
= +350 J J = 1409 J

124. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V = (+350 J) – (1.9 atm) (2.8 L – 8.3 L) =
= +350 J L atm = +350 J + (10.45 L atm)(101.3 J/1 L atm) =
= +350 J J = 1049 J
Note:

125. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V = (+350 J) – (1.9 atm) (2.8 L – 8.3 L) =
= +350 J L atm = +350 J + (10.45 L atm)(101.3 J/1 L atm) =
= +350 J J = 1049 J
Note: system compression …

126. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V = (+350 J) – (1.9 atm) (2.8 L – 8.3 L) =
= +350 J L atm = +350 J + (10.45 L atm)(101.3 J/1 L atm) =
= +350 J J = 1049 J
Note: system compression … (-) work done on system ...

127. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V = (+350 J) – (1.9 atm) (2.8 L – 8.3 L) =
= +350 J L atm = +350 J + (10.45 L atm)(101.3 J/1 L atm) =
= +350 J J = 1049 J
Note: system compression … (-) work done on system ... ∆Esys = +

128. Thermochemistry Known: @ P = k = 1.9 atm Unknown:
Piston compressed … ∆Esys = ? J
Vi = 8.3 L
Vf = 2.8 L
q = +350 J
Solution:
∆Esys = ? || q + W = q + -P∆V = (+350 J) – (1.9 atm) (2.8 L – 8.3 L) =
= +350 J L atm = +350 J + (10.45 L atm)(101.3 J/1 L atm) =
= +350 J J = 1049 J
Note: system compression … (-) work done on system ... ∆Esys = +

129. Thermochemistry
Enthalpy & Calorimetry:

130. Thermochemistry Enthalpy & Calorimetry:
Enthalpy (H) is a state function

131. Thermochemistry Enthalpy & Calorimetry:
Enthalpy (H) is a state function … ∆H is independent of pathway ...

132. ∆H = Hproducts - Hreactants
Thermochemistry
Enthalpy & Calorimetry:
Enthalpy (H) is a state function … ∆H is independent of pathway ...
∆H = Hproducts - Hreactants

133. ∆H = Hproducts - Hreactants
Thermochemistry
Enthalpy & Calorimetry:
Enthalpy (H) is a state function … ∆H is independent of pathway ...
∆H = Hproducts - Hreactants
@ P = k …

134. ∆H = Hproducts - Hreactants
Thermochemistry
Enthalpy & Calorimetry:
Enthalpy (H) is a state function … ∆H is independent of pathway ...
∆H = Hproducts - Hreactants
@ P = k …
∆H = ∆Esys + P∆V

135. Thermochemistry Enthalpy & Calorimetry:
Enthalpy (H) is a state function … ∆H is independent of pathway ...
∆H = Hproducts - Hreactants
@ P = k …
∆H = ∆Esys + P∆V = (qp + W) – W = qp

136. Thermochemistry Enthalpy & Calorimetry:
Enthalpy (H) is a state function … ∆H is independent of pathway ...
∆H = Hproducts - Hreactants
@ P = k …
∆H = ∆Esys + P∆V = (qp + W) – W = qp
P = k … ∆H = qp ...

137. Thermochemistry Enthalpy & Calorimetry:
Enthalpy (H) is a state function … ∆H is independent of pathway ...
∆H = Hproducts - Hreactants
@ P = k …
∆H = ∆Esys + P∆V = (qp + W) – W = qp
P = k … ∆H = qp ...
If ∆H > 0 … Reaction is endothermic ...

138. Thermochemistry Enthalpy & Calorimetry:
Enthalpy (H) is a state function … ∆H is independent of pathway ...
∆H = Hproducts - Hreactants
@ P = k …
∆H = ∆Esys + P∆V = (qp + W) – W = qp
P = k … ∆H = qp ...
If ∆H > 0 … Reaction is endothermic ... Heat is absorbed by the system ...

139. Thermochemistry Enthalpy & Calorimetry:
Enthalpy (H) is a state function … ∆H is independent of pathway ...
∆H = Hproducts - Hreactants
@ P = k …
∆H = ∆Esys + P∆V = (qp + W) – W = qp
P = k … ∆H = qp ...
If ∆H > 0 … Reaction is endothermic ... Heat is absorbed by the system ...
If ∆H < Reaction is exothermic ...

140. Thermochemistry Enthalpy & Calorimetry:
Enthalpy (H) is a state function … ∆H is independent of pathway ...
∆H = Hproducts - Hreactants
@ P = k …
∆H = ∆Esys + P∆V = (qp + W) – W = qp
P = k … ∆H = qp ...
If ∆H > 0 … Reaction is endothermic ... Heat is absorbed by the system ...
If ∆H < Reaction is exothermic ... Heat is released by the system ...

141. Thermochemistry
P = k Unknown:
∆Hsys = ? kJ

142. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ

143. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ;

144. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ; gfmNaOH = 40.0 gNaOH/molNaOH

145. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ; gfmNaOH = 40.0 gNaOH/molNaOH
Solution:
∆Hsys = ? kJ ||

146. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ; gfmNaOH = 40.0 gNaOH/molNaOH
Solution:
∆Hsys = ? kJ || (- 43 kJ/mol) (1 mol/40.0 g) x 14.0 g =

147. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ; gfmNaOH = 40.0 gNaOH/molNaOH
Solution:
∆Hsys = ? kJ || (- 43 kJ/mol) (1 mol/40.0 g) x 14.0 g = kJ =

148. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ; gfmNaOH = 40.0 gNaOH/molNaOH
Solution:
∆Hsys = ? kJ || (- 43 kJ/mol) (1 mol/40.0 g) x 14.0 g = kJ = ∆Hsys

149. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ; gfmNaOH = 40.0 gNaOH/molNaOH
Solution:
∆Hsys = ? kJ || (- 43 kJ/mol) (1 mol/40.0 g) x 14.0 g = kJ = ∆Hsys

150. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ; gfmNaOH = 40.0 gNaOH/molNaOH
Solution:
∆Hsys = ? kJ || (- 43 kJ/mol) (1 mol/40.0 g) x 14.0 g = kJ = ∆Hsys
∆H < 0 …

151. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ; gfmNaOH = 40.0 gNaOH/molNaOH
Solution:
∆Hsys = ? kJ || (- 43 kJ/mol) (1 mol/40.0 g) x 14.0 g = kJ = ∆Hsys
∆H < 0 … rx is exothermic ...

152. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ; gfmNaOH = 40.0 gNaOH/molNaOH
Solution:
∆Hsys = ? kJ || (- 43 kJ/mol) (1 mol/40.0 g) x 14.0 g = kJ = ∆Hsys
∆H < 0 … rx is exothermic ... heat is released …

153. Thermochemistry Known: @ P = k Unknown:
NaOH(s) ----> NaOH(aq) + 43 kJ/mol ∆Hsys = ? kJ
mNaOH = 14.0 gNaOH ; gfmNaOH = 40.0 gNaOH/molNaOH
Solution:
∆Hsys = ? kJ || (- 43 kJ/mol) (1 mol/40.0 g) x 14.0 g = kJ = ∆Hsys
∆H < 0 … rx is exothermic ... heat is released … beaker is warmer ...

154. Thermochemistry
Calorimetry ...

155. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...

156. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H

157. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H
@ V = k, qv = ∆E

158. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H
@ V = k, qv = ∆E
∆E = q + w =

159. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H
@ V = k, qv = ∆E
∆E = q + w = q + -P∆V = q + 0 = qv
In both cases ...

160. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H
@ V = k, qv = ∆E
∆E = q + w = q + -P∆V = q + 0 = qp
In both cases ... heat gain or loss is being determined …

161. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H
@ V = k, qv = ∆E
∆E = q + w = q + -P∆V = q + 0 = qp
In both cases ... heat gain or loss is being determined … amt. of heat exchanged in rx. depends upon …

162. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H
@ V = k, qv = ∆E
∆E = q + w = q + -P∆V = q + 0 = qp
In both cases ... heat gain or loss is being determined … amt. of heat exchanged in rx. depends upon …
1. ∆Trx

163. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H
@ V = k, qv = ∆E
∆E = q + w = q + -P∆V = q + 0 = qp
In both cases ... heat gain or loss is being determined … amt. of heat exchanged in rx. depends upon …
1. ∆Trx
2. masssubstance

164. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H
@ V = k, qv = ∆E
∆E = q + w = q + -P∆V = q + 0 = qp
In both cases ... heat gain or loss is being determined … amt. of heat exchanged in rx. depends upon …
1. ∆Trx
2. masssubstance
3. heat capacity (c) of substance

165. Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H
@ V = k, qv = ∆E
∆E = q + w = q + -P∆V = q + 0 = qp
In both cases ... heat gain or loss is being determined … amt. of heat exchanged in rx. depends upon …
1. ∆Trx
2. masssubstance
3. heat capacity (c) of substance
C =

166. C = heat absorbed / increase in temperature = J/’C
Thermochemistry
Calorimetry ... Experimental technique used to determine q for a reaction ...
@ P = k , qp = ∆H
@ V = k, qv = ∆E
∆E = q + w = q + -P∆V = q + 0 = qv
In both cases ... heat gain or loss is being determined … amt. of heat exchanged in rx. depends upon …
1. ∆Trx
2. masssubstance
3. heat capacity (c) of substance
C = heat absorbed / increase in temperature = J/’C

167. Thermochemistry
Calorimetry …

168. Thermochemistry
Calorimetry …
q = mc∆T

169. Thermochemistry
Calorimetry …
q = mc∆T
c = q/m∆T

170. Thermochemistry Calorimetry … q = mc∆T c = q/m∆T
Three ways to express heat capacity …

171. Thermochemistry Calorimetry … q = mc∆T c = q/m∆T
Three ways to express heat capacity …
1. Heat Capacity = J/’C

172. Thermochemistry Calorimetry … q = mc∆T c = q/m∆T
Three ways to express heat capacity …
1. Heat Capacity = J/’C
2. Specific Heat Capacity =

173. Thermochemistry Calorimetry … q = mc∆T c = q/m∆T
Three ways to express heat capacity …
1. Heat Capacity = J/’C
2. Specific Heat Capacity = heat capacity per gram of substance

174. Thermochemistry Calorimetry … q = mc∆T c = q/m∆T
Three ways to express heat capacity …
1. Heat Capacity = J/’C
2. Specific Heat Capacity = heat capacity per gram of substance = J/’C g or J/K g

175. Thermochemistry Calorimetry … q = mc∆T c = q/m∆T
Three ways to express heat capacity …
1. Heat Capacity = J/’C
2. Specific Heat Capacity = heat capacity per gram of substance = J/’C g or J/K g
3. Molar Heat Capacity = heat capacity per mole of substance = J/’C mol or J/K mol

176. What conducts heat better, water or aluminum?
Thermochemistry
Calorimetry …
What conducts heat better, water or aluminum?

177. Thermochemistry Calorimetry … Known: Unknown:
What conducts heat better,
water or aluminum?

178. Thermochemistry Calorimetry … Known: Unknown:
cwater = 4.18 J/’C g What conducts heat better,
water or aluminum?

179. Thermochemistry Calorimetry … Known: Unknown:
cwater = 4.18 J/’C g What conducts heat better,
caluminum = 0.89 J/’C x g water or aluminum?
Solution:
Heat capacity of water is 4.18 J/’C g.

180. Thermochemistry Calorimetry … Known: Unknown:
cwater = 4.18 J/’C g What conducts heat better,
caluminum = 0.89 J/’C x g water or aluminum?
Solution:
Heat capacity of water is 4.18 J/’C g ... it takes 4.18 J of energy to raise the temperature of 1 gram of water 1 ‘C ...

181. Thermochemistry Calorimetry … Known: Unknown:
cwater = 4.18 J/’C g What conducts heat better,
caluminum = 0.89 J/’C x g water or aluminum?
Solution:
Heat capacity of water is 4.18 J/’C g ... it takes 4.18 J of energy to raise the temperature of 1 gram of water 1 ‘C ...
Heat capacity of aluminum is 0.89 J/’C x g ...

182. Thermochemistry Calorimetry … Known: Unknown:
cwater = 4.18 J/’C g What conducts heat better,
caluminum = 0.89 J/’C x g water or aluminum?
Solution:
Heat capacity of water is 4.18 J/’C g ... it takes 4.18 J of energy to raise the temperature of 1 gram of water 1 ‘C ...
Heat capacity of aluminum is 0.89 J/’C x g ... It takes 0.89 J of energy to raise the temperature of 1 gram of aluminum 1 ‘C ...

183. Thermochemistry Calorimetry … Known: Unknown:
cwater = 4.18 J/’C g What conducts heat better,
caluminum = 0.89 J/’C x g water or aluminum?
Solution:
Heat capacity of water is 4.18 J/’C g ... it takes 4.18 J of energy to raise the temperature of 1 gram of water 1 ‘C ...
Heat capacity of aluminum is 0.89 J/’C x g ... It takes 0.89 J of energy to raise the temperature of 1 gram of aluminum 1 ‘C ...
It takes 5 times (4.18/0.89) to raise the temperature of an equivalent amount of water 1 ‘C.

184. Thermochemistry Calorimetry … Known: Unknown:
cwater = 4.18 J/’C g What conducts heat better,
caluminum = 0.89 J/’C x g water or aluminum?
Solution:
Heat capacity of water is 4.18 J/’C g ... it takes 4.18 J of energy to raise the temperature of 1 gram of water 1 ‘C ...
Heat capacity of aluminum is 0.89 J/’C x g ... It takes 0.89 J of energy to raise the temperature of 1 gram of aluminum 1 ‘C ...
It takes 5 times (4.18/0.89) to raise the temperature of an equivalent amount of water 1 ‘C. Al is a better conductor ... less heat causes an equal rise in temperature …

185. Thermochemistry
Constant Pressure Calorimetry ...

186. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...

187. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)

188. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:

189. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …

190. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J ||

191. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] =

192. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J

193. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...

194. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...
mass = ? g || (1.00 L)(1000 mL/1 L)(1.05 g/mL)

195. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...
mass = ? g || (1.00 L)(1000 mL/1 L)(1.05 g/mL) = 1050 g

196. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...
mass = ? g || (1.00 L)(1000 mL/1 L)(1.05 g/mL) = 1050 g

197. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...
mass = ? g || (1.00 L)(1000 mL/1 L)(1.05 g/mL) = 1050 g
3rd: Calculate the ∆T of the solution …

198. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...
mass = ? g || (1.00 L)(1000 mL/1 L)(1.05 g/mL) = 1050 g
3rd: Calculate the ∆T of the solution …
∆T = ? ‘C|| q/mc =

199. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...
mass = ? g || (1.00 L)(1000 mL/1 L)(1.05 g/mL) = 1050 g
3rd: Calculate the ∆T of the solution …
∆T = ? ‘C|| q/mc = (10,750 J)/(1050 g)(4.18 J/’C g) =

200. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...
mass = ? g || (1.00 L)(1000 mL/1 L)(1.05 g/mL) = 1050 g
3rd: Calculate the ∆T of the solution …
∆T = ? ‘C|| q/mc = (10,750 J)/(1050 g)(4.18 J/’C g) = 2.4 ‘C

201. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...
mass = ? g || (1.00 L)(1000 mL/1 L)(1.05 g/mL) = 1050 g
3rd: Calculate the ∆T of the solution …
∆T = ? ‘C|| q/mc = (10,750 J)/(1050 g)(4.18 J/’C g) = 2.4 ‘C
4th: Calculate the Tf of the reaction:

202. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...
mass = ? g || (1.00 L)(1000 mL/1 L)(1.05 g/mL) = 1050 g
3rd: Calculate the ∆T of the solution …
∆T = ? ‘C|| q/mc = (10,750 J)/(1050 g)(4.18 J/’C g) = 2.4 ‘C
4th: Calculate the Tf of the reaction:
Tf = ? ‘C || Ti + ∆T =

203. Thermochemistry
Constant Pressure Calorimetry ... Coffee Cup Calorimeter ...
P = k Known:
mNaOH = 10.0 gNaOH ; VH2O = 1 LH2O Tf = ? ‘C
cH2O = 4.18 J/’C g ; Ti = 25.0 ‘C
DH2O = 1.05 g/mL ; ∆Hrx = -43 kJ/mol
NaOH(s) ----> NaOH(aq)
Solution:
1st: Calculate the q of the reaction …
qp = J || [(-43 kJ/mol)(10.0 g x 1 mol/40.0 g)(1000 j/ 1 kJ)] = -10,750 J
2nd: Calculate the mass of the solution ...
mass = ? g || (1.00 L)(1000 mL/1 L)(1.05 g/mL) = 1050 g
3rd: Calculate the ∆T of the solution …
∆T = ? ‘C|| q/mc = (10,750 J)/(1050 g)(4.18 J/’C g) = 2.4 ‘C
4th: Calculate the Tf of the reaction:
Tf = ? ‘C || Ti + ∆T = 25.0 ‘C ‘C = 27.4 ‘C

204. Thermochemistry
Hess’s Law …

205. Thermochemistry
Hess’s Law … Enthalpy changes are state functions ...

206. Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ...

207. Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …

208. Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …
Known: Unknown:
a. 2 N2O(g) ----> O2(g) + 2N2(g) ∆Ha = -164 kJ

209. Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …
Known: Unknown:
a. 2 N2O(g) ----> O2(g) + 2N2(g) ∆Ha = -164 kJ
b. 2NH3(g) + 3N2O(g) ----> 4N2(g) + 3 H2O(l) ∆Hb = kJ

210. Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …
Known: Unknown:
a. 2 N2O(g) ----> O2(g) + 2N2(g) ∆Ha = -164 kJ ∆Hrx = ? kJ
b. 2NH3(g) + 3N2O(g) ----> 4N2(g) + 3H2O(l) ∆Hb = kJ

211. Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …
Known: Unknown:
a. 2 N2O(g) ----> O2(g) + 2N2(g) ∆Ha = -164 kJ ∆Hrx = ? kJ
b. 2NH3(g) + 3N2O(g) ----> 4N2(g) + 3H2O(l) ∆Hb = kJ
Solution:
Overall reaction NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)

212. Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …
Known: Unknown:
a. 2 N2O(g) ----> O2(g) + 2N2(g) ∆Ha = -164 kJ ∆Hrx = ? kJ
b. 2NH3(g) + 3N2O(g) ----> 4N2(g) + 3H2O(l) ∆Hb = kJ
Solution:
Overall reaction NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
1st: Multiply eqb by a factor of 2 ...

213. 4NH3(g) + 6N2O(g) ----> 8N2(g) + 6H2O(l) ∆Hb = -2024 kJ
Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …
Known: Unknown:
a. 2 N2O(g) ----> O2(g) + 2N2(g) ∆Ha = -164 kJ ∆Hrx = ? kJ
b. 2NH3(g) + 3N2O(g) ----> 4N2(g) + 3H2O(l) ∆Hb = kJ
Solution:
Overall reaction NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
1st: Multiply eqb by a factor of 2 ...
4NH3(g) + 6N2O(g) ----> 8N2(g) + 6H2O(l) ∆Hb = kJ

214. 4NH3(g) + 6N2O(g) ----> 8N2(g) + 6H2O(l) ∆Hb = -2024 kJ
Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …
Known: Unknown:
a. 2 N2O(g) ----> O2(g) + 2N2(g) ∆Ha = -164 kJ ∆Hrx = ? kJ
b. 2NH3(g) + 3N2O(g) ----> 4N2(g) + 3H2O(l) ∆Hb = kJ
Solution:
Overall reaction NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
1st: Multiply eqb by a factor of 2 ...
4NH3(g) + 6N2O(g) ----> 8N2(g) + 6H2O(l) ∆Hb = kJ
2nd: Reverse eqa ...

215. 4NH3(g) + 6N2O(g) ----> 8N2(g) + 6H2O(l) ∆Hb = -2024 kJ
Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …
Known: Unknown:
a. 2 N2O(g) ----> O2(g) + 2N2(g) ∆Ha = -164 kJ ∆Hrx = ? kJ
b. 2NH3(g) + 3N2O(g) ----> 4N2(g) + 3H2O(l) ∆Hb = kJ
Solution:
Overall reaction NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
1st: Multiply eqb by a factor of 2 ...
4NH3(g) + 6N2O(g) ----> 8N2(g) + 6H2O(l) ∆Hb = kJ
2nd: Reverse eqa ... Multiply it by 3 ...

216. Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …
Known: Unknown:
a. 2 N2O(g) ----> O2(g) + 2N2(g) ∆Ha = -164 kJ ∆Hrx = ? kJ
b. 2NH3(g) + 3N2O(g) ----> 4N2(g) + 3H2O(l) ∆Hb = kJ
Solution:
Overall reaction NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
1st: Multiply eqb by a factor of 2 ...
4NH3(g) + 6N2O(g) ----> 8N2(g) + 6H2O(l) ∆Hb = kJ
2nd: Reverse eqa ... Multiply it by 3 ...
3O2(g) + 6N2(g) ----> 6N2O(g) ∆Ha = +492kJ

217. Thermochemistry
Hess’s Law … Enthalpy changes are state functions ... ∆H may be calculated in a single step ... or a series of steps …
Known: Unknown:
a. 2 N2O(g) ----> O2(g) + 2N2(g) ∆Ha = -164 kJ ∆Hrx = ? kJ
b. 2NH3(g) + 3N2O(g) ----> 4N2(g) + 3H2O(l) ∆Hb = kJ
Solution:
Overall reaction NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
1st: Multiply eqb by a factor of 2 ...
4NH3(g) + 6N2O(g) ----> 8N2(g) + 6H2O(l) ∆Hb = kJ
2nd: Reverse eqa ... Multiply it by 3 ...
3O2(g) + 6N2(g) ----> 6N2O(g) ∆Ha = +492kJ
3rd: Add eqa + eqb ...
3O2(g) + 4NH3(g) ----> 2N2(g) + 6H2O(l) ∆Hrx = kJ

218. Thermochemistry Known:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ∆Ha = kJ

219. Thermochemistry Known:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ∆Ha = kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ

220. Thermochemistry Known:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ∆Ha = kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ

221. Thermochemistry Known:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ∆Ha = kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ
d. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ

222. Thermochemistry Known: Unknown:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ∆Ha = kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ
d. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ

223. Thermochemistry Known: Unknown:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ∆Ha = kJ ∆Hrx = ? kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ
d. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ

224. Thermochemistry Known: Unknown:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ∆Ha = kJ ∆Hrx = ? kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ
d. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
Solution:
Overall Rx … 2B(s) + 3/2O2(g) ----> B2O3(s)

225. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
Thermochemistry
Known: Unknown:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ∆Ha = kJ ∆Hrx = ? kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ
d. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
Solution:
Overall Rx … 2B(s) + 3/2O2(g) ----> B2O3(s)
1st: Leave equation d as is …
2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ

226. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
Thermochemistry
Known: Unknown:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ∆Ha = kJ ∆Hrx = ? kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ
d. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
Solution:
Overall Rx … 2B(s) + 3/2O2(g) ----> B2O3(s)
1st: Leave equation d as is …
2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
2nd: Reverse equation a … multiply by

227. Thermochemistry Known: Unknown:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3O2(g) ∆Ha = kJ ∆Hrx = ? kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ
d. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
Solution:
Overall Rx … 2B(s) + 3/2O2(g) ----> B2O3(s)
1st: Leave equation d as is …
2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
2nd: Reverse equation a … multiply by
B2H6(g) + 3O2(g) ----> B2O3(s) + 3H2O(g) ∆Ha = kJ

228. Thermochemistry Known: Unknown:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3 O2(g) ∆Ha = kJ ∆Hrx = ? kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ
d. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
Solution:
Overall Rx … 2B(s) + 3/2O2(g) ----> B2O3(s)
1st: Leave equation d as is …
2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
2nd: Reverse equation a … multiply by
B2H6(g) + 3O2(g) ----> B2O3(s) + 3H2O(g) ∆Ha = kJ
3rd: Multiply equation c by -3 …
3H2O(l) ----> 3H2(g) + 3/2O2(g) ∆Hc = +858 kJ

229. Thermochemistry Known: Unknown:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3 O2(g) ∆Ha = kJ ∆Hrx = ? kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ
d. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
Solution:
Overall Rx … 2B(s) + 3/2O2(g) ----> B2O3(s)
1st: Leave equation d as is …
2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
2nd: Reverse equation a … multiply by
B2H6(g) + 3O2(g) ----> B2O3(s) + 3H2O(g) ∆Ha = kJ
3rd: Multiply equation c by -3 …
3H2O(l) ----> 3H2(g) + 3/2O2(g) ∆Hc = +858 kJ
4th: Multiply equation b by -3/2 …
3H2O(g) ----> 3H2O(l) ∆Hb = -132 kJ

230. Thermochemistry Known: Unknown:
a. B2O3(s) + 3H2O(g) ----> B2H6(g) + 3 O2(g) ∆Ha = kJ ∆Hrx = ? kJ
b. 2H2O(l) ----> 2H2O(g) ∆Hb= +88 kJ
c. H2(g) + 1/2O2(g) ----> H2O(l) ∆Hc = -286 kJ
d. 2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
Solution:
Overall Rx … 2B(s) + 3/2O2(g) ----> B2O3(s)
1st: Leave equation d as is …
2B(s) + 3H2(g) ----> B2H6(g) ∆Hd= +36 kJ
2nd: Reverse equation a … multiply by
B2H6(g) + 3O2(g) ----> B2O3(s) + 3H2O(g) ∆Ha = kJ
3rd: Multiply equation c by -3 …
3H2O(l) ----> 3H2(g) + 3/2O2(g) ∆Hc = +858 kJ
4th: Multiply equation b by -3/2 …
3H2O(g) ----> 3H2O(l) ∆Hb = -132 kJ
5th: Cancelling we get B(s) + 3/2O2(g) ----> B2O3(s) ∆Hrx = kJ

231. Thermochemistry
Standard Enthalpy of Formation … ∆H’f ...

232. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...

233. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...
2. ∆H’f involves the formation of a compound from its elements with the
substances in their standard states ...

234. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...
2. ∆H’f involves the formation of a compound from its elements with the
substances in their standard states ...
3. Textbooks list the following standard state conditions ...

235. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...
2. ∆H’f involves the formation of a compound from its elements with the
substances in their standard states ...
3. Textbooks list the following standard state conditions ...
For an element ...

236. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...
2. ∆H’f involves the formation of a compound from its elements with the
substances in their standard states ...
3. Textbooks list the following standard state conditions ...
For an element ... It is the form which the element exists in at 25 ‘C and 1 atm ...

237. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...
2. ∆H’f involves the formation of a compound from its elements with the
substances in their standard states ...
3. Textbooks list the following standard state conditions ...
For an element ... It is the form which the element exists in at 25 ‘C and 1 atm ...
For a compound ...

238. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...
2. ∆H’f involves the formation of a compound from its elements with the
substances in their standard states ...
3. Textbooks list the following standard state conditions ...
For an element ... It is the form which the element exists in at 25 ‘C and 1 atm ...
For a compound ...
P = 1 atm ...

239. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...
2. ∆H’f involves the formation of a compound from its elements with the
substances in their standard states ...
3. Textbooks list the following standard state conditions ...
For an element ... It is the form which the element exists in at 25 ‘C and 1 atm ...
For a compound ...
P = 1 atm ...
substance in solution ...

240. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...
2. ∆H’f involves the formation of a compound from its elements with the
substances in their standard states ...
3. Textbooks list the following standard state conditions ...
For an element ... It is the form which the element exists in at 25 ‘C and 1 atm ...
For a compound ...
P = 1 atm ...
substance in solution ... It is a [1 M] ...

241. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...
2. ∆H’f involves the formation of a compound from its elements with the
substances in their standard states ...
3. Textbooks list the following standard state conditions ...
For an element ... It is the form which the element exists in at 25 ‘C and 1 atm ...
For a compound ...
P = 1 atm ...
substance in solution ... it is a [1 M] ...
pure liquid or solid ... Pure liquid or solid ...

242. Thermochemistry Standard Enthalpy of Formation … ∆H’f ...
1. ∆H’f is always given per mole of compound formed ...
2. ∆H’f involves the formation of a compound from its elements with the
substances in their standard states ...
3. Textbooks list the following standard state conditions ...
For an element ... It is the form which the element exists in at 25 ‘C and 1 atm ...
For a compound ...
P = 1 atm ...
substance in solution ... it is a [1 M] ...
pure liquid or solid ... Pure liquid or solid ...
4. ∆H’f for an element in its standard state such as Ba(s) or N2(g) equals 0 ...

243. Thermochemistry
Calculating Standard Enthalpies of Formation …

244. Thermochemistry Calculating Standard Enthalpies of Reaction … Known:
∆H’f H2O(l) = -286 kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol

245. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol

246. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H6(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)

247. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....

248. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ ||

249. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) =

250. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)

251. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol)

252. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol) = kJ

253. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol) = kJ
2nd: Calculate the ∆H’reactants ....

254. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol) = kJ
2nd: Calculate the ∆H’reactants ....
∆H’reactants = ? kJ || ∑nr∆H’f(reactants) = 2∆H’f C3H8(g) + 9∆H’f O2(g)

255. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol) = kJ
2nd: Calculate the ∆H’reactants ....
∆H’reactants = ? kJ || ∑nr∆H’f(reactants) = 2∆H’f C3H8(g) + 9∆H’f O2(g)

256. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol) = kJ
2nd: Calculate the ∆H’reactants ....
∆H’reactants = ? kJ || ∑nr∆H’f(reactants) = 2∆H’f C3H8(g) + 9∆H’f O2(g)
= (2 mol)(20.9 kJ/mol) + (9 mol)(0 kJ/mol)

257. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol) = kJ
2nd: Calculate the ∆H’reactants ....
∆H’reactants = ? kJ || ∑nr∆H’f(reactants) = 2∆H’f C3H8(g) + 9∆H’f O2(g)
= (2 mol)(20.9 kJ/mol) + (9 mol)(0 kJ/mol) = kJ

258. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol) = kJ
2nd: Calculate the ∆H’reactants ....
∆H’reactants = ? kJ || ∑nr∆H’f(reactants) = 2∆H’f C3H8(g) + 9∆H’f O2(g)
= (2 mol)(20.9 kJ/mol) + (9 mol)(0 kJ/mol) = kJ
3rd: Calculate ∆H’rx ...

259. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol) = kJ
2nd: Calculate the ∆H’reactants ....
∆H’reactants = ? kJ || ∑nr∆H’f(reactants) = 2∆H’f C3H8(g) + 9∆H’f O2(g)
= (2 mol)(20.9 kJ/mol) + (9 mol)(0 kJ/mol) = kJ
3rd: Calculate ∆H’rx ...
∆H’rx = kJ|| ∆H’f(products) - ∆H’f(reactants) =

260. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol) = kJ
2nd: Calculate the ∆H’reactants ....
∆H’reactants = ? kJ || ∑nr∆H’f(reactants) = 2∆H’f C3H8(g) + 9∆H’f O2(g)
= (2 mol)(20.9 kJ/mol) + (9 mol)(0 kJ/mol) = kJ
3rd: Calculate ∆H’rx ...
∆H’rx = kJ|| ∆H’f(products) - ∆H’f(reactants) = kJ kJ =

261. Thermochemistry Calculating Standard Enthalpies of Reaction …
Known: Unknown:
∆H’f H2O(l) = -286 kJ/mol ∆H’rx = ? kJ/mol
∆H’f CO2(g) = kJ/mol
∆H’f C3H8(g) = 20.9 kJ/mol
∆H’f O2(g) = 0 kJ/mol
Solution:
Overall Rx C3H6(g) + 9O2(g) ----> 6CO2(g) + 6H2O(l)
1st: Calculate the ∆H’products ....
∆H’products = ? kJ || ∑np∆H’f(products) = 6∆H’f H2O(l) + 6∆H’f CO2(g)
= (6 mol)(-286 kJ/mol) + (6 mol)( kJ/mol) = kJ
2nd: Calculate the ∆H’reactants ....
∆H’reactants = ? kJ || ∑nr∆H’f(reactants) = 2∆H’f C3H8(g) + 9∆H’f O2(g)
= (2 mol)(20.9 kJ/mol) - (9 mol)(0 kJ/mol) = kJ
3rd: Calculate ∆H’rx ...
∆H’rx = kJ|| ∆H’f(products) - ∆H’f(reactants) = kJ kJ = kJ