1. CHEMICAL THERMODYNAMICS
AP Chemistry
Chapter 19 Notes
CHEMICAL
THERMODYNAMICS

2. 1st Law of Thermodynamics Energy in the universe is conserved
REVIEW
1st Law of Thermodynamics
Energy in the universe is conserved

3. value depends on how process takes place (i.e. q, w)
REVIEW
Path Function
value depends on how process takes place (i.e. q, w)

4. heat
q = mCpT

5. REVIEW
State Function
value independent of path - is a defined reference or zero point (i.e. H)

6. Hf0 of elements in natural form at 250C
Enthalpy - H
zero point
Hf0 of elements in natural form at 250C

7. Enthalpy - H
types:
Hrxn = nHf0(prod) -
nHf0(react)

8. Enthalpy - H
types:
Hfus , Hvap

9. a measure of randomness or disorder of a system
ENTROPY “S”
a measure of randomness or disorder of a system

10. Entropy is NOT conserved. The universe seeks maximum disorder.

11. 2nd Law of
Thermodynamics
for a spontaneous
process, entropy
increases

12. Entropy - state function
zero reference: S = 0
for a perfect crystal at absolute zero

13. Ssolid < Sliquid << Sgas

14. S0 = standard entropy
of elements & cmpds
at P = 1 atm; T = 250C
units = J/K mol
[Appendix L, text]

15. Calculate S using
Hess’ Law
S(rxn) = nS0(prod) -
nS0(react)

16. Example:
Ca(s) + C(gr) + 3/2 O2  CaCO3(s)

17. Suniv = Ssys + Ssurr
if Suniv > 0
process spontaneous

18. Suniv = Ssys + Ssurr
if Suniv < 0
process ?

19. Suniv = Ssys + Ssurr
if Suniv = 0
process ?

20. to relate S and H
consider

21. where the water is the system & everything else is the surroundings
H2O(s)  H2O(l)
where the water is the system & everything else is the surroundings
entropy of water increases, therefore Ssys > 0
the vaporization of water is an endothermic process,
therefore heat is flowing from the surroundings to the system & the random motion of the atoms in the surroundings decrease.
thus, Ssurr < 0
so, Suniv = Ssys(+) + Ssurr(-) ; which predominates?
we know that above 1000C, water spontaneously evaporates and that below 1000C, water spontaneously condenses, thus TEMPERATURE must have an effect on relative importance of Ssys and Ssurr .

22. Temperature Dependence S = J/mol (1/T) Ssurr = -H/T
introduces central idea that the entropy changes in the surroundings are primarily determined by heat flow.
exothermic process increases the entropy of the surroundings, thus is an important driving force for spontaneity. Remember that a system tends to undergo changes that lower its energy.
The SIGN on Ssurr depends on the direction of the heat flow.
The MAGNITUDE of Ssurr depends on the temperataure.

23. Ssys Ssurr Suniv spon

24. Ssys Ssurr Suniv spon

25. Ssys Ssurr Suniv spon

26. Ssys Ssurr Suniv spon + + + Y - - - N + - ? D - +
It will be spontaneous if Ssys has a larger magnitude larger than Ssurr.

27. Ssys Ssurr Suniv spon + + + Y - - - N + - ? D - + ? D
It will be spontaneous if Ssurr has a larger magnitude larger than Ssys.

28. For a phase change:

29. GIBB’S FREE
ENERGY G

30. most abstract of thermodynamic state functions

31. w w = reversible
work w2 P V

32. G = w + PV Definition: w: reversible work PV: pressure/volume work
isothermal, reversible path

33. G = w + PV + VP
at constant P
P = 0 so VP = 0
G = w + PV

34. G = w + PV + VP
at constant P
G = w + PV
at constant V
V = 0 so PV = 0
G = w = useful work

35. must measure G over a process
G cannot be measured
must measure G over a process

36. ZERO REFERENCE
G = 0 for elements in stable form under Standard Thermodynamic Conditions
T = 25oC ; P = 1 atm

37. Gf0 = standard Free Energy of Formation from the elements
Appendix L, text

38. G follows Hess’ Law:
G0 (rxn) =
nGf0(p) - nGf0(r)

39. Summary of Laws of
Thermodynamics
Zeroth Law:
Heat Gain = Heat Loss

40. Summary of Laws of
Thermodynamics
First Law:
Law of Conservation
of Energy

41. Summary of Laws of
Thermodynamics
Second Law:
Defines Entropy

42. Summary of Laws of
Thermodynamics
Third Law:
Defines Absolute Zero

43. GIBB’S
HELMHOLTZ
EQUATION

44. combine G H T

45. G = -aT + H
G, H are state functions, thus “a” must be a state function
G = -S(T) + H

46. Gibb’s Helmholtz
Equation
G = H - TS

47. Units on the State Functions

51. thus, a process is spontaneous
if and only if
G is negative

52. enthalpy (minimum energy)
Spontaneity
controlled by
enthalpy
(minimum energy)

53. enthalpy entropy (maximum disorder)
Sponaneity
controlled by
enthalpy
entropy
(maximum disorder)

54. Sponaneity
controlled by
enthalpy
entropy
both

55. Predict Spontaneity
IF H(-) and S(+)
G = -H - T(+S)
G < 0, => spontaneous

56. Predict Spontaneity
IF H(+) and S(-)
G = +H - T(-S)
G > 0, => NOT spontan

57. Summary of Spontaneity
H S G Spont.
yes no
or ?
or ?

58. Uses of the
Gibb’s Helmholtz
Equation

59. 1. Find the molar
entropy of formation
for ammonia.

60. 2. Elemental boron, in thin fibers, can be made from a boron halide:
BCl3(g) + 3/2 H2(g) ->B(s) + 3HCl(g)

61. Calculate: H0, S0 and G0.
Spontaneous?
Driving force?

62. 3. Using thermodynamic
information, determine
the boiling point of
bromine.

63. Thermodynamic
Definition of
Equilibrium
Geq = 0

64. by definition
G = H - TS
&
H = E + PV

65. thus, G = E + PV - TS take derivative of both sides
dG = dE + PdV + VdP - TdS - SdT

66. for a reversible process
TdS = q

67. derivative used for state function while
partial derivative used for path function

68. if the only work is PV
work of expansion
PdV = w

69. First Law of
Thermodynamics
E = q - w
dE = q - w = 0

70. thus
q = w or
TdS = PdV

71. by substitution
dG = 0 + PdV + VdP
- PdV - SdT or
dG = VdP - SdT

72. let’s assume we have
a gaseous system at
equilibrium, therefore,
examine Kp at that
constant temperature

73. at constant T
SdT = 0
thus
dG = VdP

74. but

78. if “condition 2” is
at standard
thermodynamic
conditions, then
G2 = G0 and P2 = 1 atm

79. thus

80. determine G for
aA + bB  cC + dD
where all are gases

81. G = Gprod - Greact
= cGC + dGD - aGA - bGB

82. but
and likewise
for the others

83. and

84. but

85. Thus,
DG = DG0 + (RT) ln Q

86. But
aA + bB  cC + dD
G = 0

87. thus
or in general

88. THERMODYNAMICS & EQUILIBRIUM

89. information, determine
3. Using thermodynamic
information, determine
the boiling point of
bromine.

90. Thermodynamics and Keq
FACT: Product-favored systems have Keq > 1.

91. Thermodynamics and Keq
Therefore, both ∆G˚rxn and Keq are related to reaction favorability.

92. Thermodynamics and Keq
Keq is related to reaction favorability and thus to ∆Gorxn.
The larger the value of K the more negative the value of ∆Gorxn

93. Thermodynamics and Keq
∆Gorxn= - RT lnK
where R = J/K•mol

94. Thermodynamics and Keq
∆Gorxn = - RT lnK
Calculate K for the reaction
N2O4 2 NO2
∆Gorxn = +4.8 kJ
K = 0.14
When ∆G0rxn > 0, then K < 1

95. ∆G, ∆G˚, and Keq
∆G is change in free energy at non-standard conditions.
∆G is related to ∆G˚
∆G = ∆G˚ + RT ln Q where Q = reaction quotient

96. ∆G, ∆G˚, and Keq When Q < K or Q > K, reaction is spontaneous.
When Q = K reaction is at equilibrium
When ∆G = 0 reaction is at equilibrium
Therefore, ∆G˚ = - RT ln K

97. Product favored reaction
∆G, ∆G˚, and Keq
Product favored reaction
–∆Go and K > 1
In this case
∆Grxn is < ∆Gorxn , so state with both reactants and products present is MORE STABLE than complete conversion.

98. ∆G, ∆G˚, and Keq Product-favored reaction. 2 NO2  N2O4
∆Gorxn = – 4.8 kJ
Here ∆Grxn is less than ∆Gorxn , so the state with both reactants and products present is more stable than complete conversion.

99. Thermodynamics and Keq Overview
∆Gorxn is the change in free energy when reactants convert COMPLETELY to products.

100. Keq is related to reaction favorability.
When ∆Gorxn < 0, reaction moves energetically “downhill”

101. 4. For the following reaction, calculate the temperature at which the reactants are favored.

102. THERMODYNAMICS OF CHEMICAL REACTIONS

103. 5. How much useful work can be obtained from an engine fueled with 75
5. How much useful work can be obtained from an engine fueled with 75.0 L of hydrogen at 10 C at 25 atm?

104. reacting silver with water.
6. The reaction to split water into hydrogen and oxygen can be promoted by first
reacting silver with water.
2 Ag(s) + H2O(g) Ag2O(s) + H2(g)
Ag2O(s)  2 Ag(s) + 1/2 O2(g)

105. Calculate H0, S0 and G0 for each reaction.

106. Combine the reactions and calculate H0 and G0 for the combination.
Is the combination spontaneous?

107. At what temperature does the second reaction become spontaneous?

108. 7. The conversion of coal into hydrogen for fuel is:
C(s) + H2O(g)  CO(g) + H2(g)

109. Calculate G0 and Kp at 250C.
Is the reaction spontaneous?

110. At what temperature does the reaction become spontaneous?
Calculate the temperature at which K = 1.0 x 10-4.

111. 8. The generation of
nitric acid in the upper
atmosphere might
destroy the ozone
layer by:
NO(g) + O3(g)  NO2(g) + O2(g)

112. Calculate G0 (reaction)
and K at 250C.