1. Dr. Donald VanDerveer Office: Boggs 2-5 Office Hours: 8:00-9:30 MWF
chemistry.gatech.edu
Phone/voic

2. Location of Boggs 2-5
B6A

3. Equilibrium

4. Equilibrium
In order to have equilibrium for a chemical
reaction, the process must be able to move
both forwards and backwards.

5. 2 H2 + O H2O
spark

6. 2 H2 + O H2O
spark
2 H2O H2 + O2

7. Equilibrium processes include both
reactions which have chemically
different reactants and products and
phase changes.

8. Equilibrium
H2O(l)
H2O(g)
25oC

9. H2O(l)
H2O(g)

10. After a given period of time, the pressure
does not increase nor does it decrease.

11. H2O(l)
H2O(g)

12. H2O(g)
H2O(l)
H2O(g)
H2O(l)

13. As long as there is liquid water present,
there will be a liquid-gas equilibrium.

14. As long as there is liquid water present,
there will be a liquid-gas equilibrium.
At any given temperature, the equilibrium
pressure will be the same.

15. As long as there is liquid water present,
there will be a liquid-gas equilibrium.
At any given temperature, the equilibrium
pressure will be the same.
This pressure is independent of the
volume of the container.

16. Equilibrium constant:

17. Equilibrium constant:
Equilibrium constants are calculated
based on the phases of the reactants
and products.

18. Equilibrium constant:
If reactants and products are gasses, the
equilibrium constant is calculated from
partial pressures.

19. Equilibrium constant:
If reactants and products in solution, the
equilibrium constant is calculated from
concentrations.

20. Equilibrium constant:
How the value of an equilibrium
constant is determined varies,
depending on the type of
reaction.

21. Equilibrium constant:
For a liquid - gas equilibrium, the
constant, K, is equal to the vapor
pressure of the gas at equilibrium.

22. Equilibrium constant:
For a liquid - gas equilibrium, the
constant, K, is equal to the vapor
pressure of the gas at equilibrium. Px
K =
1 atm

23. H2O(l)
H2O(g)
K =
25o C
H2O(g)
H2O(l)
K =
30o C

24. Chemical equilibrium
N2O4(g)
2 NO2(g)

25. 2
17 e-

27. Place different amounts of pure NO2
in a container, equilibrate at 25oC and
measure the partial pressures of both
gasses.

28. N2O4(g)
2 NO2(g)
PNO 2

29. .
Know PN O and PNO 2 4 2

30. .
Know PN O and PNO 2 4 2
Make equilibrium constant
from an expression of
partial pressures.

31. Partial pressures
P1 + P2 + … Pn = Ptotal

32. PN O = K(PNO )2 2 4 2
PNO 2
N2O4(g)
2 NO2(g)

33. PN O = K(PNO )2 2 4 2
PNO 2
N2O4(g)
2 NO2(g)

34. PN O = K(PNO )2 2 4 2
PN O
K = 2 4
(PNO )2 2
PNO 2
N2O4(g)
2 NO2(g)

35. PN O = K(PNO )2 2 4 2
PN O
K = 2 4
(PNO )2 2
K = 25oC
PNO 2

36. Characteristics of equilibrium states
1. No changes in macroscopic states
(pressure, volume, etc.)

37. Characteristics of equilibrium states
1. No changes in macroscopic states
(pressure, volume, etc.)
2. Equilibrium is reached through
spontaneous processes.

38. Characteristics of equilibrium states
1. No changes in macroscopic states
(pressure, volume, etc.)
2. Equilibrium is reached through
spontaneous processes.
3. Dynamic balance of forward
and reverse processes.

39. Characteristics of equilibrium states
1. No changes in macroscopic states
(pressure, volume, etc.)
2. Equilibrium is reached through
spontaneous processes.
3. Dynamic balance of forward
and reverse processes.
4. Are the same regardless of
direction of approach.

40. General form of equilibrium constant
when all species are low-pressure gasses

41. General form of equilibrium constant
when all species are low-pressure gasses
Balanced chemical equation:
aA + bB cC dD c d
PCPD
= K a b
PAPB

42. c d
PCPD
= K a b
PAPB
2 NOCl NO + Cl2
PNOPCl 2
= K
PNOCl 2

43. Calculating equilibrium constants

44. Calculating equilibrium constants
Data
PNO 2

45. Calculating equilibrium constants
Data
PN O
K = 2 4
(PNO )2 2
PNO 2

46. Calculating equilibrium constants
PN O
K = 2 4
(PNO )2 2
PNO 2

47. Calculating equilibrium constants
PN O
K = 2 4
(PNO )2 2
= 5 atm
PN O 2 4
PNO 2

48. Calculating equilibrium constants
PN O
K = 2 2
(PNO )2 2
= 5 atm
PN O 2 4
= 0.75 atm
PNO 2
PNO 2

49. Calculating equilibrium constants
5 atm
K =
(0.75 atm)2
= 5 atm
PN O 2 4
= 0.75 atm
PNO 2
PNO 2

50. Calculating equilibrium constants
5 atm
K = =
(0.75 atm)2
K = 8.9
PNO 2

51. Calculating equilibrium constants
5 atm
K = =
(0.75 atm)2
K = 8.9
PNO
K = 25oC 2

52. Calculating equilibrium constants
without all partial pressures.

53. Calculating equilibrium constants
without all partial pressures.
4 NO2
2 N2O O2

54. Calculating equilibrium constants
without all partial pressures.
4 NO2
2 N2O O2
Given: starting partial pressures,
equilibrium partial pressure for NO2.

55. 4 NO2
2 N2O O2
Initial (atm)
change (atm) ? ? ?
Equilibrium ? ?

56. 4 NO2
2 N2O O2
Initial (atm)
change (atm) ? ?
Equilibrium ? ?

57. 4 NO2
2 N2O O2
Initial (atm)
change (atm) ? ?
Equilibrium ? ?
For every 4 molecules of NO2 that react,
2 molecules of N2O and 3 molecules of O2
are formed.

58. Volume changes are  coefficients
4 NO2
2 N2O O2
Initial (atm)
change (atm) -4x x x
Equilibrium ? ?

59. Volume changes are  coefficients
4 NO2
2 N2O O2
Initial (atm)
change (atm) -4x x x
Equilibrium x x

60. Volume changes are  coefficients
4 NO2
2 N2O O2
Initial (atm)
change (atm) -4x x x
Equilibrium x x
x = 2.4

61. Volume changes are  coefficients
4 NO2
2 N2O O2
Initial (atm)
change (atm) -4x x x
Equilibrium x x
x = 2.4
x = -4

62. Volume changes are  coefficients
4 NO2
2 N2O O2
Initial (atm)
change (atm) -4x x x
Equilibrium x x
x = 2.4
x =
= 0.3 atm -4

63. Volume changes are  coefficients
4 NO2
2 N2O O2
Initial (atm)
change (atm)
Equilibrium
x = 2.4
x =
= 0.3 atm -4

64. Volume changes are  coefficients
4 NO2
2 N2O O2
Initial (atm)
change (atm)
Equilibrium
(PN O)2(PO)3 2 2
(PNO )4 2

65. Volume changes are  coefficients
4 NO2
2 N2O O2
Initial (atm)
change (atm)
Equilibrium
(PN O)2(PO)3
(5.7)2(8.9)3 2 2 =
(2.4)4
(PNO )4 2

66. Volume changes are  coefficients
4 NO2
2 N2O O2
Initial (atm)
change (atm)
Equilibrium
(PN O)2(PO)3
(5.7)2(8.9)3
= 6.9 x 102 2 2 =
(PNO )4
(2.4)4 2

67. Relationships among equilibrium
expressions.

68. Relationships among equilibrium
expressions.
1. The equilibrium constant of a
reverse reaction is the reciprocal
of the forward reaction.

69. Relationships among equilibrium
expressions.
1. The equilibrium constant of a
reverse reaction is the reciprocal
of the forward reaction.
2 NO N2O4
PN O
K = 2 4
(PNO )2 2

70. Relationships among equilibrium
expressions.
1. The equilibrium constant of a
reverse reaction is the reciprocal
of the forward reaction.
2 NO2
N2O4
N2O4
2 NO2
(PNO)2
PN O
K1 = 2 4
K2 = 2
PN O
(PNO )2 2 2 4

71. 2 NO2
N2O4
N2O4
2 NO2
(PNO)2
PN O
K1 = 2 4
K2 = 2
PN O
(PNO )2 2 2 4
K1K2 =1

72. 2 NO2
N2O4
N2O4
2 NO2
(PNO)2
PN O
K1 = 2 4
K2 = 2
PN O
(PNO )2 2 2 4
K1K2 =1 1
= 1 K1 K1

73. 2. If coefficients in a balanced equation
are all multiplied by a single factor, the
equilibrium constant is raised to a power
equal to that factor.

74. 2. If coefficients in a balanced equation
are all multiplied by a single factor, the
equilibrium constant is raised to a power
equal to that factor.
2 NO2
N2O4
4 NO2
2 N2O4
PN O
(PN O )2
K1 = 2 4
K2 = 2 4
(PNO )2
(PNO )4 2 2

75. 2. If coefficients in a balanced equation
are all multiplied by a single factor, the
equilibrium constant is raised to a power
equal to that factor.
2 NO2
N2O2
4 NO2
2 N2O2
PN O
(PN O )2
K1 = 2 4
K2 = 2 4
(PNO )2
(PNO )4 2 2
K2 = K12

76. 3. Operations of addition and subtraction,
applied to chemical equations, lead to
operations of multiplication and division
of their corresponding equilibrium expressions
and constants.

77. 3. Operations of addition and subtraction,
applied to chemical equations, lead to
operations of multiplication and division
of their corresponding equilibrium expressions
and constants.
Exercise page 290
CF4 + 2 H2O CO2 + 4 HF
K1 = 5.9 x 1023

78. 3. Operations of addition and subtraction,
applied to chemical equations, lead to
operations of multiplication and division
of their corresponding equilibrium expressions
and constants.
Exercise page 290
CF4 + 2 H2O CO2 + 4 HF
K1 = 5.9 x 1023
CO + 1/2 O CO2
K2 = 1.3 x 1035

79. CF H2O CO2 + 4 HF
K1 = 5.9 x 1023
CO + 1/2 O CO2
K2 = 1.3 x 1035
2 CF4 + 4 H2O CO + 8 HF + O2

80. CF H2O CO2 + 4 HF
K1 = 5.9 x 1023
CO + 1/2 O CO2
K2 = 1.3 x 1035
2 CF4 + 4 H2O CO + 8 HF + O2

81. CF H2O CO2 + 4 HF
K1 = 5.9 x 1023
CO + 1/2 O CO2
K2 = 1.3 x 1035
2 CF4 + 4 H2O CO + 8 HF + O2

82. CF H2O CO2 + 4 HF
K1 = 5.9 x 1023
2 CO + O CO2
K2 = 1.7 x 1070
2 CF4 + 4 H2O CO + 8 HF + O2

83. 2 CF H2O CO2 + 8 HF
K1 = 34.8 x 1046
2 CO + O CO2
K2 = 1.7 x 1070
2 CF4 + 4 H2O CO + 8 HF + O2

84. 2 CF H2O CO2 + 8 HF
2 CO + O2
2 CF4 + 4 H2O CO + 8 HF + O2

85. 2 CF H2O CO2 + 8 HF
K1 = 34.8 x 1046
2 CO + O CO2
K2 = 1.7 x 1070
2 CF4 + 4 H2O CO + 8 HF + O2
K3 =
K1 x K2-1

86. 2 CF H2O CO2 + 8 HF
K1 = 34.8 x 1046
2 CO + O CO2
K2 = 1.7 x 1070
2 CF4 + 4 H2O CO + 8 HF + O2
K3 =
K1 x K2-1
34.8 x 1046
= 20.5 x 10-24
K3 =
1.7 x 1070

87. Reaction quotient c d
PCPD
Q = a b
PAPB

88. Reaction quotient c d
PCPD
Q = a b
PAPB
PA , PB, etc. not at equilibrium

89. Reaction quotient c d
PCPD
Q = a b
PAPB
Q will go to the value of K as
the partial pressures go to
equilibrium

90. Q vs K can
predict where
the reaction is
in respect to
equilibrium.

91. c d
K =
Q < K
PCPD a b
PAPB
aA + bB cC + dD c d
PCPD
Q = a b
PAPB

92. c d
K =
Q < K
PCPD a b
PAPB
aA + bB cC + dD c d
PCPD
Too much reactant,
not enough product.
Q = a b
PAPB

93. c d
K =
Q < K
PCPD a b
PAPB
aA + bB cC + dD c d
PCPD
Q > K
Q = a b
PAPB
aA + bB cC + dD

94. c d
K =
Q < K
PCPD a b
PAPB
aA + bB cC + dD c d
PCPD
Q > K
Q = a b
PAPB
aA + bB cC + dD
Too much product, not enough
reactant.

95. c d
K =
Q < K
PCPD a b
PAPB
aA + bB cC + dD c d
PCPD
Q > K
Q = a b
PAPB
aA + bB cC + dD c
Q < K a b d
PAPB
Large vs
PCPD

96. c d
PCPD
Q = a b
PAPB

97. Exercise page 291
P P2
K = 400oC
1.40 mol P4
1.25 mol P2
Volume = 25.0 L

98. P P2
(PP )2
Q = 2 PP 4

99. P P2
(PP )2
Q = 2 PP 4
T = 673 K
nRT
P =
V = 25.0 L V
R = L atm mol-1 K-1

100. P P2
(PP )2
Q = 2 PP 4
(1.4)( )(673)
nRT =
PP =
= 3.09 atm 4 V
25.0

101. P P2
(PP )2
Q = 2 PP 4
(1.4)( )(673)
nRT =
PP =
= 3.09 atm 4 V
25.0
(1.25)( )(673)
nRT =
PP =
= 2.76 atm 2 V
25.0

102. P P2
(PP )2
(2.76)2
Q = 2 = =
3.09 PP
PP =
2.76 atm 4 2
PP =
3.09 atm 4

103. P P2
(PP )2
(2.76)2
Q = 2 = =
3.09 PP
PP =
2.76 atm 4 2
PP =
3.09 atm
Q =
2.46 4
K =

104. P P2
(PP )2
(2.76)2
Q = 2 = =
3.09 PP
PP =
2.76 atm 4 2
PP =
3.09 atm
Q =
2.46 4
K =
P P2

105. Exercise page 292
CO + H2O CO2 + H2

106. Exercise page 292
CO + H2O CO2 + H2
PCO = 2.00 atm
PCO
= 0.80 atm 2
= 0.48 atm PH 2
K = @ 900 K

107. Exercise page 292
CO + H2O CO2 + H2
PCO = 2.00 atm PH
PCO
K = 2 2
PCO
PH O
PCO
= 0.80 atm 2 2
= 0.48 atm PH 2
K = @ 900 K

108. Exercise page 292
CO + H2O CO2 + H2
PCO = 2.00 atm PH
PCO
K = 2 2
PCO
PH O
PCO
= 0.80 atm 2 2 PH
PCO
= 0.48 atm PH
PH O 2 2 = 2 2
PCO K
K = @ 900 K

109. Exercise page 292
CO + H2O CO2 + H2
PCO = 2.00 atm PH
PCO
PH O 2 2 = =
PCO
= 0.80 atm 2
PCO K 2
(0.80)(0.48)
= 0.48 atm PH
= 0.30 atm 2
(2.00) (0.64)
K = @ 900 K

110. Converting between partial
pressures and concentrations.

111. Converting between partial
pressures and concentrations.
moles
Concentration = V

112. Converting between partial
pressures and concentrations.
moles
Concentration = V
PV = nRT

113. Converting between partial
pressures and concentrations.
moles
Concentration = V nA
For gas ‘A’
[A] = V
PV = nRT

114. Converting between partial
pressures and concentrations.
moles
Concentration = V PA nA
For gas ‘A’
[A] = = RT V
PV = nRT

115. Converting between partial
pressures and concentrations.
moles
Concentration = V PA nA
For gas ‘A’
[A] = = RT V
PV = nRT
PA = RT[A]

116. PA = RT[A]
N2O4(g)
2 NO2(g)

117. PA = RT[A]
N2O4(g)
2 NO2(g)
Pref = 1 atm

118. PA = RT[A]
N2O4(g)
2 NO2(g)
Pref = 1 atm
PN O
/Pref 2 4
= K
/Pref)2
(PNO 2

119. PA = RT[A]
N2O4(g)
2 NO2(g)
PN O
= RT[N2O4]
Pref = 1 atm 2 4
PN O
/Pref 2 4
= K
/Pref)2
(PNO 2

120. PA = RT[A]
N2O4(g)
2 NO2(g)
PN O
= RT[N2O4]
Pref = 1 atm 2 4
PNO
= RT[NO2]
PN O
/Pref 2 2 4
= K
/Pref)2
(PNO 2

121. ( ) N2O4(g) 2 NO2(g) PN O = RT[N2O4] PN O /Pref 2 4 2 4 = K /Pref)2-1
[N2O4] ( RT )
[N2O4](RT/Pref) x =
K =
Pref
[NO2](RT/Pref)
[NO2]

122. ( ) ( ) N2O4(g) 2 NO2(g) -1 [N2O4] RT [N2O4](RT/Pref) x = K = Pref

123. ( ) ( ) [N2O4] RT = K [NO2] Pref aA + bB cC + dD c d a+b-c-d PCPD
PAPB
[A]a[B]b

124. ( ) a+b-c-d [C]c[D]d RT = K Pref [A]a[B]b Exercise page 297
CH4 + H2O CO H2
[H2]=[CO]=[H2O]= mol L-1
K = 0.172

125. ( ) ( ) a+b-c-d [C]c[D]d RT = K Pref [A]a[B]b CH4 + H2O CO + 3 H2
[H2]=[CO]=[H2O]= mol L-1
K = 0.172 ( RT ) -2
( )4
= 0.172
Pref
[CH4]( )

126. ( ) CH4 + H2O CO + 3 H2 RT -2 (0.00642)3 = 0.172 Pref [CH4]
K(RT)-2 = x 10-5
( )3
[CH4] =
= 8.39 x 10-2 mol L-1
3.153 x 10-5