1. The reversibility of reactions
Chemical Equilibrium
The reversibility of reactions

2. Equilibrium
Many chemical reactions do not go to completion. Initially, when reactants are present, the forward reaction predominates. As the concentration of products increases, the reverse reaction begins to become significant.

3. forward reaction rate = reverse reaction rate
Equilibrium
The forward reaction rate slows down since the concentration of reactants has decreased. Since the concentration of products is significant, the reverse reaction rate gets faster.
Eventually, the
forward reaction rate = reverse reaction rate

4. Equilibrium forward reaction rate = reverse reaction rate
At this point, the reaction is in equilibrium. The equilibrium is dynamic. Both the forward and reverse reactions occur, but there is no net change in concentration.

5. Equiibrium
It is important to note that at equilibrium, the concentrations of reactants and products are not equal. Some reactions proceed far to the right, where the concentration of products is greater than that of reactants. Other reactions don’t proceed significantly to the right, and the concentration of reactants is greater than that of reactants,

6. Equiibrium
The degree to which a reaction proceeds to the right is directly related to the value of ΔG for the reaction. If the value of ΔG is large and negative, the equilibrium will favor the formation of products. If the value of ΔG is large and positive, the equilibrium will favor the formation of reactants.

7. Equilibrium: 2NO2 ↔ N2O4
The concentrations of N2O4 and NO2 level off when the system reaches equilibrium.

8. Equilibrium: 2NO2 ↔ N2O4
The system will reach equilibrium starting with either NO2 , N2O4, or a mixture of the two.

9. Equilibrium
Reactions that can reach equilibrium are indicated by double arrows (↔ or ).
The extent to which a reaction proceeds in a particular direction depends upon the reaction, temperature, initial concentrations, pressure (if gases), etc.

10. Equilibrium [C]c[D]d K = [A]a[B]b
Scientists studying many reactions at equilibrium determined that for a general reaction with coefficients a, b, c and d :
a A + b B ↔ c C + d D
K is the equilibrium constant for the reaction.
[C]c[D]d
K =
[A]a[B]b

11. Equilibrium [C]c[D]d K = [A]a[B]b a A + b B ↔ c C + d D
This is the equilibrium constant expression for the reaction. K is the equilibrium constant. The square brackets indicate the concentrations of products and reactants at equilibrium.
[C]c[D]d
K =
[A]a[B]b

12. Equilibrium Constants
The value of the equilibrium constant depends upon temperature, but does not depend upon the initial concentrations of reactants and products. It is determined experimentally.

13. Equilibrium Constants

14. Equilibrium Constants
K has no units. The square brackets in the equilibrium constant expression indicate concentration in moles/liter. Each concentration is relative to a standard molarity of exactly 1 M, so the units cancel.
Some texts will use the symbol Kc or Keq for equilibrium constants that use concentrations or molarity.

15. Equilibrium Constants
The size of the equilibrium constant gives an indication of whether a reaction proceeds to the right.

16. Equilibrium Constants
If a reaction has a small value of K, the reverse reaction will have a large value of K.
At a given temperature, K = 1.3 x 10-2 for:
N2(g) + 3 H2(g) ↔ 2 NH3(g)
If the reaction is reversed,
2 NH3(g) ↔ N2(g) + 3 H2(g)
K = 1/(1.3 x 10-2 ) = 7.7 x 101

17. Equilibrium Constants
Equilibrium constants for gas phase reactions are sometimes determined using pressures rather than concentrations. The symbol used is Kp rather than K (or Kc).
For the reaction:
N2(g) + 3 H2(g) ↔ 2 NH3(g)
Kp = (Pammonia)2
(PN2)(PH2)3

18. Kp
Kp also has not units, because the pressures, in atmospheres, are relative to a reference pressure of exactly 1 atm, so the units cancel.

19. Equilibrium Constants
The numerical value of Kp is usually different than that for K. The values are related, since P= nRT = MRT, where M = mol/liter.
For reactions involving gases:
KC = KP(RT)Δn
where n is the change in moles of gases in the balanced chemical reaction. V

20. Equilibrium Constants
The method for solving equilibrium problems with Kp or K is the same. With Kp, you use pressure in atmospheres. With K or Kc, you use concentration in moles/liter, or molarity.

21. Heterogeneous Equilibria
Many equilibrium reactions involve reactants and products where more than one phase is present. These are called heterogeneous equilibria.
An example is the equilibrium between solid calcium carbonate and calcium oxide and carbon dioxide.
CaCO3(s) ↔ CaO(s) + CO2(g)

22. Heterogeneous Equilibria
CaCO3(s) ↔ CaO(s) + CO2(g)
Experiments show that the position of the equilibrium does not depend upon the amounts of calcium carbonate or calcium oxide. That is, adding or removing some of the CaCO3(s) or CaO(s) does not disrupt or alter the concentration of carbon dioxide.

23. Heterogeneous Equilibria
CaCO3(s) ↔ CaO(s) + CO2(g)

24. Heterogeneous Equilibria
CaCO3(s) ↔ CaO(s) + CO2(g)
This is because the concentrations of pure solids (or liquids) cannot change. As a result, the equilibrium constant expression for the above reaction is:
K =[CO2] or Kp = PCO2

25. Heterogeneous Equilibria
The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

26. The Reaction Quotient
If reactants and products of a reaction are mixed together, it is possible to determine whether the reaction will proceed to the right or left. This is accomplished by comparing the composition of the initial mixture to that at equilibrium.

27. The Reaction Quotient
If the concentration of one of the reactants or products is zero, the reaction will proceed so as to make the missing component.
If all components are present initially, the reaction quotient, Q, is compared to K to determine which way the reaction will go.

28. The Reaction Quotient
Q has the same form as the equilibrium constant expression, but we use initial concentrations instead of equilibrium concentrations. Initial concentrations are usually indicated with a subscript zero.

29. The Reaction Quotient K = 2.2 x 102 for the reaction:
N2(g) H2(g) ↔ 2 NH3(g)
Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of NH3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed?

30. The Reaction Quotient K = 2.2 x 102 for the reaction:
N2(g) H2(g) ↔ 2 NH3(g)
Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of NH3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed?
1. Calculate Q and compare it to K.

31. The Reaction Quotient
A comparison of Q with K will indicate the direction the reaction will go.

32. The Reaction Quotient K = 2.2 x 102 for the reaction:
N2(g) H2(g) ↔ 2 NH3(g)
Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of NH3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed?
1. Calculate Q and compare it to K.

33. Q and ∆G
Q is used to predict the direction of a reaction, given specific starting conditions. The direction in which a reaction will proceed is also related to ∆G. Previously, ∆Go was calculated for standard conditions. For non-standard conditions, ∆G can be related to Q. (Text section 19.8)

34. ∆G for Non-Standard Conditions
The thermodynamic tables are for standard conditions. This includes having all reactants and products present initially at a temperature of 25oC. All gases are at a pressure of 1 atm, and all solutions are 1 M.

35. ∆G for Non-Standard Conditions
For non-standard temperature, concentrations or gas pressures:
∆G = ∆Go + RTlnQ
Where R = J/K-mol
T is temperature in Kelvins
Q is the reaction quotient

36. ∆G for Non-Standard Conditions
For non-standard temperature, concentrations or gas pressures:
∆G = ∆Go + RT ln Q
For Q, gas pressures are in atmospheres, and concentrations of solutions are in molarity, M.

37. ∆Go and Equilibrium
A large negative value of ∆Go indicates that the forward reaction or process is spontaneous. That is, there is a large driving force for the forward reaction. This also means that the equilibrium constant for the reaction will be large.

38. ∆Go and Equilibrium
A large positive value of ∆Go indicates that the reverse reaction or process is spontaneous. That is, there is a large driving force for the reverse reaction. This also means that the equilibrium constant for the reaction will be small.
When a reaction or process is at equilibrium, ∆Go = zero.

39. ∆Go and Equilibrium

40. ∆Go and Equilibrium ∆G = ∆Go + RT lnQ
At equilibrium, ∆G is equal to zero, and
Q = K.
0 = ∆Go + RT lnK
∆Go = - RT lnK

41. ∆Go and Equilibrium Calculate, ∆Go and K at 25oC for:
H2O(l) ↔ H2(g) + ½ O2(g)

42. ∆Go and Equilibrium Calculate, ∆Go and K at 25oC for:
H2O(l) ↔ H2(g) + ½ O2(g)
Since hydrogen and oxygen react explosively to form water, the reverse reaction should be unfavorable, and have a positive ∆Go , and a small equilibrium constant.

43. ∆Go and Equilibrium Calculate, ∆Go and K at 25oC for:
H2O(l) ↔ H2(g) + ½ O2(g)

44. ∆Go and Equilibrium Calculate, ∆Go and K at 25oC for:
H2O(l) ↔ H2(g) + ½ O2(g)
∆Go = Σ∆Gfo products –Σ∆Gfo reactants
∆Go = [(1 mol) (∆Gfo H2(g)) +(½ mol) (∆Gfo H2(g))] –[(1 mol) (∆GfoH2O(l)]
∆Go =0 –[( 1 mol)(-237.1kJ/mol)] = kJ

45. Free Energy and Equilibrium
∆Go = - RT lnK where R = J/K-mol lnK = –∆Go /RT For H2O(l) ↔ H2(g) + ½ O2(g) : lnK = –237.1 kJ(103J/kJ)/(8.314J/mol-K) (298K) ln K = –95.70 K = e–95.70 = 2.7 x 10-42

46. Free Energy and Equilibrium
In summary, an unfavorable reaction has a positive free energy change, and a small equilibrium constant.

47. Problem: Calculation of K
At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction:
N2(g) H2(g) ↔ 2 NH3(g)

48. Problem: Calculation of K
At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction:
N2(g) H2(g) ↔ 2 NH3(g)
1. Write the equilibrium constant expression.

49. Problem: Calculation of K
At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction:
N2(g) H2(g) ↔ 2 NH3(g)
1. Write the equilibrium constant expression.
K = [NH3]2/([N2][H2]3)

50. Problem: Calculation of K
At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction:
N2(g) H2(g) ↔ 2 NH3(g)
2. Make a table of concentrations.

51. Problem: Calculation of K
N2(g) + 3 H2(g) ↔ 2NH3(g)
[N2]
[H2]
[NH3]
initial
3.00mol L
change
equili-
librium
2.50mol
2.00 L

52. Problem: Calculation of K
N2(g) + 3 H2(g) ↔ 2NH3(g)
3. Complete the table.
[N2]
[H2]
[NH3]
initial
3.00mol L
change
equili-
librium
2.50mol
2.00 L

53. Problem: Calculation of K
N2(g) + 3 H2(g) ↔ 2NH3(g)
3. Complete the table.
[N2]
[H2]
[NH3]
initial
3.00mol L
change
-.50mol
2.00 L
equili-
librium
2.50mol
2.00 L

54. Problem: Calculation of K
N2(g) + 3 H2(g) ↔ 2NH3(g)
3. Complete the table.
[N2]
[H2]
[NH3]
initial
3.00mol L
change
-.50mol
2.00 L
equili-
librium
2.50mol
2.00 L

55. N2(g) + 3H2(g) ↔2NH3(g) [N2] [H2] [NH3] initial change equili- librium
3.00mol L
change
+.25mol
2.00 L
+.75mol
2.00L
-.50mol
equili-
librium
2.50mol
2.00 L

56. N2(g) + 3H2(g) ↔2NH3(g)
[N2]
[H2]
[NH3]
initial
3.00mol L
change
+.25mol
2.00 L
+.75mol
2.00L
-.50mol
equili-
librium
2.50mol
2.00 L
The equilibrium values are the sum of the initial concentrations plus any changes that occur.

57. N2(g) + 3H2(g) ↔2NH3(g)
[N2]
[H2]
[NH3]
initial
3.00mol L
change
+.25mol
2.00 L
+.75mol
2.00L
-.50mol
equili-
librium
+.13M
+.38M
+1.25M
The equilibrium values are the sum of the initial concentrations plus any changes that occur.

58. 4. Substitute and solve for K.
N2(g) + 3H2(g) ↔2NH3(g)
[N2]
[H2]
[NH3]
equili-
librium
+.13M
+.38M
+1.25M
4. Substitute and solve for K.
K = [NH3]2/([N2][H2]3)
K=(1.25)2/[(.13)(.38)3] = 2.2 x 102

59. Problem H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102
Initially, Phydrogen = Piodine = atm. Calculate the equilibrium partial pressures of all three gases.

60. Problem H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102
Initially, Phydrogen = Piodine = atm. Calculate the equilibrium partial pressures of all three gases.
1. Write the Kp expression.

61. Problem H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102
Initially, Phydrogen = Piodine = atm. Calculate the equilibrium partial pressures of all three gases.
1. Write the Kp expression.
Kp = (PHI)2
(PH2)(PI2)

62. Problem H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102
Initially, Phydrogen = Piodine = atm. Calculate the equilibrium partial pressures of all three gases.
2. Make a table of initial, change and equilibrium pressures.

63. Problem: H2(g) + I2(g) ↔ 2HI(g)
initial
.500 atm
change
equili-
brium

64. Problem: H2(g) + I2(g) ↔ 2HI(g)
initial
.500 atm
change -x
+2x
equili-
brium

65. Problem: H2(g) + I2(g) ↔ 2HI(g)
initial
.500 atm
change -x
+2x
equili-
brium
.500-x 2x

66. Problem: H2(g) + I2(g) ↔ 2HI(g)
equili-
brium
.500-x 2x
3. Substitute and solve.

67. Problem: H2(g) + I2(g) ↔ 2HI(g)
equili-
brium
.500-x 2x
Substitute and solve.
Kp= (PHI)2/(PH2)(PI2) = 1.00 x 102
(2x) = 1.00 x 102
(.500-x) (.500-x)

68. Problem: H2(g) + I2(g) ↔ 2HI(g)
3. Substitute and solve.
Kp= (PHI)2/(PH2)(PI2) = 1.00 x 102
(2x) = 1.00 x 102
(.500-x) (.500-x)
Take the square root of both sides:
(2x) = 10.0
(.500-x)

69. Problem: H2(g) + I2(g) ↔ 2HI(g)
3. Substitute and solve.
(2x) = 10.0
(.500-x)
2x = 10.0 (.500-x) = x
12.0 x = 5.00
x = .417

70. Problem: H2(g) + I2(g) ↔ 2HI(g)
4. Answer the question: Calculate the equilibrium partial pressures of all three gases.
x = .417 H2 I2 HI
equili-
brium
.500-x 2x

71. Problem: H2(g) + I2(g) ↔ 2HI(g)
4. Answer the question: Calculate the equilibrium partial pressures of all three gases.
x = .417 H2 I2 HI
equili-
brium
.500-x
0.083 atm 2x
.834 atm

72. Problem: H2(g) + I2(g) ↔ 2HI(g)
5. Check your answer, if possible. Does (PHI)2/(PH2)(PI2) = 1.00 x 102 ?
(.834)2/(.083)(.083) = (.696)/(.0069) = 1.0 x 102 H2 I2 HI
equili-
brium
.500-x
0.083 atm 2x
.834 atm

73. Problem: N2(g) + O2(g) ↔ 2NO(g)
At 2200 oC, K = for the above reaction. Initially, 1.60 mol of nitrogen and mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium.

74. Problem: N2(g) + O2(g) ↔ 2NO(g)
At 2200 oC, K = for the above reaction. Initially, 1.60 mol of nitrogen and mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium.
1. Write the equilibrium constant expression.
K = = [NO]2/[N2][O2]

75. Problem: N2(g) + O2(g) ↔ 2NO(g)
At 2200 oC, K = for the above reaction. Initially, 1.60 mol of nitrogen and mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium.
2. Make a table of initial, change and equilibrium concentrations.

76. Problem: N2(g) + O2(g) ↔ 2NO(g)
initial
1.60mol
2.00 L
.400 mol
change -x
+2x
equili-
brium x x 2x

77. Problem: N2(g) + O2(g) ↔ 2NO(g)
equili-
brium x x 2x
Substitute and solve.
K = = [NO]2/[N2][O2]

78. Approaches to Solving Problems
Some equilibrium problems require use of the quadratic equation to obtain an accurate solution. In cases where the equilibrium constant is quite small (roughly 10-5 or smaller), it is often possible to make assumptions that will simplify the mathematics.

79. Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
At 35oC, K = 1.6 x 10-5 for the above reaction. Calculate the equilibrium concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of Cl2 are placed in a 1.0 liter flask.
1. K = 1.6 x 10-5 =[NO]2[Cl2]/[NOCl]2

80. Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
At 35oC, K = 1.6 x 10-5 for the above reaction. Calculate the equilibrium concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of Cl2 are placed in a 1.0 liter flask.
2. Make a table of concentrations.

81. Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
initial
2.0mol
1.00 L
1.0 mol
change
-2x
+2x +x
equili-
brium x 2x
1.0+x

82. Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
equili-
brium x 2x
1.0+x
Substitute and solve.
K = 1.6 x 10-5 =[NO]2[Cl2]/[NOCl]2
1.6 x 10-5 =[2x]2[1.0 + x]/[2.0-2x]2

83. Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
Substitute and solve.
1.6 x 10-5 = [2x]2 [1.0 + x]
[2.0-2x]2
Since K is small, x, the amount of product formed, will be a small number. As a result, x ≈1.0, and 2.0-2x≈2.0

84. Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
Substitute and solve.
1.6 x 10-5 = [2x]2 [1.0 + x]
[2.0-2x]2
The expression simplifies to:
1.6 x 10-5 = [2x]2 [1.0]
[2.0]2

85. Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
Solve for x.
1.6 x 10-5 = [2x]2 [1.0]
[2.0]2
4x2 = 1.6 x 10-5 (4)/1 = 6.4 x 10-5
x2 = 1.6 x ; x =0.0040

86. Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
4. Check your assumption.
x =0.0040
Does x ≈1.0, and 2.0-2x≈2.0?
= 1.0
2.0 – 2(.004) = =2.0

87. Validity of the Assumption
These assumptions are generally considered valid if they cause an insignificant error. In this case, due to significant figures, no error is introduced. Usually, if the difference is within 5%, the error is considered acceptable.

88. Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
equili-
brium x 2x
1.0+x
Answer the question.
[NOCl] = 2.0M; [NO]= M;
[Cl2] = 1.0M

89. Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
Check your answer.
[NOCl] = 2.0M; [NO]= M;
[Cl2] = 1.0M
Does [NO]2[Cl2]/[NOCl]2 = 1.6 x 10-5 ?
(.0080)2(1.0)/(2.0)2 = 1.6 x 10-5

90. Le Chatelier’s Principle
If a stress is applied to a system at equilibrium, the position of the equilibrium will shift so as to counteract the stress.

91. Le Chatelier’s Principle
If a stress is applied to a system at equilibrium, the position of the equilibrium will shift so as to counteract the stress.
Types of stresses:
Addition or removal of reactants or products
Changes in pressure or volume (for gases)
Changes in temperature

92. Fe3+(aq) + SCN- ↔ FeSCN2+

93. Heat + N2O4(g) ↔ 2 NO2(g)

94. CoCl42- + 6 H2OCo(H2O)62++ 4Cl1- + heat

95. Le Chatelier’s Principle
For reactions involving gaseous products or reactants, changes in pressure of volume may shift the equilibrium. The equilibrium will shift only if there is an unequal number of moles of gas on either side of the reaction. Reactions involving liquids or solids are not significantly affected by changes in pressure or volume.

96. Le Chatelier’s Principle

97. Pressure Changes

98. Le Chatelier’s Principle
For the reaction
C(s) + 2 H2(g) ↔ CH4(g) ∆Ho=-75kJ
Assuming the reaction is initially at equilibrium, predict the shift in equilibrium (if any) for each of the following changes.
a) add carbon
b) remove methane
c) increase the temperature

99. Le Chatelier’s Principle
The value of K will change with temperature. The effect of a change in temperature on K depends on whether the reaction is exothermic or endothermic.
The easiest way to predict the result when temperature is changed is to write in heat as a product (for exothermic reactions) or a reactant (for endothermic reactions).

100. Le Chatelier’s Principle
For the reaction
C(s) + 2 H2(g) ↔ CH4(g) ∆Ho=-75kJ
Assuming the reaction is initially at equilibrium, predict the shift in equilibrium (if any) for each of the following changes.
d) add a catalyst
e) decrease the volume
f) remove some carbon
g) add hydrogen