1. Chapter 14 Chemical Equilibrium.

2. CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
The Equilibrium State
Chemical equilibrium is the state reached when the concentrations of the products and reactants remain constant over time. The mixture of reactants and products in the equilibrium state is the equilibrium mixture.
Until now, we have shown chemical reactions go to completion with a single-headed arrow ().
For many reactions, this assumption is not true. Consider
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)

3. CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
This reaction “does not go to completion”… OR
“ The reaction occurs in both directions at the same time.”

4. .
Be Careful!
The terms reactants and products are arbitrary. We must always refer to a balanced equation, as it is written, to be completely understood.
N2O4 (g)  2 NO2 (g)
reactant product
2 NO2 (g)  N2O4 (g)

5. Figure
As time progresses, the concentrations of the “reactants” decreases while the concentrations of the “products” increases. This is what we saw in the Chapter on Reaction Rates

6. Figure
At some point in time the concentrations of both the reactant and the product stop changing.
We say this system is in equilibrium.

7. CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
The reaction occurs in both directions, in the same system, at the same time.
It is reversible.
When we say
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
we really mean,
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
and CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)

8. Each reaction occurs at its own rate as defined by its own rate law.
One of the two reactions probably starts out with a faster initial rate. It is dominant.
We expect to lose the reactants and gain the products that we see in the in the dominant reaction. (Kinetics)

9. The opposite reaction in the system will be dominated.
Since the reactants of the dominated reaction are the same as the products of the dominant reaction, their concentration increases.
Since the products of the dominated reaction are the same as the reactants of the dominant reaction, their concentration decreases.

10. We’ve seen that the rate of a reaction depends on a rate law that often includes concentrations of reactants.
As the concentration of the reactants decreases, we expect the rate to decrease (the reaction slows down).
This happens to the dominant reaction!

11. We saw in Kinetics that the rate of a reaction depends on a rate law that often includes concentrations of reactants.
As the concentration of the reactants increases, we expect the rate to increase (the reaction speeds up).
This happens to the dominated reaction!

12. CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
In this experiment the forward reaction is dominant and slows down while the reverse reaction is dominated and speeds up.

13. If the dominant reaction always slows down and the dominated reaction always speeds up, then at some point in time, they must have the same rate.
What happens then?

14. CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
This is equilibrium. The concentrations of all chemicals remain constant because they are made at the same rate they are consumed.

15. At equilibrium the rates of the forward and reverse reactions are the same,
BUT are not equal to zero.
Be Careful!

16. . Be Careful! The system is in a dynamic state,
so while no visible change is occurring,
any individual molecule
may still be undergoing change
(Reactions are still happening.).

17. a A + b B  c C + d D
The concentrations of an equilibrium mixture are related to each other through the equilibrium equation.
This equilibrium constant Kc is for the specified reaction with a given balanced equation, and at a given temperature.

18. The Equilibrium Constant Kc{
All experiments occur at the same temperature!
The equilibrium concentrations will not always be the same because we start with different numbers C, O, and H atoms in each of the three experiments.
However the value of Kc is always the same!

19. N2O4 (g)  2 NO2 (g)
If we change the temperature that a reaction occurs at, the value of the equilibrium constant will change. For example,
Kc = 1.53 at 127 C.
Notice we generally choose not to show the units for equilibrium constants.

20. Reversing an equation Be Careful! c C + d D  a A + b B
a A + b B  c C + d D
REVERSE the balanced equation
c C + d D  a A + b B
Kc’ DOES NOT EQUAL Kc

21. Multiplying an equation
Be Careful!
a A + b B  c C + d D
MULTIPLY the balanced equation by 2
2a A + 2b B  2c C + 2d D
Kc” DOES NOT EQUAL Kc

22. When we add up two or more reactions, the equilibrium constant for the resultant reaction
EQUALS
the product of the equilibrium constants of the original reactions, and in general
Knew = K1 x K2 x K3 x …

23. Consider
a A + b B  c C + d D
PLUS
GIVES
2a A + 2b B  2c C + 2d D

24. Kinetics and equilibrium
Consider the reaction
N2O4 (g)  2 NO2 (g)

25. Kinetics and equilibrium
For the forward elementary reaction
N2O4 (g)  2 NO2 (g)

26. Kinetics and equilibrium
For the reverse elementary reaction
N2O4 (g)  2 NO2 (g)

27. Kinetics and equilibrium
At equilibrium, these elementary reaction rates are EQUAL

28. Calculate the equilibrium constant at 800 K.
Problem
The following equilibrium concentrations were measured at 800 K for the reaction:
2 SO2 (g) + O2 (g)  2 SO3 (g)
[SO2] = 3.0 x 10-3 molL-1
[O2] = 3.5 x 10-3 molL-1
[SO3] = 5.0 x 10-2 molL-1
Calculate the equilibrium constant at 800 K.

29. The Equilibrium Constant Kp
Gas phase equilibrium constants are often expressed in terms of partial pressures (easy to measure).
Recall, for a gas A
PAV = nART
PA = (nART) / V
PA = (nA/V) RT
What is n / V? It is moles over volume, which is concentration. So nA/V is [A] and
PA = [A] RT

30. Again, the equilibrium constant is unitless.
N2O4 (g)  2 NO2 (g)
We can express an equilibrium constant in terms of partial pressures because they are related to concentrations!
Again, the equilibrium constant is unitless.

31. a A + b B  c C + d D (all are GASES!)
Kc and Kp are related
a A + b B  c C + d D (all are GASES!)

32. We should expect that some of the RT terms might cancel out.
Kc and Kp are related
We should expect that some of the RT terms might cancel out.

33. Kc and Kp are related Be Careful!
Use R = (Latm)(Kmol)-1 because it relates molarity to pressure at a given temperature.

34. N2O4 (g)  2 NO2 (g)
Dngas = (2-1) = 1 so
At 25 C Kc = 4.64 x 10-3

35. and T is expressed in Kelvin!
N2O4 (g)  2 NO2 (g)
At 25 C Kc = 4.64 x 10-3
Kp = Kc(RT)
Kp = (4.64 x 10-3)( )(298)
note the lack of units
and T is expressed in Kelvin!
Kp = at 25 C
NOTE - If Dngas = zero then Kc = Kp

36. Problem
The chemical equation for the so-called water-gas shift reaction is
CO (g) + H2O (g)  CO2 (g) + H2 (g)
What is the value of Kp at 700 K if the partial pressures in an equilibrium mixture at 700 K are
1.31 atm of CO
10.0 atm of water
6.12 atm of carbon dioxide, and
20.3 atm of hydrogen gas?

37. Problem Nitric oxide reacts with oxygen to give nitrogen dioxide:
2 NO (g) + O2 (g)  2 NO2 (g)
If Kc = 6.9 x 105 at 227 C,
what is the value of Kp at this temperature?
If Kp = 1.3 x 10-2 at 1000 K,
what is the value of Kc at this temperature?

38. Heterogeneous Equilibria
Homogeneous equilibria occur in systems where all compounds in the equilibrium mixture are in the same state.
Heterogeneous equilibria occur in systems where some of the chemicals of the equilibrium mixture are in different states.

39. CaCO3 (s)  CaO (s) + CO2 (g)
Since one of the products is a gas, while the other two compounds are solids, this is a heterogeneous equilibrium.
Now, if we were to express the equilibrium constant for this reaction, we would probably say
What is the concentration of a solid?

40. What is the concentration of a solid or liquid?
Concentration is moles per unit volume. Also, density is mass divided by volume, and molar mass is mass per number of moles. So, for a pure substance
(mass / volume) / (mass / moles) = moles / volume
density / molar mass = (concentration)

41. What is the concentration of a solid or liquid?
density / molar mass = (concentration)
The density and molar mass of a solid or liquid doesn’t change. Therefore the concentration doesn’t change either!

42. CaCO3 (s)  CaO (s) + CO2 (g)
We choose not to include the concentrations of solids and liquids in the calculation of Kc because they don’t change even when the system IS NOT at equilibrium!
or Kp = (PCO2)

43. Problem
For each of the following reactions, write the equilibrium constant expression for Kc. a) 2 Fe (s) + 3 H2O (g)  Fe2O3 (s) + 3 H2(g) b) 2 H2O (l)  2 H2 (g) + O2 (g) c) SiCl4 (g) + 2 H2 (g)  Si (s) + 4 HCl (g) d) Hg22+ (aq) + 2 Cl- (aq)  Hg2Cl (s)

44. Using the Equilibrium Constant
Judging the extent of a reaction: The magnitude of the constant Kc (or Kp) gives an idea of the extent to which reactants are converted to products.
We can make general statements about the “completeness” of a given equilibrium reaction based on the value of the constant.

45. K much larger than 103 means the written reaction proceeds nearly to completion (all products, no reactants).
2 H2 (g) + O2 (g)  2 H2O (g)

46. K much smaller than 10-3 means the written reaction barely proceeds at all (all reactants and no products).
2 H2O (g)  2 H2 (g) + O2 (g)

47. K is between 10-3 and 103, means the written reaction achieves an equilibrium mixture with relatively significant amounts of the reactants and products.
H2 (g) + I2 (g)  2 HI (g)

48. Predicting the direction of a reaction
If we know a system IS NOT at equilibrium, what happens if we put the concentrations of the system into the equilibrium constant expression?
You would get a value that does not equal the value of the equilibrium constant.

49. When the system is not at equilibrium, then
We define the reaction quotient Qc (or Qp) in exactly the same way we define the equilibrium constant Kc (or Kp).
When the system is not at equilibrium, then
Qc  Kc

50. If Qc > Kc the reaction needs to create more reactants (and use up products) to get to equilibrium, so the reaction will be going from right to left.
If Qc < Kc the reaction needs to create more products (and use up reactants) to get to equilibrium, so the reaction will be going from left to right.

51. Figure

52. H2 (g) + I2 (g)  2 HI (g) If [H2]t = 0.80 mol/L,
[I2]t = 0.25 mol/L, and
[HI]t = 10.0 mol/L

53. Qc > Kc, the reaction will proceed from
H2 (g) + I2 (g)  2 HI (g)
Qc  Kc, so the system is
not at equilibrium.
Qc > Kc, the reaction will proceed from
right to left.

54. Problem The equilibrium constant Kc for the reaction
2 NO (g) + O2 (g)  2 NO2 (g)
is 6.9 x 105 at 500 K. A reaction vessel at this temperature was filled with M of NO, 0.20 M of O2, and 0.16 M of NO2.
a) Is the reaction mixture at equilibrium? If not, which direction does the reaction proceed?
b) What is the direction of the reaction if the initial amounts are 1.0 x 10-3 M of NO, 0.04 M of O2, and 0.80 M of NO2?

55. Calculating equilibrium concentrations
If we know an equilibrium constant (which must be referred to a balanced equation!) and all but one of the equilibrium concentrations of the species involved, we can find the unknown concentration.

56. 2 NO (g) + O2 (g)  2 NO2 (g) Kc = 6.9 x 105 at 500 K
The system is at equilibrium and we know [O2] = 1.0 mol/L and [NO2] = 0.80 mol/L so we can calculate [NO].

57. 2 NO (g) + O2 (g)  2 NO2 (g) Kc = 6.9 x 105 at 500 K
Since concentrations are always positive, we can throw out the negative answer.
The [NO] in the equilibrium mixture is
9.6 x 10-4 mol/L.
Let’s check this answer

58. Problem The reaction CO (g) + H2O (g)  CO2 (g) + H2 (g)
has an equilibrium constant Kc = 4.24 at 800 K.
Calculate the equilibrium concentrations of all species at 800 K if only CO and H2O are present initially at concentrations of mol/L.

59. ICE tables (also applies for Kp and pressures)
Using a balanced equation, we create a table of the Initial concentrations, and the Change in concentrations, (an unknown amount of change related by the stoichiometry of the balanced equation), and Equilibrium concentrations (sum of initial concentration and change in concentration). We can substitute our Equilibrium concentrations into our equilibrium constant expression.

60. Problem
Taking the square root of both sides

61. Problem
If we put both values of x back into all our Equilibrium concentration expressions, we’ll see one value of x will give at least one negative equilibrium concentration.
This isn’t physically possible!
Throw that value of x out and use the other.

62. Problem
We can check our results by inserting these equilibrium concentrations into the equilibrium equation.
Our result is (within rounding error) the equilibrium constant we were given, so our answers for the equilibrium concentrations are correct.

63. Problem The equilibrium constant Kp is 2.44 at 1000 K for the reaction
C(s) + H2O (g)  CO (g) + H2 (g)
What are the equilibrium partial pressures of H2O, CO, and H2 if the initial partial pressures are PH2O = 1.20 atm, PCO = 1.00 atm, and PH2 = 1.40 atm?

64. Problem
Since this question starts with both reactants and products in the initial mixture, it makes sense to first check the reaction quotient to see in which direction the reaction will proceed.

65. Reaction occurs left to right!

66. Problem Rearranging, we get This is a quadratic equation of the form
ax2 + bx + c = 0
which has solutions given by

67. Problem

68. Problem For x = -5.14 atm at equilibrium
Our equilibrium partial pressures must all be positive.
For x = atm at equilibrium
PH2O = (1.20 atm – [-5.14 atm]) = 6.34 atm,
PCO = (1.00 atm + [-5.14 atm]) = atm,
PH2 = (1.40 atm + [-5.14 atm]) = atm.
Two of these equilibrium pressures are not physically possible, since they are negative!

69. Problem For x = 0.30 atm at equilibrium
Our equilibrium partial pressures must all be positive.
For x = 0.30 atm at equilibrium
PH2O = (1.20 atm – 0.30 atm) = 0.90 atm,
PCO = (1.00 atm atm) = 1.30 atm,
PH2 = (1.40 atm atm) = 1.70 atm.
All these equilibrium pressures are physically real. This is the correct equilibrium mixture!

70. Problem We should check our answer:
which is the equilibrium constant we were given, within rounding errors.

71. Factors That Alter the Composition of an Equilibrium Mixture
We like to maximize a product yield for a reaction with a minimum of energy (and money) input.
If a reaction doesn’t go to near completion
we must adjust experimental conditions so the reaction proceeds as favourably as possible!

72. Le Chatelier’s Principle
Three factors can be changed to affect an equilibrium: the concentrations of the chemicals involved, the pressure and/or volume of the system, or the temperature.
We can describe the effect of a change on the system as it relates to the equilibrium through
Le Chatelier’s Principle.

73. Le Chatelier’s Principle
Le Chatelier’s Principle states that
if a stress (a change)
is applied to a reaction mixture at equilibrium,
reaction will occur in the direction
that minimizes the change.

74. Le Chatelier’s Principle
When a system at equilibrium is changed the system will respond by
re-establishing the equilibrium.
However this new equilibrium mixture will likely be different from the original equilibrium mixture.

75. Altering an Equilibrium Mixture: Changes in Concentration
N2 (g) + 3 H2 (g)  2 NH3 (g)
Kc = at 700 K.

76. Since we have added a reactant, the reaction should proceed towards products to minimize the amount of “extra” reactant in the system.

77. In general, if we increase the concentration of a reactant, the reaction will proceed from reactants to products.
If we increase the concentration of a product, the reaction will go from products to reactants.

78. In general, if we decrease the concentration of a reactant, the reaction will proceed from products to reactants.
If we decrease the concentration of a product, the reaction will go from reactants to products.

79. Fe3+ (aq) + SCN- (aq)  FeNCS2+ (aq)
a) The original equilibrium mixture for this reaction is orange, which is the added colours of pale yellow Fe3+ and the red FeNCS2+. SCN- is colorless.

80. Fe3+ (aq) + SCN- (aq)  FeNCS2+ (aq)
b) If we add FeCl3 to the solution, we see the mixture gets more red, meaning more FeNCS2+. Why?
The increase in the reactant Fe3+ concentration prompts the reaction to shift towards the products to reach a new equilibrium with more red FeNCS2+.

81. Fe3+ (aq) + SCN- (aq)  FeNCS2+ (aq)
If we add KSCN to the solution, we see the mixture gets more red, meaning more FeNCS2+. Why?
The increase in the reactant SCN- concentration prompts the reaction to shift towards the products to reach a new equilibrium with more red FeNCS2+.

82. Fe3+ (aq) + SCN- (aq)  FeNCS2+ (aq)
d) If we add H2C2O4 to the solution, we see the mixture gets more yellow, meaning less FeNCS2+. Why?
H2C2O4 removes Fe3+ by creating a new compound.
The decrease in the reactant Fe3+ concentration prompts the reaction to shift towards the reactants, meaning less red FeNCS2+ at the new equilibrium.

83. Fe3+ (aq) + SCN- (aq)  FeNCS2+ (aq)
e) If we add HgCl2 to the solution, we see the mixture gets more yellow, meaning less FeNCS2+. Why?
Hg2+ removes SCN- by creating a new compound.
The decrease in the reactant SCN- concentration prompts the system to shift towards the reactants, meaning less red FeNCS2+ at the new equilibrium.

84. CO (g) + H2O (g)  CO2 (g) + H2 (g)
Problem
Consider the equilibrium :
CO (g) + H2O (g)  CO2 (g) + H2 (g)
Use Le Chatelier’s Principle to predict how the concentration of H2 will change when the equilibrium is disturbed by:
a) Adding CO
b) Adding CO2
c) Removing H2O
d) Removing CO2

85. Altering an Equilibrium Mixture: Changes in Pressure and Volume
What happens when pressure is changed as a result of a change in volume?
N2 (g) + 3 H2 (g)  2 NH3 (g) Kc = at 700 K
Since PV = nRT then P = (nRT) / V
an increase in the volume decreases the total pressure of a system,
or a decrease in volume increases the total pressure of the system.

86. Pressure change through a change in volume
The stress on the equilibrium is the change in the total pressure.
The system will respond by minimizing the change in total pressure of the system until a new equilibrium mixture is achieved.

87. Pressure change through a change in volume
The total pressure is a direct result of the number of moles of gas in the system.
To minimize the stress of an increase in total pressure, the system shifts to the side of the reaction with less moles of gas.
To minimize the stress of an decrease in total pressure, the system shifts to the side of the reaction with more moles of gas.

88. Figure CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

89. Other facts to note
If the number of moles of gas on the reactants side of a balanced equation equals that on the products side, then a total pressure change due to volume change will not affect the equilibrium.
If we increase the total pressure by adding an inert gas to the system, there is no effect on the equilibrium of the system.
There is NO real stress on the equilibrium system in this case!

90. Problem
Does the number of moles of products increase, decrease, or remain the same when each of the following equilibria is subject to an increase in total pressure by decreasing the volume?
a) CO (g) + H2O (g)  CO2 (g) + H2 (g)
b) 2 CO (g)  C (s) + O2 (g)
c) N2O4 (g)  2 NO2 (g)

91. Altering an Equilibrium Mixture: Changes in Temperature
. Our reaction for the formation of ammonia
N2 (g) + 3 H2 (g)  2 NH3 (g) DH = kJ.
We see as the temperature
increases, the value of Kc decreases, so
the reaction
shifts towards the reactants with increasing T

92. Is there some relationship between DH and Kc?
Yes!
N2 (g) + 3 H2 (g)  2 NH3 (g) kJ
We can think of heat as a “product” in the reaction.
As we increase the temperature of the system, we increase the “concentration” of this “product”. The system will minimize this increase in “product” by shifting from right to left (towards the reactants).
The new equilibrium will have less products and more reactants, giving a smaller value of Kc.

93. Is there some relationship between DH and Kc?
In an endothermic reaction, heat will be a “reactant” so increasing the temperature will shift the reaction from the left to the right, increasing the value of Kc. Overall,
Exothermic reaction:
T  then KC 
Endothermic reaction:
T  then KC 

94. N2 (g) + O2 (g)  2 NO (g) DH = 180.5 kJ
Problem
When air is heated at very high temperatures in an engine, the air pollutant nitric oxide is produced by the reaction
N2 (g) + O2 (g)  2 NO (g) DH = kJ
How does the equilibrium amount of NO vary with an increase in temperature?

95. The Effect of Catalysis on Equilibrium
A catalyst increases the rate of the reaction, because a pathway through a lower energy transition state is available.
Lowering the activation energy increases the rate of both the forward and reverse reactions by the same amount.
Since equilibrium occurs when the rates are already the same, adding a catalyst
DOES NOT affect equilibrium.

96. .

97. Catalysis and equilibrium
Another way to think of it is that a catalyst does not appear in the overall balanced equation for a reaction and therefore it won’t appear in the equilibrium constant equation meaning no change in the equilibrium constant will be seen when you add a catalyst.

98. Catalysis and equilibrium
In our ammonia reaction we saw Kc decreases as T increases. We get less ammonia at high temperatures.
The rate of formation of ammonia increases with temperature. (Arrhenius Equation).
To make the reaction occur quickly enough to be useful requires us to run it at a temperature where we get very little of the desired product.

99. Catalysis and equilibrium
A catalyst allows us to run the reaction at a lower temperature with an acceptable rate, where we can get more of the desired product
from an exothermic reaction.

100. Problem
A platinum catalyst is used in automobile catalytic converters to hasten the oxidation of carbon monoxide:
Will the amount of CO increase, decrease, or remain the same when:
a) A platinum catalyst is added?
b) The temperature is increased?
c) The pressure is increased by decreasing the volume?
d) The pressure is increased by adding argon gas?
e) The pressure is increased by adding O2 gas?