1. Chapter 15 Principles of Chemical Equilibrium
Dr. Peter Warburton

2. 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.

3. “The reaction occurs both ways.”
N2O4 (g)  2 NO2 (g)
We have used two directional arrows () to show that this reaction does not go to completion.
“The reaction occurs both ways.”

4. Pitfall
The terms reactants and products are arbitrary. We must always refer to a balanced equation to be completely understood.
N2O4 (g)  2 NO2 (g)
reactant product
2 NO2 (g)  N2O4 (g)

5. Each reaction occurs at its own rate as defined by its rate law.
One reaction will initially have a faster rate than the other, and will initially dominate the system. The other reaction can be considered to be dominated in the system

6. This is what we will see in the dominant reaction.
We saw in Kinetics that the rate of a reaction decreases as time proceeds because the concentrations of the “reactants” decrease.
This is what we will see in the dominant reaction.

7. Its reaction rate will increase!
What will happen to the dominated reaction where the “reactant” concentrations increase?
Its reaction rate will increase!

8. This is a more correct means of defining equilibrium.
At some point in time the rate of the forward reaction is THE SAME as the rate of the reverse reaction.
This is a more correct means of defining equilibrium.

9. Initially dominant reaction “slows down”
Initially dominated reaction “speeds up”

10. Equilibrium is a dynamic process
At equilibrium the rates of the forward and reverse reactions
are the same,
BUT are not equal to zero.
While no visible change is occurring,
individual molecular events are still occurring.

11. The equilibrium constant expression
The final concentrations will not always be the same because we started with different numbers of N atoms and O atoms in the last three experiments.
But ratio of [NO2]2 / [N2O4] is always the same!

12. Experiments 1 and 2 Start first experiment with no NO2 and 0.04 M N2O4
Start first experiment with 0.08 M NO2 and no N2O4

13. a A + b B  c C + d D
The concentrations of all species in an equilibrium mixture are related to each other through the equilibrium constant equation in terms of concentration.
This equilibrium equation applies only to this specific balanced equation and the value of the equilibrium constant, Kc must always be stated at a specific temperature!

14. Notice there are no units for Kc!
N2O4 (g)  2 NO2 (g)
If we change the temperature then the equilibrium mixture will (most likely) change. This means the equilibrium constant will change. For this reaction
Kc = 1.53 at 127 C.
Notice there are no units for Kc!

15. K’c DOES NOT EQUAL Kc, but rather
Be careful!
Equilibrium constant equations (and therefore the value of K) depend on the balanced equation referenced.
A + B  C + D
C + D  A + B
K’c DOES NOT EQUAL Kc, but rather
K’c = 1/Kc

16. Be careful! A + B  C + D 2 A + 2 B  2 C + 2 D
Equilibrium constant equations (and therefore the value of K) depend on the balanced equation referenced.
A + B  C + D
2 A + 2 B  2 C + 2 D

17. Thermodynamic equilibrium constant Keq
The thermodynamic equilibrium constant (Keq) equation takes the same mathematical form as the equilibrium constant equation in terms of concentrations (Kc). However, the composition of the equilibrium mixture is expressed in terms of activities.

18. Thermodynamic equilibrium constant Keq
Activities relate effective properties (like concentration or pressure) of real substances at given conditions in comparison to the same substance acting ideally at standard conditions.
Since activities make comparisons of the same property type in ratio form, the property units cancel out and all activities are unitless.

19. Thermodynamic equilibrium constant Keq
a A + b B  c C + d D
where ax = [X] / c0
(c0 is a standard concentration of 1 M)
or ax = Px / P0 (P0 is a standard pressure of 1 bar)
Note that ax = 1 for pure solids and liquids

20. Problem Answer: [H2] = 0.263 M For the reaction
CO (g) + 2 H2 (g)  CH3OH (g)
the equilibrium concentrations of CH3OH and CO are found to be equal at 483 K. If Kc = 14.5 at 483 K, what is the equilibrium concentration of H2?
Answer: [H2] = M

21. Problem Answer: [H2] = 0.25 M For the reaction
N2 (g) + 3 H2 (g)  2 NH3 (g)
Kc = 1.8 x 104 at a certain temperature. What is the equilibrium concentration of H2 if the equilibrium concentrations of N2 and NH3 are M and 2.00 M respectively?
Answer: [H2] = 0.25 M

22. K for combined equilibria
If we can describe an overall equilibrium reaction as the sum of two or more other equilibrium processes, then the equilibrium constant for the overall reaction in the equilibrium constants of the processes multiplied together.
Krxn = K1 x K2 x K3 …

23. Complex ions and solubility
AgCl(s) + 2 NH3(aq) → [Ag(NH3)2]+(aq) + Cl-(aq)

24. Complex ions and solubility
AgCl(s)  Ag+ (aq) + Cl- (aq)
Ag+ (aq) + NH3 (aq)  [Ag(NH3)]+ (aq)
[Ag(NH3)]+ (aq) + NH3 (aq)  [Ag(NH3)2]+ (aq)
K for the first reaction is 1.8 x 10-10
K for the second reaction is 2.0 x 103
K for the third reaction is 7.9 x 103

25. Complex ions and solubility
The sum of the three reactions is
AgCl(s) + 2 NH3(aq)  [Ag(NH3)2]+(aq) + Cl-(aq)
which will have an equilibrium constant that is
Krxn = K1 x K2 x K3
Krxn = 1.8 x x 2.0 x 103 x 7.9 x 103
Krxn = 2.8 x 10-3

26. The equilibrium constant Kp
Gas phase equilibrium constants are often expressed in terms of partial pressures because they are generally very easy to measure as a function of the total pressure of the system.

27. The equilibrium constant Kp
Recall, for an ideal 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

28. 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.

29. 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!)

30. Kc and Kp are related Some of the RT terms in
will cancel each other out. In fact

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

32. note the lack of units and T is expressed in Kelvin!
N2O4 (g)  2 NO2 (g)
Dn = (2-1) = 1 so
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

33. Problem
In the industrial synthesis of hydrogen often the water-gas shift reaction is used
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?

34. Problem answer
Kp = 9.48

35. Problem In the industrial synthesis of nitric acid:
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?
Answers: Kp at 227C = 1.7 x 104
and Kc at 1000 K = 1.1

36. 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.

37. 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, though?

38. 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)

39. What is the concentration of a solid or liquid?
density / molar mass = (concentration)
Since both the density and molar mass of a pure solid or liquid substance are constant, the CONCENTRATION IS CONSTANT, and does not change in a reaction as long as some of the solid or liquid exists at all times.
This helps explain why the activities of solids and liquids are equal to one!

40. CaCO3 (s)  CaO (s) + CO2 (g)
We choose not to include the concentrations of solids and liquids in the calculation of Kc!
The concentrations of the solids are “hidden” inside the equilibrium constant.
If we look at the reaction in terms of pressure, then
Kp = (PCO2)

41. Thermodynamic equilibrium constant Keq
The activity of all pure solids and liquids is one, and so solids and liquids have no effect on the value of Keq

42. CaCO3 (s)  CaO (s) + CO2 (g)

43. Problem
For each of the following reactions, write the equilibrium constant expression for Kc. Where appropriate, do the same for Kp and give the relationship between Kc and Kp.
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 (size) of the constant K 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 size of the value of the equilibrium constant.

45. If the equilibrium constant is very large (>1000 for instance), then the forward reaction is initially very dominant and the reaction as written in the balanced equation proceeds nearly to completion before equilibrium is reached.
2 H2 (g) + O2 (g)  2 H2O (g)

46. If the equilibrium constant is very small (<10-3 for instance), then the reverse reaction is initially very dominant and the reaction as written in the balanced equation barely proceeds at all before equilibrium is reached.
2 H2O (g)  2 H2 (g) + O2 (g)

47. If the equilibrium constant is between 10-3 and 103, this means that the dominant reaction is not overpowering the other reaction and we reach equilibrium somewhere “in between” a state of “no reaction” and “completeness”. Appreciable concentrations of all species are present in the equilibrium mixture.
H2 (g) + I2 (g)  2 HI (g)

48. Predicting the direction of a reaction
If you put known concentrations of products and reactants into the equilibrium constant equation when you know the system is NOT at equilibrium you would get a value that does not equal the equilibrium constant.
Can we use this value to tell which reaction is dominant in this non-equilibrium system?

49. When the system is not at equilibrium, then
We define the reaction quotient Qc (or Qp or Qeq) in exactly the same way we define the equilibrium constant Kc (or Kp or Keq).
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, then
Qc  Kc, so the system is not at equilibrium.
Qc > Kc, the reaction will proceed from right to left.

53. 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 5.0 L reaction vessel at this temperature was filled with mol of NO, 1.0 mol of O2, and 0.80 mol 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 5.0 x 10-3 mol of NO, 0.20 mol of O2, and 4.0 mol of NO2?

54. Problem answer
a) Qc = 8.9 x System is not at equilibrium, and reaction will proceed right since Qc < Kc.
b) Qc = 1.6 x System is not at equilibrium, and reaction will proceed left since Qc > Kc.

55. CO (g) + H2O (g)  CO2 (g) + H2 (g)
Problem
In an earlier problem we saw the water-gas shift reaction
CO (g) + H2O (g)  CO2 (g) + H2 (g)
where we calculated the value of Kp at 700 K to be If we combine equal masses of all four chemicals and let the system come to equilibrium, which chemicals will have increased in quantity and which will have decreased in quantity during the reaction?

56. Problem answer
Qp  5.7 < Kp and reaction will proceed right meaning we will increase CO2 (g) and H2 (g) and decrease CO (g) and H2O (g).

57. Altering equilibrium conditions
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!

58. Le Chatalier’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.
Le Chatalier’s Principle states that if a stress is applied to a system at equilibrium, the system will react in the direction that minimizes the stress and brings the system to a NEW equilibrium.

59. Changes in concentration
N2 (g) + 3 H2 (g)  2 NH3 (g)
Kc = at 700 K.

60. The system re-established a NEW equilibrium by reacting in such a way as to decrease the stress to the system. Since we have added a reactant (this is the stress), the reaction should proceed towards products to minimize the amount of “extra” reactant added to the system.

61. In general, if we increase the concentration of a reactant, the reaction proceeds from reactants to products to decrease the stress of added reactant to our equilibrium system.
If we increase the concentration of a product, the reaction proceeds from products to reactants to decrease the stress of added product to our equilibrium system.

62. The system was at equilibrium!
In the ammonia example of slide 59, before we introduced more nitrogen (a reactant) the reaction quotient was:
The system was at equilibrium!

63. If we add 1. 00 molL-1 nitrogen (a stress
If we add 1.00 molL-1 nitrogen (a stress!) to the original equilibrium system, the reaction quotient will change and the system will no longer be at equilibrium!

64. The reaction quotient is now less than the equilibrium constant, meaning the reaction must move from left to right to re-establish equilibrium.
At the new equilibrium
[N2] = 1.31 molL-1
[H2] = 2.43 molL-1
[NH3] = 2.36 molL-1

65. Note that the [N2] in this new equilibrium mixture is now lower than the 1.50 molL-1 we changed the concentration to after adding N2 to the first equilibrium mixture.
The system has reacted to minimize the stress on the system by reducing the amount of N2 to reach a new equilibrium!

66. Fe3+ (aq) + SCN- (aq)  [FeNCS]2+ (aq)
a) An equilibrium mixture for this reaction is orange, which is the added colours of pale yellow Fe3+ and the red FeNCS2+. SCN- is colorless.

67. Fe3+ (aq) + SCN- (aq)  [FeNCS]2+ (aq)
b) If we add FeCl3 to the solution, we see the mixture gets more red, meaning more FeNCS2+. Why?

68. Fe3+ (aq) + SCN- (aq)  [FeNCS]2+ (aq)
If we add KSCN to the solution, we see the mixture gets more red, meaning more FeNCS2+. Why?

69. Fe3+ (aq) + SCN- (aq)  [FeNCS]2+ (aq)
d) If we add H2C2O4 to the solution, we see the mixture gets more yellow, meaning less FeNCS2+. Why?
H2C2O4 (aq)  2 H+ (aq) + C2O42- (aq)
Fe3+(aq) + 3 C2O42- (aq)  [Fe(C2O4)3]3- (aq)

70. Fe3+ (aq) + SCN- (aq)  [FeNCS]2+ (aq)
e) If we add HgCl2 to the solution, we see the mixture gets more yellow, meaning less FeNCS2+. Why?
HgCl2 (s)  Hg2+ (aq) + 2 Cl- (aq)
Hg2+ (aq) + 4 SCN- (aq)  [Hg(SCN)4]2- (aq)

71. CO (g) + H2O (g)  CO2 (g) + H2 (g)
Problem
Consider the equilibrium for the water-gas shift reaction:
CO (g) + H2O (g)  CO2 (g) + H2 (g)
Use Le Chatalier’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; also account for the change using the reaction quotient Qc:

72. CaCO3 (s)  CO2 (g) + CaO (s)
Problem
Calcination of limestone (decomposition of calcium carbonate) occurs through the following reaction:
CaCO3 (s)  CO2 (g) + CaO (s)
After we establish this equilibrium system in a constant volume container at a given temperature, what will be the effect on equilibrium of
a) Doubling the amount of CO2
b) Doubling the amount of CaO
c) Removing half the CaCO3
d) Removing all the CaCO3

73. Effect of 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 pressure of a system, or a decrease in volume increases the pressure of the system.

74. Pressure change due to volume change
Say we decrease the volume (and increase the pressure) of our ammonia formation equilibrium mixture. The stress on the equilibrium is the increase in pressure.
Le Chatalier’s Principle tells us the system will respond by decreasing the pressure of the system until a new equilibrium mixture is achieved.

75. Pressure change due to volume change
Since the pressure is a direct result of the number of moles of gas (more moles in a given volume means more pressure), the reaction will proceed in the direction where the number of moles of gas is decreased.

76. Pressure change through a change in volume
In general, an increase in the pressure of the system (caused by decreasing the system volume!) causes the reaction to shift to the side of the balanced equation with less total moles of gas.
In general, a decrease in the pressure of the system (caused by increasing the system volume!) causes the reaction to shift to the side of the balanced equation with more total moles of gas.

77. Why is this the case? We have doubled the concentration of all gases!
Say we reduce the volume of our ammonia equilibrium mixture by half.
We have doubled the concentration of all gases!
Reaction will shift from left to right!

78. 2 SO2 (g) + O2 (g)  2 SO3 (g)

79. Other things to note
If the total number of moles of gas on the reactants side of a balanced equation
EQUALS
the total number of moles of gas on the products side of a balanced equation, then a
pressure change due to volume change will not affect the equilibrium system.

80. We ALWAYS talked about pressure changes in terms of volume change.
Other things to note
We ALWAYS talked about pressure changes in terms of volume change.
If we increase the pressure by adding inert gas to the system, no equilibrium shift will be seen because the partial pressures of the gases in the equilibrium system have not changed!
In other words, there is no actual stress being placed on the equilibrium system

81. 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 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)

82. Changes in temperature and equilibrium
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

83. 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” and the reaction shifts from right to left (towards the reactants).
The new equilibrium will have less products and more reactants, giving a smaller value of Kc.

84. 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 

85. 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?

86. Catalysis and equilibrium
Since both the forward and reverse reactions pass through the same transition state, a catalyst reduces the activation energy for both the forward and reverse reactions, by the same amount.
This increases the rates of both the forward and reverse reactions by the same amount!

88. 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.

89. Problem
Suppose that you have a reaction vessel containing an equilibrium mixture of all three species. 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?

90. Equilibrium calculations
There are several different types of equilibrium-based calculations we can do:
T1) Find some equilibrium mixture data from K, balanced equation and other equilibrium mixture data (see slide 20)
T2) Find K from equilibrium mixture data (see slide 33)
T3) Find K from some initial and equilibrium mixture data and balanced equation
T4) Find equilibrium mixture data from initial data, K, and the balanced equation

91. 2 NO (g) + O2 (g)  2 NO2 (g) Kc = 6.9 x 105 at 500 K (T1)
Say the system is at equilibrium and
[O2] = 1.0 mol/L and [NO2] = 0.80 mol/L
We can calculate [NO]!

92. 2 NO (g) + O2 (g)  2 NO2 (g) Kc = 6.9 x 105 at 500 K (T1)
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

93. 2 NO (g) + O2 (g)  2 NO2 (g) Kc = 6.9 x 105 at 500 K (T1)
You might be asked to give your final answer in moles (which means you need to know the volume of your container), or in grams (need to know the container volume and the molar mass)

94. Problem (T2) Answer: Kc = 2.3 x 10-4
Equilibrium is established at 1405 K for the reaction
2 H2S (g)  2 H2 (g) + S2 (g)
in a 3.00 L reaction flask. If there are 0.11 mol S2, 0.22 mol H2, and 2.78 mol H2S in the flask, what is Kc for the reaction at 1405 K?
Answer: Kc = 2.3 x 10-4

95. Problem (T3) Answer: Kp = 2.2 x 10-2
0.100 mol SO2 and mol O2 are introduced into an evacuated 1.52 L flask at 900K. If the reaction is
2 SO3 (g)  2 SO2 (g) + O2 (g)
and mol of SO3 are found at equilibrium then what is Kp for the reaction at 900 K?
Answer: Kp = 2.2 x 10-2

96. Finding K

97. CO (g) + H2O (g)  CO2 (g) + H2 (g)
Problem (T4) #1
The H2/CO ratio in mixtures of carbon monoxide and hydrogen (called synthesis gas) is increased by the water-gas shift reaction
CO (g) + H2O (g)  CO2 (g) + H2 (g)
which 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.

98. Problem (T4) #1 (ICE tables)
Using a balanced equation, we create a table of
Initial concentrations,
the Change in concentrations (based on unknown quantities 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.
NOTE: If our system data are given as pressures, we do exactly the same thing, but with pressures.

99. Problem (T4) #1
Taking the square root of both sides

100. Problem (T4) #1
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.

101. Problem (T4) #1
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.

102. C(s) + H2O (g)  CO (g) + H2 (g)
Problem (T4) #2
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?

103. Problem (T4) #2
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 is going to occur to reach equilibrium
Since Q < K we expect to lose reactants and gain products to get to equilibrium. This tells us the signs of the changes that are occurring.

104. Problem (T4) #2
While this reaction quotient calculation step isn’t absolutely necessary,
if we perform this and assign the correct signs to our pressure changes, then we will find that
any negative value of x we calculate
will not be physically possible.

105. Problem (T4) #2
Rearranging, we get

106. Problem (T4) #2

107. This only occurs for x = 0.30 atm
Problem (T4) #2
Our equilibrium partial pressures must all be positive.
This only occurs for x = 0.30 atm
(the negative x value won’t work because we have used the reaction quotient to help set up our ICE table)
At equilibrium
PH2O = (1.20 atm – 0.30 atm) = 0.90 atm,
PCO = (1.00 atm atm) = 1.30 atm, and
PH2 = (1.40 atm atm) = 1.70 atm.

108. Problem (T4) #2 We should check our answer:
which is the equilibrium constant we were given, within rounding errors.

109. CH3Cl (aq) + OH- (aq)  CH3OH (aq) + Cl- (aq)
Problem (T4) #3
In a basic aqueous solution, chloromethane undergoes a substitution reaction in which Cl- is replaced by an OH-:
CH3Cl (aq) + OH- (aq)  CH3OH (aq) + Cl- (aq)
The equilibrium constant Kc is 1 x Calculate the equilibrium concentrations of all species in a solution prepared by mixing equal volumes of 0.1 mol/L CH3Cl and 0.2 mol/L NaOH.

110. Problem (T4) #3 answers [CH3OH] = 0.05 mol/L, [Cl-] = 0.05 mol/L
[CH3Cl] = 5 x mol/L.