1. Chemical Equilibrium
Chapter 15

2. Introduction
Many chemical reactions can under the proper conditions be made to go predominantly in one direction or the other.
A chemical equilibrium refers to the balance of two competitive reactions ( and ) in a reversible reaction.
In this dynamic process, the forward and reverse reactions proceed at the same rate.

3. Lets consider the catalytic methanation reaction
CO (g) + 3H2 (g) CH4 (g) + H2O (g)
Start with mol CO and mol H2 in a L vessel at 1200 K.
The rate of the reaction will depend on the concentrations of the reagents.

5. Applying Stoichiometry to an equilibrium mixture
When heated, PCl5 forms PCl3 and Cl2 as follows:
PCl5(g) PCl3(g) + Cl2(g)
When 1.00 mol PCl5 in a 1.00-L container is allowed to come to equilibrium at a certain temperature, the mixture is found to contain mol PCl3.
How many moles of each substance are present at equilibrium?

6. This problem, may conveniently be solved with the aid of the following table:
PCl5(g) PCl3(g) + Cl2(g)
Initial 
Equilibrium
We’re given that the equilibrium mixture contains mol PCl3

7. The Equilibrium Constant
Following a number of experiments it was observed that all of the equilibrium compositions for a reaction at a given temperature are related by a quantity called the equilibrium constant.

8. Definition of the equilibrium constant, Kc
Consider the general reaction :
aA + bB cC + dD
Kc =
[C]c [D]d
[A]a [B]b
The equilibrium constant is the value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted.

9. Some equilibrium compositions for the methanation reaction
CO (g) + 3H2 (g) CH4 (g) + H2O (g)
Starting Conc’s
Equilibrium Conc’s Kc
Experiment 1
M HCl
0.300 M H2
M CO
M H2
M CH4
M H2O
3.93
Experiment 2
M HCl
0.300 M H2
M CO
M H2
M CH4
M H2O
3.91
Experiment 3
M HCl
0.100 M H2
M CO
M H2
M CH4
M H2O
3.93

10. Example
Write the equilibrium-constant expression for the following reaction:
4NH3 (g) + 5O2 (g) NO (g) + 6H2O (g)

11. The Law of Mass Action
The Law of Mass Action is a relation that states:
the values of the equilibrium-constant expression Kc are constant for a particular reaction at a given temperature, whatever equilibrium concentrations are substituted.

12. The manipulation rules of equilibrium constants
Equilibrium constant for a reverse reaction: K1 = 1/K
cC + dD  aA + bB
If the reaction is halved, the new equilibrium constant is the square root of the original K.
If the reaction is doubled, then the new equilibrium constant is the square of the original K.

13. If two reversible reactions are added together, then the equilibrium constant of the resulting reaction will be the product of the two original K-values.
Equilibrium constant for two reactions added: K3 = K1 x K2
HA  H+ + A-
H+ + C  CH+
HA + C  CH+ + A-

14. Example
The equilibrium constant for reaction (1) is K. What is the equilibrium constant for equation (2)?
(1) /3 N2(g) + H2(g) /3 NH3(g)
(2) NH3(g) N2(g) + 3 H2(g)

15. Heterogeneous Equilibria
A homogeneous equilibrium is an equilibrium that involves reactants and products in a single phase only.
A heterogeneous equilibrium is an equilibrium involving reactants and products in more that one phase

16. Example Note 3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g)
Write the equilibrium-constant expression for the following reaction:
3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g)
Note
When writing the equilibrium-constant expression for heterogeneous equilibria, the concentration terms for pure solids and liquids are omitted.

17. Calculating Equilibrium Constants
See p570, 8th Ed.

18. Example
Determining the equilibrium constant given equilibrium concentrations.
Haber mixed some nitrogen and hydrogen and allowed it to react at 500 K until the mixture reached equilibrium with the product, ammonia. When he analysed the equilibrium mixture, he found it to consist of M NH3, M N2 and M H2
What is the equilibrium constant for the reaction?
N2(g) + 3H2(g) NH3(g)

19. Example
Determining the equilibrium constant given equilibrium concentrations.
Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen:
2CO2(g) CO(g) + O2(g)
At 3000 K, 2.00 mol CO2 is placed into a 1.00 L container and allowed to come to equilibrium. At equilibrium, 0.90 mol CO2 remains. What is the Kc at this temperature?

20. USING THE EQUILIBRIUM CONSTANT
So far:
we described how a chemical reaction reaches equilibrium.
how this equilibrium can be characterised by the equilibrium constant.
We now look at the following uses of the equilibrium constant:
Qualitatively interpreting the equilibrium constant.
Predicting the direction of reaction.
Calculating equilibrium concentrations.

21. Applications of Equilibrium Constants
Rule of Thumb
If the equilibrium constant is large, then the products are favoured at equilibrium.
Eg.
N2 (g) + 3H2 (g) NH3 (g)
For this reaction Kc = 4.1 x 108
Conversely, if the equilibrium constant is small then the reactants will be favoured at equilibrium.

22. Predicting the direction of Reaction.
Suppose a gaseous mixture from an industrial plant has the following composition at 1200 K:
M CO
M H2
M CH4
M H2O
Would the following reaction go forward or in reverse?
CO (g) + 3H2 (g) CH4 (g) + H2O (g)
To answer this question we need to calculate the reaction quotient, and compare its value to that of Kc

23. The reaction quotient (Qc) is an expression that has the same form as the equilibrium constant expression but whose concentration values are not necessarily those at equilibrium.
CO (g) + 3H2 (g) CH4 (g) + H2O (g)
Qc =
[CH4]i . [H2O]i
[CO]i . [H2]i3 =
( ) . ( )
(0.0200) . (0.0200)3
= 6.25
Remember that Kc = 3.93 for this reaction at 1200 K.
Thus we have that Qc > Kc
For Qc to become equal to Kc the reaction must shift to the left.

24. Calculating Equilibrium Concentrations
In general:
If Qc > Kc, the reaction will go left
If Qc < Kc, the reaction will go right
If Qc = Kc, the reaction is at equilibrium
Know This !!
Calculating Equilibrium Concentrations
Once you have determined the equilibrium constant for a reaction, you can use it to calculate the concentrations of substances in an equilibrium mixture.

25. Example Obtaining one equilibrium concentration given the others.
Nitrogen and oxygen form nitric oxide:
N2(g) + O2(g) NO(g)
If an equilibrium mixture at 25°C contains M of N2 and M of O2, what is the concentration of NO in this mixture? Kc at 25°C is 1 x

26. Example Solving an equilibrium problem (involving a linear equation)
Hydrogen iodide decomposes to hydrogen gas and iodine gas.
2HI(g) H2(g) + I2(g)
At 800 K, the equilibrium constant for this reaction is If 0.50 mol is placed in a 5.0-L flask, what will be the composition of the equilibrium mixture?

27. Three steps in solving equilibrium concentrations:
Remember
Three steps in solving equilibrium concentrations:
Set up a table of concentrations.
Substitute the expressions in x for equilibrium concentrations into the equilibrium constant expression.
Solve the equilibrium constant expression for the values of the equilibrium concentrations.

28. Changing the Reaction Conditions: Le Chatelier’s Principle
By changing the reaction conditions, you can increase or decrease the yield of product.
Three ways to alter the equilibrium composition of a gaseous reaction mixture:
Changing the concentrations by removing products or adding reactants to the reaction vessel.
Changing the partial pressure of gaseous reactants and products by changing the volume.
Changing the temperature.

29. Change in Reactant or Product Concentrations
Le Chatelier’s Principle states that:
when a system in chemical equilibrium is disturbed (by a change of temperature, pressure, or a concentration) the system shifts in equilibrium composition in a way that tends to counteract this change.

30. 8CO(g) + 17H2(g) C8H18(g) + 8H2O(g)
Applying Le Chatelier’s Principle When a Concentration is altered.
The Fischer-Tropsch process for the synthesis of gasoline consists of passing a mixture of carbon monoxide and hydrogen over an iron-cobalt catalyst.
A typical reaction that occurs in the process is as follows:
8CO(g) + 17H2(g) C8H18(g) + 8H2O(g)
Suppose the reaction mixture comes to equilibrium at 200°C, then is suddenly cooled to room temperature where octane liquifies. The remaining gases are then reheated to 200°C.
What is the direction of the reaction as equilibrium is attained?

31. Effects of Volume and Pressure Changes
In General:
If the pressure is increased by decreasing the volume of a reaction mixture, the reaction shifts in the direction of fewer moles of gas.

32. 8CO(g) + 17H2(g) C8H18(g) + 8H2O(g)
Applying Le Chatelier’s Principle When the Pressure is Altered
Lets consider the same reaction in the Fischer-Tropsch process:
8CO(g) + 17H2(g) C8H18(g) + 8H2O(g)
Would you expect more or less of the product octane, C8H18, as the pressure increases?

33. Effect of Temperature Change
For an exothermic reaction (H negative), the amounts of products are decreased at equilibrium by an increase in temperature.
A + B C + D + heat
For an endothermic reaction (H positive), the amounts of products are increased by a increase in temperature.
A + B + heat C + D

34. Example 2SO2(g) + O2(g) 2 SO3(g) H = -198 kJ
Applying Le Chatelier’s Principle When Temperature is altered.
One stage in the manufacture of sulfuric acid is the formation of sulfur trioxide by the reaction of SO2 with O2. Predict how the equilibrium composition of the reaction mixture will change when the temperature is raised.
2SO2(g) + O2(g) SO3(g) H = -198 kJ

35. The Effect of Catalyst
A catalyst is a substance that increases the rate of a reaction but is not consumed by it.
A catalyst has no effect on the equilibrium composition of a reaction mixture. A catalyst merely speeds up the reaction to achieve equilibrium.