1. Chapter 18 Chemical Equilibrium
18.1 The Nature of Chemical Equilibrium

2. Reversible Reactions
A chemical reaction in which the products re-form the reactants
Reactants ↔ Products
Le Chatelier’s Principle
When system at equilibrium is disturbed by application of a stress, it attains a new equilibrium position that minimizes the stress.

3. Chemical Equilibrium
The state where the concentrations of all reactants and products remain constant with time
All reactions carried out in a closed vessel will reach equilibrium

4. Chemical Equilibrium
Sometimes, if little product is formed, equilibrium lies far to the left (towards reactants)
Reactants Products
Sometimes, if little reactant remains, equilibrium lies far to the right (towards products)
Reactants Products

5. Dynamic Equilibrium Reactions continue to take place
Reactant molecules continue to be converted to product
Product continues to be converted to reactant (reverse reaction)
Forward and reverse reactions take place at the same rate at equilibrium

6. Causes of Equilibrium Beginning of Reaction
Only reactant molecules exist, so only reactant molecules may collide
Middle
As product concentration increases, collisions may take place that lead to the reverse reaction
At Equilibrium
Rates of forward and reverse reactions are identical

7. Suppose that compound A reacts with B to form C and D.
The symbol            is used for system at equilibrium.

8. If we add more A what would happen to the amount of C at equilibrium?
Adding A increases the amount of C and D at equilibrium.

9. If we add more C what would happen to the amount of A at equilibrium?
adding more C increases the amount of A and B at equilibrium.

10. If we add A what would happen to the amount of B at equilibrium?
A reacts with B to form C and D. Therefore, adding A decreases B.

11. Example - The Haber Process
N2 (g) + 3H2 (g) ↔ 2NH3 (g)
Hydrogen is consumed at 3x the rate of nitrogen
Ammonia is formed at 2x the rate at which nitrogen is consumed

12. Points to Remember Rxns in a closed system always reach equilibrium
When a rxn has reached equilibrium, that means the rate of the forward rxn equals the rate of the reverse rxn
A system will remain at equilibrium until the system is disturbed in some manner

13. The Equilibrium Constant
For the balanced forward equation:
jA + kB → lC + mD (j, k, l, m) are coefficients
K is a constant called the equilibrium constant
K varies depending on temperature and upon the coefficients of the balanced equation
K is determined experimentally
[X] represents concentration of chemical species at equilibrium
K = [C]l [D]m
[A]j [B]k

14. The value of K will tell us the position of equilibrium
Equilibrium Constant
4NH3(g) + 7O2(g)  4NO2(g) +6H2O(g)
K = [NO2]4[H2O]6
[NH3]4[O2]7
The value of K will tell us the position of equilibrium

15. Practice N2(g) + 3H2(g)  2NH3(g) K = [NH3]2 [N2] [H2]3 Calculate K
[equilibrium]
[NH3] = 3.1 x 10-2 M
[N2] = 8.5 x 10-1 M
[H2] = 3.1 x 10-3 M
Calculate K
K = [3.1x 10-2]2
[8.5 x 10-1] [3.1 x 10-3]3
K = 3.8 x 104
no units

16. Practice N2(g) + 3H2(g)  2NH3(g) K = 3.8 x 104 [NH3] = 3.1 x 10-2 M
[N2] = 8.5 x 10-1 M
[H2] = 3.1 x 10-3 M
The value of K will tell us the position of equilibrium
If K >1, the rxn favors products
If K <1, the rxn favors reactants
If K = 1, equilibrium lies in the middle

17. The rxn was inversed so K is inversed
Practice
N2(g) + 3H2(g)  2NH3(g)
Original rxn
2NH3(g)  N2(g) + 3H2(g)
New rxn
Original K
Same [ ]
K = 3.8 x 104
[NH3] = 3.1 x 10-2 M
[N2] = 8.5 x 10-1 M
[H2] = 3.1 x 10-3 M
Calculate new K
K =
3.8 x 104
The rxn was inversed so K is inversed
K = 2.6 x 10-5

18. The coefficients were reduced by ½
Practice
N2(g) + 3H2(g)  2NH3 (g)
Original rxn
½N2(g) + 3/2H2(g)  NH3 (g)
New rxn
Original K
Same [ ]
K = 3.8 x 104
[NH3] = 3.1 x 10-2 M
[N2] = 8.5 x 10-1 M
[H2] = 3.1 x 10-3 M
Calculate new K
K = (3.8 x 104)1/2
The coefficients were reduced by ½
K = 2.0 x 102

19. Interpreting K
If K = 1, then there are equal concentrations of reactants & products
If K is small, the forward rxn occurs slightly before equilibrium is established, reactants are favored
If K is large, reactants are mostly converted into products when equilibrium is established, products are favored

20. Heterogeneous Equilibria
The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present
If pure solids or liquids are involved in a chemical reaction, their concentrations are not included in the equilibrium expression for the reaction
Pure liquids are not the same as solutions, whose concentration can change

21. More Practice!
At 25 °C, an equilibrium mixture of gases contains 6.4 x 10-3 M PCl3, 2.5 x 10-2 M Cl2, & 4.0 x 10-3 M PCl5. What is the equilibrium constant for the following reaction?
PCl5 (g) ↔ PCl3 (g) + Cl2 (g)
2. At equilibrium, a 2.0 L vessel contains 0.36 mol of H2, 0.11 mol of Br2, & 37 mol of HBr. What is the equilibrium constant for the reaction at this temperature?
H2 (g) + Br2 (g) ↔ HBr (g)
K = 4.0 x 10-2
K = 3.5 x 104

22. Chapter 18 Chemical Equilibrium
18.2 Shifting Equilibrium

23. The Effect of Change in Pressure: Ways to Change Pressure
Add or remove a gaseous reactant or product
Add an inert gas (one not involved in the reaction)
An inert gas increases the total pressure but has no effect on the concentrations or partial pressures of the reactants or products

24. The Effect of Change in Pressure
Change the volume of the container
When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume.
This is done by decreasing the total number of gaseous molecules in the system
N2 (g) + 3H2 (g) → 2NH3 (g)
shifts to the right to decrease the total molecules of gas present

25. The Effect of Change in Pressure
When the container volume is increased, the system will shift so as to increase its volume
N2 (g) + 3H2 (g) ← 2NH3 (g)
shifts to the left to increase the total number of molecules of gas present

26. The Effect of Change in Concentration
If a reactant or product is added to a system at equilibrium, the system will shift away from the added component (it will attempt to "use up" the added component)
If a reactant or product is removed from a system at equilibrium, the system will shift toward the removed component (it will attempt to "replace" the removed component)

27. The Effect of a Change in Temperature
An increase in temperature increases the energy of the system. Le Chatelier's principle predicts that the system will shift in the direction that consumes the energy
For an exothermic rxn, energy is a product. The rxn will shift to the left to use up the excess energy
For an endothermic rxn, energy is a reactant. The rxn will shift to the right to use up the energy
A decrease in temperature will cause a system shift in the direction that "replaces" the lost energy

28. Le Chatelier’s Principle
When a stress is applied to a system at equilibrium, the system will shift in a direction that will relieve the stress
[changes] do not affect the value of K
pressure changes do not affect the value of K
temperature changes will affect the value of K

29. Which direction will the rxn shift to re-establish equilibrium if:
Practice ([change])
As4O6(s) + 6C(s)  As4(g) + 6CO(g)
Which direction will the rxn shift to re-establish equilibrium if:
Add CO?
left
right
no shift
Add or remove C or As4O6?
left
right
no shift
Remove As4?
left
right
no shift

30. Practice (change in pressure)
Which direction will the rxn shift to reestablish equilibrium if the volume is reduced?
P4(s) + 6Cl2(g)  4PCl3(l)
left
right
no shift
PCl3(g) + Cl2(g)  PCl5(g)
left
right
no shift
O2(g) + N2(g)  2NO(g)
left
right
no shift

31. Practice (change in temperature)
Which direction will the rxn shift to reestablish equilibrium if the temperature is increased?
N2(g) + O2(g)  2NO(g) ΔHo = 181kJ
left
right
no shift
2SO2(g) + O2(g)  2SO3(g) ΔHo = -198kJ
left
right
no shift
How will the change affect K?
N2(g) + O2(g)  2NO(g) ΔHo = 181kJ
increase
decrease
2SO2(g) + O2(g)  2SO3(g) ΔHo = -198kJ
increase
decrease

32. Completed Reactions
You do not need to consider equilibrium if you have a reaction that goes to completion.
Examples:
Formation of a gas in an open container
Formation of a precipitate
Formation of a slightly ionized product (polar) – neutralization

33. The Common-Ion Effect
When the addition of an ion common to two solutes brings about precipitation or reduced ionization.

34. Solubility Product and the Common Ion Effect
AgCl(s)  Ag+(aq) + Cl-(aq)
What happens if a solution of NaCl is added to this system?
Ksp = [Cl-][Ag+] = 1.8 x 10-10
solublility product
NaCl(s)  Na+(aq) + Cl-(aq)
The solubility of AgCl has actually decreased
rxn shifts 
AgCl(s)  Ag+(aq) + Cl-(aq)