1. Chemical Equilibrium
McMurray and Fay ch. 13

2. Equilibrium

3. Equilibrium
H2(g) + O2 (g) H2O(g)

4. Equilibrium
H2O(l) H2(g) + O2 (g) ?

5. Equilibrium vs. Kinetics
H2(g) + O2 (g) H2O(l)

6. Chapter 13, Figure 13.2 Rates of the forward and reverse reactions for the decomposition of N2O4 to NO2

7. Vocabulary: MASS ACTION EXPRESSION
A formula designed to express how a reaction is proceeding
For the reaction
aA + bB  cC + dD:
[C]c[D]d
[A]a[B]b
The mass action expression is:

8. Vocabulary: REACTION QUOTIENT
The value of the mass action expression at a particular point in the reaction
[C]c[D]d
[A]a[B]b
Qc =
When there are more
products than reactants…
When there are more
reactants than products…
Qc will be large
Qc will be small

9. For the reaction 2 C2H4 (g) + O2 (g)  2 CH3CHO (g)
1. Derive the mass action expression
[C]c[D]d
[A]a[B]b
Qc =

10. For the reaction 2 C2H4 (g) + O2 (g)  2 CH3CHO (g)
2. Calculate Qc when
[C2H4] = 0.3 M
[O2] = 0.2 M
[CH3CHO] = 0.1 M
Qc =
3. Calculate Qc when
[C2H4] = 0.1 M
[O2] = 0.1 M
[CH3CHO] = 0.5 M

11. The reaction quotient changes according to the concentrations of products and reactants
At equilibrium, the concentrations of products and reactants are not changing
The reaction quotient for a particular reaction at equilibrium is a constant.

12. Equilibrium constant [C]c[D]d Kc = [A]a[B]b For the reaction
aA + bB  cC + dD
The equilibrium constant, Kc,
is equal to the value of the mass action expression at equilibrium
[C]c[D]d
[A]a[B]b
Kc =

13. Qc and Kc tell us similar things
Qc will be large
when there are more
products than reactants
Qc will be small
when there are more
reactants than products
Kc will be large
when there are more
products than reactants
at equilibrium
Kc will be small
when there are more
reactants than products
at equilibrium

14. 2 H2(g) + O2 (g) 2 H2O(l) Kc = 9.1 x 1080 Kc will be large
when there are more
products than reactants
at equilibrium
Kc will be small
when there are more
reactants than products
at equilibrium
2 H2(g) + O2 (g) H2O(l)
Kc = 9.1 x 1080

15. CH3COOC2H5(aq) + OH- (aq) CH3COO- (aq) + C2H5OH(aq)
ethyl acetate hydroxide Acetate ethanol
At equilibrium, a solution contains M ethyl acetate, M OH-, 0.85 M acetate and 1.00 M ethanol.
Calculate the value of Kc.

16. Predicting the Direction of a Reaction
Reaction will proceed towards products or towards reactants based on Qc:
[C]c[D]d
[A]a[B]b
Qc =

17. Predicting the Direction of a Reaction
Reaction will proceed towards products or towards reactants based on Qc:
[C]c[D]d
[A]a[B]b
Qc =
If Qc > Kc:
There are more products than there would be equilibrium
System will get rid of excess products by creating reactants

18. Predicting the Direction of a Reaction
Reaction will proceed towards products or towards reactants based on Qc:
[C]c[D]d
[A]a[B]b
Qc =
If Qc < Kc:
There are more reactants than there would be equilibrium
System will get rid of excess reactants by creating products

19. Le Châtelier’s Principle
When a change is introduced into a system, the system will move to counteract the effects of the change

20. Le Châtelier’s Principle (simplified version)
CO + 2 H CH3OH
System gets rid of extra by
converting it to products
Add more of this
CO + 2 H CH3OH
Add more of this
System gets rid of extra by
converting it to reactants

21. Le Châtelier’s Principle from an equilibrium perspective
aA + bB cC + dD
[C]c[D]d
[A]a[B]b
Kc =
When products are added to a system,
the reaction quotient Q becomes larger than Kc.
To get back to equilibrium, the system must reduce the amount of
products and increase the number of reactants.

22. Le Châtelier’s Principle from an equilibrium perspective
aA + bB cC + dD
[C]c[D]d
[A]a[B]b
Kc =
When reactants are added to a system,
the reaction quotient Q becomes smaller than Kc.
To get back to equilibrium, the system must reduce the amount of
reactants and increase the number of products.

23. Example: Le Châtelier’s Principle from an equilibrium perspective
CH3COOC2H5(aq) + OH- (aq) CH3COO- (aq) + C2H5OH(aq)
ethyl acetate hydroxide Acetate ethanol
At equilibrium, a solution contains M ethyl acetate, M OH-, 0.85 M acetate and 1.00 M ethanol.
Calculate the value of Kc.

24. Example: Le Châtelier’s Principle from an equilibrium perspective
CH3COOC2H5(aq) + OH- (aq) CH3COO- (aq) + C2H5OH(aq)
ethyl acetate hydroxide Acetate ethanol
Equilibrium: 0.65 M ethyl acetate, M OH-, 0.85 M acetate and 1.00 M ethanol.
After the system has reached equilibrium, enough ethanol is added so that [C2H5OH] = 1.50 M. What is the reaction quotient?

25. Le Châtelier’s Principle: Changing the volume of a gas
N2O4 (g) NO2 (g)
PNO22
PN2O4
Kp =
PV = nRT

26. Changing the volume of a gas
As volume increases (at constant T), pressure decreases System counteracts the change in pressure by increasing the number of moles of gas

27. Le Châtelier’s Principle: Heat and exothermic reactions
CH4 + 2 O CO2 + 2 H2O + heat
Here you can think of heat as a product
If you add more products the system will create more reactants
If you take away products the system will create more products

28. Le Châtelier’s Principle: Heat and endothermic reactions
H2O (l) + heat H2O (g)
Here you can think of heat as a reactant
If you add more reactants the system will create more products

29. Equilibrium with Gases
Don’t usually measure concentrations of gases
Measure gases by partial pressures
PV = nRT
P = (n/V) RT
Partial pressure is related to concentration

30. Equilibrium constants with gaseous reactants can use partial pressures instead of concentrations
2 C2H4 (g) + O2 (g) CH3CHO (g)
[CH3CHO]2
[C2H4]2[O2]
Kc =
PCH3CHO2
PC2H42 PO2
Kp =

31. Relationship between Kc and Kp
[CH3CHO]2
[C2H4]2[O2]
PCH3CHO2
PC2H42 PO2
Kc =
Kp =
PV = nRT
P = (n/V) RT
Partial pressure is related to concentration

32. For any reaction: Kp = Kc (RT)Dn
Change in # moles of gas during reaction

33. H2O (g) + CH4 (g) CO (g) + 3 H2 (g)
Kp = 6.1 x 104 at 1125 K. What is Kc at 1125 K?

34. Equilibrium Laws for Heterogeneous Reactions
Basic rule: the concentration of anything that does not change does not need to be included in the equilibrium constant.
This generally applies to pure solids and pure liquids.

35. CaCO3 (s) CaO (s) + CO2 (g)

36. CO2 (g) + C(s) CO (g)

37. Hg (l) + Hg2+ (aq) Hg22+ (aq)

38. Equilibrium Calculations

39. Calculating Equilibrium Constants
If you don’t know all concentrations…
Make a table showing all reactants and products
Write in known initial concentrations
Add known equilibrium concentrations
Calculate concentration changes
Calculate unknown equilibrium concentrations

40. CH3COOH (aq) + C2H5OH(aq) CH3COOC2H5(aq) + H2O (l)
Ethyl acetate is produced by the reaction of acetic acid with ethanol: mol of acetic acid and 1.00 mol of ethanol are combined to create 1.00 L of solution. At equilibrium, the solution contains 0.65 M ethyl acetate. Calculate the value of Kc.
CH3COOH (aq) + C2H5OH(aq) CH3COOC2H5(aq) + H2O (l)
Make a table showing all reactants and products
Write in known initial concentrations
Add known equilibrium concentrations
Calculate concentration changes
Calculate unknown equilibrium concs

41. 2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)
A mixture of 0.10 mol of NO, mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established: At equilibrium, [NO] = M. Calculate Kc.
2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)
Make a table showing all reactants and products
Write in known initial concentrations
Add known equilibrium concentrations
Calculate concentration changes
Calculate unknown equilibrium concs

42. Calculating Equilibrium concentrations from initial concentrations & equilibrium constant
1. Make an ICE table showing all reactants and products and fill in known concentrations.
2. Use x to represent one concentration change
3. Using stoichiometric ratios, solve for other concentrations in terms of x
4. Write the expression for K in terms of x.
5. Reorganize expression so it is in the form ax2+bx+c=0
6. Use quadratic equation to solve for x
7. Use value of x in ICE table to solve for equilibrium concentrations.

43. Review: Quadratic Equation
A quadratic equation written in form
ax2 +bx +c = 0
has the solution
-b ± b2 – 4ac 2a
x =

44. 2 IBr (g) I2 (g) + Br2 (g)
Kp = 8.5 x 10-3 at 150 oC. If atm of IBr is placed in a 2.0 L container, what is the partial pressure of this substance after equilibrium is reached?
Make ICE table
Set change to x
Solve for concs in terms of x
Write K in terms of x.
Reorganize to form ax2+bx+c=0
Solve for x with quadratic equation
Solve for equilibrium concentrations.

45. N2 (g) + 3 H2 (g) NH3 (g)
Kp = 9.60 at 300 oC moles of pure NH3 is placed in a 1.00 L flask and allowed to reach equilibrium at this temperature. What are the
Make ICE table
Set change to x
Solve for concs in terms of x
Write K in terms of x.
Reorganize to form ax2+bx+c=0
Solve for x with quadratic equation
Solve for equilibrium concentrations.

46. (CH3)3CCl (g) (CH3)2CCH2 (g)+ HCl (g)
For the gas-phase thermal decomposition of t-butyl chloride, Kp = 3.45 at 500 K. (a) Calculate Kc at 500 K. (b) Calculate the molar concentrations of reactants and products in an equilibrium mixture obtained by heating 1.00 mol of t-butyl chloride in a 5.00 L vessel at 500 K.
Make ICE table
Set change to x
Solve for concs in terms of x
Write K in terms of x.
Reorganize to form ax2+bx+c=0
Solve for x with quadratic equation
Solve for equilibrium concentrations.