15-1 Dynamic Equilibrium 15-2 The Equilibrium Constant Expression 15-3 Relationships Involving Equilibrium Constants 15-4 The Magnitude of an Equilibrium Constant 15-5 The Reaction Quotient, Q: Predicting the Direction of a Net Change 15-6 Altering Equilibrium Conditions: Le Châtelier’s Principle 15-7 Equilibrium Calculations: Some Illustrative Examples

1. Slide 1 of 33
Chapter 15: Principles of Chemical Equilibrium

2. Slide 2 of 33
Contents
15-1 Dynamic Equilibrium
15-2 The Equilibrium Constant Expression
15-3 Relationships Involving Equilibrium Constants
15-4 The Magnitude of an Equilibrium Constant
15-5 The Reaction Quotient, Q: Predicting the Direction of a
Net Change
15-6 Altering Equilibrium Conditions:
Le Châtelier’s Principle
15-7 Equilibrium Calculations: Some Illustrative Examples

3. Slide 3 of 33
15-1 Dynamic Equilibrium
 Equilibrium – two opposing
processes taking place at
equal rates.
H2O(l) H2O(g)
I2(H2O) I2(CCl4)
NaCl(s) NaCl(aq)
H2O
CO(g) + 2 H2(g) CH3OH(g)

4. Slide 4 of 33
Dynamic Equilibrium

5. Slide 5 of 33
15-2 The Equilibrium Constant Expression
 Methanol synthesis is a reversible reaction.
CO(g) + 2 H2(g) CH3OH(g)
k1
k-1
CH3OH(g) CO(g) + 2 H2(g)
CO(g) + 2 H2(g) CH3OH(g)
k1
k-1

6. Slide 6 of 33
Three Approaches to the Equilibrium

7. Slide 7 of 33
Three Approaches to Equilibrium

8. Slide 8 of 33
Three Approaches to Equilibrium
CO(g) + 2 H2(g) CH3OH(g)
k1
k-1

9. Slide 9 of 33
The Equilibrium Constant Expression
Forward: CO(g) + 2 H2(g) → CH3OH(g)
Reverse: CH3OH(g) → CO(g) + 2 H2(g)
At Equilibrium:
Rfwrd = k1[CO][H2]2
Rrvrs = k-1[CH3OH]
Rfwrd = Rrvrs
k1[CO][H2]2 = k-1[CH3OH]
[CH3OH]
[CO][H2]2
=
k1
k-1
= Kc
CO(g) + 2 H2(g) CH3OH(g)
k1
k-1
k1
k-1

10. Slide 10 of 33
General Expressions
a A + b B …. → g G + h H ….
Equilibrium constant = Kc=
[A]m[B]n ….
[G]g[H]h ….

11. Slide 11 of 33
15-3 Relationships Involving the
Equilibrium Constant
 Reversing an equation causes inversion of K.
 Multiplying by coefficients by a common factor
raises the equilibrium constant to the
corresponding power.
 Dividing the coefficients by a common factor
causes the equilibrium constant to be taken to
that root.

12. Slide 12 of 33
Combining Equilibrium Constant
Expressions
N2O(g) + ½O2 2 NO(g) Kc= ?
Kc=
[N2O][O2]½
[NO]2
=
[N2][O2]½
[N2O]
[N2][O2]
[NO]2
Kc(2)
1
Kc(3)
= = 1.710-13
[N2][O2]
[NO]2
=
[N2][O2]½
[N2O]
=
N2(g) + ½O2 N2O(g) Kc(2)= 2.710-18
N2(g) + O2 2 NO(g) Kc(3)= 4.710-31

13. Slide 13 of 33
KP = Kc(RT)Δn
:

14. Slide 14 of 33
Pure Liquids and Solids
 Equilibrium constant expressions do not contain
concentration terms for solid or liquid phases of a
single component (that is, pure solids or liquids).
Kc =
[H2O]2
[CO][H2]
C(s) + H2O(g) CO(g) + H2(g)

15. Slide 15 of 33
 Worked Examples Follow:

16. Slide 16 of 33
15-1 Practice Example B

17. Slide 17 of 33

18. Slide 18 of 33

19. Slide 19 of 33
 CRS Questions Follow:

20. Slide 20 of 33
H2
CO
CH4 = H2O
time
moles
of
substance
0
1
2
3
f
r
k
2 4 2
k
CO(g) + 3H (g) CH (g) + H O(g)
Which of the following statements is
correct?
1. At equilibrium the reaction stops.
2. At equilibrium the rate constants
for the forward and reverse
reactions are equal.
3. At equilibrium the rates of the
forward and reverse reactions are
equal.
4. At equilibrium the rates of the
forward and reverse reactions are
zero.

21. Slide 21 of 33
H2
CO
CH4 = H2O
time
moles
of
substance
0
1
2
3
f
r
k
2 4 2
k
CO(g) + 3H (g) CH (g) + H O(g)
Which of the following statements is
correct?
1. At equilibrium the reaction stops.
2. At equilibrium the rate constants
for the forward and reverse
reactions are equal.
3. At equilibrium the rates of the
forward and reverse reactions are
equal.
4. At equilibrium the rates of the
forward and reverse reactions are
zero.