Chapter 15: Principles of Chemical Equilibrium
Slide 1 of 33
Contents
15-1
15-2
15-3
15-4
15-5
15-6
15-7
Slide 2 of 33
Dynamic Equilibrium
The Equilibrium Constant Expression
Relationships Involving Equilibrium Constants
The Magnitude of an Equilibrium Constant
The Reaction Quotient, Q: Predicting the Direction of a
Net Change
Altering Equilibrium Conditions:
Le Châtelier’s Principle
Equilibrium Calculations: Some Illustrative Examples
15-1 Dynamic Equilibrium
 Equilibrium – two opposing
processes taking place at
equal rates.
H2O(l)
NaCl(s)
H2O(g)
H2O
CO(g) + 2 H2(g)
Slide 3 of 33
NaCl(aq)
CH3OH(g)
I2(H2O)
I2(CCl4)
Dynamic Equilibrium
Slide 4 of 33
15-2 The Equilibrium Constant Expression
 Methanol synthesis is a reversible reaction.
CO(g) + 2 H2(g)
CH3OH(g)
k-1
k1
CH3OH(g)
CO(g) + 2 H2(g)
CO(g) + 2 H2(g)
Slide 5 of 33
k1
k-1
CH3OH(g)
Three Approaches to the Equilibrium
Slide 6 of 33
Three Approaches to Equilibrium
Slide 7 of 33
Three Approaches to Equilibrium
CO(g) + 2 H2(g)
Slide 8 of 33
k1
k-1
CH3OH(g)
The Equilibrium Constant Expression
k1
Forward: CO(g) + 2 H2(g) → CH3OH(g)
k-1
Reverse: CH3OH(g) → CO(g) + 2 H2(g)
At Equilibrium:
Rfwrd = Rrvrs
CO(g) + 2 H2(g)
Rfwrd = k1[CO][H2]2
Rrvrs = k-1[CH3OH]
k1
k-1
k1[CO][H2]2 = k-1[CH3OH]
k1
k-1
Slide 9 of 33
=
[CH3OH]
[CO][H2
]2
= Kc
CH3OH(g)
General Expressions
a A + b B …. → g G + h H ….
Equilibrium constant = Kc=
Slide 10 of 33
[G]g[H]h ….
[A]m[B]n ….
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.
Slide 11 of 33
Combining Equilibrium Constant
Expressions
N2O(g) + ½O2
N2(g) + ½O2
N2(g) + O2
2 NO(g) Kc= ?
N2O(g)
2 NO(g)
Kc(2)=
2.710-18
Kc(3)= 4.710-31
[N2O]
=
[N2][O2]½
[NO]2
=
[N2][O2]
[NO]2
[NO]2 [N2][O2]½
1
-13
Kc=
=
K
=
=
1.710
c(3)
[N2O][O2]½ [N2][O2] [N2O]
Kc(2)
Slide 12 of 33
KP = Kc(RT)Δn
:
Slide 13 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).
C(s) + H2O(g)
[CO][H2]
Kc =
[H2O]2
Slide 14 of 33
CO(g) + H2(g)
 Worked Examples Follow:
Slide 15 of 33
15-1 Practice Example B
Slide 16 of 33
Slide 17 of 33
Slide 18 of 33
 CRS Questions Follow:
Slide 19 of 33
Which of the following statements is
correct?
kf
CO(g) + 3H 2 (g)
kr
CH 4 (g) + H 2O(g)
3
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.
Slide 20 of 33
moles of substance
1. At equilibrium the reaction stops.
2
H2
1
CO
CH4 = H2O
0
time
Which of the following statements is
correct?
kf
CO(g) + 3H 2 (g)
kr
CH 4 (g) + H 2O(g)
3
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.
Slide 21 of 33
moles of substance
1. At equilibrium the reaction stops.
2
H2
1
CO
CH4 = H2O
0
time