Chemical equilibrium

Chapter 13
Chemical Equilibrium

Section 13.1
The Equilibrium Condition
Equilibrium is a state in which there are no observable
changes as time goes by.
Physical equilibrium
H2O (l)
Chemical equilibrium
N2O4 (g)
H2O (g)
2NO2 (g)

Section 13.1
Copyright © Cengage Learning. All rights reserved 4
 The state where the concentrations
of all reactants and products remain
constant with time.
 On the molecular level, there is frantic activity. Equilibrium
is not static, but is a highly dynamic situation.
constant
constant

Section 13.1
5
 Concentrations reach levels where the rate of the forward
reaction equals the rate of the reverse reaction.
A + 2B AB2
kf
kr
ratef = kf [A][B]2
rater = kr [AB2]
Equilibrium
ratef = rater
kf [A][B]2 = kr [AB2]
kf
kr
[AB2]
[A][B]2
= Kc =

Section 13.2
The Equilibrium Constant
Consider the following reaction at equilibrium:
eA + rB lC + yD
 A, B, C, and D = chemical species.
 Square brackets = concentrations of species at equilibrium.
 e, r, l, and y = coefficients in the balanced equation.
 K = equilibrium constant (K values are usually written without units).
e
l
r
y
[B][A]
[D][C]
K =

Section 13.2
The Equilibrium Constant
 K always has the same value at a given temperature
regardless of the amounts of reactants or products
that are present initially.
 For a reaction, at a given temperature, there are many
equilibrium positions but only one equilibrium
constant, K.
 Equilibrium position is a set of equilibrium
concentrations.

The following results were collected for two experiments
involving the reaction at 6000C between gaseous sulfur
dioxide and oxygen to form gaseous sulfur trioxide:
Show that the equilibrium constant is the same in both
cases.
For experiment 1:
For experiment 2:
Experimental errors

Section 13.4
Heterogeneous Equilibria
Homogeneous Equilibria
 Homogeneous equilibria – involve the same phase:
N2(g) + 3H2(g) 2NH3(g)
HCN(aq) H+(aq) + CN-(aq)

Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
Kc =‘
[CH3COO-][H3O+]
[CH3COOH][H2O]
[H2O] = constant
Kc =
[CH3COO-][H3O+]
[CH3COOH]
14.2

The equilibrium concentrations for the reaction between carbon
monoxide and molecular chlorine to form COCl2 (g) at 740C are
[CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate
the equilibrium constants Kc.
CO (g) + Cl2 (g) COCl2 (g)
Kc =
[COCl2]
[CO][Cl2]
=
0.14
0.012 x 0.054
= 220
14.2

Consider the reaction represented by the equation:
Fe3+(aq) + SCN-(aq) FeSCN2+(aq)
6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed at a certain
temperature and at equilibrium the concentration of
FeSCN2+(aq) is 4.00 M.
What is the value for the equilibrium constant for this reaction?
12
EXERCISE!

Section 13.5
Applications of the Equilibrium Constant
Set up ICE Table
Fe3+(aq) + SCN–(aq) FeSCN2+(aq)
Initial 6.00 10.00 0.00
Change – 4.00 – 4.00 +4.00
Equilibrium 2.00 6.00 4.00
K = 0.333Copyright © Cengage Learning. All rights reserved 13
 
  
2
3
FeSCN 4.00
= =
2.00 6.00Fe SCN

 
  
      
M
K
M M

Section 13.4
 Heterogeneous equilibria – involve more than one
phase:
2KClO3(s) 2KCl(s) + 3O2(g)
2H2O(l) 2H2(g) + O2(g)

Section 13.4
 The position of a heterogeneous equilibrium does not
depend on the amounts of pure solids or liquids present.
 The concentrations of pure liquids and solids are
constant.
2KClO3(s) 2KCl(s) + 3O2(g)
 3
2= OK
CaCO3 (s) CaO (s) + CO2 (g) K = [CO2]

Section 13.4
PCO2
= Kp
CaCO3 (s) CaO (s) + CO2 (g)
[CO2
] does not depend on the amount of CaCO3 or CaO
14.2

Kc and Position of Equilibrium (Qc)
A (aq) B (aq)
Kc= (known) Qc= (may calculate)
[B]
[A]
Then: compare Kc and Qc to determine
the direction of reaction to move toward equilibrium

The reaction quotient (Qc) is calculated by substituting the
initial concentrations of the reactants and products into the
equilibrium constant (Kc) expression.
IF:
• Qc > Kc system proceeds from right to left to reach equilibrium
• Qc = Kc the system is at equilibrium
• Qc < Kc system proceeds from left to right to reach equilibrium
14.4

a:
b:
c:
Q>K The system will shift to the left
Q=K The system is at equilibrium
Q<K The system will shift to the right

Calculating Equilibrium Concentrations
1. Express the equilibrium concentrations of all species in
terms of the initial concentrations and a single unknown
x, which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the
equilibrium concentrations. Knowing the value of the
equilibrium constant, solve for x.
3. Having solved for x, calculate the equilibrium
concentrations of all species.
14.4

At 12800C the equilibrium constant (Kc) for the reaction
Is 1.1 x 10-5. If the initial concentrations are [A2] = 0.063 M . Calculate
the concentrations of all species at equilibrium.
A2 (g) 2A (g)
Let x be the change in concentration of A2
Initial (M)
Change (M)
Equilibrium (M)
0.063 0.0
-x +2x
0.063 - x + 2x
[A]2
[A2]
Kc = Kc =
(2x)2
0.063 - x
= 1.1 x 10-5
Kc is very small therefor x can ignore and “0.063 - x” is equall 0.063
A2 (g) 2A (g)
NOTE: If Kc < 10-3 you can ignore x
x= 4.16×10-4

Section 13.3
Equilibrium Expressions Involving Pressures
 K involves concentrations.
 Kp involves pressures.
What is Kp?
In most cases
K  Kp
Example: N2(g) + 3H2(g) 2NH3(g)
 
  
3
2 2
2
NH
p 3
N H
P
=
P P
K
 
  
2
3
3
2 2
NH
=
N H
K
equilibrium pressure
equilibrium concentration

Section 13.3
Example
N2(g) + 3H2(g) 2NH3(g)
Equilibrium pressures at a certain temperature:
3
2
2
2
NH
1
N
3
H
= 2.9 10 atm
= 8.9 10 atm
= 2.9 10 atm






P
P
P

Section 13.3
Example
N2(g) + 3H2(g) 2NH3(g)
 
  
3
2 2
2
NH
p 3
N H
P
=
P P
K
 
  
22
p 31 3
2.9 10
=
8.9 10 2.9 10

 

 
K
4
p = 3.9 10K

Section 13.3
The Relationship Between K and Kp
Kp = Kc(RT)Δn
 Δn = sum of the coefficients of the gaseous products
minus the sum of the coefficients of the gaseous
reactants.
 R = 0.08206 L·atm/mol·K
 T = temperature (in Kelvin)

Section 13.3
Example
N2(g) + 3H2(g) 2NH3(g)
Using the value of Kp (3.9 × 104) from the previous
example, calculate the value of Kc at 35°C.
 
  
p
2 44
7
=
3.9 10 = 0.08206 L atm/mol K 308K
= 2.5 10


   

n
K K RT
K
K

Consider the following equilibrium at 295 K:
The partial pressure of each gas is 0.265 atm. Calculate
Kp and Kc for the reaction?
NH4HS (s) NH3 (g) + H2S (g)
Kp = PNH3 H2SP = 0.265 x 0.265 = 0.0702
Kp = Kc(RT)n
n = 2 – 0 = 2 T = 295 K
Kc = 0.0702 / (0.0821 x 295)2 = 1.20 x 10-4

If an external stress is applied to a system at equilibrium, the
system adjusts in such a way that the stress is partially offset
as the system reaches a new equilibrium position.
Le Châtelier’s Principle
• Changes in Concentration
N2 (g) + 3H2 (g) 2NH3 (g)
Add
NH3
Equilibrium
shifts left to
offset stress
14.5

• Changes in Concentration continued
Change Shifts the Equilibrium
Increase concentration of product(s) left
Decrease concentration of product(s) right
Decrease concentration of reactant(s)
Increase concentration of reactant(s) right
left
aA + bB cC + dD
AddAddRemove Remove

Section 13.7
Effects of Changes on the System
3. Pressure and Volume: for the gaseous equilibrium
a) Increase pressure shifts the equilibrium toward
the Side with fewest moles of gas
2A (g) B(g)
a) Decrease volume shifts the equilibrium toward
the side with fewer moles of gas.
2A (g) B(g)
Needs 2 volumes Needs 1 volume
1 mole of gas2 moles of gas

Heat + N2O4(g) 2NO2 (g) ΔH°= +58.0 kJ/mol (Endothermic Reaction)
2NO2(g) N2O4 (g) + Heat ΔH°= -58.0 kJ/mol (Exothermic Reaction)
2NO2(g) N2O4 (g)
Brown Colorless
In cold water:
[N2O4](Colorless) increase
In hot water:
[NO2](Brown) increase
N2O4(g) 2NO2 (g)2NO2(g) N2O4 (g)
• Changes in Temperature

uncatalyzed catalyzed
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
• Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner

Change Shift Equilibrium
Change Equilibrium
Constant
Concentration yes no
Pressure yes no
Volume yes no
Temperature yes yes
Catalyst no no
14.5

Chemistry In Action: The Haber Process
N2 (g) + 3H2 (g) 2NH3 (g) H0 = -92.6 kJ/mol

Chemical equilibrium

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