CHAPTER 6 - CHEMICAL EQUILIBRIUM

If ; % α = x x 100 is < 5%
[A]initial

assumption is allowed

If ; % α = x
[A]initial x 100% is > 5%

 assumption is NOT allowed
 use quadratic formula

101

Exercise :

Phosgene is a potent chemical warfare agent that is
now outlawed by international agreement. It
decomposed by the reaction

COCl2 (g) CO (g) + Cl2 (g)

Kc = 8.3 x 10-4 (at 360 oC)

Calculate [CO], [Cl2] and [COCl2], when the following
amounts of phosgene decompose and reach
equilibrium in a 10 L flask :

(a) 5.000 mol of COCl2
(b) 0.100 mol of COCl2

102

Reaction Quotient (Q)

• The equilibrium constant is used to predict the
direction in which a reaction mixture will
proceed to achieve equilibrium.

• For the reaction at equilibrium :

aP (g) + bQ (g) cR (g) + dS (g)

Keq  Rc Sd  at equilibrium
Pa Qb

103

• For the reaction that have not reached
equilibrium :

aP (g) + bQ (g) cR (g) + dS (g)

Q  Rc Sd  at non-equilibrium
Pa Qb conditions

• To predict the direction of reaction, compare
the values of Keq and Q.

104

• 3 possible cases for a system are as follow :

The ratio of initial concentration of products to
reactants is too large. To reach equilibrium,
i. Q > Keq products must be converted to reactants.
 [Product]  and [Reactant] 
 The system proceeds from right to left.

The initial concentrations are equilibrium
ii. Q = Keq concentration. The system is already at

equilibrium.

The ratio of initial concentrations of products
to reactants is too small. To reach equilibrium,
iii. Q < Keq reactants must be converted to products.
 [Product]  and [Reactant] 
 The system proceeds from left to right.

105

Exercise 1 :
At 199C, the initial moles of H2, N2 and NH3
are 2.0 mol, 1.0 mol and 2.0 mol respectively
was placed in a 1.0 L stainless-steel. If the Kc
for the reaction is 0.105 at this temperature,
decide whether the system is at equilibrium. If it
is not, predict which way will the reaction
proceed.

106

Answer

3H2 (g)  N2 (g)  2NH3 (g)

NHHN322   2M
  2M
  1M

QC  NH3 2
H2 3 N2 

 22

23 1

 0.5

Qc Kc
 positonof equilibriu m shift to the left.107

Exercise 2 :

The equilibrium constant, Kc for the formation of
nitrosyl chloride from nitric oxide and molecular
chloride

2NO (g) + Cl2 (g) 2NOCl (g)

is 6.5 x 106 at 35°C. In a certain experiment, 2.0 x
10-2 mol NO, 12.3 x 10-1 mol Cl2, and 6.8 mol NOCl
are mixed in a 2.0 L flask. In which direction will
the system proceed to reach equilibrium?

108

Answer

nNO  2.0 x 102 mol , NO  0.01M

nCl2  12.3 x10 1 m ol , Cl2   0.615 M
nNOCl  6.8 m ol , NOCl  3.4 M

V  2.0L

QC  NOCl2
NO 2Cl2 

 3.42

0.012 0.615

 1.88 x 105

Qc Kc 109
 positonof equilibrium shift to the right.

Exercise 3:

An equilibrium mixture in a 1.00 L vessel at 490 K contains
0.228 mol H2(g), 0.228 mol I2(g) and 1.544 mol HI(g). An
additional 0.200 mol H2(g) is then pumped into the container
and allowed to achieve to the new equilibrium.

i. Calculate the Kc of the following reaction :

H2 (g) + I2 (g) 2HI (g)

ii. Calculate the value of Q when H2(g) is added before the
new equilibrium is established.

iii. What is the effect of increasing the amount of H2(g) to the
equilibrium position ?

iv. Calculate the new equilibrium concentrations for each11g0as.

SOLUTION :

H2(g) + I2(g) 2HI(g)

nat equilibrium 0.228 0.228 1.544
[ ]at equilibrium
0.228 0.228 1.544
1
1 1
 0.228  0.228 1.544

i. KC  HI2 111
H2 I2 

49.5  1.5442
0.2282

 45.9

SOLUTION :

ii. When 0.2 mol of H2 is added

H2(g) + I2(g) 2HI(g)

n 0.228 + 0.2 = 0.228 1.544
0.428
0.228 1.544
[ ] 0.428 1 1

1  0.288 1.544
 0.428

QC  HI2 iii. Since the value of Qc < Kc ,
H2 I2  the equilibrium position shift to the right

49.5  1.5442
0.4282

 24.4

112

SOLUTION : iV

H2(g) + I2(g) 2HI(g)

n initial 0.428 0.228 1.544
n +2x
nat equilibrium -x -x 1.544 + 2x
0.428 –x 0.228-x
[ ]at equilibrium 1.544  2x
 0.428  x  0.228  x 1
1 1

113

KC  HI2
H2  I2 

49.5  1.544  2x2  x
0.428  x  0.228

x  0.0622M

HI 1.668M
HI22 
  0.166 M
 0.366 M

114

Exercise 4 :
Consider the following equilibrium process at 686°C:

CO2 (g) + H2 (g) CO (g) + H2O (g)

The equilibrium concentrations of the reacting species

are [CO] = 0.050 M, [H2] = 0.045 M, [CO2] = 0.086 M,
and [H2O] = 0.040 M.

i. Calculate Kc for the reaction at 686°C.

ii. If we add CO2 to increase its concentration to 0.50
mol L-1, what will the concentrations of all the gases
be when equilibrium is reestablished ?

115

Answer

i. KC  H2O CO
H2 CO2 

 0.04 0.05
0.0450.086

 0.5168

116

SOLUTION :

ii.

CO2 (g) + H2 (g) CO (g) + H2O (g)

[ ] initial 0.50 0.045 0.050 0.04

[ ] -x -x +x +x
[ ]at equilibrium 0.50 – x 0.045-x
0.051 + x 0.04 + x

117

SOLUTION :

ii. KC  0.52
0.52 
COH2O  X
0.52  C0.O052 HX20.04  X
0.5  X0.045

x  0.02516 M

CO2  0.5  X

 0.47 M

H2  0.045  X

 0.020 M

CO  0.05  X

 0.075 M

H2O  0.040  X

 0.065 M

118







6.3
Le Chatelier’s Principal

122

LEARNING OUTCOMES

At the end of this topic, students should be able to :
(a) state Le Chatelier’s principle.

(b) explain the effect of the following factors on a
system at equilibrium using Le Chatelier’s
principle:
i. concentration of reacting species
ii. pressure / volume
iii. addition of inert gas at constant volume at
and at constant pressure
iv. temperature
v. catalyst

123

Le Chatelier’s Principle.

The principal states that:

If a system at equilibrium is disturbed by an external
change (concentration, pressure or temperature), the
system will shift its equilibrium position to reduce the
effect of the change.

•The position of equilibrium in a system is affected by:
a) concentration
b) pressure / volume
c) temperature
d) catalyst

• The effect of each can be predicted qualitatively by Le
Chatelier’s Principle.
124

Factors Affecting Equilibrium System

• The main factors that affect the equilibrium
system are :
(a). Concentration
(b). Pressure / volume
(c). Temperature
(d). Catalyst

125

(a) Changes in Concentrations

• Adding a reactant or product shifts the
equilibrium away from the increase.

• Removing a reactant or product shifts the
equilibrium towards the decrease.

126

N2(g) + 3H2(g) 2NH3(g)

127

Example 1: H2 (g) + I2 (g) 2HI (g)

Changes in Effect to Equilibrium Position
Concentration

 The system must consume the H2
and produce product.

If H2 (reactant) is  The position of equilibrium will shift to
added to a system. the right in order to decrease [H2].

  [HI]  and [I2]  & [H2] .

If H2 (reactant) is  The position of equilibrium will shift to
removed to a
the left in order to increase [H2].
system.
  [HI]  and [I2]  & [H2] .

128

Example 1: H2 (g) + I2 (g) 2HI (g)

Changes in Effect to Equilibrium Position
Concentration

 The position of equilibrium will

shift to the left in order to

decrease [HI].
If HI (product) is

added to a system.  HI decompose to produce H2
and I2  [H2]  & [I2]  and
[HI] .

129

Example:

Consider the reaction below:

Fe 3+ + 2SCN- Fe (SCN)+2(aq)
(aq) (Blood red)

(Yellowish)

Discuss the effect of adding the following
substances to the equilibrium mixture.

i. iron(III) or Fe3+

ii. sodium hydroxide, NaOH

130

SOLUTION
At equilibrium the colour of the solution is light

blood red

i) When Fe3+ ion is added

- The system will reduce the effect of the additional
concentration of Fe3+ion. More Fe3+ ions react
with SCN- ions

- Thus more products, Fe (SCN)+2 will be formed
and the colour intensity of blood red increase

- The equilibrium position shifts forward

131

ii) When NaOH is added

- NaOH reacts with Fe3+ to form Fe(OH)3

3NaOH + Fe(NO3)3 Fe(OH)3 + 3NaNO3

- Thus decreasing the concentration of Fe3+.

- The system will overcome the change by increasing
the concentration of Fe3+.

- So the reaction shifted to the left (reaction moves
backward or reverse) to achieve a new equilibrium

.
- Therefore more Fe3+ ions will be formed and the

solution becomes more yellowish

132

b) Effect of Pressure

The pressure of a system may be changed by:
i) changing the moles of reactant or product
ii) changing the volume
iii) adding inert gas to the system

Affects only the reversible reaction involving
gaseous reactants/products

133

i) Changing moles of reactant or product

Adding or removing the gaseous reactant or
product at constant volume has the same effect as
changing the concentration

Example:
Consider the following system at equlibrium

2SO2(g) + O2(g) 2SO3(g)

• When SO2 gas is added to the system, the partial
pressure of SO2 increases (which means [SO2]
increases)

• The position of equilibrium shifts to the right to
reduce the effect of high concentration of SO2. 134

ii) Changes in Volume:

Note that : P α 1
V

(V increase) (V decrease)

Shift the position of equilibrium Shift the position of equilibrium

in the direction to increase the in the direction to decrease the

number of moles of gas. number of moles of gas.

• Example : N2 (g) + 3H2 (g) 2NH3 (g)
 when V decrease

 the position of equilibrium will shift in the

direction to the lower number of moles of gas

particles.

 the position of equilibrium will shift to the right

135

(c) Changes in Temperature

• Increasing temperature favors the endothermic
reaction

• Decreasing temperature favors the exothermic
reaction

• Kc and Kp changes in value as the T changes.

i. Exothermic Reaction
• Example :

N2 (g) + 3H2 (g) 2NH3 (g) ΔH = - 92.83 kJ

136

T increase T decrease

• Shift the position of • Shift the position of

equilibrium in the direction as equilibrium in the direction as

to decrease the T. to increase the T.

• So, the position of equilibrium • So, the position of equilibrium
will shift to the left in order to will shift to the right in order
absorbed the heat supplied. to released the heat supplied.

• More ammonia will dissociate • More ammonia will produce

•  [NH3] , [H2]  & [N2] •  [NH3] , [H2]  & [N2] 

• The value of Kc or Kp • The value of Kc or Kp

decrease increase

137

ii. Endothermic Reaction 2NO (g) ΔH = +ve
• Example : N2 (g) + O2 (g)

T increase T decrease

• Shift the position of • Shift the position of

equilibrium in the direction equilibrium in the direction

as to decrease the T. as to increase the T.

• So, the position of • So, the position of

equilibrium will shift to the equilibrium will shift to the

right in order to absorbed the left in order to released the

heat supplied. heat supplied.

• More NO will produce • More NO will dissociate

•  [NO] , [N2]  & [O2]  •  [NO]  , [N2]  & [O2] 

• The value of Kc or Kp • The value of Kc or Kp

increase decrease 138

• Note:

Increasing temperature favors the endothermic
reaction

Decreasing temperature favors the exothermic
reaction

• Adding heat (i.e. heating the vessel) favors away
from the increase:
– if H < 0, adding heat favors the reverse reaction.

• Removing heat (i.e. cooling the vessel), favors
towards the decrease:
– if H < 0, cooling favors the forward reaction.

139

Exercise :

The graph below shows the percentage of C at 100K
and 200K for reaction:

A2 (g) + B2 (g) 2C (g)

%C
100K
200K

time

i. Write the equilibrium constant expression Kp for
the reaction.

ii. Based on the graph, explain whether the forward
reaction is endothermic or exothermic.

iii. Explain whether the value of Kp is high at 100K or
200K. 140

Solution:

141

(d). Effect of Catalyst

• A catalyst increase the rate of a chemical reaction by
providing an alternative pathway with lower activation
energy, Ea.

• Activation energy, Ea is the minimum amount energy
required to initiate a chemical reaction.

• The presence of catalyst does not :
i. change the composition of equilibrium mixture
ii. affect the position of equilibrium
iii. affect the value of equilibrium constant, K

• Time taken to achieve equilibrium is shortened.

142

A (g) + B (g) C (g) + D (g)

uncatalyzed catalyzed

143

(e). Effect of Adding Inert Gases

i. At Constant Volume

• The total number of gaseous molecules increase.

• Total pressure increase of the equilibrium system
increases.

• The volume and the quantity of the reacting species
remain constant so the partial pressure of each
gases in the equilibrium system remains unchanged.

•  does not change the position of equilibrium
and equilibrium constant

144

ii. At Constant Pressure

• Adding inert gas at to the system a constant
external pressure increase the volume of the
container.

• The partial pressure for the gases in the system
are lowered.

• have the same effect as P  or V 

•  the position of equilibrium will shift in the
direction to increase the number of moles or
molecules

145

Exercise :
Consider the gas-phase reaction :

2CO (g) + O2 (g) 2CO2 (g)

Predict the shift in equilibrium position when
helium gas is added to the equilibrium mixture

i. at constant pressure
ii. at constant volume

146

Exercise :

Consider the following reaction at equilibrium :

C (s) + CO2 (g) 2CO (g) ΔH = +119.8 kJ

What will be the effect on the equilibrium of each of the
following changes ?

i. Addition of CO2
ii. Addition of C (s)
iii. An increase in temperature
iv. Decrease the volume of container
v. An increase in pressure
vi. Addition of inert gas at constant volume

147

Application of Equilibrium Principles in The Haber
Process

• To make an industrial process economically
worthwhile :

N2 (g) + 3H2 (g) 2NH3 (g) ΔH = -92.83 kJ

• Three ways to maximize the yield of ammonia :

 Decrease [NH3]
 Decrease volume (increase pressure)
 Decrease temperature

“Application of Le Chatelier’s principle”

148

STAGES IN THE HABER PROCESS FOR
SYNTHESIZING AMMONIA

149

N2 (g) + 3H2 (g) 2NH3 (g) ΔH = -92.83 kJ
• Decrease [NH3] :

Product : Ammonia

 By removing ammonia, the system will produce
more in continual drive to reattain equilibrium

 The equilibrium will shift to the right

• Decrease volume (increase pressure)

 4 mol of gas reacts to form 2 mol of gas
 Decreasing the volume will shift the equilibrium
towards fewer moles of gas
 Produces more ammonia

150