CHAPTER 6 - CHEMICAL EQUILIBRIUM

CHAPTER 6
CHEMICAL EQUILIBRIUM

CHEMISTRY UNIT
KOLEJ MATRIKULASI MELAKA

SHARED BY MISS DALINA BINTI DAUD

CHAPTER 6

Chemical Equilibrium

6.1 Dynamic Equilibrium
6.2 Equilibrium Constant
6.3 Le Chatelier’s Principal

6.1
Dynamic Equilibrium

3

LEARNING OUTCOMES
At the end of this topic, students should be able to :

(a) Explain the term reversible reaction, dynamic
equilibrium and law of mass action.

(b) State the characteristics of a system at
equilibrium.

(c) Interpret the curve of concentration reactants
and products against time for a reversible
reaction.

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Reversible and Nonreversible Reactions

(a) Non-reversible reaction

• The reaction that proceed in only one direction that is
toward the formation of product.

A+B C+D

• The sign indicates non-reversible reaction.

 A and B react to form C and D (forward reaction)

 C and D does not react to form A and B.

• Example : Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g)

5

( b ) Reversible reaction

A+B C+D

• Consider the above reversible reaction which
occurs in both forward and reverse directions.

• When molecules of reactant, A and B react,
molecules of products C and D are formed.

• As soon as some product molecules (C and D )are
formed, the reverse process (reaction) begins to
take place in which reactants, (A and B) are formed
from the products.

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A+B C+D

• The sign indicates reversible reaction.

 Forward reaction : reaction proceeds from left to
right (formation of C and D)

 Reverse reaction : reaction proceeds from right to left
( formation A and B )

• The net concentrations of A, B, C and D do not change.

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Some examples of reversible reactions are

i. The Haber Process : 2NH3 (g)
N2 (g) + 3H2 (g)

ii. Decomposition of PCl5 :

PCl5 (g) PCl3 (g) + Cl2 (g)

iii. Stage two in the Contact Process :

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

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State of Equilibrium

• Equilibrium is a state in which there are no
observable changes as time goes by.

• Chemical equilibrium – chemical composition
of the system does not change at dynamic
equilibrium

e.g : N2O4 (g) 2NO2 (g)

• For the above system, at equilibrium state, the
concentrations of the reactants, N2O4(g) and
products, NO2(g) remain constant over time
and there are no visible changes in the system

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• Physical Equilibrium : a system whose physical
state does not change
when dynamic equilibrium
is reached.

e.g :

H2O (l) H2O (g)

• Both water liquid and water vapour exist at dynamic
equilibrium

10

Characteristic of A System in Equilibrium

• A system in chemical equilibrium has the following
characteristic :
i. occurs only in a closed system

ii. forward and a reverse reaction proceed at
equal rates

iii. the reaction quotient, Q = the equilibrium
constant, K.

iv. the properties (concentrations, pressure and
colour) of the reactants and products are
constant over time.

• The point at which the rate of forward reaction
equals the rate of reverse reaction is known as
dynamic equilibrium.

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• For the following reversible reaction :

A+B C+D

Concentration Concentrations of
reactants and
products are
constant

C and D (products)

A and B (reactants)

ta Time
Figure 1 : Concentration versus Time Curve 12

• Refer to Figure 1

• As the reaction progresses :
 [C] and [D] increases with time

 [A] and [B] decreases with time
 After time, ta, there is no net change in the

concentration of products or reactants.

 The system is in the state of equilibrium
 The equilibrium is dynamic equilibrium where

after ta, the reaction did not stop.
 But, the rate of forward reaction equals to the

rate of reverse reaction.

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• Figure 2 shows the rate changes of both forward
and the reverse reactions over time.

rate of foward
reaction
decreases

Forward reaction rate

Rate A+B C+D

Reverse reaction rate

rate of reverse C+D A+B
reaction
increases

ta Time
14
Figure 2 : Rate versus Time Curve

• Figure 2 shows:

• The rate of forward reaction decrease in time
because the amount of reactant decrease and
the rate of reverse reaction increase since the
concentration of product increase.

• At first, the rate of reverse reaction zero because
no products are formed.

• As the reaction proceeds :
 [C] and [D] increase with time, hence the
rate of reaction also increase.

15

Example:

N2O4(g) 2NO2(g)

At equilibrium,

• Rate of forward reaction = Rate of reverse reaction

N2O4(g)  2NO2(g) 2NO2(g)  N2O4(g)

 [N2O4(g)] and [NO2(g)] remain constant
But!

[N2O4(g)] ≠ [NO2(g)]

16

Law of Chemical Equilibrium

The equilibrium concentration of reactants and
products are related by law of mass action / law of
chemical equilibrium.

Law of Mass Action or Equilibrium Law
The rate at which a chemical reaction takes
place at a given temperature is proportional to
the product of the active massses of the
reactants

When a reversible reaction has achieved dynamic
equilibrium at constant temperature, the ratio of
reactant to product concentrations has a constant
value, K.

Consider general reaction :

aA + bB cC + dD

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• At equilibrium ; K  Cc Dd
where : A a B b   

  
  
  

K = equilibrium constant
[ ] = concentration in the unit of mol L-1 @ M
a, b, c, d = stoichiometric coefficient for the

reacting species A, B, C, and D in
a balanced equation.

• Note : magnitude of K indicates how far the reaction
proceeds toward product at a given
temperature.

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• For a reverse reaction :

K

cC + dD K’ aA + bB

• At equilibrium ;

K  A a Bb or K 1
K


C c D d   

  
  
  

where ;

K’ - equilibrium constant for reverse reaction.

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6.2
The Equilibrium Constant

(KC and KP)

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LEARNING OUTCOMES 6.2

• At the end of this topic, students should be able to :

(a) define homogeneous and heterogeneous
equilibria.

(b) write expression for equilibrium constant in
terms of concentration, Kc and partial
pressure, Kp for homogeneous and
heterogeneous systems.

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Homogeneous Equilibrium

• In homogeneous equilibrium, all products and
reactants are in the same phase.

(a) Reaction in a Aqueous Phase

• The quantity of reactants and products are
mention in concentration unit, mol L-1 or M.

• So, the equilibrium constant expression symbol is
Kc.
aA (aq) + bB (aq) cC (aq) + dD (aq)

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K c  Cc Dd
A a B b   

  
  
  

where :
[ ] = concentration in mol L-1 @ M

Kc = equilibrium constant expression in terms
of concentration

a, b, c & d = stoichiometric coefficients

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(b). Reaction in a Gaseous Phase

• The concentration of reactant and product can be
mention in concentration unit or in partial pressure

• So, the equilibrium constant can be expressed as

Kc or Kp.

aA (g) + bB (g) cC (g) + dD (g)

Kc  Cc Dd or Kp   PC caPPDBdb
AaBb  PA
   

where :

PA = partial pressure of gas A and etc.

Kp = equilibrium constant expression in terms of
pressures
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Exercise 1 :

Write the equilibrium constant expression in terms
of concentration and pressure the following
reactions.

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

ii. 4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)

iii. CH3COOH (aq) CH3COO- (aq) + H+ (aq)

iv. C3H8 (g) + O2 (g) CO2 (g) + H2O (g)

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Answer

i. Kc  SO3 2     Kp P2
SO2 2 O2  SO 3

P P2 O2
SO 2

    Kp 
ii. Kc  NO4 H2O6 P P4 6
NH3 4 O2 5 NO H2O

P P4 5
NH3 O2

   iii.
Kc  CH3COO H Kp  

CH3COOH

iv. Kc  CO2  H2O   Kp PCO2PH2O
C3H8  O2  PC3H8 PO2

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• The equilibrium constant ( Kc and Kp ) is a
dimensionless (no unit) quantity.

• Example : PCl5 (g)
PCl3 (g) + Cl2 (g)

Kc = 1.67 (at 500 K)

Kp = 4.07 x 10-2 (at 500 K)

• NOTE:

The values of Kp and Kc will change when
temperature changes

28

Exercise 2 :

Methanol (CH3OH) is produced commercially by
the catalysed reaction of carbon monoxide and
hydrogen :

CO (g) + 2H2 (g) CH3OH (g)

An equilibrium mixture in a 2.00 L vessel is found
to contain 0.0406 mol CH3OH, 0.170 mol CO and
0.302 mol H2 at 500 K. What is the value of Kc at
this temperature?

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Answer

KC  CH3OH
COH2 2

 0.0406 
 2
 0.17  0.302 2


 2  2 

 10.47

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Exercise 3 :
Equilibrium is established at 25°C in the reaction

N2O4 (g) 2NO2 (g) Kc = 4.61 x 10-3

If [NO2] = 0.0236 M in a 2.26 L flask, how many
grams of N2O4 are also present?

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Answer

KC  NO2 2
N2O4 

4.61x 10 3  0.0236 2

N2O4 

N2O4   0.1208 M

n  MV

 0.1208 molL1 2.26 L

 0.237 mol

mass N2O4  0.273 x 92
 25.12 g

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Exercise 4 :

The equilibrium constant Kp for the reaction

2NO2 (g) 2NO (g) + O2 (g)

is 158 at 1000 K. What is the equilibrium pressure
of O2 if the PNO2 = 0.400 atm and PNO = 0.270 atm?

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Answer

   KP 
PNO 2 PO2
PNO2 2

 158
 0.272 PO2
0.42

PO2  346.78 atm

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Exercise 5 :
The following equilibrium process has been
studied at 50°C.

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

In one experiment the concentration of the reacting
species at equilibrium are found to be [H2] = 0.250
M, [NH3] = 0.050 M. calculate the [N2] if the value
of Kc at this temperature is 6.0 x 102.

35

Answer

KC  NH3 2
N2 H2 3

 0.0502

N2 0.253

N2   2.67 x104 M

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• The values of Kc or Kp are depend on the way in
which the balanced chemical equation is written .

• Example :

1. N2O4 (g) 2NO2 (g) (at 25oC)
Kc = [NO2]2 = 4.63 x 10-3

[N2O4]

2. 2NO2 (g) N2O4 (g) (at 25oC)

K’c = [N2O4] ; K’c = 1
[NO2]2
Kc = 216

3. ½N2O4 (g) NO2 (g (at 25oC)

K’’c = [NO2] ; K’’c = 1 OR K’’c = √Kc = 0.0680
[N2O4]1/2 √ K’c

37

Exercise 6 :

The value of Kc for the following reaction is
1.0 x 10-3 at 200 oC.

2NOBr (g) 2NO (g) + Br2 (g)

Calculate Kc for the following reaction at the
same temperature.

NO (g) + ½Br2 (g) NOBr (g)

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Answer

KC  1.0 x103

K'C  1
KC

1
1.0 x103

K'C  31.62

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Exercise 7 :

At 25 oC, Kc = 7.0 x 1025 for the reaction :

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

What is the value of Kc for the reaction :

SO3 (g) SO2 (g) + ½O2 (g)

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Answer

KC  7.0 x1025

K'C  1
KC

1
7.0 x1025

K'C  1.2 x1013

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Heterogeneous Equilibrium

• A heterogeneous equilibrium is achieved from
a reversible reaction involving reactants and
product that are in different phases.

• The concentration of pure substances, liquid
or solid can be neglected or constant, so its
concentration is not included in the
equilibrium expression, Kc and Kp.

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Heterogeneous equilibrium

 In a heterogeneous system, all substances are in
different phases.

Example :

In the following reaction,

CaCO (s) CaO(s) + CO (g)
3 2

The system involves a gas, CO2 in equilibrium
with two solids , CaCO3 and CaO

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• Concentration of a substance is define as amount
(in mol) per unit volume

• For pure solid or liquid, concentration correspond
to its density.

• Density of solid or liquid is almost constant at any
temperature

• Since the concentration of pure substances i.e
liquid or solid is constant, its concentration is
not included in the equilibrium expression

• So, the equilibrium constant expression, K are :

Kc = [CO2] and Kp = PCO2

44

Exercise 8:

Write the equilibrium constant expression in
terms of concentration and pressure for the
following reactions.

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

ii. FeO (s) + CO (g) Fe (s) + CO2 (g)

iii. CO2 (g) + H2 (g) CO (g) + H2O (l)

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Answer

i. Kc  CO2   Kp  PCO2

CO2   Kp
ii. Kc  CO  PCO2

PCO 

 CO PCO 
CO2  H2 
PCO2 PH2
iii. Kc    Kp 

46

Exercise 9 :

Consider the following equilibrium at 295 K :

NH4HS (s) NH3 (g) + H2S (g)

The partial pressure of each gas is 0.265 atm.
Calculate Kp for the reaction.

47

Answer

  KP  PNH3 PH2S

 0.2650.265

 0.07

48

Exercise 10 :

At equilibrium, the pressure of reacting mixture

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

is 0.105 atm at 350oC. Calculate Kp for this
reaction.

49

Answer

 KP  PCO2

 0.105

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