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Chemical Equilibrium
Definition: In Chemical reaction, the concentration of reactant and products shows no net change over the time is called as chemical equilibrium.

Types of Chemical equilibrium:

1.  Homogenous equilibrium
All reactants and products are in same phase and can not be separated
N2(g) + 3H2(g)  → 2NH3(g)

2.  Heterogeneous equilibrium
Reactants and products are not in homogenous phase and can be separated.
NH3(g) + H2O(l) → NH4OH(l)

Criteria for chemical equilibrium:
  • Reaction must be take place in closed vessel
  • Reaction must be reversible
  • Temperature, pressure and concentration of reactants remain constant.

Law of chemical equilibrium:
The measurement of equilibrium concentration is expressed as equilibrium constant. At equilibrium the rates of the forward and reverse processes are identical, but the rate constants are generally different. To determine the equilibrium constant, first consider the simple reversible reaction at constant temperature.

A  +  B → C  +  D

According to the law of active mass the rate of chemical reaction is proportional to the product of the concentration in moles in proper power of the reactants at a given temperature and pressure.

The rate of conversion of A and B is proportional to their concentration and denoted as r1

r1 = k1 × [A] × [B]

k1 is a constant known as rate constant or rate coefficient and square brackets denote the concentrations (mol/liter) of the substances enclosed within the brackets

Similarly, the rate of conversion of C and D is as below,

r2 = k2 × [C] × [D]

At equilibrium, the two rate constants will be equal, r1 = r2

k1 × [A] × [B] = k2 × [C] × [D]
k1/k2 = [C] × [D]  /  [A] × [B] = Keq

Keq is the equilibrium constant at given temperature.

Keq = [C] × [D]  /  [A] × [B]

This equation is called equation of law of chemical equilibrium. At equilibrium, the concentration of reactants is expressed as moles/lit so Keq = Kc and if it expressed as partial pressure then Keq = Kp.

Calculations:

Example 1: Two moles of hydrogen reacts with 2 moles of iodine to form 2 moles of hydrogen iodide in a 10 lit vessel at 700 K temperature. Equilibrium is extended to 60%, calculate the Kc.

H2(g) + I2(g) ↔ 2HI(g)

Reason: 
  Concentration of Reactants Concentration of products
[H] [I] [HI]
At beginning 2 moles 2 moles 0 moles
At 60% equilibrium 1.2 moles 1.2 moles 2.4 moles
Unsused 0.8 moles 0.8 moles 2.4 moles
Moles/lit at equilibrium 0.8/10 moles/lit 0.8/10 moles/lit 2.4/10 moles/lit
 
Volume = 10 lit

At equilibrium

Kc = [HI]2 / [H][I]
Kc = (0.24)2 / 0.08 × 0.08 = 9.0 moles/lit


Example 2: The value of Kc is 0.04 at 400K for the reaction given below. The concentration of hydrogen is 0.6 moles/lit and iodine is 0.8 moles/lit. Calculate the concentration of hydrogen iodide.
 
H2(g) + I2(g) ↔ 2HI(g)

Reason:
Kc = 0.04 moles/lit
[H] = 0.6 moles/lit
[I] = 0.8 moles/lit

Kc = [HI]2 / [H][I]
0.04 = [HI]2 / (0.6 × 0.8)
[HI]2 = 0.04 × 0.6 × 0.8 = 0.0192
[HI] = (0.0192)1/2 = 0.0138 moles/ lit


Example 3: Find out the concentration of sulphur dioxide and sulphur trioxide at equilibrium in a 1 lit vessel at 400K, taking one mole of each. The value of Kc is 25 for the reaction given below.

SO2(g) + NO2(g) ↔ SO3(g) + NO(g)

Reason:
 
  Concentration of Reactants Concentration of products
  [SO2] [NO2] [SO3] [NO]
At beginning 1 moles 1 moles 1 moles 1 moles
At equilibrium X moles X moles X moles X moles
Unsused 1-X moles 1-X moles 1+ X moles 1+ X moles
Moles/lit at equilibrium
(Volume = 1 lit)
 1-X
moles/lit		 1-X
moles/lit		 1+ X moles/lit 1+ X moles/lit
 
Kc = [NO][SO3] / [SO2] [NO2]
25 = {(1 + X) (1 + X)} / {(1 - X) (1 - X)}
25 = (1 + X)2 / (1 - X)2
5 = (1 + X) / (1 - X)
5 – 5X = 1 + X
X = 0.66 moles/lit

[SO3] = 1 + X = 1.66 moles/lit
[SO2] = 1 - X = 0.33 moles/lit
              
Example 4: Calculate the Kp for the reaction where in solid ammonium hydrogen sulphide was heated at a temperature 400K in which the total pressure inside the closed system was 0.16 atmosphere at the equilibrium.
 
NH4SH(s) ↔ NH3(g) + H2S(g)

Reason:
Suppose we consider P NH3  = X atmosphere
At equilibrium P H2S = P NH3  = X atmosphere

P total = P H2S + P NH3  
0.16 = X + X
0.16 = 2X
X = 0.08 atmosphere

Kp = P H2S x P NH3  
= 0.08 x 0.08
= 0.0064 atmosphere2

Example 5: 4 moles of phosphorous on heating give 2 moles of chlorine at 300K in a 500ml container. Calculate the Kc for reaction given below.

PCl5(g) → PCl3(g) + Cl2(g)

Reason:

  Concentration of Reactants Concentration of products
  [PCl5] [PCl3] [CL2]
At beginning 4 moles 0 moles 0 moles
At equilibrium 2 moles 2 moles 2 moles
Unsused 2 moles 2 moles 2 moles
Moles/lit at equilibrium
(Volume = 500 ml)		 2/0.5 = 4 moles/lit 4 moles 4 moles

Kc = [PCl3] [CL2] / [PCl5]
Kc = 4 × 4 / 4
Kc =  4 moles/lit
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