Consider the reactions (i) 2CO (g) + 2H2O (g) \underset{}{\rightleftharpoons} 2CO2 + 2H2(g) ; K1 (ii) CH4 (g) + H2O (g) \underset{}{\rightleftharpoons} CO(g) + 3H2(g) ; K2 (iii) CH4 (g) + 2H2O (g) \underset{}{\rightleftharpoons} CO2(g) + 4H2(g) ; K3 which of the following relation is correct?

Engineering

Chemistry

Equilibrium Constant and Mass Action Law

Question

Consider the reactions

(i) 2CO (g) + 2H₂O (g) $\underset{}{⇌}$ 2CO₂ + 2H₂(g) ; K₁
(ii) CH₄ (g) + H₂O (g) $\underset{}{⇌}$ CO(g) + 3H₂(g) ; K₂
(iii) CH₄ (g) + 2H₂O (g) $\underset{}{⇌}$ CO₂(g) + 4H₂(g) ; K₃
which of the following relation is correct?

$K_{3} = \frac{K_{1}}{K_{2}}$

$K_{3} = \frac{K_{1}^{2}}{K_{2}^{2}}$

$K_{3} = \sqrt{K_{1}} \times K_{2}$

K₃ = K₁K₂

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To find the relationship between the equilibrium constants, observe how the reactions combine. Reaction (iii) can be obtained by reversing reaction (i) and adding it to reaction (ii).

First, reverse (i): 2CO₂ + 2H₂ ⇌ 2CO + 2H₂O; its constant is 1/K₁.

Then add it to (ii): CH₄ + H₂O ⇌ CO + 3H₂; constant is K₂.

The sum is: CH₄ + 2H₂O ⇌ CO₂ + 4H₂, which is reaction (iii).

When adding reactions, their equilibrium constants multiply. Therefore, K₃ = (1/K₁) × K₂ = K₂/K₁.

Final Answer: $K_{3} = \frac{K_{2}}{K_{1}}$