If equilibrium constant for the reaction $A_{2} + B_{2} ⇌ 2 A B$ is $K$ , then for the backward reaction $A B ⇌ \frac{1}{2} A_{2} + \frac{1}{2} B_{2}$ , the equilibrium constant $K^{'}$ is $\frac{1}{K}$ .

True

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Solution

The correct option is B False
If equilibrium constant for the reaction $A_{2} + B_{2} ⇌ 2 A B$ is K, then for the backward reaction $A B ⇌ 1 / 2 A_{2} + 1 / 2 B_{2}$ the equilibrium constant K' is $\frac{1}{\sqrt{K}}$ .
$K = \frac{[A B^{2}]}{[A_{2}] [B_{2}]}$
$K^{'} = \frac{[A_{2}]^{1 / 2} [B_{2}]^{1 / 2}}{A B} = \frac{1}{\sqrt{\frac{[A B^{2}]}{[A_{2}] [B_{2}]}}} = \frac{1}{\sqrt{K}}$

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