The equilibrium constant for a reaction $A + 2 B ⇌ 2 C$ is 40. The equilibrium constant for reaction $C ⇌ B + 1 / 2 A$ is :

1 / 40

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(1 / 40)^{1 / 2}

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(1 / 40)^{2}

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40

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Solution

The correct option is A $(1 / 40)^{1 / 2}$
$A + 2 B ⇌ 2 C$ , .....(i) $K = 40$
and for $C ⇌ B + \frac{1}{2} A$ , ......(i) $K_{1} = ?$
$K = \frac{1}{\sqrt{K}} = \frac{1}{(40)^{1 / 2}}$
Since equation (2) is obtained by reversing and dividing by 2, the equation (1).

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2 A + B ⇌ 2 C, K = x

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