If degree of dissociation is 0.5 at equilibrium, then the equilibrium constant $K_{c}$ for the given reaction is
$2 H I ⇌ H_{2} + I_{2}$

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Solution

$2 H I ⇋ H_{2} + I_{2}$

Let initial concentration $C$ $0$ $0$

Final concentration $C (1 - α)$ $C α$ $C α$

$K_{C} = \frac{(C α) (C α)}{[C (1 - α)]^{2}} = \frac{c α^{2}}{(1 - α)^{2}} = \frac{(0.5)^{2}}{(0.5)^{2}} = 1$

$⟹ K_{C}$ (Equillibrium constant) $= 1$

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