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Q. At a certain temperature, only $50\%$ $HI$ is dissociated into $H_2$ and $I_2$ at equilibrium. The equilibrium constant is :

JEE MainJEE Main 2014Equilibrium

Solution:

$\begin{array}{l} 2 HI \rightleftharpoons H _2+ I _2 \\ \text { a } 0 \quad 0 \\ a (1-0.5) \quad 0.25 a \quad 0.25 a \\ K =\frac{(0.25 a )(0.25 a )}{(0.5 a )^2} \\ K =0.25 \\ \end{array}$