The equilibrium constant for the reaction, (1)/(2) H_ 2(g) +(1)/(2)...

The given reaction is, $\frac{1}{2} \mathrm{H}_{2(g)}+\frac{1}{2}$\mathrm{I}$_{2(g)} \rightleftharpoons \mathrm{HI}_{(g)}$
Hence, K_c=\frac{[\mathrm{HI}]}{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}
Now, reverse the eqn. (i) and multiply by 2 , we get $2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2(g)}+$\mathrm{I}$_{2(g)}$ Hence, K_c^{\prime}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}
Equating equations (ii) and (iii), we get $K_c^{\prime}=\frac{1}{\left(K_c\right)^2}