Chemistry - Chemical Equilibrium Question with Solution | TestHub

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 $K_c^{\prime }=\frac{1}{{\left(K_c\right)}^2}$