The equilibrium constant for the reaction, $\frac{1}{2} \mathrm{H}_{2(g)}+\frac{1}{2} \mathrm{I}_{2(g)} \rightleftharpoons \mathrm{HI}_{(g)}$ is $K_c$. Equilibrium constant for the reaction, $2 \mathrm{HI}_{(g)} \rightleftharpoons \mathrm{H}_{2(g)}+\mathrm{I}_{2(g)}$ will be

• A

  $1 / K_c$
• Correct answer
  B

  $1 /\left(K_c\right)^2$
• C

  $2 / K_c$
• D

  $2 /\left(K_c\right)^2$

Solution

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

Hence, the correct answer is option: 'B'.