At $$500\,K,$$  equilibrium constant, $${K_c},$$  for the following reaction is 5.
$$\frac{1}{2}{H_{2\left( g \right)}} + \frac{1}{2}{I_{2\left( g \right)}} \rightleftharpoons H{I_{\left( g \right)}}$$
What would be the equilibrium constant $${K_c}$$  for the reaction : $$2H{I_{\left( g \right)}} \rightleftharpoons {H_{2\left( g \right)}} + {I_{2\left( g \right)}}?$$

Solution :
$$\eqalign{ & \frac{1}{2}{H_{2\left( g \right)}} + \frac{1}{2}{I_{2\left( g \right)}} \rightleftharpoons H{I_{\left( g \right)}};{K_c} = 5...\left( {\text{i}} \right) \cr & {\text{Multiply}}\,\,{\text{eqn }}\left( {\text{i}} \right){\text{ by 2,}} \cr & {{\text{H}}_{2\left( g \right)}} + {I_{2\left( g \right)}} \rightleftharpoons 2H{I_{\left( g \right)}};{K_c} = {\left( 5 \right)^2}...\left( {{\text{ii}}} \right) \cr & {\text{Now, reverse the reaction}} \cr} $$
$$2H{I_{\left( g \right)}} \rightleftharpoons {H_{2\left( g \right)}} + {I_{2\left( g \right)}};$$     $${K_c} = \frac{1}{{{{\left( 5 \right)}^2}}} = \frac{1}{{25}} = 0.04$$