$\begin{array}{l}\text{ The value of }{\mathrm{K}}_{\mathrm{c}}\text{ is }64\text{ at }800\mathrm{K}\text{ for the reaction }{\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)\text{ . The value }\\ \text{ of }{\mathrm{K}}_{\mathrm{c}}\text{ for the following reaction is: }{\mathrm{NH}}_3\left(\mathrm{g}\right)\rightleftharpoons \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{3}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)\end{array}$

Given  ${\mathrm{N}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_{2\left(\mathrm{g}\right)}\rightleftharpoons 2{\mathrm{NH}}_{3\left(\mathrm{g}\right)}{\mathrm{K}}_{\mathrm{C}}=64$

If we reverse the reaction then equilibrium constant becomes $\frac{1}{{\mathrm{K}}_{\mathrm{c}}}$

$\Rightarrow 2{\mathrm{NH}}_{3\left(\mathrm{g}\right)}\rightleftharpoons {\mathrm{N}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_{2\left(\mathrm{g}\right)}{\mathrm{K}}_{\mathrm{C}}^1=\frac{1}{64}$

If the entire reaction is multiplied by ‘X’ then the equilibrium constant becomes $\left(\frac{1}{K^x}\right)$

$\begin{array}{l}{\mathrm{NH}}_{3\left(\mathrm{g}\right)}\rightleftharpoons \frac{1}{2}{\mathrm{N}}_{2\left(\mathrm{g}\right)}+\frac{3}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)\\ {\mathrm{K}}_{\mathrm{eq}}=\frac{1}{\sqrt{64}}=\frac{1}{8}\end{array}$