The equilibrium constant for the given reaction is 100. $N_2\left(g\right)+{20}_2\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ What is the equilibrium constant for the reaction given below$N{O}_2\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+O_2\left(g\right)$

$N_2+{20}_2\rightleftharpoons 2N{O}_2$

           $K_1=\frac{{\left[N{O}_2\right]}^2}{\left[N_2\right]{\left[O_2\right]}^2}$

or     $100=\frac{{\left[N{O}_2\right]}^2}{\left[N_2\right]{\left[O_2\right]}^2}$

Again, $\left[N{O}_2\right]\rightleftharpoons \frac{1}{2}{N}_2+O_2$

                   $K_2=\frac{{\left[N_2\right]}^{\frac{1}{2}}\left[O_2\right]}{\left[N{O}_2\right]}$

or                   $K_2^2=\frac{\left[N_2\right]{\left[O_2\right]}^2}{{\left[N{O}_2\right]}^2}$

Eqs. $\left(i\right)\times \left(ii\right),$ we get

               $100\times K_2^2=1$

or      $K_2^2=\frac{1}{100}or{K}_2=\frac{1}{10}=0.1$