At  298 K 
 $$\begin{gathered} N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right),K_1=4\times {10}^5 \\ {N}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right),K_2=1.6\times {10}^{12} \\ {H}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons H_{2}O\left(g\right),K_3=1.0\times {10}^{-13} \end{gathered}$$
 Based on above equilibria, the equilibrium constant of the reaction,
 $2N{H}_3\left(g\right)+\frac{5}{2}{O}_2\left(g\right)\rightleftharpoons 2NO\left(g\right)=1.0\times {10}^{-13}$
 Is _______________ $\times {10}^{-33}$ (Nearest integer)
 

$$\begin{gathered} N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right),K_1=4\times {10}^5\mathrm{...........}\left(i\right) \\ {N}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right),K_2=1.6\times {10}^{12}\mathrm{.........}\left(ii\right) \\ {H}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons H_{2}O\left(g\right),K_3=1.0\times {10}^{-13}\mathrm{..........}\left(iii\right) \end{gathered}$$

By applying  $\left(ii\right)+3\times \left(iii\right)-\left(i\right),weget$

$$\begin{gathered} 2N{H}_3\left(g\right)+\frac{5}{2}{O}_2\left(g\right)\rightleftharpoons 2NO\left(g\right)+3H_{2}O\left(g\right) \\ {k}_{eq}=\frac{k_2\times k_3^3}{k_1}=\frac{1.6\times {10}^{12}\times {\left({10}^{-13}\right)}^3}{4\times {10}^5} \\ =\frac{1.6}{4}\times {10}^{-32}=4\times {10}^{-33} \end{gathered}$$