1. We need the equilibrium constant for $\mathrm{2NH_3(g)+\frac{5}{2}O_2(g)\rightleftharpoons 2NO(g)+3H_2O(g)}$

Given:

$\mathrm{N_2+3H_2\rightleftharpoons 2NH_3},\qquad K_1=4\times 10^5$ $\mathrm{N_2+O_2\rightleftharpoons 2NO},\qquad K_2=1.6\times 10^{12}$ $\mathrm{H_2+\frac12 O_2\rightleftharpoons H_2O},\qquad K_3=1.0\times 10^{-13}$

2. Construct the required reaction.

First, reverse reaction (1): $\mathrm{2NH_3\rightleftharpoons N_2+3H_2}$ So its equilibrium constant is $K_1' = \frac{1}{K_1}$

Now add reaction (2): $\mathrm{N_2+O_2\rightleftharpoons 2NO}$ with constant $K_2$.

Also take reaction (3) three times: $\mathrm{3H_2+\frac{3}{2}O_2\rightleftharpoons 3H_2O}$ Its equilibrium constant becomes $K_3^3$

3. Add these three reactions:

• $\mathrm{2NH_3\rightleftharpoons N_2+3H_2}$
• $\mathrm{N_2+O_2\rightleftharpoons 2NO}$
• $\mathrm{3H_2+\frac32 O_2\rightleftharpoons 3H_2O}$

Cancel $\mathrm{N_2}$ and $\mathrm{3H_2}$ from both sides:

Left side: $\mathrm{2NH_3 + O_2 + \frac32 O_2 = 2NH_3 + \frac52 O_2}$

Right side: $\mathrm{2NO + 3H_2O}$

Thus we get exactly: $\mathrm{2NH_3+\frac52 O_2\rightleftharpoons 2NO+3H_2O}$

4. Therefore, $K = \frac{1}{K_1}\cdot K_2 \cdot K_3^3$

Substitute values: $K = \frac{1}{4\times 10^5}\times (1.6\times 10^{12})\times (1.0\times 10^{-13})^3$

Now, $(1.0\times 10^{-13})^3 = 1.0\times 10^{-39}$

So, $K = \frac{1.6\times 10^{12}\times 10^{-39}}{4\times 10^5}$ $K = \frac{1.6}{4}\times 10^{12-39-5}$ $K = 0.4\times 10^{-32}$ $K = 4\times 10^{-33}$

5. Hence the required number in $\underline{\hspace{2cm}}\times 10^{-33}$ is $\boxed{4}$