$\begin{array}{l}
\text {Given, } \\
\begin{aligned}
\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; K_{1} &=\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}} \\
\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons & 2 \mathrm{NO} ; K_{2}=\frac{[\mathrm{NO}]^{2}}{\left[\mathrm{N}_{2}\right]\left[\mathrm{O}_{2}\right]} \\
\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons & \mathrm{H}_{2} \mathrm{O} ; K_{3}=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]}{\left[\mathrm{H}_{2}\right]\left[\mathrm{O}_{2}\right]^{1 / 2}}
\end{aligned}
\end{array}$
The chemical equation for the oxidation of 2 mol of
$\mathrm{NH}_{3}$ to give NO is
$2 \mathrm{NH}_{3}+\frac{5}{2} \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}+3 \mathrm{H}_{2} \mathrm{O}$
(ii) and To get the equation (iv) from equation (i),
(iii) following steps are followed:
Reversing equation (i), we get
$2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{N}_{2}+3 \mathrm{H}_{2} \text { so } K^{\prime}=\frac{1}{K_{1}}$
Multiplying cquation (iii) by 3, we get
$3 \mathrm{H}_{2}+\frac{3}{2} \mathrm{O}_{2} \rightleftharpoons 3 \mathrm{H}_{2} \mathrm{O} \mathrm{so} \mathrm{K}^{\prime}=K_{3}^{3}$
Adding equation (ii) and (vi)
$\mathrm{N}_{2}+3 \mathrm{H}_{2}+\frac{5}{2} \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}+3 \mathrm{H}_{2} \mathrm{O}$
so, $K^{\prime}=K_{2} \cdot K_{3}^{3}$
On combining (v) and (vii), we get the required equation having equilibrium constant $K^{\prime}=K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$