The equilibrium constants of the following are:${\text{N}}_2+3{\text{H}}_2\rightleftharpoons 2{\text{NH}}_3(K_1)$${\text{N}}_2+{\text{O}}_2\rightleftharpoons 2\text{NO}(K_2)$${\text{H}}_2+\frac{1}{2}{\text{O}}_2\to {\text{H}}_2\text{O}(K_3)$ The equilibrium constant ($K$) of the reaction:$2{\text{NH}}_3+\frac{5}{2}{\text{O}}_2\rightleftharpoons 2\text{NO}+3{\text{H}}_2\text{O}$ will be:

We need to manipulate the given reactions to obtain the target reaction:Target: $2{\text{NH}}_3+\frac{5}{2}{\text{O}}_2\rightleftharpoons 2\text{NO}+3{\text{H}}_2\text{O}$ Reverse reaction 1: $2{\text{NH}}_3\rightleftharpoons {\text{N}}_2+3{\text{H}}_2$ ($K=\frac{1}{K_1}$)Use reaction 2 as is: ${\text{N}}_2+{\text{O}}_2\rightleftharpoons 2\text{NO}$ ($K=K_2$)Multiply reaction 3 by 3: $3{\text{H}}_2+\frac{3}{2}{\text{O}}_2\rightleftharpoons 3{\text{H}}_2\text{O}$ ($K=K_3^3$)Adding these three steps:$(2{\text{NH}}_3)+({\text{N}}_2+{\text{O}}_2)+(3{\text{H}}_2+\frac{3}{2}{\text{O}}_2)\rightleftharpoons ({\text{N}}_2+3{\text{H}}_2)+(2\text{NO})+(3{\text{H}}_2\text{O})$ Net reaction: $2{\text{NH}}_3+\frac{5}{2}{\text{O}}_2\rightleftharpoons 2\text{NO}+3{\text{H}}_2\text{O}$ Overall $K=\left(\frac{1}{K_1}\right)\times K_2\times K_3^3=\frac{K_2{K}_3^3}{K_1}$.