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(B) The target reaction is the oxidation of $2 	ext{ mol}$ of $NH_{3}$ to $NO$:$2 NH_{3} + \frac{5}{2} O_{2} ightleftharpoons 2 NO + 3 H_{2} O$Give

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$K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$

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(B) The target reaction is the oxidation of $2 	ext{ mol}$ of $NH_{3}$ to $NO$:$2 NH_{3} + \frac{5}{2} O_{2} ightleftharpoons 2 NO + 3 H_{2} O$Given equations:$(i) N_{2} + 3 H_{2} ightleftharpoons 2 NH_{3} ; K_{1}$$(ii) N_{2} + O_{2} ightleftharpoons 2 NO ; K_{2}$$(iii) H_{2} + \frac{1}{2} O_{2} ightleftharpoons H_{2} O ; K_{3}$To obtain the target reaction:$1$. Reverse equation $(i)$: $2 NH_{3} ightleftharpoons N_{2} + 3 H_{2} ; K' = \frac{1}{K_{1}}$$2$. Keep equation $(ii)$ as it is: $N_{2} + O_{2} ightleftharpoons 2 NO ; K_{2}$$3$. Multiply equation $(iii)$ by $3$: $3 H_{2} + \frac{3}{2} O_{2} ightleftharpoons 3 H_{2} O ; K'' = K_{3}^{3}$Adding these three equations:$(2 NH_{3} + N_{2} + O_{2} + 3 H_{2} + \frac{3}{2} O_{2}) ightleftharpoons (N_{2} + 3 H_{2} + 2 NO + 3 H_{2} O)$Simplifying gives: $2 NH_{3} + \frac{5}{2} O_{2} ightleftharpoons 2 NO + 3 H_{2} O$The equilibrium constant $K$ is:$K = K' \cdot K_{2} \cdot K'' = \frac{1}{K_{1}} \cdot K_{2} \cdot K_{3}^{3} = K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$

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The following equilibrium constants are given:$N_{2} + 3 H_{2} \rightleftharpoons 2 NH_{3} ; K_{1}$$N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$$H_{2} + \frac{1}{2} O_{2} \rightleftharpoons H_{2} O ; K_{3}$The equilibrium constant for the oxidation of $2 \text{ mole}$ of $NH_{3}$ to give $NO$ is