The following equilibrium are given: $\begin{array}{ll} \mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2…
The following equilibrium are given:
$\begin{array}{ll}
\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 & \mathrm{~K}_1 \\
\mathrm{~N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} & \mathrm{K}_2 \\
\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{H}_2 \mathrm{O} & \mathrm{K}_3
\end{array}$
The equilibrium constant of the reaction
$2 \mathrm{NH}_3+\frac{5}{2} \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}+3 \mathrm{H}_2 \mathrm{O}$
in terms of $\mathrm{K}_1, \mathrm{~K}_2$ and $\mathrm{K}_3$ is:
$\mathrm{K}_1 \mathrm{~K}_2 \mathrm{~K}_3$
$\mathrm{K}_1 \mathrm{~K}_2 / \mathrm{K}_3$
$\mathrm{K}_1 \mathrm{~K}_3{ }^2 / \mathrm{K}_2$
$\mathrm{K}_2 \mathrm{~K}_3^3 / \mathrm{K}_1$
Solution
The given equation can be obtained by adding eq. I and III and subtracting eq. II.
$\text { equilibrium constant }=\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_3}$