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The equilibrium constants of the following are:
$N_2 + 3H_2 \rightleftharpoons 2NH_3$, $K_1$;
$N_2 + O_2 \rightleftharpoons 2NO$, $K_2$;
$H_2 + \frac{1}{2}O_2 \rightarrow H_2O$, $K_3$.
The equilibrium constant (K) of the reaction:
$2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$, will be
$N_2 + 3H_2 \rightleftharpoons 2NH_3$, $K_1$;
$N_2 + O_2 \rightleftharpoons 2NO$, $K_2$;
$H_2 + \frac{1}{2}O_2 \rightarrow H_2O$, $K_3$.
The equilibrium constant (K) of the reaction:
$2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$, will be
Explanation
Adding reactions multiplies K; reversing inverts it; multiplying by n raises to power n.
Detailed Solution
(I) $N_2 + 3H_2 \rightleftharpoons 2NH_3$; $K_1 = \frac{[NH_3]^2}{[N_2][H_2]^3}$
(II) $N_2 + O_2 \rightleftharpoons 2NO$; $K_2 = \frac{[NO]^2}{[N_2][O_2]}$
(III) $H_2 + \frac{1}{2}O_2 \rightarrow H_2O$; $K_3 = \frac{[H_2O]}{[H_2][O_2]^{1/2}}$
(II + 3 × III – I) gives $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$
$\therefore K = K_2\times K_3^3/K_1$
(II) $N_2 + O_2 \rightleftharpoons 2NO$; $K_2 = \frac{[NO]^2}{[N_2][O_2]}$
(III) $H_2 + \frac{1}{2}O_2 \rightarrow H_2O$; $K_3 = \frac{[H_2O]}{[H_2][O_2]^{1/2}}$
(II + 3 × III – I) gives $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$
$\therefore K = K_2\times K_3^3/K_1$