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For the reaction $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$, the equilibrium constant is $K_1$. The equilibrium constant is $K_2$ for the reaction $2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$. What is K for the reaction $NO_2(g) \rightleftharpoons \frac{1}{2}N_2(g) + O_2(g)$?
A
$\frac{1}{(K_1K_2)}$
B
$\frac{1}{(2K_1K_2)}$
C
$\frac{1}{(4K_1K_2)}$
D
$\left[\frac{1}{K_1K_2}\right]^{1/2}$
Detailed Solution
(i) $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$; $K_1$
(ii) $2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$; $K_2$
Adding (i) and (ii): $N_2(g) + 2O_2(g) \rightleftharpoons 2NO_2(g)$; when reactions are added their constants multiply, so $K = K_1K_2$
Reversing this reaction: $2NO_2(g) \rightleftharpoons N_2(g) + 2O_2(g)$; the constant becomes the reciprocal, $\frac{1}{K_1K_2}$
Dividing the equation by 2: $NO_2(g) \rightleftharpoons \frac{1}{2}N_2(g) + O_2(g)$; the constant is raised to the power $\frac{1}{2}$
$K = \left[\frac{1}{K_1K_2}\right]^{1/2}$
(ii) $2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$; $K_2$
Adding (i) and (ii): $N_2(g) + 2O_2(g) \rightleftharpoons 2NO_2(g)$; when reactions are added their constants multiply, so $K = K_1K_2$
Reversing this reaction: $2NO_2(g) \rightleftharpoons N_2(g) + 2O_2(g)$; the constant becomes the reciprocal, $\frac{1}{K_1K_2}$
Dividing the equation by 2: $NO_2(g) \rightleftharpoons \frac{1}{2}N_2(g) + O_2(g)$; the constant is raised to the power $\frac{1}{2}$
$K = \left[\frac{1}{K_1K_2}\right]^{1/2}$