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Equilibrium constants $K_1$ and $K_2$ for the following equilibria :
$NO(g) + \dfrac{1}{2}O_2 \rightleftharpoons NO_2(g)$; $K_1$ and
$2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$; $K_2$
are related as -
$NO(g) + \dfrac{1}{2}O_2 \rightleftharpoons NO_2(g)$; $K_1$ and
$2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$; $K_2$
are related as -
A
$K_2 = \dfrac{1}{K_1}$
B
$K_2 = \dfrac{K_1}{2}$
C
$K_2 = \dfrac{1}{K_1^2}$
D
$K_2 = K_1^2$
Detailed Solution
$K_1 = \dfrac{[NO_2]}{[NO][O_2]^{1/2}}$
The second reaction is the first reaction reversed and multiplied by 2.
Reversing a reaction gives the reciprocal of the equilibrium constant: $\dfrac{1}{K_1}$
Multiplying the equation by 2 squares the equilibrium constant.
$K_2 = \dfrac{[NO]^2[O_2]}{[NO_2]^2} = \left(\dfrac{1}{K_1}\right)^2$
$K_2 = \dfrac{1}{K_1^2}$
The second reaction is the first reaction reversed and multiplied by 2.
Reversing a reaction gives the reciprocal of the equilibrium constant: $\dfrac{1}{K_1}$
Multiplying the equation by 2 squares the equilibrium constant.
$K_2 = \dfrac{[NO]^2[O_2]}{[NO_2]^2} = \left(\dfrac{1}{K_1}\right)^2$
$K_2 = \dfrac{1}{K_1^2}$