One mole of a compound AB reacts with one mole of a compound CD according to the equation
$large A{B_{(g)}}, + ,C{D_{(g)}}, rightleftharpoons ,A{D_{(g)}}, + ,C{B_{(g)}}$
. When equilibrium had been established it was found that $\frac{3}{4}$ mole of reactants AB and CD had been converted to AD and CB. There is no change in volume. The equilibrium constant for the reaction is

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$\frac{9}{16}$

$\frac{1}{9}$

$\frac{16}{9}$

answer is D.

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Detailed Solution

$\overset{1 mole}{AB} + \overset{1 mole}{CD} ⇌ AD + BC$

At equilibrium $\frac{3}{4}$ mole of AB and CD is reacted.

[AB]_eq = 1 - $\frac{3}{4} = \frac{1}{4}$ = [CD]_eq

[AD]_eq = [BC]_eq = $\frac{3}{4}$

K_C= $\frac{[AD] [BC]}{[AB] [CD]} = \frac{\frac{3}{4} \times \frac{3}{4}}{\frac{1}{4} \times \frac{1}{4}} = 9$

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