One mole of a compound AB reacts with 1 mole of a compound CD according to the equation
$AB + CD ⇌ AD + CB .$
When equilibrium had been established it was found that $\frac{3}{4}$ mole each of reactant 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

$AB + CD ⇌ AD + CD$

mole at t = 0 : 1 1 0 0

at equilibrium: $(1 - \frac{3}{4}) (1 - \frac{3}{4}) ⇌ (\frac{3}{4}) (\frac{3}{4})$

$0.25 0.25 ⇌ 0.75 0.75$

$K_{C} = \frac{0.75 \times 0.75}{0.25 \times 0.25} = \frac{0.5625}{0.0625} = 9$

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