One mole of a compound AB reacts with one mole of a compound CD a… - Vidyadip

MCQ

One mole of a compound $AB$ reacts with one mole of a compound $CD$ according to the equation $AB + CD$ $\rightleftharpoons$ $AD + CB$. 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

A
$\frac{9}{{16}}$
B
$\frac{1}{9}$
C
$\frac{{16}}{9}$
✓
$9$

✓

Answer

Correct option: D.

$9$

(d) $AB$ + $CD$ $ \rightleftharpoons $ $AD$ + $CD$

mole at $t = 0$ $1$ $1$ $0$ $0$

Mole at equilibrium $\left( {1 - \frac{3}{4}} \right)$ $\left( {1 - \frac{3}{4}} \right)$ $ \rightleftharpoons $ $\left( {\frac{3}{4}} \right)$ $\left( {\frac{3}{4}} \right)$

$0.25$ $0.25$ $ \rightleftharpoons $ $0.75$ $0.75$

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

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