Question

One mole of a compound $AB$ reacts with $1$ 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

Answer 1

$AB +CD \rightleftharpoons AD+CB$
mole at $t=0$
Mole at equilibrium $\underset{\overset{(1-\frac{3}{4})}{0.25}}{1} \,\,\underset{\overset{(1-\frac{3}{4})}{0.25}}{1}\,\,\, \underset{\overset{(\frac{3}{4})}{0.75}}{0} \,\,\underset{\overset{(\frac{3}{4})}{0.75}}{o}$
$K_{c} =\frac{0.75 \times 0.75}{0.25 \times0.25} =\frac{0.5625}{0.0625}=9$