One mole of a compound AB reacts with one mole of a compound CD according to the equation, A B(g)+CD(g) leftharpoons AD(g)+C B(g). When equilibrium had been established it was found that 3 / 4 mole each 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

Question

One mole of a compound AB reacts with one mole of a compound CD according to the equation, A B(g)+CD(g) leftharpoons AD(g)+C B(g). When equilibrium had been established it was found that 3 / 4 mole each 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 (A) 9 / 16 (B) 1 / 9 (C) 16 / 9 (D) 9

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/0290d8747a635edf8336d59a76209ec3-.png" /> Kc=([A D][C B]/[A B][C D])=(3 / 4 × 3 / 4/1 / 4 × 1 / 4)=9

$9/16$

$1/9$

$16/9$

$9$

$K_{c} = \frac{[ A D ] [ CB ]}{[ A B ] [ C D ]} = \frac{3/4 \times 3/4}{1/4 \times 1/4} = 9$

9/16

1/9

16/9

9

Kc​=[AB][CD][AD][CB]​=1/4×1/43/4×3/4​=9

$9/16$

$1/9$

$16/9$

$9$

$K_{c} = \frac{[ A D ] [ CB ]}{[ A B ] [ C D ]} = \frac{3/4 \times 3/4}{1/4 \times 1/4} = 9$