1 mole of HI is heated in a closed container of capacity of 2 L. At equilibrium half a mole of HI is dissociated. The equilibrium constant of the reaction is

Question

1 mole of HI is heated in a closed container of capacity of 2 L. At equilibrium half a mole of HI is dissociated. The equilibrium constant of the reaction is (A) 0.25 (B) 0.35 (C) 1 (D) 0.5

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/851acc16233b2e126da5b8753315aa37-.png" /> K C ([ H 2][ I 2]/[ HI ]2) K C = ((0.25 × 0.25/2)/(0.5 × 0.5)2) =(1/4)=0.25

Q. $1$ mole of $H I$ is heated in a closed container of capacity of $2 L$ . At equilibrium half a mole of $H I$ is dissociated. The equilibrium constant of the reaction is

0.25

0.35

1

0.5

$K_{C} \frac{[ H _{2} ] [ I _{2} ]}{[ H I ] ^{2}}$
$K_{C} = \frac{\frac{0.25 \times 0.25}{2}}{\frac{0.5 \times 0.5}{2}}$
$= \frac{1}{4} = 0.25$

Q. 1 mole of HI is heated in a closed container of capacity of 2L. At equilibrium half a mole of HI is dissociated. The equilibrium constant of the reaction is

0.25

0.35

1

0.5

KC​[HI]2[H2​][I2​]​ KC​=20.5×0.5​20.25×0.25​​ =41​=0.25

Q. $1$ mole of $H I$ is heated in a closed container of capacity of $2 L$ . At equilibrium half a mole of $H I$ is dissociated. The equilibrium constant of the reaction is

$K_{C} \frac{[ H _{2} ] [ I _{2} ]}{[ H I ] ^{2}}$
$K_{C} = \frac{\frac{0.25 \times 0.25}{2}}{\frac{0.5 \times 0.5}{2}}$
$= \frac{1}{4} = 0.25$