Equivalent amounts of H2 and I2 are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to HI, the Kc at this temperature is

Question

Equivalent amounts of H2 and I2 are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to HI, the Kc at this temperature is (A) 64 (B) 16 (C) 0.25 (D) 4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/b3185f18477933e587a74d212f8863cc-.png" /> Kc=([ HI ]2/[ H 2][ I 2]) =(1.6 × 1.6/0.2 × 0.2) Kc =64

Q. Equivalent amounts of $H_{2}$ and $I_{2}$ are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to $H I$ , the $K_{c}$ at this temperature is

64

16

0.25

4

$K_{c} = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]}$
$= \frac{1.6 \times 1.6}{0.2 \times 0.2}$
$K_{c} = 64$

Q. Equivalent amounts of H2​ and I2​ are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to HI, the Kc​ at this temperature is

64

16

0.25

4

Kc​=[H2​][I2​][HI]2​ =0.2×0.21.6×1.6​ Kc​=64

Q. Equivalent amounts of $H_{2}$ and $I_{2}$ are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to $H I$ , the $K_{c}$ at this temperature is

$K_{c} = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]}$
$= \frac{1.6 \times 1.6}{0.2 \times 0.2}$
$K_{c} = 64$