Question

Calculate the percent dissociation of h2s (g), if 0.1 mole of h2s is kept in 0.4 litre vessel at 1000k. for the reaction 2h2s(g) ↔ 2h2(g) + s2(g) the value of kc is 1.0 x 10-6 a. 5 b. 3 c. 2 d. 4

Answer 1

OPTION B. 3 IS CORRECT. .

Answer 2

Answer: The percent dissociation of $H_2S(g)$ is  $4.75\times 10^{-3}$

Explanation:

$\alpha$= ?

$concentration=\frac{\text{moles}}{\text{Volume in Litres}}=\frac{0.1}{0.4L}=0.25mol/L$

            $H_2S\rightleftharpoons H_2+\frac{1}{2}S_2$

initially           c                   0         0

eq'm     $c- c\alpha$  $c\alpha \frac{c\alpha}{2}$

So eq'm constant will be:

$K_{c}=\frac{{c\alpha}\times \frac{c\alpha}{2}^{1/2}}{c-c\alpha}$

$1.0\times 10^{-6}=\frac{{0.25\alpha}\times \frac{0.25\alpha}{2}^{1/2}}{0.25-0.25\alpha}$

$\alpha=4.75\times 10^{-5}$

$\ %\alpha=4.75\times 10^{-3}$