0.1 mol of H_2S(g) is kept in a 0.4 litre ves | vedclass

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Question

$\mathop {2{H_2}S(g)}\limits_{0.25\,[1 - \alpha ]}  \leftrightarrow \mathop {2{H_2}}\limits_{0.25\alpha }  + \mathop {{S_2}}\limits_{\frac{{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\mathop {2{H_2}S(g)}\limits_{0.25\,[1 - \alpha ]}  \leftrightarrow \mathop {2{H_2}}\limits_{0.25\alpha }  + \mathop {{S_2}}\limits_{\frac{{0.25\alpha }}{2}} $

Approx : $\alpha<<1$

$10^{-6}=\frac{(0.25 \alpha)^{2}-\left(\frac{0.25 \alpha}{2}ight)}{(0.25)^{2}}=\alpha^{3}(0.125)$

$\alpha=2 	imes 10^{-2}$

$0.1$ $mol$ of $H_2S(g)$ is kept in a $0.4$ litre vessel at $1000\,K$. For the reaction -
$2{H_2}S(g)\,\rightleftharpoons\,2{H_2}(g)\, + \,{S_2}(g)\,;\,{K_c} = {10^{ - 6}}\% $ dissociation of $H_2S$ is.......$\%$

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$0.1$ $mol$ of $H_2S(g)$ is kept in a $0.4$ litre vessel at $1000\,K$. For the reaction - $2{H_2}S(g)\,\rightleftharpoons\,2{H_2}(g)\, + \,{S_2}(g)\,;\,{K_c} = {10^{ - 6}}\% $ dissociation of $H_2S$ is.......$\%$