Question

Calculate the percent dissociation of $H_{2}S\left(g\right)$ of 0.1 mole of $H_{2}S$ is kept in 0.4 it. Vessel at 1000K. for the reaction ${2H}_{2}S\left(g\right)\rightleftharpoons {2H}_{2}S\left(g\right)+S_2\left(g\right)$ the value of $K_{c}is{1\times 10}^{-6}$.

Answer 1

Molar conc. of $H_{2}S=\frac{0.1}{0.4}mol{L}^{-1}$
$=0.25mol{L}^{-1}$
Suppose degree of dissociation of $H_{2}S=\alpha$ Then

	$H_{2}S$	$\rightleftharpoons 2H^{+}$	$+S^{2-}$
Initial conc.	$0.25M$
Conc. at eqm	$0.25(1-\alpha )$	$2\times 0.25\alpha$ $=0.5\alpha$	$00.25\alpha$

$K_c=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}$
${10}^{-6}=\frac{\left(0.5\alpha {)}^2\right(0.25\alpha )}{0.25(1-\alpha )}=\frac{0.25{\alpha }^2}{1-\alpha }$
Neglecting $\alpha$ in comparison to $1$, we get
${10}^{-6}=0.25{\alpha }^2$ or ${\alpha }^2=4\times {10}^{-6}$
or $\alpha =2\times {10}^{-3}=0.002$
$\therefore \%$ age dissociation $=0.2\%$