Question

Calculate the percent dissociation of $H_{2}S\left(g\right)$ if $0.1mole$ of $H_{2}S$ is kept in $0.4L$ vessel at $1000K$. For the reaction, the value of $K_c$ is $1.0\times {10}^{-6}$.

Answer 1

$2H_{2}S\left(s\right)\rightleftharpoons 2H_2\left(g\right)+S_2\left(g\right)$
At equi. $(0.1-x)$ $x$ $x/2$
Molar conc. $\frac{(0.1-x)}{0.4}$ $\frac{x}{0.4}$ $\frac{x}{0.8}$
$K_c=\frac{{\left[H_2\right]}^2\left[S_2\right]}{{\left[H_{2}S\right]}^2}=1.0\times {10}^{-6}$
or $\frac{x^3}{0.8{(0.1-x)}^2}=1.0\times {10}^{-6}$
as $x$ is very small
$0.1-x\to 0.1$
$x=2\times {10}^{-3}$
So, percent dissociation$=\frac{2\times {10}^{-3}}{0.1}\times 100=2.0$