Calculate the percentage dissociation of $H_{2}S\left(g\right)$ if $0.1$ mole of $H_{2}S$ is kept in $0.4$ litre vessel at $1000$ K for the reaction,
$2H_{2}S\left(g\right)\rightleftharpoons 2H_2\left(g\right)+S_2\left(g\right)$
The value of $K_c$ is $1.0\times {10}^{-6}$.

Given reaction is:
$2H_{2}S\left(g\right)\rightleftharpoons 2H_2\left(g\right)+S_2\left(g\right)$
$$\begin{array}{cccccc}\text{At equilibrium}& (0.1-x)& & x& & \frac{x}{2}\\ \text{Molar conc.}& \frac{(0.1-x)}{0.4}& & \frac{x}{0.4}& & \frac{x}{0.8}\end{array}$$
$K_c=\frac{\left[H_2{]}^2\right[S_2]}{[H_{2}S{]}^2}=\frac{{\left(\frac{x}{0.4}\right)}^2\left(\frac{x}{0.8}\right)}{{\left(\frac{0.1-x}{0.4}\right)}^2}\Rightarrow 1.0\times {10}^{-6}=\frac{x^3}{0.8(0.1-x{)}^2}$
as x is very small: $0.1-x\to 0.1\frac{x^3}{0.8\times (0.1{)}^2}=1.0\times {10}^{-6}\Rightarrow x^3=8\times {10}^{-9}\Rightarrow x=2\times {10}^{-3}\therefore \alpha =\frac{x}{0.1}=\frac{2\times {10}^{-3}}{0.1}$
So, percentage dissociation $=\frac{2\times {10}^{-3}}{0.1}\times 100=2.0$