Calculate the percent dissociation of ${\mathrm{H}}_{2_{ }}{\mathrm{S}}_{\left(\mathrm{g}\right)}$if 0.1 mole of is kept in 0.4 litter vessel at 1000 K for the reaction, The value of ${\mathrm{K}}_{\mathrm{c}}\text{ is }1.0\times {10}^{-6}$

                           $2{\mathrm{H}}_2{\mathrm{S}}_{\left(g\right)}\rightleftharpoons 2{\mathrm{H}}_{2\left(g\right)}+{\mathrm{S}}_{2\left(g\right)}$

At equilibrium  $(0.1-\mathrm{x})$                 x          x/2

                            $\frac{(0.1-\mathrm{x})}{0.4}\frac{\mathrm{x}}{0.4}\frac{\mathrm{x}}{0.8}$
Molar conc.${\mathrm{K}}_{\mathrm{c}}=\frac{{\left[{\mathrm{H}}_2\right]}^2\left[{\mathrm{S}}_2\right]}{{\left[{\mathrm{H}}_2\mathrm{S}\right]}^2}=\frac{{\left(\frac{\mathrm{x}}{0.4}\right)}^2\left(\frac{\mathrm{x}}{0.8}\right)}{{\left(\frac{0.1-\mathrm{x}}{0.4}\right)}^2}=1.0\times {10}^{-6}$
or $\frac{{\mathrm{x}}^3}{0.8(0.1-\mathrm{x}{)}^2}=1.0\times {10}^{-6}$As x is very small; $0.1-x\to 0.1$
$\frac{{\mathrm{x}}^3}{0.8\times (0.1{)}^2}=1.0\times {10}^{-6}{\mathrm{x}}^3=8\times {10}^{-9}\text{ Or }\mathrm{x}=2\times {10}^{-3}$

So, percent dissociation =$\frac{2\times {10}^{-3}}{0.1}\times 100=2.0$