Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\underset{0.2}{{\mathrm{PCl}}_3\left(\mathrm{g}\right)}+\underset{0.1}{{\mathrm{Cl}}_2\left(\mathrm{g}\right)}\underset{}{\rightleftharpoons }\underset{0.4}{{\mathrm{PCl}}_5\left(\mathrm{g}\right)}$

${\mathrm{K}}_{\mathrm{c}}=\frac{0.4}{0.2\times 0.1}=20$

If $0.2$ moles of ${\mathrm{Cl}}_2$ is added. When reactant is added the reaction takes place in forward direction.

$20={\mathrm{K}}_{\mathrm{c}}=\frac{0.4+\mathrm{X}}{\left(0.3-\mathrm{X}\right)\left(0.2-\mathrm{X}\right)}$

$\Rightarrow \left(0.4+\mathrm{X}\right)=20\left(0.3-\mathrm{X}\right)\left(0.2-\mathrm{X}\right)$

$\therefore 0.4+\mathrm{X}=20\left(0.06+{\mathrm{X}}^2-0.5\mathrm{X}\right)$

$0.4+\mathrm{X}=1.2+20{\mathrm{X}}^2-10\mathrm{X}$

$20{\mathrm{X}}^2-11\mathrm{X}+0.8=0$

$\mathrm{X}=\frac{11\pm \sqrt{121-64}}{40}$

$=\frac{11-7.55}{40}\simeq 0.08625$

$\therefore \left({\mathrm{PCl}}_5\right)=0.48625\simeq 48.625\times {10}^{-2}$ or $49\times {10}^{-2}$