The equilibrium composition for the reaction PCl _ 3 + Cl _ 2 ⇌ _...

$\mathop{{PCl}_{3}\left(g\right)}_{0 . 2} + \mathop{{Cl}_{2}\left(g\right)}_{0 . 1}\mathop{⇌}_{}\mathop{{PCl}_{5}\left(g\right)}_{0 . 4}$

${K}_{c} = \dfrac{0 . 4}{0 . 2 \times 0 . 1} = 20$

If $0 . 2$ moles of ${Cl}_{2}$ is added. When reactant is added the reaction takes place in forward direction.

$20 = {K}_{c} = \dfrac{0 . 4 + X}{\left(0 . 3 - X\right)\left(0 . 2 - X\right)}$

$⇒ \left(0 . 4 + X\right) = 20\left(0 . 3 - X\right)\left(0 . 2 - X\right)$

$∴ 0 . 4 + X = 20\left(0 . 06 + {X}^{2} - 0 . 5X\right)$

$0 . 4 + X = 1 . 2 + 20{X}^{2} - 10X$

$20{X}^{2} - 11X + 0 . 8 = 0$

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

$= \dfrac{11 - 7 . 55}{40} ≃ 0 . 08625$

$∴ \left({PCl}_{5}\right) = 0 . 48625 ≃ 48 . 625 \times {10}^{- 2}$ or $49 \times {10}^{- 2}$