The equilibrium composition for the reaction ${\mathrm{PCl}}_3+{\mathrm{Cl}}_2\rightleftharpoons {\mathrm{PCl}}_5$ at 298 K is given below ${\left[{\mathrm{PCl}}_3\right]}_{eq}=0.2{\mathrm{molL}}^{-1},{\left[{\mathrm{Cl}}_2\right]}_{eq}=0.1{\mathrm{molL}}^{-1}{\left[{\mathrm{PCl}}_5\right]}_{eq}=0.40{\mathrm{molL}}^{-1}$
If 0.2 mol of Cl₂ is added at the same temperature, the equilibrium concentrations of PCl₅ is _________x 10^-2 mol L^-1 Given: K_c for the reaction at 298 K is 20

$$\begin{gathered} \begin{array}{l} {\mathrm{PCl}}_3+{\mathrm{Cl}}_2\leftrightharpoons {\mathrm{PCl}}_5 \\ 0.2 0.1 0.4\\ 0.2-x 0.3-x 0.4+x\\ 20=\frac{(0.4+\mathrm{x})}{(0.2-\mathrm{x})(0.3-\mathrm{x})}\\ 20=\frac{0.4+\mathrm{x}}{0.06-0.2\mathrm{x}-0.3\mathrm{x}+{\mathrm{x}}^2}\\ 20=\frac{0.4+\mathrm{x}}{0.06-0.5\mathrm{x}+{\mathrm{x}}^2}\\ 20\left(0.06-0.5\mathrm{x}+{\mathrm{x}}^2\right)=0.4+\mathrm{x}\\ 1.2-10\mathrm{x}+20{\mathrm{x}}^2=0.4+\mathrm{x}\\ \begin{array}{l}1.2-10\mathrm{x}+20{\mathrm{x}}^2-0.4-\mathrm{x}=0\\ 0.8-11\mathrm{x}+20{\mathrm{x}}^2=0\\ 20{\mathrm{x}}^2-11\mathrm{x}+0.8=0\end{array}\end{array} \end{gathered}$$
$\begin{array}{l}\mathrm{x}=\frac{11\pm \sqrt{{11}^2-4\times 20\times 0.8}}{2\times 20}\\ \mathrm{x}=\frac{11\pm \sqrt{121-64}}{40}\\ =\frac{11\pm \sqrt{57}}{40}\\ =\frac{11+7.55}{40}\\ =\frac{11-7.55}{40}\\ \mathrm{x}=0.08625\\ \left[{\mathrm{PCl}}_5\right]=0.4+0.08625\\ =0.48625\\ =48.625\times {10}^{-2}\\ =49\times {10}^{-2}\end{array}$