QJEEM23Ch76da: JEE Main 2023 Chemistry - Class 11 JEEM23Ch76da Solution: The equilibrium composition for the reaction PCl 3 + Cl 2 ⇌ PCl …

The initial equilibrium concentrations are given as:

$[\mathrm{PCl}_{3}]_{\text{eq}} = 0.2 \, \mathrm{mol/L}$ , $[\mathrm{Cl}_{2}]_{\text{eq}} = 0.1 \, \mathrm{mol/L}$ , $[\mathrm{PCl}_{5}]_{\text{eq}} = 0.4 \, \mathrm{mol/L}$ .

After adding $0.2 \, \mathrm{mol}$ of $\mathrm{Cl_2}$ , the new concentration of $\mathrm{Cl_2}$ becomes $0.1 + 0.2 = 0.3 \, \mathrm{mol/L}$ .

Let the change in concentration be $x$ , so at the new equilibrium, the concentrations are:

$[\mathrm{PCl}_{3}] = 0.2 - x \, \mathrm{mol/L}$ , $[\mathrm{Cl}_{2}] = 0.3 - x \, \mathrm{mol/L}$ , $[\mathrm{PCl}_{5}] = 0.4 + x \, \mathrm{mol/L}$ .

Using the equilibrium constant expression $K_c = 20$ , we get:

$20 = \frac{0.4+x}{(0.2-x)(0.3-x)}$ .

Solving for $x$ , we find $x \approx 0.086$ .

Therefore, the new equilibrium concentration of $\mathrm{PCl}_5$ is:

$[\mathrm{PCl}_5]_{\text{new eq}} = 0.4 + 0.086 = 0.486 \, \mathrm{mol/L} = 48.6 \times 10^{-2} \, \mathrm{mol/L}$ .