JEE Main 2023 — Chemical Equilibrium Question with Solution

The initial equilibrium concentrations are given as:

$[{\mathrm{PCl}}_3{]}_{\text{eq}}=0.2mol/L$, $[{\mathrm{Cl}}_2{]}_{\text{eq}}=0.1mol/L$, $[{\mathrm{PCl}}_5{]}_{\text{eq}}=0.4mol/L$.

After adding $0.2\mathrm{mol}$ of $\mathrm{C}{\mathrm{l}}_2$, the new concentration of $\mathrm{C}{\mathrm{l}}_2$ becomes $0.1+0.2=0.3mol/L$.

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

$[{\mathrm{PCl}}_3]=0.2-xmol/L$, $[{\mathrm{Cl}}_2]=0.3-xmol/L$, $[{\mathrm{PCl}}_5]=0.4+xmol/L$.

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:

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