JEE MAIN - Chemistry (2023 - 6th April Evening Shift - No. 20)
The equilibrium composition for the reaction $$\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}$$ at $$298 \mathrm{~K}$$ is given below:
$$\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=0.2 \mathrm{~mol} \mathrm{~L}^{-1},\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=0.1 \mathrm{~mol} \mathrm{~L}^{-1},\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=0.40 \mathrm{~mol} \mathrm{~L}^{-1}$$
If $$0.2 \mathrm{~mol}$$ of $$\mathrm{Cl}_{2}$$ is added at the same temperature, the equilibrium concentrations of $$\mathrm{PCl}_{5}$$ is __________ $$\times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$$
Given : $$\mathrm{K}_{\mathrm{c}}$$ for the reaction at $$298 \mathrm{~K}$$ is 20
Explanation
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:
$[\mathrm{PCl}_5]_{\text{new eq}} = 0.4 + 0.086 = 0.486 \, \mathrm{mol/L} = 48.6 \times 10^{-2} \, \mathrm{mol/L}$.
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