# A sample of pure PCl5 was introduced into an evacuated vessel at 473 K.

Question:

A sample of pure $\mathrm{PCl}_{5}$ was introduced into an evacuated vessel at $473 \mathrm{~K}$. After equilibrium was attained, concentration of $\mathrm{PCl}_{5}$ was found to be $0.5 \times 10^{-1}$ mol $\mathrm{L}^{-1}$. If value of $K_{c}$ is $8.3 \times 10^{-3}$, what are the concentrations of $\mathrm{PCl}_{3}$ and $\mathrm{Cl}_{2}$ at equilibrium?

$\mathrm{PCl}_{5}(\mathrm{~g}) \longleftrightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$

Solution:

Let the concentrations of both $\mathrm{PCl}_{3}$ and $\mathrm{Cl}_{2}$ at equilibrium be $x \mathrm{molL}^{-1}$. The given reaction is:

Now we can write the expression for equilibrium as:

$\frac{\left[\mathrm{PCl}_{2}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=K_{\mathrm{C}}$

$\Rightarrow \frac{x \times x}{0.5 \times 10^{-1}}=8.3 \times 10^{-3}$

$\Rightarrow x^{2}=4.15 \times 10^{-4}$

$\Rightarrow x=2.04 \times 10^{-2}$

$=0.0204$

$=0.02($ approximately $)$

Therefore, at equilibrium,

$\left[\mathrm{PCl}_{3}\right]=\left[\mathrm{Cl}_{2}\right]=0.02 \mathrm{~mol} \mathrm{~L}^{-1} .$