Question

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, ${\mathrm{PCl}}_5\rightleftharpoons {\mathrm{PCl}}_3\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$. If the total pressure at the equilibrium of the reaction mixture is $\mathrm{P}$ and the degree of dissociation of ${\mathrm{PCl}}_5$​ is $\mathrm{x}$, the partial pressure of ${\mathrm{PCl}}_3$​ will be:

Answer 1

Partial pressure:

• The pressure exerted by an individual gas in a mixture of gases is termed partial pressure of the gas.
• According to Dalton's law total Pressure of a mixture is the sum of individual partial pressure of the gases in the mixture. i.e. ${\mathrm{P}}_{\mathrm{total}}={\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3+................$
• The partial pressure of a gas in the mixture can also be calculated as: ${\mathrm{P}}_1={\mathrm{x}}_1{\mathrm{P}}_{\mathrm{total}}$; where ${\mathrm{x}}_1$ is the mole fraction of gas1 and ${\mathrm{P}}_1$ is the partial pressure of gas1.

In the given reaction ${\mathrm{PCl}}_5\rightleftharpoons {\mathrm{PCl}}_3\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$, the total pressure is $\mathrm{P}$ and degree of dissociation is $\mathrm{x}$.

Let the total $1$mole of ${\mathrm{PCl}}_5$ dissociates. so we can infer that $\mathrm{x}$ mole of ${\mathrm{PCl}}_3$ and ${\mathrm{Cl}}_2$are formed.

$\underset{1-\mathrm{x}}{{\mathrm{PCl}}_5}\rightleftharpoons \underset{\mathrm{x}}{{\mathrm{PCl}}_3}+\underset{\mathrm{x}}{{\mathrm{Cl}}_2}$

The mole fraction of ${\mathrm{PCl}}_3$ can be given as: $\frac{\mathrm{moles}\mathrm{of}{\mathrm{PCl}}_3}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{moles}}=\frac{\mathrm{x}}{1-\mathrm{x}+\mathrm{x}+\mathrm{x}}=\frac{\mathrm{x}}{1+\mathrm{x}}$

Therefore, the partial pressure of ${\mathrm{PCl}}_3$ is given as the mole fraction of ${\mathrm{PCl}}_3\times {\mathrm{P}}_{\mathrm{total}}$ $\Rightarrow \frac{\mathrm{x}}{1+\mathrm{x}}\times \mathrm{P}$

Hence, the partial pressure of ${\mathrm{PCl}}_3$ is $\frac{\mathrm{x}}{1+\mathrm{x}}\times \mathrm{P}$.