Question

At temperature T K, $PC{{l}_{5}}$ is 50% dissociated at an equilibrium pressure of 4 atm. At what pressure it would dissociate to the extent of 60% at the same temperature?
A. 0.05 atm
B. 0.50 atm
C. 0.75 atm
D. 2.50 atm

Answer 1

Hint: There is a relationship between partial pressure of the gas, mole fraction and total pressure and it is as follows.
\[\text{partial pressure of the gas = mole fraction }\!\!\times\!\!\text{ total pressure}\]

Complete answer:
- In the question it is given that phosphorus pentachloride is 50% dissociated at 4 atm pressure, at what pressure 60% of the phosphorus pentachloride will dissociate.
- We have to find the pressure of 60% dissociation of the phosphorus pentachloride.
- The chemical equation of dissociation of phosphorus pentachloride is as follows.
 \[\begin{align}
  & \text{ }\underset{\text{initial}}{\mathop{{}}}\,\underset{1}{\mathop{\text{ PC}{{\text{l}}_{\text{5}}}}}\,\rightleftarrows \underset{0}{\mathop{\text{PC}{{\text{l}}_{\text{3}}}}}\,\text{+}\underset{0}{\mathop{\text{C}{{\text{l}}_{\text{2}}}}}\, \\
 & \underset{\text{At equilibrium}}{\mathop{{}}}\,\underset{1-x}{\mathop{\text{ PC}{{\text{l}}_{\text{5}}}}}\,\rightleftarrows \underset{x}{\mathop{\text{PC}{{\text{l}}_{\text{3}}}}}\,\text{+}\underset{x}{\mathop{\text{C}{{\text{l}}_{\text{2}}}}}\, \\
\end{align}\]
- Total number of moles at equilibrium = 1- x + x + x = 1 + x
- We know the relationship between partial pressure of the gas, mole fraction and total pressure.
\[\begin{align}
  & \text{partial pressure of the gas = mole fraction }\!\!\times\!\!\text{ total pressure} \\
 & \text{mole fraction = }\dfrac{\text{partial pressure of the gas}}{\text{total pressure of the gas}} \\
 & {{K}_{p}}=\dfrac{{{P}_{PC{{l}_{3}}}}\times {{P}_{C{{l}_{2}}}}}{{{P}_{PCl5}}}=\dfrac{\dfrac{x}{1+x}P\times \dfrac{x}{1+x}P}{\dfrac{1-x}{1+x}P}=\dfrac{{{x}^{2}}P}{1-{{x}^{2}}} \\
\end{align}\]
- Here x = 50% dissociation of the phosphorus pentachloride = 0.5
- Therefore substitute x = 0.5 in the above equation.
 \[\begin{align}
  & {{K}_{p}}=\dfrac{{{P}_{PC{{l}_{3}}}}\times {{P}_{C{{l}_{2}}}}}{{{P}_{PCl5}}}=\dfrac{\dfrac{x}{1+x}P\times \dfrac{x}{1+x}P}{\dfrac{1-x}{1+x}P}=\dfrac{{{x}^{2}}P}{1-{{x}^{2}}} \\
 & =\dfrac{{{(0.5)}^{2}}\times 4}{1-{{(0.5)}^{2}}}=1.33 \\
\end{align}\]
- The x value for 80% dissociation is 0.8.
- Therefore
\[\begin{align}
  & {{K}_{p}}=\dfrac{{{P}_{PC{{l}_{3}}}}\times {{P}_{C{{l}_{2}}}}}{{{P}_{PCl5}}}=\dfrac{\dfrac{x}{1+x}P\times \dfrac{x}{1+x}P}{\dfrac{1-x}{1+x}P}=\dfrac{{{x}^{2}}P}{1-{{x}^{2}}} \\
 & \dfrac{{{(0.8)}^{2}}\times P}{1-{{(0.8)}^{2}}}=1.33 \\
 & P=0.75atm. \\
\end{align}\]
- Therefore at 0.75 atm of pressure 80% of the dissociation of the phosphorus pentachloride occurs.

So, the correct option is C.

Note:
First we have to calculate the mole fraction of 50 % of the dissociation of the phosphorus pentachloride. Later substitute the mole fraction value of 50% dissociation of the phosphorus pentachloride to get the pressure exerted by the dissociation of the 80% of the phosphorus pentachloride.