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

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.

& \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{aligned}$$ \- Total number of moles at equilibrium = 1- x + x + x = 1 + x \- We know the relationship between partial pressure of the gas, mole fraction and total pressure. $$\begin{aligned} & \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{aligned}$$ \- Here x = 50% dissociation of the phosphorus pentachloride = 0.5 \- Therefore substitute x = 0.5 in the above equation. $$\begin{aligned} & {{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{aligned}$$ \- The x value for 80% dissociation is 0.8. \- Therefore $$\begin{aligned} & {{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{aligned}$$ \- 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.