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.
 $$\begin{array}{rl}& \underset{\text{initial}}{}\underset{1}{\text{ PC}{\text{l}}_{\text{5}}}\rightleftarrows \underset{0}{\text{PC}{\text{l}}_{\text{3}}}\text{+}\underset{0}{\text{C}{\text{l}}_{\text{2}}}\\ & \underset{\text{At equilibrium}}{}\underset{1-x}{\text{ PC}{\text{l}}_{\text{5}}}\rightleftarrows \underset{x}{\text{PC}{\text{l}}_{\text{3}}}\text{+}\underset{x}{\text{C}{\text{l}}_{\text{2}}}\end{array}$$
- 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.
$$\begin{array}{rl}& \text{partial pressure of the gas = mole fraction }\times \text{ total pressure}\\ & \text{mole fraction = }{\displaystyle \frac{\text{partial pressure of the gas}}{\text{total pressure of the gas}}}\\ & K_p={\displaystyle \frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PCl5}}}={\displaystyle \frac{{\displaystyle \frac{x}{1+x}}P\times {\displaystyle \frac{x}{1+x}}P}{{\displaystyle \frac{1-x}{1+x}}P}}={\displaystyle \frac{x^{2}P}{1-x^2}}\end{array}$$
- Here x = 50% dissociation of the phosphorus pentachloride = 0.5
- Therefore substitute x = 0.5 in the above equation.
 $$\begin{array}{rl}& K_p={\displaystyle \frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PCl5}}}={\displaystyle \frac{{\displaystyle \frac{x}{1+x}}P\times {\displaystyle \frac{x}{1+x}}P}{{\displaystyle \frac{1-x}{1+x}}P}}={\displaystyle \frac{x^{2}P}{1-x^2}}\\ & ={\displaystyle \frac{{(0.5)}^2\times 4}{1-{(0.5)}^2}}=1.33\end{array}$$
- The x value for 80% dissociation is 0.8.
- Therefore
$$\begin{array}{rl}& K_p={\displaystyle \frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PCl5}}}={\displaystyle \frac{{\displaystyle \frac{x}{1+x}}P\times {\displaystyle \frac{x}{1+x}}P}{{\displaystyle \frac{1-x}{1+x}}P}}={\displaystyle \frac{x^{2}P}{1-x^2}}\\ & {\displaystyle \frac{{(0.8)}^2\times P}{1-{(0.8)}^2}}=1.33\\ & P=0.75atm.\end{array}$$
- 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.