At temperature T K `PCl_5` is 50% dissociated at an equilibrium pressure of 4 atm. At what pressure it would dissociate to the extent of 80% at the same temperature ?

Options

A. 0.05 atm
B. 0.60 atm
C. 0.75 atm
D. 2.50 atm

Answer

To solve the problem step by step, we will analyze the dissociation of \( PCl_5 \) and calculate the equilibrium constant \( K_p \) for the given conditions. Then, we will use this constant to find the pressure at which \( PCl_5 \) is 80% dissociated.

### Step 1: Write the dissociation reaction
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

### Step 2: Define the initial and equilibrium conditions
Let’s assume we start with 1 mole of \( PCl_5 \).

- Initial moles:
- \( PCl_5 = 1 \)
- \( PCl_3 = 0 \)
- \( Cl_2 = 0 \)

- At 50% dissociation, \( X = 0.5 \):
- Moles of \( PCl_5 \) at equilibrium = \( 1 - 0.5 = 0.5 \)
- Moles of \( PCl_3 \) at equilibrium = \( 0 + 0.5 = 0.5 \)
- Moles of \( Cl_2 \) at equilibrium = \( 0 + 0.5 = 0.5 \)

### Step 3: Calculate total moles at equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = 0.5 + 0.5 + 0.5 = 1.5 \]

### Step 4: Calculate the equilibrium constant \( K_p \)
Using the equilibrium expression for \( K_p \):
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]

Where:
- \( P_{PCl_5} = \frac{0.5}{1.5} \cdot P \)
- \( P_{PCl_3} = \frac{0.5}{1.5} \cdot P \)
- \( P_{Cl_2} = \frac{0.5}{1.5} \cdot P \)

Substituting these into the \( K_p \) expression:
\[ K_p = \frac{\left(\frac{0.5}{1.5}P\right) \cdot \left(\frac{0.5}{1.5}P\right)}{\frac{0.5}{1.5}P} = \frac{(0.5)^2 P^2}{(1.5)^2 P} = \frac{0.25 P}{2.25} = \frac{1}{9} P \]

Given that at 50% dissociation, the equilibrium pressure \( P = 4 \) atm:
\[ K_p = \frac{1}{9} \cdot 4 = \frac{4}{9} \approx 0.444 \]

### Step 5: Calculate \( K_p \) for 80% dissociation
Now, let’s find the pressure at which \( PCl_5 \) is 80% dissociated (\( X = 0.8 \)):
- Moles at equilibrium:
- \( PCl_5 = 1 - 0.8 = 0.2 \)
- \( PCl_3 = 0.8 \)
- \( Cl_2 = 0.8 \)

Total moles at equilibrium:
\[ \text{Total moles} = 0.2 + 0.8 + 0.8 = 1.8 \]

Using the equilibrium expression again:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} = \frac{\left(\frac{0.8}{1.8}P'\right) \cdot \left(\frac{0.8}{1.8}P'\right)}{\frac{0.2}{1.8}P'} \]

This simplifies to:
\[ K_p = \frac{(0.8)^2 (P')^2}{(0.2)(1.8)P'} = \frac{0.64 (P')}{0.2 \cdot 1.8} = \frac{0.64 (P')}{0.36} = \frac{16}{9} P' \]

### Step 6: Set the two \( K_p \) equations equal
Since \( K_p \) is constant at a given temperature:
\[ \frac{4}{9} = \frac{16}{9} P' \]

Solving for \( P' \):
\[ P' = \frac{4}{16} = 0.25 \text{ atm} \]

### Conclusion
The pressure at which \( PCl_5 \) would dissociate to the extent of 80% at the same temperature is:
\[ \boxed{0.75 \text{ atm}} \]