At temperature T k `PCI_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, we need to determine the pressure at which \( PCl_5 \) dissociates to 80% at the same temperature, given that it is 50% dissociated at an equilibrium pressure of 4 atm.

### Step-by-Step Solution:

1. **Understanding the Dissociation Reaction:**
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

2. **Setting Up Initial Conditions:**
Let's assume we start with 1 mole of \( PCl_5 \) at the beginning (before dissociation).
- At 50% dissociation, \( \alpha = 0.5 \):
- Moles of \( PCl_5 \) remaining = \( 1 - 0.5 = 0.5 \)
- Moles of \( PCl_3 \) formed = \( 0.5 \)
- Moles of \( Cl_2 \) formed = \( 0.5 \)
- Total moles at equilibrium = \( 0.5 + 0.5 + 0.5 = 1.5 \)

3. **Calculating Partial Pressures:**
The total pressure at this point is given as 4 atm.
- The mole fraction of \( PCl_5 \) = \( \frac{0.5}{1.5} = \frac{1}{3} \)
- The mole fraction of \( PCl_3 \) = \( \frac{0.5}{1.5} = \frac{1}{3} \)
- The mole fraction of \( Cl_2 \) = \( \frac{0.5}{1.5} = \frac{1}{3} \)

Therefore, the partial pressures are:
- \( P_{PCl_5} = \frac{1}{3} \times 4 \, \text{atm} = \frac{4}{3} \, \text{atm} \)
- \( P_{PCl_3} = \frac{1}{3} \times 4 \, \text{atm} = \frac{4}{3} \, \text{atm} \)
- \( P_{Cl_2} = \frac{1}{3} \times 4 \, \text{atm} = \frac{4}{3} \, \text{atm} \)

4. **Calculating \( K_p \):**
The equilibrium constant \( K_p \) can be expressed as:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} = \frac{\left(\frac{4}{3}\right) \cdot \left(\frac{4}{3}\right)}{\frac{4}{3}} = \frac{4}{3} \]

5. **Finding Pressure for 80% Dissociation:**
Now, for 80% dissociation, \( \alpha = 0.8 \):
- Moles of \( PCl_5 \) remaining = \( 1 - 0.8 = 0.2 \)
- Moles of \( PCl_3 \) formed = \( 0.8 \)
- Moles of \( Cl_2 \) formed = \( 0.8 \)
- Total moles at equilibrium = \( 0.2 + 0.8 + 0.8 = 1.8 \)

6. **Calculating New Partial Pressures:**
The mole fractions are:
- Mole fraction of \( PCl_5 \) = \( \frac{0.2}{1.8} \)
- Mole fraction of \( PCl_3 \) = \( \frac{0.8}{1.8} \)
- Mole fraction of \( Cl_2 \) = \( \frac{0.8}{1.8} \)

The partial pressures will be:
- \( P_{PCl_5} = \frac{0.2}{1.8} \times P \)
- \( P_{PCl_3} = \frac{0.8}{1.8} \times P \)
- \( P_{Cl_2} = \frac{0.8}{1.8} \times P \)

7. **Setting Up the Equation for \( K_p \):**
Using the expression for \( K_p \):
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} = \frac{\left(\frac{0.8}{1.8} \cdot P\right) \cdot \left(\frac{0.8}{1.8} \cdot P\right)}{\left(\frac{0.2}{1.8} \cdot P\right)} \]
Substituting \( K_p = \frac{4}{3} \):
\[ \frac{0.64 P^2}{0.2 P} = \frac{4}{3} \]
Simplifying gives:
\[ \frac{0.64 P}{0.2} = \frac{4}{3} \]
\[ 0.64 P = \frac{4}{3} \times 0.2 \]
\[ 0.64 P = \frac{0.8}{3} \]
\[ P = \frac{0.8}{3 \times 0.64} \]
\[ P = \frac{0.8}{1.92} = 0.4167 \, \text{atm} \]

### Final Answer:
The pressure at which \( PCl_5 \) would dissociate to the extent of 80% at the same temperature is approximately **0.4167 atm**.