`PCl_(5)` is `50%` dissociated into `PCl_(3) and Cl_(2)` at 1 atmosphere pressure. It will be `40%` dissociated at:

Options

A. 1.75 atm
B. 1.84 atm
C. 2.00 atm
D. 1.25 atm

Explanation

To solve the problem, we need to analyze the dissociation of \( PCl_5 \) into \( PCl_3 \) and \( Cl_2 \) under two different conditions of dissociation.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- We know that \( PCl_5 \) is 50% dissociated at 1 atmosphere pressure.
- The degree of dissociation (\( \alpha \)) is given as 50%, which can be expressed as:
\[ \alpha = 0.50 \]

2. **Setting Up the Reaction**:
- The dissociation reaction can be written as:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]
- Let the initial number of moles of \( PCl_5 \) be 1 mole.

3. **Calculating Moles at Equilibrium**:
- At equilibrium, if 50% of \( PCl_5 \) dissociates:
- Moles of \( PCl_5 \) left = \( 1 - 0.50 = 0.50 \)
- Moles of \( PCl_3 \) formed = \( 0.50 \)
- Moles of \( Cl_2 \) formed = \( 0.50 \)
- Total moles at equilibrium = \( 0.50 + 0.50 + 0.50 = 1.50 \)

4. **Calculating Partial Pressures**:
- The total pressure is given as 1 atmosphere.
- The partial pressures can be calculated using the mole fraction:
- \( P_{PCl_5} = \frac{0.50}{1.50} \times 1 = \frac{1}{3} \) atm
- \( P_{PCl_3} = \frac{0.50}{1.50} \times 1 = \frac{1}{3} \) atm
- \( P_{Cl_2} = \frac{0.50}{1.50} \times 1 = \frac{1}{3} \) atm

5. **Setting Up for 40% Dissociation**:
- Now, we need to find the pressure when \( PCl_5 \) is 40% dissociated.
- Here, \( \alpha' = 0.40 \).
- Moles of \( PCl_5 \) left = \( 1 - 0.40 = 0.60 \)
- Moles of \( PCl_3 \) formed = \( 0.40 \)
- Moles of \( Cl_2 \) formed = \( 0.40 \)
- Total moles at equilibrium = \( 0.60 + 0.40 + 0.40 = 1.40 \)

6. **Equating Equilibrium Constants**:
- The equilibrium constant \( K_p \) remains the same for both cases since the temperature is constant.
- For the first case (50% dissociation):
\[ K_p = \frac{(P_{PCl_3})(P_{Cl_2})}{(P_{PCl_5})} = \frac{\left(\frac{1}{3}\right) \left(\frac{1}{3}\right)}{\left(\frac{1}{3}\right)} = \frac{1}{3} \]
- For the second case (40% dissociation):
\[ K_p = \frac{(P_{PCl_3})(P_{Cl_2})}{(P_{PCl_5})} = \frac{(0.40)(0.40)}{(0.60)} \cdot P \]

7. **Solving for Total Pressure \( P \)**:
- Setting the two \( K_p \) expressions equal:
\[ \frac{(0.40)(0.40)}{(0.60)} \cdot P = \frac{1}{3} \]
- Rearranging gives:
\[ P = \frac{1/3 \cdot 0.60}{0.40 \cdot 0.40} = \frac{0.20}{0.16} = 1.25 \text{ atm} \]

8. **Final Result**:
- The pressure at which \( PCl_5 \) will be 40% dissociated is **1.25 atm**.