The equilibrium pressure necessary to obtain `50%` dissociation of `PCl_5` at `250^@C` is numerically________of `K_p`.

Options

A. six times
B. four times
C. five times
D. three times

Explanation

To solve the problem, we need to determine the relationship between the equilibrium pressure necessary to obtain 50% dissociation of \( PCl_5 \) at \( 250^\circ C \) and the equilibrium constant \( K_p \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The dissociation of phosphorus pentachloride can be represented as:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

2. **Initial Moles**:
Let's assume we start with 1 mole of \( PCl_5 \).

3. **Degree of Dissociation**:
Given that 50% of \( PCl_5 \) dissociates, this means:
\[ x = 0.5 \text{ moles of } PCl_5 \text{ dissociate.} \]
Therefore, at equilibrium:
- Moles of \( PCl_5 \) remaining = \( 1 - x = 1 - 0.5 = 0.5 \)
- Moles of \( PCl_3 \) formed = \( x = 0.5 \)
- Moles of \( Cl_2 \) formed = \( x = 0.5 \)

4. **Total Moles at Equilibrium**:
The total number of moles at equilibrium is:
\[ \text{Total moles} = \text{Moles of } PCl_5 + \text{Moles of } PCl_3 + \text{Moles of } Cl_2 = 0.5 + 0.5 + 0.5 = 1.5 \text{ moles} \]

5. **Calculating Partial Pressures**:
Assuming the total pressure at equilibrium is \( P \):
- The partial pressure of \( PCl_5 \):
\[ P_{PCl_5} = \frac{0.5}{1.5} \times P = \frac{1}{3} P \]
- The partial pressure of \( PCl_3 \):
\[ P_{PCl_3} = \frac{0.5}{1.5} \times P = \frac{1}{3} P \]
- The partial pressure of \( Cl_2 \):
\[ P_{Cl_2} = \frac{0.5}{1.5} \times P = \frac{1}{3} P \]

6. **Expression for \( K_p \)**:
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]
Substituting the partial pressures:
\[ K_p = \frac{\left(\frac{1}{3} P\right) \cdot \left(\frac{1}{3} P\right)}{\frac{1}{3} P} = \frac{\frac{1}{9} P^2}{\frac{1}{3} P} = \frac{1}{9} P \cdot 3 = \frac{1}{3} P \]

7. **Finding the Relationship**:
From the above expression, we can rearrange to find \( P \):
\[ P = 3 K_p \]

### Final Answer:
The equilibrium pressure necessary to obtain 50% dissociation of \( PCl_5 \) at \( 250^\circ C \) is numerically **3 times** \( K_p \).