`PCl_(5)` s dissociating 50% at `250^(@)C` at a total pressure of P atm. If equilibrium constant is `K_(p)`, then which of the following relation is numerically correct -

Options

A. `K_(p)=3P`
B. `P=3K_(p)`
C. `P=(2K_(p))/(3)`
D. `K_(p)=(2P)/(3)`

Answer

To solve the problem, we need to analyze the dissociation of \( PCl_5 \) and derive the relationship between the equilibrium constant \( K_p \) and the total pressure \( P \) at equilibrium.

### Step-by-Step Solution:

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

2. **Initial Moles**:
Let the initial number of moles of \( PCl_5 \) be \( A \). Since it dissociates 50%, the change in moles at equilibrium will be:
- Moles of \( PCl_5 \) at equilibrium = \( A - A \cdot 0.5 = A \cdot 0.5 \)
- Moles of \( PCl_3 \) at equilibrium = \( A \cdot 0.5 \)
- Moles of \( Cl_2 \) at equilibrium = \( A \cdot 0.5 \)

3. **Total Moles at Equilibrium**:
The total number of moles at equilibrium \( N_{total} \) is:
\[ N_{total} = (A \cdot 0.5) + (A \cdot 0.5) + (A \cdot 0.5) = A \cdot 1.5 \]

4. **Calculating \( K_p \)**:
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{PCl_3})(P_{Cl_2})}{(P_{PCl_5})} \]
In terms of moles, we can express the partial pressures as:
\[ K_p = \frac{(A \cdot 0.5)(A \cdot 0.5)}{A \cdot 0.5} \cdot \frac{P}{N_{total}} = \frac{(A \cdot 0.5)^2}{A \cdot 0.5} \cdot \frac{P}{A \cdot 1.5} \]

5. **Simplifying \( K_p \)**:
Simplifying the expression:
\[ K_p = \frac{0.25 A^2}{0.5 A} \cdot \frac{P}{1.5 A} = \frac{0.25 A}{0.5} \cdot \frac{P}{1.5} = \frac{0.5 P}{1.5} = \frac{P}{3} \]

6. **Final Relationship**:
Rearranging gives us:
\[ P = 3 K_p \]

### Conclusion:
The correct numerical relation derived from the dissociation of \( PCl_5 \) at 250°C and total pressure \( P \) is:
\[ P = 3 K_p \]