In the thermal dissociation fo `PCl_(5).` the partical pressure in the gaseous equlibrium mixture is 1.0 atmosphere when half of `PCl_(5)` is found to dissociate. The equlibrium constant of the reaction `(K_(p))` in atomosphere is

Options

A. `0.25`
B. `0.50`
C. `1.00`
D. `0.3`

Explanation

To solve the problem of finding the equilibrium constant \( K_p \) for the thermal dissociation of \( PCl_5 \), we can follow these steps:

### Step-by-Step Solution:

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

2. **Initial Conditions**:
Let's assume we start with 1 mole of \( PCl_5 \). According to the problem, half of \( PCl_5 \) dissociates. Therefore, the initial amounts are:
- \( PCl_5 \): 1 mole
- \( PCl_3 \): 0 moles
- \( Cl_2 \): 0 moles

3. **Change in Moles**:
When half of \( PCl_5 \) dissociates, the change in moles will be:
- \( PCl_5 \): \( 1 - 0.5 = 0.5 \) moles
- \( PCl_3 \): \( 0 + 0.5 = 0.5 \) moles
- \( Cl_2 \): \( 0 + 0.5 = 0.5 \) moles

4. **Equilibrium Moles**:
At equilibrium, we have:
- \( PCl_5 \): 0.5 moles
- \( PCl_3 \): 0.5 moles
- \( Cl_2 \): 0.5 moles

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

6. **Partial Pressures**:
Given that the total pressure at equilibrium is 1.0 atmosphere, we can find the partial pressures:
- Partial pressure of \( PCl_5 \):
\[ P_{PCl_5} = \frac{0.5}{1.5} \times 1 = \frac{1}{3} \text{ atm} \]
- Partial pressure of \( PCl_3 \):
\[ P_{PCl_3} = \frac{0.5}{1.5} \times 1 = \frac{1}{3} \text{ atm} \]
- Partial pressure of \( Cl_2 \):
\[ P_{Cl_2} = \frac{0.5}{1.5} \times 1 = \frac{1}{3} \text{ atm} \]

7. **Equilibrium Constant \( K_p \)**:
The expression for 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:
\[ K_p = \frac{\left(\frac{1}{3}\right) \cdot \left(\frac{1}{3}\right)}{\frac{1}{3}} = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{9} \times 3 = \frac{1}{3} \]

### Final Answer:
Thus, the equilibrium constant \( K_p \) is:
\[ K_p = \frac{1}{3} \text{ atm} \]