A certain temperature, the degree of dissociationof `PCI_(3)` was found to be 0.25 under a total pressure of 15 atm. The value of `K_(p)` for the dissociation of `PCl_(5)` is

Options

A. 1
B. 0.25
C. 0.5
D. 0.75

Answer

To find the value of \( K_p \) for the dissociation of \( PCl_5 \), we will follow these steps:

### Step 1: Write the equilibrium reaction
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \]

### Step 2: Define the degree of dissociation
Let \( \alpha \) be the degree of dissociation of \( PCl_5 \). Given that \( \alpha = 0.25 \), this means that 25% of \( PCl_5 \) has dissociated.

### Step 3: Set up initial and equilibrium pressures
Assuming we start with 1 atm of \( PCl_5 \):
- Initial pressure of \( PCl_5 = 1 \, atm \)
- Initial pressure of \( PCl_3 = 0 \, atm \)
- Initial pressure of \( Cl_2 = 0 \, atm \)

At equilibrium, the pressures will be:
- Pressure of \( PCl_5 = 1 - \alpha = 1 - 0.25 = 0.75 \, atm \)
- Pressure of \( PCl_3 = \alpha = 0.25 \, atm \)
- Pressure of \( Cl_2 = \alpha = 0.25 \, atm \)

### Step 4: Calculate total pressure
The total pressure \( P_{total} \) at equilibrium is given by:
\[ P_{total} = P_{PCl_5} + P_{PCl_3} + P_{Cl_2} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha = 1 + 0.25 = 1.25 \, atm \]

However, we are given that the total pressure is 15 atm, which means we need to scale our pressures accordingly. The pressures can be expressed in terms of the total pressure:
- \( P_{PCl_5} = (1 - \alpha) \cdot P_{total} = 0.75 \cdot 15 = 11.25 \, atm \)
- \( P_{PCl_3} = \alpha \cdot P_{total} = 0.25 \cdot 15 = 3.75 \, atm \)
- \( P_{Cl_2} = \alpha \cdot P_{total} = 0.25 \cdot 15 = 3.75 \, atm \)

### Step 5: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]

### Step 6: Substitute the equilibrium pressures into the \( K_p \) expression
Substituting the pressures we found:
\[ K_p = \frac{(3.75)(3.75)}{11.25} \]

### Step 7: Calculate \( K_p \)
Calculating the above expression:
\[ K_p = \frac{(3.75)^2}{11.25} = \frac{14.0625}{11.25} = 1.25 \]

### Final Answer
Thus, the value of \( K_p \) for the dissociation of \( PCl_5 \) is:
\[ \boxed{1.25} \]