The volume percentage of `Cl_2` at equilibrium in the dissociation of `PCl_5` under a total pressure of 1.5atm is (Kp = 0.202) ,

Options

A. 74.5
B. 36.5
C. 63.5
D. 26.6

Answer

To solve the problem of finding the volume percentage of \( Cl_2 \) at equilibrium in the dissociation of \( PCl_5 \) under a total pressure of 1.5 atm, we can follow these steps:

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

### Step 2: Set up initial conditions
Assume we start with 1 mole of \( PCl_5 \) and no products at time \( t = 0 \):
- Initial moles of \( PCl_5 \) = 1
- Initial moles of \( PCl_3 \) = 0
- Initial moles of \( Cl_2 \) = 0

### Step 3: Define changes at equilibrium
Let \( \alpha \) be the degree of dissociation of \( PCl_5 \). At equilibrium:
- Moles of \( PCl_5 \) = \( 1 - \alpha \)
- Moles of \( PCl_3 \) = \( \alpha \)
- Moles of \( Cl_2 \) = \( \alpha \)

### Step 4: Calculate total moles at equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 5: Use the total pressure
Given that the total pressure \( P \) is 1.5 atm, we can express the partial pressures as:
\[ P_{PCl_5} = \frac{(1 - \alpha)}{(1 + \alpha)} \cdot P \]
\[ P_{PCl_3} = \frac{\alpha}{(1 + \alpha)} \cdot P \]
\[ P_{Cl_2} = \frac{\alpha}{(1 + \alpha)} \cdot P \]

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} = \frac{\left(\frac{\alpha}{1 + \alpha} \cdot P\right) \cdot \left(\frac{\alpha}{1 + \alpha} \cdot P\right)}{\frac{(1 - \alpha)}{(1 + \alpha)} \cdot P} \]
This simplifies to:
\[ K_p = \frac{\alpha^2 \cdot P}{(1 - \alpha)(1 + \alpha)} \]

### Step 7: Substitute known values
We know \( K_p = 0.202 \) and \( P = 1.5 \) atm:
\[ 0.202 = \frac{\alpha^2 \cdot 1.5}{(1 - \alpha)(1 + \alpha)} \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ 0.202(1 - \alpha)(1 + \alpha) = 1.5\alpha^2 \]
Expanding and rearranging leads to a quadratic equation in terms of \( \alpha \).

### Step 9: Solve for \( \alpha \)
After solving the quadratic equation, we find:
\[ \alpha^2 = 0.11868 \implies \alpha \approx 0.345 \]

### Step 10: Calculate volume percentage of \( Cl_2 \)
The volume percentage of \( Cl_2 \) is given by:
\[ \text{Volume percentage of } Cl_2 = \frac{\alpha}{1 + \alpha} \times 100 = \frac{0.345}{1 + 0.345} \times 100 \approx 25.6\% \]

### Step 11: Final answer
Comparing with the given options, the closest value is:
\[ \text{Volume percentage of } Cl_2 \approx 26.6\% \]