3 mole of `PCl_(5)` were heated in a vessel of three litre capacity. At equilibrium `40%` of `PCl_(5)` was found to be dissociated. What will be the equilibrium constant for disscoiation of `PCl_(5)`?

Options

A. 0.14
B. 0.24
C. 0.267
D. 0.35

Answer

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

### Step 1: Write the balanced chemical equation
The dissociation of \( PCl_5 \) can be represented by the following equation:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 2: Determine initial moles and changes
Initially, we have:
- Moles of \( PCl_5 \) = 3 moles
- Moles of \( PCl_3 \) = 0 moles
- Moles of \( Cl_2 \) = 0 moles

At equilibrium, we know that 40% of \( PCl_5 \) has dissociated. Therefore:
\[ \text{Moles of } PCl_5 \text{ dissociated} = 0.40 \times 3 = 1.2 \text{ moles} \]

Thus, at equilibrium:
- Moles of \( PCl_5 \) = \( 3 - 1.2 = 1.8 \) moles
- Moles of \( PCl_3 \) = \( 1.2 \) moles (produced)
- Moles of \( Cl_2 \) = \( 1.2 \) moles (produced)

### Step 3: Calculate concentrations at equilibrium
The volume of the vessel is given as 3 liters. The concentration (C) is calculated using the formula:
\[ C = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Calculating the concentrations:
- Concentration of \( PCl_5 \):
\[ [C_{PCl_5}] = \frac{1.8 \text{ moles}}{3 \text{ L}} = 0.6 \text{ M} \]
- Concentration of \( PCl_3 \):
\[ [C_{PCl_3}] = \frac{1.2 \text{ moles}}{3 \text{ L}} = 0.4 \text{ M} \]
- Concentration of \( Cl_2 \):
\[ [C_{Cl_2}] = \frac{1.2 \text{ moles}}{3 \text{ L}} = 0.4 \text{ M} \]

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 5: Substitute the concentrations into the equilibrium constant expression
Substituting the calculated concentrations:
\[ K_c = \frac{(0.4)(0.4)}{0.6} \]

### Step 6: Calculate \( K_c \)
Calculating the value:
\[ K_c = \frac{0.16}{0.6} = 0.267 \]

### Final Answer
The equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \) is:
\[ K_c = 0.267 \]