`2 "mole"` of `PCl_(5)` were heated in a closed vessel of `2 litre` capacity. At equilibrium `40%` of `PCl_(5)` dissociated into `PCl_(3)` and `Cl_(2)`. The value of the equilibrium constant is:

Options

A. 0.267
B. 0.53
C. 2.63
D. 5.3

Answer

To find the equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \) into \( PCl_3 \) and \( Cl_2 \), we will follow these steps:

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

### Step 2: Determine initial moles and changes at equilibrium
Initially, we have 2 moles of \( PCl_5 \) in a closed vessel.

- Initial moles of \( PCl_5 = 2 \) moles
- At equilibrium, 40% of \( PCl_5 \) dissociates.

Calculating the amount that dissociates:
\[ \text{Amount dissociated} = 0.4 \times 2 = 0.8 \text{ moles} \]

Thus, the moles at equilibrium will be:
- Moles of \( PCl_5 \) left = \( 2 - 0.8 = 1.2 \) moles
- Moles of \( PCl_3 \) formed = \( 0.8 \) moles
- Moles of \( Cl_2 \) formed = \( 0.8 \) moles

### Step 3: Calculate concentrations at equilibrium
The volume of the vessel is 2 liters. Therefore, the concentrations of the species at equilibrium can be calculated as follows:

- Concentration of \( PCl_5 \):
\[ \text{Concentration of } PCl_5 = \frac{1.2 \text{ moles}}{2 \text{ L}} = 0.6 \text{ M} \]

- Concentration of \( PCl_3 \):
\[ \text{Concentration of } PCl_3 = \frac{0.8 \text{ moles}}{2 \text{ L}} = 0.4 \text{ M} \]

- Concentration of \( Cl_2 \):
\[ \text{Concentration of } Cl_2 = \frac{0.8 \text{ moles}}{2 \text{ L}} = 0.4 \text{ M} \]

### 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 \( K_c \) expression
Substituting the equilibrium concentrations into the equation:
\[ K_c = \frac{(0.4)(0.4)}{0.6} \]

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

### Final Answer
The value of the equilibrium constant \( K_c \) is:
\[ \boxed{0.267} \]