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

Answer

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

### Step 1: Determine Initial Concentration
We start with 2 moles of \( PCl_5 \) in a 2-liter vessel. The initial concentration (molarity) of \( PCl_5 \) can be calculated using the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Substituting the values:

\[ \text{Concentration of } PCl_5 = \frac{2 \text{ moles}}{2 \text{ liters}} = 1 \text{ M} \]

### Step 2: Write the Dissociation Reaction
The dissociation reaction of \( PCl_5 \) is:

\[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \]

### Step 3: Determine the Change in Concentration
Given that 40% of \( PCl_5 \) dissociates, we can calculate how many moles dissociate:

\[ \text{Moles of } PCl_5 \text{ dissociated} = 0.4 \times 2 \text{ moles} = 0.8 \text{ moles} \]

### Step 4: Calculate Equilibrium Concentrations
At equilibrium, we can find the concentrations of each species:

- **For \( PCl_5 \)**:
\[ \text{Concentration of } PCl_5 = 1 - 0.8 = 0.2 \text{ M} \]

- **For \( PCl_3 \)** and **\( Cl_2 \)** (both produced in equal amounts):
\[ \text{Concentration of } PCl_3 = 0.8 \text{ M} \]
\[ \text{Concentration of } Cl_2 = 0.8 \text{ M} \]

### Step 5: Write the Expression for Equilibrium Constant
The equilibrium constant \( K_c \) for the reaction is given by:

\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]

Substituting the equilibrium concentrations:

\[ K_c = \frac{(0.8)(0.8)}{0.2} \]

### Step 6: Calculate \( K_c \)
Now, calculate the value:

\[ K_c = \frac{0.64}{0.2} = 3.2 \]

### Final Answer
The value of the equilibrium constant \( K_c \) is **3.2**.

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