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 can follow these steps:

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

### Step 2: Determine the initial moles and volume
We are given that initially there are 3 moles of \( PCl_5 \) in a 3-liter vessel.

### Step 3: Calculate the moles dissociated
At equilibrium, 40% of \( PCl_5 \) is dissociated. Therefore, the moles of \( PCl_5 \) that dissociate can be calculated as:
\[ \text{Moles dissociated} = 0.4 \times 3 = 1.2 \text{ moles} \]

### Step 4: Calculate the moles at equilibrium
- Moles of \( PCl_5 \) at equilibrium:
\[ \text{Moles of } PCl_5 = 3 - 1.2 = 1.8 \text{ moles} \]
- Moles of \( PCl_3 \) formed:
\[ \text{Moles of } PCl_3 = 1.2 \text{ moles} \]
- Moles of \( Cl_2 \) formed:
\[ \text{Moles of } Cl_2 = 1.2 \text{ moles} \]

### Step 5: Calculate concentrations at equilibrium
Concentration is calculated using the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
- 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 6: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 7: Substitute the concentrations into the equilibrium expression
Substituting the values we calculated:
\[ K_c = \frac{(0.4)(0.4)}{0.6} \]

### Step 8: Calculate \( K_c \)
\[ K_c = \frac{0.16}{0.6} = 0.267 \]

Thus, the equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \) is approximately **0.267**.

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