A quantity of `PCI _5` was heated in a 2 litre vessel at 525 K . It dissociates as `PCI_5 (g) hArr PCI _3 (g) + CI_2 (g)` At equilibrium 0.2 mol each of `PCI_5 , PCl_3and Cl_2` is found in the reaction mixture. The equilibrium constant `K_c` for the reaction is -

Options

A. 0.2
B. 0.5
C. 0.1
D. 0.05

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \) into \( PCl_3 \) and \( Cl_2 \) at equilibrium.

### Step-by-Step Solution:

1. **Write the Balanced Equation:**
The dissociation reaction is:
\[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \]

2. **Identify the Initial and Equilibrium Concentrations:**
At equilibrium, we are given that:
- Moles of \( PCl_5 = 0.2 \) mol
- Moles of \( PCl_3 = 0.2 \) mol
- Moles of \( Cl_2 = 0.2 \) mol

3. **Calculate the Volume of the Vessel:**
The volume of the vessel is given as \( 2 \) liters.

4. **Calculate the Concentrations:**
Concentration (\( C \)) is calculated using the formula:
\[ C = \frac{\text{moles}}{\text{volume (L)}} \]
- For \( PCl_5 \):
\[ C_{PCl_5} = \frac{0.2 \, \text{mol}}{2 \, \text{L}} = 0.1 \, \text{M} \]
- For \( PCl_3 \):
\[ C_{PCl_3} = \frac{0.2 \, \text{mol}}{2 \, \text{L}} = 0.1 \, \text{M} \]
- For \( Cl_2 \):
\[ C_{Cl_2} = \frac{0.2 \, \text{mol}}{2 \, \text{L}} = 0.1 \, \text{M} \]

5. **Write the Expression for the Equilibrium Constant \( K_c \):**
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

6. **Substitute the Concentrations into the \( K_c \) Expression:**
\[ K_c = \frac{(0.1)(0.1)}{0.1} \]

7. **Calculate \( K_c \):**
Simplifying the expression:
\[ K_c = \frac{0.01}{0.1} = 0.1 \]

8. **Conclusion:**
The equilibrium constant \( K_c \) for the reaction is \( 0.1 \).

### Final Answer:
The equilibrium constant \( K_c \) for the reaction is \( 0.1 \).

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