5 moles of `PCl_(5)` are heated in a closed vessel of 5 litre capacity. At equilibrium `40%` of `PCl_(5)` is found to be dissociated. What is the value of `K_(c)` ?

Options

A. 0.266 M
B. 0.133 M
C. 2.5 M
D. 0.20 M

Explanation

To solve the problem step by step, we need to follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of phosphorus pentachloride (PCl₅) can be represented as:
\[ \text{PCl}_5 (g) \rightleftharpoons \text{PCl}_3 (g) + \text{Cl}_2 (g) \]

### Step 2: Determine the initial moles and concentration of PCl₅
We are given that there are 5 moles of PCl₅ in a closed vessel of 5 liters.

Initial concentration of PCl₅:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} = \frac{5 \, \text{moles}}{5 \, \text{liters}} = 1 \, \text{M} \]

### Step 3: Calculate the moles of PCl₅ that dissociate
We know that 40% of PCl₅ is dissociated at equilibrium.

Moles of PCl₅ that dissociate:
\[ \text{Dissociated moles} = 0.40 \times 5 \, \text{moles} = 2 \, \text{moles} \]

### Step 4: Calculate the moles of PCl₅, PCl₃, and Cl₂ at equilibrium
- Moles of PCl₅ remaining:
\[ \text{Remaining moles of PCl}_5 = 5 - 2 = 3 \, \text{moles} \]

- Moles of PCl₃ produced:
Since 2 moles of PCl₅ dissociate, 2 moles of PCl₃ are produced.

- Moles of Cl₂ produced:
Since 2 moles of PCl₅ dissociate, 2 moles of Cl₂ are produced.

### Step 5: Calculate the equilibrium concentrations
Now, we can calculate the equilibrium concentrations of each species in the 5-liter vessel.

- Concentration of PCl₅:
\[ [\text{PCl}_5] = \frac{3 \, \text{moles}}{5 \, \text{liters}} = 0.6 \, \text{M} \]

- Concentration of PCl₃:
\[ [\text{PCl}_3] = \frac{2 \, \text{moles}}{5 \, \text{liters}} = 0.4 \, \text{M} \]

- Concentration of Cl₂:
\[ [\text{Cl}_2] = \frac{2 \, \text{moles}}{5 \, \text{liters}} = 0.4 \, \text{M} \]

### Step 6: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]

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

### Step 8: Calculate \( K_c \)
\[ K_c = \frac{0.16}{0.6} = 0.2667 \, \text{M} \]

### Final Answer
Thus, the value of \( K_c \) is approximately \( 0.267 \, \text{M} \).

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