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. (a) 0.266 M
B. (b) 0.133 M
C. (c) 2.5 M
D. (d) 0.20 M

Explanation

To solve the problem step by step, we will follow the reaction and the calculations involved:

### Step 1: Write the Reaction
The dissociation reaction of phosphorus pentachloride (PCl₅) is:
\[ \text{PCl}_{5(g)} \rightleftharpoons \text{PCl}_{3(g)} + \text{Cl}_{2(g)} \]

### Step 2: Determine Initial Moles and Volume
We are given:
- Initial moles of PCl₅ = 5 moles
- Volume of the vessel = 5 liters

### Step 3: Calculate Moles Dissociated
We are told that 40% of PCl₅ dissociates. Therefore, the number of moles that dissociate is:
\[ \text{Moles dissociated} = 40\% \text{ of } 5 \text{ moles} = 0.4 \times 5 = 2 \text{ moles} \]

### Step 4: Calculate Moles at Equilibrium
At equilibrium:
- Moles of PCl₅ remaining = Initial moles - Moles dissociated
\[ \text{Moles of PCl}_{5} = 5 - 2 = 3 \text{ moles} \]
- Moles of PCl₃ formed = 2 moles (since 1 mole of PCl₅ produces 1 mole of PCl₃)
- Moles of Cl₂ formed = 2 moles (since 1 mole of PCl₅ produces 1 mole of Cl₂)

### Step 5: Calculate Concentrations at Equilibrium
To find the concentrations, we divide the number of moles by the volume (5 liters):
- Concentration of PCl₅:
\[ [\text{PCl}_{5}] = \frac{3 \text{ moles}}{5 \text{ L}} = 0.6 \text{ M} \]
- Concentration of PCl₃:
\[ [\text{PCl}_{3}] = \frac{2 \text{ moles}}{5 \text{ L}} = 0.4 \text{ M} \]
- Concentration of Cl₂:
\[ [\text{Cl}_{2}] = \frac{2 \text{ moles}}{5 \text{ L}} = 0.4 \text{ M} \]

### Step 6: Write the Expression for Kc
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 Concentrations into the Kc Expression
Substituting the values we calculated:
\[ K_c = \frac{(0.4)(0.4)}{0.6} \]

### Step 8: Calculate Kc
Calculating the above expression:
\[ K_c = \frac{0.16}{0.6} = 0.2667 \]

### Final Answer
Thus, the value of \( K_c \) is approximately:
\[ K_c \approx 0.266 \]

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