`1.5 mol` of `PCl_(5)` are heated at constant temperature in a closed vessel of `4 L` capacity. At the equilibrium point, `PCl_(5)` is `35%` dissociated into `PCl_(3)` and `Cl_(2)`. Calculate the equilibrium constant.

Explanation

To solve the problem of calculating the equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \), we will follow these steps:

### Step 1: Write the balanced chemical equation
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 the change at equilibrium
We start with \( 1.5 \) moles of \( PCl_5 \) in a \( 4 \) L container. At equilibrium, \( 35\% \) of \( PCl_5 \) is dissociated.

- Initial moles of \( PCl_5 \) = \( 1.5 \) moles
- Moles of \( PCl_5 \) dissociated = \( 0.35 \times 1.5 = 0.525 \) moles

### Step 3: Calculate moles at equilibrium
At equilibrium, the moles of each species will be:
- Moles of \( PCl_5 \) at equilibrium = \( 1.5 - 0.525 = 0.975 \) moles
- Moles of \( PCl_3 \) at equilibrium = \( 0 + 0.525 = 0.525 \) moles
- Moles of \( Cl_2 \) at equilibrium = \( 0 + 0.525 = 0.525 \) moles

### Step 4: Calculate concentrations at equilibrium
To find the concentrations, we divide the number of moles by the volume (4 L):
- Concentration of \( PCl_5 \) = \( \frac{0.975 \text{ moles}}{4 \text{ L}} = 0.24375 \, \text{M} \)
- Concentration of \( PCl_3 \) = \( \frac{0.525 \text{ moles}}{4 \text{ L}} = 0.13125 \, \text{M} \)
- Concentration of \( Cl_2 \) = \( \frac{0.525 \text{ moles}}{4 \text{ L}} = 0.13125 \, \text{M} \)

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 6: Substitute the concentrations into the \( K_c \) expression
Substituting the concentrations we calculated:
\[ K_c = \frac{(0.13125)(0.13125)}{0.24375} \]

### Step 7: Calculate \( K_c \)
Calculating the above expression:
\[ K_c = \frac{0.017265625}{0.24375} \approx 0.0708 \]

### Final Answer
The equilibrium constant \( K_c \) is approximately \( 0.0708 \).

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