Two moles of `PCl_(5)` were heated to `327^(@)C` in a closed two-litre vessel, and when equilibrium was achieved, `PCl_(5)` was found to be `40%` dissociated into `PCl_(3)` and `Cl_(2)`. Calculate the equilibrium constant `K_(p)` and `K_(c)` for this reaction.

Explanation

To solve the problem of calculating the equilibrium constants \( K_p \) and \( K_c \) for the dissociation of \( PCl_5 \), we can follow these steps:

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

### Step 2: Set up the initial conditions
We start with 2 moles of \( PCl_5 \) in a 2-liter vessel. Therefore, the initial concentrations are:
- \( [PCl_5] = \frac{2 \text{ moles}}{2 \text{ L}} = 1 \text{ M} \)
- \( [PCl_3] = 0 \text{ M} \)
- \( [Cl_2] = 0 \text{ M} \)

### Step 3: Determine the change in concentration at equilibrium
Given that \( PCl_5 \) is 40% dissociated, we can calculate the moles that dissociate:
\[ \text{Moles dissociated} = 0.4 \times 2 = 0.8 \text{ moles} \]
At equilibrium:
- Moles of \( PCl_5 \) remaining = \( 2 - 0.8 = 1.2 \) moles
- Moles of \( PCl_3 \) formed = \( 0.8 \) moles
- Moles of \( Cl_2 \) formed = \( 0.8 \) moles

### Step 4: Calculate the equilibrium concentrations
Now we can find the equilibrium concentrations:
- \( [PCl_5] = \frac{1.2 \text{ moles}}{2 \text{ L}} = 0.6 \text{ M} \)
- \( [PCl_3] = \frac{0.8 \text{ moles}}{2 \text{ L}} = 0.4 \text{ M} \)
- \( [Cl_2] = \frac{0.8 \text{ moles}}{2 \text{ L}} = 0.4 \text{ M} \)

### Step 5: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.4)(0.4)}{0.6} = \frac{0.16}{0.6} \approx 0.267 \text{ M} \]

### Step 6: Calculate \( K_p \)
To find \( K_p \), we use the relation:
\[ K_p = K_c \cdot R T^{\Delta n} \]
Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 327 \, \text{C} + 273 = 600 \, \text{K} \)
- \( \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - 1 = 1 \)

Substituting the values:
\[ K_p = 0.267 \cdot 0.0821 \cdot (600)^1 \]
Calculating:
\[ K_p = 0.267 \cdot 0.0821 \cdot 600 \approx 13.15 \text{ atm} \]

### Final Answers
- \( K_c \approx 0.267 \, \text{M} \)
- \( K_p \approx 13.15 \, \text{atm} \)