1 mole of `N_2` and 3 moles of `PCl_5` are placed in a 100 litre vessels heated at ` 227^(@) C ` the equilibrium pressure is 2.05 atm Assuming ideal behaviour,Calculate degree of dissociation of `PCl_5 and K_p ` for the reaction
` PCl_5 (g) hArr PCl_3 (g) + Cl_2(g)`

Explanation

To solve the problem, we need to calculate the degree of dissociation of \( PCl_5 \) and the equilibrium constant \( K_p \) for the reaction:

\[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \]

### Step 1: Initial Moles and Changes at Equilibrium

Initially, we have:
- Moles of \( PCl_5 = 3 \)
- Moles of \( PCl_3 = 0 \)
- Moles of \( Cl_2 = 0 \)

Let \( x \) be the degree of dissociation of \( PCl_5 \). At equilibrium, the moles will change as follows:
- Moles of \( PCl_5 = 3 - x \)
- Moles of \( PCl_3 = x \)
- Moles of \( Cl_2 = x \)

### Step 2: Total Moles at Equilibrium

The total number of moles at equilibrium will be:
\[ \text{Total moles} = (3 - x) + x + x + 1 = 4 + x \]
(Note: We add 1 for the 1 mole of \( N_2 \) present.)

### Step 3: Using Ideal Gas Law to Find Total Moles

We know the equilibrium pressure \( P = 2.05 \, \text{atm} \) and the volume \( V = 100 \, \text{L} \). Using the ideal gas law \( PV = nRT \), we can find the total number of moles \( n \):

\[ n = \frac{PV}{RT} \]

Where:
- \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- Temperature \( T = 227 + 273 = 500 \, \text{K} \)

Substituting the values:
\[ n = \frac{2.05 \times 100}{0.082 \times 500} = \frac{205}{41} = 5 \, \text{moles} \]

### Step 4: Setting Up the Equation

Now, we can set up the equation for the total moles at equilibrium:
\[ 4 + x = 5 \]

### Step 5: Solving for \( x \)

Solving for \( x \):
\[ x = 5 - 4 = 1 \]

### Step 6: Degree of Dissociation

The degree of dissociation \( \alpha \) is given as:
\[ \alpha = \frac{x}{3} = \frac{1}{3} \approx 0.333 \]

### Step 7: Calculate \( K_p \)

To find \( K_p \), we can use the expression:
\[ K_p = \frac{(P_{PCl_3})(P_{Cl_2})}{(P_{PCl_5})} \]

Where:
- \( P_{PCl_5} = \frac{(3 - x)}{(4 + x)} \times P \)
- \( P_{PCl_3} = \frac{x}{(4 + x)} \times P \)
- \( P_{Cl_2} = \frac{x}{(4 + x)} \times P \)

Substituting \( x = 1 \):
- \( P_{PCl_5} = \frac{(3 - 1)}{(4 + 1)} \times 2.05 = \frac{2}{5} \times 2.05 = 0.82 \, \text{atm} \)
- \( P_{PCl_3} = \frac{1}{5} \times 2.05 = 0.41 \, \text{atm} \)
- \( P_{Cl_2} = \frac{1}{5} \times 2.05 = 0.41 \, \text{atm} \)

Now substituting into \( K_p \):
\[ K_p = \frac{(0.41)(0.41)}{0.82} = \frac{0.1681}{0.82} \approx 0.204 \, \text{atm} \]

### Final Answers
- Degree of dissociation \( \alpha = 0.333 \)
- Equilibrium constant \( K_p = 0.204 \, \text{atm} \)