`1`mol of `N_(2)` and `3` mol of `PCl_(5)` are placed in a `100 L` vessel heated to `227^(@)C`. The equilibrium pressure is `2.05` atm. Assuming ideal behaviour, calculate the degree of dissociation for `PCl_(5)` and `K_(p)` for the reaction.
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`

Answer

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: Determine Initial Moles
We start with:
- 1 mole of \( N_2 \)
- 3 moles of \( PCl_5 \)

### Step 2: Set Up the Change in Moles at Equilibrium
Let \( x \) be the degree of dissociation of \( PCl_5 \). At equilibrium:
- Moles of \( PCl_5 \) = \( 3 - x \)
- Moles of \( PCl_3 \) = \( x \)
- Moles of \( Cl_2 \) = \( x \)

### Step 3: Calculate Total Moles at Equilibrium
The total moles at equilibrium will be:
\[ \text{Total moles} = 1 + (3 - x) + x + x = 4 + x \]

### Step 4: Use Ideal Gas Law to Relate Pressure and Moles
Using the ideal gas law \( PV = nRT \):
- Given \( P = 2.05 \) atm
- Volume \( V = 100 \) L
- \( R = 0.0821 \) L·atm/(K·mol)
- Temperature \( T = 227 + 273 = 500 \) K

We can rearrange the ideal gas law to find the total number of moles at equilibrium:
\[ n = \frac{PV}{RT} \]
Substituting in the values:
\[ n = \frac{2.05 \times 100}{0.0821 \times 500} = \frac{205}{41.05} \approx 4.99 \text{ moles} \]

### Step 5: Set Up the Equation for Degree of Dissociation
From the total moles at equilibrium:
\[ 4 + x = 4.99 \]
Solving for \( x \):
\[ x = 4.99 - 4 = 0.99 \]

### Step 6: Calculate the Degree of Dissociation
The degree of dissociation \( \alpha \) is given by:
\[ \alpha = \frac{x}{\text{initial moles of } PCl_5} = \frac{0.99}{3} \approx 0.33 \]
To express this as a percentage:
\[ \text{Percentage dissociation} = \alpha \times 100 = 33.3\% \]

### Step 7: Calculate \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]
Where:
- \( P_{PCl_5} = \frac{(3 - x) \cdot (2.05)}{4.99} \)
- \( P_{PCl_3} = \frac{x \cdot (2.05)}{4.99} \)
- \( P_{Cl_2} = \frac{x \cdot (2.05)}{4.99} \)

Substituting \( x = 0.99 \):
\[ P_{PCl_5} = \frac{(3 - 0.99) \cdot 2.05}{4.99} = \frac{2.01 \cdot 2.05}{4.99} \approx 0.81 \text{ atm} \]
\[ P_{PCl_3} = P_{Cl_2} = \frac{0.99 \cdot 2.05}{4.99} \approx 0.40 \text{ atm} \]

Now substituting into the \( K_p \) expression:
\[ K_p = \frac{(0.40)(0.40)}{0.81} \approx \frac{0.16}{0.81} \approx 0.1975 \approx 0.20 \]

### Final Answers
- Degree of dissociation of \( PCl_5 \) = 33.3%
- \( K_p \approx 0.20 \)