At 273 K and 1atm , 10 litre of `N_(2)O_(4)` decompose to `NO_(4)` decompoes to ` NO_(2)` according to equation
`N_(2)O_(4)(g)hArr2NO_(@)(G)`
What is degree of dissociation `(alpha)` when the original volume is 25% less then that os existing volume?

Options

A. `0.25`
B. `0.33`
C. `0.66`
D. `0.5`

Answer

To solve the problem, we need to determine the degree of dissociation (α) of \( N_2O_4 \) when the original volume is 25% less than the existing volume after decomposition.

### Step-by-Step Solution:

1. **Write the Reaction**:
The decomposition reaction is given as:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

2. **Initial Moles**:
Let's assume the initial number of moles of \( N_2O_4 \) is \( a \). Given that the volume of \( N_2O_4 \) is 10 liters at 1 atm and 273 K, we can calculate \( a \) using the ideal gas law:
\[ PV = nRT \implies n = \frac{PV}{RT} \]
Where:
- \( P = 1 \, \text{atm} \)
- \( V = 10 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 273 \, \text{K} \)

Calculating \( n \):
\[ n = \frac{1 \times 10}{0.0821 \times 273} \approx 0.446 \, \text{moles} \]
Thus, \( a \approx 0.446 \).

3. **Change in Moles**:
If \( \alpha \) is the degree of dissociation, then at equilibrium:
- Moles of \( N_2O_4 \) remaining = \( a(1 - \alpha) \)
- Moles of \( NO_2 \) produced = \( 2a\alpha \)

Therefore, the total moles at equilibrium:
\[ n_{\text{total}} = a(1 - \alpha) + 2a\alpha = a(1 + \alpha) \]

4. **Volume Relationship**:
We know that the original volume is 25% less than the existing volume. Let \( V_{\text{equilibrium}} \) be the volume at equilibrium. Then:
\[ V_{\text{original}} = 0.75 V_{\text{equilibrium}} \]
Since volume is proportional to the number of moles at constant temperature and pressure, we can write:
\[ n_{\text{original}} = 0.75 n_{\text{total}} \]

5. **Set Up the Equation**:
Initially, \( n_{\text{original}} = a \), and substituting for \( n_{\text{total}} \):
\[ a = 0.75 \cdot a(1 + \alpha) \]

6. **Simplifying the Equation**:
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ 1 = 0.75(1 + \alpha) \]
Expanding this gives:
\[ 1 = 0.75 + 0.75\alpha \]
Rearranging:
\[ 0.25 = 0.75\alpha \]
Solving for \( \alpha \):
\[ \alpha = \frac{0.25}{0.75} = \frac{1}{3} \approx 0.33 \]

### Final Answer:
The degree of dissociation \( \alpha \) is approximately **0.33**.