For the following equilibrium in gaseous phase,
`N_(2)O_(4)hArr 2NO_(2)`
`NO_(2)` is `50%` of the total volume, when equilibrium is set up. Hence, percent dissociation of `N_(2)O_(4)` is :

Options

A. `50%`
B. `25%`
C. `66.66%`
D. `33.33%`

Answer

To solve the problem, we need to determine the percent dissociation of \( N_2O_4 \) at equilibrium, given that \( NO_2 \) constitutes 50% of the total volume.

### Step-by-Step Solution:

1. **Define Initial Conditions**:
Let the initial number of moles of \( N_2O_4 \) be \( N \). Initially, there are no moles of \( NO_2 \).

2. **Set Up the Equilibrium Expression**:
At equilibrium, let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). Then:
- Moles of \( N_2O_4 \) at equilibrium = \( N - N\alpha = N(1 - \alpha) \)
- Moles of \( NO_2 \) produced = \( 2N\alpha \)

3. **Total Moles at Equilibrium**:
The total number of moles at equilibrium is:
\[ \text{Total moles} = N(1 - \alpha) + 2N\alpha = N(1 + \alpha) \]

4. **Mole Fraction of \( NO_2 \)**:
The mole fraction of \( NO_2 \) at equilibrium is given by:
\[ \text{Mole fraction of } NO_2 = \frac{2N\alpha}{N(1 + \alpha)} = \frac{2\alpha}{1 + \alpha} \]

5. **Given Condition**:
We know that the mole fraction of \( NO_2 \) is 0.5 (since it occupies 50% of the total volume):
\[ \frac{2\alpha}{1 + \alpha} = 0.5 \]

6. **Solve for \( \alpha \)**:
Cross-multiplying gives:
\[ 2\alpha = 0.5(1 + \alpha) \]
Expanding and rearranging:
\[ 2\alpha = 0.5 + 0.5\alpha \]
\[ 2\alpha - 0.5\alpha = 0.5 \]
\[ 1.5\alpha = 0.5 \]
\[ \alpha = \frac{0.5}{1.5} = \frac{1}{3} \]

7. **Calculate Percent Dissociation**:
The percent dissociation of \( N_2O_4 \) is given by:
\[ \text{Percent dissociation} = \alpha \times 100 = \frac{1}{3} \times 100 = 33.33\% \]

### Final Answer:
The percent dissociation of \( N_2O_4 \) is **33.33%**.