At `298 K`, the equilibrium between `N_(2)O_(4)` and `NO_(2)` is represented as : `N_(2)O_(4(g))hArr2NO_(2(g))`. If the total pressure of the equilibrium mixture is `P` and the degree of dissociation of `N_(2)O_(4(g))` at `298 K` is `x`, the partial pressure of `NO_(2(g))` under these conditions is:

Options

A. `((2x)/(1+x))P`
B. `2xp`
C. `((2x)/(1-x))P`
D. `(2)/(3)xp`

Answer

To find the partial pressure of \( NO_2 \) at equilibrium given the dissociation of \( N_2O_4 \) at 298 K, we can follow these steps:

### Step-by-Step Solution

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

2. **Define Initial Conditions**:
Let's assume we start with 1 mole of \( N_2O_4 \) and no \( NO_2 \) at time \( t = 0 \):
- Initial moles of \( N_2O_4 = 1 \)
- Initial moles of \( NO_2 = 0 \)

3. **Define Degree of Dissociation**:
Let \( x \) be the degree of dissociation of \( N_2O_4 \). This means that at equilibrium:
- Moles of \( N_2O_4 \) that dissociate = \( x \)
- Moles of \( NO_2 \) formed = \( 2x \)

4. **Calculate Moles at Equilibrium**:
At equilibrium, the moles of each species will be:
- Moles of \( N_2O_4 = 1 - x \)
- Moles of \( NO_2 = 2x \)

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

6. **Calculate Mole Fraction of \( NO_2 \)**:
The mole fraction of \( NO_2 \) is given by:
\[ \text{Mole fraction of } NO_2 = \frac{\text{Moles of } NO_2}{\text{Total moles}} = \frac{2x}{1 + x} \]

7. **Calculate Partial Pressure of \( NO_2 \)**:
The partial pressure of \( NO_2 \) can be calculated using the total pressure \( P \):
\[ P_{NO_2} = \text{Mole fraction of } NO_2 \times P = \frac{2x}{1 + x} \times P \]

### Final Expression
Thus, the partial pressure of \( NO_2 \) at equilibrium is:
\[ P_{NO_2} = \frac{2xP}{1 + x} \]