If in the reaction `N_2O_4 hArr 2NO_(2),alpha` is that part of `N_2O_4` which dissociates, then the number of moles at equilibrium will be

Options

A. 3
B. 1
C. `(1- alpha)^2`
D. `(1+ alpha)`

Explanation

To solve the problem, we need to analyze the equilibrium of the reaction:

\[ N_2O_4 \rightleftharpoons 2NO_2 \]

Let's denote the initial number of moles of \( N_2O_4 \) as \( 1 \) mole (for simplicity). The variable \( \alpha \) represents the fraction of \( N_2O_4 \) that dissociates at equilibrium.

### Step-by-Step Solution:

1. **Initial Moles**:
- At the start (T = 0), we have:
- Moles of \( N_2O_4 \) = 1
- Moles of \( NO_2 \) = 0

2. **Change in Moles**:
- When \( \alpha \) moles of \( N_2O_4 \) dissociate, the changes in the moles will be:
- Moles of \( N_2O_4 \) that dissociate = \( \alpha \)
- Moles of \( NO_2 \) produced = \( 2\alpha \) (since 1 mole of \( N_2O_4 \) produces 2 moles of \( NO_2 \))

3. **Moles at Equilibrium**:
- At equilibrium, the moles will be:
- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \)

4. **Total Moles at Equilibrium**:
- To find the total number of moles at equilibrium, we add the moles of both species:
\[ \text{Total moles} = \text{Moles of } N_2O_4 + \text{Moles of } NO_2 \]
\[ \text{Total moles} = (1 - \alpha) + (2\alpha) = 1 + \alpha \]

5. **Final Answer**:
- Therefore, the total number of moles at equilibrium is \( 1 + \alpha \).