If in the reaction `N_(2)O_(4) = 2N_(2)` , `alpha` is that part of `N_(2)O_(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 dissociation of \( N_2O_4 \) into \( N_2 \) and determine the number of moles at equilibrium.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Let's assume we start with 1 mole of \( N_2O_4 \) at the beginning of the reaction.

2. **Define Alpha:**
- Let \( \alpha \) be the fraction of \( N_2O_4 \) that dissociates. This means that \( \alpha \) moles of \( N_2O_4 \) will dissociate.

3. **Calculate Remaining Moles of \( N_2O_4 \):**
- The amount of \( N_2O_4 \) remaining after dissociation will be:
\[ \text{Remaining } N_2O_4 = 1 - \alpha \]

4. **Calculate Moles of \( N_2 \) Produced:**
- According to the reaction \( N_2O_4 \rightarrow 2N_2 \), for every mole of \( N_2O_4 \) that dissociates, 2 moles of \( N_2 \) are produced. Therefore, the moles of \( N_2 \) produced will be:
\[ \text{Produced } N_2 = 2\alpha \]

5. **Total Moles at Equilibrium:**
- The total number of moles at equilibrium can be calculated by adding the remaining moles of \( N_2O_4 \) and the produced moles of \( N_2 \):
\[ \text{Total moles} = \text{Remaining } N_2O_4 + \text{Produced } N_2 \]
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

6. **Final Result:**
- Therefore, the total number of moles at equilibrium is:
\[ \text{Total moles} = 1 + \alpha \]