If in the reaction, `N_(2)O_(4)(g)hArr2NO_(2)(g), alpha` is the degree of dissociation of `N_(2)O_(4)`, then the number of moles at equilibrium will be

Options

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

Answer

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

\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

Let’s denote:
- The initial number of moles of \( N_2O_4 \) as \( C \).
- The degree of dissociation of \( N_2O_4 \) as \( \alpha \).

### Step 1: Initial Moles
At the start of the reaction (before any dissociation occurs), we have:
- Moles of \( N_2O_4 \) = \( C \)
- Moles of \( NO_2 \) = \( 0 \)

### Step 2: Change in Moles
As the reaction proceeds, \( N_2O_4 \) dissociates. The degree of dissociation \( \alpha \) indicates the fraction of \( N_2O_4 \) that has dissociated. Therefore, at equilibrium:
- Moles of \( N_2O_4 \) that dissociate = \( C \alpha \)
- Remaining moles of \( N_2O_4 \) = \( C - C\alpha = C(1 - \alpha) \)

Since each mole of \( N_2O_4 \) produces 2 moles of \( NO_2 \), the moles of \( NO_2 \) produced will be:
- Moles of \( NO_2 \) = \( 2 \times C\alpha = 2C\alpha \)

### Step 3: Total Moles at Equilibrium
Now, we can calculate the total number of moles at equilibrium by adding the moles of both species:
- Total moles at equilibrium = Moles of \( N_2O_4 \) + Moles of \( NO_2 \)
- Total moles at equilibrium = \( C(1 - \alpha) + 2C\alpha \)

### Step 4: Simplifying the Expression
Now, we simplify the expression:
- Total moles at equilibrium = \( C(1 - \alpha) + 2C\alpha \)
- = \( C - C\alpha + 2C\alpha \)
- = \( C + C\alpha \)
- = \( C(1 + \alpha) \)

### Conclusion
Thus, the total number of moles at equilibrium is:
\[ \text{Total moles at equilibrium} = C(1 + \alpha) \]

### Final Answer
The number of moles at equilibrium will be \( C(1 + \alpha) \).

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