`A_(3)(g)hArr 3A(g)`, In the above reaction, the initial conc. Of `A_(3)` is 'a' moles/lit. If x is degree of dissociation of `A_(3)`. The total number of moles at equilibrium will be

Options

A. `a-(ax)/(3)`
B. `(a)/(3)-x`
C. `(a-ax)/(2)`
D. None

Answer

To solve the problem, we need to determine the total number of moles at equilibrium for the reaction:

\[ A_3(g) \rightleftharpoons 3A(g) \]

Given:
- Initial concentration of \( A_3 \) = \( a \) moles/liter
- Degree of dissociation of \( A_3 \) = \( x \)

### Step-by-Step Solution:

1. **Initial Moles Calculation**:
- Initially, we have \( a \) moles of \( A_3 \) and 0 moles of \( A \).
- So, at \( t = 0 \):
- Moles of \( A_3 \) = \( a \)
- Moles of \( A \) = 0

2. **Change in Moles Due to Dissociation**:
- The degree of dissociation \( x \) means that \( x \) fraction of \( A_3 \) will dissociate.
- Therefore, the amount of \( A_3 \) that dissociates = \( ax \).
- Consequently, the amount of \( A \) produced from this dissociation = \( 3 \times (ax) = 3ax \).

3. **Moles at Equilibrium**:
- At equilibrium, the moles of \( A_3 \) remaining will be:
\[ \text{Moles of } A_3 = a - ax = a(1 - x) \]
- The moles of \( A \) formed will be:
\[ \text{Moles of } A = 3ax \]

4. **Total Moles at Equilibrium**:
- The total number of moles at equilibrium is the sum of the moles of \( A_3 \) and \( A \):
\[ \text{Total moles} = \text{Moles of } A_3 + \text{Moles of } A \]
\[ \text{Total moles} = a(1 - x) + 3ax \]
\[ \text{Total moles} = a(1 - x + 3x) = a(1 + 2x) \]

Thus, the total number of moles at equilibrium is:

\[ \text{Total moles} = a(1 + 2x) \]