In the following reaction, `3A (g)+B(g) hArr 2C(g) +D(g)`, Initial moles of `B` is double at A . At equilibrium, moles of A and C are equal. Hence `%` dissociation is `:`

Options

A. `10%`
B. `20%`
C. `40%`
D. `5%`

Answer

To solve the problem step by step, we will analyze the given reaction and the information provided:

**Given Reaction:**
\[ 3A(g) + B(g) \rightleftharpoons 2C(g) + D(g) \]

**Step 1: Define Initial Moles**
Let the initial moles of \( A \) be \( A \) (we'll use \( A \) as a variable). According to the problem, the initial moles of \( B \) are double that of \( A \). Therefore:
- Initial moles of \( A = A \)
- Initial moles of \( B = 2A \)
- Initial moles of \( C = 0 \)
- Initial moles of \( D = 0 \)

**Step 2: Define Change in Moles at Equilibrium**
Let \( x \) be the amount of \( A \) that dissociates at equilibrium. The changes in moles can be represented as follows:
- Moles of \( A \) at equilibrium: \( A - 3x \)
- Moles of \( B \) at equilibrium: \( 2A - x \)
- Moles of \( C \) at equilibrium: \( 2x \)
- Moles of \( D \) at equilibrium: \( x \)

**Step 3: Set Up the Equation Based on Given Condition**
We know that at equilibrium, the moles of \( A \) and \( C \) are equal:
\[ A - 3x = 2x \]

**Step 4: Solve for \( x \)**
Rearranging the equation:
\[ A = 5x \]
Thus, we can express \( x \) in terms of \( A \):
\[ x = \frac{A}{5} \]

**Step 5: Calculate the Percentage Dissociation**
The percentage dissociation is given by the formula:
\[ \text{Percentage Dissociation} = \left( \frac{x}{\text{Initial moles of } A} \right) \times 100 \]
Substituting the values we have:
\[ \text{Percentage Dissociation} = \left( \frac{\frac{A}{5}}{A} \right) \times 100 \]
\[ = \left( \frac{1}{5} \right) \times 100 \]
\[ = 20\% \]

**Final Answer:**
The percentage dissociation is \( 20\% \).

---