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 chemical reaction and the conditions stated in the question.

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ 3A (g) + B(g) \rightleftharpoons 2C(g) + D(g) \]

### Step 2: Define initial moles of A and B
Let the initial moles of A be \( x \). According to the problem, the initial moles of B are double that of A. Therefore, we can write:
- Initial moles of A = \( x \)
- Initial moles of B = \( 2x \)
- Initial moles of C = 0
- Initial moles of D = 0

### Step 3: Set up the changes in moles at equilibrium
Let \( y \) be the amount of A that dissociates at equilibrium. The changes in moles for each substance can be expressed as follows:
- Moles of A at equilibrium = \( x - 3y \)
- Moles of B at equilibrium = \( 2x - y \)
- Moles of C at equilibrium = \( 2y \)
- Moles of D at equilibrium = \( y \)

### Step 4: Use the given condition at equilibrium
According to the problem, at equilibrium, the moles of A and C are equal:
\[ x - 3y = 2y \]

### Step 5: Solve for \( x \) in terms of \( y \)
Rearranging the equation gives:
\[ x - 3y = 2y \]
\[ x = 5y \]

### Step 6: Substitute \( x \) back to find moles at equilibrium
Now, substituting \( x = 5y \) back into the expressions for moles at equilibrium:
- Moles of A at equilibrium = \( 5y - 3y = 2y \)
- Moles of B at equilibrium = \( 2(5y) - y = 10y - y = 9y \)
- Moles of C at equilibrium = \( 2y \)
- Moles of D at equilibrium = \( y \)

### Step 7: Calculate the percentage of dissociation
The percentage of dissociation is calculated based on the initial and final moles of B:
\[ \text{Percentage of dissociation} = \frac{\text{Initial moles of B} - \text{Final moles of B}}{\text{Initial moles of B}} \times 100 \]

Substituting the values:
- Initial moles of B = \( 2x = 2(5y) = 10y \)
- Final moles of B = \( 9y \)

Now, substituting these into the formula:
\[ \text{Percentage of dissociation} = \frac{10y - 9y}{10y} \times 100 = \frac{y}{10y} \times 100 = 10\% \]

### Final Answer
The percentage of dissociation is \( 10\% \).

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