For the reaction `A+3B hArr 2C+D` initial mole of A is twice that of B . If at equilibrium moles of B and C are equal , then percent of B reacted is

Options

A. 10
B. 20
C. 40
D. 60

Answer

To solve the problem step by step, let's analyze the given reaction and the conditions provided:

### Given Reaction:
\[ A + 3B \rightleftharpoons 2C + D \]

### Initial Conditions:
- Let the initial moles of \( B \) be \( x \).
- Therefore, the initial moles of \( A \) will be \( 2x \) (since the initial moles of \( A \) are twice that of \( B \)).

### At Equilibrium:
Let \( y \) be the moles of \( A \) that reacted. According to the stoichiometry of the reaction:
- Moles of \( A \) at equilibrium = \( 2x - y \)
- Moles of \( B \) at equilibrium = \( x - 3y \) (since 3 moles of \( B \) are consumed for every mole of \( A \) that reacts)
- Moles of \( C \) at equilibrium = \( 2y \) (since 2 moles of \( C \) are produced for every mole of \( A \) that reacts)
- Moles of \( D \) at equilibrium = \( y \) (since 1 mole of \( D \) is produced for every mole of \( A \) that reacts)

### Given Condition:
At equilibrium, the moles of \( B \) and \( C \) are equal:
\[ x - 3y = 2y \]

### Solving for \( y \):
1. Rearranging the equation:
\[ x - 3y = 2y \]
\[ x = 5y \]

### Finding Percent of \( B \) Reacted:
1. Initially, the moles of \( B \) = \( x = 5y \).
2. At equilibrium, the moles of \( B \) = \( x - 3y = 2y \).
3. The moles of \( B \) that reacted = Initial moles of \( B \) - Final moles of \( B \):
\[ \text{Moles of } B \text{ reacted} = 5y - 2y = 3y \]

4. Percent of \( B \) reacted:
\[ \text{Percent of } B \text{ reacted} = \left( \frac{\text{Moles of } B \text{ reacted}}{\text{Initial moles of } B} \right) \times 100 \]
\[ = \left( \frac{3y}{5y} \right) \times 100 = \frac{3}{5} \times 100 = 60\% \]

### Final Answer:
The percent of \( B \) reacted is **60%**.