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. 0.1
B. 0.2
C. 0.4
D. 0.6

Answer

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

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

### Step 2: Define initial moles
Let the initial moles of B be \( x \). According to the problem, the initial moles of A are twice that of B, so:
\[ \text{Initial moles of A} = 2x \]
\[ \text{Initial moles of B} = x \]

### Step 3: Define changes at equilibrium
Let \( y \) be the amount of A that reacts at equilibrium. Since 1 mole of A reacts with 3 moles of B, the change in moles of B will be \( 3y \) and the change in moles of C will be \( 2y \).

At equilibrium:
- Moles of A = \( 2x - y \)
- Moles of B = \( x - 3y \)
- Moles of C = \( 2y \)
- Moles of D = \( y \) (not needed for this problem)

### Step 4: Set up the equation based on the condition given
We are told that at equilibrium, the moles of B and C are equal:
\[ x - 3y = 2y \]

### Step 5: Solve the equation
Rearranging the equation:
\[ x - 3y = 2y \]
\[ x = 5y \]

### Step 6: Find the value of \( y \)
From the equation \( x = 5y \), we can express \( y \) in terms of \( x \):
\[ y = \frac{x}{5} \]

### Step 7: Calculate the amount of B that reacted
The amount of B that reacted is given by:
\[ \text{Amount of B reacted} = 3y = 3 \left(\frac{x}{5}\right) = \frac{3x}{5} \]

### Step 8: Calculate the percentage of B that reacted
The percentage of B that reacted is calculated as follows:
\[ \text{Percentage of B reacted} = \left(\frac{\text{Amount of B reacted}}{\text{Initial amount of B}}\right) \times 100 \]
Substituting the values:
\[ \text{Percentage of B reacted} = \left(\frac{\frac{3x}{5}}{x}\right) \times 100 = \left(\frac{3}{5}\right) \times 100 = 60\% \]

### Final Answer
The percentage of B that reacted is **60%**.