The conversion of `A rarr B` follows second-order kinetics. Doubling the concentration of `A` will increase the rate of formation of `B` by a factor

Options

A. `1//4`
B. 2
C. `1//2`
D. 4

Explanation

To solve the problem, we need to analyze how the rate of the reaction changes when the concentration of the reactant \( A \) is doubled, given that the reaction follows second-order kinetics.

### Step-by-Step Solution:

1. **Understanding the Rate Law**:
The rate of a reaction can be expressed using the rate law. For a second-order reaction with respect to \( A \), the rate \( R \) is given by:
\[ R = k[A]^2 \]
where \( k \) is the rate constant and \( [A] \) is the concentration of reactant \( A \).

2. **Initial Rate Calculation**:
Let’s denote the initial concentration of \( A \) as \( [A] = A \). Therefore, the initial rate \( R \) can be expressed as:
\[ R = k[A]^2 = kA^2 \]

3. **Doubling the Concentration**:
Now, if we double the concentration of \( A \), the new concentration becomes \( [A] = 2A \). We can calculate the new rate \( R' \) using this new concentration:
\[ R' = k(2A)^2 \]

4. **Calculating the New Rate**:
Expanding the expression for \( R' \):
\[ R' = k(4A^2) = 4kA^2 \]

5. **Relating the New Rate to the Initial Rate**:
Since the initial rate \( R \) is \( kA^2 \), we can express \( R' \) in terms of \( R \):
\[ R' = 4R \]

6. **Conclusion**:
Therefore, when the concentration of \( A \) is doubled, the rate of formation of \( B \) increases by a factor of 4.

### Final Answer:
Doubling the concentration of \( A \) will increase the rate of formation of \( B \) by a factor of **4**.

---