The conversion of A `to` B, follows second order kinetics. tripling the concentration of A will increase the rate of reaction by a factor of :

Options

A. `1//9`
B. 3
C. `1//3`
D. 9

Answer

To solve the problem, we need to analyze how the rate of a reaction changes when the concentration of a reactant is altered, specifically in the context of second-order kinetics.

### Step-by-Step Solution:

1. **Identify the Rate Law Expression**:
The reaction given is the conversion of A to B, which follows second-order kinetics. The rate law for a second-order reaction can be expressed as:
\[ \text{Rate} (r) = k [A]^2 \]
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 written as:
\[ r = k a^2 \]

3. **Change in Concentration**:
According to the problem, the concentration of A is tripled. Thus, the new concentration of A becomes:
\[ [A] = 3a \]

4. **New Rate Calculation**:
We can now calculate the new rate \(r'\) when the concentration is tripled:
\[ r' = k (3a)^2 \]
Simplifying this gives:
\[ r' = k \cdot 9a^2 = 9k a^2 \]

5. **Determine the Factor of Increase in Rate**:
To find out how much the rate has increased, we can take the ratio of the new rate \(r'\) to the initial rate \(r\):
\[ \frac{r'}{r} = \frac{9k a^2}{k a^2} \]
Here, \(k a^2\) cancels out:
\[ \frac{r'}{r} = 9 \]

6. **Conclusion**:
This means that tripling the concentration of A increases the rate of reaction by a factor of 9. Therefore, the correct answer is:
\[ \text{The rate of reaction increases by a factor of } 9. \]

### Final Answer:
The correct option is 9.