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 of how tripling the concentration of A affects the rate of the reaction A to B, which follows second-order kinetics, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Rate Law for Second Order Kinetics**:
For a reaction that follows second-order kinetics, the rate of reaction (R) can be expressed as:
\[ R = k[A]^2 \]
where \( k \) is the rate constant and \([A]\) is the concentration of reactant A.

2. **Initial Rate Expression**:
Let's denote the initial concentration of A as \([A]\). Therefore, the initial rate of reaction can be written as:
\[ R = k[A]^2 \]

3. **Change in Concentration**:
If the concentration of A is tripled, the new concentration will be:
\[ [A'] = 3[A] \]

4. **New Rate Expression**:
Substitute the new concentration into the rate law:
\[ R' = k[A']^2 = k(3[A])^2 \]
Simplifying this gives:
\[ R' = k \cdot 9[A]^2 \]

5. **Relate New Rate to Initial Rate**:
Now, we can relate the new rate \( R' \) to the initial rate \( R \):
\[ R' = 9k[A]^2 = 9R \]

6. **Conclusion**:
Therefore, tripling the concentration of A increases the rate of reaction by a factor of 9.

### Final Answer:
The rate of reaction increases by a factor of **9**.

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