For a chemical reaction rate law is, rate `= k{A]^(2)[B]`. If [A] is doubled at constant [B], the rate of reaction

Options

A. increases by a factor of 8
B. increases by a factor of 4
C. incr-eases by a factor of 3
D. increases by a factor of 2

Explanation

To solve the problem, we need to analyze how the rate of reaction changes when the concentration of reactant A is doubled while keeping the concentration of reactant B constant.

### Step-by-Step Solution:

1. **Understand the Rate Law Expression**:
The rate law is given as:
\[ \text{Rate} = k[A]^2[B] \]
Here, \(k\) is the rate constant, \([A]\) is the concentration of reactant A, and \([B]\) is the concentration of reactant B.

2. **Define Initial Rate**:
Let the initial concentration of A be \([A]\) and the concentration of B be \([B]\). The initial rate of the reaction can be expressed as:
\[ R = k[A]^2[B] \]

3. **Change the Concentration of A**:
If the concentration of A is doubled, then the new concentration of A becomes \(2[A]\). The concentration of B remains constant at \([B]\).

4. **Calculate the New Rate**:
Substitute \(2[A]\) into the rate law to find the new rate \(R'\):
\[ R' = k(2[A])^2[B] \]

5. **Simplify the Expression**:
Now, simplify the expression for \(R'\):
\[ R' = k(4[A]^2)[B] = 4k[A]^2[B] \]

6. **Relate New Rate to Initial Rate**:
Since \(R = k[A]^2[B]\), we can substitute this into the equation for \(R'\):
\[ R' = 4R \]

7. **Conclusion**:
Therefore, when the concentration of A is doubled, the rate of reaction increases by a factor of 4.

### Final Answer:
The rate of reaction increases by a factor of 4.