A reaction is first order in A and second order in B : How is the rate affected when the concentrations of both A and B are doubled?

Answer

To solve the problem, we need to analyze how the rate of the reaction changes when the concentrations of both reactants A and B are doubled.

### Step-by-Step Solution:

1. **Understand the Rate Law**:
The rate of a reaction can be expressed using the rate law. For a reaction that is first order in A and second order in B, the rate law is given by:
\[ \text{Rate} = k[A]^1[B]^2 \]
where \( k \) is the rate constant, \( [A] \) is the concentration of A, and \( [B] \) is the concentration of B.

2. **Initial Concentrations**:
Let the initial concentrations of A and B be \( [A] \) and \( [B] \) respectively. The initial rate of the reaction (let's call it \( R \)) can be expressed as:
\[ R = k[A][B]^2 \]

3. **Doubling the Concentrations**:
When the concentrations of both A and B are doubled, the new concentrations become \( 2[A] \) and \( 2[B] \).

4. **Calculate the New Rate**:
Substitute the new concentrations into the rate law:
\[ R' = k[2A][2B]^2 \]
Simplifying this, we get:
\[ R' = k(2[A])(2[B])^2 = k(2[A])(4[B]^2) = 8k[A][B]^2 \]

5. **Relate the New Rate to the Initial Rate**:
Since the initial rate \( R = k[A][B]^2 \), we can substitute this into the equation for \( R' \):
\[ R' = 8R \]

6. **Conclusion**:
Therefore, when the concentrations of both A and B are doubled, the rate of the reaction increases by a factor of 8.

### Final Answer:
The rate of the reaction increases by a factor of 8 when the concentrations of both A and B are doubled.