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 determine how the rate of a reaction changes when the concentrations of both reactants A and B are doubled. The reaction is first order with respect to A and second order with respect to B.

### Step-by-Step Solution:

1. **Understand the Rate Law**:
The rate law for the reaction can be expressed as:
\[ \text{Rate} = k [A]^1 [B]^2 \]
where \( k \) is the rate constant, \( [A] \) is the concentration of reactant A, and \( [B] \) is the concentration of reactant B.

2. **Initial Rate Expression**:
Let the initial concentrations of A and B be \( [A] \) and \( [B] \). The initial rate (denoted as \( r \)) can be written as:
\[ r = k [A]^1 [B]^2 \]

3. **Doubling the Concentrations**:
If we double the concentrations of both A and B, the new concentrations will be \( 2[A] \) and \( 2[B] \).

4. **New Rate Expression**:
The new rate (denoted as \( r' \)) with the doubled concentrations can be expressed as:
\[ r' = k (2[A])^1 (2[B])^2 \]

5. **Calculating the New Rate**:
Now, substituting the doubled concentrations into the rate expression:
\[ r' = k (2[A])^1 (2[B])^2 = k (2[A]) (4[B]^2) = k \cdot 2[A] \cdot 4[B]^2 \]
Simplifying this gives:
\[ r' = 8 k [A]^1 [B]^2 \]

6. **Relating New Rate to Initial Rate**:
Since \( r = k [A]^1 [B]^2 \), we can substitute \( r \) into the equation:
\[ r' = 8r \]

7. **Conclusion**:
The new rate \( r' \) is 8 times the initial rate \( r \). Therefore, when the concentrations of both A and B are doubled, the rate of the reaction increases by a factor of 8.

### Final Answer:
When the concentrations of both A and B are doubled, the rate of the reaction becomes 8 times the original rate.