A reaction is 2nd order w.r.t A. and 1st order w.r.t B. When the concentrations of both A and B are doubled the rate of reaction will become

8 times

The correct answer is option 4. 8 times.

To determine how the rate of the reaction changes when the concentrations of both A and B are doubled, we can use the rate equation for a second-order reaction with respect to A and a first-order reaction with respect to B. The rate equation for this reaction can be written as:

\(\text{Rate} = k[A]^2[B]\)

Where:

\( k \) is the rate constant.

\( [A] \) and \( [B] \) are the concentrations of A and B, respectively.

When both \( [A] \) and \( [B] \) are doubled, the new rate of the reaction can be calculated as follows:

\( \text{New Rate} = k(2[A])^2(2[B]) \)

\( = k \times 2^2[A]^2 \times 2[B] \)

\( = 4k[A]^2 \times 2[B] \)

\( = 8k[A]^2[B] \)

So, the new rate of the reaction is 8 times the original rate. Therefore, the correct answer is option 4. 8 times.