For a second-order reaction, the rate law can be expressed as:Rate = k[A]^m[B]^nwhere Rate is the reaction rate, k is the rate constant 0.002 M^-1s^-1 in this case , [A] and [B] are the concentrations of reactants A and B, and m and n are the orders of the reaction with respect to A and B.Since the reaction is second-order, the sum of m and n is 2. We are given that the concentration of reactant B is fixed, so we can assume that the reaction is second-order with respect to reactant A m = 2 and zero-order with respect to reactant B n = 0 .Now, the rate law becomes:Rate = k[A]^2Let's consider the initial rate when the concentration of A is doubled:Rate' = k 2[A] ^2Rate' = k 4[A]^2 Rate' = 4k[A]^2Now, let's compare the new rate Rate' to the initial rate Rate :Rate'/Rate = 4k[A]^2 / k[A]^2 The k[A]^2 terms cancel out:Rate'/Rate = 4So, when the concentration of reactant A is doubled, the initial rate of the second-order reaction increases by a factor of 4.