The reaction $ A + 2B \to products $ , was found to have the rate law, $ rate = k\left[ A \right]{\left[ B \right]^2} $ . Predict by what factor the rate of reaction will increase when the concentration of A and B is doubled?

Hint: Given reaction has the rate $ rate = k\left[ A \right]{\left[ B \right]^2} $ , which means the rate is directly proportional to the concentration of reactant A and square of the concentration of reactant B. When the concentration of both the reactants are doubles, substitute the double of the concentration of the reactants gives the new rate.

Complete Step By Step Answer:
Given reaction is $ A + 2B \to products $ which means the reactant A and B are reacted to give products. The rate law for the above reaction is $ rate = k\left[ A \right]{\left[ B \right]^2} $
From the rate law, it was clear that the rate is directly proportional to the concentration of reactant A and square of the concentration of reactant B.
Let us consider concentration of A is x and concentration of B is y then the rate will be $ rate = k.x.{y^2} $
When the concentration of A and B are doubled, substitute the value of double concentration in the above equation which means $ x = 2x $ and $ y = 2y $ $ rate = k.2x.{\left( {2y} \right)^2} = k.8.x{y^2} $
By comparing the two-rate equation, it was clear that the rate increases by a factor $ 8 $
The rate of reaction will increase by a factor $ 8 $ when the concentration of A and B is doubled.

Note:
The rate law can be written from the concentration of the reactants. Thus, when the concentration is doubled the concentration of both the reactants should be doubled and should be substituted in the rate equation. The factor $ 2 $ should be multiplied for the two reactants concentrations.