A reaction has the following rate law: Rate = k[A][B]₂
In experiment 1, the concentrations of A and B are both 0.10 mol L^-1; in experiment 2, the concentrations are both 0.30 mol L^-1. If the temperature stays constant, what is the value of the ratio, Rate(2) /Rate(1) ?

Question

Quizplus · Accepted Answer

The ratio of rates is directly proportional to the ratio of [A][B]. In this case, the ratio of [A][B] in experiment 2 compared to experiment 1 is 3, since both concentrations tripled. Therefore, the ratio of rates is also 3^(1+1+2+0) = 27. So the answer is E.

A Reaction Has the Following Rate Law: Rate = K[A][B]2

A Reaction Has the Following Rate Law: Rate = K[A][B]₂