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 given by: (rate2)/(rate1) = k[A]2[B]2² / k[A]1[B]1² Canceling out the rate constant and substituting the given concentrations, we get: (rate2)/(rate1) = (0.30 mol/L)2(0.30 mol/L)² / (0.10 mol/L)2(0.10 mol/L)² Simplifying the expression gives: (rate2)/(rate1) = 27 Therefore, the answer is E) 27.

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

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