The rate law for the reaction 2 A + B -> C is experimentally determined to be Rate = k [A]^2 [B]. If the concentration of A is tripled and the concentration of B is doubled, by what factor does the initial reaction rate increase?
Answer and explanation
Correct answer: C. 18 times
Substitute the concentration changes into the rate law: Rate_new / Rate_old = ([3A]^2 [2B]) / ([A]^2 [B]) = 3^2 2 = 9 2 = 18. The reaction rate increases by a factor of 18.
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