Question #37a30

1 Answer
Aug 19, 2016

See below.

Explanation:

Let us assume that the reaction is under the following general form:

A+B->CA+BC

The rate law is determined experimentally to be:

R=k[A][B]^2R=k[A][B]2

The rate law with the first concentrations could be written as:

R=k[A_1][B_1]^2R=k[A1][B1]2

If the concentration of AA is doubled, this means that the new concentration is: [A_2]=2xx[A_1][A2]=2×[A1]

and the concentration of BB is tripled: [B_2]=3xx[B_1][B2]=3×[B1]

Replacing the new concentration in the expression of the rate law: R'=k[A_2][B_2]^2R'=k[A2][B2]2

Therefore, we have:
R'=k[A_2][B_2]^2=k(2xx[A_1])(3xx[B_1])^2R'=k[A2][B2]2=k(2×[A1])(3×[B1])2

=>R'=(2)(3)^2xxunderbrace(k[A_1][B_1]^2)_R

=>R'=(2)(3)^2xxR

=>R'=2xx9xxR

=>R'=18xxR

Which indicates that the rate of the reaction will increase by a factor of 18.

Here is a complete lesson on reaction rate:
Chemical Kinetics | Reaction Rates & Rate Law.