Rate Laws and Method of Initial Rates

Question

Consider the following rate law:

How does the rate change when the concentration of [BrO₃^-] is doubled (all other concentrations remain the same)?

I am confused on this problem, as when I try this out with sample concentrations, doubling the concentration of BrO3 gives me 1/2rate, however, the answer sheet says it is 2x rate. How does doubling the concentration of one of the reactants affect the rate?

Answer 1

Hello,

Here's an explanation for the rate doubling

Rate = k[A] ^m [B]^n

where k is the rate constant [A] is the concentration of a reactant and [B] is the concentration of another reactant.

m + n is the order of the reaction

in this case [BrO3 ^-] is first order

now if we double the concentration of [A] then as we see the rate also double in the equation.

k 2[A][B]

2Rate = k2[A][B]

to make it simpler for you say that:

k[A][B] = 10

thus Rate =10

however, if we multiply 10 by two (by doubling concentration of B)

Rate=2(10)

Rate =20

This is because being a first order reaction Rate is directly proportional to Concentration.

Hope this helps.