For the rate law expression rate=k[A][B]^2, what happens to the rate if the concentration of B is increased by a factor of two ?

1 Answer
May 16, 2018

Let's find out...


Given the rate law

r_1(t) = k[A][B]^2,

doubling the concentration of B results in a new rate of...

r_2(t) = k[A] (2[B])^2

= 4 cdot k[A][B]^2

And since r_1(t) = k[A][B]^2,

r_2(t) = 4 cdot r_1(t)

Thus, the rate quadruples. What is the order with respect to reactant B?