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 ?

Question

721 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-16T18:21:02

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 $B$ ?