The reaction A + 2 B → C has the rate law rate = k[A][B]. By what factor does the rate of reaction increase when both [A] and [B] are doubled?

Answer 1

Hi Alex T.!

We can use the rate law to help us answer this question:

rate = k[A][B]

The order of this reaction is 1st order with respect to reactant A, 1st order with reactant B and 2nd order overall.

One way to help visualize this to consider the original concentrations as equal to 1 M:

rate = k[A][B]

=> [A]_o = 1M

=> [B]_o = 1M

rate_o = k(1)(1)

rate_o = k

Now, let's imagine that we doubled both concentrations:

rate = k[A][B]

=> [A] = 2[A]_o = 2M

=> [B] = 2[B]_o = 2M

rate = k(2)(2)

rate = 4k = 4rate₀

We can see that the rate of the reaction is 4X higher than the original rate when we double both concentrations.

Hope this helps!

Cheers