The gas-phase reaction, A2 + B2 --> 2AB, proceeds by bimolecular collisions between A2 and B2 molecules. If the concentrations of both A2 and B2 are doubled, the reaction rate will change by a factor of.....?

Question

Will you please explain instead of just giving me an answer. thanks

Answer 1

Hey! amarillo science on youtube does a great job at explaining it:
basically, the rate eq'n is: rate = k [A][B]

Note: we know that k will always stay the same since it's a constant

so, if the concentrations were, for example, 1 M for each A and B, our eq'n would be:

rate = [1][1]
1 = 1 x 1
(i've omitted k here for now for simplicity's sake since it would stay the same regardless of what we did to the concentrations)

now, if we doubled both the rates as stated in the question, they would each become 2 M:

rate = 2 x 2
4 = 2 x 2

thus, the reaction rate will change by a factor of 4.