Rate of reaction question?

Question

If the rate law for a reaction is rate = k[A][B] (both first order), and they both have a concentration of .02 M, the rate after 25% of B has reacted is what percent of the inital rate?

Could you explain how you got the answer? thanks (my main focus is trying to learn how to do it :) )

Answer 1

The initial concentration is .02M - yes?

Is that's the case it's just a matter of writing it out and dividing

rate(i) = k[0.02][0.02]
rate(t)= k[(0.02-(.25)*0.02))]
[(0.02-(0.25)*0.02))] = k[0.015][0.015]

SO
rate(t)/rate(i)=(0.015)^2
/(0.02)^2 *100% = 56.25%

Answer 2

Note that rate constant is a CONSTANT.

for the initial rate r1 = k (.02)(.02) or 0.0004 k

after 25% of B has reacted means
1. 25 % of A also has reacted being first order each. Every molecule of A has to react with every molecule of B

2. Thus, the remaining concentration of A or B is equal to 75% of original.

rate 2 = r2 = k (0.015)(0.015)= 0.000225

Now relate r2 with initial r1.

% = r2/r1 = (0.000225/0.0004) x100 = 56.25 % or just 56 % of the initial rate