Chemistry wizzes want a challenge?

Question

What is the half-life of a first-order reaction if it takes 4.8x10^-3 seconds for the concentration to decrease from 1.33M to 0.51 M?
I think the answer is 3.5x10^-3 s, which is one of the options but I am not getting exactly that, I am a little off.
Please explain!
Any chemistry advice welcomed!!!

Answer 1

ln ([A]o/[A]) = kt
k = ln (1.33/0.51)/4.8*10^-3 = 0.959/4.8*10^-3 = 199.8

t 1/2 = 0.693/k = 0.693/199.8 = 3.47 * 10^-3

With rounding, and the fact that most of the parameters have two significant figures, it would be 3.5 * 10^-3.

Answer 2

If you are a little off, I wouldn't complain. The answer looks about right. I don't know how much math you have had, but to keep it simple, in a half-life, the concentration of a first-order reactant will decrease to half the concentration at the start of the time period. In 2 half-lives, it will decrease to 1/4th its initial conc, and so on.

In the problem, the concentration decreased to 0.51 from 1.33, or a reduction to about 1/2.2. This is slightly more than a half-life reduction; hence the half-life should be somewhat less than the 0.0048 seconds.