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15.35

Posted: Mon Mar 05, 2018 12:14 am
by Merzia Subhan 1L
Not sure why I'm not getting the right answer

The half-life for the second-order reaction of a substance A is 50.5 s when [A]0 = 0.84 mol*L^-1. Calculate the time needed for the concentration of A to decrease to (a) one-sixteenth....of its original value

I used the the half life 2nd order reaction to get k:
t(1/2)=1/k*[A]0
50.5=1/k*0.84
k=0.0236

ln(1/16)=-0.0236(t)
t=118s

but the answer is 7.4 x 10^2 s.
If someone can tell me what I'm doing wrong, I would greatly appreciate it.

Re: 15.35

Posted: Mon Mar 05, 2018 5:46 am
by RyanS2J
Since it's a second order reaction, you would use 1/[A]t = 1/[A]o + kt for the final step rather than ln[A]t = ln[A]o - kt, which only applies for first order reactions.

Re: 15.35

Posted: Mon Mar 05, 2018 3:29 pm
by zanekoch1A
So for second order reactions, unlike first order, we have to use two equations not just the the one half life equation?