## 15.35

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

Merzia Subhan 1L
Posts: 33
Joined: Thu Jul 13, 2017 3:00 am

### 15.35

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.

RyanS2J
Posts: 32
Joined: Thu Jul 27, 2017 3:00 am

### Re: 15.35

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.

zanekoch1A
Posts: 25
Joined: Thu Jul 13, 2017 3:00 am

### Re: 15.35

So for second order reactions, unlike first order, we have to use two equations not just the the one half life equation?