## 15.35

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

Christina_F_3F
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

### 15.35

"The half-life for the second order reaction of a substance A is 50.5 s when [A]naught=0.84 M. Calculate the time needed for concentration of A to decrease to a) one sixteenth, b) one-fourth, c) one-fifth it's original value."
For this question, after looking at the solutions manual, I could pretty easily see the pattern of what you're supposed to do. For example, for a) you would write:
T= (16/[A]naught - 1/[A]naught) / k
What I don't understand, however, is why this equation works, and why you would write the equation this way.
If anyone could explain this I would much appreciate it.
Thank you!

Anna_Kim_2E
Posts: 31
Joined: Wed Sep 21, 2016 2:56 pm

### Re: 15.35

For second order, -d[A]/dt = k[A]^2 . Integration of both sides results in an equation 1/[A] = kt + 1/[A]0. Hence, in order to find out for the half life, we use this equation when it is in second order.