## Course Reader 2nd Order Rxn

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

004643111
Posts: 14
Joined: Fri Sep 25, 2015 3:00 am

### Course Reader 2nd Order Rxn

In the course reader on page 67, when integrating, how does -(d[A])/([A]^2)=kdt turn into positive (1/[A])=kt+b? I understand the integration but not why the negative disappears.

Edwin Ng 1G
Posts: 20
Joined: Fri Sep 25, 2015 3:00 am

### Re: Course Reader 2nd Order Rxn

When you integrate (1/[A]2) it becomes (-1/[A]), and so the two negatives cancel

Bryan Trieu 1F
Posts: 11
Joined: Fri Sep 25, 2015 3:00 am

### Re: Course Reader 2nd Order Rxn

Yes, the integration is no longer based off of natural log rules, just basic exponential integration. I believe that 1st order reactions depend on natural logs when integrating but the others are based off of basic exponential integration .