## Hw 15.27 and 15.35

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

Jaret Nishikawa
Posts: 15
Joined: Wed Sep 21, 2016 2:57 pm

### Hw 15.27 and 15.35

In problem 15.27 it asks to find the time need for a first order concentration of reactant to decrease by 1/8 of its original value and then 1/4 and then 15%.

Also, in 15.35 it asks to find the time needed, but in second order, of a concentration of reactant to decrease by 1/16, 1/4, and 1/5.

I was wondering why we can't use the same process used in 15.27 solutions manual to solve for 15.35? Thanks in advance

Manpreet Singh 1N
Posts: 41
Joined: Wed Sep 21, 2016 2:59 pm
Been upvoted: 1 time

### Re: Hw 15.27 and 15.35

Hello,

We cant use the same process because it 15.27 the reaction is in 1st order but it 35 it is in second order. So it 27, the equation we used is ln[A]=-kt+ln[A0]. But for a second order reaction the equation would be 1/[A]=kt+1/[A0].

Jessica Chern 1H
Posts: 22
Joined: Sat Jul 09, 2016 3:00 am

### Re: Hw 15.27 and 15.35

Also another note, since I got tripped up here, with a first-order reaction, t = ln[A0/A] / k. If A = 1/8 A0, the equation would be t = ln[A0 / (1/8 * A0)] / k, so the A0 would cancel out. With a second-order reaction however, you need to include the initial concentration in your calculation because they don't cancel out, since it would be t = (1/A - 1/A0) / k.