## 7b.3c - where did the ln come from?

Sue Bin Park 2I
Posts: 52
Joined: Mon Jun 17, 2019 7:24 am

### 7b.3c - where did the ln come from?

i got up to the new concentration of A to be .085, but i don't understand why we are applying the natural log to (.153/.085) before dividing by time.

Jordan Young 2J
Posts: 102
Joined: Thu Jul 25, 2019 12:17 am
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### Re: 7b.3c - where did the ln come from?

You plug the concentrations into the first order integrated rate law equation to find k
ln[A]=-kt+ln[Ainitial]

Astrid Lunde 1I
Posts: 103
Joined: Sat Sep 07, 2019 12:16 am

### Re: 7b.3c - where did the ln come from?

It is a first order rate law so use lnA = -kt + lnA

Junwei Sun 4I
Posts: 125
Joined: Wed Oct 02, 2019 12:16 am

### Re: 7b.3c - where did the ln come from?

The natural log comes from rearranging the first order rate law: ln[A] = -kt + ln[A]initial

RRahimtoola1I
Posts: 102
Joined: Fri Aug 09, 2019 12:15 am

### Re: 7b.3c - where did the ln come from?

If you use the equation ln[A] = -kt + ln[A]0 you can see this will happen. You should be subtracting over the ln[A]0 before dividing and with laws of natural logs, if you subtract two logs you are basically taking the natural log of the two divided so that's why you are dividing 0.085/0.153 and then taking the natural log.

Tanmay Singhal 1H
Posts: 143
Joined: Sat Jul 20, 2019 12:16 am

### Re: 7b.3c - where did the ln come from?

rate law equation asks for it

Myka G 1l
Posts: 100
Joined: Fri Aug 30, 2019 12:17 am

### Re: 7b.3c - where did the ln come from?

You could also use the equation [A]t= [A]o e^-kt