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

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

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

Postby Sue Bin Park 2I » Tue Mar 10, 2020 9:42 pm

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

Postby Jordan Young 2J » Tue Mar 10, 2020 9:48 pm

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

Astrid Lunde 1I
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Re: 7b.3c - where did the ln come from?

Postby Astrid Lunde 1I » Tue Mar 10, 2020 10:08 pm

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

Junwei Sun 4I
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Re: 7b.3c - where did the ln come from?

Postby Junwei Sun 4I » Thu Mar 12, 2020 10:49 am

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

RRahimtoola1I
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Re: 7b.3c - where did the ln come from?

Postby RRahimtoola1I » Thu Mar 12, 2020 11:41 am

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

Postby Tanmay Singhal 1H » Thu Mar 12, 2020 11:42 am

rate law equation asks for it

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

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

Postby Myka G 1l » Thu Mar 12, 2020 11:45 am

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


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