Equation variations


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ClaireHW
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Equation variations

Postby ClaireHW » Wed Mar 07, 2018 8:20 pm

For 1st order reactions the equation is typically ln[A]f = -kt + ln[A]0
I found a variation used in the textbooks: ln([A]0/[A]f) = kt
Can these equations be used interchangeably and produce the same answer?

Thanks!

(Claire Woolson Dis 1K)

Andrea ORiordan 1L
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Re: Equation variations

Postby Andrea ORiordan 1L » Wed Mar 07, 2018 8:49 pm

Rearrange ln[A]f = -kt + ln[A]0 and you will get kt= ln[A]0-ln[A]f. When you subtract ln[x]-ln[y] you can use log rules to write the expression as ln[x/y]. So ln([A]0/[A]f) = kt is the exact same equation, just written differently. You will get the same answer.

Erica Nagase 1H
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Joined: Sat Jul 22, 2017 3:00 am

Re: Equation variations

Postby Erica Nagase 1H » Thu Mar 08, 2018 10:16 am

These equations are the same, just rearranged using log rules, so it does not matter which you use. They are just written differently to make it easier to solve for different terms.

William Satyadi 2A
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Joined: Sat Jul 22, 2017 3:00 am

Re: Equation variations

Postby William Satyadi 2A » Thu Mar 08, 2018 1:19 pm

Yes, the two equations you provided can be used interchangeably, as they are the same, just rearranged differently using logarithm rules. From the first equation you provided, we can rearrange it so that . Dividing by -1 gives us . Using log rules, , so this gives us the second equation, meaning they are equivalent.

Rohan Chaudhari- 1K
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Re: Equation variations

Postby Rohan Chaudhari- 1K » Thu Mar 08, 2018 3:57 pm

You can aso use this equation for first order:

[A]=[Ainitial]e^(-k*t)

Jennie Fox 1D
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Re: Equation variations

Postby Jennie Fox 1D » Thu Mar 08, 2018 4:19 pm

Yes, all they did in the book was switch the ln[A]f to the right side, and moved -kt to the left side. ln[A]o-ln[A]f is the same as ln([A]o/[A]f)

Jennie Fox 1D
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Re: Equation variations

Postby Jennie Fox 1D » Thu Mar 08, 2018 4:19 pm

^ They just rearranged the equation. Both versions should yield the same answer

Joanne Guan 1B
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Re: Equation variations

Postby Joanne Guan 1B » Sun Mar 11, 2018 10:24 pm

The equations will give you the same thing; the second is just rearranged using log rules.

Jaewoo Jo 2L
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Re: Equation variations

Postby Jaewoo Jo 2L » Mon Mar 12, 2018 12:00 am

the second is just rearranged with log rules, so it's the same thing

Emma Miltenberger 2I
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Re: Equation variations

Postby Emma Miltenberger 2I » Mon Mar 12, 2018 1:08 pm

Yes, this equation yields the same answer. The book uses logarithm rules to rearrange the equation to make kt positive. ln(Af/Ao)=-kt is the same as ln(Ao/Af)=kt.


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