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)
Equation variations
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Re: Equation variations
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.
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Re: Equation variations
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.
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Re: Equation variations
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.
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Re: Equation variations
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)
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Re: Equation variations
^ They just rearranged the equation. Both versions should yield the same answer
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Re: Equation variations
The equations will give you the same thing; the second is just rearranged using log rules.
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Re: Equation variations
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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