## Equation variations

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Moderators: Chem_Mod, Chem_Admin

ClaireHW
Posts: 60
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

### Equation variations

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
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

### 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.

Erica Nagase 1H
Posts: 30
Joined: Sat Jul 22, 2017 3:00 am

### 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.

William Satyadi 2A
Posts: 31
Joined: Sat Jul 22, 2017 3:00 am

### 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 $ln[A]_f - ln[A]_0 = -kt$. Dividing by -1 gives us $ln[A]_0 - ln[A]_f = kt$. Using log rules, $ln[A]_0 - ln[A]_f = ln\frac{[A]_0}{[A]_f}$, so this gives us the second equation, meaning they are equivalent. $ln\frac{[A]_0}{[A]_f} = kt$

Rohan Chaudhari- 1K
Posts: 32
Joined: Fri Sep 29, 2017 7:06 am

### Re: Equation variations

You can aso use this equation for first order:

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

Jennie Fox 1D
Posts: 66
Joined: Sat Jul 22, 2017 3:01 am

### 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)

Jennie Fox 1D
Posts: 66
Joined: Sat Jul 22, 2017 3:01 am

### Re: Equation variations

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

Joanne Guan 1B
Posts: 30
Joined: Sat Jul 22, 2017 3:01 am

### Re: Equation variations

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

Jaewoo Jo 2L
Posts: 31
Joined: Fri Sep 29, 2017 7:06 am

### Re: Equation variations

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

Emma Miltenberger 2I
Posts: 51
Joined: Thu Jul 27, 2017 3:00 am

### 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.

Return to “First Order Reactions”

### Who is online

Users browsing this forum: No registered users and 1 guest