Negative sign in ln [A]t = -k t + ln [A]o

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Annie Chantasirivisal_4G
Posts: 114
Joined: Wed Sep 18, 2019 12:21 am

Negative sign in ln [A]t = -k t + ln [A]o

Postby Annie Chantasirivisal_4G » Sat Mar 07, 2020 8:35 pm

For the formula ln [A]t = -k t + ln [A]o,

why is there a negative sign when t=-(1/k) ln[A]t/ln[A]o
but not when t=(1/k) ln[A]o/ln [A]t?

Sarah Zhari 1D
Posts: 103
Joined: Sat Sep 14, 2019 12:16 am

Re: Negative sign in ln [A]t = -k t + ln [A]o

Postby Sarah Zhari 1D » Sat Mar 07, 2020 9:09 pm

When you rearrange the equation ln [A]t = -k t + ln [A]o, you get ln [A]t-ln [A]o= -k t. Following the log rules, ln[A]t-ln[A]o is equal to ln([A]t/[A]o). Flipping the fraction to get ln([A]o/[A]t) is simply the negative, which explains the change in sign when you flip the fraction.

Adam Kramer 1A
Posts: 103
Joined: Sat Aug 24, 2019 12:15 am

Re: Negative sign in ln [A]t = -k t + ln [A]o

Postby Adam Kramer 1A » Sat Mar 07, 2020 10:16 pm

It has to do with the log rules, if it is organized in a way that the negatives cancel when subtracting the ln[A]t over, then you get t=(1/k) ln[A]o/ln [A]t. If you change it, you can rearrange the equation so t=-(1/k) ln[A]t/ln[A]o, and the ln[A]t is divided by the ln[A]o.

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