## problem 15.23B

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

Amy Zheng 2l
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

### problem 15.23B

For part B,how come we use In(Ainitial/Afinal)=kt? shouldn't it be -kt?

Anna Goldberg 2I
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

### Re: problem 15.23B

ln(At)/(Ainitial)=-kt

When you do ln(A initial)/(At), the swap of the numerator and denominator makes ln(A initial)/(At)=kt

Hena Sihota 1L
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 2 times

### Re: problem 15.23B

ln[A]t-ln[A]0=-kt If you divide both sides of this equation by -1 then you get -[A]t+[A]0=kt which can be rewritten as [A]0-[A]t=kt. According to log rules, this equation can be rewritten as ln[A]0/ln[A]t=kt

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

### Re: problem 15.23B

You can also just use [A]=[A]initial * e^-kt, which is just a simplified verision of the ln[A]=-kt+ln[A]initial equation