## Half Life for First Order

Victoria Luu - 1C
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

### Half Life for First Order

When determining the half-life of a first order reaction, why is it that ln2 is in the numerator of the equation? How did we get that?

Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

### Re: Half Life for First Order

From page 628 of the 6th edition textbook

ln2 will simplify to 0.693, which is the equation given on the provided formula sheet on tests.
Attachments Timothy_Yueh_4L
Posts: 57
Joined: Fri Sep 28, 2018 12:28 am

### Re: Half Life for First Order

The integrated rate law of a first order reaction is ln[A] = -kt + ln[A]o. In order to determine the half life of a first order reaction, we set ln[A] = ln[0.5A]o and solve for t. Overall we get the unsimplified version: (ln[0.5A]o - ln[A]o)/-k = t. According to log rules when you subtract 2 logs, you divide the two inputs to each other and when there's a coefficient in front of the log, you raise the input to that power, and by knowing this, first simplify the equation to ln[0.5]/-k = t, then bring the negative in the front to exponent of the log's input therefore getting ln/k = t.

Ronald Thompson 1F
Posts: 59
Joined: Fri Sep 28, 2018 12:25 am

### Re: Half Life for First Order

isolating t1/2 and halving A should get you the half life with some calculus.