## Half Life for First Order

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

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.
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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[2]/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.

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