Half Life for First Order

[Victoria Luu - 1C]

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]

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.

[Timothy_Yueh_4L]

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]

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