Half Life for First Order
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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?
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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.
ln2 will simplify to 0.693, which is the equation given on the provided formula sheet on tests.
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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.
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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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