## half life

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

skyeblee2F
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

### half life

Why is the half-life for first order reactions (t1/2=(ln2/k) independent of the initial reactant concentration [A]0, while the half-life of zero order and second order reactions are dependent? ([A]0 is included in both of their t1/2 formulas)

Luc Lorain 1L
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

### Re: half life

The value of the half life is ultimately dependent on the order of the reaction. For first order equations, d[A]= -kdt, in which the reactant is conspicuously absent.

In reality, we just have to accept that it just be like that sometimes.
(Sorry fam, I just need those fat Chemistry community points)

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
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### Re: half life

The half-life equations are derived using the integrated rate law formulas by setting the final concentration equal to half the initial concentration and solving for time (t). When deriving the half-life equation for the 1st order reactions, the initial concentration terms will cancel out. However, they will not cancel out when deriving the half-life equations for the 0th and 2nd order reactions.

Hope this helps! :)

Jack Hewitt 2H
Posts: 67
Joined: Fri Sep 28, 2018 12:27 am

### Re: half life

Because when the first order half-life equation is derived the initial concentration terms will cancel out.