Question

How do you calculate half life decay?

Answer 1

It depends on the order of the reaction with respect to the reactant undergoing half-life decay.

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The common half-life expressions for one reactant `A` are:

`t_"1/2" = ([A]_0)/(2k)` (zero order)

`t_"1/2" = (ln2)/k` (first order)

`t_"1/2" = 1/(k[A]_0)` (second order)

`t_"1/2" = 3/(2k[A]_0^2)` (third order)

where `[A]_0` is the initial concentration in `"M"` and `k` is the rate constant in the appropriate units to obtain time units of `"s"`.

If you wish to derive them, that requires some calculus. I can do first order as an example, as that is the hardest one. For the general rate law

`r(t) = k[A] = -(d[A])/(dt)`,

of the first-order reaction

`A -> B`,

where the negative sign indicates disappearance of reactant,

separation of variables gives:

`-kdt = 1/([A]) d[A]`

Now, integrate from `0` to `t` on the left-hand side and `[A]_0` to `[A]` for the right-hand side.

`-int_(0)^(t) kdt = int_([A]_0)^([A]) 1/([A])d[A]`

The integral of `1/x` is `ln|x|`, and `[A] >= 0`, so:

`-kt = ln[A] - ln[A]_0`

For a half-life, `[A] = 1/2[A]_0`, i.e. the concentration of `A` has halved. So:

`-kt_"1/2" = ln(1/2 [A]_0) - ln[A]_0`

`kt_"1/2" = -ln(1/2 cancel(([A]_0)/([A]_0)))`

`=> color(blue)(t_"1/2 " ("1st order") = (ln2)/k)`