How do you calculate half life decay?

1 Answer
Jun 16, 2017

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


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)#