Is the half-life of a first-order decay dependent on temperature?

1 Answer
Dec 13, 2016

Of course. As rate constants depend on temperature, so does the first-order half-life.

For a first-order process, let us consider simply a reactant #A#, whose rate law is:

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

where #[A]# is the concentration of #A#, and #-(d[A])/(dt)# is the rate of disappearance of #A# as time passes.

To determine the change in concentration over time, we consider an interval of time from #0# seconds to some general time #t#.

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

Integrate from #t = 0# to #t = t# on the left side, and from the initial concentration #[A]_0# to the final concentration #[A]# on the right side.

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

(the integral of #1/x# is #ln|x|#.)

#kt = -ln[A] - (-ln[A]_0)#

#= -{ln[A] - ln[A]_0}#

#= -ln\frac([A])([A]_0)#

If we let #t_"1/2"# be the half-life time duration, then at #t = t_"1/2"#, we have that #[A] = 1/2[A]_0#. Thus, we get:

#kt_"1/2" = -ln\frac(1/2cancel([A]_0))(cancel([A]_0))#

#= -ln(1/2) = ln2#

So...

#color(blue)(t_"1/2") = (ln2)/k ~~ color(blue)(0.693/k)#

Although the first-order half-life depends on the rate constant #k# and not the initial concentration #[A]_0#, #bb(k)# is a function of temperature, and increases with increasing temperature.

Note the Arrhenius equation:

#ln(k_2/k_1) = -(E_a)/R[1/T_2 - 1/T_1]#

where clearly, #k# is related to #T#.

For instance, if #T_2 > T_1#, then #1/T_2 - 1/T_1 < 0#, and thus, #ln(k_2/k_1) > 0#, meaning that #k_2 > k_1#. Therefore, we've shown that #k# increases with increasing temperature.