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

##### 1 Answer

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

#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

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

Integrate from

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

(the integral of

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

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

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

If we let

#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 **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,

For instance, if