Question

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

Answer 1

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.