Question

What is the order of the reaction if doubling the concentration of `A` divides the half-life in 4?

Answer 1

This is third order. See the derivation below for why... but note that third-order decompositions are quite rare...

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The rate law for a third-order reaction

`A -> B`

is:

`r(t) = k[A]^3 = -(Delta[A])/(Deltat)`

For an infinitesimally small change, we write:

`k[A]^3 = -(d[A])/(dt)`

And we rearrange to get:

`-kdt = 1/([A]^3)d[A]`

The integral then gives:

`-int_(0)^(t) kdt = int_([A]_0)^([A]) 1/([A]^3)d[A]`

`-kt = -1/(2[A]^2) - (-1/(2[A]_0^2))`

`kt = 1/(2[A]^2) - 1/(2[A]_0^2)`

Thus, the third-order integrated rate law is:

`barul(|stackrel(" ")(" "1/(2[A]^2) = kt + 1/(2[A]_0^2)" ")|)`

A half-life is when `[A] = 1/2[A]_0`, i.e. the concentration of the reactant halves. Thus, we define the half-life `t_"1/2"` as:

`1/(2 cdot (1/2[A]_0)^2) = kt_"1/2" + 1/(2[A]_0^2)`

`2/([A]_0^2) = kt_"1/2" + (1/2)/([A]_0^2)`

`3/(2[A]_0^2) = kt_"1/2"`

Therefore, the third-order half-life is given by:

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

Given that the half-life changed by a factor of `1/4`, you can convince yourself that doubling the concentration of `A` quarters the half-life.