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

##### 1 Answer

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

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

#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