# What is first order half life derivation?

##### 1 Answer

Well, let us begin from the rate law... Is the FIRST-order half-life dependent on concentration?

The **first-order rate law** for the reaction

#A -> B#

is

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

#r(t)# is therateas a function of time, which we take to be the initial rate in#"M/s"# #k# is therate constant.What are the units? You should know this.#[A]# is theconcentrationof a reactant#A# in#"M"# .#(d[A])/(dt)# is therate of disappearanceof reactant#A# , a NEGATIVE value. But the negative sign forces the rate to be positive, fitting for a FORWARD reaction.

By separation of variables:

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

Integrate on the left from time zero to time

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

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

Therefore, the **first-order integrated rate law** is:

#color(green)(ln[A] = -kt + ln[A]_0)#

As we should recognize, the **half-life** is when the concentration drops by half. Hence, we set

#ln0.5[A]_0 = -kt_(1//2) + ln[A]_0#

#ln0.5[A]_0 - ln[A]_0 = -kt_(1//2)#

#ln\frac(0.5cancel([A]_0))(cancel([A]_0)) = -kt_(1//2)#

#-ln0.5 = ln2 = kt_(1//2)#

As a result, the half-life is given by...

#color(blue)(barul|stackrel(" ")(" "t_(1//2) = (ln2)/k" ")|)#