# How is the half-life of a first-order reaction affected by change in concentration?

##### 1 Answer

It's not. The half-life of a reactant is intrinsic to the identity of the reactant, which has its own rate constant.

A first-order reaction **rate law**, for the reaction

#A -> B# ,

would be written as:

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

#r(t)# is therate of reactionin#"M/s"# ,#k# is therate constantin#"s"^(-1)# , and#(d[A])/(dt)# is the change in the concentration#[A]# of#A# over the change in time#t# (otherwise known as therate of disappearance).

When you separate variables by multiplying by

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

To observe the process over time, we will be integrating from

We get:

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

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

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

which is just a rearrangement of the integrated rate law for a first-order reaction.

Now, if we say that we want to know how much time it takes for **half-life**,

Therefore:

#ln((1/2cancel([A]_0))/(cancel([A]_0))) = -kt_"1/2"#

Since

#-ln2 = -kt_"1/2"#

#ln2 = kt_"1/2"#

#=> color(blue)(t_"1/2" = ln2/k)#

*So, the half-life depends only on the rate constant for the first-order reaction that is occurring.*