# Question #f4729

##### 1 Answer

Okay, first of all, I'll need to figure out how you got that equation.

**Half-life reactions** are generally *first-order* decompositions, meaning that they do **NOT** depend on the concentration of the starting reactant. That means we should not expect to find

First, let's start from the **rate law**, which consists of the rate *time*, the rate constant

#r(t) = k[A]#

But we're missing something that'll help. Recall that in *any* reaction, a reactant is *consumed*. Thus, its rate for *any* reaction is *negative*, and is represented by the **derivative of concentration with respect to time**:

#\mathbf(r(t) = k[A] = stackrel("derivative of conc. with respect to time")overbrace(-(d[A])/(dt)))#

That is, it is the rate of change of concentration of **half-life equation**.

#[A]kdt = -d[A]#

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

Now we can start integrating it, from time

#int_(t_0)^(t) kdt = -int_([A]_0)^([A]_t) 1/([A])d[A]#

To keep it simple, let

#|[kt]|_(0)^(t) = -|[ln[A]]|_([A]_0)^([A]_t)#

#kt - cancel(k*0) = -(ln[A]_t - ln[A]_0)#

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

As a side note, this is known as the **integrated rate law**:

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

Surely enough, if I use the properties of logarithms, I get what you meant to give me (you meant an equal sign instead of a minus sign):

#-kt = ln\frac([A]_t)([A]_0)#

Next, since we are going for a **half-life equation**, we consider **half** of the previous concentration is left after **each** half-life that passes.

Thus, we have:

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

#= -ln\frac(1)(2)#

#= -ln2^(-1)#

#= ln2#

So, our final result is:

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

There we go; we have the half-life, and indeed, it does ** not** require one to know the concentration of the reactant. In other words, it does

**NOT**depend on the concentration of the reactant.