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
#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
#[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
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.