Question

Question #f4729

Answer 1

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 `[A]` in the final equation.

First, let's start from the rate law, which consists of the rate `r(t)` of reaction as a function of time, the rate constant `k`, and the concentration `[A]` of reactant `A`.

`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` over time. Now we're ready to derive the half-life equation.

`[A]kdt = -d[A]`

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

Now we can start integrating it, from time `t_0` to `t`, and from the initial (`[A]_0`) to the final (`[A]_t`) concentrations.

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

To keep it simple, let `t_0 = "0 s"`. Then we have:

`|[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 `[A]_t` to be equal to `1/2[A]_0`, given that 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.