# Question #f4729

May 9, 2016

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 $\left[A\right]$ in the final equation.

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

$r \left(t\right) = k \left[A\right]$

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:

$\setminus m a t h b f \left(r \left(t\right) = k \left[A\right] = \stackrel{\text{derivative of conc. with respect to time}}{\overbrace{- \frac{d \left[A\right]}{\mathrm{dt}}}}\right)$

That is, it is the rate of change of concentration of $A$ over time. Now we're ready to derive the half-life equation.

$\left[A\right] k \mathrm{dt} = - d \left[A\right]$

$k \mathrm{dt} = - \frac{1}{\left[A\right]} d \left[A\right]$

Now we can start integrating it, from time ${t}_{0}$ to $t$, and from the initial (${\left[A\right]}_{0}$) to the final (${\left[A\right]}_{t}$) concentrations.

${\int}_{{t}_{0}}^{t} k \mathrm{dt} = - {\int}_{{\left[A\right]}_{0}}^{{\left[A\right]}_{t}} \frac{1}{\left[A\right]} d \left[A\right]$

To keep it simple, let ${t}_{0} = \text{0 s}$. Then we have:

$| \left[k t\right] {|}_{0}^{t} = - | \left[\ln \left[A\right]\right] {|}_{{\left[A\right]}_{0}}^{{\left[A\right]}_{t}}$

$k t - \cancel{k \cdot 0} = - \left(\ln {\left[A\right]}_{t} - \ln {\left[A\right]}_{0}\right)$

$k t = - \left(\ln {\left[A\right]}_{t} - \ln {\left[A\right]}_{0}\right)$

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

$\textcolor{g r e e n}{\ln {\left[A\right]}_{t} = \ln {\left[A\right]}_{0} - k t}$

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):

$- k t = \ln \setminus \frac{{\left[A\right]}_{t}}{{\left[A\right]}_{0}}$

Next, since we are going for a half-life equation, we consider ${\left[A\right]}_{t}$ to be equal to $\frac{1}{2} {\left[A\right]}_{0}$, given that half of the previous concentration is left after each half-life that passes.

Thus, we have:

$k {t}_{\text{1/2}} = - \ln \setminus \frac{\frac{1}{2} \cancel{{\left[A\right]}_{0}}}{\cancel{{\left[A\right]}_{0}}}$

$= - \ln \setminus \frac{1}{2}$

$= - \ln {2}^{- 1}$

$= \ln 2$

So, our final result is:

$\textcolor{b l u e}{{t}_{\text{1/2}} = \ln \frac{2}{k}}$

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.