# What is a first order half life?

Jun 17, 2016

See explanation.

#### Explanation:

The half-life of a chemical reaction, regardless of its order, is simply the time needed for half of an initial concentration of a reactant to be consumed by the reaction.

Now, a first-order reaction is characterized by the fact that the rate of the reaction depends linearly on the concentration of one reactant.

For a first-order reaction

$\text{A " -> " products}$

the differential rate law allows you to express the rate of the reaction in terms of the change in the concentration of the reactant (or of a product)

"rate" = -(d["A"])/dt = k * ["A"]

The integrated rate law, which establishes a relationship between the change in the concentration of the reactant and time, can be obtained by integrating the differential rate law.

The integrated rate law for a first-order reaction looks like this -- mind you, I will not do the derivation here.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\ln {\left[\text{A"] = ln["A}\right]}_{0} - k \cdot t} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$\left[\text{A}\right]$ - the concentration of reactant $\text{A}$ after a time $t$
${\left[\text{A}\right]}_{0}$ - the initial concentration of the reactant $\left[\text{A}\right]$
$k$ - the rate constant of the reaction

So, the half-life of the reaction, ${t}_{\text{1/2}}$, corresponds to the period of time in which the concentration of the reactant goes from ${\left[\text{A}\right]}_{0}$ to ${\left[\text{A"] = ["A}\right]}_{0} / 2$.

You can thus say that

ln(["A"]_0/2) = ln["A"]_0 - k * t_"1/2"

This can be rearranged to get

ln(["A"]_0/2) - ln["A"]_0 = - k * t_"1/2"

ln(color(red)(cancel(color(black)(["A"]_0)))/2 * 1/color(red)(cancel(color(black)(["A"]_0)))) = -k * t_"1/2"

$\ln \left(\frac{1}{2}\right) = - k \cdot {t}_{\text{1/2}}$

You can rewrite this as

${\overbrace{\ln \left(1\right)}}^{\textcolor{b l u e}{= 0}} - \ln \left(2\right) = - k \cdot {t}_{\text{1/2}}$

$- \ln \left(2\right) = - k \cdot {t}_{\text{1/2}}$

The half-life of a first-order reaction will thus be equal to

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{t}_{\text{1/2}} = \ln \frac{2}{k}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

An important thing to notice here is that the half-life of a first-order reaction depends exclusively on the rate constant of the reaction.

In other words, the initial concentration of the reactant has no influence on the half-life of the reaction, i.e. the half-life is constant regardless of the concentration of the reactant.