ChemTalk

Half-Life

half-life of carbon-14

Core Concepts

In this article, you will learn what a half-life is, in 0th. 1st, and 2nd order reactions. You will also learn what the half-life formula is and how to use it to find the age of a substance.

Topics Covered in Other Articles

What is a Half-Life?

A half-life (t1/2) is the time it takes half the concentration of a sample to decay. For example, if the starting concentration of a sample is 1M, then the half-life will be the time it takes for the concentration of the sample to decay to 0.5M.

First Order Half-Life

First-order reactions, including radioactive decay, are reactions where the rate of reaction is only dependent on the concentration of one reactant. First-order reactions have constant half-lives. We can prove this using the integrated first-order rate law:

    \begin{gather*} ln[A] = -kt + ln[A]_0 \end{gather*}

[A] is the concentration of a reactant at time t, [A]₀ is the initial concentration of the reactant, and k is the rate constant. Here, we can move all the natural logs to the left side of the equation:

    \begin{gather*} {\text{ln}\frac{[A]} {[A]_{\text{0}}}} = -kt \end{gather*}

Since we are trying to find the half-life of the reactant, we can set t as the half-life and [A] as half of the initial concentration:

    \begin{gather*} {t} = {t_{\text{1/2}}} \end{gather*}

    \begin{gather*} [A] = \frac {[A]_0}{2} \end{gather*}

Substituting those values into the equation gives us the following:

    \begin{gather*} {\text{ln}\frac{[A]_0/2} {[A]_{\text{0}}}} = -kt_{1/2} \end{gather*}

Finally, dividing the k and the negative gives us an equation for the half-life of a first-order reaction:

    \begin{gather*} {t_{\text{1/2}} = \frac{ln2}{k}} \end{gather*}

Radioactive decay of Carbon-14, which has a half-life of 5730 years.
Graph showing the radioactive decay of Carbon-14. Carbon-14 has a half-life of 5730 years, so the amount of Carbon-14 is cut in half after every 5730 years. Geologists often use carbon dating to find the age of rocks and organisms. Source.

Second Order Half-Life

Second-order reactions are reactions where the rate of reaction is dependent on the concentrations of 2 reactants. The half-lives of second-order reactions depend on the reactant’s initial concentration. The integrated rate law for second-order reactions is as follows:

    \begin{gather*} \frac{1}{[A]} = kt + \frac{1}{[A]_0} \end{gather*}

Since we are trying to find the half-life of the reactant, we can set t as the half-life and [A] as half of the initial concentration:

    \begin{gather*} {t} = {t_{\text{1/2}}} \end{gather*}

    \begin{gather*} [A] = \frac {[A]_0}{2} \end{gather*}

Substituting those values into the equation gives us the following:

    \begin{gather*} \frac{2}{[A]_0} = kt_{1/2} + \frac{1}{[A]_0} \end{gather*}

We can then isolate the t to solve for the half-life of a second-order reaction:

    \begin{gather*} \frac{1}{[A]_0} = kt_{1/2} \end{gather*}

    \begin{gather*} t_{1/2} = \frac{1}{k[A]_0} \end{gather*}

Zero Order Half-Life

In zero-order reactions, the rate of reaction does not depend on the reactant’s concentration. The half-lives of zero-order reactions, however, do depend on the initial concentration of the reactant. The integrated rate law for zero-order reactions is as follows:

    \begin{gather*} [A] = [A]_0 - kt \end{gather*}

Since we are trying to find the half-life of the reactant, we can set t as the half-life and [A] as half of the initial concentration:

    \begin{gather*} {t} = {t_{\text{1/2}}} \end{gather*}

    \begin{gather*} [A] = \frac {[A]_0}{2} \end{gather*}

Substituting those values into the equation gives us the following:

    \begin{gather*} \frac{[A]_0}{2} = [A]_0 - kt_{1/2} \end{gather*}

We can then isolate the t to solve for the half-life of a zero-order reaction:

    \begin{gather*} kt_{1/2} = \frac{[A]_0}{2} \end{gather*}

    \begin{gather*} t_{1/2} = \frac{[A]_0}{k2} \end{gather*}

Practice Problems

  1. The half-life for a first-order reaction is 2768 years. Starting with a concentration of 0.345M, what will the concentration be after 11072 years?
  2. What is the half-life of a compound if 75 percent of a given sample of the compound decomposes in 60 min? Assume first-order kinetics.
  3. A substance hydrolyzes in water with a rate constant of 2.0 × 10−3 min−1. Calculate the t1/2 for the hydrolysis reaction. Assume first-order kinetics.

Solutions

  1. 0.0216M. First, find out how many half-lives have passed. By dividing 11072 by 2768, we find that 4 half-lives have passed. Therefore, we can find the final concentration by dividing 0.345M by 2 four times. This leaves us with 0.0216M.
  2. 30 mins. 25 percent of the sample is left, meaning 2 half-lives have passed. 60 mins divided by 2 is 30 mins, which is the half-life.
  3. 350 mins. Using the equation for the half-life of a first-order reaction, we can replace k with 2.0 × 10−3 min−1.

        \begin{gather*} {t_{\text{1/2}} = \frac{ln2}{2.0 \times 10^{-3}}} \end{gather*}

    We can then conclude that the half-life is 350 mins.

Further Reading