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A reaction is first order with respect to a reactant $A$ if the rate is directly proportional to the concentration of $A,$ i.e. $$rate = k[A],$$ where $k$ is a constant that does not depend on $[A].$ The quantity $k$ is called the rate constant, and the above equation is called the rate law for the reaction.

We can solve for $[A]$ as a function of time $t$ and the initial concentration $[A]_0$ by solving the differential equation expressed by the rate law: $$\frac{d[A]}{dt} = -k[A]$$ $$\frac{d[A]}{[A]} = -k dt$$ $$\int_{[A]_0}^{[A]}\frac{d[A]}{[A]} = \int_{0}^{t}k dt$$ $$\ln[A] - \ln[A]_0 = -kt$$ $$[A] = [A]_0e^{-kt}.$$

The half-life of a reaction with respect to a particular reactant is the amount of time required for half of the reactant to participate in the reaction. One can see from the above equation that, since $[A]$ depends exponentially, the half-life of a first-order reaction is a constant dependent only on $k.$ One can easily solve for the half-life $t_{1/2}$ as follows: $$[A] = \frac{1}{2}[A]_0$$ $$e^{-kt_{1/2}} = \frac{1}{2}$$ $$-kt_{1/2} = -\ln 2$$ $$t_{1/2} = \frac{\ln 2}{k}.$$

(Video) Half-Life | Chemical Kinetics

by Ben's Chem Videos

In this video, Ben explains half-life. He starts with the definition. Then he derives the half-life equations for zero, first, and second-order reactions. In the end, he shows you an example problem. Are you looking for help understanding half-life? This video can help!

We're exploring chemical kinetics. It's a subset of physical chemistry that studies reaction rates. We want to know how fast a chemical reaction happens. What does that mean? We want to see how our product and reactant concentrations change per unit of time. We've explored reaction rates, reaction order, and writing rate laws. What's the result? If we know the molarity (moles per liter), we can calculate the rate.

Often want to know how concentration changes with time. For that, we use the integrated rate laws. The equations for the integrated rate laws required some calculus to derive. But if you don't know calculus, ignore the derivation. You can use the final equations. In general chemistry classes, you typically won't have to derive the equations.

Today, we'll explore half-life. It's the time required for the reactant's concentration to decrease by half. It's a relationship between time and concentration. So, we'll have to use the integrated rate equations.

Zero Order Half-Life

The integrated rate law equation for a zero-order reaction is $[A]=[A]_{o}-kt$. Let's start by solving the equation for $t$.

$t=\frac{[A]-[A]_{o}}{-k}$

By definition, the half-life occurs when $[A]=\frac{1}{2}[A]_{o}$. Let's plug that into our equation!

$t=\frac{[A]-[A]_{o}}{-k}\rightarrow t_{\frac{1}{2}}=\frac{\frac{1}{2}[A]_{o}-[A]_{o}}{-k} \rightarrow t_{\frac{1}{2}}=\frac{[A]_{o}}{2k}$.

What does this tell us? For a zero-order reaction, the half-life depends on the initial concentration and the rate constant. So, the subsequent half-lives will be shorter.

First Order Half-Life

Let's repeat the process for a first-order reaction. The integrated rate law was $\ln{[A]}= \ln{[A]_{o}}-kt$.

Again, let's solve for $t$.

$t=\frac{\ln{[A]} - \ln{[A]_{o}}}{-k}$

We can take advantage of the properties of logarithms to clean up the equation.

$t=\frac{\ln{\frac{[A]}{[A]_{o}}}}{-k}$

Again, by definition, the half-life occurs when $[A]=\frac{1}{2}[A]_{o}$. Again, we'll take advantage of the properties of logarithms to tidy our result.

$t_{\frac{1}{2}}=\frac{\ln{\frac{\frac{1}{2}[A]_{o}}{[A]_{o}}}}{-k} \rightarrow \frac{\ln{\frac{1}{2}}}{-k} \rightarrow \frac{\ln{2}}{k}\rightarrow \frac{0.693}{k}$

What does this tell us? For first-order reactions, the half-life is only affected by the rate constant. In general chemistry, we assume the rate constant doesn't change during a chemical reaction. So, the half-life is constant. That means that each subsequent half-life is the same.

In future lessons, we'll study nuclear chemistry. Many of the lanthanides and actinides are radioactive elements. That means they have unstable nuclei. So, they undergo various methods of radioactive decay. But every type of decay only depends on one component. That's the unstable nucleus. So, all types of radioactive decay are first-order! We use the half-life to calculate the rate of radioactive decay. We use the radioactive decay rate in radioactive dating. For example, many scientists use the isotope carbon-14 for archeological dating. Its half-life is 5715 years. So, it's perfect for determining the age of ancient artifacts.

If you want to learn more about radioactive decay, look at our lessons on alpha decay, beta decay, gamma decay, and electron capture.

Second Order Half-Life

One last time, let's solve the integrated rate equation for second-order reactions.

$\frac{1}{[A]}= kt + \frac{1}{[A]_{o}}$

This time, let's plug in the half-life definition first.

$\frac{1}{\frac{1}{2}[A]_{o}}= kt_{\frac{1}{2}} + \frac{1}{[A]_{o}}$

Now we can rearrange the equation. Both fractions have the same denominator. So, the subtraction is simple.

$\frac{2}{[A]_{o}}-\frac{1}{[A]_{o}}=kt_{\frac{1}{2}}$

Now, all we have to do is divide both sides by the rate constant.

$\frac{1}{k[A]_{o}}=t_{\frac{1}{2}}$

In this case, the half-life depends on the rate constant and the initial concentration. But this time, both are in the denominator. So, subsequent half-lives take longer.

Types of Problems You Could Encounter

Many problems will ask you to determine the half-life of a reaction. You'll need to know...

• The reaction order.

• You might have to calculate this based on the method of initial rates.

• Use the reaction order to determine which equation to use.

• The rate constant.

• You might have to calculate it from the Arrhenius equation or the method of initial rates.

• If the reaction is zero or second-order, you'll need the initial concentration.

Another variety of problems involves converting the half-life to a rate constant. You'll need to know...

• The half-life.

• The reaction order. Use the reaction order to determine which equation to use.

• If the reaction is zero or second-order, you'll need the initial concentration.

Chemistry in the World

Why do we care about the half-life? It's a bit arbitrary. We could just as easily use a quarter-life. But, it's a convenient measure of concentration decrease. For example, in college, I studied environmental chemistry. When environmental chemists track a pollutant through a system, they often think of half-life. If a contaminant has a rapid half-life, it may only require containment. But a long half-life might require complete remediation.

We also consider the half-life in medicine. Let's say you take a medication. Often, the drug gets converted into its active form in the liver. The liver is also the primary site where chemicals get broken down. The liver uses various enzymes to remove foreign chemicals. Enzymes are essentially biological catalysts. So, they break down molecules into smaller and more soluble forms. The molecules then get removed from the body. The effectiveness of a medication expires after about five half-lives. For some drugs, the half-life is remarkably fast. So, you may have to take multiple doses throughout the day. But sometimes, it can take many months for a medicine to undergo five half-lives. So, you may receive treatment only periodically.