Integrated Rate Laws — Overview & Calculations

We've started to study kinetics! We're beginning to answer some interesting questions. How fast does a chemical reaction happen? What factors influence a reaction's progress? Today, we'll look at equations that compare the change in concentration versus time. Arriving at the equations does require some calculus. If you don't know calculus, skip the derivation sections.

What We've Learned about Kinetics So Far

Before we look at the equations, let's remind ourselves of what we've already studied. We need to know these to help us understand the equations.

Reaction rates measure the change in concentration over time.
The reaction orders tell us how a reactant affects the reaction rate.

We find the reaction orders for each reactant experimentally.

The total reaction order is the sum of the reactants' reaction orders.

The rate law is an equation that describes the reaction rate.

It's written as $rate=k[A]^{n}[B]^{m}=\frac{\Delta\,[reactants]}{\Delta\,time}$ .

K is the rate constant. We can calculate it from the initial rates method or the Arrhenius Equation.

[A] and [B] are the molar concentrations (moles per liter) of the reactants. If we had more reactants, we would add more terms.

n and m are the reaction order for each reactant. The total reaction order is $n+m$ .

A reactant's half-life is the time it takes for a reactant's concentration to halve.
We can derive the equations for half-life from the rate laws.

Two Forms of Rate Laws

There are two types of rate laws. The differential rate law is the one we've already seen. It expresses the reaction rate as a function of the concentration change over a period of time. The differential equation helps us know what is happening on a molecular level. We use it to determine reaction mechanisms.

The second type is called the integrated rate law. We derive it from the differential, using calculus. The result of the integration is a function that tells us the concentration at a specific time. We use it to determine the reaction order and constant from experimental data. The integrated law gives us a macroscopic understanding. We can track how our reactant concentrations change with time.

In the following sections, we'll derive both types of rate laws.

Differential Rate Law

Say we have a generic balanced chemical equation. We'll use a decomposition reaction.

$A \rightarrow B + C$

How would we write the rate law?

$rate=-\frac{\Delta [A]}{\Delta [t]} = k[A]^{n}$

If you know calculus, you know we can write the changes in concentration over time as a differential.

$rate=- \frac{d[A]}{dt}= k[A]^{n}$

We now have the generic form of the derived rate law.

Deriving the Integrated Rate Laws

Zero Order:

If we have a zero-order reaction, we know that $n=0$ . So our derived equation is equal to $rate= -\frac{d[A]}{dt}= k[A]^{0}$ .

What happens if we integrate the derived rate law? First, let's set our integration boundaries.The initial concentration of A will be $[A]_{o}$ at time, $t_o$ . Our second boundary will be a later concentration of A, $[A]$ at time $t$ .

$- \frac{d[A]}{dt}= k[A]^{0}$ .

$-d[A]=k dt$

$-\int_{[A]_{o}}^{[A]} d[A]=k\int_{t_{o}}^{t}dt$

$(-1)([A]-[A]_{o})=k(t-t_{0})$

$[A]=-k(t-t_{o})+[A]_{o}$

If we assume that $t_{o}=0$ , we get $[A]=-k(t)+[A]_{o}$

First Order:

For first-order reactions, we know that n=1.

$-\frac{d[A]}{dt}= k[A]^{1}$ .

$\frac{d[A]}{[A]}=-k dt$

$\int_{[A]_{o}}^{[A]}\frac{d[A]}{[A]}=-k\int_{t_{o}}^{t}dt$

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

Again, if we assume that $t_{o}=0$ , we get $\ln{[A]}=-kt+\ln{[A]_{o}}$ .

Second Order:

For second-order reactions, we know that n=2.

$-\frac{d[A]}{dt}= k[A]^{2}$ .

$\frac{d[A]}{[A]^{2}}=-k dt$

$\int_{[A]_{o}}^{[A]}\frac{d[A]}{[A]^{2}}=-k\int_{t_{o}}^{t}dt$

To help with the integration, we can write $\frac{1}{[A]^{2}}$ as $[A]^{-2}$ .

$\int_{[A]_{o}}^{[A]}[A]^{-2}d[A]=-k\int_{t_{o}}^{t}dt$

$-[A]^{-1} + -[A]_{o}^{-1}= -k(t-t_{0})$

Again, if we assume $t_{0}=0$ then, $\frac{1}{[A]}= kt + \frac{1}{[A]_{o}}$ .

Integrated Rates Laws and What They Tell Us

Zero order: $[A]=-k(t)+[A]_{o}$ .

Notice that the equation is in the form y=mx+b. If we plotted a zero-order reaction, the slope would be -k.
Because the slope is negative, eventually, the concentration would become negative. So, a zero order reaction model only works over a specific range.

First order: $\ln{[A]}=-kt+\ln{[A]_{o}}$ .

This equation is also in the form y=mx+b. Again, the slope of the graph is -k.

Second order: $\frac{1}{[A]}= kt + \frac{1}{[A]_{o}}$ .

The second order equation also has the form y=mx+b. However, this time the slope is positive.
The slope is positive because we have the inverse of the concentration. So, at higher concentrations of A, $\frac{1}{[A]}$ is smaller. But, as the reactant depletes, $\frac{1}{[A]}$ becomes larger.

Solving the Integrated Rate Laws for Half-Life

For all the equations, the half-life occurs when $[A]=\frac{1}{2}[A]_{o}$ and $t = t_\frac{1}{2}$ .

Zero Order:

$[A]=-k(t)+[A]_{o}$

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

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

The zero order half-life depends on the initial concentration and the rate constant.
As the initial concentrations become smaller, so does the half-life. So, the time between half-lives shrinks as the reaction proceeds.

First Order:

$\ln{[A]}=-kt+\ln{[A]_{o}}$

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

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

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

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

$\frac{-\ln{\frac{1}{2}}}{k}=t_{\frac{1}{2}}$

$\frac{ln{2}}{k}=t_{\frac{1}{2}}$

The half-life only depends on the rate constant. As the reaction proceeds, the half-life occurs at regular intervals.

Second Order:

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

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

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

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

The half-life depends on the rate constant and the inversed initial concentration.
As the reaction proceeds, the time between half-lives increases.

Because the half-life increases, second-order chemicals can persist for a long time. Their persistence is a challenge in environmental chemistry. For example, traces of a pollutant may remain in a system for a long time. Even small concentrations of a toxin can cause unbelievable damage!

With high concentrations, there is a greater probability of the molecules colliding. So, the half-life is faster.

Chemistry In The World: Uranium in the Water

We are most familiar with uranium as a radioactive element. All of uranium's isotopes have unstable nuclei. So, they are all radioactive. That makes uranium a toxic metal. The United States began uranium mining in the Navajo Nation in the 1940s. From World War II through the Cold War, the U.S. extracted over 30 million tons of uranium ore. Sadly, many of the uranium deposits were near underground aquifers. The aquafers provide drinking water for the Navajo Nation. Today, many of the aquafers have high concentrations of uranium. The result is that many Navajo people have experienced chronic uranium exposure.

What does this have to do with half-lives? The problem is that all of uranium's isotopes have long half-lives. The shortest is uranium-234, which has a half-life of 250,000 years. Even when uranium decays, it undergoes a decay series in which many of the products are also toxic. The good news is that uranium has very low solubility in groundwater. So, it does not spread easily. But, it's still a massive problem that will take decades of remediation work.