Reaction Order — Definition & Overview

The reaction order refers to the powers that the rate is dependent on. Consider the reaction below, $$A \rightarrow B$$ Let's say this reaction has the corresponding rate law. $$\text{rate}=k[A]$$ where $k$ is a constant. This reaction is said to be a first order reaction since $[A]$ is raised to the first power. If the rate law were given as $\text{rate}=k[A]^{2}$, then the reaction would be second order.

What happens if the rate is dependent on multiple chemical species? Let's consider the reaction below, and the associated rate law. $$A+B \rightarrow C$$ $$\text{rate}=k[A]^{2}[B]$$ This reaction is said to be third order overall. That is the sum of all the exponents. But the reaction is said to be second order with respect to $A$ and first order with respect to $B$.

It is possible for a reaction to be zeroth order. That is to say, the reaction is independent of the concentration of any species. These reactions always proceed at the same rate. Some biological processes, such as the liver oxidizing ethanol, are zeroth order reactions.

(Video) Kinetics: Initial Rates and Integrated Rate Laws

by Professor Dave Explains

In this video, professor Dave discusses many aspects of kinetics. You will learn about how reaction rate, reaction order, and rate law are all related. You'll also have a chance to practice what you've learned with a problem at the end.

We've spent a lot of time investigating the various chemical reactions. We even incorporated some math when we learned stoichiometry. But so far, we haven't talked about how chemical reactions happen. What goes on during a chemical reaction?

Now, we're beginning our study of chemical kinetics. It's a subject within physical chemistry. It investigates how quickly reactions occur and the factors that affect them. We started by discussing the reaction rates. They show us how our reactant or product concentrations change with time. So, our units are typically molarity per second. Today, we'll study reaction order. It tells us how the concentration changes affect the rate.

Finding Reaction Order

Remember, chemistry is an experimental science. Sometimes, we can get away with theoretical work. But often, we have to go into the lab and perform experiments. We can only determine the reaction order from experimental data. So, for any example problems, you'll be given a data set.

Let's assume we have a reaction that only depends on [X]. Let's look at various experimental scenarios.

Scenario 1:

Let's say that we performed two experimental trials. For the first trial, our concentration was [X], and the initial rate was $t$. For the second, we doubled the concentration. Let's say our initial rate also doubled. So, we had 2[X] and $2t$. What observation can we make? Our rate is directly proportional to the concentration of X. When we doubled X, the rate also doubled. How could we write this mathematically? We could write $rate\propto [X]$. As a rate law, it would be $rate=k[X]^{1}$. $k$ is the rate constant. It varies from reaction to reaction. But it's just a proportionality. Because our exponent is one, this reaction is first order.

Scenario 2:

Again, let's say that we performed two experimental trials. For the first trial, our concentration was [X], and the initial rate was $t$. For the second, we doubled the concentration. But this time, our initial rate quadrupled. So, we had 2[X] and $4t$. What observation can we make? The rate of our reaction is directly proportional to $[X]^{2}$. So, we could write $rate\propto [X]^{2}$. As a rate law, it would be $rate=k[X]^{2}$. Because our exponent is two, this reaction is second order.

Scenario 3:

Let's look at one final scenario. Again, we perform two experimental trials. For the first trial, our concentration was [X], and the initial rate was $t$. For the second, we doubled the concentration. But this time, our initial rate stayed the same. So, we had 2[X] and $1t$. What observation can we make? The rate of our reaction is directly proportional to $[X]^{0}$. The rate does not depend on the reactant concentration! So, we could write $rate\propto [X]^{0}$. As a rate law, it would be $rate=k[X]^{0}$. Anything raised to the zero-power is one. So, our rate equals our rate constant. Because our exponent is zero, this reaction is zero order.

How can a chemical reaction not depend on concentration? Zero-order reactions depend on an outside factor. The most common is thermal decomposition. It typically occurs when we have massive temperatures. The system has a lot of heat energy. What's the result? The molecular vibrations overcome the bond energy. The molecule decomposes. What's another example? Catalysis! Remember, catalysts lower a reaction's activation energy. So, they increase the reaction rate. In catalysis, the catalyst's surface area is the driving factor.

Reaction Order with Multiple Reactants

In each of the above scenarios, we only had one reactant. What if we have two or more reactants? We have to find the reaction order for each individual reactant. The total order for the reaction is the sum of the exponents. So, let's say we had a reaction with two reactants. Both reactants are second order. What is the overall order? It's fourth order! Why? Two plus two is four.

Practice Problem Walkthrough

Let's consider the redox reaction between solutions of ammonium and nitrite ions. They react to form nitrogen gas and water.

Let's start by writing the balanced chemical equation.

$NH^{+}_{4(aq)} + NO^{-}_{2(aq)} \rightarrow N_{2(g)} + 2H_{2}O_{(l)}$.

Let's start by writing a generic rate law based on the reactant concentrations.

$rate=K[NH^{+}_{4(aq)}]^{X}[NO^{-}_{2(aq)}]^{Y}$

Notice that each concentration has exponents (X and Y). That's because we don't know the reaction order for each reactant. We cannot determine it from the chemical equation! We must you experimental data! Let's look at the table below.

$$\begin{array}{r|c|l} [NH^{+}_{4(aq)}] & [NO^{-}_{2(aq)}] & Rate \\ \hline 1 & 1 & 1 \\ 2 & 1 & 2 \\ 1 & 2 & 2 \end{array}$$

Please know that these numbers are very artificial. But they highlight how we'll work with more complex examples. We see that when $[NH^{+}_{4(aq)}]$ doubled, the rate also doubled. That's a first-order situation (scenario 1). The same is also true for $[NO^{-}_{2(aq)}]$. So both X and Y equal 1. Each reactant is first order. Our overall reaction order is second ($X+Y=1+1=2$).