Integrated Rate Laws

integrated rate laws graph

Core Concepts

In this article, we learn about integrated rate laws, including their chemical meaning, mathematical derivation, and usefulness in studying chemical kinetics.

Topics Covered in Other Articles

Rate Laws and Integrated Rate Laws

Chemical reactions can happen at many different speeds depending on the specifics of the reaction. As a reaction progresses, you can expect to see the concentration of a reactant decrease over time as products form. If you need an introduction to kinetics and rates of reaction, check out this article!

Integrated rate laws are mathematical equations that represent the concentration of a reactant as a function of time. These equations provide a lot of insight to chemists, who use integrated rate laws in order to closely model reactions or systems of reactions. As the name suggests, integrated rate laws come from using calculus to integrate what chemists call rate laws.

Rate laws illustrate the mathematical relationship between reactant concentration and reaction rate. 

    \begin{gather*} {aA+bB\rightarrow cC+dD} \\ {\text{Rate}=k[A]^{n}[B]^{m}} \end{gather*}

As you may have noticed, the concentrations of the reactants are raised by some power (n and m) in the rate law. These values are the reaction order with respect to each reactant. Put simply, reactants often don’t have a one-to-one effect on reaction rate. Doubling the concentration of a reactant could double the reaction rate, or quadruple it, or have no effect at all. Each of these cases implies a different reaction order with respect to that reactant. To learn more about the kinetics of reaction order, check out this article.

Additionally, rate laws include an additional parameter (k). Chemists all this value the rate constant, and it mathematically accounts for other factors affecting the reaction rate. Some of these factors include temperature or the presence of catalysts. The mathematical impact of these factors on the rate constant are illustrated through the Arrhenius equation. If you’d like to learn more about the Arrhenius equation, check out this article.

Integrating Rate Laws

As mentioned before, to yield an integrated rate law, we need to do some calculus on a rate law expression. In a rate law, “reaction rate” means the rate of product production. Put differently, reaction rate is the change of product concentration over time, adjusted for stoichiometry

    \begin{gather*} {aA+bB\rightarrow cC+dD} \\ {\text{Rate}=\frac{d[C]}{dt}\left(\frac{1}{c}\right)} \end{gather*}

When accounting for differences in stoichiometry, products of the same reaction are produced at the same rate. Additionally, the rate of product formation always equals reactant consumption or the negative change in reactant concentration.

    \begin{gather*} {\text{Rate}=\frac{d[C]}{dt}\left(\frac{1}{c}\right)=\frac{d[D]}{dt}\left(\frac{1}{d*}\right)=\frac{-d[A]}{dt}\left(\frac{1}{a}\right)=\frac{-d[B]}{dt}\left(\frac{1}{b}\right)} \end{gather*}

(Note: d* indicates the stoichiometric coefficient in the reaction equation)

In order to integrate rate laws, we need to use the version of reaction rate that includes the concentration of the reactant in the rate:

    \begin{gather*} {\frac{-d[A]}{dt}\left(\frac{1}{a}\right)=k[A]^{n}} \end{gather*}

Depending on the reaction order, the exact steps to integrate a rate law differ. However, the basic method of integration followed by solving for [A] remains the same, as we’ll see.

Integrated Zero-Order Rate Law

Take the following zero-order rate law:

    \begin{gather*} {A\rightarrow B} \\ {\text{Rate}=\frac{-d[A]}{dt}=k} \end{gather*}

Using calculus, we can integrate the rate law according to the following proof. Note that [A] indicates the initial concentration of A at the start of the reaction.

    \begin{align*} {\frac{-d[A]}{dt}&=k} \\ {-d[A]&=kdt\vphantom{\frac12}} \\ {-\int_{[A]_{0}}^{[A]}d[A]&=k\int_{0}^{t}dt} \\ {-\left([A]-[A]_{0}\right)&=k\left(t-0\right)\vphantom{\frac12}} \\ {[A]_{0}-[A]&=kt\vphantom{\frac12}} \end{align*}

Following the rules of integration, we have a fairly straightforward, linear expression. We can solve for [A] and therefore end up with a y=mx+b style expression. This is our integrated zero-order rate law.

    \begin{gather*} {[A]=-kt+[A]_{0}} \end{gather*}

Integrated First-Order Rate Law

Take the following reaction and corresponding first-order rate law:

    \begin{gather*} {A\rightarrow B} \\ {\text{Rate}=\frac{-d[A]}{dt}=k[A]} \end{gather*}

Using calculus, we can again integrate the rate law according to the same method:

    \begin{align*} {\frac{-d[A]}{dt}&=k[A]} \\ {\frac{-d[A]}{[A]}&=kdt} \\ {-\int_{[A]_{0}}^{[A]}\frac{-d[A]}{[A]}&=k\int_{0}^{t}dt} \\ {-\left(ln[A]-ln[A]_{0}\right)&=k\left(t-0\right)\vphantom{\frac12}} \\ {ln[A]_{0}-ln[A]&=kt\vphantom{\frac12}} \end{align*}

Interestingly, following the rules of integration, we have our concentrations of A in natural logs. We can solve for ln[A] and therefore end up with a y=mx+b style expression. We can also take that same expression, using the properties of natural logs, and end up with an exponential expression. Both are valid (and useful) expressions of [A] as a function of time.

    \begin{align*} {ln[A]&=-kt+ln[A]_{0}} \\ {[A]&=[A]_{0}e^{-kt}} \end{align*}

Integrated Second-Order Rate Law

Take the following reaction and corresponding second-order rate law:

    \begin{gather*} {A\rightarrow B} \\ {\text{Rate}=\frac{-d[A]}{dt}=k[A]^{2}} \end{gather*}

Using the same method, we can integrate the rate law according to the following proof:

begin{align*} {\frac{-d[A]}{dt}&=k[A]^{2}} \\ {\frac{-d[A]}{[A]^{2}}&=kdt} \\ {-\int_{[A]_{0}}^{[A]^{2}}\frac{-d[A]}{[A]^{2}}&=k\int_{0}^{t}dt} \\ {-\left(\frac{1}{[A]_{0}}-\frac{1}{[A]}\right)&=k\left(t-0\right)} \\ {\frac{1}{[A]}-\frac{1}{[A]_{0}}&=kt} \end{align*}

Interestingly, following the rules of integration, we have inverses of our concentrations of A. We can solve for 1/[A] in order to once again end up with a  y=mx+b style expression. We can once again take that same expression, using simple algebra, and end up with an (also useful) inverse expression.

    \begin{align*} {\frac{1}{[A]}&=kt+\frac{1}{[A]_{0}}} \\ {[A]&=\frac{[A]_{0}}{[A]_{0} kt+1}} \end{align*}

Integrated Rate Law Graphs

Between the zero-, first-, and second-order integrated rate laws, each is a different mathematical class of function. This then means that each reaction order graph ([A] vs. time) looks noticeably different.

graphed data of different orders

Additionally, we’ve previously identified  y=mx+b forms of the first- and second-order integrated rate laws. This consequently means that we can display first-order reaction data on a ln[A] vs time graph and generate a linear graph. We can also do the same with second-order data on a 1/[A] vs. time graph and also get a linear graph.

graphed linearized integrated rate laws

The following table summarizes the behavior of a kinetic graph, given data of a certain reaction order. The most important detail to look at is concavity:

 [A] vs. timeln[A] vs. time1/[A] vs. time
Zero-order dataLinearConcave-DownConcave-Up
First-order dataConcave-UpLinearConcave-Up
Second-order dataConcave-UpConcave-UpLinear
all combinations of graphs and integrated rate laws

Application of Integrated Rate Laws: Finding Reaction Order

As mentioned before, these integrated rate laws are useful tools for chemists studying the kinetics of chemical reactions. Specifically, a chemist can take empirical kinetic data from a chemical reaction they can determine the reaction’s order according to which integrated rate law best models the kinetic data. 

Unlike other ways of determining reaction order, this method does not require directly measuring reaction rate nor multiple trials. Instead, integrated rate laws allow chemists to find reaction order by monitoring the concentration of a reactant over the course of one trial. 

Let’s take a look at an example. You decide to study a reaction that has the following stoichiometry:

    \begin{align*} {A\rightarrow B+C} \end{align*}

You use spectrophotometry in order to continually measure the concentration of A over the course of multiple minutes. After using the Beer-Lambert Law to convert absorbance to concentration, you have the following measurements:

Time (min)[A] (M)

As we discussed in the previous section, kinetic data like this looks different depending on reaction order once graphed. Thus, you then take your data and make three graphs: [A] vs. time, ln[A] vs. time, and 1/[A] vs. time:

graphed data using integrated rate law models

Interestingly, the ln[A] vs. time graph has a linear shape while the other two have a subtle concave. This means your reaction must be first-order since it follows the first-order integrated rate law:

    \begin{align*} {ln[A]&=-kt+ln[A]_{0} \rightarrow \text{Rate}=k[A]} \end{align*}

Also, as a neat bonus, the slope of the graph corresponds to the reaction constant. The ln[A] vs. time has a slope of  -0.1012 , which thus means our reaction constant is 0.1012 \text{s}^{-1} :

rate constant derived from integrated rate law

Integrated Rate Laws Practice Problems

Problem 1

Given the following first-order reaction and an initial reactant concentration of  [A]_{0} = 0.300\text{M}, how much reaction time will pass until the concentration of A halves? (Solve for t when [A]_{t} = \frac{[A]_{0}}{2} = 0.150\text{M}) ( k = 3.43\text{s}^{-1} )

     \begin{gather*} A \rightarrow 2B \end{gather*}

Problem 2

Given the following second-order rate law:

     \begin{gather*} {2A \rightarrow B + C} \\ {Rate = k[A]^{2}} \\ {k = 0.0246\text{M}^{-1}\text{s}^{-1}} \end{gather*}

You measure the concentration of A after 30 seconds is  [A]_{t=30} = 0.750\text{M} . What was the initial concentration of A?

Integrated Rate Laws Practice Problem Solutions

1:  t = 0.202\text{s}

2:  [A]_{0} = 1.68\text{M}