Using calculus to integrate rate equations

Rates of all kinds should remind us of calculus, and indeed, calculus – differentiation and integration of functions – is an important part of analyzing the kinetics of a reaction.

In this section we'll work through a few examples of how to integrate rate laws to come up with useful equations that help us analyze what's going on kinetically in chemical processes.

Zeroth-order reaction

Consider a simple chemical process

that obeys a zero-order rate law (meaning that the exponent of [A] is zero:

We can re-express the word "Rate" as the change in the concentration of A over time, or equivalently as the change in B over time. We'll choose the former and make sure that the rate is negative because we expect to be losing A:

Now integrating this simple differential equation is easy

The indefinite integral is

Now at time t = 0, we'll call the concentration of A "[A]o". Plugging in t = 0 and [A] = [A]o, we get

The integrated rate expression is

The concentration of A plotted vs. time will yield a linear graph with intercept [A]o and slope -k.

So a zero-order reaction might be identified by plotting [A] vs. time and checking for a linear graph. If it is linear, then the slope of that graph will be the rate constant. An example of a zero-order reaction is the decomposition of ammonia:

2NH3 (g) → N2 (g) + 3 H2 (g)

First-order reaction

Sticking with our simple model equation, A → B, but now assuming that this is a first-order reaction, that is, it obeys the rate law

we can substitute a derivative for "Rate."

This separable differential equation can be separated like this:

Then we can integrate both sides. Notice that the integral on the left suggests a log function solution.

The integrated equation is

Now if we let [A]o be the concentration of A at time t = 0, we get the integrated equation for this first-order reaction:

This suggests that if we suspect that such a reaction is first order, and we plot ln[A] vs. time (t), we should obtain a linear graph with slope = -k and y-intercept ln[Ao].

Second-order reaction

Keeping with our simple model equation, A → B, but this assuming that this is a second-order reaction, that is, it obeys the rate law

The differential equation is

It is separable, and the variable-separated form is

The integral is also a simple one

The general solution is

And once again if we let [A]o be the concentration of A at time t = 0, we get the integrated equation for this second-order reaction:

This equation suggests that if we suspect a rate equation is second order in one component, a plot of 1/[A] vs. t should yield a linear graph with a slope of k and a y-intercept of 1/[A]o.

Example 1

Find the rate law and calculate the rate constant from the data.

Sucrose (table sugar), C12H22O11, will decompose in weakly-acidic solutions to form two smaller sugars, fructose and glucose (both have the formula C6H12O6). These smaller sugars are structural isomers. The reaction is

C12H22O11 (aq) + H2O (l) → 2 C6H12O6 (aq)

The following concentration measurements were made at the time intervals shown, at 23˚C and in a weak HCl solution:

Time (min.) [C12H22O11]
0 0.315
40 0.275
80 0.235
140 0.190
210 0.145

In a problem like this, we have concentration vs. time data. We don't know whether it's zero, first or second order kinetically – or perhaps even more complicated. We'll start by assuming that the rate law looks pretty much like this:

Now we can make good use of the fact that zero- first- and second-order data can be plotted in different ways, and the correct model should yield a linear graph. We'll begin by looking at the possibility that this is a zero-order reaction, and plotting [A] vs time:

This graph isn't too curved. In fact, the R2 value of a linear fit (blue line) yields a correlation coefficient of 0.995. Still, it's clear from the magenta line that there is a slight upward curvature. Maybe we can do better.

If this reaction is first order, a plot of 1/[A] vs. t should yield a linear graph:

Well, that's a little worse, with a clear upward curvature and a lower linear correlation, so we'll rule out first order.

How about second order? A plot of ln[A] vs time has the highest linear correlation and just by looking there is little if any curvature.

From this data set, we'd have to conclude that this is a second-order reaction. That means that the detailed mechanism likely depends upon some sort of collision and interaction between two sucrose molecules before they break apart into fructose and glucose.

One thing you've probably noticed is that the distinction of reaction orders was pretty subtle here. It's often more obvious, but kinetic measurements and data can be tricky. That's the way it is. It often takes a great deal of data of various types to fully understand the kinetic details of a reaction.

The rate constant, k, is the slope of our line, so k = 3.6 × 10-3, and the units of rate must be molarity/second, so the units of k are M-1·s-1

Example 2

Determine the order of the reaction 2N2O5 (g) → 4NO2 (g) + O2 (g) from the data.

The data, concentration measurements of N2O5 taken over time are:

[N2O5] (M) Time (s)
0.100 0
0.709 50
0.051 100
0.249 200
0.126 300
0.0063 400

We'll follow the same procedure as above, plotting graphs to determine whether this is a zero-, first- or second-order reaction.

Zero order

From this data we can clearly see that the plot of [N2O5] vs. time is non-linear, so we can rule the zero-order case out.

First order

The plot of ln[N2O5] vs time is linear, with a correlation coefficient very near 1. This looks like first-order kinetic data.


The plot of the reciprocal of [N2O5] vs time is also clearly non-linear, so now we can be sure that this is a first order reaction.

The reaction is first order and the rate constant comes from the slope of the ln[N2O5] vs. time graph, k = 0.0069 s-1 (different units than the last example!)

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