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. More specifically, chemical rate equations are (usually simple) differential equations, which we have the tools to solve.

Zero^th-order reaction

Consider a simple chemical process

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

$$Rate = k[A]^o$$

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:

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

Now integrating this simple differential equation is easy

$$\int d[A] = \int k \, dt$$

The indefinite integral is

$$[A] = -kt + C$$

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

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

$$Rate = k[A]$$

we can substitute a derivative for "Rate."

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

This separable differential equation can be separated like this:

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

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

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

The integrated equation is

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

$$Rate = k[A]^2$$

The differential equation is

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

It is separable, and the variable-separated form is

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

The integral is also a simple one

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

The general solution is

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

Example 1

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

Solution: 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

C₁₂H₂₂O_{11 (aq)} + H₂O _(l) → 2 C₆H₁₂O_{6 (aq)}

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

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.

Example 2

Determine the order of the reaction 2N₂O_{5 (g)} → 4NO_{2 (g)} + O_{2 (g)} from the data.

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

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 half life

Very often when measuring data on the rate of a reaction, it is more convenient to measure the time it takes for the concentration of a reactant to be reduced to one-half it's starting concentration — and then to ¼ ⅛ and so on. The time that it takes for such a concentration reduction is called the half life.

Zero^th-order half life

the zero^th-order rate law is

$$[A] = [A]_o - kt$$

The trick to finding the half-life expression is, first, to solve for time, t, and then to replace [A] with [A]_o/2. First solve for t:

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

Then substitute [A]_o/2 for [A] and simplify to get the half life formula:

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

Notice that the half-life for a zero-order reaction is dependent on the initial concentration.

First-order half life

We follow the same procedure as above for determining the half-life for a first-order reaction. Fist, the integrated rate law (one version of it) is:

$$ln[A] = -kt + ln[A]_o$$

Isolating the time gives

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

Now we'll use the property of logs that ln(x) - ln(y) = ln(x/y) to rearrange:

$$\frac{1}{-k} \cdot \frac{ln[A]}{ln[A]_o} = t$$

Now substitute t_1/2 for t, and at that time, [A] = [A]_o/2:

$$\frac{1}{-k} \cdot \frac{ln([A]_o/2}{ln[A]_o} = t$$

Dividing the [A]_o gives a simple log of a constant in the numerator, which we can approximate:

$$\frac{ln\left( \frac{1}{2} \right)}{-k} \approx \frac{0.693}{k} = t_{1/2}$$

Second-order half life

Finally, let's find an expression for the half life of a second-order reaction. We begin again with the integrated rate law:

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

Isolating the kt term gives

$$kt = \frac{1}{[A]} - \frac{1}{[A]_o}$$

This is a convenient step to convert to half life:

$$kt_{1/2} = \frac{1}{[A]_o/2} - \frac{1}{[A]_o}$$

Rearranging the first fraction gives a nice expression for kt_1/2:

$$kt_{1/2} = \frac{2}{[A]_o} - \frac{1}{[A]_o} = \frac{1}{[A]_o}$$

And dividing by k gives us our result:

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