Integrated Rate Laws

Objectives

By the end of this section, you will be able to:

Explain the form and function of an integrated rate law
Perform integrated rate law calculations for zero-, first-, and second-order reactions
Define half-life and carry out related calculations
Identify the order of a reaction from concentration/time data

The rate laws we have seen thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.

Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. At the first-year level, you are not generally expected to perform these derivations, only to work with their results.

Click here to see the derivation of the first-order integrated rate law.

A first order rate law is generally written as: $$rate = k [A]$$ where A is the species we are measuring and $[A]$ is its concentration in mol/L.
Rate in this case is defined as the change in $[A]$ with time, or $\frac{\Delta[A]}{\Delta t}$. The rate law can also be represented as: $$\frac{\Delta[A]}{\Delta t} = k [A] $$ If we take the change measured by the rate from a “bulk” change ($\Delta $) to an “infinitesmal” change ($d$), this becomes: $$\frac{d[A]}{d t} = k [A] $$ This (hopefully) is a more recognizable (from calculus) form – and why the rate law is also called the differential rate law. A small rearrangement to isolate the variables (put all A’s on one side and all t‘s on the other) gives us: $$\frac{\mathrm{d} [A]}{[A]} = -k\, \mathrm{d}t$$ Taking the integral from time 0 (start of reaction) to time t: $$ \int_{[A]_0}^{[A]_t} \! \frac{\mathrm{d} [A]}{[A]} = \int_{0}^{t} \! -k\, \mathrm{d}t \\ $$ Since $\int \frac{1}{x}\, \mathrm{d}x = ln(x)$ and $\int k\, \mathrm{d}x = k\, x$, we will obtain: $$ln \left( \frac{[A]_t}{[A]_0]} \right)= -k\, t$$ Which is our first order integrated rate law. You will also see it rearranged as: $$ ln {[A]_t}= -k\, t + ln {[A]_0} \\ [A]_t = [A]_0 e^{-kt}$$

Since integrated rate laws are integrated with respect to time, they are most useful in situations where you need to relate concentrations and times for a reaction. For example:

Determining how long it will take a reaction to reach a certain point (e.g. 99% complete)
Predicting how much reactant will remain (or how much product will be produced) after a certain amount of time

We will focus on the integrated rate laws for first-, second-, and zero-order reactions, since these are the most common orders and the integrated rate laws give linearizable equations as results. Each of these orders is discussed on the following pages.