## Determining rate order from concentration vs. time data

As a reaction proceeds, we can measure the concentration of a reactant at various times. Then we use this data to determine the reaction order. For example, here’s the data for some arbitrary reaction during its first 30 seconds:

 Time (s) [A] (M) 0 0.100 5 0.0632 10 0.0400 15 0.0253 20 0.0160 25 0.0101 30 0.0064

To find the reaction order, we create three graphs. One is [A] vs. time, a second is ln [A] vs. time and the third is 1/[A] vs. time. If the first graph is linear, the reaction is zeroth order with respect to [A]. If the ln of concentration graph is linear, the reaction is first order, and if the reciprocal concentration graph is linear, the reaction is second order. In the old days (before calculators), we would calculate all the numbers and graph the data by hand and hope that one was obviously linear. Here, I used my graphing calculator to quickly generate three scatter graphs, shown below:

It should be pretty obvious that the red graph (ln [A] vs time) is the only linear graph, and that means the reaction is first order with respect to [A]. The rate equation is thus: rate = k[A] and the absolute value of the slope of the line (found using a linear regression) is the rate constant k.

What if the data has a bit of variability in it and the data doesn’t fit perfectly on a straight line on any of the three graphs? With your graphing calculator, create the three scatter graphs as I did above, then perform a linear regression on each graph. The graph with r2 closest to 1.0 will be the correct graph.

## Determining rate order from initial reaction rate

When you are studying chemical kinetics, a typical problem will look something like this:

Compounds A and B react together according to the following reaction:

A + 2 B → C + D

The following initial reaction rate data was collected. Determine the rate equation for this reaction.

 Rxn [A] [B] Initial rxn rate 1 0.0500 M 0.0500 M 1.2 x 10-4 M/s 2 0.100 M 0.0500 M 2.4 x 10-4 M/s 3 0.100 M 0.100 M 9.6 x 10-4 M/s

These problems are pretty easy to solve if you have learned exponential equations in algebra 2. The trick is to compare two different reactions where only one compound changes its concentration. For example, if you compare reaction 1 with reaction 2, you will see that [A] changes and [B] is the same. If you compare reactions 2 and 3, you see that [B] changes and [A] is held constant. However, when you compare reactions 1 and 3, both [A] and [B] change.  So do not compare reactions 1 and 3.

Rate = k [A]m [B]n

Therefore, your job is to determine the values of m, n, and k. You do this by taking the values from the table and plugging them into this equation.

Let’s find the value of m first. You need two equations where [A] changes and [B] does not—that is, equations 1 and 2.

$\begin{matrix} (2) \quad & 2.4 \cdot 10^{-4} = k(0.100)^m(0.0500)^n \\ (1) \quad & 1.2 \cdot 10^{-4} = k(0.0500)^m(0.0500)^n \end{matrix}$

Now divide the first equation by the second. The magic here is that k and the term with the n exponent both cancel out, leaving you with a simple expression:

$2.0 = 2^m$

It should be pretty obvious that m = 1. What if the numbers don’t divide evenly and you end up with an equation such as

$2.03 = 1.982^m?$

Unless told otherwise, you can assume that the exponent is supposed to be an integer, and therefore m= 1 in this problem too.

Now, let’s solve for n. We need two equations where [B] changes and [A] does not. That would be equations 2 and 3. Set it up the same way as before:

$\begin{matrix} (3) \quad & 9.6 \cdot 10^{-4} = k(0.100)^m(0.100)^n \\ (2) \quad & 2.4 \cdot 10^{-4} = k(0.100)^m(0.0500)^n \end{matrix}$

And divide as before:

$4.0 = 2^n$

Thus n = 2. Now we can solve for k by taking any of the three original equations and plugging in all the given values. Let’s use equation 3.

$9.6 \cdot 10^{-4} = k(0.100)^1(0.100)^2$

Solving this equation gives k = 0.96. Therefore, our final answer is

Rate = 0.96 [A] [B]2

Now here’s an additional trick I give you at no extra charge: If you are pretty good with exponents and scientific notation, you should see that you can often solve for the values of m and n in your head. In our first step, the rate doubled when [A] doubled, so the rate with respect to A is first order (m = 1). In our second step, the rate quadrupled when [B] doubled, so the rate with respect to B is second order (n = 2). (That was certainly a lot easier than the work we did above!) Then plug the known values into the rate equation to solve for k.

## Differential rate law vs. integrated rate law

When we study rates of chemical reactions, we look for an equation that describes the reaction rate. The equation can be in one of two forms, known as the differential rate law and the integrated rate law.

If you have taken calculus, you’ll recognize these names and you even know how to derive the integrated law from the differential law. If you haven’t taken calculus, don’t worry! You probably won’t be asked to derive the equations. However, you should recognize each form of the law for zeroth, first and second order reactions.

The differential rate law gives the reaction rate as a function of the concentration of the reactants. The integrated rate law gives the concentration as a function of time. Both equations are useful, and you will probably use both in your chemistry class.

The table below shows the two rate laws for each reaction order in terms of a single reactant A.

 Order Differential rate law Integrated rate law Zeroth Rate = k $[A] = -kt + [A]_0$ First Rate = k[A] $\ln{[A]} = -kt + \ln{[A]_0}$ Second Rate = k[A]2 $\dfrac{1}{[A]} = kt + \dfrac{1}{[A]_0}$

You might have noticed that each of the integrated law equations is in the form y = mx + b, the equation for a line. In the old days (before graphing calculators), this is the method that chemists used to determine the reaction order. If the graph of concentration vs. time is a straight line, the reaction is zeroth order. If the graph of ln(concentration) vs. time is a straight line, the reaction is first order. And if the graph of 1/concentration vs. time is a straight line, the reaction is second order.

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