## 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.

Chemistry
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