Rate laws can be confusing for students, but they don’t need to be. Let’s try to walk through some problems that will hopefully clarify what these problems are and how to approach them.
Rate laws appear in the following form:
Rate laws can only be determined via experimentation, never from the chemical equation alone. Take for instance the following reaction:
We might be inclined to write the rate law for this reaction as
However, this would be incorrect. Instead we would have to look at a table like the following:
This table contains everything we need to find for rate laws: the order of reaction and the rate constant, k. Let’s take a look at how we can accomplish this.
First we want to look at two rows that have one of the reactants with constant concentrations. Notice how the bolded and italicized concentrations are the same across two rows.
Now let’s look at the first two columns. We hold B constant, which means any change in rate has to be attributed to changes in A. Notice that [A] doubles, from 0.10 to 0.20, and it caused the rate to go from 0.0021 to 0.0082. Now, there are two ways to go about finding the order. The formal way is to use a bit of math to solve for the order, which can be very useful if you want a sure way to get the correct order. The other method is a bit more intuitive. We’ll explore both!
Option 1: Math solves everything!
[A]1 and Rate 1 = the first row, and [A]2 and Rate 2 = the second row values. Notice that the k’s cancel! This will make things easier for us.
We are now going to be able to solve for n from here. We need to use some properties of logs to do so, so if this is fuzzy material be sure to review those rules. Continuing on,
Finally we get n = 1.963, which we can safely round up to n = 2. Hurray! We just found that the order of [A] is 2! Repeating this process (don’t worry, it’s not as long as it looks!) will allow us to find the order of B. At the end we should arrive at this rate law:
To finish out, let’s look at the second option.
Option 2: Lets guesstimate!
The overall idea of this process is basically the same as the math approach, but we don’t do any math with logs. First we look at what we are doing to the concentration of A: we’re doubling it. Now we see what is happening to the rate: it roughly quadruples. So in order to get from a 2x ([A]) increase to a 4x increase (in rate) we know we have to square the 2x. Thus, since we squared, that means that [A] must be second order. If we doubled [A] and the rate also doubled, that would be a 1:1 increase, so [A] would be first order. And if we doubled [A] and got no change in the rate, we would know that [A] would be zero order. A bit tricky, but it can save a lot of time!
I hope this helps make sense of rate laws and how we arrive at these equations. If you have other questions don’t hesitate to come visit CAPS on the 3rd floor of Zimmerman Library! Our hours and locations can be found here. Have fun with kinetics!