5.7 Reaction Mechanisms and Elementary Steps

Reaction mechanisms go more in-depth regarding chemical reactions than just a net equation. For example, a reaction between an acid and a base must happen in the presence of water, yet water is not included as a reactant. The actual reaction happens in two steps. A mechanism takes a full reaction and splits it up into elementary steps, essentially what actually happens in a reaction. Here's an example for the decomposition of hydrogen peroxide (H2O2):

As you can see, each of these steps are individual reactions, and when added up and spectators are cancelled out, we end up with the total reaction. Why is this important, you may ask? Well, if we find the total reaction by adding the elementary steps, we see that IO- and I- cancel out to leave 2H2O2 --> 2H2O + O2. So then why are they there? Well, as we'll see in future lessons, iodide acts as a catalyst for this reaction, speeding it up. It goes in and comes out the exact same way and thus does not impact the reaction products, but rather impacts the mechanism by changing the way the reaction occurs. The natural decomposition of H2O2 actually occurs, but it is incredibly slow. By changing the steps in the reaction, iodide creates IO-, which is called an intermediate. It is not a product, nor a reactant, but takes place in the reaction.

Often when dealing with mechanisms, you will see one elementary step labeled "slow" and the others labelled "fast". The slow step is also called the rate-determining step as it defines the rate law. Let's take a look at an example:

Step one is to find the overall reaction for this mechanism, which comes out to H2 + 2ICl --> I2 + 2HCl. We now know what the overall reaction is. To find the rate law for this reaction, we must look at the rate limiting step. When dealing with elementary steps (AND ONLY ELEMENTARY STEPS), we can use the stoichiometric coefficients to tell us the order for each reactant. Thus, looking at the slow reaction, we know that the rate law is: R = k[H2][ICl]

An important thing to note is that sometimes the rate-determining step will have an intermediate in it. In this case, you will need to use some math to make a substitution since you cannot have an intermediate in your rate law. This math involves a topic in chemistry that you most likely learned called equilibrium. If you have, what you do is you essentially use the Keq of one of the elementary steps (typically one will be in "fast equilibrium") and use some substitutions. To see what this looks like, check out the 2019 Scoring Guidelines, Page 14.

We will consider the following mechanism:

To find the rate law for this
mechanism, we look to the slow, rate-limiting step. We find the rate-limiting step to be Step 1. By using the stoichiometric coefficients (which again, we can ONLY do with elementary steps), we find the rate law to be R = k[NO2][NO2] = k[NO2]^2.