   1,000s of Fiveable Community students are already finding study help, meeting new friends, and sharing tons of opportunities among other students around the world! Don't miss out!  🎉  # 5.8 Writing Rate Laws Using Mechanisms dylan black

⏱️ April 29, 2020

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🎥 Watch: Rate Determining Steps

## Quick Review of Mechanisms

Before we start learning about how to use mechanisms to find something out about the rate of a reaction, let's do a quick refresher on what exactly a mechanism is. A mechanism describes the steps of a reaction as it takes place from reactants to products. It's important to note that mechanisms tell us exactly what happens in a reaction, as the vast majority of reactions happen in many steps. When you see a reaction A --> B, there are typically multiple steps that lead to A yielding B. This could mean A reacting with an intermediate, or the reaction being catalyzed. Let's take a look at an example: This is the mechanism for the reaction: O3 + 2I- + H2O --> O2 + I2 + 2OH-. How did we find that out? Well, if you add up all of the steps of a mechanism, it MUST (and I mean must) add up to the overall reaction.

## Rate Determining Steps

Now that we've been refreshed on what mechanisms are, let's start discussing rate laws and how to write the rate law using a mechanism. When writing the rate law of a mechanism, we need to find the rate determining step. The rate-determining step is the slowest step of a mechanism. This step essentially constrains how fast the reaction can go. Then, we can use the stoichiometric coefficients to find the rate law. It is important to note that besides mechanisms, rate laws MUST be determined experimentally. In fact, problems often are formatted along the lines of: the mechanism for a reaction is {mechanism}. If {rate law} is the experimentally determined rate law, is this mechanism valid? Since the overall rate law of a reaction must be found experimentally, it can be used to verify mechanisms. Let's look at an FRQ that shows this:

### Example FRQ

Suppose a reaction with an experimentally determined rate law of R = k[NO2]^2 and the following mechanism: i) Is this mechanism consistent with the rate law experimentally determined?

Using the mechanism above, we see that the slow step is Step 1. Then, we find the rate law to be R = k[NO2][NO2] = k[NO2]^2, which matches the experimentally determined rate law.

### Experimental Rate Laws vs. Mechanisms

The reason we look at verifying rate laws rather than finding rate laws directly from mechanisms is that the rate law for a reaction can only be determined experimentally as we saw in previous chapters. Rather, an experimentally determined rate law can help chemists verify a mechanism. If the proposed mechanism does not match the experimental rate law, we know that that mechanism cannot be true!

## Using Kc to Find the Rate Law

Let's take a look at the next part of the FRQ we were working on. The rate law is still R = k[NO2]^2. Using our rules from before, we find the slow step and write out our rate law as R = k[N2O4]. Boom, done. Wait... let's take a second look there. If we look at the overall reaction (this was given originally in the FRQ) it is 2NO2 --> 2NO + O2. Therefore, R = k[N2O4] is NOT a part of the rate law! This can also be noted by seeing that N2O4 is an intermediate for this mechanism. So then, how do we go about finding the rate law? Well, first of all, note that for Step 1, K = [N2O4]/[NO2]^2. Thus, [N2O4] = K[NO2]^2. This means that the rate law is: R = kKc[NO2]^2, which is consistent with the rate law before. Here's the CollegeBoard's answer:    