Did you know that you can compare a chemical reaction with your daily experience in the lunch line?
Did you know that you can compare a chemical reaction with your daily experience in the lunch line? Episode 24 recaps the reaction mechanism and how it relates to the rate law. It starts with a brief recap of factors affecting the rate of a reaction (1:28). A reaction itself can have several elementary steps that make up the overall reaction (2:00). The rate law of the reaction is determined by the slowest elementary step (3:55). Using a general example, episode 24 shows why you cannot use the overall stoichiometry to determine the rate law (4:18) and why termolecular reactions are rare (5:24).
Question: (6:26) Catalysts can be part of elementary steps. How can you identify a catalyst?
A. Being produced and used up.
B. Shows up on reactant and product side.
C. Shows up on reactant side.
D. Shows up on product side.
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Hi and welcome to the APsolute Recap: Chemistry Edition. Before we start this topic, I want to let you know that Episodes 26 and 28 are Listener’s choice episodes! If you have a topic, skill, or question you want a recap of - please let us know. You can contact us through our website, theapsoluterecap.com, or reach out to us on Instagram, Facebook or Twitter! Today’s episode will recap Reaction Mechanism and Rate Law.
Lets Zoom Out:
Unit 5 - Kinetics
Topic 5. 7 and 5.8 - Reaction Mechanism and Rate Law
Big idea - Transformations
Many of you are experiencing this every day: Long lines to get your lunch! First, you have to get in line to pick up the food, then you have to stand in line to pay for it. Similar to a two-step lunch procedure, a chemical reaction can also consist of several reactions. And the rate of our reaction depends on these individual steps as much as your lunch depends on how fast you get though the two lines. Therefore, today’s episode recaps the reaction mechanism and rate law.
Let’s zoom in:
In episode 23 we described the rate law. It shows us how the rate depends on the reactant concentrations and temperature. Let’s recap: In order for a chemical reaction to happen, we have discussed that chemical bonds must be broken, which takes energy, and new bonds are being formed - which releases energy. For bonds to break and form, particles have to collide. As mentioned in episode 23, we need sufficient energy and orientation for a successful collision.
These bond-breaking and bond-forming steps can be represented in the reaction mechanism, which shows the individual steps for the reaction. The chemical reaction can occur in one step or in several steps. If a reaction occurs via several steps, we can usually identify intermediate products. These are products that are formed in one step and used up in a subsequent step.
Each single step, no matter how many there are, is called an elementary step. When combining all elementary steps, you should receive the overall reaction.
Let’s go back to our analogy of your lunch line: The step of picking up the food is one elementary step. The second elementary step in the “get lunch” reaction is to pay for the lunch. Depending on staffing and preparation, we often see a long line at either one of them and the other one goes rather quickly. So your rate of getting lunch depends on how fast you make it through the slower of the two lines.
This is similar in chemistry: The rate law is determined by the slower step. You will see two different types of problems: Either you will be given the elementary steps and it will be indicated which of the elementary steps is the slow step or you will be given the rate law and you determine which of the given elementary steps is the slow step. When given the slow step and asked for the rate law you can use the slow steps stoichiometry.
Let’s look at an example: The first elementary reaction is A + A yields B. This is the slow step. The second elementary reaction is B + C yields D and this step is fast. The overall reaction for this is 2A + C yields D. The rate law for this reaction is: k times the concentration of A times A - or A squared or: this reaction is 2nd order with respect to the concentration of A. The rate law does NOT match the overall equation, but it matches the slow step. Because this is the one that determines the rate of the reaction. In this example, B is an intermediate. It is being produced in the first elementary reaction and used up in the second elementary reaction. This example shows WHY you CANNOT use the overall equation to determine the rate law - because the rate law is determined by the slow step. Unless your reaction consists only of one step, the rate law and the overall reaction are different!
This example also illustrates another point: If this was a reaction that happens in one step, you would need two A molecules and one C molecule to collide with the right amount of energy and orientation. Chemists call this termolecular. These reactions are very, rare - and slow, if happening at all. Most chemical reactions are either unimolecular - involving one molecule - or bimolecular - involving two molecules.
Back to our rate law: What if the second step is the slow step? Intermediates are never part of the rate law! You will need to use a steady-state approximation, but that is a topic for another episode!
To recap…
Bond-breaking and bond-forming steps of a reaction can be represented in the reaction mechanism, which shows the elementary steps for the reaction. The elementary steps of a reaction add up to the overall balanced equation. The slow step of the reaction mechanism is rate-limiting and therefore determines the rate law of the reaction.
Coming up next on the APsolute RecAP Chemistry Edition: Catalysis
Today’s Question of the day is about Catalysts
Question: Catalysts can be part of elementary steps. How can you identify a catalyst?
A. Being produced and used up.
B.Shows up on reactant and product side.
C.Shows up on reactant side.
D. Shows up on product side.