Reaction Mechanism - Video Tutorials & Practice Problems
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A Reaction Mechanism is a step-by-step sequence of elementary steps by which an overall chemical change occurs.
Understanding Reaction Mechanism
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concept
Reaction Mechanism Concept 1
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3m
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A reaction mechanism is a step by step sequence of elementary steps by which an overall chemical change occurs. Now when we use the term elementary step, an elementary step is just one step in a series of reactions that show the progress of a reaction at the molecular level. Now here, we're gonna have a reaction mechanism overview. So we're gonna say here step 1 and step 2 represent our 2 elementary steps. So this would be elementary step 1 and elementary step 2. In elementary step 1, we have c l gas plus o 3 gas, giving us c l o gas, plus o 2 gas. In elementary step 2, we have c l o gas plus c l o gas gives me c l gas plus c l o two gas. Now both elementary steps together, well that represents our reaction mechanism. So these 2 elementary steps together our reaction mechanism, and we can say here that our overall reaction can be created by canceling out what we call our catalyst and reaction intermediates. So if we take a look here, we're going to say that a catalyst is a compound shown as a reactant in the first step, and then as a product in the final step. So here the first step is just elementary step 1, final step is elementary step 2. Now if this mechanism of another mechanism we see has, let's say, 4 elementary steps, then the first step would still be the initial step, and it will be the very last one, step 4, that would be our final step. Okay? But in this case, since there's only 2, this is the first step, this is the second step. Now our reaction intermediate, this is a compound that first appears as a product in one elementary step, and then later as a reactant in another elementary step. Now these are canceled out, and by canceling them out what's left behind comes down to form our overall reaction. So if we looked at our overall reaction, this would come down here, this came down here, this came down here, and then this came down over here. So keep this in mind when you're looking at a reaction mechanism. A reaction mechanism is just all of the elementary steps together. If we were to cancel out our catalyst and our reaction or our reaction intermediates, that should give you the overall, reaction for the equation. K? So just keep this in mind, these key terms when it comes to any type of reaction mechanism and elementary step.
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example
Reaction Mechanism Example 1
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Consider the following reaction mechanism for the formation of Nitrobenzene. Alright. So identify the reaction intermediate or intermediates and catalyst or catalysts. So here, if we take a look, all of this together represents our reaction mechanism. Each individual reaction represents our elementary step. Right? So here this would be elementary step 1, this would be elementary step 2, and elementary step 3. Right? So remember, a reaction intermediate appears first as a product, then later as a reactant. So if we take a look here, let's try to find them. Let's see. So here, h two n o three appears as a product, and then later we see it appear as a reactant. So that would definitely be a reaction intermediate. But is that the only one? We would say, no, that's not the only one. Who else does that? Here's another one. C 6 H6 n o two plus here, c 6 H6 n o two plus here. That is also a reaction intermediate. It appears first as a product and later as a reactant. Do we see any others? Yes. We see this, h s o four minus appears as a product, and then later as a reactant. Now do we have any catalysts? If we look, we do. Remember, a catalyst will appear in the very first step as a reactant, then it will appear in the very last elementary step as a product. So here this would be our catalyst, and that'd be our only catalyst involved in this entire reaction mechanism. So just keep in mind, a reaction intermediate appears first as a product and then later as a reactant. A catalyst appears as one of the first reactants, and then later in the very last elementary step, it appears as one of the products. So as long as you can remember that, you'll know the difference between a reaction intermediate and a catalyst.
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Reaction Mechanism Concept 2
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Now molecularity is associated with the number of moles for reactant molecules within an elementary step. Here, if we take a look, we have 1 mole of carbon dioxide gas reacting with 1 mole of water liquid to produce 1 mole of carbonic acid. If we were to take a look here, we'd say that we have a total of 2 moles of reactants. Now the number of moles of reactants gives us different names for the molecularity. If we had only 1 mole of a reactant, then the molecularity would be called unimolecular. If there are 2 moles of reactants like we have in this example, then that would be called bimolecular. Now I know when we're naming covalent compounds in the past, when we say 2, we'd say di, like, carbon dioxide. But for molecularity, we don't use the prefix di, we use the prefix bi. So 2 moles of reactants equals bimolecular. If we had 3 moles of reactants total, then that would be ter termolecular. When we talked about numerical prefixes in the past, 3 would be tri-, but there's no such thing as trimolecular. So again, when it's 3 moles of reactants, it's termolecular. K? So when we talk about molecularity, just count the total number of reactive molecules, and you'll be able to determine the molecularity of any elementary step.
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example
Reaction Mechanism Example 2
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The elementary reaction of 2 moles of nitrogen dioxide reacting with 1 mole of foreign gas to produce 2 moles of n o two f gas is an example of a blank type of reaction. Alright. So here we're talking about molecularity, so we need to determine the total number of reactant moles involved. So here we have 2 moles of nitrogen dioxide and 1 mole of fluorine gas. Together, that totals 3 moles of reactants. So unimolecular means we only have 1 mole of of reactants, bimolecular would mean 2, Tetra would mean 4. So the answer is either b or d. So it's either trimolecular or termolecular. Remember from our earlier videos, when it comes to molecularity, 3 moles of reactants is referred to as termolecular. Trimolecular does not exist. So here, the answer would be option d.
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concept
Reaction Mechanism Concept 3
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We're going to say here in a reaction mechanism, one of the elementary steps is classified as a slow step. And when we say slow step, we're gonna say that is the rate determining step that limits the overall rate of a chemical reaction, because remember you're only as fast as your slowest step. Now we're gonna say for a slow step, we're gonna say that the coefficients of the reactants are equal to the reaction orders of the rate law. Remember, we talked about calculating rate laws beforehand by using a chart and a list of values. But now we can look at the reaction mechanism, locate our slow step, use its coefficients, and that it can give us the reaction orders. Right? So this is just another way of connecting rate law to our chemical reactions. In this case, our chemical reactions are elementary steps that comprise a reaction mechanism.
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example
Reaction Mechanism Example 3
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4m
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Here it says, consider the following elementary steps. What is the rate law of the reaction mechanism? So, in this reaction mechanism, we have 3 elementary steps. The first one is a fast step, the second one is a slow step and the last one, the third step is also a fast step. Now if we take a look, it says step 1, we need to locate the slow step of the reaction mechanism. So the slow step is this right here. We're gonna say as long as the reactants are not intermediates, their coefficients equal the reaction orders of the rate law. So looking at the reactants here, we'd say k rate equals k n o 2, the coefficient here is a 1 to the 1, and then n 205 to the 1. But we have an issue. 1 of our reactants is an intermediate. This intermediate, this intermediate. Remember, intermediates cancel out. So we could have stopped at step 1 if the reactants were not intermediates, but because they are, we have to continue to step 2. Now if a reactant is an intermediate, which we saw, You're going to cancel out with the product intermediate and we did that with the fast step before. Now, we're going to say step 3. If step 2 happens, use the elementary step possessing the product intermediate. Here, we're going to say for that elementary step, the coefficients of the reactants still equal the reaction orders. And for that elementary step, the coefficients of the products equal the inverse. The inverse of the reaction orders. So, let's see what that means. Alright. So if we come back up here, my NO2 reacting got cancelled out by this NO2 product here and we kind of have to make up for what we just lost. So this here took away the reactant that I needed. So, I'm going to take both of its reactant and its product to make up for that loss. So if we if we calculate if we look at everything, what do I have now? Well, here I have an n to o 5 that I originally had as a reactant and now I have a newly gained n 2O5 also as a reactant. So how many total moles of N205 do I have? 2. 1 from here that I just gained and one here that I originally had. So in essence, their coefficient is 2 because I have 2 moles of it. So coming down here, we'd say that our new rate law is equal to kn205 to the 2 because, again, I have 2 moles of it, one that I originally had plus one that I gained later on. And then let's look at what else we took. We also took this 1 mole of n o 3, but here it is not a reactant, it is a product. We're gonna say here, bringing it down to our rate law, if it were a reactant, it'd be n03 to the 1. But because it is a product, we said that it equals the inverse. So instead of being 1, it's to the negative one. This can be a bit tricky. This happens anytime one of our reactants is an intermediate. We have to do this making up process what we just lost. So here the rate law would be rate equals k n0205 to the 2 and n03 to negative one. Right. So this would be our new and official rate law.
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Problem
Problem
Consider the following elementary steps:
NO2 (g) + NO2 (g) → NO3 (g) + NO (g) [SLOW]
NO (g) + CO3 (g) → N (g) + CO4 (g) [FAST]
What is the rate law of the reaction mechanism?
A
rate = k [NO2]
B
rate = k [NO2]2
C
rate = k [NO][CO3]
D
rate = k [CO3]
E
rate = k [NO4]
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Problem
Problem
The following reaction of 2 Br2 (g) + 2 NO (g) → N2 (g) + 2 Br2O (g) has the following rate law:Rate = k [Br2][NO]2. The proposed mechanism for the reaction is:
Br2 (g) + NO2 (g) → N (g) + Br2O (g) [SLOW] N (g) + NO (g) → N2 (g) + O (g) [FAST] O (g) + Br2 (g) → Br2O (g) [FAST]
Which of the following statements is/are false?
a) The rate determining step is bimolecular. b) There are three elementary steps in the reaction mechanism. c) The mechanism possesses a catalyst. d) O is the only reaction intermediate in this reaction mechanism. e) This is not a valid mechanism for the reaction.