15. Chemical Kinetics
Reaction Mechanism
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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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