 Okay, how are you guys? Everybody have a good weekend? No? Why are you guys wearing ties? You have to? Why do you have to? Pledges, what are you pledging? Oh, okay, you look fantastic. All right, so we are, when I say we, I mean Steven and Mark are grading the midterm, I think it's about done and they're going to post scores later on today. There's a key that keeps getting updated. The current version of that's on the web already. It's been updated a few times. So rumor has it that you did better than on midterm one but I haven't seen the results yet so don't put too much weight on that. I'm hopeful that you all did. So, yes, yes, keys posted. Scores are going to be posted today. Exams are going to be returned as PDFs via rapid return and I think you know that that's a misnomer. It ain't rapid. It takes, you know, four or five days or so. Okay, but that's, we'll get our return as fast as we can. I'm going to post a new how am I doing score today. I'll drop your lowest two quizzes and let you know how you're doing in the class. You'll be able to look at that score and see what kind of grade you have going into the final exam, okay? There's one quiz left though. There's a quiz Friday and it's really going to be on the stuff that we talk about today and also the stuff we talked about way last Monday. The steady state approximation, all the kinetic stuff that we've talked about really is covered by this quiz. Okay, we've already had one quiz that had kinetic stuff on it. Okay, so that's the very last quiz. Some of you probably have done well enough so you don't even need to take it because you can drop two quizzes. So, if you've got five really good ones already, take the first 15 minutes off tomorrow. The final exam is going to be like this. It's comprehensive but it's going to emphasize the kinetic stuff that we're doing here at the end of the quarter because the kinetic stuff hasn't been on a midterm exam yet. So, half of the final exam is going to be kinetics, 25% thermal, 25% stat mech. But I'll break it down for you problem by problem, all right, next week so don't worry about that. But I just want to say that the stuff that we're talking about now is going to be worth sort of 100 points on the final exam so it's rather important. Okay, so we're going to review the steady state approximation because I know that's not foremost in your minds anymore. You've been studying for midterm too. It was all about thermal. We're going to do an example and then we're going to talk about the Lindemann Hinchlewood mechanism. We started to talk about this last Monday but I'm sure that that's sort of a vague memory for you at this point. So we'll go back and look at it carefully. Okay, so the whole idea in the steady state approximation is that we want to simplify the mathematical expressions for consecutive reactions, right? Reactions where here's a reaction, here's another reaction that follows that one, there could be another one that follows this one, all right? And in general, the reactions are going to be more complicated than this. I'll show you an example later on, all right? But the basic idea is that the mechanism for sequential reactions like this, the mathematical expressions for the integrated rate laws get exponentially more mathematically complex as the mechanism gets larger. They don't increase linearly with the size of the mechanism, they increase exponentially. So we need a mathematical tool that allows us to simplify what these expressions look like and that tool is the steady state approximation. We're going to use it to simplify several different types of reaction mechanisms including the Lindemann-Hinshelwood mechanism and enzyme kinetic expressions, right? Which we're about to start talking about on Friday. So the basic idea is that we're going to set the time rate of change for all intermediates to zero, right? What's an intermediate? Well, an intermediate is something that shows up in a sequential reaction, but it is not a product and it's not a reactant, right? So in this case, the intermediate is obviously B. We're going to set the time rate of change of B to zero and then we're going to solve the simplified kinetic expressions that result from making this simplification. So for example, we've set the time rate of change of B equal to zero, but what is that? Well, there's a rate of formation for B because B is formed at a rate equal to K1 times A and there's a rate of consumption of B because B is turned into C at a rate times of K2 times B so there's going to be a minus K2 times B and a plus K1 times A, right? And that difference has to equal zero. So if that's the case, then obviously that has to equal that. That's what we're showing right here. And so I can then solve for B. This should say B steady state, should be a steady state subscript here. The steady state concentration of B is equal to K1 over K2 times A. Now, one of the things that we showed way last Monday, a week ago Monday, is that the steady state approximation is really only going to work when K1 is much, much less than K2, right? That if you plot what the concentration of the intermediate is doing, you can convince yourself that only in this limit is the concentration of the intermediate going to be quasi-constant, right? Even in this limit, it's not perfectly constant but it's quasi-constant, all right? And so consequently, if we make K, here we're saying the time rate of change of B is zero, right? In other words, B is not changing. Its concentration is not changing at all as a function of time. That's what we're assuming, right? In order for this expression to approximate this expression, K1 has to be much, much less than K2. Can everyone see that? K1 has to be, K1 over K2 has to approximate zero, okay? In order for this expression to make sense. Keep in mind that A is always going to go from its initial concentration to zero. According to this mechanism here, right? A is going to change a lot, all right? So in order for B not to change, K1 over K2 has got to be small, very small. Everybody with me on that? All right, that's the assumption that we're always making somewhere in the steady-state approximation, okay? So here's the case, here's the mechanism, here are the rigorous equations that describe the concentrations of A. Here's what's happening to A, just like I said, it starts off at some initial concentration and then goes to zero. Here's C building up and here's the concentration of the intermediate, it's quasi-constant. Why? Because K1 over K2 is just 0.02, it's tiny. This is the limit where the steady-state approximation is going to work pretty darn well for us. Now we're going to solve the simplified equations that result, so we said the steady-state concentration of B is K1 over K2 times A and so we can then plug that in to the rate at which C is produced. So here's the rate of the reaction in terms of C. It's K2 times B, all right, and now I can just plug this expression in for B, boom, all right, and the K2's are going to cancel and so the rate of this reaction is just going to be equal to K1 times A. And we know what the integrated rate expression is for A, all right, it's just a first order reaction with rate constant K1, this is what it's equal to if I work it out and so if I want to know what the rate at which the product is produced, obviously this is a decaying exponential, it's doing that. That's their integrated rate law for A and now I'm going to integrate that to get the concentration of C as a function of time and this is what that integral gives me right here, okay, so the C as a function of time is going to be given by that equation, we get really simplified integrated rate laws from the steady-state approximation compared to the rigorous expression and the thing to keep in mind is this is the best case scenario, this is the simplest possible sequential reaction mechanism, all right, two first order reactions in sequence where there's only one reactant and one product in each, all right, and I think you can see there's already a pretty big difference in complexity between these equations and these equations, especially look at C, all right, that mathematical expression is a lot simpler than this one and as we make this mechanism more and more complex these equations blow up exponentially, okay, that's why we need the steady-state approximation, how well does it work, here K1 over K2 is tiny again, look how well it works, the dashed line is a steady-state approximation, the solid line is the perfect or the rigorous equations for the concentration, look at A, look at C, C is what we really care about, all right, C is telling us what the rate of the reaction is, steady-state approximations also predicting B, all right, and you can see B is quasi-steady state, concentration doesn't change that much, it's changing but it doesn't change that much, if I make K1 over K2 bigger, all right, things should get worse and you can sort of see that it's getting worse, all right, if I make it bigger yet, now to .9 you can't really see what's going on unless I blow this up but when I blow it up you can see that it's not doing a very good job of predicting B anymore, all right, here's what the steady-state approximation is doing, the steady-state approximation says B is some fraction of A and A is changing a lot and B is really not doing what the steady-state approximation is assuming, now look at the difference between C, here's the steady-state approximation and the actual, all right, so it's starting to break down, we expect it to break down here, it's not surprising that it does that and if we make this bigger yet, we get a complete train wreck, here's the steady-state approximation for C, this is the dashed line, here's the solid line, all right, big difference between the steady-state approximation and what the concentration of C actually is, so we want to keep in mind the steady-state approximation doesn't always work, all right, this is even worse, this is even worse than that, that's the steady-state approximation, that's what's actually happening, no, sorry, bad idea to use it, okay, so does everyone understand the steady-state approximation? Well, let's do an example, all right, here's a real-world example for the steady-state approximation, here's the reaction mechanism, right, three reactions, if I wrote down the actual kinetic rate expressions for this mechanism, it would take three screens, all right, it's a nightmare of enormous proportions, let's see if we can use the steady-state approximation to simplify what's going on here, first of all, what do we got to do, we have to find intermediates, if there are no intermediates, there's no point in using the steady-state approximation, all right, the steady-state approximation only assumes that the time rate of change of intermediates is zero, identify the intermediates, so look at this guy right here in pink, he's produced by this reaction, consumed by this reaction, he does not appear as a product, he is by definition an intermediate, right, what about this guy? That's another intermediate, so there's two intermediates in this case, wait a second, here's bromine, bromine showing up here and here is bromine an intermediate? No, he is a catalyst, right, it's consumed that drives the reaction forward but it's regenerated as a product, without B this reaction, without Br minus, this reaction grinds to a halt, with it the reaction occurs at whatever rate is characteristic of the reaction, all right, but Br minus is not consumed by the reaction because it's regenerated as a product even though it's consumed by the second step of the reaction, so Br minus is a catalyst, you have to be able to recognize that, it's not an intermediate and you can't apply the steady-state approximation to its concentration, okay, what about Br minus is no, it's a catalyst, okay, so apply the steady-state approximation to the following mechanism, the rate of formation of the product, this is the product we care about right here by the way, all right, its rate of formation is going to be equal to K3 times the concentration of this stuff times the concentration of this stuff, everyone agree with that? Okay, so to apply the steady-state approximation the first thing we do is we write down this rate right here, the rate of the reaction is the rate of the last step, then we look at these two concentrations and we ask ourselves, is that an intermediate or is that an intermediate, that's not an intermediate, that's a rate constant, are either of these species intermediates, yes, that guy is, we agree, that guy is an intermediate, there he is and there he is, and so he's getting generated by this step and consumed by this step, so ONBR is definitely an intermediate, let's apply the steady-state to its concentration, steady-state approximation, to do that we set the time rate of change of that species to zero, then we write a rate expression for it, all right, we look at the mechanism, here's the mechanism for the reaction and we ask ourselves, how is ONBR generated? Well, where's ONBR, it's generated from this reaction right here, so I've got H2NO2, that concentration times BR minus with a rate constant of K2 is generating ONBR, boom, that's what this term is right here, it's the rate at which this guy is getting generated by what? By this step right here, what's this guy, this is the rate at which the ONBR is getting consumed by this reaction right here, okay, so that rate is the steady-state concentration of ONBR times the concentration of this stuff times K3, so there's a minus sign in front of this guy and a plus sign in front of this guy, that's the generation rate, that's the consumption rate, everyone see that, okay, we're going to set that and that have to be equal to one another if the steady-state approximation is valid, okay, so then I'm just going to solve for ONBR, the steady-state concentration of ONBR from this expression right here, all right, I can move this guy over to the left-hand side, divide through by K3 and the concentration of this stuff, boom, that's the kinetic expression for my steady-state concentration of ONBR and then I just plug that in, we already said the rate of the reaction is equal to this expression right here, I can plug that into this expression and I've got a steady-state approximation expression for the rate of the reaction, so I just took that whole thing there, stuck it in here and so now I've got K3, that's the K3 right there, times this whole mess, that's right here, times the concentration of our product, that's this, boom, that's our steady-state approximation integrated rate law, differential rate law, sorry, now I'm looking at this and I'm asking myself can I simplify it further, we've applied the steady-state approximation once, looking at this guy, are there any intermediates among these species here, I guess I could have canceled that and that, the same, all right, is BR an intermediate? No, we agreed it was a catalyst, is H2NO2 plus an intermediate? Oh, I did cancel, look at that, good, okay, now, are either one of these two guys intermediates, this guy could be, right, where is he, he's right there and he's right there so he is an intermediate, so we can apply the steady-state approximation to his concentration too, we already applied it to this guy, now we can apply it to this guy, okay, good, so let's ask again, are there any, yes, yes, yes, so we said it's time rate of change to zero, this is the rate at which it's generated, there are two processes that consume it, let's look back and so okay, it's generated by the reaction of protons with NHO2, with the rate constant K1, what did I do? Oh, wait, sorry, protons, NHO2 with the rate constant K1, this is the rate at which this stuff is generated and then there's two processes that remove it from the system, all right, one of them is the back reaction here, all right, so the rate of that is just given by H2NO2 plus times K to the minus one, that's this guy right here and the other one is this forward reaction here, H2NO2 plus reacting with BR minus with the rate constant K2 and that's this guy right here, okay, so these are consumption rates, this is the generation rate for this species, all right, and then what we're going to do is we're going to solve for this guy, all right, and if you do that, this is the expression that you get and then what we do is we plug this into our original steady state expression that we got after we applied the steady state approximation the first time, now we've applied it a second time, so I'm going to plug this mess in for this species here, okay, there it is and is there some cancellation? No, that's it, that's our steady state rate law, it's a differential rate law, all right, now there's going to be two limiting experimentally observed rate laws, looking at this guy, all right, it's possible that this reaction will appear to have two different rate laws associated with it because of this denominator here, all right, if this term dominates, we're going to see one thing and if this term dominates relative to this, we're going to see something else. Let's think about that for a second, consider first the case where this guy K2 times Br minus is much larger than K to the minus 1, all right, if that's true, then we can neglect K to the minus 1 in this denominator because there's an addition here, all right, if one term is much, much larger than the other, then this will have an insignificant effect on the total size of the denominator, so what we're looking for are addition operations within the rate law, all right, there's one right there, that tells us that there's the potential to have two distinguishable rate behaviors, all right, if K2 Br is much, much bigger than K to the minus 1, then this guy simplifies to this, all right, we just leave out the K to the minus 1, the K2's cancel, the Br minus cancels and we end up with a rate, all right, that's the rate right there, equal to K1 times the proton concentration times NH02, that's the first thing that can happen, we would predict that this would be observed, for example, at high bromide concentrations, if I make the bromide concentration higher and higher and higher, at some point the reaction should stop depending on bromide, shouldn't see any dependence at all anymore and then in that limit this rate law should apply to the reaction, the other possibility is that K2 Br is much, much less than K to the minus 1, if that's true then I leave this guy out and I only keep the K to the minus 1, okay, and in this limit the reaction does depend on Br minus and I get an effective rate constant that bundles together these three rate constants right here, this K effective would be K1 times K2 divided by K to the minus 1, everyone see that? So I would expect to see a transition from this behavior at low Br concentrations to this behavior at high Br concentrations, all right, we should see the reaction rate stop depending on Br minus in the limit where this rate law applies, everyone see that? This will always happen when there's an addition operation somewhere in your rate law, there will always be the possibility for limiting behaviors, okay, now we're going to look at a different application of the steady state approximation, we're going to talk about the mechanism of this reaction right here, unimolecular reaction of some species, call it A. Now, if you look at this reaction right here, A plus B gives you products, it's completely intuitive how this reaction might occur, imagine for example, A and B are gas phase species, you've got A zooming around in the gas phase, you've got B zooming around in the gas phase, A and B collide, generate an activated complex and then form products, all right, there's a collision of A with B to form some activated complex that breaks up to produce products. Bimolecular reactions have an obvious mechanism in the gas phase, A collides with B to generate some kind of activated complex of A and B and then this breaks up to give products that have some A character and some B character associated with them, this collision between A and B generates an energized transition state, for now these are just words, we're going to make, we're going to talk about transition state theory later on, it'll be very important, but right now it's just conceptual, okay? Now, the basic idea is that this transition state here is located on this reaction coordinate here between substrate which are these guys and product which are these guys, it's located halfway in between, it's located in fact right at the peak of this energy level diagram, but once again, we're going to make this more quantitative later, right now it's just conceptual. All right, that's how this happens, that makes perfect sense to us, A collides with B to give products, how does that happen? All right, there isn't any B, there's only A. What kind of reactions conform to this reaction mechanism? What kind of reactions do this? A just reacts. Well, there's two main kinds, unibolecular reactions are either isomerizations, here's resveratrol, remember that, the stuff that's found in red wine, it's so delicious, impedes the growth of cancer in your body and has all kinds of other benefits for you, all right, if you ingest kilograms of it, all right, and decomposition reactions. Here's A, it falls apart into B and C, all right. A just falls apart, that's a decomposition reaction, and that's unibolecular, all right, the reaction happens just to this molecule by itself, it falls apart, this molecule here oscillates between two different chemical states because of isomerization around that double bond, right there, right, these are the two primary types of unibolecular reactions. How do they happen? How do they occur? How do we understand what's going on? We need a reaction mechanism and these two guys proposed one in the early 1900s. Lindemann was an English guy, a physicist, who did a lot of stuff for the British government during World War II, he was the science advisor to Winston Churchill, and he was a really arrogant guy who most people hated, but one of the things that he did is he worked out this reaction mechanism and then somebody who he didn't even know, Hinchelwood, another English guy, came along and worked out all the mathematics and made it quantitative. Lindemann cooked up the mechanism, Hinchelwood refined it, Hinchelwood ended up getting a Nobel Prize, not only for this, I mean Hinchelwood did other stuff too, but this is one of the things. Okay, so for this reaction, so the Lindemann Hinchelwood mechanism basically postulates that this reaction occurs in three steps. One, A collides with itself. When that happens, you generate activated A and ground state A, non-activated. This reaction can go back, that's what this is right here. Activated A can collide again with another A. So here's A colliding with A to produce activated A. Activated A can collide with A again to become deactivated. That's what this is, all right? Here's a collision between activated A and ground state A. I get two ground state A's. There's no more activated A. And finally, activated A can fall apart into B or can turn into B. This looks like a decomposition mechanism. A's decomposing into two B's, okay? So we can apply the steady state approximation to this guy, the basic idea and the most important assumption of the Lindemann Hinchelwood mechanism and it's assumption that turns out to be wrong, all right? The assumption is that A star possesses an internal energy that exceeds the activation energy. In other words, if there's a bond that has to break, in A to form two B's, A star has enough energy within it to break that bond. That turns out to be a bad assumption because not all collisions are going to generate an activated A that has enough energy to break up necessarily, right? You can imagine there's going to be a wide distribution of collision energies because A can hit itself at a small angle and transfer very little energy to itself, all right? That's one way in which A star could have a smaller amount of energy than it needs to react. This is called the strong collision assumption of Lindemann Hinchelwood theory. The strong collision assumption is that these collisions that occur here, right, generate an activated A that has enough internal energy to go on and do the reaction that you care about. It's either decomposition or an isomerization. Now, can we apply the steady state approximation to this mechanism right here? Yes, we can because there's an intermediate right there, all right? The activated A is an intermediate, right? Here's our reactants, here's our products. A is formed and then consumed. So it only exists transiently within the mechanism. That's the definition of an intermediate, okay? So what are our rate expressions going to be like? While the rate at which A star is formed, here's the generation rate and there's two processes that consume it. The generation rate, of course, is that, right? A squared times K1 and then how is A star consumed? There's two things that consume it. There's this reaction here, A star times A with a rate constant K minus 1, that's just the reverse of this, of course, and A star can react with a rate constant K2, all right? And so both of these two processes use up A star and so they have a minus sign in front of them. Minus sign here because A star is used up by that guy, minus sign here because A star is used up by that guy, plus sign here because A star is generated by this guy, okay? So you've got to be able to write these rate expressions by looking at the mechanism. Then all we do is set this equal to zero. Oh, well, these are the rate expressions for the other species, A and B, all right? And then we set this guy equal to zero, don't we? That's what the steady state approximation is. We set this guy equal to zero and then we solve for A star. So if this is equal to zero, then I can set all the positive terms equal to all the negative terms, all right? Now I'm being more careful. I'm putting this little steady state subscript on the A star concentration because now we're talking in particularly about the steady state concentration of A star and I can just solve for A star in this expression here to get this guy right here, okay? And so once I've got that, then I can just plug A star into the expression for the rate at which B is produced, okay? So I just plug this whole guy in for A star, boom, there it is, all right? And that's my rate law for the Lindemann-Hinshelwood mechanism. What does it predict? So look at this guy, all right? You see how there's an addition operation in the denominator, that means there's going to be two limiting behaviors depending on which one of those two terms is bigger or smaller, all right? That's the first thing to notice. So we're going to talk about those two limits. If A is big, in other words K minus 1 times A is much larger than K2, all right? This would be in the limit of high pressures of A, all right? At high pressure, remember there's only one reactant in this thing, you've got a vessel with A in it in the gas phase, if you put a lot of pressure of A, that's the limit that you're in, this guy right here. This guy is much, much bigger than K2 in that limit and one of these A's cancel with one of these A's. So I'm left with one A in the numerator, all right? No A's in the denominator, all right? And if you look at this, all right, it's just I can cluster together these rate constants in that two to get an effective rate constant times A, okay? So the rate of this reaction, if it conforms to the Lindemann-Hinshelwood mechanism, it's going to look like this. It's going to be apparently first order in A, all right? That's what you expect just looking at the mechanism. It's a unimolecular reaction. You'd expect it to be naively, you'd expect it to be first order in A and it is at high A pressures. Now, what does this mechanism mean mechanistically? Well, what does it mean for K minus 1A to be much, much bigger than K2? It means that this reaction is fast compared to this reaction here, all right? The rate of deactivation at high pressures, in other words, you create an excited A but it collides right away with a ground state A to get deactivated, all right? In that limit, you get a reaction that is first order in A. Now, that's true for large A. This is what we get. What happens if A is small? What happens if small pressures, low pressures for A? Well, obviously, this guy is going to be much smaller than that so I can get rid of him. Where is he? I've just got K2 now in the denominator. I don't have any A in the denominator canceled so that's still A squared and so this K2 here cancels with that K2 right there so I'm going to have 2K1 times A squared, it's going to act like a second order reaction. At low pressures, it's going to look like a second order reaction. Bizarre, all right? It's just A falling apart into products or a summarizing, all right, but that reaction is going to act like a second order reaction. Weird. What does this mean mechanistically? Well, if that is much less than that, all right, that means that this reaction is fast, this reaction is slow. The rate of deactivation at low pressures, all right, the rate at which if you form an excited A, it doesn't get deactivated very, it's not very likely to get deactivated, there aren't very many collisions to deactivate it, it's more likely to fall apart. Once you form this guy, it's more likely to undergo this reaction than it is the deactivation reaction at low pressures, that's what it means. So this plot should make sense to us, right? Here's what the Lindemann-Hinshelwood mechanism predicts. Let's look at this and make sure that we understand this because this is really everything in a nutshell here. All right, what's on this axis? This is the pressure of A, log pressure of A. All right? What's on this axis? This is the log of the rate constant, the effective first order rate constant. All right, what do I mean by that? If I pretend that the reaction has this form, all right, then this rate constant here, if I apply it to this case right here, it's going to be 2 times K1 times A is going to be equal to K effective. Can everyone see that? In other words, if I assume that the reaction always has this form, the reaction rate always has this form, then I can write an expression for K effective. All right, K effective in this case is going to be equal to 2 times K1 times A because that's what it would take if K effective would have to be equal to that in order for K effective times A to give me this, okay? So this is the log of the rate constant. This is log of the pressure of A and what this plot says is, hey, at low pressures of A, the effective rate constant depends on A. It goes up as a function of the pressure of A until it gets to a point where there's no dependence of the effective rate constant on A at all and we just arrive these two cases, all right? Here the effective rate constant is 2 times K1 times A. It does depend on A. It goes up linearly with A. That's what this shows right here, all right? And this rate constant here doesn't depend on A at all. It's completely A independent. Now, of course, the rate of the reaction does depend on A up here because the rate is K1 times A, all right? But if I take the rate constant, all right, and I plot, I'm only plotting the rate constant here, all right? I'm not plotting the total rate, all right? Here the rate constant becomes constant. It's equal to this collection of rate constants from those individual steps. Here the rate constant depends on A. Okay? So this is classical behavior that's modeled by the Lindemann-Hinshelwood mechanism. At low pressures, the reaction rate constant, the effective rate constant depends on A linearly and at high pressures, it does not, all right? Here the reaction is acting like a first-order reaction. Here the reaction is acting like a second-order reaction. The Lindemann-Hinshelwood mechanism explains that. It doesn't get it exactly right, but it comes close. Okay, did everyone get that? Now, here's our Lindemann-Hinshelwood rate. Here's the expression that we derived earlier. Let's recast this equation, all right? If I assume the rate looks like this, in other words, I'm defining an effective first-order rate constant, all right? Just like I did for this, exactly the same way I did for this plot, all right? I can then say that K effective, if this is true and this is true, then that K effective has got to be equal to all this nonsense, including one of these A's. See how that's A squared? I'm going to take one of those A's. The other one is right here, so if I take this, if this is K effective and I multiply by A, I get that, okay? So this is my effective Lindemann-Hinshelwood first-order rate constant. And I can just take the reciprocal of this. So I can take 1 over K effective. I move these two guys into the numerator. There they are, all right? I can split this into two terms. I can put the K minus 1 here, and if I do that, the A's are canceling, and I can put the K2 here, and if I do that, the K2's cancel, all right? So this is my simplified expression if I take the reciprocal of this guy. Everyone see that? So this now is my Lindemann-Hinshelwood equation, if you will. What it says is if I plot 1 over K effective, versus 1 over A, either the concentration of A or the pressure of A, the partial pressure of A, I should get a straight line, all right? I'm plotting that, versus 1 over A, I should get a straight line with the slope of 1 over 2 times K1, and a positive intercept. Everyone see that? So that's how you tell whether your reaction is conforming to the Lindemann-Hinshelwood mechanism. You make that plot. What is it? 1 over K versus 1 over A. 1 over K versus 1 over A. This is 1 over the partial pressure, all right? How well does it work? Well, not that great. I mean, okay, it's working pretty well at low pressure. All right? This is 1 over pressure, so if it's working well here, that's low pressure, right? The Lindemann-Hinshelwood theory predicts a rate that is too low. Here's what we predict, all right? This is 1 over the rate, all right? It predicts a rate that is too low at high A. This is high A because this is 1 over A. This is very confusing. All right? This is high A, low A. High rate, low rate because it's all reciprocal. It's a totally reciprocal plot. All right? So the Lindemann-Hinshelwood theory predicts a rate that is too low at high A. There's a good reason for that. All right? We'll talk about it later on. All right? But where we expect it to work is at low A. All right? We expect a positive intercept. No one see that? And we are looking for linear behavior down here, all right? We expect it to roll off here. Lindemann-Hinshelwood theory has a well-known defect that we're going to understand in detail later, but for now expect this guy to roll off like this at high A or low 1 over A, okay? So all of this is based on our application of the steady state approximation to the Lindemann-Hinshelwood mechanism, right? Actually, the Lindemann-Hinshelwood mechanism uses the steady state approximation whether we want to or not. Okay? It's kind of deep and confusing. So that's what this is. Now, we're going to talk about one more case, all right? So what were we just talking about? We were talking about unimolecular reaction mechanisms. All right? How do unimolecular reactions occur? You've got a molecule, it's just falling apart, all right? What's the mechanism for that? It's this Lindemann-Hinshelwood mechanism. We can understand that mechanism in terms of the steady state approximation. Derive equations. What other reactions are going to be useful to look at with a steady state approximation? It turns out that enzyme reactions are another case where that's true and that's why reactions of this form are important, all right? Reactions where a pre-equilibrium is established within the reaction mechanism. Pre-equilibrium, what am I talking about? Look at this. A reacts with B to give some complex of A and B. That's going to be the enzyme substrate complex. A and B can fall apart to give A and B again, all right? So A reacts with B to form complex A, B. Complex A, B falls apart to give A and B separately again. All right? There can be an equilibrium that involves this forward step and this reverse step. That's the pre-equilibrium. Then A, B can fall apart to give products. This looks like an enzyme reaction, doesn't it? All right? Substrate reacts with enzyme to form enzyme substrate complex. Enzyme substrate complex falls apart to give enzyme and substrate or a reaction occurs and generates product. But there's lots of other reaction mechanisms that also adhere to this pre-equilibrium model. Oh, that slide is out of, yeah, this is a problem where you're supposed to plot the data and find out whether the reaction conforms to the Lindemann-Hinshelwood mechanism. All right? So what is this? This is the pressure of some gas, all right? We call it A. And this is the effective rate constant that we're measuring for the, the effective first order rate constant that we're measuring for the reaction of this gas to form products. Could be an isomerization reaction. Could be a decomposition reaction. Some unimolecular reaction. So the question is, does this data conform to the Lindemann-Hinshelwood mechanism? How do you tell? Well, you have to make a plot of 1 over K effective versus 1 over partial pressure. And see if there's any linearity in that plot at low pressures. Remember? And so you take 1 over P. You take 1 over K effective. All right? So you take each one of these guys, take the reciprocal, and now you plot them. Here are these, these data points are the plot. Does this look like Lindemann-Hinshelwood mechanistic behavior? Sort of. I mean, it's really ugly. All right? It's sort of linear at low A. All right? And it has this deviation that we're expecting. All right? We would hate to see this on a test because it's sort of nebulous whether this is really conforming to the Lindemann-Hinshelwood mechanism or not. It's not very linear. All right? I should really concoct a data set that looks better than this. But that's what you would do. All right? The point is that you make this plot. You take these reciprocals. You plot this versus this. And you look for linear behavior over here. All right? You're not getting really good linear behavior in this case. This is actual, this is real data actually. That's the part of the reason. One thing you'll learn if you ever have to do experiments is that experiments never or very rarely adhere perfectly to the theory that you're trying to apply to them. So this is a case of that. All right? So getting back to the pre-equilibrium, we can use the steady-state approximation again. And we'll do that more on Friday. Okay? So that's what quiz seven is going to be about steady-state approximation.