 Okay, so we posted the results for quiz six this morning. You guys did well. The mean was a little lower than last week. We're pretty high. A's, B lower grades. Okay, midterm is Friday and we're going to have one more quiz. A week from Friday, quiz seven. Then we're done. We're winding down. What we'll do is we're going to review in detail what's going to be on the midterm on Wednesday. I'll walk you through it, problem by problem and review that material for you. What will not be on it for sure are these two topics from chapter 19 right at the end. Alright, polymerization and photochemistry, we're not going to get to those two topics. But I think everything else in chapter 19 will be included. Alright, we're going to get to, we're going to get right up to section 19.8 by the end of the lecture today. Okay, so we're in the final stretch. There's just seven lectures including this one left and we've just got one quiz left. But most of the points in this class are still to be decided. We have a 200 point final. We have a 100 point midterm and we still have one quiz left. So this is the, these next two weeks would be good weeks to focus on doing well in the class because it could make a huge difference to you. I mean so far we've basically decided for most of you what your quiz grade is going to be for the quarter. Just got one quiz left, alright. And since you can drop two, you've pretty much, you've come close to locking in your quiz grade. You can still affect it with quiz seven a little bit. And you've taken midterm one, alright. But we've got 300 points and exams left in the course and so it's tempting to think that we're coasting into the summer vacation period but in reality now is the time to do well. Alright, so we're going to certainly do a good job of reviewing this midterm two for you on Friday so you know exactly what's going to be on it. So what we're going to do today is we're going to talk a little bit about Arrhenius. Arrhenius developed a phenomenological equation that we find very useful in chemistry. It's used virtually all the time by people doing all kinds of chemistry and so it's very important to us but it's really a phenomenological equation. Hopefully by the end of the quarter we'll understand where it comes from. Now we're going to talk a little bit about consecutive reactions and the steady state approximation. This is very important. This is right in chapter 19 and there will definitely be a problem on midterm two that involves this topic because it's one of my favorite topics. And then we're going to talk about the Langmuir Hinchelwood mechanism which is a complicated mechanism for unimolecular reactions that exploits the steady state approximation. The Langmuir Hinchelwood mechanism. Okay, so Arrhenius is a Swedish guy and we already know he created a definition of acids and bases for us that involves proton donation and hydroxide accepting. Yup, but for the purpose of kinetics he recognized the relationship between temperature and reaction rate. He recognized a trend in the scientific literature for the temperature dependence of reaction rates that other people hadn't noticed. And so we'll say more about that in a second but he also discovered the greenhouse effect. Most people don't appreciate that. Here's the Arrhenius equation that relates the rate constant for a reaction to the temperature but he also figured out the greenhouse effect. In 1896, and this is the paper that he wrote, a great deal has been written on the influence of the absorption of the atmosphere upon the climate. So he understood climate change in 1896. Tindall in particular has pointed out the enormous importance of this question to him most seriously, is the mean temperature of the ground in any way influenced by the presence of heat absorbing gases in the atmosphere. The first guy to think about this is actually a guy named Fourier. Yes, that Fourier but in this paper actually Arrhenius works out the expected temperature change for the known concentrations of things like CO2 in the atmosphere and he gets a decent answer for the amount of warming that should occur. It's just really amazing that you could even make that calculation in 1896. Carbonic, on the influence of carbonic acid in the air upon the temperature of the ground. Amazing. All right, that doesn't really have anything to do with kinetics but this equation does. All right, this is the famous Arrhenius equation and so what we have here, this EA here, that's something called the activation energy and we haven't talked about that yet but we will before the end of the quarter. And A is a pre-exponential factor, obviously. It's before the exponential, all right, but it's also called the frequency factor. Its units have to match those of K, obviously. All right, but what this equation predicts, so first of all, this activation energy, there's some product state, there's some reactant state and in between the product and the reactant state, there's an energy barrier that has to be surmounted. And up at the top here is something called the transition state. It's the state of the reactants when they're halfway to becoming products, all right, that's the so-called transition state and that's a high energy state for the system. In fact, it's the highest energy state for the system as it progresses from reactants to products. You have to get over this peak and at the very top of this peak is the transition state by definition. And so the height of this energy barrier here with respect to the reactant state, that's the activation energy. Now we're not going to really say more about it right now but I just wanted you to know where that comes from. So if you go to the lab and you carry out a measurement of a reaction rate as a function of temperature, what you notice and everybody in this room already knows this is the rate of most chemical reaction goes up as you increase the temperature. And so here's what that data might look like. This is supposed to be the rate constant on this vertical axis here. This is the temperature on this horizontal axis and as we increase the temperature, it turns out that we're talking about this reaction right here, all right, the dissociation of NO2, right, the reaction rate goes up and up and up and up, all right. And what Arrhenius noticed is that if you replot this data as log of the rate constant versus 1 over temperature, these data fall almost on a straight line. And he noticed that this was true for a wide variety of different types of reactions, reactions that occur in solution, reactions that occur in the gas phase. This was a very general observation that if you recast rate versus temperature data as log rate versus 1 over temperature data, you get a straight line. And this equation explains why that would be the case, right, if chemical reactions adhere to this functional form. Because if I just take the log of both sides of this equation, I get log K equals log A minus EA over RT and so you would expect a plot of log K versus 1 over RT to be a straight line with a negative slope and the slope is going to be minus EA over R and the intercept is going to be log A. And so this would explain phenomenologically only though why the data would adhere to this functional form. He didn't explain what this activation energy was in detail and he didn't explain what A was in detail. We have to understand those things yet. And we're going to talk about something called transition state theory that allows us to understand what these parameters are in more detail but suffice it to say this is a phenomenological reaction, equation rather, that is very important to us. So chemists are constantly varying the temperature of the reaction, measuring the reaction rate as a function of temperature, calculating an activation energy using that equation right there. All kinds of chemists for all kinds of reactions. Okay, so we're not going to say more about that for now. We're only pointing out that there is this parameterization of the rate constant with temperature from which we can extract some parameters A and E sub A rather that have relevance in terms of the kinetic mechanism but we haven't said what the relevance is yet. We're going to come back to it. Okay, so consecutive reactions are sequences of reactions which the product of the reaction serves a reaction. So many times in chemistry what we observe is a reaction occurs and then the product of the reaction serves as the reactant for a following reaction and then the product of that following reaction serves a reactant for another reaction so you get sequences of reactions occurring. All right, not a single reaction in isolation of others which is what we always talk about in kinetics. It turns out that's relatively unusual what we often see are cascades of reactions occurring. This is an example of a case where there's two reactions that occur in sequence. All right, we've got this methyl bromide reacting with whatever the heck this thing is. All right, an ethyl ester. All right, an intermediate is formed by that reaction. That's this guy right here which is an ester onto which we've appended a methyl group and then an hydrolysis occurs. All right, and we end up with this guy, all right, where this ethyl group has fallen off. I should actually have a plus ethyl group here. All right, this is malonic acid and this is the synthesis malonic acid from this methyl ethyl ester. So the point is it's important to understand consecutive reactions because consecutive reactions are sort of the rule, not the exception, right? Reactions are usually consecutive. There's usually more than one thing happening in sequence rather than just having something happening in isolation of other chemical reactions. All right, because they teach us how the concentration of intermediate is, yes. So what we're going to talk about first is the simplest consecutive reaction, all right? One where we've got irreversible conversion of A to B and irreversible conversion of B to C with these two ray constants right here. No reversibility in either one of these two reactions. All right, the simplest possible case of two consecutive reactions, the minimum number of reactants and products. Now, without going through and deriving these equations, they're derived for you in Chapter 19. This is equation 19.27B. This is equation 19.27C. This equation here, we did derive. All right, this is just the first order reaction of A going to B and we derive an integrated rate law like three times, okay? So these are just the integrated rate laws for A, B and C for this reaction mechanism right here. That's not really what we're interested in, all right, as I'm going to show you. What we want to find are simpler expressions for these integrated rate laws, all right? Because in general, the consecutive reactions that we're going to care about are going to be much more complex than this, all right? And I think you'll agree that these equations are already getting pretty complex even for the simplest case where we've stripped down the number of reactants and products to the absolute minimum, we're already getting pretty complex mathematical expressions here. Imagine if this was A plus B goes to C plus D plus E goes to F plus G plus, all right, and if there's twos and if the stoichiometric coefficients are not all one, I think you can see this is going to start to be a nightmare scenario for us. So what we would like to do is look at these complex equations, see if there's a way to strip down the complexity, all right, and generate a simpler way to think about these reactions. That's what the steady-state approximation does for us, all right, it allows us to simplify these integrated rate laws for reactions that are consecutive. Okay, so let's first of all look at what these rigorous equations predict and see if we can see something generalizable from what they predict that will help us in terms of figuring out what this steady-state approximation is. What do these equations predict about B versus time? B is the intermediate here, right, and intermediate is not a reactant and not a product, all right? By definition, it's something that is formed transiently in the middle of this reaction, so what we may see in the lab is we put A into a beaker, C comes out, but B was there transiently, all right, it was formed, and then it was consumed. It was formed by this reaction, and then it was consumed by this reaction, but what we want to know is what happens to its concentration as a function of time, all right? Let's consider the case first where K1 is faster than K2. That's fast, that's slow, all right? Qualitatively, what do we expect to happen? If this is fast and this is slow, B is going to build up, right? Because we're going to form B rapidly, but then it's only going to decay to C slowly, and so what we're going to expect qualitatively is the larger of the disparity between K1 and K2 in this direction, where this is fast and this is slow, the higher the concentration of B is going to be. So I calculated the concentrations of A, B, and C versus time, all right, for the case where K1 over K2 is 5. Remember, K1 is faster than K2, so the ratio between K1 and K2 is going to be bigger than 1, right? All right, so let's say we make it 5. If that's the case, we're going to start off with A equals 1 molar, okay, on this axis, and this is just a time axis here, and so you can see A just decays like it would in any unimolecular reaction. This is just an exponential decay of A, not very interesting. But look what happens to B. B increases, reaches a peak, and then decreases as a function of time, so there is a big spike in B as a consequence of the fact that K1 is faster, we form B, and then it slowly reacts to give us C, all right? We form B rapidly, and then it slowly reacts to give us C, and you can see, here's the concentration of C as a function of time. It makes sense. Now, if we change this ratio to make it bigger, now I've made K1 faster relative to K2, you can see B gets bigger. Here's where it was, here's where it is now, all right? The concentration of the intermediate builds up more if I make K1 versus K2 faster, and this is K1 over K2 equals 20, bigger yet, here's 10, here's 20, okay? So qualitatively, these equations tell us exactly what we expect intuitively. We expect as we make K1 faster versus K2, we expect the build up of this intermediate to be more and more prominent. We expect the peak concentration of the intermediate to go up and up and up as we make that number bigger. Oh, that's 20, that's 50, I think you get the idea. Now, let's look at the other possibility. If we make K1 slower than K2, what's going to happen? We slowly form B and then B reacts quickly. Under those conditions, we don't expect B to build up very much. We slowly form B and then as soon as we form B, it reacts fast, that's what that limit means right there. K2 is faster than K1, so this is slow, this is fast, and if we look at, if we make that ratio 1 to 10, in other words, K1 over K2 is 0.1, look, B forms, but it doesn't, we never get very much B and then it leaks away, all right, but the concentration of B never gets very large, compare that to the case where K1 over K2 is 10, all right, look how big B gets, enormous difference, okay? If we increase this, so now I've made the ratio 1 to 20, in other words, 0.05, we've made the K1 even slower compared to K2, B gets smaller. There's B when that's 0.1, B when it's 0.05, B when it's 0.02, smaller yet, all right? So what's interesting about this is if you look at this now, it looks like B is becoming quasi-constant as a function of time, all right? It's concentrations hardly changing, all right? The concentration here is not quite zero, all right? It's becoming almost invariant as a function of time. That's the key thing that we want to focus on, all right? So there's an assumption that we can use, that we can bring to bear on reactions like this that allows us to simplify the mathematics enormously. And that's the steady-state approximation. Basically what we're going to do is we're going to assume for all of the intermediates in the reaction that the time rate of change of those intermediates, the time rate of change of their concentration is going to be zero. So here, the time rate of change of B would be zero. That's the steady-state approximation. What does that do for us? Well, it's going to allow us to derive simplified equations for the rate at which the product appears. Mainly, that's what we care about. How fast does the product appear as a function of time? What's the integrated rate law for the product? That's what we really care about here. So what we want to do is apply the steady-state approximation, derive a new integrated rate law for the product that is simpler than what it would be if we had to work out all of the complex mathematics. So we're going to solve a simplified equation that results. So for this reaction mechanism right here, the time rate of change of B is going to be K1 times A because that's the rate of B is formed. So it would be plus K1 times A. And then B is consumed by this reaction so there will be minus K2 times B. So that's the rate at which B is consumed. That's the rate at which B is formed. And what we're saying in the steady-state approximation is that since that time rate of change is zero, those two things have to be equal to one another. The rate at which it's formed and the rate at which it's consumed are going to be equal to one another. And if that's true, then B sub-steady-state, right, in the special case that we're talking about the steady-state approximation, B is going to be equal to K1 over K2 times A. Very simple, much simpler than the whole truth about the reaction, than the rigorous equation that we talked about on an earlier slide. And I'll show it to you again in a second. We always have to keep one thing in mind. It's tempting to mindlessly apply the steady-state approximation to every reaction mechanism that we come in contact with. But it's important to understand that this reaction, this assumption, the steady-state approximation is an approximation and this approximation fails badly if K1 is faster than K2. If K1 is faster than K2, we're going to get build-up of B. Its concentration is not going to be constant as a function of time, not even close. What we've learned by looking back and calculated in concentration of B is the steady-state approximation can be expected to work when K2 is faster than K1. When the intermediate is used up rapidly in subsequent chemical steps but formed, yeah. In other words, this is the case. When K1 over K2 is small, K2 is fast and K1 is slow. Okay. So to reconcile this equation with this equation right here, we insist that K1 be much, much less than K2. All right? If that ratio approaches zero, why these two results can be reconciled with one another? Right? This one says the time rate of change of B is zero. This one says the time rate of change of B matches the time rate of change of A just ratioed by K1 over K2, right? But if K1 over K2 is zero, then the concentration of B is going to be very low at all times. All right? That's really what we're assuming in the steady-state approximation. And when we deviate from that assumption, the steady-state approximation is not going to work very well. Let me show you that. Yes, I just pointed that out. Okay. So this suggested an expedient method for dealing with such reactions. We call this steady-state approximation self-sufficiency. Yes. Okay. So here is the integrated rate law for C, all right? If this is the concentration of B, all right, then we know C is just equal to K2 times B, right? And consequently, if I just plug in for B from the steady-state approximation, all right, so this should really be B sub-steady state. And this should really be B sub-steady state. And so when I plug in for B, K2's cancel, I just get K1 times A. That is the differential rate law for C. In other words, DCDT is K1 times A. So I can then solve that just by integrating in the normal way to derive an integrated rate law for C. And when I work through the integral, I find that the concentration of C is equal to the initial concentration of A times 1 over this EXP term here, right? It depends on K1. It doesn't even depend on K2, all right? So this is the new integrated rate law subject to the assumption that we're applying the steady-state approximation, all right? So here are the integrated rate laws that we have from the steady-state approximation. This is the one that we derive directly by thinking about the steady-state of B by setting DBDT equal to 0. We derive this equation directly. And then this one has never changed. It's just the first order integrated rate law for a unimolecular reaction of any kind. That one's always the same. Same as it is here, same as it is here. This one we derive by just making a substitution of B into the differential rate law for C. Which is what we did on the previous slide. I submit to you that these equations are a lot simpler than these equations, but the degree of simplification here is tiny compared to what it normally is, all right? Because we're talking about the most elemental sequential reaction mechanism possible right now, right? As you make that A goes to B goes to C. As you make that incrementally more complex, these mathematics get exponentially more complex, all right? And so the degree of simplification that you get becomes very significant, which is why we use the steady-state approximation so much in chemistry. It's used a lot. It's overused. How good is it? Let's look at some data. How well does the steady-state approximation work? Let's first examine a case where we expect it will work well. All right? What case is that? K1 over K2 small, all right? In other words, we're forming the intermediate and it's reacting fast, all right? We looked at this data before. The dashed lines are the steady-state approximation. The solid lines are the full and correct answer, right? For the concentration as a function of time, all right? This is just the concentration of these three guys as a function of time as before, all right? Check this out. This dashed line here almost falls on top of that solid line. We've got excellent agreement. This green data right here, by the way, is the one that we care about most. We want to know what the product is doing as a function of time, right? The product concentration. We're going to use the steady-state approximation to get at that and here we're going to do a great job, all right? We've got an excellent fit of the steady-state approximation to the true answer. Look at what the steady-state approximation predicts for the intermediate. It's killing it. It's right on top of it, all right? And, of course, it's doing a good job for the reactant as well, all right? So the steady-state approximation works great here. It works great because that's much less than one. Now let's make it worse, all right? Let's make it a little bit bigger. Now we're not killing B quite as well. It's a little bit hard to see. A still looks great. C looks pretty good, but not quite as good as it did here. It's kind of hard to see that there's a difference, but it's still doing a pretty good job, don't you think? All right? We care mostly about this green curve and the dashed line here is almost right on top of that green line. Not quite now. You can see a little bit of daylight there, but not much. All right? So the steady-state approximation is still working pretty well here. Now what if we make this 10? In other words, we're going to start forming the intermediate faster than it gets consumed, right? This is a case where we expect the steady-state approximation to fail, all right? And we're going to remember that even though nobody else does. Even though other people use the steady-state approximation indiscriminately, we will not do that. We will always remember that this is a case where the steady-state approximation really shouldn't work very well. Doesn't work very well. What am I talking about? Check this out. Here's the dashed line from the steady-state approximation for the product. Here's what the product concentration is really doing. That's a pretty big delta. All right? That doesn't look like it fits that very well at all. All right? And the intermediate, terrible. All right? Except down here, actually not quite as terrible as you might think, but it's still, A, is still doing a pretty good job. All right? So it's doing a poor job when K1 over K2 is 10. If I make K1 over K2 50, it gets downright nasty. All right? Look at this. Here's the dashed green line. Here's the solid green line. There's miles between these two curves here. All right? And many people will apply the steady-state approximation without any qualms at all to cases that conform to this set of ray constants right here. All right? So you have to be careful that the intermediate is not building up because in that case, you doesn't make any sense to apply the steady-state approximation. All right? So for the steady-state approximation, what do we do? All right? If we want to apply the steady-state approximation to some kinetic mechanism, right, for a sequential reaction, first of all we write the rate law for the process. I'm talking about the process that generates product usually. It's usually what we want to know. In other words, in this case, DCDT is K2 times B, right? That's the rate at which C is formed. Then what we do is we ask are there intermediates in this rate law? Identify them and then apply the steady-state approximation to all of them. Here we can find an intermediate. It's right there, all right? And it shows up in the differential rate law for the formation of product as well. Perfect. This is a great candidate for the application of the steady-state approximation. Then what we do is set the time rate of change for these intermediates to zero. In this case, DBDT is equal to zero. And then we write equations for the creation and consumption of the intermediate, equate these and solve for the steady-state concentration of the intermediate. That's what we did here. All right, so I set the time rate of change of the intermediate to zero. And then I said what that means for this reaction mechanism is that the rate at which B is formed has to equal the rate at which it's consumed, right? These two rates have to be equal to one another. Now I just solve for the concentration of B, right? You solve for the concentration of the intermediate, all right? And inevitably, the concentration of the intermediate appears in the rate expression for the concentration of the product. And so the next thing to do is just plug this in to the differential rate law for the product, all right? Substitute the suppression of the rate law for the product. In other words, DBDT is K2 times B, all right? And now, instead of being K2 times B, it's going to be K2 times B, steady-state. And we just plug in what we said is the steady-state concentration of B is K1 over K2. And in this case, the two K2's cancel. And that then becomes our differential rate law for C. And we can integrate that to get an integrated rate law if we want to. Remember, this works if and only if K1 is less than K2. Did I say that already? I know I did. Okay. Now we're going to take a little conceptual break. Allow that information to sink in deep into our cortexes. And we're going to talk about Irving Langmuir. Irving Langmuir was one of the first great American chemists. And with G. N. Lewis, he invented Lewis dot structures. He, Lewis called them Langmuir dot structures. That's not true. I just made that up. He's the father of surface chemistry. He revolutionized the light bulb. He discovered how to manipulate monomolecular layers on water. Who cares? Well, that turns out to be rather interesting to do, it turns out. He revolutionized the light bulb? Yeah. The incandescent light bulb. I mean, there's been several revolutions in the meantime. This is a page from, in case you're interested, G. N. Lewis's notebook. He actually made Lewis dot structures. That's why they call them Lewis dot structures. I don't know if this is a notebook or a cocktail napkin, but pretty cool. And he wasn't very neat. You see, these aren't, so Langmuir figured out how to make the light bulb work properly. Lots of people had worked on the light bulb. Edison's famous for the light bulb, right? But he wasn't the first one to make an incandescent light bulb, it turns out. There were other light bulbs before his work better. When you think about it, this is an enormously unlikely way to make light. What is this? That is a tungsten filament that is almost so tiny in diameter that you need a microscope to see it. And tungsten, what do we know about tungsten? Is tungsten a very stable metal? How does it compare to platinum? Not as stable as platinum. It's less noble than platinum, right? Does it react in air? When you buy a wedding ring, do you want a tungsten wedding ring? Is anyone going to be impressed by that? The problem with tungsten is it reacts rapidly and exothermically with oxygen to form tungsten oxide. And so, you know, if you were going to choose a metal to make light with, you might not choose tungsten. And in fact, Edison didn't choose tungsten. He chose carbon. The original light bulbs were carbon. They only lasted 40 hours because carbon filaments don't like oxygen either. And then a guy named Hiram Webster figured out how to make really, really good carbon filaments, a general electric back in like 1897. And he made carbon filament light bulbs that lasted 100 hours. These were the king of carbon filament light bulbs. But they only lasted 100 hours and 100 hours really wasn't good enough. I mean, that was like a couple of months at most and your light bulb had to be replaced. And at that time, these were not cheap. Until 1906, all light bulbs had carbon filaments. So it turns out, Lamier was not the guy who figured out that you should use tungsten for this filament. There was another guy, a general electric in 1908 who did it first. He made tungsten filaments. Why tungsten? Especially in view of this. What makes tungsten special? Anybody know? Tungsten turns out is the best at something. No. Heat capacity? That's a good, I like that answer, but that's not the correct. What problem are you going to have when you think about this? I'm going to put 100 volts across this filament that's the width of a human hair. All right, what problem am I going to have? I mean, the filament's going to get hot, right? Melting. Did you say melting? Melting. What else is going to happen? What happens to solids when you heat them up and they get really hot, even before they melt? Let's say that it doesn't decompose somehow. Because, yes, that's the first thing that you'd expect. All right, but let's say it doesn't decompose, it could still fall apart. How? What? What about sublimation? What's sublimation? Direct transition from the solid to the gas phase, right? Metal's sublime. All right, so it turns out, Tungsten does have the highest melting point of every metal in the periodic table, all right? And it has an extremely low rate of sublimation, all right? And so the problem with Tungsten was that when you turn on the light bulb, as soon as you turn on the light bulb, the Tungsten would start to sublime onto these glass surfaces here and it would turn this light black, all right? So the day you buy this light bulb, you screw it in, it's clear, two weeks later it's starting to turn gray, two weeks after that it's getting darker, and so the amount of light that comes out of your light bulb just decays exponentially as the inside surfaces of this light bulb get coated with that, all right? Because what happened is the Tungsten would evaporate onto the light bulb and then this Tungsten filament, this Tungsten film would react with trace oxygen and water, all right? And it would coat the inside of this light bulb turn of black. So Lehmier solved that problem, all right? He solved the light bulb turns black problem and he solved the Tungsten, so when he solved the light bulb turns black problem, he created another problem, right? His filaments didn't last very long and he had to solve that problem too. So just to make a long story short, what he did is he filled the inside of this bulb with nitrogen and it turns out that the reason these Tungsten light bulbs were turning black is it's because of this reaction here, but the reason that the Tungsten was sublime under the surface of this glass was because it was reacting with trace water that was inside the light bulb, all right? And that formed the Tungsten oxide and it's actually the Tungsten oxide that got sublime under the surface of this glass, all right? Not the Tungsten directly. The Tungsten oxide actually got sublime. Metal oxides, it turns out, have much higher vapor pressures than parent metals, usually, all right? Metal oxides, you wouldn't necessarily expect that. So he found out that if he filled this whole thing with nitrogen, got rid of all the water, not only did the Tungsten filament last longer, that the surfaces of this glass thing didn't turn black anymore, he had a patent for that. So the way that he made light bulbs is the same way that they're made today, all right? He essentially solved the incandescent light bulb problem. Of course, this is completely obsolete today. Okay. That was supposed to rest your thought process so we can go back to thinking about kinetics. We've got eight minutes. Let me just tell you, he figured out something about how really simple reactions occur. So even a reaction that looks like this can be stoichiometric. In other words, not elementary. Even a reaction that's as simple as this as this can have multiple steps, all right? How is that possible? Well, as an example, if there's a barrier for the reaction to occur, all right? If in the process of going from A to B, A has to surmount an energetic barrier, all right? It can be necessary for there to be an activation of A that has to occur before the reaction can happen. How could that happen? Well, let's say A is in the gas phase. A can collide with itself, and if that collision is energetic enough, you could generate an excited A that has a lot of kinetic energy or a lot of internal energy, and then that activated A could give rise to products, all right? So there's several steps here. A reacts with itself to give an activated A, that's this orange thing, then, of course, it can deactivate as well, all right? So there's a back reaction. This is just K minus one. So this is just this going back to this. I've just written a whole new reaction for it here. The activated A collides with another A and loses its energy. That could also happen, and finally, this activated A could give rise to B products. That is the Langmuir-Hinshelwood mechanism for unimolecular reactions. Turns out that lots of reactions that look as simple as this occur by this mechanism right here, all right? This activation of the reactant by itself, all right? It's activating itself, has to occur before the reaction can happen, all right? The Langmuir-Hinshelwood mechanism. Now, where A possesses enough internal energy to fall apart to give B, yes, said that. Why do we bring this up right after the steady-state approximation? I think you know the answer. We're going to apply the steady-state approximation, okay? So, let's think about what happens to the time rate of change of the intermediate here. Does everyone agree that the intermediate is A star? How do I know A star is an intermediate? I don't see A star here, right? It's not a product and it's not a reactant. It's formed during the reaction, and then it's consumed in the course of forming the product, isn't it? All right, so the first thing we have to be able to do is spot the intermediate in the reaction mechanism. That can be non-trivial, depending on how complex the reaction mechanism is. Okay, then what we want to do is write a differential rate law for it, all right? What's the time rate of change of A? Looking at this reaction right here, I can see A is formed by a second-order reaction of A with itself with the rate constant of K1, and so I'm going to write K1 times A squared. That's the rate at which A is formed. Now, can everyone see that I'm going to form, sorry, A star? Can everyone see I'm going to form one A star as a consequence of this reaction? What else can happen? Well, A star can react with A, and it can get consumed by this reaction right here, right? So I have to write a term for that. So I'm going to have K minus 1. I'm going to have a minus sign, first of all, because this makes A star smaller, not bigger. This one makes A star bigger. And what is that rate? It's A star times A with a rate constant of K minus 1. And finally, A star gets consumed by this reaction right here, all right? And so I've got K2 times A star, and that's a minus sign, too. That's a minus sign because A star gets smaller. That's a minus sign because A star gets smaller, and that's a plus sign because A star gets bigger, all right? So this is my differential rate law. I've got to be able to write that for the intermediate, because the next thing I'm going to do in the steady-state approximation is set at equal to zero. So this is just analogous rate law for A. A is consumed by this reaction, formed by this reaction. Those are the only two that matter, so I only need two terms. Why is this minus K1A squared and not 2K1A squared? Because this pertains to this reaction right here. Think about that, and if the answer doesn't arrive, I mean the short answer is you're only consuming 1A in this reaction. It looks like two A's are getting consumed, all right, but you're forming one, and so in reality you're only losing one A. That's why that's a 1 and not a 2, okay? And the differential rate law for B is just this. Everyone can see that. Yes, yes, yes. Now, what we're going to do with the steady-state approximation is set this guy equal to zero, here he's not zero, here he is, and then we're going to work out what the kinetics of the Lindemann-Hinshelwood reaction are, okay? So once I set this guy equal to zero, he's still equal to that, now I'm just going to rearrange this equation. I'm going to move the negative terms to the left-hand side and the positive terms to the right-hand side and set them equal to one another, and then I'm going to solve for A star, and then I'm going to plug A star into the expression for dB dt, and this is going to give me an integrated rate law for the Lindemann-Hinshelwood reaction, and I'm going to tell you more about that on Wednesday.