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.  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?  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?  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?  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.


The BRICS bloc is expanding and it will have worldwide implications. What lies ahead for the grouping? Israeli Interior Minister Itamir Ben-Guir has once again made extremely offensive comments about Palestinians. What do they say about the current government of Israel? And Australia and the Philippines have conducted joint military drills in the South China Sea.  This is the daily debrief. These are our stories for the day. And before you go any further, if you are watching this on YouTube, don't forget to hit that subscribe button. The 15th summit of BRICS leaders concluded in Johannesburg on Thursday and it lived up to its billing as one of the most significant in recent times. The highlight was the expansion of the bloc. Six new countries have been invited to join from January 2024. The leaders also released a statement which had several interesting pronouncements on the world order. To find out more, we have with us Prashant. So Prashant, the BRICS summit has concluded with some interesting developments, including the expansion of the bloc to six new countries. Can you tell us about this? Right. So I think prior to the BRICS summit, this was a single biggest question on everyone's minds. Is the BRICS bloc going to expand? Are new countries going to come in? If so, who are going to come in? And we know that, you know, about 40 countries at various points of express interest, I believe at least two dozen countries officially applied. And I think even among the members at the same time, there was a lot of discussion on, you know, how do you sort of expand a bloc like BRICS while keeping its character, keeping its goals, and at the same time, trying to sort of increase the footprint it has, increase the kind of relevance it has. And so finally, they've chosen a very interesting set of countries to begin with. I mean, if you take each one of them, you have Argentina, the country going through a grave crisis. In fact, a country which is in some senses a model of what, how a country can be destroyed by the IMF, by international financial institutions, and a country which is also very vital for Latin America. So the fact that Argentina has been brought in itself is a very interesting signal. You look at say another country like Ethiopia, which actually went through a prolonged civil war, which in which the Ethiopian government faced off against a group, the TPLF, which was believed to have strong backing from the United States and the West. And, you know, the Ethiopian government trying to sort of rebuild after the TPLF had ruled Ethiopia before, and they had a very different vision of Ethiopia as a fragmented, tribally split country, whereas the new government at IBM was trying to build a different kind of Ethiopia. And that was also one of the reasons for civil war. So again, Ethiopia being brought in very interesting in that sense. Iran no need to talk about it. We know why Iran faced so many sanctions. So definitely a very interesting move. And then you have three other very interesting countries that is actually interesting. The United Arab Emirates, Saudi Arabia and Egypt, all of which are actually very close to the West in various ways. We know that Saudi Arabia for long the linchpin of the petrodollar strategy, long decades and decades of relationships with the United States, the UAE also very similar. Egypt, you know, under Abdul Fateh LCC has received a huge amount of weapons from the United States. But also these countries in recent times have also, I think, been considering the possibility of expanding their traditional roles, expanding the kind of cooperation they have with other groups as well. And you know, for instance, the fact that Egypt, Saudi Arabia and India, you all have had good relationships with India, for instance, with China, for instance, and for that matter with Russia. So the expansion, I think it's a very, you know, I think it's a very carefully chosen group of six countries. On the one hand, many of them are very resource rich. Iran, UAE and Saudi Arabia together bringing a huge amount of, you know, oil reserves. And BRICS has now become one of the leading blocks in terms of oil production if you look at it, I believe it's a substantial amount. But also in terms of GDP, also the focus, for instance, on Africa, you have Egypt, you have Ethiopia, South Africa is already there. And, you know, there's a lot of focus. South Africa is considered one of the most important regions in the future for humanity as a whole. So definitely a lot of focus over there. So I think it's a very calculated, very interesting set of choices. Also in some senses, I think blunting the Western propaganda that you could see in many media houses in the days prior to the summit that there was too much disagreement between the BRICS countries, the fact that the BRICS countries, you know, they were just not going to do it because some were more pro-US, some were very anti-US. And these dissensions, they were stopped at India and Brazil were far more closer to the US. So they would not agree to this. So all that, I think they figured out a way, I think, whatever differences they may have had, whatever disagreements they may have had, they've sort of negotiated. And I think that really says a lot about BRICS. The fact that BRICS is not some kind of a perfect alliance. It's not the G7, where everyone is subsumed under US hegemony. It's not one of those blocks. BRICS is a block where different countries often have different agendas, but also are pushing for certain similar agenda points. So I think a lot of hope, I think, from this expansion process. Right. Can you also tell us about some of the other key outcomes of this summit and what implications will they have for building a multilateral world? Look at the statement, two of the important things to note. One, there was a lot of talk about the international financial system. And that was addressed. Now, again, important to note that I think some of the Western sources really focused on what was called de-dollarization. And they pitched or de-dollarization as the ultimate aim, which was the idea that there would be a new currency. And if there was no new currency, it would probably mean a defeat of the BRICS summit. Whereas I don't think the BRICS countries ever, at least at this point, are really considering and ordinates the dollar properly as a kid. Right now, I think the main focus seems to be on trading in their own currencies or mutual trade within these currencies and the building of infrastructure for that. I think these countries do understand that there are certain risks that trading with the dollar poses, including the fact that sanctions are an ever-present reality before all of them. The Russia faces them. China faces them. India, when it actually bought goods from Russia, faced a lot of criticism from the US for that. So I think many of these countries have realized that there are certain risks in completely trading in the dollar. They want to consider the option of mutual trade. They have also already initiated some of these experiments. But this will take time. This will take political will. This will take building a particular kind of infrastructure. And I think this summit has kind of set the stage for a lot of that to happen. There's been a definite push as far as that is concerned. Secondly, I think it's very interesting that they have talked about the United Nations, both Brazil and India, very keen, and South Africa for that matter, all very keen on getting permanent seats in the United Nations Security Council. I think there's a recognition of the fact that this way the system is structured right now is kind of really unfair towards the Global South. So they're really pointed out. Interestingly, they've pointed out to the WTO as well, which at some point of time, the United States and its allies really pushed. And now they are the ones really, you know, because they don't want to go into those arbitration mechanisms. So that's also very much on the agenda. So I think overall, a very interesting and progressive statement, which kind of brings the possibility of multiplicity in terms of both economic structures in terms of political resolution of various conflicts in terms of the kind of agenda that is being said is mentioned of climate change also. So I think a lot of these issues definitely breaks taking a very good stand. Now, the important point again, like I said in my last question, the fact is not that all these countries need to agree on everything. I don't think that's really at all part of the agenda at this point in time. I think what's important to note that there is space for negotiation. There is space for poaching common alliances. There is space for taking a common stand. And there is space for pushing back against some of the, you know, some of the points that the Global North has been pushing, what you would call the ex-colonial and settler colonial countries have been pushing on the Global South. There is a possibility of pushing back against some of that. And I think that is what BRICS represents at this point of time. And considering that ambition, considering that agenda, I think this has been quite a successful summit. Absolutely. Thanks Prashant for this update. Our next story is from Palestine, where Israeli Interior Minister Itamir Ben-Guy said that his right to move around is more important than those of Palestinians. Now this might sound like a textbook definition of a pathide and it has, in fact, caused a huge controversy. But this is nothing new for Ben-Guy, who has made such outrageous comments on a number of occasions and has followed it up with outrageous actions too, especially targeted at Palestinian prisoners. We go to Abdul for more. So, Abdul, can you start by telling us about the recent comments made by Ben-Guy and also give us some context and background into who he is and all the controversies surrounding him? Well, Ben-Guy is quite famous now. Ever since he became the Minister of Police, Interior Minister, Netanyahu's Cabinet last year, he has been making comments, the comments he made other day, similar comments he has made several other times. What he said on that day is nothing in a way new. The only thing which surprised the people that one of the ministers in the Israeli government is openly accepting something which is considered to be a taboo yet and Israel has objected about being an apartheid state, recognized being an apartheid state in the past. But Ben-Guy said was primarily basically defines Israel as an apartheid state. Ben-Guy said during a press conference with the Palestinian journalist, of course, that his right to move across the occupied territories, of course, whether it is West Bank, Israel, or any other Palestinian territory is superior, more important than the right to movement of Palestinians living there. When he said he, of course, it meant that the settler colonial citizens recognized as the citizens of Israel, the Zionist set of people. So that is exactly what he said. And that basically, again, you can, as I said before, he has said such things before also. For example, when while talking about Palestinian resistance, he often calls them terrorists and even talks about having a dead penalty for those who are involved in murder of quote unquote Israeli citizens. Only the Palestinians, by the way, not the other Israeli citizens. So if an Israeli citizen's motives and Israeli citizen, there is no need of a death penalty. But if a Palestinians is found involved in such a crime, then there should be a death penalty. This is a similar thing. Then he has also talked about how Palestinians need to be expelled from the occupied territories and should be exiled, should be sent into other parts of the world primarily sticking to the traditional Zionist notion of Jordan being Palestine. So he basically wants to expel all the Palestinians out of historic Palestine and send it to other parts of the world. So what he said basically is a repetition of what he believes in him being an extremist. Some of the people have identified him as a racist. He kind of a person who has also openly talked about mass ethnic cleansing of the occupied territories. So what he said other day is not something which is surprising in any way. Right. As you said, this is not the first time he's made such statements. So, you know, what kind of response has he gotten this time and in the past both within and outside the region? Well, there is already a very well-known debate in the global arena. Primarily, the progressive sections of the profile, you can say the people who stand with the Palestinian struggle have known this for quite some time that Israel is an apartheid state. The policies it is pursuing inside the occupied territories are basically the policies which were followed by South Africa when it had apartheid system before 1994. So the discrimination, the restrictions on the movement of Palestinians, the treatment of Palestinians vis-à-vis having different set of policies, then the passing of the Jewish national law which basically, even within the Israel, there is creates two kinds of different citizens, one Palestinians, Arabs and the Israelis and claiming that Israel is a Zionist state, completely denouncing the right of the Palestinians. So the kind of policies which the Palestinians have been witnessing both inside 1948 Israel and inside the occupied territories have basically made it very clear that Israel is an apartheid state and increasingly becoming so, much more pronounced apartheid state, particularly after the coming of Benjamin Netanyahu's government in power last year in November, December. So when he said those things, some of the only thing which is noticeable is some of those who so far have restrained themselves have not to say Israel as an apartheid state have not agreed to the idea of that. They had some disputes saying that okay, there are discriminations, there is an occupation, but it is not an occupied apartheid state yet. Even some of them have started questioning this notion and they have made clearly clear cut statements on Twitter and other social media platforms. If you see that more and more people are now are kind of willing to see Israel as an apartheid state. Of course, this is not a decisive shift. Of course, as I said before, this has been going on for a quite long time, but that has been the case that a large number of people are now much more in clear to see what Israel stands for and particularly under this particular regime. This also comes in the context of the increasing violence in the occupied territories. One should not forget that already according to the UN data, this is just the eight month of 2023 has been the deadliest year on record in the occupied territories, particularly in West Bank and more than 200 people Palestinians have been killed in the repeated raids. And then there are settlements are increasing on a regular basis and the settler violence has been allowed to increase. They have been unleashed. You can say there is complete impunity. No settler has been found guilty or punished for what they did and they are doing it increasingly. People are get Palestinians are getting killed and so on and so forth. So, it seems that what Ben Guir said on that day has basically made it quite clear for those who did not want to see it all this while that what is what Israel actually stands for and particularly at this point of time. Right, Abdul. Thanks for this update. And finally, Australia and the Philippines conducted joint military exercises on Friday, August 25 near disputed territories in the South China Sea. The drill was held near the Scarborough Shoal and island controlled by China and disputed by the Philippines. The joint military exercises is the first of its kind between the two countries and comes just weeks after recent tensions between China and Philippines at the second Thomas Shoal in the region. Australia's involvement is the latest in the Philippines attempt to involve foreign players in a bilateral dispute with similar exercises and patrols conducted in collaboration with the US and Japan. We speak to Anish for more. So Anish, this is the first time that Philippines and Australia are holding military exercises of this sort. Can you give us a bit of context and background into this and what does this mean for the tensions in the region between China and Philippines? Yes, I think we should start off by pointing out that very recently Philippines and China actually went into a standoff near the Scarborough Shoal, which is one of the disputed set of islands between Philippines and China in the South China Sea. So this was quite contentious, quite tense. It created a situation which was like very closely monitored by everybody around the globe at the time. And this happened very just a couple of weeks ago. And a military exercise of the sort pretty much especially initiated by two countries in the region is creating sort of like it is only aggravating the tension at this point. And we need to understand that this is a more multilateral kind of dispute that we're looking at when we talk about the South China Sea and it's not just the Philippines, but also closed a couple of other countries in the region. But the islands in dispute pretty much can be and should be ideally dealt with through resolution, something that both China and Philippines can do through diplomacy. But that is not happening. One of the primary reasons why this is not happening is because the current administration of Marcos Jr in the Philippines is pretty much taking this very anti-China, very strongly anti-China and very strongly prone United States. That's the more important part, like the fact that they are more very strongly aligned to US foreign policy creates a situation where tensions are going to keep burning in the region. And that is pretty much part of this whole joint military exercise. And the fact that Australia is also getting to become a part of this as a major pivot in the Southern Pacific region is also speaking volumes of it. Like this is essentially in many ways part of the US extension of US foreign policy of trying to encircle China. And we have spoken about this quite a lot in this show as well. Like this is pretty much an extension of that policy and that is only going to create more tensions in the region at this point. Can you also tell us more about the role being played by Australia in this encirclement of China? Yeah, it is quite an interesting position that Australia is in right now because obviously there has been a lot of back and forth. It is not as ideologically committed to the central policy of creating tensions with China, provocating or for that matter even confronting China on pretty much every single issue that we can think of. Because obviously China is the biggest trading partner for Australia and that is something that is quite irreplaceable at this point in time. No matter how much the Australian elite tries to make sure that tries to replace or make a sort of US leaning foreign policy, it is still not going to undermine the kind of economic significance that relation between these two countries have. Nevertheless there is definitely a significant section within the Australian ruling class that does believe that China is a threat and that is something that they have taken as a given. And we have seen not hints of that like very blatant pronouncements of that in the recent years. We have seen defence papers very recently talking very specifically about it pretty much parrots US line on what the Indo-Pacific policy should be and that is pretty much what Australia has taken up. We have also seen very recently Australia trying to sign a security pact with Pacific Islands in the region. Vanuatu currently is going through an entire political crisis pretty much because a deal of this kind was signed by the Australia and initiated obviously by the Australian government and all of that being done with China in mind. Australia is pretty much trying to become and we have seen a very similar kind of policy which says South Korea and Japan specifically in the region and so in the South Pacific you have Australia doing the same kind of thing and that is pretty much creating an extension of US foreign policy and pretty much jeopardises Australia's own national interest at this point and it is the same thing with Philippines. The fact is that Philippines depends a lot on China for a whole host of consumer products and not just that they also look for China for a lot of their products to be sold there like a massive market that cannot be replaced and so despite that you have governments primarily pushed by a certain kind of ideological inclination pushing cold war rhetoric and also furthering tensions that need not be there at this point in time and we have seen in the past that these tensions could have at least been kept at bay rather than kept burning the manner in which it does so this is pretty much the situation that we are at right now. Right, thank you Anish for talking to us. And this is all we have in this episode of the daily debrief for more details on these stories and for other such stories visit our website peoplecispatch.org and our social media pages on Facebook, Twitter and Instagram for more video updates visit our YouTube page. Thank you for watching.


Here's how easy it is to install the Facebook Marketplace dropshipping helper extension and enjoy price monitoring, stock monitoring, quick product importing and more business automation. First things first, make sure that you are logged in to your Facebook account as well as to your AutoDS account. Then proceed to install the Marketplace dropshipping helper extension. Once it's installed, be sure to pin it onto your Chrome browser. So I'm going to click on this puzzle icon on the top right extensions and click on the pin icon next to the Marketplace dropshipping helper. Now I see it up here, left click on it and log in using your AutoDS credentials. Once you fill it in, click on log in. Then you'll get the screen, welcome to Facebook monitoring and choose your Facebook store from the drop down menu. Now I only have one Facebook store called my Facebook store. That is why you need to be logged in to your Facebook account with that Facebook store inside. So I'm going to choose that store and click on start sync. Besides that, you also have the logs to see all of the changes that happened on the AutoDS Marketplace extension since you activated it for the first time. Now let's see how fast and simple it is to import products to your Facebook store. Let's go with this product. For example, first things first, copy the products URL, then head to the AutoDS platform, click on add products. And in this case, we're going with single product. Now paste the URL of the product that you just copied. Next, click on edit now. Now all of the products details are being added and it'll end up in the draft section of your store. From here, you can optimize the product's title, the automation settings, the product's description, the images and so forth. Once you're done, click on save. Now the product is in the draft section of our store. Now whether you optimize the product on the drafts page beforehand doesn't really matter because you can always do it on the Facebook Marketplace. But as long as the product is on the draft section of our store, we can move on to the next step, which is heading back to the Facebook Marketplace. Once there on the left side, we'll click on create new listing. And here in the middle on the choose listing type, we're going to choose item for sale. Now you can see a drop down menu. And here are the products that we have on the draft section of our Facebook stores on AutoDS that haven't been published yet. We're going to click on paste info from AutoDS. And now as you can see all of the info from AutoDS from the draft section of this product got imported to our Facebook Marketplace listing, where here we can continue optimizing the product listing, including the categories, all of the texts, all of the tags. And once we are done with the optimization process, we're going to click on next and publish once we are done. Now what's happening in the background is this product is now being published on my Facebook Marketplace store. And also on AutoDS simultaneously, it is moving automatically from the draft section to the product section of our store. And now you can see that it is listed on my Facebook store. It is a top listing that you can see right here. And if I go to create new listing and item for sale, and I click on the drop down menu, you don't see that item here anymore, because that's done on the draft section of AutoDS. Here you can see it being imported to the product section of our store. It's not in the draft section. And that is because we published that product. Also, you can see the Facebook ID right here. And as soon as you click on it, it takes you to the Facebook ID of that product. That of course is a different product, but I also wanted to show you that new feature. And that is how simple it is to download and install the Facebook Marketplace dropshipping helper extension and begin enjoying business automation.


Any Mendoza here with RIT head coach Jack Coon and coach congrats on the big win obviously knocking off a tough opponent and amhurst. What do you credit the fast start I mean an 8-0 lead in the first quarter that it's pretty unheard of. Yeah we had a couple of balances go our way I think the first two goals are rebound goals and you know give our guys credit for staying with it but you know we kind of got off to a you know a lucky start really with a couple of rebounds but that kind of got our confidence going our momentum going and then from there I think we we had great energy I was a little worried coming off of yesterday's grind that we might not have the energy we needed but I give the guys credit for for stepping up and coming coming ready to go yeah and obviously the they made a pretty furious come back there in the third quarter notice that you switch goalkeepers was that injury related or was that just kind of momentum related trying to get a spark for your team yeah trying to get a spark Walker's been just incredible all year for us and he struggled down here I'm not quite sure why you know he was struggling the second half wasn't seeing the ball well so we have great confidence in our goalies and I thought let's let's give hole in a try here and see what happens and he came in and had four or five nice saves for us so you know you know Walker and holding both a great goalies and I gotta have the confidence to go with our next guy if we need to so you know Walker's a great kid he he's supporting holding when he's in there and he's a team guy so and what do you credit the the ability really to kind of respond to their response it's tough when when a team has got so much momentum behind them yeah and they're they're making the comeback that they did you guys were able to really answer goal for goal in that fourth quarter yeah I had a couple times there what I thought about calling a timeout just to slow the momentum down and and we're talking about on the sidelines to you know keep your composure and you know sometimes you got to let them grind it out and figure it out and you know yesterday I thought one of the keys to our win today was just a composure that we showed whereas yesterday I thought we're you know we're little jittery and a little you know just pushing the ball a little too fast instead of settling down and getting composed so our leadership was much better today and obviously undefeated on the season still how do you take these two wins from this weekend and kind of continue the momentum for the rest of the season yeah I mean I think it's gonna give us some confidence being a team like Amherst they're a great great group you know I think we just got to keep keep doing what we've been doing and you know keep grinding keep working you know we got some formidable opponents coming up too so we've got to you know focus on the next one I think we have Geneseo next and that's the next one we got to focus on thanks so much for the time coach good luck the rest of the season appreciate the time thank you Andy appreciate it for more coverage on the Mustangs Classic you can go to GoMustangsSports.com