 Okay, so we're talking about Chapter 19, and we're going to talk about the stuff that we talked about on Wednesday, and then we're going to talk very briefly about reversible reactions. We started to talk about those on Wednesday, and then we're going to say a few words about Arrhenius and his equation. It's a very exciting lecture today. Very exciting. Okay, so we mentioned on Wednesday that we can, you know, we're used to looking at chemical reactions that are written down like this, and we classified these as two different types. They're either stoichiometric or elementary. If they're stoichiometric, what this reaction represents is the overall stoichiometry. Two moles of B reacts overall with one mole of A to give one mole of C. That's the only information that's conveyed by this equation, the overall stoichiometry of the reaction. There's no kinetic information conveyed by it at all. Okay, no mechanistic information. Reaction order is conveyed. In other words, A does not necessarily ever interact with 2B directly. Right? It happens through some indirect mechanism that has multiple steps in it that we don't know about. All we know from the stoichiometric reaction is that this stoichiometry here, the reaction rate will have this general form, but we don't know what alpha and beta are. We don't know anything about them from this. In particular, beta is not 2 and alpha is not 1 necessarily. They could be, but they're not necessarily equal to these stoichiometric coefficients here. Okay? Now, you can't tell by looking at this reaction whether it's stoichiometric or elementary. You have to be told, right? You can't tell just by looking at it. It's another important thing to know about. Elementary reactions describe a single chemical event or reactive encounter, right? If this was an elementary reaction instead of a stoichiometric reaction, we could write this rate law just by looking at the reaction. We could write the rate law down, right? These exponents here are going to correspond to the stoichiometric coefficients for the reactants. So if we know it's an elementary reaction, we can write down the differential rate law right away just by looking at it. Okay? So there's two types of reactions. Stoichiometric that don't convey any kinetic information. They tell us about the stoichiometry of reactants and products and the elementary reaction. For elementary reactions, the rate law can be written down immediately upon inspection. If this is our stoichiometric, if this is our elementary reaction rather, it's a unimolecular reaction. All that means is that there's a one in front of this A. We call it unimolecular. All right? The differential rate law will be this where there's a one exponent on the A because there's a one in front of the A. If this is our elementary reaction, then the differential rate law is going to look like this and there's a one exponent on the A and the B because there's a one stoichiometric coefficient in front of the A and the B. You get the idea. All right? I can look at this reaction and write down this differential rate law without any problem. This is a bimolecular reaction. When I say bimolecular reaction, I mean the overall order of the reaction is two. One plus one. Right? This is also a bimolecular reaction in which the overall order is two. In this case, there's just two A instead of an A and a B and this is what the differential rate law would look like for that. Now, notice how we normalize the rate using this stoichiometric coefficient. Here I'm expressing the rate in terms of A and in terms of B. Right? I think you can see that the rate at the apparent rate is going to be twice as fast for A as it is for B because two A's are reacting to give B in this case. And so to compensate for that, we multiply the rate with respect to A by one half. All right? And in general, we multiply by one over whatever that stoichiometric coefficient is to get a rate that is invariant, independent of which reactant or product we decide to use to express the rate. Make sense? Remember, the stoichiometric coefficient is negative for reactants. That's why there's a minus sign here. All right? So these two rates, whether I express the rate with respect to B or I express the rate with respect to A, I'm going to get the same rate. If I don't get the same rate, I've somehow done that rate calculation wrong. So let me emphasize this is a differential rate law. It does not allow for a comparison of A and B as a function of time. I think you can see that. This is the time rate of change of B. This is the time rate of change of A. All right? And this is what that's equal to. I can't create a plot of A as a function of time or B as a function of time directly from this equation, can I? All right? It's a differential rate law, not an integrated rate law. If I integrate it, then I can. Okay? So if we perform the integration, let's say we choose to use A, I move this one half over to the right-hand side and I evaluate what that integral looks like. It's just 1 over A minus 1 over A0. That's going to equal minus 2 times kT because I'm integrating from 0 here. And so this is going to be my integrated rate law, right? Well, this is not the one that's in your book. All right? If you look up this reaction in your book, it's reaction 19.13B and it looks like this. It doesn't look like this. And the reason is that this 2 by convention is always rolled up together with the k. All right? That 2 is always incorporated into the k. The 2 gets devoured by the k. All right? The constant includes that 2. Okay? So it's not there. When we talk about the rate constant, we're talking about this k that includes this 2 if you want to think about it that way. Now, we have to be able as chemists to experimentally determine what this rate law is. We've got a reaction vessel. We take a sample from it. We inject that onto a GCMS. We get some peaks. It tells us that there are three reactants in our reaction vessel at the beginning of the reaction before we initiate the reaction somehow, maybe with a temperature jump, maybe by stirring it, whatever. All right? We know there's three reactants, but we don't know what the stoichiometric coefficients are. We want to determine what the rate law is for the reaction that's occurring between these three reactants, A, B and C. Let's say those stoichiometric coefficients are alpha, beta and gamma and they're forming some products here. All right? How do we as chemists figure out what the actual differential rate law is? All right? What's the stoichiometry with which these reactants are reacting with one another? This is a very important issue for us as chemists. We know the differential rate law is going to have this form, yes, or assuming this is an elementary reaction. And I've indicated here that it is because you can't tell by looking at it. Okay. So, there's three methods for doing this. The first is called the method of initial rates. The basic idea is very simple. We want to design this reaction so that the rate, overall rate of the reaction only depends on one reactant at a time. That's the key. So, to do that, we're going to make two of these other reactants large, much larger than A. How do we know when it's much larger than A? Let's say that A is one millimolar. We might start off by making B and C one molar, a factor of a thousand larger. Then what do we do? We measure the reaction rate as a function of time. And then we change B and C. All right? If we go from one molar to 1.1 molar or from one molar to 0.9 molar, there shouldn't be any change in the reaction rate because the reaction rate should only depend on A. Okay? So experimentally, we can tell whether we've isolated A or not. Right? If we haven't isolated A, we need to make the concentrations of B and C two molar or five molar, however large we can go before we're out of solubility limit, possibly. Maybe it's impossible to isolate A. Right? We won't know until we start to do some experiments. All right? But we can tell in the laboratory whether we've succeeded if we change the concentrations of B and C and the reaction rate does not change, we know we're isolated with respect to A. That's the limit that we have to be in. This will isolate reactant A. Okay? So once we've done that, there's a new effective rate constant, K prime, that includes B and C. B and C are not going to change appreciably as A reacts. Right? Of course, B and C are reacting too. Their concentrations are going to be going down. But the change in the reactant, the change in the concentrations of B and C are going to be so small that the overall concentrations of B and C are insignificantly affected by this change. Okay? So we've got one millimolar of A. All of the A could react and we're going to have what? Right? 0.999 molar of B if we started off with one molar. Right? So B is going to be the same to within three sig figs. It's not going to be affected by this reaction and we can assume it's constant and we can roll it up into this pseudo alpha order rate constant. Okay? So you see this K prime? Once I make these guys big, I can roll them up together with K to form this K prime and now my apparent reaction rate looks like this. This is my apparent differential rate equation. Okay? Then what do we do? So, no, before isolation, the overall reaction order was alpha plus beta plus gamma but after isolation, the apparent, apparent overall reaction order is now just alpha. Here it's just alpha. Here it's alpha plus beta plus gamma. Am I beating this to death? All right, then we're going to measure alpha. To do this, we measure the reaction rate as a function of alpha zero, the initial concentration of A. It's function of alpha zero, A zero rather. Okay? So if I take the logarithm of this guy, I get log of the rate. That's log of K prime plus alpha times log of A. And so all I'm going to do is I'm going to change the concentration of A instead of using one molar, one millimolar, I'm going to use two millimolar and three millimolar and four millimolar. All of those concentrations are still small compared to one molar, B and C, right? So I've still isolated A, presumably. And in fact, you see how this is A and this is A zero? The reason I'm going to use the initial concentration of A is so that in case there's a reverse reaction that occurs, I will not be influenced by that. Because at the beginning of the experiment, there's no product to cause there to be a reverse reaction, right? And so if there's no product, I'm only going to measure the rate of the forward reaction. And if there is a reverse reaction that I don't know about, perhaps, it won't influence what I'm trying to determine. I'm trying to determine the stoichiometric coefficient of A, right? That's why we're going to use, that's why it's the initial rate method, okay? We measure the initial rates of the possible influence of a reverse reaction in case it's there, we don't have to think about it. Okay, so here's what that data looks like. Here's four millimolar, three millimolar, two millimolar, and one millimolar of A, all right? And I'm measuring the initial rate, which is the slope of this red line that I've drawn here, all right? We don't know what happens to A during the rest of this reaction. We don't have an integrated rate law for A, all right? All we're going to determine is the initial rate for different concentrations of A. Yes, that's what the slope of this line is, and then we're going to plot that log of the rate versus log of A0. We should get a straight line, and the slope of the straight line is alpha. It's beautifully simple. It can't fail. It works, hesitate to say it works every time, all right? It's a very simple and straightforward way to determine what the stoichiometric coefficient of each reactant is. The problem is you've got to do a lot of experiments. You've got to do four experiments here, right? That's just to get the stoichiometric coefficient of A, and then if there's B and C, a lot of experiments are involved. Okay, that's the first method, method of initial rates. The second method is to calculate the integrated rate law, integrate the differential rate law. When you do that, you get an integrated rate law that now explicitly predicts what A will be as a function of time. The integrated rate law allows you to model the concentration of any reactant or product as a function of time in your computer. You can just put the equation into Excel and boom. You can fit your experimental data using the rate constant and the initial concentration, all right? And when you do that, you should be able to match up your predicted A versus time curve with your experimental data, and that tells you that whatever differential rate law you assumed for A, first order, second order, zero order, right? Whatever differential rate law you assumed is correct. If you can fit your experimental data as a function of time and your predicted curve overlays your experimental data, why you've got it, right? Whatever molecularity, first order, second order, zero order, third order that you've assumed is what you actually have in your experiment. So, in this case, we're plotting log of A on this axis versus time and the intercept is obviously going to be log of A0 and slope is going to be minus K. You don't have to create a linearized form of this equation the way that these textbooks do that were written 23 years ago, 25 years ago, okay it's not necessary to do that anymore necessarily although it is convenient to be able to just get the K from the slope of this equation, see how linear this is. This gives you a good feeling for whether the fit of your first order reaction, whether your first order reaction fits the data or not, but you can just plot it directly also and fit it. Most graphing programs will do that automatically now for you, okay? So, the integrated rate law is the second way to determine what the differential rate equation is. It's better than the initial rate law in the sense that you're fitting over a range of times and over a range of reactant and product concentrations and so if you get a good fit it really tells you unequivocally that the reaction is first order over a range of times and a range of concentrations for your reactants and that's not true for the initial rate law, the initial slope method, the initial method of initial rates, sorry. Okay, so there is a third method, we already talked about it on Wednesday and that is once you've got your integrated rate law there's a shortcut that you can take which is not to do any fitting, it's just to determine what the half-life of the reaction is and you determine that by plugging A0 over 2 in for A. Here I've just plugged A0 over 2 in for A and when I do that the A0 is cancelled and I get 1 over 2 and that's just minus log of 2 and so the half-life of the reaction is just going to be log 2 over K. The half-life is the time when the concentration is dropped to half of its initial value, obviously. And so this is yet a third way, it just involves taking the integrated rate law, substituting A0 over 2 for A. That time is now called the half-life and you can solve for that and see if what you measure conforms to what you expect. All right, so this is the half-life or first order reaction and the half-life stays the same. The second half-life is the same as the first, the third half-life is the same as the second. What you're interested in here is the progression of half-lives as a function of time, the first half-life. How does that compare to the second half-life? How does that compare to the third half-life? To do that you've essentially got to map out this green curve here. So it's not really a time saver. But it does allow you to just look at the data and say, hey, this is obviously first order data. All right, without having a computer handy to fit it, you can just look at it and say, all right, I started off at point two, here's one half-life, now I'm at point zero five, that's the second half-life, yep, it's first order. So it's handy to be able to do that without having to get into your computer and get into Excel and transfer the plus slope and do everything else that you would normally need to do. Okay, so there are two types of second order reactions. We already mentioned this. There's two way goes to products. That's a reaction that is overall second order and we can work out the integrated rate law for that. We already did that. Keeping in mind that this two gets rolled up into the K, the half-life in this case is different than it is for a first order reaction. It's proportional to one over A zero. And so now the half-life doubles. Every success of half-life is twice as long as the one before it for a second order reaction. All right, I measure this half-life and then I call that zero and I measure a second half-life and it's twice as long. Now I measure the, okay, so it's inversely proportional to A zero. If this goes down by a factor of two, the half-life will double and then quadruple. Everyone see that? So once again, if you're taking this data and you're looking at your computer screen, you can see immediately, if you're starting at point two, that's your first half-life. Your second one's twice as long. Looks like it could be a second order reaction with respect to A. Now there's a second type of second order reaction isn't there that has this differential rate law and the integrated rate law in this case is a little more complicated. It's worked out for you on page 802 of your book. All right, but suffice it to say there is no simple T1 half for a second order reaction that involves two reactants. All right, there is no simple closed form expression for T to the one half for this case. All right, it really only works for two-way goes to products. So here you would actually have to fit your integrated rate law, you'd have to fit your experimental data with your integrated rate equation to see whether it's really second order in A and B. Okay, we've mentioned first and second order reactions. These are the most common times, but we left out one other important reaction that we talked about in your quiz today, which is the zero order reaction in which A goes to products. And a zero order reaction is characterized by a differential rate law that doesn't depend on the concentration of A. That's a little odd. All right, the rate doesn't depend on A. If I make A small, I get a rate, and if I make A big, I get the same rate. That's odd. Something similar happens in the Haber process. The rate doesn't depend necessarily on the hydrogen and nitrogen concentrations. How can that happen? That's right, the rate of the reaction is independent of A. Well, if you work out the math, this is the integrated rate law that you get. A is proportional, or I should say decreases linearly as a function of time. And so if I work out the half-life, I just plug in A over 2 for A and simplify this equation. The half-life is equal to A0 over 2k. So if I look at what that looks like, now the half-life gets shorter as a function of time. The reaction rate is linear for a zero order reaction. So if you're looking at the data on your computer, you can see immediately if it's a zero order reaction because your reacting concentrations are decreasing on a straight line, you can see it immediately. Only a zero order reaction does that. But the effect is for the half-life, the second half-life is half as long as the first one, the third half-life is half as long as the second one, and so on, the half-life gets shorter as a function of time for a zero order reaction. What kind of reaction does this? Catalyzed reactions have zero order. For example, 2nOx reacts to give xO2 plus n2, we have catalytic converters that eat up the nOx and the CO and the hydrocarbons by catalyzing these reactions to these benign products, or we used to think CO2 was a benign product. The catalyst is in your catalytic converter. The rate of that reaction doesn't depend on the concentrations of these guys. What it depends on is how much room there is on the catalyst particle. The catalyst particle is where this reaction is occurring. Your reactant, let's say nOx comes in, it sticks to this catalyst particle right here, and then the chemistry occurs. The only thing that matters is how much room there is on this catalyst particle. That will determine the rate, not the concentration of nOx, there's plenty of nOx, even at really low concentrations to saturate the surface coverage of these tiny metal particles. What you observe experimentally is the rate at which the nOx is getting eaten up by the catalyst doesn't depend on what the nOx concentration is. It does depend on how much of the catalyst you have. More catalyst, faster rate. Makes sense, right? Something similar happens in the Haber process. Same idea, in the Haber process you have an iron particle, right? In the case of these reactions here, in general the particles are made of a platinum alloy. So they're kind of expensive. Trying to figure out how to get rid of the platinum in these catalysts. Okay? So there are some integrated rate laws that are very common, zero order, right? The rate of any reactant in the zero order reaction just increases linearly as a function of time, boom, straight line. See a straight line, you know immediately it's a zero order reaction. Log A minus log A0 equals minus KT, that's a first order integrated rate law. Here the log of A decreases linearly as a function of time, not A directly. And in the second order reaction we have 1 over A equals 1 over A0 plus KT. So it's 1 over A as a function of time that increases now linearly because it's 1 over A, all right? Third order reactions, that'd be a good test question for midterm two. Okay, so in reality we have three methods for classifying a reaction with respect to its order. Method one, method of initial rates, isolate one reactant, make the other two big, measure the initial rate as a function of the initial concentration of the isolated reactant, plot your data, determine what the stoichiometric coefficient is. Method two, derive the integrated rate equation, fit your data over a wide range of times with different integrated rate equations to find out which one fits your data. Then that tells you it's first order, it's second order, it's third order because that's the one that fits. It's tedious to do that but that's what you do. There isn't a shortcut. And finally, reaction half-life, you take the integrated rate equation. In that integrated rate equation you plug in A0 over 2 for A, simplify the equation for the half-life and that's the half-life equation. And you compare that with what you observe for your experimental data as a function of time over multiple half-lives, that's the key. You've got to see if the half-life stays the same, if it gets shorter or if it gets longer. You have to look at the data over multiple half-lives to see how the half-life evolves as the reaction is proceeding. This shows what that looks like, zero order reaction, straight line, half-life gets shorter according to this equation right here. First order reaction, second order reaction, look the same qualitatively, don't they? It's just a curve, all right? You can't tell if it's first order or second order just by looking at it but if you look at the half-lives, here the half-life's not changing, here the half-life is doubling, all right? That's a second order reaction, that's a first order reaction, that's a zero order reaction. So you can tell by looking at the half-lives. So SO2Cl2, what is that? I have no idea. It gives SO2 plus Cl2 with this rate constant right here at 593 degree Kelvin, what percent will decompose after one hour assuming this is an elementary reaction that I've written down here? In other words, it's really unimolecular, all right? It's really first order. If that's true, here's the integrated rate law. Okay, so it's basically plug and chug from that point. I want to solve for A over A0. I plug in the rate constant. Here's the amount of time that we're talking about one hour is 3,600 seconds. I get 92.4 percent is left, so the amount that decomposes is 100 minus that, right? How long for half of it to decompose at this temperature right here? Well, I can solve for the half-life. This is the normal equation for the half-life of a first-order reaction. That's the rate constant, log 2, 31500 seconds. That's eight hours and 45 minutes. That's the half-life of the reaction. Now, we talked a little bit about reversible reactions on Wednesday, and what we said was, look, you can treat a reversible reaction like an irreversible reaction if you are able to remove the product as it's formed. All right, if I remove this HI as it's formed, then the rate of this reverse reaction here will be zero because there won't be any reactant present for this reverse reaction if I get rid of this. In this case, if we study the reaction from left to right, we remove it as, so the rate law would just be the rate law that you would expect for the forward reaction, assuming this is an elementary reaction. But in general, you don't remove the reactant. You don't remove the product, rather, as it's formed. You need to know what the reaction rate is with the product in there. So what about this more general case? Let's look at the simplest example of this. All right, A forms B and it's a reversible reaction. There's a forward rate constant and the reverse rate constant for this equilibrium. It's an equilibrium, isn't it? Okay, so we can write a differential rate law with respect to A, for example. There's going to be the forward rate, right, KF times A minus the reverse rate, all right, which is KB times B. That's going to represent the overall rate at which A disappears on the left-hand side. So let's say for the sake of argument that we allow the reaction to go until we don't measure any change in the concentrations of A and B anymore. We allow the reaction to go all the way to equilibrium. Then the time rate of change of A is zero because it's a flat line. It's not changing anymore. And if that's the case, then this expression is equal to zero but we have to remember we're talking now about particular concentrations of A and B, the equilibrium concentrations, right? Okay, and so in other words, that has to be equal to that at equilibrium. And so if I derive the equilibrium constant by putting B over A, that's the expression for our equilibrium constant, it becomes apparent that equilibrium constant is also equal to the ratio of these two rate constants, all right? KF over KB. This is an important thing to remember. So the equilibrium constant is the ratio of the forward and the reverse rate constants, yes. The higher the forward rate constant, the larger the K equilibrium, the greater the extent. Products are favored with respect to reactants. It all makes intuitive sense. Okay, I'll just continue this on Monday. All right, have a good weekend.