 Okay, how are you guys doing? Good. All right, I'm glad to hear that. I'm doing awesome too. All right, we're going to talk about the final exam today. Let me remind you that it's Tuesday, not Monday. It's at 1.30 to 3.30. Okay, in this room. Now, I understand, is Stephen here? Maybe you know Jean-Marc. When the review session, I just checked Facebook and I didn't, right. So if you haven't weighed in on when you want the review to be, I think he's already settled on it being on Monday, but I don't think he's got the room nailed down yet and the time, all right? So if you haven't made your opinion known, you might want to do that, all right? And just stay tuned. It'll be posted, I presume, by the end of the day to day. Right, that, yes, thank you, it's wrong. The review is not on Saturday. Right now, the version of this lecture that's posted to the website says Saturday, it's not on Saturday. It's on Monday. Okay? So there's going to be, I wrote the exam yesterday. I know exactly what's on it, so I can tell you. There's four problems, 50 points. I don't know why. Okay, and here's what they are. There's a problem on Stack Mac that, I looked at this problem, I went and found it, it's similar to stuff that is in Chapter 13, it's stuff that's in Chapter 13. There's a thermal problem that's from Chapters 14 and 15, not so different from one that you saw on midterm one and then on midterm two and the quiz material. So you are really familiar with problems one and two. Problem three is enzyme connection. These last two problems are kinetics. We haven't had kinetics on a midterm exam yet, so we're going to focus this review on these last two problems and make sure that we review some of this stuff. All right, there's two extra credit problems that are worth a total of 20 points, each one. The first one is on transition state theory and solution. We barely talked about that. That was Wednesday, at the end of the lecture, we were rushing. All right, so we're going to go back and look at that again. We're going to try and get you this 10 points. To get this 10 points, you've got to get the answer exactly right. There isn't going to be any partial credit on these extra credit problems. There's also a Stack Mac problem that you've seen multiple times in different forms. I'm asking it in a slightly different way. This will not be something that catches you unprepared if you do any preparation. Okay, so that's what it's going to be. I don't think it'll take you a whole two hours to do this exam, honestly, but you'll have a whole two hours to do it. All right, you can relax, make sure you get everything right, double check your work. All right, there should be plenty of time to do this exam. Now, normal rules apply, open book, open notes. There will be no equations on the exam, no constants, no conversion factors. As usual, you need to be able to dig those out of your book. It's not open anything else. Computers, phones, iPads, solutions manuals, other books, no. All right, part of the problem is that this workspace gets kind of congested if everybody drags in a library worth of stuff. All right, so you've got your book. Any notes that you want are fine. Okay, that's it. No other books. I don't think we've enforced this on the previous midterms, but we're going to enforce it on Tuesday. Okay, any questions? Bring a ruler or some sort of straight edge. You're going to need to draw a straight line through data points on a graph. That's a hint. Okay, so what we want to do is we want to go back and look at the stuff that I threw at you in such haste on Wednesday. All right, because this is 10 points for everybody in this room if you can get this right. I put this on the exam because I think it's important. All right, transition state for ionic reactions in solution. Here are the three things that we want to understand. First of all, the first thing that we want to understand is that equilibrium itself in solution depends on whether ions are present even if those ions are not directly participating in the reaction. Okay, equilibrium is affected by the ionic strength of the solution. It's affected by ions that are not even reactants or products in the reaction that we care about. All right, that's the first thing to understand. The second thing to understand is transition state theory. So in transition state theory, one of the most important postulates is that there's an equilibrium between the reactants and a transition state. All right, and so that equilibrium, like every other equilibrium that involves ions, is perturbed by the presence of ions in the solution. All right, it's not a special case. All right, it's just another equilibrium, the transition state for equilibrium strength. All right, this is why the presence of ions can influence the rate of a reaction. So here we're talking about the equilibrium constant. Here we're talking about the rate of a reaction and the reason the rate is affected is because the equilibrium between reactants and the transition state, that equilibrium is perturbed. And if the transition state is favored, if the ionic content of the solution favors the transition state, you accelerate the reaction. It makes a perfect intuitive sense once you understand it that way. What did I write here? The rate of the reaction, so does this depend on the favor of the transition, so the rate is to accelerate, yes. Okay, so the main point is that we've got charge reactants. The charge on A is Z of A, charge on B is Z of B. The total charge on the transition state then is the sum of Z A plus Z B. All right? And what we're going to postulate is that this transition state exists and that it's an equilibrium with these reactants right here. Okay, and once you form this transition state, it's just a unimolecular reaction with rate constant K double dagger, not two pluses, to form products. So we're just going to call this thing AB double dagger. The rate of the reaction is just equal to that rate constant times the concentration of the transition state, yes. And this is the normal way that we would write the rate of the reaction, right? We would just, I don't know why I called it K2, but this is just the rate constant times the concentration of A times the concentration of B. This is just the normal way that we'd write the differential rate law for this reaction, isn't it? Okay, so we've got this. We're going to come back to it in a second. What we need to understand first, that first bullet that we have to understand is why ions in solution affect equilibrium, right? Why is that the case and how do we understand that? We talked briefly, I'll admit it, about activities, right? The activity of some molecule A, which can be an ion or a neutral, is given by this expression right here where that's the concentration of A and this is the activity coefficient, right? The activity coefficient. Now, obviously, if the activity coefficient is 1, the activity just equals the concentration. What we're interested in is cases where the activity coefficient is not equal to 1. And that turns out to be the case if A is charged. If A is charged, in general, the activity coefficient will not be equal to 1. It'll be less. So that's the activity. It just has units of concentration. That's the activity coefficient. It's got no units. It's dimensionless. Okay, so this activity coefficient, there's some math that we can use to calculate it. This is the simplest equation. There are equations that take up a whole screen, right? We're not going to pay attention to those. This is the only one we care about. That A is just a constant that's equal to 0.509 in water. That's the charge on the ion that we're calculating the activity coefficient for. And that is the ionic strength. The ionic strength is just a concentration of each ion in the solution multiplied by its charge squared. So if you had sulfate with a 2 minus charge, you'd be multiplying by 4. If you have sodium with a 1 plus charge, you'd be multiplying by 1. That's just a concentration. There's a term in this summation for each and every ion in the solution. You just take all of the ions that are in the solution, take their concentration, multiply by their charge, line them up, multiply by 1 half. That's the ionic strength. That's how it's defined. Okay, so this is called the Debye Huckle limiting law and it allows us to calculate this gamma. If you look at this equation, you can have zero intuition about gamma, right? This equation doesn't, it's not obvious what gamma is doing. But here let me show you what this equation predicts. What this equation, so what am I plotting? This is gamma on this axis right here. Notice that all of these curves are starting up here at 1 and on this axis right here, that's the ionic strength. Right, so at infinite dilution, we're here in pure water and gamma for everything, no matter what the ion is, gamma is equal to 1. Now as the ionic strength starts to increase, gamma becomes less than 1. All right, what are these different curves? This curve right here is for an ion, a cation or an anion that has a charge of 1, plus 1 or minus 1. All right, it's got a size of 9, right? That's its hydrated radius. We don't need to know in detail about that. But the main point is that bigger ions show larger gammas. See how this gamma over a whole range of ionic strength is higher than for this blue guy. This blue guy is a smaller ion that also has a charge of 1. All right, so there's a negative deviation from 1 as the ionic strength increases and if the charge goes up, this is 1, this is 1, this is 3, there's a huge effect. All right, there's a huge deviation from 1 that happens for highly charged ions. All right, di-cations or anions, things like sulfate, phosphate has a 3 minus charge, magnesium has a 2 plus charge. These ions have large activity effects if they are highly charged. Okay, so what do we want to know about this plot right here? Basically, the activity coefficient is always less than or equal to 1. It's never more than 1. As the ionic strength approaches 0, this should really be I because that's what we're calling it in this lecture. Gamma approaches 1 for any ion, yes. As the ionic strength approaches 0, gamma approaches 1. Okay, for neutrals, if the molecule has no charge, by definition, its gamma is equal to 1. All right, so whatever gamma is describing is an electrostatic effect, neutrals don't participate in it. Gamma decreases deviating more from 1 as the ion gets smaller. Yes, that's this difference right here. See how that's lower than that? That's what I'm talking about. The smaller the ion, the greater the negative deviation from 1. Usually you can tell just by looking at an equilibrium which direction will be favored by the addition of an inert salt. In other words, if you change the ion content of the solution, you change the ionic strength, what we as chemists want to be able to understand qualitatively is what's going to happen to some equilibrium that's in solution. Is the equilibrium going to shift to the right towards products as we add salt, for example? Is it going to shift to the left towards reactants when we add salt? As chemists, we need to have that intuition. Okay, so let's look, take this equilibrium right here. We already know that we can write the equilibrium constant for this equilibrium in terms of activities. If we do that, oops, I didn't write it here, that would be the activity of C, the activity of A, the activity of B. We take gamma to the C power, C to the C power, B to the B power, and so on. All right, we write each activity as an activity coefficient and a concentration. We don't forget these exponents. We can pull all of these gammas to one side, pull them out of the expression. Now we've just got concentrations here, and we can use this collection of gammas to write a concentration equilibrium constant. So there's two kinds of equilibrium constants, it turns out. What you look up in the back of your book is a thermodynamic equilibrium constant, this K right here. It doesn't depend on the ion concentration in the solution, it's constant, it really is constant. Then there's a K prime, that's a concentration equilibrium constant. That equilibrium constant applies specifically to a certain concentration of ions in the solution, a particular ionic strength. All right, and it includes the activity effects. So this is called the concentration equilibrium constant because what you've done is you've stripped all the gammas out, now your equilibrium constants written out in terms of concentrations. Well that's how we normally write it anyway, isn't it? But it turns out it's only strictly correct to do this. If you pull out all these gammas, then you take the thermodynamic equilibrium constant, multiply by these gammas, and you get the concentration equilibrium constant. Okay, so by comparing the thermodynamic equilibrium constant with the concentration equilibrium constant, you can tell what will happen if you change the ionic strength. All right, if K prime is bigger than K, if the concentration equilibrium constant is bigger than the thermodynamic equilibrium constant, that means that as you add salt, equilibrium shifts towards products. If the concentration equilibrium constant is less than the thermodynamic equilibrium constant, that means if you add salt, everything shifts towards reactants. Let's do an example. All we're trying to do here is qualitatively get the answer right. All right, I want to know, here's acetic acid. We're going to add salt to an acetic acid solution. I want to know if the pH is going to go up or down. You should be able to predict that with 100% accuracy every time. All right, there's no doubt about what will happen. It will not be a small change. It will be a change that you can measure with your pH meter. Your pH meter was going to, the digits are going to click as you add salt to this solution. So it's not a rounding error. All right, here's the equilibrium constant. Now I've pulled all these gammas out. They're gamma for hydronium. There's the gamma for acetate. There's the gamma for acetic acid and the gamma for water. Now I'm going to calculate the concentration equilibrium constant for this equilibrium. How do I do that? I just pull, so I take my thermodynamic equilibrium constant and I multiply by these two guys in the numerator. There they are, and I divide by these two guys. There they are. Now that's my concentration equilibrium constant. All right, and then I look at these gammas here and I can already guess how big they are. All right, notice these two guys in the numerator, what's special about them? They're neutral. All right, so their gammas are one by definition. Boom and boom. What about these bottom guys? They're charged. All right, their gammas are going to be less than one by definition, right? Okay, and so just this really crude thinking here, those are one, those are less than one. That means that this thing in the red box here is going to be greater than one. That means that the concentration equilibrium constant is greater than the thermodynamic equilibrium constant. That means if I add salt to this solution and these guys get smaller, right, when I add salt, these gammas are going to get smaller. Remember, because if we're here and we add more salt, we go down here. These gammas are getting smaller when we add salt in this direction right here. Okay, these two gammas are smaller when we add salt. So K prime we could look. All right, now we can go through and mathematically figure this out using this approach. For any equilibrium, you can do this. You can do the analogous thing. All right, now what happens if you have ions on both sides while you can't tell? In other words, if these guys were charged in the numerator, this would be less than one and this would be less than one, so you'd have less than one, less than one, less than one, less than one, you can't just do a back of the envelope prediction about what will happen. You'd have to actually calculate the numerical gammas using the Debye Huckle limiting law, then you could figure out what's happening. All right, here we're just doing sort of a meatball estimation. All right, which works fine, a lot of the time it turns out. Okay, so what we're going to predict is if we add salt to this solution, it's going to shift to the right. Solution is going to become more acidic. The hydronium concentration is going to go up. The pH is going to go down. That's what actually happens. What about this guy? Now notice something. I'm looking at this equilibrium here. What we just said is it's going to shift to the right. Look at these guys and look at these guys. Which state of the system is more ionic? Reactants or products? Products. The products are charged. The reactants are neutral. All right, it turns out that adding salt favors the more ionic state of the system. Here products are more ionic than reactants. And so if I add salt, equilibrium is going to shift to the right. If you just remember that, you'll get the right answer every time. All right, equilibrium favors the addition of salt to a system in equilibrium favors the most ionic state of the system. Yes? Yep. If it's the reactants, is it more basic? Yes. Look at this guy. All right, we're talking about the solubility of lead sulfide. Lead dissolves to give lead 2 plus sulfur, 2 minus. Which state of this system is more ionic? Products again. That's totally neutral. All right, what's going to happen if I add sodium chloride to this electrolyte? Amazingly, the lead sulfide is going to become more soluble. The solubility of the lead sulfide goes up by quite a bit. It's not a small effect. We neglect it in CAM 1, in O-CAM, in inorganic chemistry. We totally sweep these effects under the carpet, but they're not small effects. How do I know that that's true? I can work it out mathematically. Here are the two gammas for those two ions. Here's the KSP expression. All right? So that concentration equilibrium constant is going to be equal to the thermodynamic equilibrium constant divided by these two gammas. How big are those two gammas? Well, they're both going to be less than 1, because both of these species are charged. And so when I calculate, that means that this whole thing here is going to be greater than 1, and when I multiply, wait a second, right. This is greater than, sorry, brain lock. All right? So the concentration equilibrium constant is going to be greater than the thermodynamic equilibrium constant. Yes? And the solubility is going to go up. We call this salting in. All right? Hardly anybody calls it that anymore. All right? But old-time analytical chemists still know what this means. Salting in means you add an inert salt to a solution where you're trying to get some sparingly soluble salt that's on the bottom of your Erlemeyer. You're trying to get that into solution. It's not working. What are you going to do? If you can heat the solution up, of course, that's one option. Another option is just to add an inert salt. When you add an inert salt, these gammas go down. They become smaller. The solubility, this concentration equilibrium constant goes up. The solubility goes up. All right? It's a trick. That every old-timer knows. Absolutely. Yeah. Saturation concentration goes up. Yeah. That's actually what it means to say that the solubility goes up. Why? Why does this happen? Is this magic? All right? This is in some ways the most important thing to understand. All right? This is what Peter Debye understood when he derived the Debye-Huckl equation. Coulomb's law. What is this? This is the energy acting between two ions in the solution. This is the distance these two ions are from one another. They're center to center distance. Okay? As they get further and further apart, the energy goes to zero. Zero's up here at the top. All right? So as these ions move far apart, they have no Coulomb interaction. But as they come closer and closer together, the energy goes down. These two ions stabilize one another if they have an opposite charge. If they have the same charge, then there's a curve like this that goes up. All right? But the point is in solution, you've got ions. There's no barrier to these ions moving around. They can go wherever they want to. Wherever the energy is the lowest is where they choose to go. How do they arrange themselves? They automatically arrange themselves like this. Right? You see how cations are close to anions? Anions are close to cations. All right? There's few cases where anions are close to one another. They don't have to be so they're not. Now a solution is very flexional. Ions are moving around all over the place. But on average, if you take a snapshot, you'll see this loose ionic lattice exists even in liquid salt solutions. All right? There's an ionic lattice that exists even in liquid salt solutions. It's counterintuitive. All right? How could there be a lattice of ions in a liquid? All right? Well, it's not a precise lattice of ions, all right? But it's a very loose. So you see this, you don't see this. All right? The solution doesn't look like that. Otherwise, it would be electrostatically destabilized. Okay? So these physics exert an influence on every ionic reaction because ions want to join this ionic lattice. Why? Because they stabilize the whole system. The energy of the whole system goes down when they join the lattice. You add more ions, the lattice gets reinforced, all right? The energy goes down even more. That's why gamma keeps going down. All right? There's a greater and greater electrostatic effect acting as you increase the concentration because the ions are closer to one another. The solution looks like this, not like this. All right? If it looked like this, you'd get electrostatic destabilization, all right? So the ions, this is completely counterintuitive to me, all right? You've got these ions that can be anywhere they want to but they form a lattice, they organize themselves so that they lower their electrostatic energy, all right? And what that does is it changes the driving force for every reaction that involves ions. That's what we're trying to understand with these gammas, all right? The gammas allow us to predict the influence of this ionic lattice on equilibrium. Okay. So coming back to this picture, we can apply this logic now to this equilibrium, all right? So far we've just been talking about generic equilibria for reactions and solution. Now we can come back to the rate of this reaction which is going to be influenced by this equilibrium here, right? Because in transition state theory we postulate there's an equilibrium between the transition state and the reactants, okay? And so now I can write an expression for this equilibrium constant. Here's the activity of the transition state. Here's the activity of A and B. I can write each activity in terms of gamma times concentration. I can roll all these gammas up into a K gamma, this is what your book calls it, all right? So K gamma is the equilibrium constant for the activity coefficients, if you will. Very strange notation to use but this is what they chose to use. So that is just that collection of gammas there, okay? And if I solve for the transition state concentration here, all right, this is the expression that I get, this is the equilibrium constant here, this is that collection of gammas, all right? And then I've got A times B. And so the rate of the reaction is now going to be given by this expression right here, okay? This is your equation 2049A. So what this says, notice, very important, as this equilibrium constant gets larger, the reaction accelerates. What is that equilibrium constant? It's the equilibrium between the transition state and the reactants. As that gets bigger, the reaction goes faster. That's what transition state theory says, okay? And by putting this gamma in the denominator, we include now the effect of salt in the solution on this equilibrium. Now, we're not going to go through and derive all of these other equations here, but this turns out to be an important one, all right? What this says is the effective second-order rate constant of the reaction, the thing that we've been calling the rate constant all along, all right? That's equal to the rate constant at infinite dilution of salt, ionic strength equals zero. That's what that zero means. Minus log of K gamma and K gamma is this collection of, it's that, all right? That's K gamma. Okay, this is the master equation that describes this transition state theory in solution. What does this predict? Oh, yeah, that's those guys, yes, we can calculate them, yes, we can do some algebra, yes, we can derive this equation from that equation, this is a better equation and it actually explicitly says what the effect of K is on the ionic strength, that's 2051. Here's what's predicted, all right? This plot is kind of hard to understand. Let's make sure we understand it. What is on this axis? This is the rate constant K2, all right? Reaction going faster in this direction, slower in this direction. This is the square root of the ionic strength. So this is infinite dilution here and now I'm adding salt, adding salt, adding salt as I go in this direction. The point of this plot is to understand what happens to the reaction rate as we increase the ionic strength. All right, what happens to the reaction rate? It's a little counterintuitive, I think. Here's the reaction rate at infinite dilution. For every reaction, all right? If the reactant and the product have the same charge, the reaction rate gets accelerated. That's true here, that's true here, that's true here. The amount of acceleration depends on how charged the reactants are. The more charged they are, the more the reaction gets accelerated by adding salt. If they have an opposite charge, the reaction slows down, right? As I add salt, I slow the reaction down. That's bizarre. All right, the reason this happens, adding salt, is because in this case right here, look at this one, reactants both have plus 2 charge. The transition state, that will have a plus 4 charge, right? That plus 4 charge is a higher charge state than the 2 plus 2s. You might think plus 4 equals 2 plus 2. So isn't the amount of charge the same? No, 4 trumps 2 2s. All right, a higher magnitude has much stronger coupling with the ionic network in the solution than 2 2 plus ions. Okay? And so that plus 4 gets stabilized and the reaction gets accelerated. So the more salt I add, the faster and faster and faster the reaction goes. If I look at this guy down here, this is a plus 2 and a minus 2. What's the charge on the transition state here? Zero. So now you're creating, you've got ions that are strongly coupled to this ionic network. Now you're creating a transition state that doesn't want to be in the network. It doesn't care. It's neutral. All right, the reaction gets slowed down by the addition of salt. Why? Because the reactants are stabilized. So equilibrium is shifting in the reactant direction. They are stabilized by the salt. The transition state is not. The reaction doesn't get accelerated. So if you can understand this plot qualitatively, you should be able to get 10 extra points on the final. Questions? Yes? 4 plus and a 2 minus. 4 plus and a 2 minus. Great question. Let's think about that. You've got reactants that are 4 plus and 2 minus. All right, what's going to happen to that reaction? Accelerated, deaccelerated or no effect? 2 plus, 4 minus, or 4 plus, 2 minus. All right, those are your reactants. 4 plus, 2 minus. How many people think that if I add salt to that solution, I will accelerate the reaction rate? How many people think if I add salt to that solution, I will deaccelerate the reaction rate? Right. 4 plus and 2 minus is 6 total charges in the reactants, the transition state is going to have a charge of what? 2 plus or minus. Right, 2, so you've got 6 total charges in the reactants and 2 for the transition state, so the reactants are going to be favored. The reactants are more ionic than the transition state. Okay, if the reactants have the same charge, then it's a little confusing because if you have 2 and 2 and they're both pluses, then the transition state is 4 and what you have to remember is 4 trumps 2 and 2. If they have opposite charges, that won't be the case. It'll be obvious, which one has, okay, everyone going to get those 10 points for a total score of 210. All right, so there's going to be a problem not dissimilar from this on your exam. Big piece of graph paper, concentration of some substrate velocity and we have to be able to figure out what's going on. We need to know, be able to calculate all of the relevant parameters, Km, Vmax, turnover number, catalytic efficiency, no, we didn't talk about it. No, that's not on your exam. All right, this picture keeps coming up in connection with everything that we do. We can't get away from it. All right, in this case, this is not the transition state, it's the enzyme substrate complex. This is the cartoon that applies. This is the rate constant for the reaction, the enzymatic reaction. All right, it's also the turnover number. It's also a constant called K subcat. All right, so if you're reading the literature on enzymes, they may say Kcat is this. They're just talking about that rate constant right there. All of those guys have units of seconds to the minus 1 because they're first order rate constants, first order in the enzyme substrate complex. Okay, so we worked out the mathematics for this using the steady-state approximation. This big Km here is the McHale's constant. It's this connection of rate constants right here related to those guys. This is what the generic enzyme reaction does. All right, this is the reaction rate on this axis. This is the substrate concentration here, not a time axis. This is the concentration of the substrate. Of course, if the substrate is zero, the reaction rate is zero. Now if I add substrate, the reaction looks first order at first, okay, but eventually when I fill up all the enzyme with substrate, I reach a Vmax. All right, this dashed line here represents what that is. Vmax is equal to K2 times the initial concentration of enzyme that I added at the beginning of the experiment, the total amount of enzyme that's present in the beaker. All right, that's its concentration there, so I can calculate Vmax if I know K2. We derived this equation, which is not called the line weaver-berk equation, but the plot that we make from it mysteriously is called the line weaver-berk plot. What is it? It's a plot of one over substrate versus one over the reaction rate. Okay? So if I'm given substrate and V information, I've got to calculate one over V and one over S before I can make a line weaver-berk plot. Once I've made it, this intercept here on the one over V axis is going to be one over Vmax. This intercept is one over, actually minus one over Km. All right, so I get Vmax and Km from those two intercepts right away. It's really easy. It's helpful if I can draw a straight line. That's why you need that ruler. Here's what your plot should look like, qualitatively. Okay? You've got to get these intercepts right. You've got to read them off the graph. You've got to be able to make a graph that looks, the graph is actually worth a lot of points. It's important to get it right, quantitatively right. So you'll have plenty of time to make it, don't worry. And then once you've got it, you can read 0.44, that's one over Vmax, 0.90, that's one over Km, K2 is Vmax over the concentration of enzyme. All right, I can calculate all this stuff. Make sure you get the units right. Units are going to be worth points, as always. Okay, here's an old exam. Here's an old exam question. Consider the following reaction mechanism, blah, blah, blah, assuming the steady-state approximation means you've got to derive an expression for the reaction rate. This is like the hardest steady-state problem I've ever asked on an exam. If you can do this one, you can do the one that's on the exam on Tuesday, I guarantee it. The one that could be on the exam Tuesday, I should say. All right, proton nitric acid, nitrous acid, sorry, reacts to protonate that. There's a plus sign there that's hard to see here. So that's H2NO2 plus, H2NO2 plus reacts with bromide to give ONBr, right here are the rate constants. Okay, so what do we have to know to get this right? First of all, we have to be able to pick out all the intermediates. What's an intermediate? Well, that guy looks like an intermediate. Look, he's here and here and he's not here. So he's definitely an intermediate. What else? ONBr, he's here and here and he's not here. Okay, so he's an intermediate. That's why this is hard. You've got to apply the steady-state approximation twice. You've got to apply it to every intermediate. Okay, so let's go through and do that. What do you do? I don't remember who my teaching assistant was in 2010, but they didn't have very good penmanship. This is not my handwriting. My handwriting is immaculate. So what do you do? The time rate of change of this guy, that's what this says right here. I know you can't read that. Time rate of change of H2NO2 plus, that's this. You set that equal to zero. That's the first thing that we're looking for. And then you've got to have each source and sink for that species represented by a term, right? This is the build-up of the H2NO2 plus from here, all right? Protons, HNO2, boom, boom, K1, boom. Then you've got K minus 1. You've got the second reaction here, boom. All right, so you've got three terms and now you solve for the concentration of the intermediate. You should really say H2NO2 plus steady-state to distinguish it from the general concentration for that species. That's what it's going to be equal to right there. All right, six points for getting that guy right. Then we've got to do the same thing with this guy, ONBR, here he is, set it equal to zero. You do exactly the same thing, source, sink of ONBR. In this case there's only two terms, one for that, one for that, okay? But when I write it out, this guy appears in this expression. And so once I've got this expression, I can substitute this whole mess here in for the H2NO2 plus. I get this monster equation here which happens to be correct, it is kind of monstrous. You should see what it looks like if you don't apply the steady-state approximation, total nightmare. So once I've got this guy, notice that that guy is the only one that appears in the rate expression for the product. That's the product we care about, that's the guy that appears in the rate expression, not this guy, but this guy appears indirectly because he appears in the expression for this guy, right there, okay? And so if I want to know the rate of the reaction, it's K3 times ONBR times C6H5 NH2, okay? And there it is, plug everything in, that's what the differential rate law is for this reaction in the steady-state approximation. Okay, now I can ask questions about this. I can say, look, what happens to the apparent rate law as a function of the bromide concentration? Well, there's bromide here and bromide here, right? If I increase, if I make bromide large, we expect this term in the denominator to dominate K minus 1, right? If that dominates, the bromide here will cancel, all right? And so I'm going to end up with, so at high BR, these BRs are going to cancel for us because this term K2BR dominates over K minus 1, BRs cancel, I get a reaction that has overall second order that doesn't depend on BR minus. At high BR minus, we shouldn't have any BR minus sensitivity to the rate, that makes sense. We've effectively isolated the reaction with respect to BR minus. You make BR minus big enough, all right, and the reaction doesn't care about it anymore. It cares about the ones that are not very, the reactants that are not very big. And at low BR minus, what happens, oops, at low BR minus, K minus 1 dominates, this term goes away, all right, we have a collection of rate constants here where we've got 1, 2, 3 concentrations, okay, and so the reaction has overall third order and it does depend on BR minus. So there's two limiting cases. You can tell there's going to be two limiting cases immediately because there's a plus sign in the denominator. Okay, so what else could we talk about, possibly? What other kinetic issues could be broached on this exam? I don't really know. I mean, it's, you know, you have to be able to write the reaction rate, all right, here are the rules for doing that from an earlier lecture. You have to be able to figure out what the units are on the rate constant, all right, you do that by just putting the units, writing the, first of all, writing the differential rate expression and then putting the units on the concentrations, putting the units on the reaction rate, which is always molar per second, and then just solving for the units of K. No matter what this thing looks like, you can always get the right units of K by doing that. Here's an example. Here's another example. We have three methods for determining what the reaction rate, what the differential reaction rate is, what the rate law is, we have three methods for doing that. One of them is the method of initial rates. You make everything, all the reactants in the mixture big except one, and if you do that, you've isolated the reaction with respect to that reactant. Let's say in this case it's A, and if you work out the math for doing that, this equation applies, log of the reaction rate equals log of the K prime, which includes the K and all the other concentrations, so you make everything big except one reactant, in this case A, did I just say that or did I say that wrong? Everything big except one reactant, you've isolated that one reactant in this case, it would be A, and then you can plot log of the reaction rate as function of log of A, and the slope is just the molecularity of the reaction with respect to A, right, the number of A's that are participating in the reaction. And you can pick the reaction mechanism apart reactant by reactant by doing that. Okay, what other options do you have? Well, you can use an integrated rate law. We talked about that, and you can fit your data to an integrated rate law, and there's the half-life. If you can measure the half-life of your reaction, here's an example of working out the integrated rate law for a reaction that has ordered three halves, that would be bizarre, but you could work it out if you had to, here's what that looks like. All right, and here's the half-life, and the half-life behaves, the half-life itself tells you absolutely nothing, but successive half-lives tell you a lot. All right, the successive half-life goes down by a factor of two, the reaction is first order. Half-life doesn't change, oh sorry, zero order. Half-life doesn't change, it's first order. Half-life gets doubled on each successive half-life, the reaction is second order, and so on. And you can derive the T1 half expression for more complicated mechanisms if you have to. Okay, and finally there's the Arrhenius equation that tells you about the temperature dependence of the reaction rate, all right, you have to know what the activation energy is, the temperature, and you have to know what this pre-exponential factor is, or you can determine the pre-exponential factor if you want from the data that you have because the pre-exponential factor is the intercept of this Arrhenius plot of log K versus 1 over T. I think that's everything. Now, I've located a website that you're going to be interested in, you guys are going to want this bumper sticker here, you're going to want a mug like this one, or it's possible, I don't think you're going to want it, but if, and I notice also that they have many of these in different colors and sizes in case you should want such, this is a terrible thing to say, but, okay, so everyone's going to do well on Tuesday, good luck.